Engineering Mechanics.Statics

- l )(i cos fI + j sin lJ)

) 5 - 4 sin

- - y

Prablem 2/110

Prablem 2/ 109

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Ar t icle 2 /8

2 /8

Momen t a n d Coupl e

73

MOMENT AND COUPLE

In two-dimensional analyses it is often conveni ent to determin e a moment magni tude by scalar multi plication using the moment-arm rule. In three dimensions , however, the determin ation of the perpendicular dist ance betw een a point or line and the line of action of the force can be a tedious computa tion. A vector approach with cross-product multi plicati on th en becomes advantageo us . Moments in Three Dimensions

Cons ider a force F with a given line of act ion act ing on a body, Fig. 2/2 10, an d any point 0 not on this line. Point 0 a nd the lin e of F establish a plan e A. Th e mom en t Mo of F about a n axis throu gh 0 normal to the plan e has the magnitude M o = Fd , wher e d is the perpendi cular distance from 0 to the line of F . Th is moment is also refe r red to as th e moment of F about the point O. The vecto r M o is normal to the plan e and is directed along the axis through O. We can describe both the magnitude a nd th e direct ion of Mo by the vector cross-product relati on introduced in Art. 2/4. (Refer to item 7 in Art. C/ 7 of Appe ndix C.) Th e vector r runs from 0 to any point on t he line of action of F . As describ ed in Art. 2/4, t he cross prod uct of r an d F is written r x F a nd ha s the magn itude tr sin CtlF, which is the same as Fd , t he magnitud e of Mo . The correct direction and sense of t he momen t are esta blished by t he rig ht- ha nd rule, describ ed pr eviously in Arts. 2/4 a nd 2/5 . Thus, wit h r and F treated as free vectors emanating from 0 , Fig. 2/2 1b, th e thu mb point s in the direction of Mo if the fingers of th e rig ht hand curl in t he direction of rotat ion from r to F thro ugh the a ngle Ct. Th er efore, we may write the moment of F about the axis through 0 as

---- ~

.-. r----a) :.:u-----~/ / a / d

F la l

(2 / 14 )

The order r x F of the vectors must be mainta ined because F x r would produ ce a vector with a sense opposite to th at of Mo; that is, F x r = - Mo .

The cross-product express ion for 1\1 0 may be written in the determinant form j

k (2 / 15)

(Refer to item 7 in Art. C/ 7 of Appendix C if you are not a lready famili ar with th e determinant representation of the cross product.) Not e the symmetry and order of the te r ms , and not e that a right-handed coordinate syste m must be used. Expansion of the determinant gives MO

=

(ryF z - rzF , l j + (rzFx - rxFzlj

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(hi Figure 2/21

Evaluating the Cross Produd

+

(rxFy - ryFxlk

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...... /

~

/

//

74

Chap ter 2

Force Systems

To gain more confidence in the cross-product relationship, exam ine th e three compone nts of the moment of a force about a point as obtaine d from Fig. 2/ 22. Th is figu re shows the t hree components of a force F act ing at a point A located relat ive to 0 by the vecto r r . The scalar magnitudes of th e mom ents of these forces about th e positi ve X - , yo, and z-axes through 0 can be obtained from the moment -arm rule, and are

z I Mz

CD I

I I

r-

I I

a

/ /

which agree with the respective terms in the determinant expansion for the cross product r x F. Moment about an Arbitrary Axis Figure

2/22

We can now obtain an expression for the moment M,\ of F about any axis A through 0 , as shown in Fig. 2/ 23. If n is a unit vector in the

F

A-direction, then we can use the dot-product expression for the componen t of a vector as described in Art. 2/ 7 to obtain M o ' n , the comp onen t of M o in th e direction of A. Thi s scalar is the magn itude of the mom en t M, of F about A. To obtai n the vector expres sion for the momen t M, of F about A, multiply the magnitude by th e dir ecti onal unit vecto r n to obtain

( M,

\

\ Figure

2/23

=

(r x F -n)n )

(2/16)

where r x F replaces Mo . The expression r x F· n is known as a triple scalar product (see it em 8 in Art. C/ 7, Append ix C). It nee d not be written (r x F ) -n becau se a cross pr odu ct cannot be form ed by a vecto r and a scalar. Thus, the association r x (F · n ) would have no meaning. Th e tr iple scalar product may be represen ted by the determinant

IM,I = M ,

where

a,

(2/17)

f3, yare the direction cosines of the unit vector n .

Varignon 's Theorem in Three Dimensions

In Art. 2/4 we introduced Vari gnori's t heorem in two dimensions. Th e theorem is easily exte nded to t hree dim ensions. Figure 2/ 24 shows a sys tem of concurrent forces F 11 F 2' F 3' . . . . The sum of the moment s about 0 of these forces is

r x F 1 + r x F 2 + r x F 3 + . . . = r x (F , + F 2 + F 3 + . . .) Figure

2/24

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r x

~F

Articl e 2 /8

where we have used the distributive law for cross products. Using the symbol 1\1 0 to represen t th e sum of the moments on th e left side of the above equation, we have ( 1\10

= ~(r

x F)

=

r x

R)

(2/18)

This equation states that the sum of the moment s of a syste m of concurrent forces about a given point equals the moment of their sum about t he sa me point. As ment ioned in Art. 2/4 , t his pr inciple has many application s in mechanics.

Couples in Three Dimensions Th e concept of th e couple was introduced in Art. 2/5 an d is eas ily exte nded to three dimen sions. Figu re 2/ 25 shows two equa l a nd opposite forces F a nd - F acting on a body. Th e vector I' runs from any point B on the line of action of - F to any point A on the line of actio n of F . Points A and B are located by position vectors rA and rn from any point O. The combined moment of the two forces about 0 is

However, rA - ru = r-, so that all reference to the moment center 0 disapp ea rs, and t he momen t of t he couple becomes

(2/ 19) Thus, the moment of a couple is the same about all points . The magnitude of 1\1 is M = Fd , where d is th e per pendicular distan ce betw een th e lines of act ion of the two forces, as descr ibed in Art. 2/ 5. The moment of a couple is a free vector, whereas the moment of a force about a point (which is a lso t he momen t about a defined axis through the point) is a sliding vector whose direction is along the axis through the point. As in the case of two dimensions, a couple tends to prod uce a pure rotat ion of the body about an axis norm al to th e plan e of the forces which const itute the couple.

Figure 2/2 5 Marwan and Wa see m A l-lraqi

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Moment and Cou pl e

75

76

Ch apter 2

Force Systems

Figure 2/26

Couple vectors obey all of the ru les which govern vector qu antities. Thus, in Fig. 2/ 26 t he couple vecto r M, due to F , and - F , may be added as sho wn to the coupl e vecto r M2 du e to F 2 an d - F 2 t o pr oduc e the couple M , which, in turn, can be produced by F and - F . In Art . 2/5 we lea rn ed how to repl ace a force by its equivalent forcecouple syst em. You should also be able to carry out t his replacem en t in three dimensions. Th e procedure is represented in Fig. 2/27, where the force F acti ng on a rigid body at point A is replaced by a n equal force at poin t B and t he couple M = r x F. By adding the equal and opposite forces F and - F at B , we obtain the couple composed of - F and t he origi nal F . Thus , we see that th e couple vector is simply the moment of the or iginal force about t he point t o whi ch the force is bein g moved. We emphasize that r is a vector which runs from B to any point on the line of action of the origin al force passing through A . M =r xF

Figure 2/27

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Article 2/8

Moment and Couple

Sample Problem 1 /10

77

y

I

A tension T of magni tude 10 kN is applied to th e cable attached to th e top A of the rigid mast and secu red to t he gro und at B. Determine t he mom ent M z of T abou t the a-axis pass ing t hrough th e base O.

r-.- I A

[ 15m

CD

T

10 [

12i - 15j + 9k

~

TnAS

~

10(O.566i - 0.707j + 0.424k ) kN

~

IM o

~

r x F] Mo

/

/

'>> , 9m

B ]

Hel pfu l Hints

/ ([ 2)2 + (- 15)2 + (9)2

CD We could also use the vector from 0 to B for r and obtain the same result, but using vector OA is simpler.

From Eq. 2/ 14, ~ ~

15j X IOW.566i - 0.707j + 0.424k ) 150(-0.566k + 0.424iJ kfc-m

It is always hel pful to accompany your

The valu e M z of the desired moment is th e scalar component of M o in th e zd irection or M z = Mo· k. Th erefore, M, = 150(- 0.566k + 0.4241) ' k = - 84.9 kN· m



T = 10 kN

SoluUon (a). Th e required moment may be obtai ned by finding t he compone nt along t he z-ax is of the moment 1\10 of T about point O. T he vector 1\10 is normal to th e plane defined by T an d point 0 , as shown in th e accompa nying figure . In t he use of Eq. 2/ 14 to find M o , the vect or r is any vecto r fro m point 0 to the line of action of T . The simplest choice is the vecto r from 0 to A, which is written as r = l 5j m. Th e vector expr ession for T is

An s.

Th e minus sign indicates that the vector Mz is in the negati ve z-direction. Expressed as 8 vector, t he moment is Mz = - 84.9k kN . m.

vecto r operations with a sketch of the vectors so as to retain a clear pictu re of th e geometry of th e problem.

@ Sketch the

x-y view of th e problem

and show d . )'

I

A

@

Solution (6). Th e force of magnitude T is r esolved into compone nts T z and Txy in th e x-y plan e. Sinc e T z is parallel to t he a-axis, it can exert no moment about thi s axis . Th e momen t M z is, th en , du e on ly to Txy and is Alz = T~d, where d is t he perpendicu lar distance from Txy to O. Th e cosin e of th e angle betwe en T and Try is JI 52 + 122 / JI 52 + 122 + 9 2 = 0.906, and th er efor e, T xy

~

10W.906 )

= 9.06 kN

The momen t arm d equals OA multiplied by th e sine of th e angle between Txy and OA, or

d = 15

12

-, -

B

= 9.37 m

JI 22 + 152

y

I

Hen ce, the moment of T about th e a-axis has the magnitude M,

~

9.06(9.37)

~

84 .9 kN · m

and is clockwise when viewed in t he x -y plan e.

T,.\• -

Solution (e). Th e component Txy is further r esolved into its components Tx and TY' It is clear that T;y exerts no momen t abou t th e a-axis since it passes through it , so th at th e required mom ent is du e to T x alone . The direct ion cosine of T with respect to t he .r-axis is 121 J92 + 122 + 15 2 = 0.566 60 that T. = 10(0.566 ) = 5.66 kN . Thu s, M, ~ 5.66(15) = 84.9 kN · m Ans. Marwan and Waseem AI-Iraqi

www. gigapedia.c om

A

Tx

I \~ ~»..r I \ I \

Ans.

t

\\ I

~~ T I T/ xy t 15 m

II / 0).. - \ L--/

/

z

- -,

_

- _ _

\,I --...'l._ '"

12 '";i,- -

"' 9 m B

X

78

Chapter 2

Forc e Sys t ems

Sample Problem 2/11

30N

30 N

Determine the magnitude an d direction of t he couple M which will re place the two given couples and st ill produce th e sa me exte rn al effect on th e block. Specify th e two forces F and - F, applied in the two faces of th e block parallel to the y -z plan e, which may replace the four given forces. The 30-N forces act parall el to the y -z plane.

x,

100 mm

... ...

50m m

/

Solution. The couple du e to t he 30-N forces has the magn itude M I = 30(0.06) =

25N

1.80 N· m. Th e direction of M I is normal to th e plane defined by t he two forces , and t he sense, shown in the figure , is esta blished by t he right-ha nd convention. The couple due to the 25-N forces has the magnitude M 2 = 25(0.10) = 2.50 N· m with th e direction and se nse shown in th e sa me figure. Th e two couple vectors combine to give th e compone nts

I

z

... M

My = 1.80 sin 60' = 1.559 N . m M, = - 2.50 + 1.80 cos 60' = - 1.600 N · m

CD

AI = / (1.5591' + (- 1.600)2 = 2.23 N· m

Thus,

8 ~ tan ' 11. 559 ~ tan -' 09i4 ~ 44 3'

with

1.600

.

An s.

F = 2.23 0.10

= Fd ]

J

All = 1.8 N'm

I

z

Ans.

Q) Bear in mind that the cou ple vectors are free vectors an d th er efore have no u nique lines of actio n.

z I I

A force of 40 lb is appl ied at A to th e handle of the control lever which is atta ched to the fixed shaft OB. In determi ning t he effect of th e force on th e shaft at a cross section such as that at 0 , we may replace th e force by an equ ivalen t force at 0 and a coup le. Describe th is couple as a vector M .

'""---- S' I --'- - - =r-~

3'

-J........ kif-"::::::":~' /

Solution. Th e cou ple may be express ed in vector notation as M = r x F , where r ~ OA ~ 8j + 5k in . and F - 40i lb. Thu s,

+ 5k )

X I - 40iJ = - 200j

+ 320k

lb-in .

Alternatively we see th at moving th e 40-lb force through a distance d = / 5 2 + 8 2 = 9.43 in. to a parallel position through 0 requires the addi tion of a couple M whose magnitude is ~

.-

I

Sample Problem 2 /12

M

/ /

Helpful Hint 22.3 N

and the direction 0 = 44.3°.

M = 18j

z-v

J.. -- . . . . . . ,

8 -F

I

Ans.

.

I

x .....

---.,..-y

Th e forces F and - F lie in a plan e norm al to the cou ple 1\1, and thei r momen t arm as see n from the right-hand figure is 100 mm. Thus, each force has th e magnitude

1M

,

8, '

/~

Fd

~

40(9.431 ~ 3ii Ib-in .

Ans.

The couple vecto r is perpe ndicular to the plan e in which the force is shifted , and its se nse is that of t he moment of the given force about O. The direction of M in the y-z plane is given by

e~

ta n"!

Marwan and Waseem AI-Iraqi

~~

Ans.

32.0' www .gigapedia.com

/

X

/

I

B

40lb

Art icl e 2 /8

PROBLEMS

Pro ble ms

79

2/113 The two force s act ing on the handles of the pipe

wrenches constitute a cou ple M . Express the couple as a vecto r. An s. M ~ - 751 + 22.5j N -m

Intro ductory Problems 1/111 T he weigh t of the comp uter syste m is 80 Jb wit h cente r of gr avity at point G. Deter mine th e moment M o of this weight about point 0 on th e horizontal table to p. Find the magnitude of M o Ail s. M o ~ - 320i + 160j Ib-ft , M o ~ 3581b-ft

150 N

.

~ I 1 50 m~ 250 mm ~~

<,

' x

150 N

Problem 2/113

Problem 2/111

1 /1 12 Determine the moments of force F about point 0 a nd about point A.

1/114 T he helicopter of Prob. 2/58 is redra wn here with cer tain three-dimension a l geom et ry given . Du ri ng a gr oun d te st, a 400·N a ero dynamic force is applied to the ta il rot or at P as shown. Determine th e moment of thi s force about point 0 of the airframe . Z I

z I I

I I I

0)/ /

I

b 'II-------£..

a

c

:~6m

A

I

_

,--",,,,,,.. ~. .

I I I

I

F

_

-- - - -

/

/ - - - - - - j / /

1.2 m

X

Problem 2/112 Problem 2/114

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80

Chap ter 2

Forc e Systems

1 /11 S T he struct u re shown is cons t ructed of circ ular rod whi ch has a mass of 7 kg per met er of len gth. Determi ne the mom en t M o ab out 0 caused by th e we ight of the st ru ctu re. Find the magn itude of 1\1 0 , A ils . M a ~ - 192.6; - 27.5j N · m Ala = 194.6N-m

<>

2m

-<'

2/117 T he right- an gle pipe DAB of Prob. 2/ 99 is shown aga in here. Repl ace th e 750-N te nsil e force whic h th e cable exerts on poin t B by a force-cou ple sys te m at point O. Ans. R = - 598; + 4 11j + 189.5k N M a ~ - 36 li - 718j + 419k N - m

>-

1.6 01

1.2m

__________

/"

0.8 m

7"-

/ 0.8 m

-x

Problem 2/117

x-

"' y

Problem 2/11 5 1 /116 The tu rn bu ckle is tightened unti l t he te nsio n in ca ble AB is 1.2 kN. Ca lcu late the magn itu de of the moment about point 0 of the force act ing un point A .

1 /118 In an a tt empt to pull down a nearly sa wn- through branch. th e tree su rgeo n exerts a 400 -N pull on the line which is looped around th e branch a t A. Det ermine the momen t about point C of th e force exerted on th e branch a nd sta te the magnit ude of th is mo me nt . z J J

12 m 2m--:-_ 4 m "

~ I

- l~

11 m

-,...------2m

l

o

...... -_L... .../

r" "

.

<,

v

--->.. .,.

»

I

1.5 m

J--.

I

8m ---::'iI1,-"..'"

>

:

-

-~~'( IJ 611~ ~

2m B

Problem 2/116 Problem 2/118

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Article 2 / 8 2 /119 The figure of Prob. 2/101 is shown again her e. If th e magnitude of the moment of F a bout line CD is 50 N· m , determine th e magnitude of F. Ans. F = 228 N B I I

r

I

Prob le m s

81

2/121 A 50-lb force is a pplied to th e cont rol ped al as shown. The force lies in a plane parallel to th e x -z plan e a nd is perpendicular to BC. Det ermine th e moments of this force about point 0 and about th e shaft OA. Ans. M o = - 90.6i - 690j - 338k Ib-in . M O, \ ~ - 690 Ib-in.

/

O.2m

,Ie------- _

~

0.2 m

-

D

A

O.2m

Problem 2/119

Representative Problems 2 /120 Two 1.2-lb thrusters on th e non rotating sa tellite are simu ltaneo us ly fired as sho.....n. Compute th e mome nt associated with this couple an d state about wh ich satellite a xes rotations will begin to occur.

Problem 2/121 2/122 A mom ent (torque ) M applied to t he shaft a nd at tach ed a rm causes a te nsio n T of 120 Ib a pplied to A by t he restraini ng cab le AB. Det ermi ne the moment 1\1 0 of the ten sion about point O.

I- - 20·:L;'-I

.--y 25" I

/ X

-I..

/

1.21b

Problem 2 / 120

Problem 2/122

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82

Ch ap te r 2

Forc e Sy st ems

2 /125 A 50- N hor izontal force is a pplied to th e handle of th e industrial water valve as shown. The force is perpendicular to the vertical pla ne containing lin e OA of th e ha nd le. Deter mine th e equivalent forcecoup le sys tem at point O. Ail s. R = - 38.3i - 32.1j N 1\I" = 643 i - 766j + 6250 k N . mm

2/123 A 300-N force is applied to the handle of the winch as shown. The for ce lies in a plane which is parallel to th e y -s: plane and is perpendicu lar to line AB of th e handle. Determine th e momen ts of this force about point 0 a nd about the x-axis. Ails. M() = - 98.7i + 17.25j + 29.9k Nr m M x ~ - 98.7 N r m

~

1$i'R?i~i::i

-

~ 40 mm

" "75 mm

~75 mm

B

/=--=:~

p =' Problem 2/125 2/126 A space shuttle orbit er is subj ected to th ru st s from five of t he e ngi nes of it s reaction cont rol system. Four of th e thru st s a re show n in th e figu re; th e fifth is an 850-N upw ard thru st a t the righ t rear, symmetric to the 850-N t hru st shown on the left rear. Com pute t he mome nt of these force s a bout point G a nd show that t he forces have th e sa me moment ab out a ll points.

Problem 2/123

2/124 Compute the mom en t l\1() of t he 250-lb force about the axi s 0 -0 . I

o

1 2" ~ r-------------I

------.""-

250 lb

I

~m

)'

I 8"

j

o

I

1700 N

1700 N 1700 N

<, ~

Prob lem 2/124

.r

Problem 2/126

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Arti cl e 2 / 8 2 /127 The specialty wrench shown in t he figure is designed for access to t he hold-down bolt on certai n automobile distributors. For the configu ra t ion shown where the wr enc h lies in a vertical plan e a nd a horizon tal 200-N force is a pplied at A perpendicu lar to the han dle, ca lcu la te the momen t M o applied to th e bolt a t O. For wh at value of the distance d would the a-com ponen t of M o be zero'? Ans. M() ~ - 48.6j - 9.49k N · m, d = 74.5 m m

Probl em s

83

2/129 Usin g the pr inc iples to be developed in Chapter 3 on equili brium , one can determine that t he tension in cab le An is 103.6 N. Determi ne the moment about th e .r-axis of this ten sio n force acting on point A. Cornpar e your resu lt to the momen t or the weight Waf th e 15-kguniform plate about th e r -ax is. What is t he moment of the tension force act ing at A abo u t lin e OB? A n s. M x = 3 1.1 Nv m, (M-T) w = - 3 1.1 Nrm M OB = 0

A~ d = 125 mm

,.

:>

2()() N

-,

O.4 m

- y

200mm

~j

.r " 70 c01- n,-

-- y

/

Prob lem 2/129

Problem 2/127 2/128 In pickin g up a load from position B, a cable te ns ion T of magnitude 24 kN is develop ed. Ca lcu la te t he moment wh ich T produces ab out the base 0 of the cons tructi on crane.

2 / 130 T he specia l-pur pose mill ing cutter is subjected to th e force of 1200 N a nd a coupl e of 240 N · m as shown. Deter mi ne the mom ent of t his system a bou t point O.

Z

1-.-. I

18 01

--..,

I

/'

<, ;

~x

250 111m

: I I

1

I I

130 m

I/\I/"~_ x ~

~

/ -7>,.-

I

. I) :'!::s1.

i

1200 N

I

I,

6m

Ic- /

Problem 2/128 Marwan and Waseem AI-Iraqi

240 N 01 I

I/ ~ /

~~ - y 5m

A

Problem 2/130 www.gigapedia.com

84

Ch ap ter 2

Fo rc e Sy st ems

1 /131 The rigid pole and cross-arm assembly ofProb. 2/95 is shown again here. Determine the vecto r expres sion for th e moment of th e 1.2·kN te nsion (0) a bout point 0 a nd (b ) a bout th e pole z-axis. Find eac h momen t in t wo different ways. All ,' , lal M o = - 2,89i - O,962k kN · m Ibl M, ~ - O.962k kN · m

2/133 A 1.8-lb vertical force is a pplied to th e kn ob of th e window-opener mecha nism when th e crank Be is ho rizontal. Deter mine the moment of th e force about poin t A a nd abo ut line AB . Ails. M A ~ - 5.40i + 4.68j lb-in . M A IJ ~ - 4.05i - 2,34k Ib-in .

z' I I l e././

z I

B

I \ " ",~

'

.......... <,

y'

Problem 2/1 JJ 1 /134 Determine the vect or express ion for th e mom ent .1\1 0 of the 600-N force a bout point O. The design specificat ion for the bolt at 0 would req uire t his result.

Problem 2/1 J 1 1 /132 The ISO-lb force is applied a t point A of th e bracket. Dete rmine t he momen ts of t his force a bout point B , about point C, and about the line BC.

, ,

6OOi':

Problem 2/1 J4

Problem 2/1 J2

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'x

Art icl e 2 /9

2 /9

RESULTANTS

In Art. 2/6 we defined the res ulta nt as the simplest force combination which can replace a given sys tem of forces without altering the external effect on the rigid body on which th e forces act . We found the magnitude and direction of the resultant force for the two-dimensional force sys tem by a vector sum ma tion of forces , Eq. 2/9, an d we located th e line of action of th e re sulta nt force by applying t he pr inciple of moments, Eq. 2/ 10. These same principles can be extended to three dimensions. In the previous article we showed that a force could be moved to a parall el position by adding a corres ponding couple. T hu s, for th e syste m of forces F) , F 2 , F" . . . acting on a rigid body in Fig. 2/ 280, we may move each of them in turn to the arbitrary point 0 , provided we also introd uce a couple for each force tran sferred. Thus , for example, we may move force F I to 0, provided we introduce the couple 1\1} = r l x F 1> where r l is a vector from 0 to any point on the line of action of F l' When all forces are shifted to 0 in this manner , we hav e a syste m of concurrent forces at 0 and a syste m of couple vectors, as represent ed in part b of th e figur e. Th e concur rent forces may then be added vectorially to produ ce a resultant force R, and the couples may also be add ed to produce a resulta nt coupl e M , Fig. 2/2&. T he gen era l force system , then, is reduced to R

= F1 +

M = M)

=

F2 + F3 +

+ M2 + M3 +

~F

=

(2 /20) ~( r

x F)

T he couple vectors are shown through point 0 , but becau se they are free vectors, they may be represen ted in any parallel positions. The magnitudes of the resu ltant s and their component s are

n, = ss,

Ry

= ~Fy

R,

= ~F,

R = J( ~Fx )2 + (~Fy)2 + (~F,l2 Mx

= ~ (r

M" = ~ ( r x

x F ),

M

F )y

M,

= ~ (r

(2/ 2 1) x F),

= "1Mx 2 + My 2 + M z2

/. ' a)

' bJ

(e )

Figure 2/28 Marwan and Wa see m AI- Iraqi

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Result ant s

85

86

Chapter 2

For ce Systems

Th e point 0 selected as th e point of concur re ncy for th e forces is arbitrary , and th e magnitude and directi on of M depend on the particular point 0 selected, Th e magnitude and direct ion of R, however , are the same no matter which point is selected. In general, any system of forces may be replaced by its resu lt ant force R and th e resultant couple M. In dyn ami cs we usually select the mass cente r as the reference point. The change in the linear motion of the body is determin ed by th e resultant force, and th e change in th e a ngular motion of th e body is det ermined by th e resul tan t couple. In statics, the body is in complete equilibrium when the res ultant force R is zero and the resultant couple M is also zero. Thus, the determination of resultants is ess ential in both statics and dyn ami cs. \Ve now examine the result ants for several special force systems. Concurrent Forces. \Vhen forces are concurrent at a point, only the first of Eqs. 2/20 needs to be used becau se there are no moment s about the point of concurrency.

Parallel Farces. For a syste m of par allel forces not a ll in t he same plan e, th e magn itu de of the para llel resu ltant force R is simply the magnitude of t he algebra ic sum of the given for ces. Th e position of its line of action is obtai ned from the prin ciple of moments by requiring that r x R = Mo . Here r is a posit ion vector extending from the forcecouple refe re nce point 0 to the fina l lin e of actio n of R, a nd M o is the sum of the moments of th e ind ividual forces about O. See Sample P roblem 2/ 14 for a n exa mple of parall el-force systems. Coplanar Forces. Article 2/6 was devot ed to this force syste m. Wrench Resultant. When the resul tan t couple vecto r M is parallel to th e res ultant force R, as shown in Fig. 2/29, the resultant is called a wrench. By definition a wrench is posit ive if t he couple and force vectors point in the same direction and negative if they point in opposite directions. A common example of a positive wrench is found wit h the application of a screwdriver, to drive a right-handed screw. Any gen eral force system may be represen ted by a wrenc h applied alon g a un ique line of act ion. Thi s reductio n is illustrat ed in Fig. 2/ 30, where part a of t he figure represent s, for the general force system, the resultant force R act ing at some poin t 0 a nd the correspond ing resu lta nt coup le M. Although 1\1 is a free vector, for convenie nce we represent it as acting th rough O. In part b of the figure , M is resolved into components M, along t he direct ion of R and M 2 norma l to H. In par t c of t he figu re, the coup le

Pos iti ve wrench

Negative wrench Figure 2/29

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Articl e 2/9 M

I

;

t 7"1.---'

II

----!C, '-iIiIII (a)

fbi

fe)

fd)

- II

Figure 2/30

M 2 is repl aced by its equivalent of two forces R and - R se parated by a distan ce d = M 2 /R with - R appli ed at a to cancel th e or iginal R. This ste p leaves th e resul tant R, which act s a long a new and unique line of actio n, and the parallel couple M i , which is a free vector, as shown in part d of the figure. Th us, the resultants of the origi na l general force sys tem have been transformed into a wrench (positive in this illustration ) with its un ique axis defined by th e new posit ion of R. We see from Fig. 2/ 30 that the axis of the wrench resu ltant lies in a plan e th roug h a nor mal to the plan e defined by R and M. Th e wrench is th e simplest form in which th e resultant of a gen er al force system may be expres sed. Thi s form of the resultant, however, has limited a pplication , because it is us ually more conve nient to use as the reference point some point 0 such as the mass center of the body or another convenient origin of coordinates not on the wrench axis.

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Resultants

87

88

Ch ap t e r 2

Fo rc e Syst ems

Sample Problem 2 /13

z I I I

Determine th e resul tan t of the force and couple syste m which acts on th e rectangular solid.

50lb

1000I b·in .

cb Solution. We choose point 0 as a convenient reference point for t he initi al step

'--,.

of reducing th e given forces to a force-c ouple syste m. Th e resultant force is

Q)

I

R ~ ~F ~ (80 - 80li + (100 - 100)j + (50 - 50)k ~ 0 Ib

12"

The sum of the momen ts about 0 is l'tI o = [50(16 ) - 700J; + [80(12) - 960ij + [100(10) - 1000]k lb-in . = 1001 Ib-in .

/

/'

Hence, t he resu lta nt consist s of a couple, which of course may be applied at any point on the body or the body extended.

10" 100lb <, <,

...... y

Helpful Hints

CD Since the force summation is zero. we conclude that the resultant. if it exists, must be a couple.

@ The moments associated with the force pairs are easily obtained by using the At = Fd rule and assigning the unit-vector direction by inspection. In many three-dimensional problems. this may be simp ler than the 1\1 = r x F approach.

Sample Problem 2 /14

50N

Determine the resultant of the syste m of parallel forces wh ich act on th e plate. Solve with a vector ap proach.

0.5 m

z :-< 1/ O.5m ............

Solution. Transfer of all forces to point 0 results in the force-couple syste m R l'tI o

:i:F

~

I 'J-O

y /1 0.35 m

/ :.. . . . / 1

(200 + 500 - 300 - 50)j = 350j N

150(0.35) - 300(0.35)]1 + [- 50 (0.50) - 200(O.50)]k - 87.5; - 125k N · m

1

/ 500 N~ /

~oiN

y/ /

0.35m

...,........J.

300N

Th e placemen t of R so that it alone represen ts th e above force-c oup le syste m is determ ined by the principle of moments in vector form r x R =Mo (x i

+

.n

+ zk l x 350j

- 87.5;

125k

35O:ck - 350.1

- 87.5;

125k

From th e one vector equatio n we may obtain th e two scalar equatio ns 35O:c

CD

~

- 125

and

- 350z

~

- 87.5

Hence, x = - 0.357 m and z = 0.250 m are the coordinates thro ugh which the line of act ion of R mu st pass. Th e value of y may. of course , be any valu e, as permitted by th e principle of transmissibility. Thus, as expected , th e variable y drops out of the above vector analysis. MalWan and Waseem Al-l raq!

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Helpful Hint

CD You should also carry out a scalar solution to this problem.

Art icle 2 /9

Resultants

89

Sample Problem 2 /15 500 X

Replace the two forces a nd the negative wre nc h by a single force R applied at A a nd t he cor res ponding coup le M.

(

Solution. The resul tant force has th e compone nts

= 500 si n 40" +

2F-t"J

R,

[R,

~F,l

R, = 600 + 500 cos 40" cos 45° = 8 71 N

[Rz

~

R,

~ FzJ

~

700 si n 60"

Thus.

R = 928 i

/

.. ,-=:......._

120 mm . --.",./

// . - - 1...

- y

+ 87 1j + 621 k N

R = )1 9 28 )2 + (871)2 + (621)2 ~ 1416 N

a nd

BO m m

l ~~ ~ "/~-'T4-50_ _~ -tOm m600 x

mm

700 cos 60" + 500 cos 40" si n 45° = 62 1 N

u fO 'd--'-, // z

Im m I 1 50 m m

100

= 928 N

[R,

~

Ans.

Th e cou ple to be added as a resu lt of moving th e 500-N force is

CD

[M ~ r x F ]

M 500 ~ (O. 08i + 0.12j + 0.05k ) x 500l i sin 40° + j cos 40° cos 45° + k cos 400 sin 45°)

where r is the vecto r from A to B . The term -by-term, or de terminant, expansion gives M 500

~

18 .95i - 5.59j - 16.90k N . m

@ Th e mom ent of t he

600-N force ab out A is written by inspecti on of its x- a nd zcomponents, whi ch gives

CD Suggestion : Chec k t he cross-product results by eva lua ting t he mome nts about A of the compone nts of the 500-N force directly from the sketch.

M 600 = (600 )(0. 060)i + (600)(0.0401k ~ 36.0 i + 24 .0k N · m

The moment of t he 700-N force about A is easi ly obtai ned from the mom en ts of the x- and z-compcnents of th e force. The result becomes M , OO = (700 cos 60")(O.0301i - [(700 sin 60")(0 .060 ) + (700 cos 60")(0.10 0 »)j - (700 si n 60")(0.030)k = 1O.5i - 71.4j -

Helpful Hi nts

18 .19k N · m

@ For the 600-N and 700-N forc es it is eas ier to obtain th e components of th eir moments about the coordina te directions throu gh A by inspection of the figure th an it is to set up th e cro ss-product relations.

@ Also, the couple of th e give n wre nch may be wr itten M'

@ T he

25.0(- i sin 40° - j cos 40° cos 45° - k cos 40° sin 45°) - 16.0 7i - 13.54j - 13.54k N· m

Therefore, th e resultant coup le on adding toget he r the i-, j-, a nd k -ter ms of th e four 1\1' 5 is M = 49.41 - 90.5j - 24 .6k N . m and

M

) (49.4)2 + (90.5)2 + (24.6)2 ~ 106.0 N vm

Ans.

25-N ' m coup le vector of th e wrench poin ts in the direction opp osite to that of the 500·N force, a nd we mu st resolve it into its .r-, y -, and z-com pone nts to be added to th e ot her cou ple-vecto r components .

@ Alth ough th e resulta nt couple vect or 1\1 in the sketch of the resultants is shown through A , we recogni ze th at

a couple vector is a free vector and therefo re has no specified line of act ion .

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90

Chapte r 2

For ce Systems

Sample Problem 2 /16

z

Determine the wrench re sultant of th e three forces acting on the brac ket. Calculate the coordinates of th e point P in the x-y plane t hrough which the res ulta nt force of th e wrench acts. Also find the magnitude of the couple M of the wrench .

1 I

4"

5"

CD

Solution. Th e directi on cosines of th e cou ple 1\1 of th e wrench mu st be t he sa me as those of th e resu ltant force R, assuming that the wre nch is positi ve. The resultant force is R

= 20i

<,

...... x

40lb

+ 40j + 40k Ib

and its direct ion cosines are cos 8x = 20/60 = 1/ 3 cos 8v = 40/ 60 = 2/ 3 cos 8x = 40/ 60 = 2/ 3 The momen t of the wre nch couple must equal th e sum of th e moments of t he given forces about point P through which R passes. Th e moments about P of the t hree forces are (lU )n.

20yk lb-in.

(Mi n.• = - 40(3li - 40xk lb-in. (M) n.

= 40(4 - y li - 40(5 - xlj lb-in.

and t he total momen t is

Helpful Hint

CD We assu me initia lly that the wrench

1I1 = (40 - 40yli + (- 200 + 4Ox )j + (- 40x + 2Oy)k lb-in.

is positive. If 1\1 turns out to be negative, then the direction of the cou ple vector is opposite to that of the resultant force.

Th e directi on cosines of M a re (40 - 4Oy)/ M (- 200 + 4Oxl/M

cos 0)'

cos 8x = (- 4Ox + 2Oyl/M where M is th e magnitude of M. Equ ati ng th e direction cosines of R an d 1\1 gives M 40 -4Oy = "3

2M 3

- 200 + 40x = -

- 4Ox + 20y = 2M 3

Solu tion of th e three equations gives M = - 120 lb-in .

x = 3 in .

y = 2 in.

Ans.

We see that M tu rned out to be negative, which mean s t hat the couple vector is pointing in the direction opposite to R , which makes th e wren ch negative.

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Article 2 /9

PROBLEMS Introductory Problems 2 /135 A baseball is thrown with spin so th at th ree concu rrent Iorces act on it as shown in th e figur e. Th e weight \V is 5 OZ , th e drag D is 1.7 OZ , a nd th e lift L is perpendicul ar to th e velocity v of the ball. If it is kn own that th e j-compone nt of the resu ltant is - 5.5 0 2 and t he a-componen t is - 0.866 oz, determine L , 0, and R. A il s. L = 1 OZ , H = 3D", R = 5.82 0 2

Problems

91

1 /137 T he thi n rectangu lar plat e is subjected to the four force s shown. Deter mine th e equivalent forcecoup le sys tem a t O. Is R perpendicula r to M o '! AilS. It = FW.5j - 0.1340 k l M o ~ Fb! 1.866 i + 0.268j + k }, yes z I I

F

o

b

- y

Probl em 2/137

Problem 2/135

2/136 Three equal forces ar e exerted on the equilate ral pla te as shown. Reduc e the force sys tem to an equ iva lent force- couple syste m at point O. Show that R is perpen dicular to Mo .

1 /138 T he spacecraft of Prob 2/ 120 is rep eated he re . T he plan is to fire four L2- lb thrusters as shown in ord er to spin up t he spacec raft abo ut its a-axis. but the t hru ster at A fails. Deter mine the eq uivalent forcecouple sys te m at G for th e remai ni ng th ree th rusters.

F

F

b

z I I 0 '

t

b

2

1.2 1b / /

/

b

2

b

/

F

/

t

I 20· "

-'1-

25"

/

"'A~'f"-r;-k-_ J.

• • • •1 ~ :

Problem 2/136

1.21b

I

Probl em 2/1 38

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. --y

92

Chapter 2

Force Systems

2 /139 Th e pull ey a nd gea r a re subjected to the loads shown. For t hese forces, determine th e equi valent force-couple sys te m at point O. Ail s. H = 792i + 1182j N M o = 2S0i - 504j + 28.Sk N . m

,

2 /141 Two up wa rd loads a re exerted on th e sma ll threedim en siona l tru ss. Reduc e t hese two load s to a single force- couple sys tem at point O. Show that R is perpend icul ar to Mo . T hen determine the point in the .r-z plan e through whic h t he resultant passes. Ails. H = SOOj lb, M o = 1200 i + 4800 k lb-ft x = 8 ft , Z = - 2 ft

330 / 11

, / mm v

/'

.:--- mm 220 / I I I I

I

3' ;/",__

'~~~'--... \100 mrn 800 N

75 mm

3'

r;o__

z/A

- x

2001'

J1.-'

10'

1200 N Problem 2/139 Problem 2/141

Representative Problems 2 /140 The comme rcial a irl iner of P rob . 2/78 is redrawn he re with three-di me ns iona l inform ation supplied. If engin e 3 suddenly fails, determine the resu lta nt of th e three remai nin g engi ne thrust vectors, eac h of which has a magn itu de of 90 kN. Spec ify th e y a nd z-coordina tes of the poin t through wh ich t he line of action of the resulta nt passes. T his information wou ld be critical to the design cr ite r ia of pe rfor ma nce wit h engine failu re .

2 /141 Represen t t he resultant of t he force syste m a ct ing on th e pipe assembly by a single forc e R at A a nd a coup le M.

120 N

lOO r-.:

x _ <, -

~;iOi"''"'AA~O N ~ ... ... y

JJ -t;I-S" 0N ', ~~~~r<

200mm

~ 25'

Y Problem 2/142

Problem 2/140

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Art icl e 2 /9

Probl em s

93

1 /143 Replace th e two forces and single couple by an equiva lent force-couple syste m at point A. Ans. H ~ - 20i - 37 .9j + 12.65k kN M = 45.3j + 40.9k kN . m 35 kjc-m ,-"~?~ l m

1m

- y

1m

................y

I I I I

Z

Problem 2/143

Problem 2/ 145

1 / 144 Determine th e x-an d y -coordinates of a point

1 /146 In tighte ning a bolt whose cente r is at point 0 , a

through which the resulta nt of th e par allel for ces passes.

person exerts a 40-1b force on the rat chet ha ndle wit h his right hand. In addition, with his left hand he exerts a 20-lb force as shown in order to secu re th e socket onto th e bolt head . Deter mine th e equ ivalent force- couple syste m at O. Then find the point in the x·y plane through which the line of action of t he single-force resu ltant passes.

y I I I

701b 301b

fiOl h

60~--r ,- -8"_

,

_ ,...-- 10" ~ 1O" -

z ..... -- -- "'j"

80 Ib

Probl em 2/ 144 2 /145 Th e motor mounted on the bracket is act ed on by

its 160-N weight , and its shaft resist s th e 120-N t hru st and 25·N - m couple appli ed to it. Det ermine th e resul ta nt of th e for ce syste m shown in term s of a force R at A an d a coup le 1\1. Ans. H ~ - 120i - 160k N M ~ - 7i + 9j + 24k N 'm

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20 lb

,

..... y

Problem 2/146

94

Chapter 2

Force Syst ems

1 /147 Rep lace the two for ces acti ng on th e pole by a wr en ch . Write the moment 1\1 associated with th e wrenc h as a vect or a nd specify th e coordinates of th e point P in the y-z plane through which the line of action of th e wre nc h passes. - aT o, z 7a A ns. M - - (i + j ), y 2 2

The resulta nt of th e two forces a nd couple may be re present ed by a wr en ch . Det ermine t he vector expr ession for the moment 1\1 of the wrench and find t he coordinates of th e point P in t he x -z plane through which th e resultant force of th e wr en ch passes. Ans. l\1 = lOi x

z

+

lOj N vm

= z = 0.1 m

----_...........

I I

-..... -. T

I

a

400 mm

---y 3a

o7 /

__

,,

I -

-

.I.. r

:»

/

Problem 2/149

/

r

,

/

/ X

Problem 2/147

.... 1 /148 Replace th e two forces actin g on t he rectangular solid by a wre nch. Wri te th e moment 1\1 ass ociated with th e wre nc h as a vector a nd specify th e coor dinates of the point P in the x -y plan e th rou gh which the line of action of the wre nc h passes.

A ns. M = ' : x = a

(i -

+ c, y

.... 1 /150 Repl ace the sys te m of two forces and coup le shown in Prob. 2/ 143 by a wre nch . Determine th e ma gn it ude of t he mom ent 1\1 of th e wr ench, th e magnitude of the force R of th e wre nc h, an d th e coordina tes of the point P in the x -y plan e through whi ch R passes. A ns. M ~ 26.9 kN . m , R = 44.7 kN x ~ 0.22 1 m, y ~ - 0.950 m

k)

= b/ 2

z I I I

Problem 2/148

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Chapter Review

CHAPTER REVIEW In Chapte r 2 we have es tablished the properties of for ces, mom ents, and cou ples , an d the correct procedures for rep resenting th eir effects , Mastery of this materi al is essential for our study of equilibrium in the chapters whi ch follow. Failure t o correctl y use the procedures of Chapter 2 is a common cause of errors in applyin g the principles of equilibrium. When d ifficul ties ar ise, you should refer to thi s chapter to be sure that the forces, moments, and couples are correctly represented. Forces

There is frequent need to represent forces as vectors, to resolve a single force into components along desired directions, and to combine two or more concurrent forces into an equivalent resultant force. Specifically, you shoul d be able to: I. Resolve a given force vector into its compone nts al ong given directions, and express the vector in terms of the unit vectors along a given set of axes.

2. Express a force as a vector when given its magnit ude and information about its lin e of action . This infor mation may be in the form of two points along the line of action or a ngles whi ch or ient th e lin e of action . 3. Use the dot produ ct to compute th e proj ectio n of a vect or onto a specified lin e and the angle betw een tw o vect ors.

4. Compute the resultant of two or more forces concurrent at a point . Moments The tende ncy of a force to rotate a body about an axis is described by a moment (or to rque ), which is a vecto r quantity. We have see n that find in g the moment of a for ce is oft en facilitated by combining the moment s of the components of the force. When working with moment vectors you sho uld be able to: 1. Determine a moment by using the moment-arm rule.

2. Use the vector cross product to compute a moment vector in terms of a force vector and a posit ion vector locating the line of action of the force. 3. Utilize Va rign on 's theo rem to simplify th e calcu lation of moments, in both scalar and vector forms. 4. Use the triple sca lar product to compute the momen t of a force vector about a given axis through a given point. Couples A cou ple is the combined mo ment of two equal, oppos ite , and noncollinear forces. The un iqu e efTect of a cou ple is to produce a pure twist or rotation regardl ess of whe re the for ces are located. The cou ple is useMarwan and Waseem Al-lra qi

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95

96

Chapter 2

Force Sy s t em s

ful in rep lacing a force act ing at a point by a force-couple system at a different point. To solve problems involvin g couples you should be able to: 1. Compute th e moment of a couple, given th e couple forces and either their separation distance or any position vectors locating their lines of action.

2. Replace a given force by an equivale nt force-couple system, and vice versa.

Resultants We can reduce an arbitrary system of forces and couples to a si ngle resultant force applied at an arbitrary point, a nd a corresponding re sultant couple. We can further combine this resultant force and couple into a wrench to give a single result ant force along a unique line of action, togeth er with a parallel couple vector . To solve 'problems involving resultants you should be able to: 1. Compute th e magnitud e, dir ection , and line of action of t he resultant of a system of coplanar forces if that resul tant is a force; otherwise , compute th e moment of th e res ultant couple.

2. Apply th e principle of moments to simplify the calculation of the moment of a system of coplanar forces about a given point. 3. Replace a given genera l force system by a wren ch along a given line of action.

Equilibrium You will use th e preceding concepts a nd meth ods when you study equilibrium in the following cha pters. Let us summarize t he concept of equilibrium: 1. When th e resulta nt force on a body is zero (~ F = 0), the body is in translational equilibrium. Th is means that its center of mass is eit her at rest or moving in a st raight line with constant velocity.

2. When th e resultant couple is zero ( ~M = 0), the body is in rotational equilibrium, either having no rotati onal motion or rotating with a constant angular velocity. 3. When both resulta nts are zero, t he body is in complete equili bri um.

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Rev i e w Prob lems

REVIEW PROBLEMS

97

2 /151 Th e blades of t he portable fan gene ra te a l -lb th ru st T as shown. Compute th e moment At 0 of thi s force

1 /15 1 Using t he pri ncipl es of equilibriu m to be developed

about th e rear su pport point O. For comparison, determ ine the moment about 0 du e to th e weight of the mot or- fan un it AB , whose weight of 9 lb acts at G. Ans. M o ~ 13.14 Ib-in. CCW M 0 w = 26.1 Ib-in . CW

in Cha pte r 3. you will soon be able to verify tha t the ten sion in cable AB is 85.81H of the weight of t he cylinder of mass m . while th e ten sion in cable AC is 55.5% of th e sus pended weight. Write each tension force acting on point A as a vector if the ma ss m is 60 kg. Ans. T AB = -2BO i + 420j N T AC ~ 2BOi + 16B. lj N -'~~~- 2 m -~~-

c 1.2 m

U

y I I

L_- x

Problem 2/151 1 /151 A die is being used to cut threads on a rod. If I5-lb

forces are ap plied as shown, determ ine the magn itude P of the equa l forces exerted on th e t-in. rod by each of the four cutt ing su rfaces so that their extern al effect on the rod is equ ivalent to that of th e two I5-lb for ces. , I"

- 14

@

Problem 2/153 2/154 For th e angu lar posit ion (I = 60° of th e crank OA. the gas pressure on t he pist on induces a co m pressive for ce P in the conn ecting rod along its centerline AB . If t his force produces a moment of 720 N vm about th e cra nk axis 0 , calcu late P.

15 1b

~5"

I I CIA: = 125 rom

I AB =300mm o Problem 2/ 152

Problem 2/154

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98

Chapter 2

Force Sy s t ems

2 /155 Ca lculate the mom ent M o of t he 250 -N force a bout the ba se po int 0 of the robot. Ail s. M 0 = I S9.6 N · m CCW

2 /158 A force F ac ts along t he line AB ins ide t he right circ u lar cylindri cal she ll as show n. The qu antities r, h , H, and F a re known . Usin g the X-, y- , an d z-coord ina tes shown, express F as a vector.

-" I

>

300 111m

/

- x Probl em 2/155 2 {156 Rep resen t t he resultant of t he t hree forces a nd cou ple by a force-cou ple system loca ted a t point A . 3 kN

- - - - - - -A

-

- - 2 111

5 kN 4kN

Pro blem 2/156 2{157 Reduce t he given loadi ng syst em to a force-couple sys tem at point A . Then det ermine the dist an ce .r to the righ t of point A at which the resul tant of t he t hree forc es acts. A il s. R ~ SO Ib down, M A ~ 1240 Ib-in . CW .r = 15 .5 in .

Prob lem 2/ 158 2{159 The direct ion s of rotat ion of t he inpu t sha ft A a nd out put sha ft B of t he wor m-gea r reduce r are ind icated by the curved da shed arrows . An inpu t torque (cou ple) of 80 N · m is applied to sha ft A in the di rection of ro tation . The output shaft B su pplies a to rque of 320 N . m to t he mac hine whic h it drives (not shown). The shaft of t he d riven machine exerts a n equal a nd opposite reacting torque on t he output sha ft of t he red ucer . Determ ine the resu lt a nt M of the t wo couples which act on t he reducer u nit a nd calcul at e t he directi on cosine of M with respec t t o t he x-axis. Ans. M ~ - 320i - SOj N · m cos fI% = - 0. 970 r

IS0lb

200 lb

20"

S"

B)(

-:

A

IS"

150 111111......,"'-

.i-:

x

3001b

Pro ble m 2/157 Problem 2/159

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Rev iew Prob l e m s

2 /160 Replac e th e force P a pplied at point A by an equ iv-

alent force- coupl e system at point B.

,--

z

I I AI

I I

P

I

~ ' II _ 3

99

2/162 When the pole OA is in the position shown , the te nsion in cab leAB is 600 lb. (0 ) Wr ite th e tens ion force exe rte d on point A as a vector usin g th e coordi nat es shown. (b) Dete rmi ne the momen t of this force about point 0 a nd state t he moments abo ut the X - , y- , and z-axes. (c ) Det ermin e the projection of thi s te ns ion force onto line AD .

--J

OA = 50' OB = 40' OC = 30'

b b

b X /

/

- y

B

40'

Problem 2/ 160

- .r

2(161 Th ree couples a re form ed by t he three pairs of equal a nd opposit e forces. Det ermine the resu lta nt M of the th ree couples. Ails. M = - 201 - 6.77j - 37.2k N· m

2/163 The combined action of the three forces on the base at 0 may be obtained by esta blishing their resultant through O. Determine the magn it ude s of R and the accompa nying coup le M. Ails. R ~ 1093 lb, M = 9730 lb-tt

~<>~{I -, 7<78 0N -:

100 N lOO N

Problem 2/162

>;~=_-:'" 120

N

Pro blem 2/161

Problem 2/163

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100

Ch ap t er 2

Forc e Systems

1 /164 During a drilling operation, the small robotic device is subjected to a n 800·N force a t point C as shown. Replace this force by an equivalent force-cou ple systern at poin t O.

-1 / 166 The trap door OA is raised by th e cabl e AB wh ich passes over the sma ll guid e pull eys at R. The ten sia n everywhere in t he cable is T , and this ten sion applied at A cau ses a momen t Al 0 a bout t he hi nge . . . Al() at O. P lot th e qu an tity T ' wh ich I S t he momen t a rm rela tive to 0 of the ten sio n a pplied a t A , as a fun ction of t he door a ngle (J for the range 0 es II :=:; 90~ . Determin e th e maximum a nd mi nim um valu es of th e moment arm over thi s ra nge of O.

-0

•' C

I 250mm

T ~-i'r.i"::"-'-

\ •

L

I

0.4

;;00\ F = 800 N

III

1

0.5

III

Problem 2/ 166 Problem 2/ 164

'!

' Computer-Oriented Problems

' 2/ 165 Four force s are exerted on t he eyebolt as shown. If the net effect on the bolt is a direct pull of 600 Ib in the y -di rect ion, det e rmine the necessa ry values of T a nd fl. All", T ~ 204 lb, /I = 21.70 240lb

T

I

\

' 2 / 167 Wit h th e 300-lb cylindrical part P in it s gr ip, th e rob otic a r m pivot s ab out 0 through th e range - 4 5~ :=:; (I :=:; 45° with th e an gle at A locked at 120°. Determine a nd plot (as a fun ction of m th e mom ent at 0 due to the combined effect s of t he 300-lb part P, t he 120·lb weight of member OA (mass cente r a t G , l, a nd the 50·lb weight of memb er AB (mass cent e r a t G 2 l. Th e en d gri p is include d as a pa rt of member AB. T he lengt hs 1.. 1 a nd 1.. 2 a re 3 It and 2 Ft, respectively. Wha t is th e maxim um value of Al() and at what value of Ii does this maximum occur? Ail S. M a : : : 1230 cos lJ + 650 cosl60° - OJ lb-It (M U )m ll x ::::: 1654 lb-Ft at tJ ::::: 19.90<)

360 II,!

- =r:¢~

Probl em 2/165 Probl em 2/1 67

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Review Pro blem s

·1 /168 A flagpole with attached light t riangu lar fram e is shown here for a n a rbit ra ry position during its raising. Th e 75-N ten sion in the erect ing cab le remains cons ta nt. Determ ine and plot the momen t a bout the pivot 0 of t he 75-N force for t he ran ge O :s O:s 90"'. Determine the maximum value of th is mom ent an d the elevation a ngle at which it occurs ; comment on th e physical sign ifica nce of the latter . T he effects of th e diam eter of the drum a t D may be neglecte d.

101

·1/170 As pa rt of the design pr ocess for a larger mechanism , t he portion shown in th e figu re is considered . The spring of modulu s k = 200 N/m is a ttached to the fixed point 0 a nd to th e slider A which moves along the slot. T he uns t retched len gth of the spring is 150 mm , and the force in th e spring is the cons ta nt k times t he deflection of the spring. P lot the X - , y -, and a-compone nt s of the spr ing force as app lied to A as the slider moves in t he range - 200 :s x :s 200 mm .

c

,, \

)

\ \

1

\

\

a

I

150 0101

---oJ

- - - 6 01- - --1

100 0101 -- -<'--

Problem 2/168

· 2/ 169 Th e recta ngu la r plat e is til ted about it s lower edge by a cabl e tensioned a t a constant 600 N. Deter mine and plot the momen t of thi s ten sion abou t the lower edge AB of th e plate for th e range a :s a « 90"'. 7200 sin (/ N vm •• A ns . l YlAJJ = ) 4 1 + 24 cos (J

z I I I I

I I I

c A

'"--__ o~ -I

3 01 / /

5 01

4m

Problem 2/169

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Problem 2/ 170

-- x

In many applications of mechanics. the sum of the forces acting on a body is zero, and a state of equilibrium exists. What are the two prima ry forces acting on each slowly movin g balloon? MalWan and Waseem Al-l raqi

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Chapter

EQUILIBRIUM CHAPTER OUTLINE 3/1

Introduction

SECTION A Equilibrium in Two Dimens ions

3/2 3 /3

System Isolation and the Free-Body Diagram Equilibrium Conditions

SECTION B. Equilibrium in Three Dimensions 3/ 4 Equilibrium Conditions

Chapte r Review

3/ 1

I NTRO DUCTIO N

Sta tics deals primarily with the description of t he force conditions necessary a nd su fficient to maintain th e equilibrium of engineeri ng str uct ures. Th is cha pte r on equilibrium, th erefore, cons t itutes the most importa nt part of sta tics , an d th e pr ocedur es developed her e form th e basis for solving problems in both stat ics a nd dyn ami cs. We will ma ke continua l use of t he concepts developed in Cha pte r 2 involving forces, momen ts, couples, a nd resultant s as we apply the pri nciples of equ ilibrium. When a body is in equ ilibrium, the res ultant of all forces acting on it is zero . Th us , the resultant force R an d the resu lta nt coup le M are both zero, and we have the equilibrium equations ( R = :!:F = O

(3 /1)

Th ese requ irem ents are both necessary a nd sufficient conditions for equi libri um . All physical bodies are t hree-dimensional , but we can treat many of them as two-dimensional when the forces to whi ch they a re subjected act in a single plan e or ca n be project ed onto a single plane. \Vhen this simplification is not possible, the problem mu st be treated as threeMarwan and Waseem AI-Iraqi

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103

104

Chapt er 3

Equilibrium

dimensional. We will follow t he ar ra ngeme nt used in Cha pte r 2, a nd discuss in Sect ion A the equilibrium of bodies subjected to two-dimen siona l force systems a nd in Section B the equilibrium of bodies subjected to three-dimensional force sys tems.

SECTION A. EQUILIBRIUM IN TWO DIME NS IO NS 3 /2 SYSTEM ISOLATION AND THE FREE-BODY DIAGRAM Before we apply Eqs. 3/ 1, we must define un a mbiguously t he particular body or mechanical syste m to be a na lyzed and represent clearly and complete ly all forces acti ng an the body. Omiss ion of a force which acts on the body in qu estion, or inclusion of a force which does not act on the body, will give erroneous result s. A mechanical system is defined as a body or gr oup of bodies which can be conceptually isolat ed from all othe r bodies. A system may be a single body or a combination of connecte d bodies. Th e bodies may be r igid or nonrigid. Th e syste m may also be an identi fiable fluid mass, either liquid or gas, or a combination of fluids and solids. In sta tics we st udy primarily forces which act on rigid bodies at rest, althoug h we a lso study forces acting on fluids in equilibrium. Once we decide which body or combina tion of bodies to a na lyze, we th en trea t th is body or combination as a single body isolated from all sur rounding bodies. This isolation is accomp lished by means of t he fr eebody diagra m , which is a diagrammatic representa tion of the isolated system treated as a single body. Th e diagram shows all forces appli ed to the syste m by mechanical contact with other bodies, which a re imagined to be removed. If appreciable body forces are present , such as gravita tional or magneti c attraction, then these forces must also be shown on th e free-body diagram of the isolat ed syste m. Only afte r such a diagram has been carefully dr awn shou ld the equilibri um equations be written. Because of its critical importance, we emphasize here that the free-body diagram is the most important single step in the solution of problems in mechanics. Before attem pting to draw a free -body diagr am, we mu st recall the basic characteristics of force. These characteristics were described in Art. 2/2, with primary atte ntion focused on the vector properti es of force. Force s can be applied either by direct physical contact or by remote action . Forces can be either internal or external to the syst em under consideration. Application of force is accompanied by reactive force, and both applied and reactive forces may be either concentrated or dist r ibuted. Th e pr inciple of transmissibili ty per mits the treatment of force as a sliding vecto r as far as its exte rnal effect s on a r igid body are concerned. MarNan and Waseem AI-Iraqi

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Articl e 3/2

Syst em Isolation and the Free -Body Diagram

\Ve will now use th ese force characteri sti cs to develop conceptual models of isolated mechanical systems. Th ese models enable us to write th e appropriate equa t ions of equilibrium, which ca n th en be a na lyzed.

Modeling the Adion of Forces Figure 3/ 1 s hows the common types of force application on mecha nica l systems for ana lysis in two dimensions. Each examp le shows the force exer ted 011 th e body to be isolated, by th e body to be removed . Ne vvrton 's third law, which notes the existe nce of a n equa l a nd opposite reactio n to every action , mu st be ca refully observed. Th e force exerted 0 11 the body in question by a contacting or supporti ng member is always in the sen se to oppose t he movement of th e isolated body which would occur if th e con tac ti ng or support ing body were re moved.

MODELING THE ACTION OF FORCES IN TWO·DIMENS IONAL ANALYSIS Type of Contact and Force Origi n Act ion on Body to Be Isolated 1. Flexible cable, belt . cha in. or rope

~

Weight of ca ble ne gligible

~ e

Weight of cable not negligible 2. Smoot h s u rfaces

~

...

<.

~ T

pC T

,

:/<'

3. Hou gh s u rfaces

~

W

R'

/

'

/

,

' v' N

4. Roller s upport

~

SI::!.

If: k 0 ,

:f

5. Freely s lidi ng gu ide

JfifJnl_

For ce exerted by a flexible ca ble is a lways a te nsi on away from the body in the direct ion of th e ca ble.

~

J.~ N

,

Cont act force is com pressive a nd is norm al to t he s u rface.

Rough s u rfaces are ca pable of s uppor t ing a tange nt ia l compo nen t F (fr ict iona l force ) as well as a no rm al component N of the resultant contact force R.

Roller, rocker. or hall su pport transmits a compressive force normal to t he s upport ing s urface.

Colla r or s lider free to move along s moot h guides; can s upport force normal to guide onl y.

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105

106

Chapt e r 3

Equilibr ium

MODELI NG TH E ACTION OF FORCES IN TWO· DIMENSIONAL ANALYSIS (co nt.) Typ e of Contact a nd Force Origin Acti on on Body to Be Isolated Pin

Pin not free to turn to turn

6. Pin conn ection

free

~~I

k 7. Built-i n or fixed s upport

""""'I

A

F~C=

~2 weld

A

or

I ~j

n,

R.~.

V

8. Gra vitational attraction

r

m

111111 9. Spri ng act ion Ne utral pos iti on

} ,,,,v~1 vvl..o-+-F

W= mR

Linear

Non lin ear

F I I F = kx

F I Ha rdenin g

~~l;ening ~.T ~

y

I

A freel y hin ged pin conn ection is capabl e of s up porti ng a force in a ny direct ion in th e plane normal to th e axis; usu ally sh own as two com pone nts R, an d By. A pin not free to tum may a lso s u pport a couple .\l. A buil t-in or fixed su pport is ca pa ble of sup porting a n axial force F, a tran sverse force V tsh ear for ce ), and a couple M (be ndi ng mom ent) to pr event rotation . Th e resu lt a nt of gravit at ional attraction on all elements of a body of mass 111 is t he we ight W = mR a nd net s toward the cen te r of th e ea rth through t he cente r mass G. Spri ng force is tens ile if s pr ing is s t retched a nd compressiv e if com pressed . For a linea rly clastic s pr ing the st iffness Il is th e force requ ired to deform th e s pr ing a u nit di st an ce.

Figure 3/1 . continued

In Fig. 3/ 1, Example 1 depicts the action of a flexible cab le, belt, rope, or cha in on the body to which it is attached. Beca use of its flexibility, a rope or cable is un abl e to ofTer any resistance to ben di ng, shea r, or compressio n an d therefore exert s only a te nsion force in a direction ta ngent to th e cable at its point of attachment. The force exerted by the cable on th e body to which it is attac hed is always away from the body. When the tension T is large compared with t he weight of the cable, we may ass ume that th e cable form s a straigh t line. When th e cable weight is not negligible compared with its tension, the sag of t he cable becomes importa nt , a nd the ten sion in th e cable changes directi on and magn itude along its length. When the smooth surfaces of two bodies are in contact, as in Example 2, the force exerted by one on the other is normal to th e tan gent MarNan and Waseem AI-Iraqi

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A rticl e 3 /2

Syst em Iso lation and th e Fr ee -Body Dia gr am

to the surfaces and is compressive . Although no actual surfaces are perfectly smooth, we can assume thi s to be so for practical purposes in many instan ces. When mating sur faces of contacting bodies are rough , as in Example 3, the force of contact is not necessarily normal to the tangent to the surfaces, but may be resolved into a tangential or frictional component F and a normal component N . Example 4 illustrates a number of forms of mechani cal support which effec tively eliminate tangential friction forces. In these cases the net reaction is normal to the supporting surface. Example 5 shows th e action of a smooth gu ide on th e body it supports. Th er e ca nnot be any resistance parallel to th e gu ide. Example 6 illustrates the action of a pin connection . Such a connecti on can support force in any direction normal to the axis of the pin. \Ve usually represent thi s action in terms of two rectangul ar components. The correct se nse of these components in a specific problem depend s on how the member is loaded. When not otherwi se init ially known , the se nse is arbitrarily assigned and the equilibrium equations are th en written . If the soluti on of these equation s yields a positive algebraic sign for the force component , the assigned sense is correct. A negat ive sign indicates the sens e is opposite to that initially assigned. If the joint is free to turn about the pin, the connect ion can support only the force R . If the joint is not free to turn, the connect ion can also support a resist ing couple M . The sense of M is arbitrarily shown here, but the tru e sense depen ds on how the memb er is loaded. Example 7 shows the resu ltants of the rather complex distribut ion of force over the cross section of a slender bar or beam at a built-in or fixed support. Th e sense of the reactions F a nd V a nd th e bending coup le M in a given problem depends, of course, on how the member is loaded. One of the most common forces is that due to gr avitational att raction, Exa mple 8. This force affect s all eleme nts of mass in a body a nd is, ther efore, distributed throughout it. T he resu ltant of the gravi ta tional forces on all eleme nts is the weight IV = mg of th e body, which passes through th e center of mass G and is directed toward the center of th e eart h for ea rt hbound stru ctures. Th e location of G is frequently obviou s from th e geometry of the body, particularly where th ere is symmetry . When th e location is not readily ap parent, it mu st be determined by experiment or calculations. Simil ar remarks apply to the remote action of magnetic and electric forces. These forces of remote action have the same overall effect on a rigid body as forces of equa l magnitude a nd directi on applied by direct external contact. Example 9 illustrates the action of a linear elastic spring and of a nonlin ear spring with either hardenin g or softenin g characte ristics. The force exerted by a linear spring, in tension or compression , is given by F = kx , where k is th e stiffness of the spr ing a nd x is its deformation measured from the neutral or undeformed position . The represent ation s in Fig. 3/ 1 are 1I0t free-body diagrams , but are merely elements used to constru ct free-body diagrams. Study these nine conditions and identify them in the pro blem work so that you can draw th e correct free-body diagrams. MarNan and W aseem AI-Iraqi

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107

lOB

Chapt e r 1

Eq u ili b ri um

Construction of Free-Body Diagrams

Th e full procedure for dr awing a free-body diagr am which isolat es a body or syste m consists of the following steps. Step I. Decide which system to isolate . The syste m chose n should us ually involve one or more of the desired unknown quantities. Step 2. Next isolat e the chosen system by drawing a diagram which rep resents its compl ete external boundary. This boundary defines the isolation of the system from all othe r attracting or contacting bodies, which are considered removed. This step is often the most crucial of all. Make certain that you have completely isolated the system before proceeding wit h the next step. Step J . Ident ify all forces which act on the isolated syst em as applied by th e removed contacting and attracting bodies , an d repre sent them in their pr oper positio ns on th e diagra m of the isolated system. Make a systematic traverse of the entire boundary to identify all contact forces . Include body forces such as weights, where appreciable. Represe nt all known forces by vector arrows, each with its proper magnitude , direction, and se nse indicated. Each unknown force should be represe nted by a vector arrow with the unknown magnitude or direction indicated by symbol. If th e sense of the vector is also unknown , you must arbitra rily assi gn a sense. Th e subseque nt calculatio ns with the equilibrium equat ions will yield a positive quantity if the correct se nse was assumed and a negative quanti ty if the incorrect sense was assumed. It is necessary to be consistent with the assigned characteristics of unkn own forces throughout all of the calculations. If you a re consiste nt , the solutio n of the equilibrium equations will reveal the correct se nses. Step 4. Show the choice of coordina te axes dir ectly on th e diagram. Pertinent dimensions may also be represented for convenience . Note, however, that th e free-body diagram serves th e purpose of focusin g attention on the actio n of the external forces, and th erefore the diagram shou ld not be cluttered with excessive extraneous information. Clearly distinguish force arrows from arrows represe nting quantities othe r than forces. For this purpose a colored penc il may be used .

Complet ion of t he foregoing four steps will produce a cor rect freebody diagram to use in applying th e governing equations , both in statics an d in dynamics. Be careful not to omit from t he free-body diagr am certain forces which may not appear at first glance to be needed in the calcu lation s. It is only through complete isolation a nd a syste matic rep rese ntatio n of all external forces th at a reliable accoun t ing of th e efTects of all a pplied an d react ive forces can be made. Very ofte n a force which at first glance may not appear to influence a desired result does indeed ha ve a n influe nce. Thus, th e only safe procedure is to include on th e free-body diagram all forces whose magnitudes are not obvious ly negligible. Th e free-body method is ext remely important in mecha nics because it ensures an accurate definition of a mechan ical syste m and focuses MarNan and W aseem AI-Iraqi

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Articl e 1/2

Syst em Isola tion and the Fre e -Body Diagram

atte ntion on the exact meanin g and application of the force laws of statics and dynamics. Review the foregoing four steps for constructi ng a free-body diagram while stu dying t he sam ple free-body diagrams shown in Fig. 3/2 a nd the Sa mple Problems which appea r at t he end of th e next article . Examples of Free-Body Diagrams

Figure 3/2 gives four exam ples of mechan isms and st ruc tures together with thei r correct free-body diagrams. Dimensions and magnitudes are omitted for clarity. In each case we treat the entire system as a single body, so that the internal forces are not sho wn. The cha rac te ristics of the various types of contact forces illustrated in Fig. 3/ 1 are used in the four exa mples as th ey apply. SA.\ \ PLE FREE-BODY DIAGRAMS

Mechanical System

Free-Body Diagram of Isolated Body

1. Plane trus s Weight of t ru ss assumed negligible compared with P k ---)(

p

A. 2. Cantileve r beam

'\' IA

F

!

2

V

FJ

Mas s m

~ .II

'\"

F

t

W = mg

3. Beam Sm ooth s urface contac t a t A. Mass m

p

!

2

FJ

y I I

L __ x

~" ~I \.~ Bx tw = mg

I

L __ .r

B.vT 4 . Rigid sy stem of in terconn ected bodies analyze d as a single un it

p

~

We ight of mec hanism neglected

m A

y

8

n fA . 8,-:.1

Figure 1/2 MarNan and Waseem AI-Iraqi

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L __ x

W=mg

109

110

Chapte r 3

Equilibrium

In Example 1 t he truss is composed of structu ra l elements wh ich, taken all togeth er, constitute a rigid framework. Thus, we may remove the entire truss from its supporting foundation and treat it as a single rigid body. In addition to the applied external load P, the free -body diagram must include the reactions on the truss at A and B. The rocker at B can support a vertical force only, and th is force is transmitte d to the st ruc t ure at B (Example 4 of Fig. 3/1). Th e pin connecti on at A (Example 6 of Fig. 3/l) is capa ble of su pplying both a hori zon tal a nd a verti ca l force component to th e truss. If t he total weight of th e truss members is apprecia ble compared wit h P and t he forces at A and B, then t he weig ht s of the mem bers mu st be includ ed on t he free -body d iagr am as external forces. In this r elatively simple exa mple it is clear that th e ver tica l component A v must be directed down to prevent the truss from rotatin g clockwi se' about B. Also, the hori zontal component Ax will be to th e left to keep the truss from movin g to the righ t under th e influe nce of th e hori zontal component of P. Thus, in cons truct ing t he free -body diagr a m for this simple truss , we can eas ily perceive the correct se nse of each of the comp onen t s of force exe rted on t he t russ by the fou ndation at A and can, therefore, represent its correct physical sense on the diagram. Wh en the correct physical se nse of a force or its component is not eas ily rec ognized by direct observation, it must be assigne d arbitrarily, and the correctnes s of or error in the ass ignment is determined by the algebraic sign of its calculated val ue. In Example 2 the cantilever beam is secure d to the wall and su bj ect ed to three applied loads. When we isolate th at pa rt of the beam to th e r ight of the sectio n at A , we must include t he re act ive forces applied to th e beam by the wall. The resu lt ants of th ese react ive forc es are shown acting on th e sect ion of th e beam (Example 7 of Fig. 3/ 1). A ver tica l force V to counteract the excess of downward applied force is shown, and a tension F to balance th e excess of applied force to t he r igh t must also be included. Then, to prevent the beam from rotat ing about A, a coun terclockwise cou ple M is a lso required, The weight mg of t he beam must be re presented through the mass cen ter (Example 8 of Fig. 3/1). In th e free-body diagr am of Example 2, we have represent ed the somewhat complex system of forces which actually act on the cut section of the beam by the equ ivalent for ce-couple sys te m in whi ch the force is broken down into its verti cal compon ent V (shear for ce) and its ho rizo nta l compone nt F (te ns ile force). The coup le M is t he bending mom en t in the beam . The free-body diagram is now complete and shows th e beam in equilibrium under the actio n of six forces and one couple . In Exa mple 3 the weigh t W = mg is shown acti ng t hrou gh th e center of mass of the beam, whose location is ass umed known (Example 8 of Fig. 3/ 1). The force exe rte d by the corner A on th e beam is normal to th e smoot h surface of th e beam (Example 2 of Fig. 3/1). T o perceive this action more clearly, visualize an enlargeme nt of the contact point A, which would appear somewhat rounded, and cons ider the force exerted by this ro un ded corner on th e straight surface of the beam , which is ass umed to be smoo th. If the contacting surfaces at the corner were not smooth, a tangential frictional component of force could exist. In addition to the applied force P a nd cou ple M , t here is t he pin connection Ma rwan and W aseem AI-Iraqi

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Articl e 3 / 2

System I so lati on an d t h e Fr e e-Bod y D i a g ra m

at B, which exerts both an x- and a y-component of force on the beam. The positive sense s of these components are assigned arbitrarily. In Example 4 t he free -body diagram of t he ent ire isolated mechanism contains three unknown forces if the loads mg and P are known . Anyone of many internal configurations for securing the cable leading from the mass m would be possible without affecting the ext ernal response of t he mechanism as a who le, and t h is fact is brough t ou t by th e free-body diagram . This hypot heti cal exa mple is used to show t hat t he forces internal to a rigid assembly of members do not influence the values of the external react ions. We use t he fr ee-body diagr am in writing t he equ ilib riu m equatio ns, which are discussed in the next article. When these equations are solved, some of the calculated force magnitudes may be zero. This would ind icate t hat t he assu med force does not exist. In Example 1 of Fig. 3/2, any of the reactions Ax, Av' or B v can be zero for specific values of the truss geome t ry and of t he m agn itude, direction, and sense of th e applied load P. A zero react ion force is oft en difficu lt to iden t ify by in spection , bu t ca n be det ermined by solving t he equilibriu m equation s. Simila r comments apply to calculate d for ce magnitud es which are negative. Such a result indicates that the actual sense is the opposit e of the assu med se nse. The assumed positive sen ses of Bx and By in Example 3 and B v in Example 4 are shown on the free-body diagrams. The correctness of these ass umptions is proved or disproved according to whether the algebraic signs of the computed forces are plus or minu s when the calculations are carried out in an actual problem . The isolation of the mechanical system under consideration is a crucial step in t he formulation of t he mathematical model. Th e most important asp ect to t he cor rect const ruction of t he a ll-important free-body diagram is the clear-cut and unambiguous decision as to what is included and what is excluded. This decision becomes unambiguous only when t he bou nda ry of the fre e-body diagram repre se nts a compl ete t raverse of th e body or sys te m of bodie s to be isolated , start ing at some arbit ra ry point on t he bou ndary and returning to t ha t sa me point. T he system withi n t his closed boundary is t he isolated fre e body, and a ll contact forces a nd a ll body forces tran smi tted to th e sys te m acr oss t he bounda ry must be accounted for. The following exercises provide practice with drawing free-body dia grams. This practice is helpful before using suc h diagr ams in the applicat ion of the principles of force equilibrium in the next article.

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111

112

Chapter 1

Equili brium

FREE-BODY DIAGRAM EXERCISES 3 jA In each of the five following exam ples, th e body to be isolated is shown in t he left-hand diagram , and an incomplete free-body diagram (FBD) of the isolated body is shown on the right. Add whatever forces are nee-

essary in each case to form a complete free -body diagram. Th e weights of t he bodies are negligible unless otherwise indicated. Dimen sions and numerical values are omitted for simplicity.

Incomplet e FBD

Body 1. Bell crank

su pporting ma ss m with pin s upport atA . Pu ll P 2. Control lever a pplying torque t o sha ft at O.

3. Boom OA. of negligib le mass compared with mass m . Boom hinged at 0 and suppo rted by hoisti ng cable at B .

T~

mg

o

4. Un iform crat e of

ma ss m lean ing agains t smoot h verti cal wall and s upport ed on a rough horizonta l surface .

B

5. Load ed br acket su pported by pin connection at A and fixed pin in smo ot h slot at B.

A

Figure l /A

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BY\L

Fr e e-Body Diag ram Exerci se s

11 3

[

3/B In eac h of the five followin g examples , the body to be isolated is sh own in the left -ha nd diagram, and either a wrong or an incomp lete free-body diagram (F BD) is shown on the right. Mak e wh atever ch anges or addi-

tions are necessary in each case to form a correc t a nd comp let e free-bo dy diagr am . T he weig hts of the bodie s are negligible un less otherwise indicated. Dimen sions and numerical values are omitted for simplicity.

Wrong or Incomplete FBD

Body

~P

~--

1. Lawn roller of

ma ss m being pushed up inclin e B.

__

mg

N

2. Prybar lifting body A ha ving smoot h hori zont al sur face. Ba r rests on horizont al rough surface. 3. Uniform pole of mass m being hoisted into position by winch. Horizontal support ing surface notched to prevent slipping of pole.

Notch / F

4. Support ing angle bracket for fra me; Pin joints.

r----\",I B

A • F

5. Bent rod welded to support at A and subjected to two forces an d couple.

A y

I I

M

~

L __ x P Figure 3/B

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F

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114

l Ie

Chapt er 3

Equilibrium lab eled . (N ote: The se nse of so me reac t ion com pon ents ca n not a lways be det ermined without numerical

Draw a complete and correct free-bod y diagram of each of the bodies desi gn ated in the state me nts . The weights of the bodi es a re sign ifica nt onl y if the mass is sta ted. All for ces, kn own a nd unknown , shou ld be

1. Uniform horizontal bar of mass m sus pended by vert ical ca ble at A and

supporte d by rough inclin ed surface at B .

2. Wheel of mass "I on verge of bein g rolled over curb by pull P .

P

calculatio n.I

5. Uniform grooved whe el of mass ", su pported by a rough surface a nd by act ion of horizont al ca ble.

6. Bar , init ially horizont a l bu t deflected und er load L. Pinn ed to rigid su pport at eac h end .

E

3. Load ed truss supported by pin joint at A a nd by ca ble at B.

7. Uniform heavy plat e of mass m supported in vert ical plane by cable C a nd hinge A .

B

A

4. Uniform bar of mass m and roller of mass "1 0 ta ken toget her . Subjected to couple At a nd su pport ed as shown. Roller is free to tum .

8. Entire fra me, pull eys, a nd contact ing cable to be isolated as a single un it.

Tn.

A •

Figure 3/C

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Tn

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Arti cle 3/3

3 /3

EQUILIBRIUM CONDITIONS

In Art. 3/ 1 we defined equilibrium as the condition in which the result ant of all forces and moments acting on a body is zero. Stated in a nother way, a body is in equi librium if all for ces and mo ments applied to it are in balance. These requirements are contained in the vecto r equati ons of equilibrium, Eq s. 3/1, which in two dime nsion s may be written in scalar form as "YM o =

0)

(3 /2 )

The thi rd equation represe nts the zero sum of the moments of all forces about a ny point 0 on or off the body. Equations 3/2 are t he nece ssary and sufficient conditions for complete equilibrium in two dimensions. They are necessary conditions because, if they are not satisfied, there can be no force or moment balance. They are sufficient because once they are satisfied, there can be no imbalance, and equilibrium is assured. The equat ions relating force and acceleration for rigid-body motion are developed in Vol. 2 Dynamics from Newton 's second law of motion. T hese equatio ns show that the acceleration of the mass center of a body is proportional to the resultant force ~ F acting on the body. Conse quently , if a body moves with constant velocity (zero acceleration), t he resu ltant force on it mu st be zero, and the body may be t reated as in a state of translational equilibrium. Forcomplet e equilibrium in two dim ension s, all threeofEqs. 3/ 2 must hold. However, these conditions are independent requirement s, and one may hold without another. Take, for example, a body which slides along a horizonta l surface with increasing velocit y under the action of applied forces. The force-equilibrium equations will be satisfied in the vertical direction where the acceleratio n is zero, but not in the horizontal direction. Also, a body, such as a flywheel, whic h r otates about its fixed mass center with increasing angular speed is not in rotational equilibrium, but t he two force-equil ibrium equations will be sati sfied. Categories of Equilibrium

App lications of Eqs. 3/2 fall naturally into a numb er of categories which are eas ily ide ntified. T he categories of force systems acting on bodies in two-dime nsio nal equilibrium are summarized in Fig. 3/3 and are explained further as follows . Ca tegory 1, equilibrium of collinear forces, clearly requires only the one force equation in the direction of the forces (r -dire ction), since all other equatio ns are automat ically satisfied. Ca tegory 2, eq ui libri um of forces whic h lie in a plane (x-y pla ne) and are concurrent at a point 0 , requires the two force equations only, since the mom ent sum abo ut 0 , that is, about a z-ax is through 0 , is necessarily zer o. In cluded in thi s category is the case of the equ ilibri um of a particle. Ca tegory 3, equilibr ium of parallel for ces in a plan e, req uire s th e one force equatio n in the direction of the forces (r -direct ion ) and one moment equatio n about an axis (a-axis ) norm al to the pla ne of the for ces. Marwan and Waseem AI-Iraqi

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Equ i libr iu m Con d it ion s

115

116

Chapter 1

Equil ibr ium

CATEGORI ES OF EQUI LIBRIUM IN TWO DIM ENSIONS Force Sys tem 1. Collinea r

Free-Body Diagram

Independent Equations

~--

-"

LFx = 0

F~ 2. Concurrent at a point

n",",

F, 1' .

-,

-,

, .>

/f'

y

I I

F,

L _ _x

X

F. ~ Fa

y

3. Parallel

_ F, _ a F _-

4. General

y

I

Ft

LFx = 0 'L Fy = 0

LFx = 0

LMz

=0

LFx = 0

LAll

=0

I

L _ _x F,

/

F, Fa Y I I

M)

t

'LFy = 0

L __ x

F,

Figure 1/1

Ca tegory' 4, equilibrium of a general system of forces in a plane (r -y ), requi res th e two force equations in the plane and on e momen t equa t ion about an axis (z-axis) normal to the plan e. Two- and Three-Force Members p

-p

-p Two-force mem bers Figure 1/4

Marw an and W aseem AI-I raqi

You shou ld be alert to two frequ en t ly occur ring equilibr iu m situa tions. The first situation is the equilibrium of a body under the act ion of two forces on ly. Two examples are shown in Fig. 3/ 4, and we see that for such a two-force member to be in equilibrium, the forces must be equal, opposite, a nd collinear . The s hape of t he memb er does not affect this simple requiremen t. In the illustrations cited, we cons ider the weights of the members to be negligible compared wit h th e a pplied forces. The second situation is a three-force member , which is a body under the action of three forces, Fig. 3/00. We see that equil ibrium requires the lines of action of the t hree forces to be concurrent. If t hey were not concurrent, then one of the forces would exert a resultant moment about th e point of intersection of the ot her two, which would violate the requirement of zero moment about every point. The only exception occurs when the three forces are para llel. In th is case we may consider the point of concurrency to be at infinity . www .gigapedia.c om

Article 3 / 3

Equ i librium Cond itio n s

Th e pri nciple of t he concurrency of three forces in equilibrium is of cons iderable use in carrying out a graphical solution of the force equation s. In this case th e polygon of forces is drawn and made to close, as shown in Fig. 3/ 5b. Frequently, a body in equilibrium under the action of more than three forces may be reduced to a three-force mem ber by a combination of two or more of the know n forces. Alternative Equilibrium Equations

F3

In addition to Eqs. 3/2, t here are two ot her ways to express t he general condi tions for th e equilibrium of forces in two dimen sions. Th e first way is illustrated in Fig. 3/ 6, parts (a) a nd (b ). For the body shown in Fig. 3/6a, if"i.MA = 0, the n t he resul tan t, if it still exists , cannot be a couple, but must be a force R passing through A . If now th e equation ";;;F x = 0 holds, where the x-direction is arbitrary, it follows from Fig. 3/ 6b that th e resultant force R, if it st ill exists, not only must pass through A , but also must be perpendicular to the x-direction as shown . Now, if ";;;AlB = 0, where B is any point such th at th e line AB is not perpendicular to th e x-direction, we see that R must be zero, and thus th e body is in equilibrium. Therefore, an alternative set of equilibrium equations is "i.MA

=

0

where the two poin ts A a nd B must not lie on a line perp endi cular to t he x-direction. A third formu lat ion of the equilibrium conditions may be made for a coplanar force syste m. Thi s is illustrated in Fig. 3/ 6, parts (c ) a nd (d ). Again, if ";;;M A = 0 for any body such as th at shown in Fig. 3/6c, th e resulta nt , if a ny, must be a force R through A . In additio n, if "i.MB = 0, th e resultant, if one still exists, must pass through B as shown in Fig. 3/6d . Such a force cannot exist, however, if "i.Mc = 0, where C is not

I .\fA = 0 sa tis fied

--- ~ •

__ x

B

A

(a)

(b)

l:MA=O} . fi d LAIn = 0 satls Ie

LAtA = 0 s atis fied

1 (e)

(d)

Figure3/6 Marwan and Wa see m AI- Iraqi

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(u )

(b )

Three-force member

Clas ed polygaa satis fies I F = 0

Figure3/5

117

118

Chapt er:}

Equ ilibrium

collinea r with A and B. Thus , we may write the equations of equilibrium as ":::M c = 0 where A , B , and C ar e an y three points not on the same stra ight lin e. When equilibrium equations are written which are not independent, redundant information is obtained, and a correct solution of the equat ions will yield 0 = O. For example, for a genera l problem in two d imens ions with three unknowns, three moment equations written about three points which lie on the same straight line are not indepe ndent. Such equations will conta in duplicated information, and solution of tw o of th em can at best det erm ine two of t he u nknowns , with the third equat ion merely ver ifying the identity 0 = o.

Constraints and Statical Determinacy The equilibr ium equations developed in this arti cle a re both necessa ry a nd su fficient condit ions to establish the equilibr ium of a body. However, they do not necessarily provide all the informat ion required to calculate all the unknown forces whi ch may act on a body in equ ilibrium. Whether the equatio ns are adequate to det ermine a ll th e unknowns depends on the characteristics of the constra ints against possible movement of the body provided by its su pports. By constraint we mean the restriction of movement. In Example 4 of Fig. 3/ 1 the roller , ba ll, and rocker provide constraint normal to the surface of contact, but none tangent to the surface. Thus, a tangential for ce cannot be su pported. For the collar a nd slide r of Example 5, cons traint exis ts only norm al to the guide. In Example 6 the fixed-pin connection provides constraint in both directions, but offers no resistance to rotation about the pin unless the pin is not free to turn. T he fixed su pport of Example 7, however, offers constraint agai nst rotation as well as lateral movement. If th e rocker which supports t he truss of Example 1 in Fig. 3/2 were replaced by a pin joint, as at A , th ere would be on e additiona l constraint beyond those required to support an equilibrium configuration with no freedom of movement. The three scalar conditions of equilibrium, Eq s. 3/ 2, would not pro vide sufficient information to det ermine all four unkn owns, since Ax and Bx could not be solved for se parate ly; only their sum could be determined. These two components of force would be de pen den t on th e defor mation of th e membe rs of t he t russ as influenced by their corresponding stiffness properties. The horizonta l reactions Ax and Bx would also depend on any initial deformation required to fit th e dimensions of th e stru cture to those of t he foundation between A and B. Thus, we ca nnot det ermine Ax and Bx by a rigid -body analysis. Again referring to Fig. 3/2, we see that if the pin B in Examp le 3 were not free to turn , the support could tra nsmit a couple to the beam t h ro ugh th e pin . Therefor e, th ere would be four unknown sup port ing reactions acting on the beam, namely, the force at A, the two componen ts of force at B, an d the cou ple at B. Conseq ue ntly th e t hree indeMa rwan and W aseem AI-I raqi

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Articl e 3 / 3

pendent scalar equations of equilibrium would not provide enough information to compute all four unknowns. A rigid body, or rigid combi nat ion of eleme nts treated as a single body, which possesses more external supports 0 1' const raints than are necessary to maintain an equilibrium position is called statically indeterminate. Supports which can be removed without destroying the equilibrium condition of the body are said to be redunda nt . The number of redundant supporting elements present corresponds to the degree of statical indeterminacy and equals the total number of unk nown external forces, minus the number of available independent equat ions of equilibd um. On th e oth er hand, bodies which are su pported by the mi nim um number of constraints necessary to ensure an equilibrium configuration are called stati cally determinate , and for such bodies the equilibrium equat ions are sufficient to determine the unknown external forces. T he prob lem s on equilibr ium in this article and throughout Vol. 1 S tatics are gene ra lly rest r ict ed to statically deter minate bodies where the const raints are just sufficient to ensure a stable equilibrium configuration and where the unknown supporting forces can be completely determ ined by the avai lable inde pen den t equations of equilibrium . We must be aware of the nature of the constraints before we atte mpt to solve an equilibr ium problem . A body can be recogni zed as stat ically indeterminate when there are more unknown exte rnal reactions than there are available independent equilibrium equat ions for the force sys tem involved. It is always well to coun t the nu mber of u nk now n va riables on a given body an d to be cert ain that an equal numb er of independent equations can be written; otherwise, effort might be wasted in attempt ing a n impo ssible solution with the aid of the equilibriu m equations only. The unknown variables may be forces, couples, distan ces, or angles.

Equil ibr ium Conditio n s

(u )

Complet e fixity Adequa te constraints

Adequacy of Constraints In discussing the relationship between const raints and equilibrium, we should look further at the question of t he ade quacy of con stra ints. The existence of three constraints for a two-dimensional problem does not always guarantee an equilibrium configuration. Figure 3/ 7 shows fou r differe nt types of cons tr aints. In part a of t he figu re, point A of th e r igid body is fixed by the two links and can not move, a nd the thi rd link prevents any rotat ion about A . Thu s, this body is completely fixed with three adequ ate (prope r ) const rai nts. In part b of the figu re, the third link is positioned so t ha t th e force tra nsmitted by it passes t h rough point A where t he other tw o cons t raint forces act. Thus, this configuration of constraints can offer no initia l resistance to rotatio n about A, which would occur when exte rnal loads were applied to th e body. We conclude, the refore , tha t this body is incompletely fixed under pa rtial const raints. The configuration in part c of the figure gives us a similar condition of incomplete fixity because the three para llel lin ks could offer no initial resistance to a small vertical movement of the body as a result of external loads app lied to it in t his dire ct ion . T he constra ints in thes e tw o examples are often termed imp roper . Marwan and Waseem AI-Iraqi

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(bl Incomplete fixity Partial constraints

(c)

Incomplete fixity Partial cons traints

(dJ Excessive fixity

Redundant cons traint

Figure 3/7

11 9

120

Chapter 3

Equ il ibrium

In part d of Fig. 3/ 7 we have a condition of complete fixity, with link 4 acting as a fourth constraint which is unnecessary to maintain a fixed positi on. Link 4, the n, is a redu ndant constraint , and th e body is statically indeter minate. Ai; in the four examples of Fig. 3/7, it is genera lly possible by dir ect observation to conclude whethe r the constraints on a body in twodimensional equilibrium are adequate (proper), partial (improper), or redun dan t . Ai; indicated pr eviously, the vast majority of problems in this book are statically det ermi nate with adequa te (proper) cons t raints.

Approach to Solving Problems The sam ple pr oblems at the end of this article illus trate th e application of free-body diagrams and the equations of equilibrium to typical statics pr oblems. Th ese solutions should be studied th oroughly. In th e problem work of this cha pter an d throu ghout mechanics, it is important to develop a logical a nd syste mat ic approach which includes the following steps: 1. Ident ify clearly th e quan tities which are known a nd u nkn own.

2. Mak e an un ambiguous choice of the body (or syste m of connected bodies treat ed as a single body) to be isolated and dr aw its complete free -body diagram , labelin g all exte rnal known and unknown but identifi able forces and couples which act on it. 3. Choose a convenient set of reference axes , always usin g righthanded axes when vector cross products are employed. Choose momen t centers with a view t o simplifying the calculatio ns. Gen erally the best choice is one through which as many u nknown forces pass as possible. Simultaneous solut ions of equilibriu m equations are frequen tly necessary, bu t can be minimized or avoid ed by a careful choice of reference axes and moment centers. 4. Identify and state the appli cable force and moment principles or equa tions which govern th e equilibrium conditio ns of th e problem . In th e following sample pr oblems these relations are shown in brackets an d pr ecede each major calcu lati on. 5. Mat ch the number of independent equations with th e number of unknowns in each problem. 6. Carry out th e solution and check the results. In many problems engineering judgment can be developed by first mak ing a reasonable guess or estimate of the result pri or to th e calculat ion and th en comparing th e estimate with the calculated value.

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Article 3/3

Equil ib rium Conditions

Sample Problem 3 /1

y I

Determine the magnitudes of the forces C and T, which, along with the other three forces shown, act on the brid ge.tru ss joint.

CD

I

c

Solution. Th e given sketch constitutes the free-body diagram of the isolated

y~

section of th e join t in que stion and shows th e five forces which are in equi librium.

I

,

T -,

16kN

Solution I (scalar algebra). For the x-y axes as shown we have [LFx

= 0]

8 + T cos 40· + C sin 20· - 16

(a )

T sin 40" - C cos 20" - 3 = 0 0.643T - 0.940C

Simultaneous solut ion of Eqs. (a) and

T

~

9.09 kN

(b )

(b )

produces

Ans.

C = 3.03 kN

axes x' ~y ' with the firs t su mma tion in th e y ' -direction to elimi nate refe rence to

@ T. Thus,

0]

II

I

\\11 1

/

/

\ 1,1"/

/

- ~~--

40°

8 kN

--x

- C cos 20" - 3 cos 40" - 8 sin 400 + 16 sin 40" = 0 C

~

T

+ 8 cos 40" - 16 cos 40" - 3 sin 40" - 3.03 sin 20°

Helpful Hints

CD Since this is a prob lem of concu rrent

=3

Solution II (scalar algebra). To avoid a simultan eous solu tion, we may use

[LFy ' = 0]

\

=0

0.766T + 0.342C = 8

OJ

121

3.03 kN

forces, no necessary.

momen t

equation

is

@ The select ion of reference axes to Iacilitate computati on is always an important consideration. Alternatively in this exa mple we could tak e a set of axes along and normal to the direction of C and emp loy a force summati on normal to C to eliminat e it.

An s. = 0

Ans.

T = 9.09 kN

Solution "' (vedor algebra). With unit vectors i and j in t he x - an d y -dir ectio ns , the zero summation of forces for equilibrium yields th e vector equ ation [LF = OJ

8i + (T cos 40·)i + (T sin 40·)j - 3j + (C sin 20·)i - (C cos 20· )j - 16i

o

Equ ating th e coefficients of the i· and j -ter ms to zero gives 8 + T eas 40· + C sin 20· - 16

0

T sin40· - 3 - Ccos20·

0

which are the sa me, of course , as Eqs. (a) and (b) , wh ich we solved above.

@

Solution IV(geometric). Th e polygon representing the zero vector sum of the five forces is shown. Equations (a) and (b) are seen imm ediately to give the projection s of th e vecto rs onto the ,X- and y-directions. Simil arly, projections onto the x '- and y ' -directions give the alte rn at ive equations in Solution II. A graphical solution is easily obtained. The known vectors are laid off headto-tail to some conven ient scale, and the directions of T and C are then drawn to close t he polygon . The r esult ing inte rsec tion at point P completes the solut ion, thu s enabling us to measure th e magnitu des of T and C directly from the drawing to whatever degree of accuracy we incorporate in th e construction. Marwan and Waseem AI-Iraqi

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8 kN 3 kN

@ Th e known vectors may be add ed in any order desired , but they mu st be added before the u nknown vectors .

122

Chapte r 3

Equ ilibrium

Sample Problem 3 /2

T

~ = 30'

Calculate the ten sion T in th e cable which supports t he lOOO·lb load with the pulley arra ngement shown. Each pulley is free to rotate about its beari ng, and t he weights of all parts ar e small compared with the load. Find th e magn itu de of the total force on t he bearing of pulley C.

B

Solution. The free -body diagram of ea ch pull ey is drawn in it s re lative position to t he othe rs. We begin with pulley A, which includes th e only known force. With the unspecified pulley rad ius designated by r, th e equilibr iu m of momen ts abou t its cen ter 0 and the equilibriu m of forces in the vertical direction requ ire

= 01

cW A Fx

T 1r - T 2r = 0

[LMo = OJ [ ~Fy

1000 1b

T, + T 2

1000

-

~

1
0

From the example of pulley A we may write the equilibr ium of forces on pulley B by inspection as

To = T.

~

'~;:

T 2 / 2 = 250 Ib

For pulley C the angle (} = 3OC' in no way affects the moment of T about th e cente r of t he pulley, so th at momen t equilibri u m requires T

~

Ts

or

T

~

250 1b

TI~T, a

An s .

IL __ - x

A

Equilibrium of the pu lley in the x· an d y-direcrions requires

[YF, [F

0]

250 cos 30" - Fx = 0

OJ

Fy + 250 sin 30" - 250 = 0

~ jr=F"/ ;-+- "Fy"2" j

F

~

10001b

217 1b

Fy

) (217)2 + (125)2

Helpful Hint

125 1b

250 1b

Ans.

CD Clearly th e radi us r does not influence t he resu lts. Once we have analyzed a simple pulley, the res u lts should be perfectly clear by inspection .

Sample Problem 3 /3 Th e uni for m IOO-kg I-beam is supported initially by its end roller s on the hori zontal surface at A and B . By mean s of the cable at C it is desired to elevate end B to a position 3 m above end A . Determine the required ten sion P, the reaction at A , an d the angle 8 made by the beam with the horizon tal in the elevated positio n.

P~ 6m

2m B

A

Solution. In constructing t he free-body diagram , we note th at th e reaction on the roller at A and the weight are vertical forces. Consequentl y, in the absence of ot her horizontal forces, P mu st also be vert ical. From Sample Problem 3/2 we see imm ediate ly that the ten sion P in the cable equals the ten sion P applied to the beam at C. Moment equilibrium about A elim inates force R and gives P(6 cos 8) - 981( 4 cos 8) = 0

P = 654 N

Ans.

Equ ilibrium of vertical forces requires

(YFy = OJ

654

+ R - 981 = 0

Ma rwan and Waseem A1·l raqi

- - -8- - - .100 - (9.81) - - -N- - -

A

:r

R

R = 327 N

Ans.

The angle 0 depends only on th e specified geometry and is sin 8 = 3/8

I3m

8 = 22.0'

CD Clearly the equilibriu m of thi s paraIAns.

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Helpful Hint lei force system is independent of 8.

Article 3 /3

Equilibrium Condi tions

123

Sample Problem 3 /4 Det er min e th e magn itude T of th e ten sion in th e supporti ng cable and th e magnitude of th e force on th e pin at A for the jib crane shown. Th e bea m AB is a sta nda rd O.5-m l-beam with a mass of 95 kg per meter oflengt h.

Algebraic solution. Th e syste m is symmetrical about the vertical x-y plan e

CD

through the center of the beam, so th e probl em may be analyzed as th e equilibrium of a coplanar force syste m. The free-body diagram of the beam is shown in th e figure wit h th e pin reaction at A repr ese nt ed in terms of its two rectangu lar compo nents. The weight of th e beam is 95(10 . 3 )(5)9.81 ~ 4.66 kN and acts through its center . Note th at there are three unknowns Ax, Ay , and T which may be found from th e three equations of equilibrium. We begin with a moment equation about A , which eliminates two of th e three unknowns from th e equ atio n. In applying th e moment equation about A, it is simpler to cons ider th e moments of the x- and y-componen ts of T tha n it is to compute th e perpe ndicular dis tance fro m T to A . Hence, with the counterclockwise sense as positiv e we write (T

10 kN

1-- - - - 5 m - - - --J

cos 25°)0.25 + (T sin 25°)(5 - 0.12) - 10(5 - 1.5 - 0.12)

4.66(2.5 - 0.12 1 ~ 0 Free-body diagram

T

from which

~

19.61 kN

Ans. Helpful Hints

Equ ating t he sums of forces in the .r- and y-direct ions to zero gives [~Fx

Ax - 19.61 cos 25°

= 0]

[ ~Fy ~

OJ

Ay

+ 19.61 sin 25° - 4.66 - 10

~

0

= 0

CD Th e just ification for th is step is Varig-

Ax = 17.77 kN A y = 6.37 kN

@ [A = J A/ + A/I A ~ J (l 7.77)2 + (6.37)2 ~ 18.88 kN

Ans.

Graphical solution. Th e principle that thr ee forces in equilibrium mu st be concu rrent is util ized for a gra phical solutio n by combin ing the two known ver tical forces of 4.66 and 10 kN into a single 14.66-kN force, located as shown on the modified free-body diagram of the beam in th e lower figur e. Th e position of this res ultant load may easily be determ ined gra phically or alge braically. Th e intersect ion of t he 14.66-kN force with th e line of action of th e unknown ten sion T defines th e point of concurre ncy 0 through which th e pin reaction A mu st pass. Th e un kn own magnitudes of T and A may now be found by adding t he forces head-to-tail to form th e closed equilibriu m polygon of forces, thus satisfying th eir zer o vector su m. After t he known vert ical load is laid off to a convenient scale, as shown in the lower part of th e figure , a line representing th e given direction of th e ten sion T is drawn through the tip of the 14.66-kN vector. Likewise a line representing the direction of the pin reaction A, deter mined fro m t he concur rency established with the free -body diagram, is drawn thro ug h the ta il of th e 14.66-kN vector. Th e intersect ion of the lines representing vectors T and A esta blishes th e magnitudes T and A necessary to make the vector sum of the for ces equal to zero. Th ese magnitudes are scaled from the diagr am . The x - and y -compo ne nts of A may be constructed on the force polygon if desired .

@

@

non' s theorem , explained in Art. 2/4. Be prepar ed to take full advantage of th is pri ncip le frequen tly. Th e calcu lation of moments in twodimen sional problems is generally handled more simply by scala r algebra th an by the vector cross product r x F . In th ree dim en sions, as we will see later, the reve rse is often the case . Th edi rection of t he force atA could be easi ly calcu lated if desired . However , in designing th e pin A or in checking its strength, it is only th e magnitude of t he force th at ma tters.

T Il' A...:"' "

_

Graphical solution Marwan and Wa seem AI-Iraqi

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124

Chapter 3

Equ ilib rium

PROBLEMS Introductory Problems 3 /1 The mass center G of the 1400-kg rear-engine car is

3/3 A carpenter holds a 12-lb g-in. by 4-in . board as shown. If he exerts vertical forces on the board, determine the forces at A and B . Ans. N A ~ 121b down , N B ~ 24 lb up

located as shown in the figure. Determine the normal force under each tire when the car is in equilibrium. State any assumptions. Ans. N r ~ 2820 N, N , ~ 4050 N B

6.---J Problem 3/1

Problem 3/3

3 /1 A carpente r carries a 12·1b 2-in. by 4-in. board as

shown. What downward force does he feel on his shoulder at A?

i s'

3/4 The 450-kg uniform I-beam supports the load shown. Determine the reactions at the supports.

f--- - - 5.6 m - - --+- 2.4 m

--1

A •

B

220 kg

Problem 3/4 1 /5 The 20-kg homogeneous smooth sphere rests on the

two inclines as shown. Determin e the contact forces at A and B .

Ans. N A

~

101.6 N, N B

Problem 3/2

Problem 3/5

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~

196.2 N

Articl e 3 /3

3/6 With wha t force magni tud e T must th e per son pull on th e cabl e in order to caus e the scale A to read 500 lb? The weights of the pulleys and cables are negligible. State any assu mpt ions.

Problem s

12 5

3/8 The 600-lb drum is being hoisted by th e lifting device which hooks over the end lips of th e drum. Det ermine t he ten sion T in each of t he equal -lengt h rods which form t he two U-shaped member s of th e device.

A

Problem 3/8

Problem 3/6

3/7 What horizontal force P must a worker exert on the rope to posit ion the 50-kg crate directly over th e trailer? Ans. P ~ 126.6 N

3/9 What fracti on n of the weight W of a jet airp lan e is the net thrust (nozzle thrust T minu s air resistance R ) in ord er for th e airplane to climb wit h a constant speed at an angle () with the hor izontal? A ns. n = sin fI R

1 ---+

4m

~

~

\

,

/

.......

2m

Iffii~1 ~

•

~B

Problem 3/9

3/10 Determine the force magnitud e P requi red to lift one end of t he 250.kg crate with th e lever dolly as shown. State any assu mpti ons.

QL 1m

Problem 3/7

B

Problem 3/10 Marwan and Waseem AI-Iraqi

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126

Chapte r 3

Equilib r ium

3 /11 Find th e a ngle of tilt tJ with the hori zont al so that th e contact force at B will be one-ha lf th at at A for th e smooth cylinder. A ns. II = 18 .43°

Problem 3/13 Prob le m 3/ 11

3/12 Determin e the magnitude P of the vert ical force required to lift the whe elbarrow free of th e gr ound at point B. The combined weight of th e wheelbarrow and its load is 240 Ib with center of gra vity at G.

3/14 T hree cables a re j oined at th e ju nct ion r ing C. Deter mine the ten sions in ca bles AC a nd Be caused by t he weight of t he 3D-kg cylinder. A

p

24"

30 kg

30'

j

B

19"

21" Problem 3/12

Problem 3/14

1 /13 To facilitate shift ing th e position of a lifting hook when it is not under load, th e sliding han ger shown is u sed . The project ion s at A and B e ngage t he fla nges of a box beam when a loa d is supported, an d the hook projec ts through a hori zon tal slot in t he beam . Compute th e forces at A a nd B when the hook

3/15 Th e 100-kg whee l rests on a rough su rface a nd bea rs aga ins t the roller A when the couple J.\J is applied . If M = 60 N . m a nd the whee l does not slip, compute th e rea ct ion a ll t he roller A. A il S. FA = 23 1 N

,,

s u ppo r ts a 300-kg mass.

Ans. A = 4.91 kN . B = 1.962 kN

, I

_' A

Prob lem 3/15

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Articl e 1 / 1 3/16 Th e un iform beam has a mass of 50 kg per meter of length . Compute th e rea ction s at the su pport O. The force loads shown lie in a vert ical plane.

Pr oble m s

127

Representative Prablems 3 /18 Determine the magn itude P of the force which the

man mu st exert perpendi cular to the handle of the high-pressure wash er in ord er to cause loss of contact at the front suppor t B. Note that the operator prevents movement of the whee l wit h his left foot. The 60-kg machine has its mass center at point G, Tr eat th e problem as two-dim ensional.

1.4 kN

c

o

Problem 1/16 3/17 To accommodate th e r ise and fall of the t ide, a walkway fro m a pier to a float is supported by two rollers as shown. If t he mass center of the 300-kg walkw ay is at G, calculate t he te nsion T in the hor izontal cable which is att ached to the cleat and find the force unde r t he roller a t A. A ns. T ~ 850 N. A ~ 1472 N mm mm

T

_ _

Problem 1/1 B 1/19 If the screw B of the wood clam p is tightened so that th e two blocks are under a comp ression of 500 N, determine the for ce in screw A. (No te: Th e force supported by eac h screw may be taken in the dir ectio n of th e scre w.) Ans. A ~ 1250 N

Problem 1/17

150 mm

100 mm

Problem 3/19

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128

Chaple r 3

Equ ilibrium

3/20 The u nifo rm 15-m pole has a mass of 150 kg and is supported by it s smoot h end s against t he vertical walls and by the tension T in t he vertical cable. Comput e the reactions at A a nd B.

-B

/~

180 m m

p

Problem 3/22 }<---

-

12 m - - - - I

Problem 3/20 3/21 Determine t he force P required to begi n rollin g th e uniform cylinder of mass m over the obstruction of height h. ~ mg hrh - h 2 Ans. P r h

3/23 The spring of modulus k ~ 3.5 kN /m is st re tc hed 10 mm when the disk center 0 is in the leftm ost position x = O. Determine t he ten sion T required to positio n t he disk cente r at x = 150 mm . At that positio n, what force N is exerted on the hori zontal slotted guide? The mass of the disk is 3 kg. Ans. T ~ 328 N, N ~ 203 N up

p

k = 3.5 kN/m

.h

I I

t

f- %-1

Problem 3/21 Problem 3/n 3/12 The elements of a heavy -duty fluid valve ar e shown in the figure . When t he memb er DB rotates clockwise abou t t he fixed pivot 0 u nder the actio n of the force P, th e element S slides free ly upward in its slot, releasing th e flow. If an int ernal torsional spring exerts a mom en t M = 20 N· m as shown, det ermine th e force P requ ired to open th e valve. Neglect all fr icti on.

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Articl e J /J 3/24 A block placed under th e head of th e claw hammer as shown gr eatly facilit at es the extraction of the nail. If a 50·1b pu ll on t he handle is required to pull th e nail , calculate th e tension T in th e nail and t he magnitude A of the force exerted by the hammer head on the block. The cont acting su rfaces at A are su fficiently rough to prevent slipping.

Pro blems

129

3/26 To test th e validity of aerodyna mic ass u mptions mad e in th e design of the aircraft , its mode l is being tested in a wind tu nnel. Th e sup port bracket is connected to a force and mom ent ba lanc e, which is zeroed when there is no airflow. Und er test condition s, t he lift L, drag D, an d pitching momen t MG act as shown. Th e force balance records th e lift, drag, an d a momen t M p . Det ermine M G in te rms of L, D , and M p .

- - d---j ~ ~ ~

Airflow

=:::: ~ ~ ~

•

-'---------r; . ~ •

•

Problem 3/26

Problem J/2 4 3/25 Th e indicat ed location of the center of gravity of th e 3600-1b pickup truck is for th e un laden condition. If a load whose center of gravity is x = 16 in. behind th e rear axle is added to the truck, determine the load weight WL for which th e normal forces under the front and rear wheels are equal. Ans. WL = 550 1h

1 /27 In a procedure to evaluate the str engt h of the triceps mu scle, a person pushes down on a load cell wit h the palm of his hand as ind icated in the figure. If the load -cell reading is 35 Ib, determine t he vertical te nsile force F gene ra ted by th e t riceps mu scle. Th e lower arm weighs 3.2 lb with mass center at G. Sta te an y ass umptions . Ans. F = 401 Ih

---'e-

-

Humeru s Tr iceps

Ulna Hand

G

~1"

Problem 3/25

Problem J /27

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130

Chapter 3

Equilib rium

3/28 A person is performing slow arm curl s with a 20-1b

weight as indicat ed in the figure. The brachia lis mu scle gro u p (consisting of the biceps and brn chiali s mu scles) is th e major factor in th is exercise. Deter mine the magnitude F of the brach ialis-muscle-group force a nd the magnitude E of th e elbow joint react ion at point E for th e forea rm posit ion shown in t he figur e. Take th e dim ensions s hown to locat e th e effective points of applicat ion of th e two mu scle groups; th ese points are 8 in . directly abov e E and 2 in. directly to the righ t of E. Include the 3.2-lb for earm weight which acts at point G. State any assumptions.

.I'

I

L __

r - - - - 25 Bra ch iali s

,/

I

.4"U

3/30 With his weigh t W equally distributed on both feet,

a man begin s to slowly rise from a squatt ing- position as indi cat ed in th e figure. Deter mine the ten sile force F in th e patellar tendon and the magn it ude of the force react ion at point 0 , which is the contac t area between th e tibi a and th e femur. Not e th at th e line of acti on of the pat ellar tendon force is along its midline. Neglect th e weight of th e lower leg.

I

I

lVH

Problem 3/29

Biceps

8"

r

\VI.

f-- - - 16.5"

-

S ib

I

Ulna ( Radius

G E

~- 4"=.j

_ _ _ _

~/ _::

Quadriceps muscle

~ 14" ----I Problem 3/28 3 /29 A woman is holding an 8-lb weight in her hand with

th e entire arm held hori zon tally as shown in th e figur e. A tensile force in th e delt oid mu scle prevent s th e arm from rotati ng about the shoulder joint 0 ; th is force acts at the 210 angle shown. Determ ine the force exerted by the deltoid mu scle on the upp er arm at A and t he .r- and y -ccmponents of th e force reaction at th e shou lder joint O. The weight of the upp er arm is \V U = 4.1 Ib, th e weight of t he lower ar m is 11', = 2.4 Ib, and the weight of th e hand is 11'" = 0.9 lb; all th ese weight s act at th e locations shown in the figure . An s. FD ~ 160.2 lb, Ox = 149.5 Ib, Oy = - 42.2 lb

Marwan and W aseem AI-Iraqi

Problem 3/30 3/31 For the design of th e belt -ten sioning device, det ermine the dimen sion I if t he mass m maint ain s a specified tension T in the belt for the position shown. Neglect th e mass of th e arm and central pulley compared with m. Also determine th e magnitude R of the force supported by th e pin at O. Tb /3 Ans. I = --'- R = .J3T 2 + m 2Ef mg '

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Article 3 /3

Pr o b le m s

Bl

Problem 3/33

m

3/34 Ca lcu late the magn itud e of t he force sup ported by th e pin at C un der t he action of t he 900-N load applied to the bracket. Neg lect frict ion in th e slot.

T

Problem 3/31

3/31 The u niform l.Svkg ba r OA is held in the posit ion shown by the smooth pin at 0 a nd the cable AB. Determi ne t he te nsio n T in the cable a nd th e magn itude and di rection of the exte rnal pin reac tio n at O.

Problem 3{34 3/35 A un iform r ing of mas s m a nd rad ius r ca rries a n eccent r ic mass /no at a ra dius b a nd is in a n equi librium positio n on th e inclin e, which ma kes an a ngle a wit h t he horizon ta l. If t he contact in g su rfaces are ro ugh enough to preven t slipping , write t he expression for t he ang le (J which defines the equi librium

B/ C!'J\ - - - - - - ---CC/
posit ion. -

-

-

- 1.2 m- -

An s.

(J

~ sin - I [~ (1+ I~~) sin u ]

- -i

Problem 3/32 l /ll T he exercise machi ne is designed wit h a ligh tweight cart wh ich is mou nted on sma ll rollers so t hat it is free to move along the inclined ra mp. Two cables a re attached to the cart- one for each ha nd. If the hands a re together so that t he cables a re parallel an d if each cable lies esse ntially in a vertical plan e, determine t he force P which each ha nd must exert on its cable in order to mainta in a n equi libr ium positio n. The ma ss of the person is 70 kg, t he ramp a ngle (} is 15°, a nd t he angle f3 is 18°. In additi on, calculate t he force R wh ich the ra mp exerts on the cart. Ans. P = 45.5 N, R = 69 1 N

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m

- _u_ --j Problem 3{35

132

Ch a p t e r 3

Eq u il i b r i u m

3/36 Th e concrete hopper and its load have a combined mass of 4 met ric tons (l met ric ton equals 1000 kg ) with mass center at G and is being elevated at co nsta nt velocity along its ver tical guide by the cable tension T . Th e design calls for two sets of gu ide rollers at A , one on each side of th e hopper, and two sets at B. Determine the force su pported by each of th e two pins at A a nd by each of t he two pins at B . T

3/38 Th e elements ofa wall-mounted s wing-a way stool are shown in the figure. The hi nge pin P fits loosely th rough the frame tu be, and th e fra me tube ha s a slight clearance bet ween the supports A and B. Determ ine the reactions on the fra me tube at A and B associat ed with the weight L of an SO-kg per son . Also, calcu late the changes in th e horizont al react ions at C and D du e to th e same load L . St at e any assu mpt ions.

10'

-jf- -- j

65 mm

1

300mm B

~-- 350 mm -

I

G o

-:'''--1'' ,85 mm

600 mm

~_

Problem 3{38

Problem 3{36

3/17 Dur ing an engine test on th e gr ound . a propeller thru st T = 3000 N is generated on th e ISOO-kg airplan e with mass center at G. Th e main wheels at B are locked and do not skid ; th e sma ll tail whee l at A has no brake. Comput e th e perce nt cha nge 11 in the normal forces at A an d B as compar ed wit h their " engine-off' values. A ns. lI .-\ = -32.67" lin = 2.2S7c

3/39 Th e hook wrench or pin span ner is used to turn shafts and collars . If a moment of SO N · m is required to turn the 200-m m·diameter colla r about its center o u nder the action of th e applied force P, determ ine th e contact force R on the s mooth sur face at A . Engagement of the pin at B may be considered to occu r at t he per iphery of the colla r. A ns. R = 1047 N p

Problem 3/39

I A_~ r4 m-

- -

3/40 In sa iling at a const an t speed with th e wind, the sa ilboat is driv en by a 1000-lb force against its mainsail an d a 400·lb force against its staysail as shown. Th e tota l resist ance du e to fluid frictio n thro ugh the water is the force R. Determine the resultant of the latera l forces perpendicular to moti on applied to th e hull by the water.

Problem 3/37

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Article 3/3

Wind

-+-

-+-+~

-+-+-+-+-+-+-+-

Problem 1/ 40

3/41 A portion of t he shifte r mecha nism for a manual car transm issio n is shown in the figure . For the 4·lb force exe rted on the shift knob, det ermine the correspond in g force P exert ed by the shift link BC on t hetransm ission (not shown >. Neglect friction in th e ba ll-a nd - socket joi nt at 0 , in the joint at B , a nd in the slip tube near support D. Note tha t a soft rubber bushing at D a llows the slip t ube to self-a lign wit h link BC. An s. P ~ 13.14 Ib

Problems

133

3/43 T he car complet e wit h driver weig hs 1700 lb and wit hout the t wo airfoils ha s a 50%-50% fron t-rear weigh t dist ribu tion at a certain speed at whic h there is no lift on t he car. It is est ima te d t hat at th is speed eac h of the airfoils AI and A 2 will gene ra te 400 lb of downward force Land 50 Ib of drag force D on the car. Specify the vertica l reactions N A and N B un der t he two pai rs of wh eels at that speed wh en t he airfoils a re added . Assu me that th e addit ion of the a irfoils does not affect t he dra g a nd zero-lift conditions of t he car body it self a nd that t he engine has sufficient power for equilibriu m at t hat speed. The weigh t of the airfoils may be neg lect ed. Ans. N A ~ 12011 b (48.0%), N n ~ 12991b (52.0%)

rr-= Detail

--

~

Airflow

A2

1-C1--12 "

-

-

120 "-

-

-

---!

Problem 1/41

15°

3/44 Determine th e exte rnal reactions a t A a nd F for th e roof t russ loaded as show n. The vert ica l loa ds represent th e effect of th e suppor te d roofing materials, while the 400·N force represents a win d load.

D

500N

J 3/42 A torque (moment) of2 4 N · m is requ ired to tu rn t he bolt about it s axis. Det erm ine P an d th e force s betwee n th e smoot h ha rde ned jaws of t he wre nch and t he cor ners A a nd B of t he he xagonal head . Assume that t he wr ench fits easi ly on the bolt so that cont act is ma de at corners A a nd B on ly. -

-

-

120 mm -----~

I l-

Problem 3/42 Marwan and Waseem AI-Iraqi

250 N E

30' 1 - -- - -- 10 m -

-

Problem 1/44

p

14 mm

500N

D

Prob lem 1/41

1 - --

S OON

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30' -

-

-

-I

134

Chapter 3

Equilibrium

3/45 Calculate th e normal forces associated with the front and rear wheel pa irs of th e 1600·kg front-wheel-drive va n. Th en repeat the calcu lations when th e van (a) climb s a lO·percent grade and (b) descends a 10percent grade, both at constant speed. Compute the percent changes " Il and nn in the norm al forces compared with the nominal valu es. Be su re to recognize that pro pulsive and braking forces are present for case s (a) and (bl. AilS. N A ~ 9420 N, NlJ = 6280 N (a ) N A ~ 9030 N (-4.14'7< 1, N n ~ 6590 N (+ 4.98'7< ) (b) N A ~ 971 0 N (+ 3.15'7<), NlJ ~ 5900 N (- 5.97%)

1 /47 Th e man pushes the lawn mower at a steady speed

with a force P that is parallel to the incli ne. Th e mas s of th e mower with attached grass bag is 50 kg with mass center at G. If 0 = 15°, det erm ine th e norma) forces N nand N c under each pair of wheels Band C. Neglect friction . Compare with th e nor mal forces for the condition s of tJ = 0 and P = O. Ans. NlJ = 214 N, N c ~ 260 N Wit h II = P ~ 0: NlJ ~ 350 N , N c = 140.1 N

Problem 3/47

Problem 3/45 3/46 It is desired that a per son be able to begin closing the va n hatch from the open posit ion shown wit h a 10· Ib vertical force P. As a design exercise, determ ine the necessary for ce in each of th e two hydraul ic st ru ts AB. Th e mass center of t he 90· lb door is 1.5 in. directly below point A. Treat th e problem as twodimensional.

3/48 Th e small cran e is mounted on one side of the bed of a pickup truck. For th e positi on (J = 40°, determ ine the magnitude of th e force supported by th e pin at 0 and the oil pressur e p aga inst the 50-m m-diameter pisto n of the hyd raulic cylinder BC. A

A

o• Hinge axis

IB

360

7"

mm I

Strut detail

Problem 3/48

1 p

Problem 3/ 46

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Articl e 3 /3

3/49 The pin A , wh ich connects th e 200-kg stee l beam with center of gravi ty at G to the vert ical colu mn. is welded both to th e beam an d to th e column. To test t he weld. th e SO-kg man loads th e beam by exert ing a 300-N force on the rope wh ich passes th rough a hole in the beam as shown. Calculate th e torque (cou ple) AI supported by th e pio. An s. AI ~ 4.94 kN · m CCW

135

3/51 Th e ca rgo door for an ai rplane of circular fuselage sectio n cons ists of th e u niform qu arter-circular segment AB of mass m . A det ent in th e hinge at A holds the door open in th e positio n shown. Deter mine the mome nt exerted by the hin ge on th e door . A ilS. Al A = 0.709mgr CCW

Welded / ' pin A

Probl ems

. '

.

-)

C.

~-o-

/

~ -1200 mm --l 60o

I

300 mm

..,""

B

I

mm

Closed position of B Problem 1/51

1 /52 Th e rubber-tired tra ctor shown has a mass of 13.5

Problem 1/49

Mg wit h cente r of mass at G a nd is used for pushing or pulling heavy load s. Determine th e load P wh ich the tractor can pu ll at a con stant speed of 5 km/ h up the 15-percent gr ade if th e dri ving force exer ted by th e gro und on each of its fou r wheels is SO per cen t of the norm al force under t hat wheel. Also find th e total nor mal reaction N[j und er the rear pair of wheels at B.

1 /50 The cargo door for an ai rp lane of circu lar fuselage sectio n consists of th e uniform semicircu lar cowling AB of mass m. Determine th e compress ion C in the hori zontal stru t at B to hold the door open in t he position shown. Also find a n expression for th e total force supported by th e hinge at A. (Consu lt Table D/3 of Appendi x D for th e position of the centro id or mass cen ter of the cowling.)

I

p

I S C:--= 100

I / / / /

/

Problem 1/52

Closed position of R Problem 1/50

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Chapter 3

Equilibriu m

3/53 Pu lley A deliver s a steady torque (moment) of900 lbin . to a pump through its shaft at C. Th e tension in th e lower side of th e belt is 150 lb. Th e drivin g motor B weighs 200 Ib and rotat es clockwise. As a design considerat ion, determine the magnitude R of th e force on the supporting pin at O. Ans. R ~ 287 Ib

3/55 A slender rod of mass Inl is welded to th e hor izontal edge of a un iform semicylindrical shell of mass "taDeter mine an expression for th e angle 0 wit h t he horizontal mad e by th e diam eter of the shell t hro ugh m 1(Consu lt Tab le 0 / 3 in Appendix D to locate the center of gravity of the semicircula r section. ) A lIs . H = tan "! 7Trnl 2m 2

A

I

8"

J~ Problem 3/55

Problem 3/53 3/54 The receiving uni t for a wire less microphone system, exclusive of t he ante nna, has a mass of 1100 gram s wit h ma ss center at G. A single 375-g half-wave anten na with mass cente r at C is mou nt ed to the receiver at point 0 as shown. Plot the reacti on forces at A and B and th eir sum as fun ctions of the antenna angle fJ over th e range 0 == 0 == 90°. Physically inter pret your plot. Treat the problem as twodim ens ional.

3/56 When setti ng the anchor so that it will dig into th e sandy bottom, th e engi ne of th e 80,000·lb cruiser wit h center of gravity at G is ru n in reverse to produce a ho rizontal t hru st T of 500 lb. If the an cho r cha in makes an angle of 60° wit h t he horizon tal , deter mine t he forward shift b of the cente r of buoyancy from its posit ion when th e boat is float ing free. T he center of buoyan cy is th e poin t thro ugh which th e resulta nt of t he buoyant forces passes.

3'

IT

1----24"---.:

Problem 3/56 ,

30

mm

75

mm

,

60 15 30 mm mmmm

Problem 3/54

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Article 3 /3 3/57 Th e uniform 400-kg dru m is mounted on a line of rollers at A a nd a line of roller s at R. An 80.kg man moves slowly a distance of 700 mm from th e vert ica l cente rl ine before t he dru m begins to rota te. All rollers are perfectly free to rotate, except one of them at B which must overcome a ppreciable fricti on in its bearing. Calcu late th e friction force F exert ed by th at one roller tange nt to the dru m an d find th e magn itude R of the force exerted by all ro llers at A on the drum for this condition. Ans. F ~ 305 N, R = 3 770 N

Probl ems

137

center G of th e jig by calculat ing ra nd o. Note that the mass cente r of t he pipe section is at O. Ail S. r = 36 7 mm , II = 79.8°

Horiz.

T

Probl em 3/ 59

Problem 3/57

1 /58 Th e pipe bend er consists of two grooved pulleys mounted an d free to tu rn on a fixed fram e. The pipe is bent into th e sha pe shown by a force P = 60 lb. Ca lculate the forces support ed by the bearings of the pu lleys.

pr I

.. 1 /60 T he lu mbar port ion of the hum an spine supports the enti re weight of the up per torso and th e force load imposed on it. We cons ider here the disk (shaded red ) betwee n th e lowest verteb ra of the lumbar region (L s ) and the up per most verte bra of the sa cru m region. (u) For th e case L = O. det er mine the compressive force C and th e shea r force S supporte d by thi s disk in terms of the body weight W. Th e weight W/I of' th e upp er torso (above the disk in question ) is esrh' of the tota l body weight Wand act s at G l ' Th e vertica l force F which the rectus mu scles of' t he back exert on the upper torso acts as s hown in t he figu re. (6) Repeat for t he case when the person holds a weight of magni tu de L = W/ 3 as s hown. State any ass umptions. AilS. (01 C ~ 0.77 01V, S = 0.669 \V (bl C = 2.53 1V, S ~ 2.20\V

I I

20 '-I,~'- - - Problem 3/ 58

.. l jS9 A special jig is designed to positio n large concrete

pipe sectio ns (shown in gray) and consists of an 80· Mg secto r mou nt ed on a line of roller s at A and a line of rollers at B . One of the rollers at B is a gear which mesh es with n ring of gear teet h on th e secto r so as to tur n th e secto r about its geometric center O. Wh en u = 0, a counte rclockwise torque of 2460 N . m must be applied to the gear at B to keep th e asse mb ly from ro t at ing. When n = 30°, a clockwise torque of 4680 N . m is required to prevent rotation. Locat e th e mass

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Probl em 3/60

1 ~8

Chapter 3

Equilibrium

SECTION B. EQUILIBRIUM IN THREE DIMENSIONS 3 /4

EQUILIBRIUM CONDITIONS

We now exte nd our principles and meth ods developed for two-dimensiona l equilibrium to the case of three -dimensional equilibrium. In Art. 3/ 1 the general conditions for the equilibr iu m of a body were sta ted in Eqs. 3/ 1, which req uire that th e resultant force a nd resulta nt couple on a body in equilibrium be zero . Th ese tw o vecto r equa tio ns of equilibrium and their scalar components may be written as

r

F = 0

IF

0

or

ITy,

0

IFz = 0

(3 / 3)

0

IM = 0

or

{ IM, IM = 0 uJz = 0

The first three scalar equations state that there is no resu ltant force acting on a body in eq uilibrium in a ny of the three coordinate dire ctions. Th e second three scala r equations expre ss the further equilibrium requiremen t that t here be no re sultant moment acting on th e body about any of the coordinate axes or about axes parallel to the coordinate axe s. These six equations are both necessary and sufficient condition s for complete equilibrium. The reference axe s may be chosen arbitrarily as a matter of convenience, the only rest riction being that a right -handed coordinate sys te m should be chose n when vector notation is used . The six sca lar re lationsh ips of Eqs. 3/ 3 a re independ ent condit ions because any of them can be valid without the ot he rs . For example, for a car which accelerates on a straight and level road in the .r-dire ct ion, Newton's second law tells us that the resultant force on the car equals 0, but the remaining two its mass times its acceleration. Thu s, ":5:.Fx force-equilibrium equations are satisfied because all other acce leration compone nts are zero. Simil arly, if th e flywheel of t he engine of the accelerating car is rotatin g with increasi ng angular speed about the x-axis, it is not in rot ational equilibrium about thi s axis. Thus, for t he flywh eel 0 a long with IF, 0, bu t th e rem aini ng fou r equilibri um alone, IM, equa t ions for the flywheel would be sat isfied for it s mass-center axes . In applying the vector form of Eqs. 3/ 3, we firs t express each of th e forces in terms of the coordinate un it vecto rs i, J. and k. For the first equa tion, I F = 0, th e vector sum will be zero only if the coefficien ts of i , J. and k in the express ion are, respectively, zero. These three sums, when each is se t equal to zero, yield precisely the t hree scala r equa tio ns of equilibrium, ':5:.Fx = 0, ':5:.Fy = 0, and ~ Fz = o. For the seco nd equation, ~M = 0, where the moment sum may be taken about any convenient point 0, we express the moment of each force as the cross product r x F , where r is the position vector

'*

*

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*

Article '5 / 4

from 0 to any point on t he line of ac t ion of t he force F. T h us ~ M ~ ~ ( r X F ) = O. When th e coefficients of i , j , and k in t he res ulting moment equation are set equal to zero, respectively, we obtain the three scalar moment equat ions ~x 0, ~)' = 0, and ~[z = o. Free-BDdy Diagrams

T he summation s in Eqs. 3/3 inclu de th e effects of all forces on the body under considerat ion. We learned in the previous article that the free-body diagram is t he on ly r eliabl e meth od for disclosing all forces and moments whic h should be included in our equilibriu m equations . In three dimensions the free-body diagram se rves the same essential purpose as it does in two dimensions and should always be drawn. \Ve have ou r choice eit he r of drawing a pictori a l view of t he isolated body wit h a ll external forces represented or of drawin g the ort hogo na l projec t ions of t he free-body d iagram. Bot h representations are illustrated in th e sa mple problem s at the end of this a rticle . The correct represen tation of forces on th e free- body diagram requires a knowledge of the characteristics of contacting surfaces. These characteristics we re described in Fig. 3/ 1 for two-dimensional problems, and their exte nsio n to three-dimensional problems is represented in Fig. 3/8 for the most common situations of force transmission. The represe ntations in both Figs. 3/ 1 an d 3/8 will be used in t hree-dimen sional analys is. The essen ti al pu rp ose of t he fre e-body d iagr am is to develop a reliable pict u re of t he physical action of a ll forces (a nd coupl es if any ) acting on a body. So it is helpfu l to represen t t he forces in t heir cor rect physical se nse whenever possible. In this way, the free-body diagram becomes a closer model to th e actua l physical problem t han it would be if t he forces were arbitrarily ass igned or always ass igned in the same mathematical se nse as that of the ass igned coordinate axis. For exa mple, in part 4 of Fig. 3/8, t he correct se nse of th e un kn own s R , and R~. may be known or perceived to be in the se nse opposite to those of t he assigned coord inate axes. Si milar cond itio ns apply to t he se nse of coup le vecto rs , parts 5 and 6, where th eir se nse by th e righ thand rule may be ass igned opposite to that of the respective coordinate direction . By this time, you should recognize that a negative answer for an unk nown force or couple vector merely indicates that its physical action is in t he se nse opposite to that assign ed on t he free-body diagr am . Frequently, of course, the correct physical se nse is not known initially, so that an arbitrary ass ignment on the free-body diagram becomes necessary . CategDries Df Equilibrium

App lication of Eqs , 3/3 fa lls into fou r ca tegor ies which we ide ntify wit h t he a id of Fig. 3/9. These catego ries d iffer in th e num ber a nd type (force or moment ) of independent equilibrium equations required to so lve t he prob lem. Ca tegory' I , equ ilib r iu m of forces all concurren t at point 0 , requires all three force equations , but no momen t equations because the moment of the forces about any axis through 0 is zero. Marwan and Waseem AI-Iraqi

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Equ ilib riu m Co nd it io n s

139

14 0

Chapt e r 3

Equilibr ium

MOD E LI N G T HE A C TIO N O F FORCES IN THREE.DIMENSIONAL ANALYSIS Type of Contact a nd Foree Ori gin

Action on Body to Be Isolated

1. Member in contact wit h s mooth surface, or ball -supported member z z I I I I I I

z I

~ /

,

-,

,

x ./

' y

' y

/

2. Member in contact with rough su rface

<,

For ce m ust be normal to th e s urface and directed to wa rd th e member.

I

' '

N

' y

Th e possi bility exists for a foree F tangent to th e s urface (friction force) to act on the member, as we ll as a normal foreeN.

, ' y

3. Roller or wh eel s upport with latera l const ra int

A lateral force P exerted by the guide on t he wheel can exi st, in addit ion to th e normal foree N .

4. Ba ll-a nd -socket joint A ball-and -sock et joint free to pivot a bout the cente r of th e ball ca n s uppor t a force R wit h a ll t h ree com ponent s.

z I

I

5. F ixed conn ection (embedded or welded )

In add it ion to three components of force, a fixed connec t ion ca n su pport a cou ple M represen ted by its three compone nts.

6. Thrust-bearing s upport Thrust bea ri ng is ca pa ble of su pport ing ax ia l force Ry as well as radial forces Rx. and R:. Couples Mx. and M: mu st , in some case s, be assume d zero in order to provide statica l determinacy.

z "

I

Figure3/8

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Art icle 3 /4

CATEGORIES OF EQUILI BRIUM IN THREE DIMENSIONS Force System 1. Concurrent

at a point

Free-Body Diagram

F t" - .

/

Independent Equations

Y I I

F,

~

_x

1.: : ,-

' ~O

F:?'~ /S

' z

LFz = 0 W y= O

LFl

F.

=0

F, 2. Concurrent with a line

F'j ~

). ---\ ' , I

,

t

'

' z

F.

LFz =O

rMy =O

LFy=O

rMl

LFl

=0

=0

F. Y I

r F,

-,

F,

~ Ft

3 . Parall el

_x

1.: : -

'"

I I ____ __ /~ .....

'"

-

IY I

~

F' 1

.: «

F.

' z

1.: : ,-

/

=0 r.Mz = 0

LFz =0

IMy

LFx = 0

r.Mx = 0

Wy = 0

rMy =O

F,~ 4. General

F1 \

M

F' /

,)1

Y I

I

1;: -

_x

-,

' z

~

LFl

F,

\

=0

rMz

=0

F, Figure 3/9

Category 2, equilibrium of forces which are concurrent with a line, requires all equations except the moment equation about that line , which is auto ma tically satisfied. Category :1. equ ilibrium of parallel forces, requires only one force equation, the one in the direction of the forces (x-direction as shown), and two mom ent equations abou t the axes (y an d z) which are nor mal to the direction of th e forces . Ca tegory' 4, equilibrium of a general system of forces, requires all three force equations and all three moment equations. The observations cont ained in these statements are generally quite evident when a given pr oblem is being solved.

Constraints and Statical Determinacy The six scalar relations of Eqs. 3/3, although necessary and sufficient conditions to establish equilibrium, do not necessarily pr ovide all of the inform ation requ ired to calculate th e unknown forces acti ng in a Ma rwan and W aseem AI-Iraqi

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Equil ibrium Co nd i tions

141

142

(a)

Chapt e r 3

Equilib rium

Comp lete fixity Adequat e constrai nts

lbl lncomplete fixity Part ial constraints

( e)

three-dimensional equilibrium situation. Again, as we found with two dimensions, the questio n of adequacy of information is decided by the cha ra cter istics of t he constra ints provided by the supports . An a nalytical criterion for determining the adequacy of constraints is available, but it is beyond th e scope of this treatment. ' In Fig. 3/ 10, however, we cite four examples of constraint conditions to alert the reader to the problem. Part a of Fig. 3/ 10 shows a rigid body whose corner point A is complet ely fixed by th e links 1, 2, and 3. Links 4, 5, a nd 6 prevent rotations about th e axes of links 1, 2, a nd 3, respecti vely, so t ha t th e body is completely fixed a nd the const ra ints are sa id to be adequat e. Part b of th e figu re shows th e sa me num ber of constraints, but we see that t hey provide no resistance to a moment which might be applied about axis AB. Here th e body is incompletely fixed and on ly partially constrained. Similarly, in Fig. 3/ 1Oc the cons traints pr ovide no resista nce to an unbalanced force in the y-direct ion, so here also is a case of incomplete fixity with partial cons traints. In Fig. 3/ lOd, if a seven th const ra ini ng link were imposed on a system of six constraints placed properly for compl ete fixity, more supports would be pr ovided th an would be necessary to esta blish the equ ilibrium position , and link 7 would be redun dant. Th e body wou ld t hen he statically indeterm inate with such a sevent h lin k in place. With only a few exceptio ns, th e su pporting constraints for rigid bodies in equilibrium in this book are adequate . and t he bodies are stat ica lly det er minat e.

Incomplete fixity Partial constra ints

fd) Excessiv e fixit y Redundan t const rai nts Figure 3/10

"See the first autho r's Statics, 2nd Edition SI Version, 1975 . Art. 16. MarNan and Waseem AI-Iraqi

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Art icle 3/4

Equilibrium Conditions

143

Sample Problem 3 /5 The uniform 7-m steel shaft has a mass of 200 kg and is supported by a bal land-socket joint at A in the horizon tal floor. Th e ball end B rests against the smooth vertical walls as shown . Compute th e forces exerted by th e wall s and the floor on th e end s of the shaft.

SolutIon. The free-body diagram of the shaft is first drawn where the contact

CD

forces acting on the shaft at B are shown normal to the wall surfaces. In addition to the weight W = mg = 200(9.81) = 1962 N, t he force exerted by the floor on the ball joint at A is represe nted by its X- , y -, and a-component s. These components are shown in their correct physical sen se, as should be evident from the requirement th at A be held in place . The vertical position of B is found from 7 = ,/22 + 62 + h 2 , h = 3 m. Right-handed coordinate axes are assigned as shown.

Vedor solution. We will use A as a moment center to eliminate reference to th e forces at A. The position vectors needed to compute th e moments about A ar e r AG = l i - 3j + 1.5km and rAE = - 2i - 6j + 3k m -r

where th e mass center G is locate d halfway between A and B . Th e vector moment equation gives

=

[ ~MA

rAE x

OJ

+ By) + rAG x W

(Ex

=

0

(- 2i - 6j + 3k l x (B) + By j ) + (- i - 3j + l.5k ) x (- 1962k ) = 0 j

k

j

-2 -6 3

+ - I -3

1.5

0

- 1962

s,

B,

0

0

Helpful Hints

k

CD We could , of course, assign all of the = 0

(- 3B, + 5890 )1 + (3B x 1962Jj + (- W y + 6Bx)k = 0 Equating the coefficients of i , j , and k to zero an d solving give

B x = 654 N

and

Ans.

By = 1962 N

The forces at A are easily determ ined by [~ F

= 01

and

\654 - A x)i + (1962 - A, lj + (- 1962 + Axlk Ax

A

Finally

= 654 N

A,

=

1962 N

o

@ Note t hat th e thi rd equa tion - 2By + BBx = 0 merely checks the re sults of

A x = 1962 N

,IA/ + Ay2 + A/ = J (654)2 + (1962 ,2 + (1962 :2 = 2850 N

An s.

Scalar solution. Eval uat ing the scalar moment equat ions about axes t hrough A parallel, res pectively, to the x - and y-axes, gives [~

0)

A.

OJ

1962(3) - 3B y

l ~lA

OJ

- 1962(1) + 3B x

.-

=0 =0

By

1962 N

Bx

654N

Th e force equat ions give, simply, I~Fx

OJ

- A x + 654

0

Ax

654 N

I~Fy

0)

- A.,. + 1962

0

Ay

1962 N

I~Fx

= OJ

1962

0

A,

1962 N

Ax

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unknown components of force in the positi ve mathematical sense, in which case A x and A y would turn ou t to be negati ve upon computa tion. Th e freebody diagr am describ es th e phy sical situa t ion, so it is genera lly preferable to show th e forces in th eir correct physical se nses wher ever possibl e.

th e first two equations. This res ult could be anticipa ted from th e fact that an equi libri um syste m of forces concurrent with a line requires only two moment equations (Category 2 u nde r Categories of Equilibrium ).

@ We observe th at a moment sum abo ut an axis thro ug h A parallel to t he a-axis merely gives us 6B x 2By = 0, which serves only as a check as noted previously. Alternativ ely we could have first obtained A z fro m ""iF z = 0 and t he n ta ken ou r momen t equa tions abo ut axes th rou gh B to obtai n Ax and A.v.

144

Chapter 3

Equ ilibrium

Sample Problem 3 /6 A 200-N force is a pplied to th e handle of the hoist in the direction shown. Th e bearing A supports the thru st (force in th e direction of t he shaft axis), while bea ring B su pports only rad ial load (load norm al to t he shaft axis ). Determ ine the mass m which can be supported and the total rad ial force exerted on t he shaft by each bearing. Assu me neith er bear ing to be capable of supporti ng a mome nt abou t a line nor mal to t he shaft axis.

(

Rad ia l l <, bearing Thru st bearin g

,

Z

m Dime nsion s in millimet ers

Solution. The syste m is clearl y th ree-dimen sional with no lines or plan es of

CD

symmetry, and th erefore the pro blem must be ana lyzed as a general space system of forces. A scalar solutio n is used here to illust rate th is approach, althoug h a solution us ing vecto r nota tion wou ld also be sa tisfactory. Th e free-body diagr am of the shaft, lever, and dru m consid ered a single body cou ld be shown by a space view if desired , but is represen ted her e by its three orthogona l projec tions. Th e 200-N force is resolved into its three components, an d each of the three views shows two of these components. The correct directions of Ax and Bx may be see n by inspection by observing t ha t the line of action of the res ultant of th e two 70.7-N forces pass es between A and B. The correct sense of the forces Ay and By cannot be determined until the magnitudes of the moments are obtained , so th ey are arbitrarily as signed. The x-y projection of the bearing forces is show n in te rm s of the su ms of the un kno wn x- and y -ccmponents. Th e addi tion of Al' an d th e weight lV = mg completes th e free -body diagrams. It should be noted tha t the th ree views represen t three two-d imensional prob lems related by th e corres ponding comp one nts of the forces.

=

o

m

44.1 kg

150B, + 175(70.7) - 250 (70.7)

o

e,

35.4 N

Ax

35 .4 N

OJ

100(9.81m ) - 250 (173.2)

Ans.

From the x-z projection I ~IA = I~F,

OJ

= OJ

Ax + 35.4 - 70.7 = 0

Q) The y -z view gives 1 ~\1A I~F, I ~Fx

= OJ

520 N 86.8 N

= 70.7 N

= JAr' + it,2J

IB = J Bx 2 + B, 2]

A, -- z

m~

= 9.81m

Helpful Hint s graph ic project ion are not entirely familiar, th en review and practice t hem. Visualize the three views as the images of the body projected onto the front, top, and end surfaces of a clear plastic box placed over and aligned with the body . x.z

project ion ra ther than with the .r-y projection.

Q) Th e y-z view could have followed im-

Ans.

mediately after th e x-y view since the det erm ina tion of A y and B.y may be made aft er Tn is foun d.

Ans.

@ With ou t t he ass u mpt ion of zero mo-

The total radial forces on the beari ngs become lA,

70.7 N

@ We could have started with the

= OJ 150B, + 175(173.2) - 250 (44.1)(9.81) = 0 B, it, = OJ it, + 520 - 173.2 - (44.1)(9.8 1) = 0 A,

: f 07N

CD If the standard th ree views of or tho-

From t he x-y projection

l:ufo

x

A, = J (35.4 )2 B

=

+ (86.8)2

=

93.5 N

J (35.4 )2 + (520)2 = 52 1 N

ment supported by eac h bearing about a line normal to the shaft axis, the prob lem wou ld be statically indeterminate.

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Article 3/4

Equ ilibrium Cond it ion s

145

Sample Problem 3 /7 Th e welded tubular fram e is secured to t he horizontal x -y plane by a balland-socket joint at A and receives support from the loose-fitting ring at B. Under th e act ion of th e 2-kN load, rotation about a line from A to B is prevented. by th e cable CD , and th e frame is stable in th e position shown . Neglect th e weight of th e fram e compared wit h th e applied. load and determine th e tension T in th e cable, the react ion at the ring, an d the reaction components at A.

Solution. The system is d early three-dimensional with no lines or plan es of sym metry. an d t herefore the problem must be analyzed as a general space syste m of forces. The free-body diagram is drawn, wher e th e rin g reactio n is shown in te rms of its two components. All unknowns except T may be eliminated by a moment su m about th e line AB . The direction of AB is specified. by th e uni t


1 (4.5j + 6k ) ~ ! (3j + 4k ). The moment of T about AB J 6 2 + 4.52 is th e component in t he direction of AB of the vector moment about th e point A and equals r l x T ·n. Similarly t he moment of th e applied. load F about AB is r a x F · n. With CD = J 46.2 m, th e vector expressions for T, F , r Io and r 2 are

vector n =

T =

~ (2i +

2.5j - 6k )

F = 2j kN

)46.2

rl = - i

+ 2.5j m

r2

2.5i

+ 6k

m

The moment equat ion now becomes f~AB ~

OJ

(- i

+ 2.5j ) x _ T_ (2i + 2.5j - 6k H (3j + 4k ) J 46.2

+ (2.51 + 6k ) x (2j H(3j + 4k )

o

Completion of t he vector opera tions gives

--

48T

- + 20 = 0

T = 2.83 kN

Ms.

J 46.2

and th e components of T become

T, = 0.833 kN

Ty

Helpful Hints ~

1.042 kN

T, = - 2.50 kN

CD The

We may find the re mai ning unknowns by mome nt and force summations as follows: fL\t , =

OJ

2(2.5) - 4.5B, - 1.042(3)

~

0

B,

f ~,

OJ

4.5B, - 2(6) - 1.042(6)

~

0

B, = 4.06 kN

Ms.

=

lU, =

~

0.417 kN

Ms.

OJ

A, + 0.417 + 0.833 = 0

A, = - 1.250 kN

Ans.

A, + 2 + 1.042

A, = - 3.04 kN

Ms.

A, = - 1.556 kN

Ans.

@ fUy

~

OJ

fU ,

~

0)

~

0

A, + 4.06 - 2.50 = 0

advantage of using vector notation in this problem is the freedom to take moments directly about any axis. In this problem this freedom permits the choice of an axis that eliminates fiveofthe unknowns.

Q) Recall that the vector r in the expressian r x F for the moment of a force is a vector from the moment center to any point on the line of action of the force. Instead of rl . an equally simple choice would be the vector AC.

@ The negative signs associated with the A-components indicate that they are in the opposite direction to those shown on the free-body diagram. Marwan and Waseem AI-Iraqi

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146

Chapte r 3

Equ il ibrium

PROBLEMS

1/63 Dete rmine th e te nsions in cables AB, ACt and AD. Ans. T AB = 47.8 lb. T AC = 47.8 lb. T AO = 31.2lb

Introductory Problems

C

3/6 1 A force of magn itude P = 40 lb is ap plied to the sta tionary machine handle as shown. Write th e force and moment reactions at 0 as vect ors. Neglect t he weight of the handle assembly. Ans. R = - 38.6i - 10.35k lb M = - 103.5i - 193.2j + 386k lb-in.

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Problem 3/63

3/64 Th e uniform I-beam has a mass of 60 kg per meter of its length. Determine th e ten sion in the two supporting cables and the reaction at D.

Prob lem 3/ 61

3/61 Thr ee workers are carrying a 4· ft by 8· ft pan el in th e horizontal position shown. If the homogeneou s pan el weighs 100 lb, estimate th e lifting force exerted by each wor ker.

Prob lem 3/64

Problem 3/62

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3/65 The vertical mast support s t he 4-kN force an d is constrained by the two fixed cables Be and BD and by a ball -and-socket connection at A. Calcu late the tension T I in BD . Can th is be accomp lished by using on ly one equ at ion of equilibrium? Ans. T 1 = 4.90 kN

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Articl e 3/4 z

Problems

147

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Problem 3/65 3/66 An over hea d view of a car is shown in th e figure. Tw o different locations C a nd D are consi de red for a single jac k. ]0 eac h case, the entire right side of the car is lifted ju st ofT th e gro u nd . Determine the normal reaction forces a t A a nd B a nd th e vertica l jacking for ce required for the case of each ja cking locati on. Con-

sider th e 1600-kg car to be rigid. Th e mass center G is on the midline of the car. B

A

Problem 3/67 3/68 Th e indust rial door is a uniform rectangular panel weighing 1200 lb and rolls alon g t he fixed rail D on its hanger-mounted wheels A and B . Th e door is maintained in a vertical plane by th e floor-mounted guide roller C, which bears against th e bottom edge. For t he posit ion shown compute the hori zontal side thru st on each of the whee ls A and B , which mu st be accounted for in the design of the brackets.

1575 mm

1.- 1400 4 + 1120 -.j mm

280

mm

mm Problem 3/66

Detail of door hanger

3/67 Th e light r ight-a ngle boom which supports th e 400kg cylinder is supported by three cables and a ba lland-socket joint at 0 att ached to the vert ical x -y su rface. Determine th e reacti ons at 0 and the cable ten sions. Ans. Ox ~ 1962 N, O; = 0, Ox = 6540 N TA C = 4810 N, T Bn ~ 2770 N , T BE ~ 654 N

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Problem 3/68

148

Chapter 3

Equilib rium

3/69 A three-legged stool is subjec ted to the load L as sh own . Determine th e vertical force reaction under eac h leg. Neglect the weight of the stool. Ans. N A = O.533L , N B = N c ~ O.233L 100

75

75

Representative Prablems 3 /71 The three-wheel truck is used to ca rry th e 100-kgbox as shown . Calculate t he cha nges in th e normal force reactions a t the three whee ls due to th e weight of th e box. Ans. !!.NA ~ 66.1 N,!!.NB = 393 N !!.Nc ~ 522 N

~~~~375mm

325

,4 Dimensions in millimeters

Problem 3/69 3 /70 Ca lcu la te t he compression P in each leg of the equilatera l spreader fram e ABC , whic h e ns ures the ap plica tion of equal vertical forces to the ri m of the concrete hopper . The tota l load L is 840 lb. Note th at determination of P wou ld be a first ste p in the design of th e fra me.

550mm

Problem 3/71

3/72 On e of th e vertical walls su pporting end B of the 200· kg uniform shaft of Sample P roblem 3/5 is turned throu gh a 30° a ng le as shown here. End A is still supported by the ball -an d-socket connection in th e hor izontal x-y plan e. Ca lcu late th e ma gn it ud es of th e forces P and R exe rted on t he ball e nd B of th e shaft by th e vertical wa lls C a nd D, re spectively .

Problem 3/ 72

Problem 3/70

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Art icl e 3 /4 3/13 Th e smooth homogeneous sphere rests in the 1200 groove a nd bea rs aga inst t he end plate which is normal to the di rection of the groove. Det ermine t he angle 0, measured from th e hori zont al, for whic h t he reaction on ea ch side of t he groove equals the force suppor te d by t he end plate. A ns. (} = 30 0

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Pro b lems

3/15 As part of a check on its design , a lower A-ar m (pa rt of a n au tomobile suspension) is supported by bearings at A a nd B a nd subject ed to th e pa ir of 900-N forces a t C a nd D. T he sus pension spr ing, not sh own for clarity, exer ts a force Fs at E as sh own , where E is in pla ne ABCD . Determin e t he magn it ude F s of the spring force a nd th e magni t ude s FA a nd Fa of the bea ring forces at A a nd B which ar e perpendi cular to th e hi nge axi s AB. AilS. F s = 3950 N, F,\ ~ 437 N, F lJ ~ 2450 N

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or V-groove

Problem 3/73 3/14 T he mass cent er of the 3D-kg door is in th e cente r of the panel. If the weig ht of the door is supported en tirely by the lower hinge A , ca lculate th e magn itu de of the to tal force supported by the hin ge at B.

900 N

Problem 3/75 3/16 Determine th e magni t ud es of th e force R an d coup le M exerted by t he nut and bolt on t he loaded brack et at 0 to ma intain equilibriu m.

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1500 mm

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Problem 3/74

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Problem 3/76

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150

Ch a pter 3

Equi li br ium

3 /77 The square steel plat e ha s a mass of 1800 kg wit h

3/79 One of th e three landing pad s for a propo sed Mars

mass center at its center G. Calculate the tension in eac h of the three cables wit h which th e plat e is lifted wh ile rema ining horizontal. Ails. T A = T il = 5.41 kN , Tc = 9.87 kN

lander is shown in the figure . As part of a design check on th e dist rib ut ion of force in the landing struts, compute th e force in eac h of th e stru ts AC. BC, an d CD when t he lan der is rest ing on a hor izonta l sur face on Mars. The arrangement is symmet ical with respect to the x-z plan e. The mass of th e lander is 600 kg. (Assume eq ua l support by th e pad s and consult Table D/ 2 in Appendix D as needed.I An s. FAC = Fell = 240 N ten sion F eD = 1046 N compression

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Problem 1/77 3 /78 During a test , th e left engine of th e twin-en gine air-

plan e is revved up a nd a 500-lb thrust is generat ed. Th e main whee ls at B and C are braked in order to prevent moti on. Determine th e change (com pared with the nominal values with both engi nes in the nor mal react ion forces at A, B , and C.

am

Dimensions in millimeters

Problem 3/79 3 /80 The spring of modulus k = 900 N/m is stre tched a distance Ii = 60 mm when the mechan ism is in th e position shown. Calcu late th e force P min required to

init iate rotation about the h inge axis Be. and deter min e the corresponding magnitudes of the bearing forces which are perpendicular to BC. Wh at is th e normal reacti on force at D if P = Pmin/ 2?

Problem 3/78

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Art icl e 3 / 4

3 /81 A smooth homogeneous sphere of mass III and ra dius r is suspended by a wire AB of lengt h 2,. from point B on th e line of inte rsect ion of the two smooth ver-

Pro b l e ms

151

3 /83 Th e shaft, lever , and ha ndl e are welded togeth er and

const itut e a single rigid body. Th eir combined ma ss is 28 kg wit h mass center at G. The assembl y is mounted in bear ings A and B , and rot ation is prevented by link CD . Deter mine t he forces exerted on the shaft by beari ngs A and B wh ile the 30-N . m cou ple is applied to the hand le as shown. Would these for ces change if the coup le were applied to the shaft AB rather than to th e handle'? Ail s. A = 167.9 N , 8 = 117,1 N

tical walls at right an gles to one anot her . Deter mine the react ion R of each wall against the sphere. Ans. R = mg / ,/'7

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Problem 3/81

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1 /82 The u nifor m 15-kg plate is welded to the vertical shaft, which is suppor ted by bear ings A and B. Cal-

culate the magn it ude of the force supported by bearing B during a pplication of the 120-N ' m couple to t he shaft. Th e cable from C to D prevents t he plate and shaft From turn ing, and the weight of the assembly is carr ied entire ly by bear ing A.

Problem 3/83 3/84 Consider the ru dder assembly of a rad io-controlled

model airpl an e. For t he 15° posit ion shown in t he figure , th e net pressur e act ing on th e left side of th e recta ngu lar rudder area is p ;:: 4 (1 0 ~ 5 ) N/ mm 2 . Determi ne t he required force P in th e cont rol rod DE and the hor izon tal componen ts of the reacti ons at hinges A and B wh ich are parallel to the rudder surface. Assu me t he ae rodynamic pressu re to be u niform . I

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Problem 3/82

Problem 3/84 Marwan and Waseem AI-Iraqi

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152

Chap t er 3

Equ ilib rium z'

3 /85 Th e u nit ABCD of the radia l-arm saw weighs 40 lb with cente r of gravity at G. If a hor izont allO-lb force is applied to the cont rol handle in sawing t he board , ca lculat e t he corresponding bending moment act ing on th e colu mn at A (total momen t about a hor izontal axis th rough A ). The reacti on oft he wood on th e sa w teeth is 15 lb in th e direct ion sho wn, and to a close approxi ma tion. its point of application may be ta ken as E. What ju stification exists for treat ing t he saw as being in equi librium? Ail s. At = 711 lb-in.

Problem 3{86

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Problem 3{85

3/87 Th e up per ends of t he vertical coil spr ings in the stoc k racecar ca n be moved up and down by means of a sere' v• mechani sm not shown. Thi s adjust ment permits a change in th e downward force at eac h wheel as an opti mum ha ndling set up is sought. Initially, scales indicate the normal forces to be 800 lb. 800 Ib, 1000 Ib, a nd 1000 Ib at A, B , C, and D , respect ively. If t he top of th e right rear spr ing at A is lowered so th at the sca le at A read s an additio na l 100 lb, determi ne the corresponding changes in t he normal forces at B, C, and D. Neglect the efTects of the small attitude cha nges (pitch an d ro ll angles) caus ed by th e spring adju stm ent. The fron t wheels are the sa me distance apart as the rea r whee ls. A ns. .1N 1l = - 100 Ib, .1N c = 100 Ib .1N" = - 100 1b

3/86 T he rigid pole and cross-arms of Prob. 2/ 95 are shown again here. Determine the tensions TAE.' and TG/.. in the two support ing cables result ing from th e 1.2-kN ten sion in cable CD . Assu me t he abse nce of any resisti ng moments on the base of the pole at a about th e .r- an d y-axes, bu t not about th e a-axis.

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Article 3/4

Problem s

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Problem 3/89 Problem 3/87

.... 3/90 A rectangul ar sign over a store has a mas s of 100 kg,

3 /88 Th e un iform 30- by 40-in . tra p door weighs 200 lb

and is propp ed open by the light st ru t AB at t he a ngle o = ta n - 1 (4/ 3), Calculate the compress ion F B in the strut a nd the force supported by the hinge D nor mal to t he hinge axis. Assume th at the hinges act at the ext re me ends of th e lower edge .

with the center of ma ss in the center of th e rec ta ng le. T he su ppor t agai nst th e wall at point C may be treated as a ba ll-and -socket jo int. At corne r D su pport is pro vided in t he y-direct ion on ly. Calcul ate th e ten sion s T 1 a nd T 2 in t he supporting wires, th e total force suppor ted at C, and the later al force R su pported at D. Ans. T, ~ 347 N, T 2 ~ 431 N R ~ 63.1 N, C ~ 768 N

Problem 3/88 3 / 8 9 T he boom AB lies in the vertical y -z pla ne an d is sup-

ported by a ball-a nd-socket join t a t B a nd by the tw o cables at A . Calculate the te nsion in each cable resulting from the 20-kN force acting in the horizontal pla ne a nd a pplied at the midpoint M of th e boom. Neglect th e weight of the boom. An• . T, = 33.0 kN, T 2 ~ 22.8 kN

Problem 3/90

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154

Chapter 3

Equ il ibrium

... 3/91 The awning windo w is temporarily held open in the 50° pos it ion shown by a wooden prop CD until a cra nk-type opening mechani sm can be installed . If a = 0.8 m a nd b = 1.2 m a nd th e mas s of the wind ow is 50 kg with mas s center at its geometric center , determi ne the compressive force F CD in th e prop and all components of the forces exerted by the hinges A and B on th e window. Assume that A is a th rustbearing hi nge but t hat hinge B is not . An s. Ax ~ - 140.9 N, A" = 118.2 N, A, = -92.0 N B; = - 47.0 N, By ~ 285 N, FeD ~ 227 N

1200 mm

B

Problem 3/92

"'3/93 Under the action of the 40-N ' m torque (couple) applied to th e vertical shaft, the restraining cable AC limits th e rota tion of th e arm OA and attached shaft to an angle of 60° measured from the y-ax is . Th e collar D fastened to th e shaft prevents downw ard motion of th e shaft in its bearing. Calculate the bending momen t M, th e compression P, and the shear force V in the shaft at section B . (N ote: Bending moment, expressed as a vector, is normal to t he shaft axis, and shear force is also normal to the shaft axis. ) Ans. M 47.7 Nvm, P = 320 N, V = 274 N

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Problem 3/91

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.... 3/91 The un iform rectangular panel ABeD has a mass of 40 kg and is hinged at its cor ners A and B to th e fixed vertical sur face. A wir e from E to D keeps edges Be and AD hor izontal. Hin ge A can support thrust along the hinge axis AB whereas hinge B supports force norm al to the hinge axis only. Find th e te ns ion T in the wire and the magnitude B of t he force su pported by hi nge B. Ans. T = 277 N, B = 169.9 N

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Problem 3/93

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r-

200 mm

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Art icle 3 / 4

Problems

... 3/ 94 Th e porta ble reel is used to wind up a nd store a n air hose. Th e tension in the hose is 100 N a nd a vertical 200·N force is applied to the ha ndle in or der to ste ady the reel fram e. Determine the minim um force P which mu st be a pplied perpendicula r to the handle DE a nd the vertical components of th e force reactions at the feet A, B, and C. T he diam eter of the coil of reeled hose is 300 mm , and the weight of the load ed

reel and its frame may be neglected. State any assumptions. Ans. P ~ 50 N, N A ~ 108.6 N N B = 32.4 N, N c = 58.1 N

OD =300mm

Dimen sions in millim et er s

Problem 3/94

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155

156

Chapter 3

Equ il ib rium

CHAPTER REVIEW In Chapter 3 we have applied our knowledge of the properties of forces, momen ts, a nd couples studied in Chapte r 2 to solve problems involving rigid bodies in equilibriu m. Complete equ ilibrium of a body requires that the vector resultant of all forces acting on it be zero ( ~ F = 0) an d t he vector resultant of all moments on the body about a point (or axis) also be zero ( ~ M = 0). We are guided in all of our solutions by these two requirements, which are easily comprehended physically. Frequ ently, it is not th e t heory bu t its application which presents th e difficulty. The cruc ial steps in applying our pri nciples of equ ilibrium should be quite fam iliar by now. They ar e: 1. Make an un equivocal decision as to which system (a body or collection of bodies) in equilibrium is to be an alyzed.

2. Isolate th e syste m in question from all contacting bodies by drawing its free-body diag ram showing all forces and couples acting on the isolated syste m fro m exte rnal sources. 3. Observe the principle of actio n and reaction (Newlon 's third law) when ass igning the sense of each force. 4. Label all forces and couples, kn own and unknown. 5. Choose and label reference axes, always choosing a right-handed set when vector not ation is used (which is usually th e case for t hreedimensional analysis). 6. Check th e adequacy of the constrai nts (supports) and match th e number of un knowns with the number of avai lable indepe ndent equations of equ ilibri um. When solving an equilibrium probl em , we should first check to see t hat th e body is statically det erminate. If th ere are more su pports than are necessary to hold th e body in place , th e body is statically inde terminate, and th e equations of equilibriu m by themselves will not enable us to solve for all of the extern al reacti ons. In applying the equat ions of equ ilibrium , we choose scalar algeb ra , vector algebra , or graphical analysis according to both prefere nce an d experience ; vector algebra is part icularly useful for many three-dimensional problems. Th e algebra of a solution can be simplified by the choice of a moment axis which eliminates as many unknowns as possible or by the choice of a direction for a force summation which avoids reference to certai n unkn owns. A few moments of th ought to take advan tage of these simplificat ions can save appreciable time and effort. Th e principles an d methods covered in Chapters 2 and 3 constitute the most basic part of statics. They lay the foundation for what follows not only in statics but in dynam ics as well.

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Re vi ew Problems

REVIEW PROBLEMS

157

3/97 A 50· kg acroba t pedal s her u nicycle across t le ta ut but slight ly elast ic ca ble. If t he deflect ion a t the cen-

3/95 Th e device shown in the figure is usefu l for lifting drywall panels int o position prior to faste ning to t he st ud wall. Esti mate t he magnitud e P of the force required to lift the 25-kg pa nel. State a ny as su mptions. An s. P = 351 N

ter of the 18-m span is 75 mm , determine the tension in the cable. Neglect t he effect s of the weig ht s of t he cab le and u nicycle. Ans. T ~ 29.4 kN 75 mm

t

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Problem 3/97 3/98 The uniform bar with end roller s weighs 60 lb a nd is supported by t he hor izontal a nd ver tica l sur faces a nd by the wire AC. Ca lcu la te the tensio n T in the wire a nd the reacti ons aga ins t the rollers a t A a nd at B .

Problem 3/95 3/96 The light bracket ABC is freely hin ged at A and is constrai ned by the fixed pin in the smooth slot at B . Ca lcu late the magn itude R of t he for ce supported by th e pin at A under th e act ion of the 8 0 ~ N ' m applied coupl e.

3'

Problem 3/98

Problem 3/96

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Ch a pt e r 1

Equ il ib r i um

n is supported by t he single mast as shown. T he sign . support ing fra mewo rk , a nd mast toget her weigh 600 lb. wit h cent e r of gr av ity 10 ft away from th e vert ica l centerline of th e ma st. When t he sign is subjected to th e direct blast of a 75 ml/b r wind, a n ave rage pressu re diffe renc e of 17.5 Ib/ ft 2 is developed bet ween the fron t a nd ba ck sides of the sign wit h t he resu ltant of the wind-pressu re forces acti ng at the cent er of the sign. Deter mine the magn itudes of the force an d mome nt rea ction s a t th e base of th e mast. Such resu lts \..-oul d be inst ru me nta l in t he design of th e base. A ns. R = 1396 lb. M = 25,600 lb-Ft

3/99 A free way sign measu ri ng 12 ft by 6

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Problem 3/100

I 6'

3/101 A vert ical force P on t he foot pedal of the bell cran k is required to produce a tension T of 400 N in t he vertical control rod. Determ ine t he cor res ponding bea ring react ions a t A a nd B. Aus. A = 183.9 N up . B = 424 N up

I

14'

7'"

120 mm

- x

Problem 3/99 3/100 Magne tic ta pe under a te nsion of 10 N at D passes a round the gu ide pu lleys a nd through th e erasing head at C at constant speed. As a res u lt of a small a mount of frict ion in the bea rin gs of th e pulleys, the tape at E is under a ten sion of 11 1';. Det ermine th e ten sion T in the supporting spring a t B. T he pla te lies in a hori zontal plan e a nd is mou nt ed on a precision needle beari ng a t A .

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200 mm

r/

T = 400 N

Problem 3/101

....... x

Revi ew Problem s

3/1 02 Explain why the 50-kg u niform circu lar rod cannot he in stat ic equ ilibr iu m when in the indicated position.

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Problem 3/102

3/1 03 The power uni t of the pest-hole digger supplies a torq ue of 4000 lb-in. to t he auger. The ar m B is free to slide in the su pporting sleeve C but if not free to rotat e about t he horizontal axis of C. If the unit is free to swivel about the vertica l axis of the mou nt D , determine the force exerted against the right rear whee l by the block A (or A ' ), which prevents the unbraked truck from ro lling. (H i n t : View the syste m from above.) An s. A ' = 41.7 lb

Problem 3/104

3/105 Eac h of th e three u nifor m 1200-mm bars has a mass of 20 kg. Th e ba rs are welded togeth er into the configura tion shown and suspended by t hree verti cal wires. Bar s AB and BC lie in th e hor izonta l x-y plane, an d the third bar lies in a plan e parallel to the x -z plane. Compute t he ten sion in each wire . Ans. T A = 147,2 N, T B = 245 N, T c = 196.2 N Z I

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Problem 3/103

3/104 The joints at a an d A are verti cal hin ges which can support force in all th ree coordinate directions and moment about th e x- and a-axes. The mass of arm OA is 2 kg, th at of arm AB is 2.5 kg, and th at of tray C is 4 kg. Th e mass cente r of each memb er is located at its geometr ic center. Determine the reaction at a for the configurat ion shown.

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Problem 3/105

160

Chapter 3

Equ ilib rium

3/106 A large sym met rical drum for dryin g sa nd is ope rated by t he gea red motor dri ve shown. If the mass of th e sa nd is 750 kg an d an average gea r-toot h force of 2.6 kN is supplied by t he motor pinion A to th e drum gea r normal to th e con ta ct ing su rfaces at B. ca lcu lat e the average offset j of the center of mass G of the sa nd from t he vertica l cen ter line. Neglect a ll frict ion in th e sup port ing rollers .

3/108 Three identical stee l ba lls, eac h of mass til , are placed in the cylindrica l ri ng which rests on a ho rizonta l su rface a nd whose height is slightly greater t han the radius of t he ba lls . The diam et er of t he ring is such t hat the balls a re virt ua lly touch in g one a not he r. A four th iden t ica l ball is th en placed on top of t he three ball s. Determine th e force P exerted by t he ring on each of t he three lower ba lls .

Detail of contact at B //

ci/ Problem 3/108

Probl em 3/ 106

1 /107 The 900-lb boom with center of gravity at G is held in t he posit ion shown by a ball-an d-socket jo int a t o an d the two cab les AB a nd A G. Determi ne the two cable tensions a nd the X- , y o, a nd a-com ponents of the force rea ctio n a t O. Ans . T A il ~ 4541b, T A C ~ 554 lb Ox = - 489 lb. 0 ,. = -582 lb, O. = 1445 Ib

3/109 T he drum a nd shaft a re welded togeth er a nd have a mass of 50 kg wit h mass cen te r at G. The shaft is subjecte d to a torq ue (couple) of 120 N - m, and the drum is prevented from rotating by th e cord wrap ped secure ly a round it a nd a ttached to point C. Calculate t he magn itudes of t he forces su pported by bear in gs A a nd B. An". A = 6 10 N, H = 656 N

C _ -J/

B DA

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Dime ns ions in mill im et ers

-;

...... x

50'

Prob lem 3/109

0 0 = 30' DB = D C = 40'

Probl em 3/1 07

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Review Problems

3/11 0 Th e L-shap ed bar is su pported by a ball-and- socket joint at 0 [case (a) ] and th e two cables as shown . Explain why this configu ration is improperly con strained. (b ) Th e ball-and-socket joint is now replaced by th e u niversal joint which can suppor t, in addition to th ree force reac tions , a moment about th e y-ax is but no moments about th e x- and z-axes. Plot the two cable tensions, the magnitu de of the force reaction at 0 , and th e momen t reaction at 0 as fu nct ions of t he position x of th e 100-lb cylinder over the rang e 0.5 :S x :S 4.5 ft . Explain any u nusual characteristics of these plots. Neglect the weight of the bar t hroughout.

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i

f. 2"~ Pawl C

D

Problem 3/111

I

'Computer-Oriented Problems

· 3/ 112 The jib cran e is designed for a maximu m capacity of 10 kN, a nd its u niform I-beam has a mass of200 kg. (a) Plot t he magnitude R of the force on the pin at A as a funct ion of x through its operating range of x :::: 0.2 m to x :::: 3.8 m. On the sa me set of axes, plot t he x- an d y-components of the pin reaction atA . (b ) Determine the minim um value of R an d the corresponding value of x . (c ) For what value of R should the pin at A be designed? (Use g ~ 10 rn/ s2.)

100Ib Problem 3/ 1 10 y I

3/111 Th e device shown in section can su pport the load L at vario us heights by resetting the pawl C in another toot h at the desired height on the fixed vertical column D. Determine the distan ce b at which th e load should be positio ned in order for the two rollers A and B to support equal forces. The weight of th e device is negligible compared with L . A ns. b = 10.33 in.

I A

Problem 3/112

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I

L __ - x

162

Chapter 3

Equi libr ium

-3 /113 Determine and plot as a function of 8 th e tension T which must be produced by th e winch in order to steadily rotate th e stru ctural member about th e fixed pivot at O . Use the range 0 :s; H :s; Hmax ' where 0mmc is the value of (J at which T goes to zero. The stru ctu ral member has a mass of 35 kg per meter of its length. Ans. T = [ 2750 cos 9 .+ 687 coste + 60' )] [8 + 4 sm (H + 60°)

-] / 115 Two traffic signa ls are attached to t he 36·ft su pport cable at equal inte rva ls as shown. Determi ne the equilibrium configurat ion angles a, /3, and y, as well as th e tension in each cable segme nt. A ns . a = 14.44' , {3 = 3.57 ', y = 18.16' T AB ~ 529 lb, T BC ~ 5 13 lb, T CD = 539 lb

- - - - 35'- - - - 1

I

8 cos(9 + 60' )]' 12'

(Jm ax = 79.1°

••

100lb

C

200 lb

Problem 3{115 - 3/ 116 Th e two traffic signals of Pr ob. 3/ 115 are now repositioned so th at segme nt BC of the 36· ft su pport cable is 10 ft in length and is horizontal. Specify the necessa ry lengt hs AB and CD and the tensions in all three cab le segme nts.

It - - 2 m

35'- - - - 1

Problem 3{113 -]/114 The 50·kg cylinder is sus pended from a clamping collar at C which can be positioned at any horizonta l positi on x between th e fixed supports at A and B . Th e cable is 11 m in len gth . Determine and plot the ten sions in cable segme nts A C and BC as functions of x over the int erval 0 :S X :s; 10 m. What is the maximum value of each tension and for what value of x does it occur?

~ 10 m

1

C 10'

•

•

100lb

Problem 3{116

~ A ---]-------~B

c

50 kg

Problem 3{114

Marwan and W aseem AI-Iraqi

B

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\l 200 lb

Rev iew Problems

163

-1 /117 The horizonta l boom is supported by the cablesAB

and CD and by a ba ll-and -socket joint at O. To determine the influen ce on the reaction at 0 of th e position of th e vertical load L along th e boom, we may neglect th e weight of the boom . If R represents the magnitude of the total force at 0 , determin e by calculus the minimum ratio R IL and th e corresponding value of .r, Then write a computer program for R IL and plot the results for 0 < x < 6 m as a check on your calcur.1a;:t;:io"n;::s::;.",..._--,;:-:-:Ans. R/L ~ J 47x' / 162 - x / 3 + 1 (R / L)min = 0.951 at x = 0.574 m

Problem 3/ 118 '3/119 The mass center of the 1.5-kg link OC is located at G, and the spri ng of constant k = 25 N / m is u nstretched when 0 = O. Pl ot the ten sion T required for static equilibriu m over the range 0 S (J S 90° an d state t he values of T for (J = 45 ° an d (J = 90°, Ans. T = 5.23 N at e ~ 45' T = 8.22 N at e = 90' T

<,

Problem 3/117 -1 /118 Th e basic features of a small backhoe are shown

in th e illustration . Member BE (complete with hydr auli c cylinder CD and bucket-control links DF and DE ) weighs 500 lb with mas s center at G t . Th e buck et and its load of clay weigh 350 lb with mas s cen ter at G2 • To disclose th e operational design cha rac te rist ics of the backh oe, determine and plot th e force T in the hydraulic cylinder A.B as a fun ction of th e angu lar position 0 of member BE over the ran ge 0 S 0 s 90°. For what valu e of 0 is th e force T equal to zero? Memb er OB is fixed for th is exercise ; note th at its controlling hydraulic cylinder (h idden) extends from near point 0 to pin 1. Simil arly, t he bucket-control hydrauli c cylinder CD is held at a fixed length.

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00 = 160 mm DB = BC = 240 mm

Problem 3/119

The Skydome and eN Tower in Toronto are distinctly different structures. In both cases. however . the eng ine e rs had to ca lculat e the force supported by each major component of the ove ra ll str uct ure. MalWan and Waseem At-Iraqi

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STRUCTURES CHAPTER OUTLINE 4 /1 4/2 4/3 4/4 4/5 4 /6

Introduction Plane Trusses Method of Joints Method of Sections Space Trusses Frames and Machines

Chapter Review

4 /1

INTRODUCTION

In Cha pter 3 we studied th e equilihr ium of a single rigid body or a syste m of connected memb ers t reated as a single rigid body. We first d rew a free-body diagram of t he body showing a ll forces external to t he isolated body a nd then we applied the force a nd moment equatio ns of equilibrium. In Chapter 4 we focu s on the determination of the forces internal to a st ructure , that is, forces of act ion and reaction between t he connected memb ers. An engineering struct u re is any connect ed sys te m of members built to su pport or transfer forces and to sa fely withstand the loads applied to it. To determine the forces internal to a n engi nee ring structure , we must d ismember th e st ruc t ure a nd ana lyze se parate free -body diagrams of individual members or combinations of members. This analysis req uir es ca re ful application of Newton's th ird law, whi ch states th at eac h action is accompa nied by a n equa l and oppos ite reaction . In Cha pte r 4 we analyze the internal for ces acting in several types of struct ures, nam ely, trusses, frames, and machines. In this treatment we consider only statically determinate stru ctures, whi ch do not hav e mor e su pporti ng cons train ts than are necessary to maintain an equilibr ium configu ration . Thus, as we hav e a lready seen, the equations of equilibrium are ade quate to determine all u nknown reactions. Th e analysis of trusses, fram es a nd ma chines, a nd beams under concentrated load s cons tit utes a straightforwa rd applicati on of th e mat erial developed in th e previous two chapters. The ba sic procedure developed in Cha pt er 3 for isolating a body by construct ing a cor rect free-body diagr am is essential for the analysis of statica lly deter min ate st ructure s. MalWan and Waseem AI-Iraqi

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165

166

Ch ap t er 4

Str uct ures

Stringer Cross beam

\

~~~

L

Figure 4/ 1

Pra tt

Howe

Warren

Common ly Used Brid ge Trusses

Common ly Used Roof Tru sses

Figure 4/2 Marwa n and W aseem AI-Iraqi

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Articl e 4 /2

4 /2

Plane Trusses

B

PLANE TRUSSES

A framework composed of members joined at thei r ends to form a rigid st ru ct ure is called a tru ss. Bridges, roof supports, derricks, an d other such structures are common examples of trusses. Structura l members commonly used are l-bearn s, channels, angles, bars, and special shapes which are faste ned together at t heir ends by welding, r iveted connecti ons, or large bolts or pins. \Vhen the members of the truss lie essentially in a single plane, the truss is called a plane truss . For bridges and similar structures, plane trusses are commonly utilized in pairs with one t ru ss assembl y placed on eac h side of the st ruct ure. A sect ion of a ty pical brid ge structu re is shown in Fig. 4/1. Th e combined weight of th e roadway and vehicles is tra nsfer red to the longit udina l str ingers, t hen to the cross bea ms, a nd fina lly, wit h the weights of the stringe rs and cross beams accounted for, to the upper joints of the two plane trusses which form the vertical sides of the structure. A simplified mode l of th e truss stru cture is indicated at t he left side of the illust ra tion; the forces L re prese nt the joint loadings. Several examples of commonly used trusses which can be ana lyzed as plane trusses are shown in Fig. 4/2. Simple Trusses

A 'a )

[)

A

' bJ

F

The basic element of a plan e truss is t he t ria ngle. Three bars joined by pins at th eir ends, Fig. 4/3a , cons titute a r igid frame. Th e ter m rigid is used to mean noncollapsible and also to mean that deformation of the members due to induced internal strains is negligible. On the other hand, four or more bars pin-jointed to form a polygon of as many sides constitute a nonrigid frame. \Ve can make the nonrigid frame in Fig. 4/3b rigid, or stable, by adding a diagonal bar joining A a nd D or B and C and t hereby for ming two triangles. We can exte nd the st ructure by add ing additional unit s of two end-connected bars, suc h as DE an d CE or AF and DF , Fig. 4/ 3c, which are pin ned to two fixed join ts. In t his way the entire structure will remain rigid. Structures built from a basic triangle in the manner described are known as simple trusses. \Vhen more members are present than are needed to prevent collapse, t he truss is stat ically ind eterminate. A statically indetermina te truss cannot be analyzed by th e equa tions of equilibrium alone. Additional members or supports which are not necessary for maintaining the equilibrium configuration are called redunda nt. To design a truss we must first determine the forces in the various members and then se lect appropriate sizes and structural shapes to with stan d the forces. Several ass umptions are made in the force analysis of simple trusses. First, we assume all members to be two-force members. A two-force member is one in equilibrium under the action of two forces only, as defined in genera l te rms with Fig. 3/4 in Ar t. 3/3. Each member of a truss is normally a straight lin k joi ning the two points of application of force. The two forces are applied at the ends of the member and a re necessarily equal, opposite, and collinear for equilibrium. The member may be in tension or compression, as shown in Fig. 4/4. \Vhen we represent the equilibriu m of a portion of a two-force member, the tension T or compress ion C acting on the cut section is the same Marwan and Waseem AI-Iraqi

167

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e

A (e )

Figure 4/3

T e

IT

Ie e

Tension

Compression

Two-Force Members Figure 4/4

16 8

Ch ap t e r 4

S truct u res

,

for all sections. We assum e here that t he weight of th e memb er is s mall compared with the force it supports. If it is not, or if we must account for th e s mall effect of th e weight, we ca n replace the weight W of the memb er by two forces, each W/ 2 if the memb er is uniform , with one force acting at each end of the member. These forces, in effect, are treated as loads exte rnally applied to the pin connections . Accounting for th e weight of a memb er in thi s way gives t he cor rect result for the average tension or compress ion along the member but will not accoun t for the effect of bendin g of th e member .

,

};~'0--~':"""""/",~.\ I

l

Truss Connections and Supports

Figure 4/5

F

E

D

L (a)

\Vhen welded or riveted connections are used to jom st ructural members, we may usually ass ume that the connectio n is a pin joint if the centerlines of the members are concu rrent at the joint as in Fig. 4/5. \Ve also ass ume in the analysis of simple trusses that all external forces are applied at the pin conn ection s. This condition is satis fied in most trusses. In br idge t russes th e deck is usu ally laid on cross beam s which are supported at th e jo int s, as shown in Fig. 4/1. For large trusses , a roller, rocker, or some kind of slip joint is used at one of the supports to provide for expansion and contraction due to temperature changes and for deformation from applied loads. Trusses and frames in which no such provision is made are statically indeterminate, as explain ed in Art. 3/ 3. Fig. 3/ 1 shows exa mples of such joints. Two methods for the force analysis of simple trusses will be given. Each met hod will be expla ined for t he simple tru ss shown in Fig. 4/60 . The free-body diagram of the t russ as a whole is shown in Fig. 4/6 b. The exte rn al reac tio ns are usua lly determined first , by applying the equilibrium equation s to the truss as a whole. Then the force analysis of the remainder of t he t ru ss is performed.

4 /3 L

R, (b)

Figure 4/6

y I I I I

r

AF

AB

'-'=---_ ...1

~- - - x

Tension

RI

Figure 4/7 MarNan and Waseem AI-Iraqi

METHOD OF JOIN TS

This method for finding the forces in the members of a truss cons ists of satisfying the condit ions of equilibr iu m for the forces act ing on the connecting pin of eac h joint. Th e method ther efore deal s with th e equ ilibrium of concurrent forces, and only two independent equilibrium equations are involved. We begin th e ana lysis with an y joint whe re at least one known load exists and where not more than two unkn own forces are present. The solut ion may be started with th e pin at th e lell end. It s free- body diagram is shown in Fig. 4/ 7. Wit h th e joints ind icated by letter s, we usually designat e th e force in each memb er by the two letter s definin g the ends of t he memb er . Th e prop er directions of t he forces s hould be eviden t by inspecti on for thi s simple case. Th e free-body diagram s of portions of memb ers AF and AB are also shown to clearl y indicat e the mechanism of the action and reaction . The member AB actually makes contact on th e left side of t he pin, although t he force AB is drawn from th e right side a nd is shown acting away from the pin. Thus, if we consistently draw the force arrows on the same side of the pin as the member , then ten sion (such as AB ) will a lways be ind icat ed by an ar row away www.gigapedia.com

Article 4/3

fro m the pin , an d compression (such as AF) will always be indicated by an arrow toward t he pin. Th e magnitude of AF is obtained from the equa tion ':::.Fy = 0 an d AB is t he n found from ':::.Fx = O. Joint F may be analyzed next, since it now contains only two unknow ns , EF and BF. Proceeding to the next jo int ha ving no more than two unknowns, we subsequently analyze joints B, C, E, and D in that orde r. Fig. 4( 8 shows the free-body diagram of each joint and its corresponding force polygon, which represents graphically the two equilibrium conditions 'i..Fx = 0 and 'iFy = O. The numbers indicate the order in which t he joints are analyzed . We not e that , when joint D is finally reached, th e compute d reaction R 2 must be in equ ilibriu m wit h t he forces in members CD and ED, which were determined previously from the two neighb oring jo ints . This requi remen t provides a check on the cor rect nes s of our work. Note that isolation of join t C shows tha t the force in CE is zero when the equa tio n ':::.Fy = 0 is applied. The force in

2

zl

EF

EF

{l

BF Joint F

AB

4

t CE =O

R1

BC

Joint A

E

)

CD

Joint C

3

BF BE

AB

BC

5

BE

BC

~E DE BE

L

EF

BF

Joint E 6

AB CD L

Joint B

~

CD

D~R,

R, Joint D

y I I

f->--~---I----'>; D

R1

L - - -

x

R,

L Figure 4/8

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Method of Joi nts

169

170

Chapter 4

Struct u re s

th is member would not be zero, of course, if an exte rnal vertical load were ap plied at C. It is often convenien t to indicate the tension T and compression C of t he various members directly on t he origi na l truss diagram by dra wing arrows away from the pins for ten sion and toward th e pins for compression. This designation is illustrated at th e bottom of Fig. 4/8. Sometimes we cannot initially ass ign the correct direction of one or both of the unknown forces acting on a given pin. If so, we may make an arbitrary assignment. A negat ive computed force value indicates that the initially assumed direction is incorrect.

Internal and External Redundancy If a plan e t rus s has more exte rn al supports than are necessa ry to ensure a stable equilibrium configuration, th e truss as a whole is statically indeterminate, and the extra supports constitute external redundancy. If a truss has more internal members than are necessary to pr event collapse whe n t he truss is removed from its supports, then the extra members constitute internal redundancy and the truss is again statically indet erminat e. For a truss which is stat ically determinate externally, there is a definit e relation between th e number of its members a nd th e number of its joints necessary for inte rnal sta bility without redundan cy. Becau se we ca n specify th e equilibrium of each joint by two scalar force equat ions, th ere are in all 2j such equ ations for a truss withj joints. For th e entire truss composed of m two-force mem bers and having the maximum of three unknown support reactions, there are in all m + 3 unkn owns (m tension or compression forces and three reactions). Thus, for any plan e truss, th e equ ation m + 3 = 2j will be sa tisfied if the truss is sta tically determinate internally. A simple plan e truss, formed by starting with a tri angle and add ing two new members to locate each new joint with respect to the exist ing struc ture, satis fies th e relation auto matica lly. Th e condition holds for t he initial tri angl e, where m = j = 3, and m increases by 2 for each added joint while j increases by 1. Some othe r (nonsimple) statically determinate trusses, such as th e K-truss in Fig. 4/2, are arranged differently, but can be see n to satisfy the same relation. This equa tion is a necessary condit ion for stability but it is not a sufficient conditio n, since one or more of the m members can be arran ged in such a way as not to contribute to a sta ble configu ra tion of t he ent ire truss. If m + 3 > 2j, there a re more members t ha n independent equations, a nd the tru ss is statically indeter minate intern ally wit h redundant members present. If m + 3 < 2j , there is a deficiency of internal members, and the truss is unstab le an d will collapse under load. Special Conditions We often encounter seve ral special conditions in the analys is of trusses . When two collinear members are under compression, as indicated in Fig. 4/9a, it is necessary to add a t hird member to main tain Marwa n and Waseem AI- Iraqi

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Ar t icl e 4 /J

/

h,

171

x·

X / /

M ethod of Jo int s

~ x -r - -

1/ L

'F.F;r = 0 require s F1 = }'i rF;r' = 0 require s FJ = F4

~ zr; = 0 requires F 1 = 0 r F;r' = 0 requires F2 = 0

rFy = 0 requires F3 = 0 ! Fx = 0 requires F1 = Iii (c )

(bJ

(c)

Figure 4/9

alignment of th e two members and prevent buckling. We see from a force su mmation in they-direction that the force F3 in the third member mus t be zero and from the x-direct ion that F 1 = F 2' Th is conclusion holds regardless of the a ngle 0 and holds also if th e collinear me mbers are in te nsion. If an externa l force with a component in the y-direction were a pplied to the join t, then F 3 would no longer be zero . Wh en two non ccllinear members are joined as shown in Fig. 4/ 9b, th en in t he absence of an exte rnally applied load at t his joint, th e forces in both members must be zero, as we can see from the two force summations. When two pairs of collinear members are joined as shown in Fig. 4/ 9c, the forces in each pair must be equa l and opposite . This conclusi on follows from the force summations indicated in the figure. Tru ss pan els are frequently cross-braced as shown in Fig. 4/ 10a . Suc h a panel is statically indeter minate if eac h brace can support eit he r tension or compression . However, whe n the braces are flexible members incapable of supporting compression, as are cables, then only the tension member acts and we can disregard the other member. It is usually evident from the asymmetry of th e loading how th e pan el will deflect. If the deflection is as indica ted in Fig. 4/lOb, th en member AB should be retained a nd CD disregarded. When this choice cannot be made by inspection, we may arbitrarily se lect the member to be retained. If the assumed te nsion turns out to be positive upon calculation, then the choice was correct. If the assumed tension force tu rns out to be negat ive, t hen the opposit e member mu st be ret ained and th e calculation redon e. We can avoid simultaneous solution of t he equilibrium equat ions for two unkn own forces at a joint by a careful choice of reference axes . Thus, for the jo int indicated schematically in Fig. 4/ 11 where L is known a nd F1 and F2 are unknown, a force summation in the x-direction eliminates reference to F1 and a force summation in the x ' -direction eliminates reference to F 2 . When the angles in volved are not easily found , then a simultaneous solution of the equations using one set of reference direct ions for both u nknowns may be pr eferab le. Ma rwan and W aseem AI- Iraqi

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D

B

D

B

wN c

A

c

A

(b)

(a)

Figure 4/10

Figure 4/1 1

172

Cha pt e r 4

St ructures

Sample Problem 4 /1 Compute th e force in each member of the loaded cantilever truss by the method of joints .

Solution. If it were not desired to calculate the exte rnal reactions at D an d E , the analysis for a cantilever truss could begin with t he joint at th e loaded end. However, this truss will be analyzed completely, so t he first ste p will be to compute th e exter nal forces at D and E from the free-body diagram of the truss as a whole. Th e equa tions of equilibrium give [~E = [~Fx =

OJ

[~Fy

OJ

=

5T - 20(5) - 30(10)

OJ

0

Ex = 69.3 kN

80 sin 30° + Ey - 20 - 30

0

Ey = 10 kN

80 cos 30

-

30 kN

20kN

T = 80kN

0

Ex

0

A

T y I I

Next we draw free-body diagrams showing th e forces acting on each of t he connect ing pins. Th e correc tn ess of the assigned directions of the forces is verified when each joint is considered in sequence. Th ere should be no question about th e cor rect direction of th e forces on joint A. Equilibrium requires

CD

OJ

0.866AB - 30

0]

AC - 0.5(34.6 )

o o

AB

~

34.6 kN T

Ans.

AC

~

17.32 kN C

Ans.

where T sta nds for tension and C stands for compre ssion. Joint B must be analyzed next , since there are more th an two unknown forces on joint C. The force BC must provide an upward componen t, in which case BD mu st balance the force to the left . Again th e forces are obtained from

OJ

0.866BC - 0.866 (34.6)

0]

BD - 2(0 .5)(34.6)

o o

= 34.6 kN

C

Ans.

BD = 34.6 kN T

Ans.

BC

L __

OJ

0.866CD - 0.866 (34 .6) - 20 CD

OJ

~

57.7 kN T

Ans.

CE - 17.32 - 0.5(34 .6) - 0.5(57.7)

o Ans.

Finally, from joint E th er e results 0.866DE

~

10

DE

x

5m

5m

30 kN

20kN

Ey

y I\B

I I

I I 600 _ AC - - x

BD

I\B =~ OO

34.6 kN 600

BC

30kN

J oint A

Joi ntB

Helpful Hint

o

CE = 63.5 kN C

I:EFy = 0]

5m

Ex

J oint C now contai ns only two unknowns, and these are found in the same way as before : [~Fy =

30 0 60 0

5m

11.55 kN C

an d t he equation '2:.F:r: = 0 checks.

A ns.

CD It

should be stressed th a t t he te nsion/com pression designa tion refers to th e member , not the joint . Note that we dr aw th e force arrow on the same side of th e joint as the member which exer ts th e force. In thi s way ten sion (ar row away from th e joint) is disti ngu ished from compress ion (ar row toward the joint). BC =

34.6 kN

DE 60 AC= 17.32 kN

CE

Joint C www .gigapedia.com

69.3 kN

CE =

63.5 kN 20kN

Marwan and Waseem AI-Iraqi

0

10kN

Joint E

Art icle 4/3

PROBL EMS

173

Problems

4/4 Calculate the forces in members BE and BD of the loaded trus s .

Introductory Problems 4/1 Dete rmine the force in each member of the simple equilateral truss. An s. AB ~ 736 N T , AC ~ 368 N T , BC = 736 N C

B

2

A

8' 2m

2m

~2

C

A'

8'

8'

8'

E

'B 10001b

2m

Problem 4/4 75

k~

4/5 Determine the force in each member of the loaded

C

truss.

An. , AB = 12 kN T , AE BC = 5.20 kN T , BD BE ~ 5.20 kN C, CD ~ DE

Problem 4/ 1 4/2 Dete rmine the force in each member of the loaded truss. Discuss the e lTects of varying the angle of the 450 suppor t sur face at C. A

3 kN C 6 kN T 6 kN C

A

B

6'

B

C 3D'"

E

2.5'

1001b

3 kN

D

C

Problem 4/5 45°

- - - -

4 /6 Calculate th e force in eac h memb er of th e loaded tru ss. E "'- - - "'-D -2 kN

Problem 4/2

4/3 Dete rmine the force in each member of the truss. Note the presence of any zero-force members.

An". AB = 5 kN T, BC = 5/2 kN C CD ~ 15 kN C, A C ~ 5 ,/5 kN T, AD ~ 0

3m

3m B Problem 4/6

Prob lem 4/ 3

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174

Ch apter 4

Structures

4/7 Determine the force in each member of the loaded truss. Make use of the symmetry of the truss and of the loading. Ans. AB ~ DE ~ 96.0 kN C All ~ EF = 75 kN T, BC = CD = 75 kN C BH ~ CG ~ DF = 60 kN T CH ~ CF ~ 48.0 kN C, GH ~ FG = 112.5 kN T

c

B

c

D

'i---- - - ---->:D

4m

5m

5m

E

5m

G

F

60 kN

30 kN

4 kN

2 kN Problem 4/ 9

30 kN

Problem 4/7

Representative Problems

4/8 Determine the force in each member of the loaded truss. All triangles are isosceles. 1 300 1

I

lO kI'

I

4m

4/10 Solve for the forces in members BE and BD of the truss which supports the load L. All interior angles are 60° or 120°.

A

B

~

c c

4m

B

D

F

t

G

Problem 4/ 8 4/9 Determine the force in each member of the loaded truss. All triangles are equilateral. Ans. AB ~ 9,13 kN C, AE = 5 ,13 kN T kN C, BD = 3 ,13 kN C, BE = kN C BC = kN T, DE ~ kN T CD ~

,?J3

lfJ3

E

L

Problem 4/ 10

iJ3

¥J3

4/11 Determine the force in member AC of the loaded truss. The two quarter-circular members act as twoforce members . Ans. A C = T 2

!::

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Article 4 /1

Prob l em s

175

L

B

Problem 4/11

4/12 Calcul at e the force s in members CG a nd CF for th e truss shown.

2 kip s

Problem 4/14

2 kN

4 /15 Th e equiangular t russ is loaded a nd su ppo rted as shown. Determine the for ces in a ll members in terms of th e horizontal load L . Ans. AB BC = L T , AF = EF = L C DE = CD = L/2 T, BF = DF = BD = 0

F

Problem 4/12

4/13 Each member of the truss is a un iform 20·ft bar weighing 40 0 lb. Ca lcu late the average ten sion or com pr ession in each membe r du e t o the weights of

t he members.

Ans. AB AE BD ED

=

=

2000/ ,13 Ib C BE = 800/ ,13 Ib T 1400/ ,13 Ib C =

B

Problem 4/15

4 /16 Determ ine t he force s in members BI , CI, a nd HI for t he loaded truss. All a ng les are 30°, 60°, or 90°.

D

E

20'

1000/ ,/ 3 1b T

BC

= CD

20'

Problem 4/13

4/14 A drawbridge is bein g raised by a cable £ 1. Th e four joi nt loadings shown result from the weight of t he roadway. Det ermine th e forces in members EFt DE, DF, CD, and FG.

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Problem 4/16

176

Ch ap ter 4

Sto uc tur es

4/17 A snow load transfers the forces shown to the upper joints of a Pratt roof truss. Neglect any horizontal reactions at the supports and solve for the forces in all members. Ails. AB ~ DE = BC ~ CD = 3.35 kN C AH = EF = 3 kN T , BH ~ DF ~ 1 kN C CF ~ CH ~ 1.414 kN T , FG = GH = 2 kN T

4/20 Determine the force in each member of the pair of trusses which support the 5000-lb load at their common joint C.

1 kN

~c

1 kl'i

1 kN 1 1<.."

D

F

2m

Problem 4/17

4/18 The loading of Prob. 4/ 17 is shown applied to a Howe roof truss . Neglect any horizontal reactions at the supports and solve for the forces in all members. Compare with the resul ts of Prob. 4/17.

1 kN

1 kN 1 kN

B

1 kN

D

2m

Problem 4/20

4/21 The rectangular frame is composed of four perimeter two-force members and two cables AC and BD which are incapable of supporting compression. Determine the forces in all members due to the load L in position (0) and then in position (b) . A il S. (0) AB = AD ~ BD = 0, BC ~ I. C 5L 4L A C = '3 T, CD ~ '3 C

Ibl AB 2m

H

2m

G

2m

F

~

AD

~

BC

BD

~

A C ~ 51. T CD ~ 41. C

2m

3

Problem 4/ 18

'

3

L

4/19 Calculate the forces in members CF , CG, and EF of the loaded. truss. Ans . CF ~ 15381b C, CG = 41 70 lb T , EF = 0

( a)

4d

20001b

10' D

3d 10'

20001b C

nl'----------~ l

+

pe-----"'----~B

t

L

10'

f---

-

- 26' -

-

-

Probl em 4/21

-..j

Probl em 4/ 19

Marw an and W aseem AI-Iraqi

C

H b)

10'

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~

0

Article 4/3

4/22 Determine the forces in members AB , CG, and DE of th e loaded truss. U2

L

L

L

U2 E

Problems

177

4/24 Ver ify th e fact that each of the trusses contains one or more eleme nt s of redundancy and propose two separate cha nges, either one of which would remove the redundancy and produce complete stat ical determinacy . All members can support compression as well as ten sion .

/ A

F D

Problem 4/22 (b)

(0 )

4/23 T he movable ga ntry is used to erect and pre pare a 500·Mg rocket for firing. The primary structure of th e ga ntry is approx imated by t he symmetrical plan e t russ shown, which is sta tically indeterm inate. As the gantry is positi oning a 60-Mg section of th e rocket suspended from A , strain-gage measurements indio cate a compressive for ce of 50 kN in memb er AB and a ten sile force of 120 kN in memb er CD du e to th e 60-Mg load . Calcu late th e corres ponding forces in memb er s BF and EF . Ails. BF ~ 188.4 k N C, EF ~ 120 kN T

H

C

D

I

16 m

E

'"

F

D

C

B

(d l

(e)

Problem 4/24

4/25 Analysis of the wind act ing on a small Hawaiian churc h, which wit hstood th e 165-mi/h r winds of Hu rricane Iniki in 1992, showed th e forces t ra nsmit ted to each roof truss pa nel to be as shown. Treat the structure as a symmet rical simple truss and neglect a ny horizontal component of th e support reaction at A . Ident ify the truss mem ber which supports th e largest force, te nsion or compression , and calcu lat e this force. Ans. FD ~ 24,500 Ib T 2450 lb

16'

Problem 4/ 23

f"------- :2 4 '----~r

Probl em 4/25

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178

Chapt er 4

Struc tur e s

4 /26 The 240· ft struct ure is used to pr ovide va rious support services to launch veh icles pri or to liftoff. In a test , a re-t e» weight is sus pende d from j oints F and G, wit h its weight equa lly divided. between the two joints. Determine t he forces in members GJ a nd GI. Wha t would be you r pa th of jo in t analysis for mem be rs in t he vertica l tower, such as AB or KL?

5 pa nels at 3 m H

I' 3m I

D

J

----i E IA

I '

T- IJ 1--'~-7:~

1.8 kN

3 m 15°

-j - -

3m

i

5m

J_ - .J5._ . A. Problem 4/27 .... 4/28 Find t he force s in members EF, KL , an d GL for th e Fink truss shown . Ans. EF = 75.1 kN C, KL 40 kN T GL 20 kN T 10 kN

I

20kN

E

20kN

Problem 4{26

IO kN

.. 4/27 The tower for a tra nsmi ssion line is modeled by th e truss show n. The crosse d members in the cen ter sections of t he t russ may be ass umed to be capable of supporting tension on ly. For the loads of 1.8 kN applied in the vertical plan e, compute th e forces indu ced in me mbers AB, DB, a nd CD. Ans. AB ~ 3.89 kN C, DB ~ 0, CD ~ 0.932 kN C

Marwan and Waseem Al-l raqi

20kN

1-- - - - - 6 panels at 5 m - - -- - -

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Problem 4{28

Articl e 4 / 4

4 /4

Me th o d of S ecti on s

179

METHOD OF SECTIONS

When ana lyzing plane tr usses by the method of joint s, we need only two of the three equilibrium equations because the procedures involve concur re nt forces at each joint. We can tak e adva nta ge of the th ird or moment equation of equilibrium by selecting an entire section of the truss for the free body in equilibrium under the action of a nonconcurrent sys tem of forces. This method of sections has the basic advantage that the force in a lmost a ny desired member may be found dir ectly from a n ana lysis of a section which has cut th at memb er . Thus, it is not necessary to proceed with th e calculation from joint to joint until t he member in question has been reached. In choosing a sect ion of the truss, we note that, in general, not more than three members whose forces are unknown should be cut , since t here are on ly three available ind epend ent equilibrium relations.

Illustration of the Method Th e met hod of sect ions will now be illustrated for th e tr uss in Fig. 4/ 6, which was used in th e explanation of t he met hod of joints. The truss is shown again in Fig. 4/ 12a for ready reference. The external reactions are first computed as with the method of joints, by cons idering the truss as a whole. Let us determine the force in the member BE, for example. An irnaginary section, ind icat ed by th e dashed line, is passed through th e truss, cutting it in to t wo parts, Fig. 4/ 12b. Th is sect ion has cut three memb ers whose forces are initially unknown. In order for the portion of the truss on each side of the section to remain in equilibrium, it is necessary to apply to each cut member t he force which was exerted on it by the member cut away. For simple trusses composed of two-force members, these forces, either tensile or compressive, will always be in the dire ctions of th e respec tive members. Th e left-hand section is in equilibrium under the action of t he ap plied load L, t he end reaction R 1> and th e t hr ee forces exerte d on the cut members by t he right-hand sect ion which has been removed. We can usually draw the forces with their proper senses by a visual approximation of the equilibrium requirements. Thus , in balancing the momen ts about point B for th e left-hand section, t he force EF is clearly to the left, which makes it compressive, because it acts toward the cut section of member EF. The load L is greater than the reaction R 11 so that the for ce BE must be up a nd to t he right to supply the needed upward component for vertical equilibrium. Force BE is therefore tensile, since it acts away from the cut section. With the approximate magni tudes of R , and L in mind we see th at th e balan ce of moments about point E requires that BC be to the right. A casua l glance at the truss should lead to t he sa me conclusion whe n it is reali zed that t he lower horizontal memb er will st retch under the tension cau sed by bending. Th e equat ion of momen ts ab out joint B eliminates three forces from the relation, and EF can be determined directly. The force BE is calculated fr om the equilibrium equa t ion for the y-direction. Fin ally, we deter mine BC by bala ncing moments about point Marwan and W aseem AI-Iraqi

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F

A

\

E

\

R,

C

r

R,

L ( a)

F

E / EF

E

I I / I RE 1/

C

R,

L

8,

y I I

L __ x (b )

Figure 4/12

18 0

Chapter 4

Structure s

E. In this way each of th e three u nknown s has been determined independen tly of the ot he r two. The rig ht-han d section of t he tru ss, Fig. 4/1 26, is in equilibrium und er the action of R 2 and the same three forces in the cut memb ers applied in the direction s opposite to t hose for the left sect ion . The pr oper sense for the horizontal forces can easily be seen from the bala nce of moments about points B and E .

Additional Considerations It is essen tial to understand t hat in t he method of sections an entire portion of t he tru ss is considered a single body in equilibr ium. Thus, the forces in members internal to the section are not invo lved in the ana lysis of the sect ion as a whole. To clarify the free body and the forces acting exte rnally on it, th e cutt ing sect ion is prefera bly passed th rough the members and not the join ts. \Ve may use either portion of a truss for the calcula tions, but the one involvin g the smaller number of forces will usu ally yield th e simpler solution. In some cases the methods of sections and joints can be combined for an efficient solution. For example, suppose we wish to find the force in a central member of a large truss. Furthermore, suppose that it is not possible to pass a sectio n through thi s mem ber without passin g through at least four unknown mem bers. It may be possible to determine t he forces in nearby memb ers by the method of sections a nd th en progress to th e unknown memb er by the method of joints. Such a combination of the two met hods may be more expedient than exclus ive use of either meth od. The moment equations are used to great advantage in the meth od of sections . One should choose a moment center, either on or ofT the section, through which as many unknown forces as possible pass. It is not always poss ible to assign the proper sense of an unknown force when the free-body diagram of a section is ini tially drawn . Once an arbitrary assignment is made, a positive answer will verify the assumed sense and a negative result will indicate that the force is in the se nse opposite to that assumed. An alternative notation preferred by some is to ass ign all unknown forces arbitrarily as positi ve in the te nsion directi on (away from the sectio n) and let the algebraic sign of th e an swer disti nguish betw een tension and compression. Thus, a plus sign would sign ify te nsion and a minu s sign compr ession. On th e othe r hand , the advantage of assigning forces in their correct sen se on the free-body diagram of a section wherever possible is that doing so emphasizes the physica l act ion of the forces more directly, and th is pra ctice is t he one which is pr eferred here.

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Arl icle 4/ 4

Mel h o d of Sec tio n s

181

Sample Problem 4 /2 Calculate the forces induced in memb er s KL, CL , and CR by the 20-too load on the cantilever tru ss.

CD

Solution. Although the vertical components of th e re actions at A a nd M are stat ically ind et ermi nate wit h the tw o fixed supports, all memb ers oth er than AM are statically determinate. We may pass a section dir ectly th rough member s KL, CL , and CR and ana lyze the porti on of th e truss to the left of this section as a statically determinate rigid body. The free -body diagram of the portion of the truss to the left of the section is shown. A momen t su m about L qu ickly ver ifies the assignment of CR as compr ession, and a moment sum about C quickly discloses that KL is in ten sion . The direction of CL is no t quite so obviou s un til we observe that KL and CB inter sect at a point P to the right of G. A moment sum about P elimina tes re fere nce to KL an d CB a nd shows that CL mu st be compressive to balance the moment of the 20-ton force about P . With these consi dera tions in mind th e solution becomes straight forwa rd, as we now see how to solve for each of the three unknown s inde pend ently of the other two. Su mming moments abou t L require s finding the mom ent arm BL = 16 + (26 - 16)/ 2 21 ft. Thus, 20(5)(12) - CB (21)

~

0

CB

~

57.1 Ions C

Ans.

Next we tak e moments about C, wh ich requires a calculation of cos 9. From the given dimensions we see 9 = tan - 1{5/12) so that cos 8 = 12/13. Th er efore, [LAIc =

OJ

20(4)(12) - ¥SKL(16) = 0

KL = 65.0 Ions T

Ans.

H

11 16' i-G f---.'F:---:E~-D!':--C~,--+-+---,.·t:J -

-

- 6 pa nels at 12'-

-

-

20 tons 8

KL L ~-~""7 ~ ""'ZfCL

y I

I G fl

Ill' I -....:------'--"~_ p

C

- .r

CB

20 to ns

Helpful Hinls

CD We note tha t a na lysis by the method of j oints wou ld necessit ate working wit h eight j oints in order to calcu lat e t he t hree forces in question . Th us, the me thod of sections offers a consider able advantage in t his case.

@ We could have sta rted with mo ments Final ly, we may find CL by a moment sum about P, whose distance from C is give n by PC/16 ~ 24/ (26 - 16) or PC ~ 38.4 ft. We also need fl, which is given by fl ~ ta n - 1(CB/ BL) ~ lan - 1(12/ 21) ~ 29.7' and cos fl ~ 0.868 . We now ha ve

@

[LAIp =

OJ

20(48 - 38.4 ) - CL
o

CL = 5.76 Ions C

Marwan and Waseem AI-Iraqi

Ans.

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about C or P ju st as well.

@ We could also ha ve det ermined CL by a force summa tion in either the x or y -direc tio n.

182

Chapter 4

Structur e s

Sample Problem 4 /3

10 kN 1 lO kN

Calculate the force in member DJ of the Howe roof truss illustrated. Neglect any horizontal components of force at the supports.

2

'I V

I

c B A

Solution. It is not possible to pass a sect ion through DJ without cutting four

CD

members whose forces are un known. Alth ough three of these cut by section 2 are concu rrent at J an d therefore the moment equati on about J could be used to obtain DE, the force in DJ can not be obtained from t he remaining two equilibrium pri nciples. It is necessary to consider first the adjacent sect ion 1 before analyzing section 2. The free- body diagr am for section 1 is drawn an d includes the reaction of 18.33 kN at A , which is previously calcu lated from th e equilibri um of the truss as a whole. In ass igning th e proper directions for th e forces acting on the th ree cut mem bers, we see that a balan ce of moments about A eliminates the effects of CD and JK an d clearly require s th at CJ be up and to th e left . A balan ce of moments about C eliminat es th e effect of th e three forces concurrent at C and indicates that JK must be to the right to supply sufficient count erclockwise moment. Again it should be fairly obvious that th e lower chord is under ten sion because of the bending tendency of the truss. Alth ough it should also be apparent th at the top chord is und er compress ion, for purposes of illust rat ion th e force in CD wi ll be arbitrarily ass igned as tension. By t he ana lysis of sect ion 1, CJ is obtained from 0.707CJ(12) - 10(4 ) - 10(8) = 0

CJ = 14.14 kN C

In thi s equation t he momen t of CJ is calculate d by considering its horizontal and vertical components acting at point J . Equilibrium of moments about J requires 0.894CD (6) + 18.33(12 ) - 10(4 ) - 10(8) CD

o

= - 18.63 k N

The moment of CD abou t J is calculated here by considering its two componen ts

@ as act ing through D. Th e minu s sign indicates that CD was assigned in th e wrong

_

~

J

J

H 10 kN

6 panels at 4 m - ---I

l OkN

Section 1

JK 18.33 kN

H elpful Hin ts

CD There is no harm in ass igni ng one or more of th e forces in th e wrong directi on as long as th e calculati ons are consiste nt with the ass umpt ion. A negati ve answer will show the need for reversing th e dir ectio n of the force.

CV If desired, the direction of CD

may be cha nged on the free-body dia gram and the algebrai c sign of CD reversed in the calculat ions , or else th e wor k may be left as it stands with a not e stating th e proper direction.

direction . Hence,

K I I

lO kN lOkN

CD = 18.63 k N C

From th e free-body diagram of section 2, which now includ es th e known value of CJ , a balance of moments about G is seen to elimina te DE and JK. Thu s, 12DJ + 10(16 ) + 10(20) - 18.33(24 ) - 14.14<0.707 )(12 ) = 0 DJ = 16.6 7 kN T

Ans.

Again th e moment of CJ is determi ned from its compone nts cons idered to be act ing at J . T he answer for DJ is positive. so that the ass umed te nsile direction is correct. An alter na tive approach to the entire problem is to u tilize section 1 to deter mine CD and th en use the method of joints applied at D to determine DJ .

Marwan and Waseem AI-Iraqi

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<,

Sect ion 2 <,

14.14 kN . . . . . . . . . . . .

~-~~~~'~ J - - - - - - - ~G

18.33 kN

@ Observe th at a sect ion th rough mem her s CD, D.J, an d DE could be taken which would cut on ly three unknown members. However, since th e forces in t hese three member s are all con current at D, a mome nt equation about D would yield no information about th em . Th e remaini ng two force equations would not be sufficient to solve for t he three unknowns.

Articl e 4 /4

PROBLEMS

J

Probl ems

4m

H

4m

G

4m

C

4m

D

183

Introductory Problems 4m 4 /29 Determine the force in member CG . Ans. CG

H

10'

F

10'

G

14.14 kips T

10'

E

4m

B

50kN

Problem 4/31 10'

4/32 Determine t he force in mem ber DO of the loaded

truss. A

c

B

5 kips

5 kips

r

5 kips

4'

Problem 4/29

L

L

E

C

LV

3' F

A

4 /30 Det er mine the forces in members BC, CF, and EF of

l

5 pan els at 4'

I

Dl

th e loaded t russ.

12'

A

B

B

Problem 4/32 4 /33 Deter mine the forces in memb er s BC, BE, and BF.

Th e tr ian gles are equilateral.

An s. BC

~

BE

c

B

A

Problem 4/10 4 /31 Deter mine th e forces in members OH and CO for

t he truss loaded and supported as shown. Does the sta tical ind et erminacy of the supports affect your calculation? An s. CG ~ 70.7 kN T , GH ~ 100 kN T , No

i}--------
G

E

F

L

Problem 4/33

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184

Chapt e r 4

Structu r es

Representative Problems 4/34 Determ ine th e forces in members DE and DL. 2m

6'

2m

-~"----x--~ F

8kN

H

Problem 4{36

4/37 Th e tru ss is composed of equilate ral t ria ngles of side a and is supported and loaded as shown. Determine t he forces in members BC an d Ans. BC ~ CG ~ 1-{3 T

ca.

Problem 4{34

4/35 Calcu late th e forces in mem bers BC, BE , and EF. Solve for each force from an equilibrium equation wh ich contains that force as th e only unknown . Ans. BC ~ 21 kN T , BE ~ 8.41 kN T EF = 29.5 kN C

L

G

a

14 kN

F

a

B a

E

»:

C

30'

D

Problem 4{37

Problem 4/35 4/36 Determine th e forces in members BC and FG of the loaded sym metrical tru ss. Show t ha t thi s calculation ca n be accomplished by using one section and two equa tions, each of which contains only one of the two unknowns. Are th e resul ts affected by the statical indete rminacy of the supports at t he base?

4/38 The tru ss shown is comp osed of 45° right triangles. The crossed memb ers in th e cente r two pan els are slende r tie rods incap able of supporting compression. Retain the two rods which are und er ten sion and compute the magnitudes of t he ir ten sions. Also find th e force in member MN. J

H

G

K

L

E

F

D

80 kN

A

N

100 kN Problem 4{38

Marw an and W aseem AI-Iraqi

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o

Articl e 4 / 4

Pro ble ms

18 5

4/39 Det ermine th e force in member BF.

Ans. BF

~

2.66 kip s C D

C

B L

A

LI2

Problem 4/41 9 ki ps

FI -

3 kips i

1

4 /42 Determine the forces in members CD, CJ , and DJ.

EI

r-

r-- 5'- j - ' - - 5'- - + - - 5' ---4 Problem 4(39

MI

4 /40 T he members CJ and CF of the loaded truss cros s bu t a re not con necte d to memb ers BI and DG. Compu te the force s in members BC , CJ , CI , and HI.

L

6 panels at 3' J

K

5'

B

6 kN

C

E

D

F

G

A 3m A

3m

C

L

3m

3m

B

D

L

E

L

L

L

L

Problem 4(42

4m J~--+'---I--:-:----->:r:::--....::,q F I H G

4(43 Compute the force in memb er GM of th e loaded t ru ss. A ns. GM = 0 L L

4 kN

10 kN

8 kN

L

L L 2

Problem4/40 4 /41 Th e t russ su pports a ramp (shown with a dash ed line) which extends from a fixed approach level near jo in t F to a fixed exit level nea r J . Th e loa ds shown rep resen t th e weight of th e ramp. Determine t he forces in members BH a nd CD. Ails. BH ~ 0.683L T, CD ~ 1.932L C

L F

L

H

D

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L 2

1-- - - - - 8 panels at 3 m - -- - - -4 Problem 4(43

Marwan and Waseem AI-Iraqi

L

G

E

186

Chapte r 4

Struc tur es

4 /44 Com pute th e force in member HN of the loaded

truss. Compare your answer wit h the stated resu lt of Prob. 4/ 43. L

J

L L

L

4/47 Determine the forces in members DE, EI, FI, and HI of the arc hed roof tru ss. An s. DE = 297 kN C, EI 26.4 kN T Fl = 205 kN T , HI 75.9 kN T

L L L 3

C

F

1--- - -- -

K

8 panels at 3 m - - - - - Problem 4/44

Dimen sion s in met ers

4/45 Determine th e forces in memb ers DJ and EJ of the loaded t russ. Ails. DJ = 0.45L T , EJ 0.360L T

L

1--- - - -

c

D

E

L

L

L

Problem 4/47

4/48 Find the force in member JQ for t he Bal tim or e truss wher e all angl es are 30°, 60°, 90°, or 120°.

L

6 panels at 8 m ----~ Problem 4/45 100

4/46 Determine th e force in membe r HP of the loaded truss. Members FP and GQ cross without tou chin g and are incapable of support ing compression. L 2

L

L

L

L

L

L

L

L

L 2

"''I 100 kN

Problem 4/48

.... 4/49 Deter mine the force in member DK of the loaded over head sign tru ss. All s. DK = 1 kip T 6 pan els at 8'

I A 0

N 9 panels at 20'

B

p

C Q

M

E

t t L

2 kip s

4 kips

Problem 4/49 www .gigapedia.com

S

T

II J

u

Marwan and W aseem AI-Iraq i

lc- ,

F

R

1 kip

Problem 4/46

D

I

j'

5'

Article 4/4 ~

4/50 In the traveling bridge crane shown all crosse d m em be rs are slender tie rods incapable of su pporting com pression . Determine the forces in members DF and EF and find the horizontal reaction on the truss at A. Show that if CF = 0, DE = 0 also. Ans. DF = 768 kN C, EF = 364 kN C AI = 101.1 kN

.;-- - - - 5 panels at 8 m -

-

-

-

-jI

4m l4m 14 m , 4 m 14 m 14 m I

6m

122m

J

·I c-+--+-+-----.I K~ I 6m

H

N

60 0 /

....

C

0

60'

j I

6m I

A

I!-

.

25 kN

25 kN

- x

6m

1-

I I

6m

)'

1-

Problem 4/50

6m ... 4/51 Det ermine the force in member DC of the compound truss. The joints all lie on radial lines subtending angles of 15° as indicated, and the curved members act as two-force members. Distance DC = OA = DB = R. Ans. DC = 0.569L C

Problem 4/51

MarNan and W aseem AI-Iraqi

187

... 4/51 A design model for a transm ission-line tower is shown in the figure. Members GR , FG , OP, and NO are insu lated cables; all other mem bers are stee l bars. For the loading shown, compute the forces in members FI, FJ , EJ . EK , and ER. Use a combinati on of methods if desired. Ans. FJ = ER = 0, FJ = 7.8 1 kN T EJ = 3.6 1 kN C, EK = 22.4 kN C

--.- ----<;<;------;i/c----;>;
200 kN

Problems

www.gigepedia.com

A

5.4 m

Problem 4/52

18 8

Chapt er 4

Structur es

4 /5 D

(a )

D

B ~-- I '--7 C

A

(b)

P,

G

SPACE TRUSSES

A space tr uss is the three-dimensional counte rpart of the plane truss described in the three previous art icles . The idealized space truss consists of rigid links connecte d at t heir ends by bal l-and-socket jo ints (such a joint is illustrated in Fig. 3/8 in Art. 3/4 ). Whereas a t riangle of pinconnected bars form s the basic noncollap sible un it for the pla ne truss, a space truss, on the other hand, requires six bars joined at their ends to for m th e edges of a tetrahed ron as th e basic no ncollapsibl e unit. In Fig. 4/ 130 the two bars AD a nd BD join ed at D r equire a t hi rd su pport CD to keep the triangle ADB from rotat ing about AB. In Fig. 4/ 13b the suppor ting base is replaced by thr ee more ba rs AB , BC, and AC to form a tetrah ed ron not dependen t on the foundation for its own rigidity. We may form a new rigid un it to exte nd t he st ructure with thre e additional concurrent bars whose ends are attached to three fixed joints on the existing struc ture. Th us, in Fig. 4/ 130 the bars AF, BF. an d CF a re attached to the foundati on and therefore fix point F in space. Likewise point R is fixed in space by t he bars AH, DR, a nd CR . Th e th ree addit iona l bars CG, FG, and R G are attached to the t hree fixed points C, F , and R and t herefore fix G in space. T he fixed point E is simila rly create d. We see now that the st ruc ture is entire ly r igid. T he tw o applied loads shown will resu lt in forces in all of the members. A space truss for med in t his way is called a simple space t russ. Ideally th ere must be point support, such as t hat given by a balIand-socket joint, at the connections of a space truss to prevent bending in the members. As in riveted and welded connections for plane trusses, if the centerlines of joined members intersect at a point, we can justify the assumption of two-force members under simple tension and compression.

Statically Determinate Space Trusses

(c)

Figure 4/13

M arwan and W aseem AI-Iraqi

When a space truss is su pported exte rnally so that it is statically determinate as an entire unit, a relationship exists between the number of its joints and the number of its members necessary for internal stability witho ut re du nda ncy. Becau se th e eq uilibrium of each joint is specified by three scalar force equations, there are in all 3} such equations for a space truss with} joints. For the entire truss composed of m members there are m unknowns (the te nsile or compressive forces in the members) plus six unknown support react ions in the general case of a statically determinate space structure. Thus, for any space truss, the equat ion m + 6 = 3i will be satisfied if t he truss is st atica lly determinate internally. A s imple space truss satisfies this relation automatically. St arting with th e ini tial tetrah edron, for which t he equation holds, t he structure is extended by adding three members and one joint at a time, thu s preserving the equa lity . As in the case of the plane truss, this relation is a necessary condition for stability, but it is not a sufficient condition, since one or more of the m members can be arranged in such a way as not to contribute to a stable configu ra tio n of t he enti r e tr uss. If m + 6 > 3i , th ere are more members than there are independent equations, and the truss is www .gigapedia.com

Art icl e 4 / 5

st atically indeterm inate intern ally wit h red un dant members present. If m + 6 < 3j , there is a deficiency of interna l member s, and the truss is unstable a nd subject to collapse un der load . Th is relationship between t he number of joint s and t he number of member s is very helpfu l in the prelim inary design of a stable space truss, since the configuration is not as obvious as wit h a pla ne t russ, wher e th e geome try for statica l det er min acy is gen era lly qui te apparent. Method of Joints for Space Trusses

T he method of joints developed in Art. 4/3 for plane trusses ma y be exte nded direct ly to space trusses by satisfying th e complete vector equation ~F

(4/ 0

= 0

for each joint. We norma lly begin the ana lysis at a join t wh ere at least one kno wn force acts and not more t ha n three unknown forces are present. Adjacent joints on which not more t han three unknown forces act may t hen be analyzed in turn . Th is step-by-step joint technique tend s to minimize t he number of simultaneous equa t ions to be solved when we must determine th e forces in a ll member s of the space truss. For thi s reason , although it is not read ily redu ced to a routine, such an approach is recomm end ed. As an alte r native procedure, however , we may simply wri te 3j joint equa tions by applying Eq. 4/ 1 to all joints of th e space frame. Th e number of unknown s will be m + 6 if t he st ruct ure is non collapsible when remo ved from its support s and tho se supports provide six ext ernal reactions. If the number of equations (3j) equals the number of unknowns (ni + 6), then th e entir e sys tem of equations may be solved sim ultaneously for the unknowns. Because of the large number of coup led equat ions, a computer solution is usually required. With this latter approach, it is not necessary to begin at a joint where at leas t one kn own and no mor e th an three unknown forces act. Method of Sedions for Space Trusses

Th e met hod of sections deve loped in th e previous article may also be applied to space tr usses. The two vector equa tions ::':F

=

0

and

::': M

~

0

mu st be satisfied for an y sectio n of th e truss, wher e th e zero mom ent su m will hold for all moment axes . Becau se the two vecto r equa tions are equiva lent to six sca lar equa t ions, we conclude that a sect ion should in ge neral not be passed through more th an six memb er s whose forces are unknown . Th e method of sections for space trusses is not widely used , however , becau se a momen t axis ca n seldom be found wh ich elimina te s all bu t one u nk nown, as in the case of plane trusses. Vector notation for expressing the te rms in the force a nd momen t equations for space tru sse s is of cons ide ra ble advantage a nd is used in the sample problem which follows. Marwan and Waseem AI-Iraqi

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Spac e Tru s se s

189

190

Chap t er 4

Structu r es

Sample Problem 4 /4 Th e space truss consists of the rigid tetra hedron ABeD a nchored. by a balland-socket con nectio n at A a nd preven ted from any rotat ion about the X·, y., or a-axes by th e res pective lin ks I , 2, and 3. Th e load L is applied to joint E , which is r igidly fixed to th e tetrahedron by the th ree addit ional links. Solve for t he forces in the mem ber s at joint E an d indicat e the procedu re for the deter min ation of t he forces in the remaining members of the truss.

CD @

Solution. We note first that the truss is su pported wit h six properly placed constraints, which are the three at A and the links I, 2, and 3. Also, with m = 9 members an d j = 5 joi nts , the condit ion m + 6 = 3j for a su fficiency of mem bers to provide a noncollapsi ble st ru ctu re is satisfied. The exter na l reactions at A, B, and D can be calculated eas ily as a first step, alt houg h their values will be determine d from t he solution of all forces on each of th e joints in succession. We mu st start with a joint on which at least one known force and not more than t hree unknown forces act , which in this case is joint E. The free-body diagram of joint E is shown with all force vectors arbitrarily as su med in the ir positive tension directions (away from the joi nt>. Th e vector expressions for the th ree unknown for ces are FEB ( - 1, F EB = -r.;

,2

-

FEC ( - 3'1 - 4k l , F ED ~ ""5 FED (- 3)' - 4 k 1 Jn, F EC =""5

= OJ

CD S uggestion:

Draw a free-body diagr am of th e truss as a whole and verify th at th e exte rnal forces acting on the truss are A, : Li. A, ~ Lj , A, ~ (4L/ 3Jk, B, 0, D, ,; - Lj , D, : -(4L/ 3Jk.

Q) With th is ass u mptio n, a nega tive nu mer ical value for a force wou ld indicate compression.

Equ ilibriu m of join t E requi res I ~F

He lpfu l Hints

L + F ElJ + F EC + F ED = 0

or

z

- L i + FEB (_ i _ j l + FEC (_ 3i + 4k l + FED (- 3j - 4k l /2 5 5

I

o

B

Rearranging terms gives

(

.;!! _ 3F'!C) i + (_ FEB _ 3FED - L _ FE , '2 0 ./2 5

)j + ( _ 4FEC _ 4FED ) k ~ 0 5

4m

5

Equating th e coefficients of th e i-, j -, and k -u nit vectors to zero gives th e three equations

3F + - -ED 5

FE I) + 3FEC = - L /2 5

o

o

Solving the equations gives us FEC

~

- 5L/ 6

Ans.

Thus, we conclude that FEB and FEC are compressi ve forces and F ED is tension . Un less we have computed the exter nal reactions firs t, we mu st next analyze joi nt C wit h t he known value of F EC and the t hree unknowns F CB , F CA' and Fen . Th e procedur e is iden t ical with tha t used for joint E. J oints B, D, and A are then ana lyzed in t he sa me way and in th at order, which limits the unknowns to three for each joint . Th e exter na l reactions compu ted from these ana lyses must, of cour se, agree with t he values wh ich ca n be determ ined initi ally fro m an analysis of th e truss as a whole.

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Article 4 /5

PROBLEMS

P roblems

191

800 lb

(In the followin g pro blems, use plus for tens ion and minu s for compression. ) 4 /53 Determ ine th e forces in members AR, AC, and AD. Ans . AB ~ - 4.46 kN , AC ~ - 1.521 kN AD = 1.194 kN

Problem 4/54

4 /55 Th e rectangular space truss 16 m in heigh t is ere cte d

B!G:::

on a hor izontal squa re base 12 m on a side. Guy wires ar e attached to t he st r uct u re at E a nd G as shown and are tightened until the ten sion T in each wir e is 9 kN. Calculate the force F in each of the diagon al memb ers. Ans . F ~ - 3.72 kN

"":"::J/

Problem 4/53

H

4 /54 Th e base of an au tomobile jac kstand form s an equi-

lateral triangle of side lengt h 10 in. and is centered und er th e collar A. Model the stru ct ure as one with a ball a nd socke t at ea ch joint a nd det ermine the forces in memb ers Be , RD , and CD. Neglect any hor izontal reaction compo nents under th e feet B, C, and D.

Problem 4/55

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192

Chap t e r 4

Struc tur es

4 /56 The tetrahedral space truss has a horizontal base ABC in th e form of an isoscele s triangle and legs AD , BIJ , and CD which support t he mass m from point D. Each ver tex of th e base is suspended by a ver tical wire from over head supports . Calcu late the forces induced in memb ers A C and AB .

4/58 The space truss is shown in an intermediate stage of design. The external constraints indicated are su fficient to maintain exte rn al equ ilibr ium. How many additional mem bers are needed to preven t inter nal instabi lity an d wher e can they be placed? G

~ P

t--, B

/

v

L

--- . 3m

-\ L- -+- - - - /I( E

c

,\: - - , ---{ ~F B

Problem 4/58 4/59 Deter mine the force in member BD of th e regular pyramid with square base . An s. DB = - 2.00L Problem 4/56 4/57 For the space truss shown, check t he sufficiency of th e supports an d also the num ber and arrangement of the memb ers to ens ure statical determ inacy, bot h extern al and int ernal. By inspection determine th e forces in members DC, CR. and C/o'. Calcu late the force in memb er AF and the x-component of t he reaction on th e tru ss at D. P An s. FA F = ,' 13 P D

3/2" '

x

3./2 D

50

Problem 4/59

Problem 4/57

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t.

Articl e 4 / 5

4/60 The space truss shown is secu red to th e fixed su pports at A , B, and E and is loaded by the force L which has equal x - and y-components but no vertical a-compone nt. Show that there is a sufficient number of mem bers to provide intern al stability and that their place ment is adequate for t his purpose. Nex t det ermine the forces in members CD , BG, and CEo

Prob lems

193

.. 4 / 62 A space t russ is constructed in t he form of a cub e with six diagonal members shown. Verify that t he truss is internally sta ble. If the truss is subjecte d to the compressi ve force s P applied at F and D along the diagonal FD, det erm ine the forces in members FE a nd EG.

z I I IE

p

c Prob lem 4/62 .. 4 / 6 3 The lengthy boom of an overhead cons t r uction crane, a por ti on of which is shown, is a n exa mple of a periodic st ru ct ure-one which is composed of rep eated a nd iden tic al st ruct ura l uni ts. Use the method of sectio ns to find th e forc es in members FJ and GJ. Ans. FJ ~ O. GJ ~ - 70.8 kN

Pro blem 4/60 4 /61 The pyramidal truss section BCDEF is symmet r ic about t he vertica l x-z plane as sho wn. Cabl es AE ,AF, a nd AB support a 5-k N load. Det ermine the force in member BE. Ans. TB E = -2.36 kN

-,

360

"

Il}Il}

~~

300mm

I

-"----

--- /r --t r

!

360 mm _---

I

D

I I /

V

360mm

I I F I

E

440 mm

I I :

L - -~/

I

J(

---

5000 kg

B Pro blem 4/63

:

I I I I

A -----x 5 kN

Problem 4/61

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194

Chapte r 4

Structures

.... 4/64 The space truss supports th e structure of an amuse ment park rid e (not shown) which rotates about a vert ical axis. Th e eight footpads form a regu lar octagon, a nd AB CDE is a pyramid with a fi-Ft-squa re base BCDE and vert ex A , which is 1 ft above th e base. Th e plane of figur e BCDE is 4 ft above th e plan e of th e footp ads. Th e diagonals of th e trapezoida l faces such as BCGF cross with out touchin g. If the vertical load L is trans mitt ed to point A an d if instrumentation indi cates a ten sile force of O.3L in memb er BC, determi ne the forces in members CF and CG. (Hi nt: Begin your analysis at point A and mak e full use of sy m met ry.)

Ails. CF

~

O.051L , CG

~

- O.3 12L

Problem 4/ 64

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Ar ticl e 4/6

4 /6

FRAMES AND MACHINES

A structure is called a frame or machine if at least one of its individual members is a mult iforce member . A multiforce member is defined as one with three or more forces acting on it, or one with two or more forces and one or more couples acting on it. Frames are structures which a re designed to support applied loads and are usually fixed in position . Machines are structures which contain moving parts and are designed to transmit input forces or couples to output forces or couples. Because frames and machines contain multiforce members, the forces in these member s in general will not be in th e directi ons of th e members. Therefore, we cannot analyze these structures by the methods developed in Ar ts. 4/3, 4/4, a nd 4/ 5 becau se these methods apply to simple trusses composed of two-force members where the forces are in the directions of the members. Interconnected Rigid Bodies with Multiforce Members

In Chapte r 3 we discussed th e equ ilibrium of multiforce bodies, but we conce ntrate d on the equilibr ium of a single rigid body. In this pr esent article we focus on the equilibrium of interconnected rigid bodies which in clude multiforce memb ers. Although most such bodies may be analyzed as two-dimensional sys tems, there are numerous examples of frames and machines which are three-dimensional. The forces acting on each member of a connected system are found by isolatin g the memb er wit h a free-body diagram an d applyin g the equations of equilibrium. The principle of action and reaction must be carefully observed when we represent the forces of interaction on the se parate free-body diagrams. If the structure contains more members or supports than are necessary to prevent collapse, then, as in the case of t russes, the problem is sta tically indeterminat e, and the principles of equili br ium, alt hough necessary, are not su fficient for solution. Although many frames and machines are statically indeterminate, we will consider in this article on ly those which are statically det erminat e. If the frame or machine constitutes a rigid unit by itself when removed from its supports, like th e A-fram e in Fig. 4/14a , the analysis is best begu n by establishing a ll t he forces exte rn al to the st r ucture treat ed as a single r igid body. We then dismember the stru ctu re and consider the equilibr iu m of each part separate ly. Th e equilibr ium equations for the seve ra l parts will be related through the terms involvin g t he forces

Rigid noncollapsib le

(ul

(bJ Figure 4/14

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Nonrigid collapsible

Frames and Machin e s

19 5

196

Chap t er 4

St ructur es

of int eraction. If the struc ture is not a rigid unit by itself but depends on its exte rnal supports for rigidity, as illustrated in Fig. 4/14b , th en th e calculation of the ext ernal support reacti ons cannot be complete d until the structure is dismembered and the individual parts are analyzed. Force Representation and Free-Body Diagrams

(b )

Figure 4/15

A~ ~.~ V I Vector

.<

~.

notation Figure 4/16

Ma rwan and Waseem AI-I raqi

In most cases the analysis of frames and machi nes is facilitated by representing the forces in terms of their rectangular components. This is particularly so when th e dimen sions of the parts are given in mutually perp endicular directi ons. Th e advantage of th is represen tation is th at the calculation of moment arms is simplified. In some three-dimensional problems, particularly when moments are evaluated about axes which are not parallel to the coordinate axes, use of vector notation is advanta geous. It is not always possible to assign the proper se nse to every force or its compone nts when drawing th e free -body diagrams, and it becom es necessary to make an arbitrary assignm ent. In any event, it is absolutely necessary that a force be consis tently represent ed on the diagrams for interacting bodies which involve the force in question. Thus, for two bodies connected by t he pin A , Fig. 4/1Sa, the force compone nts mu st be consistently represented in opposite directions on the se parate freebody dia grams. For a ball-and-socket connection betwe en members of a space frame, we must apply the action-and-reaction principle to all three components as shown in Fig. 4/ 15b. Th e assigned directions may prove to be wro ng when the algebraic signs of the component s are determined upon calculation. If A " for inst an ce, should turn out to be negative, it is actually act ing in th e direction opposit e to that or igina lly represented . Accordingly, we would need to reverse the direction of the force on both members and to reverse the sign of its force terms in the equations . Or we may leave the representation as originally made, and the proper sense of t he force will be under st ood from th e negati ve sign . If we choose to use vector notation in labeling the forces, then we must be careful to use a plus sign for an action and a minus sign for the corresponding reaction, as shown in Fig. 4/16. \Ve may occasiona lly need to solve two or more equat ions simultaneously in order to separate the unknowns. In most instances , however, we can avoid simultaneous solutions by careful choice of the member or grou p of memb er s for the free-body diagr am a nd by a ca reful choice of moment axes which will eliminate undesired terms from the equations . The method of solution descr ibed in the foregoing paragraphs is illustrated in the following sample pr oblems.

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Article 4 /6

Fra mes and Mach in e s

Sample Problem 4/5

1--

The frame supports t he 400.k g load in the manner show n. Neglect t he weights of the members compared wit h the forces induced by the load and cornpute the ho rizontal and vertical components of all forces acting on each of t he members.

.--- . A

3m-

.!.- 2 m -.j

I

I

i

1

1.5 m

197

I•

O .T-5m-B ~~===~~;;:=~lfj"--O.5 E

1.5 m

, C ~===::J +--

mR

F

1

CD

SoluUon . We observe first that the three supporting members which constitute

1.5 m

the fram e form a rigid assembly that can be analyzed as a single unit. We also obse rve that the arrangement of the external supports makes the frame statically det erminate. From th e free-body diagram of the entire frame we det erm ine the external reactions. Thus,

L

[:!:Fr

OJ

Ar - 4.32

0

Ar

= 4.32 kN = 4.32 kN

I~Fy

OJ

A, -

0

A,

~

[ ~A ~


OJ

5.5(0.4)(9.81) - 5D

0

D

3.92

OJ

H elpful Hints

CD We

see th at th e fram e corresponds to t he ca tegory illustrat ed in Fig. 4/1 40 .


3.92 kN

Next we disme mber t he fra me and draw a separate free-body diagr am of each member. The diagrams are arranged in the ir a pproximate relative positions to aid in keeping track of the common forces of inte raction. Th e exte rnal reactions ju st obtained are ente red onto th e diagram for AD . Other known forces are the 3.92·kN forces exerted by the shaft of th e pull ey on th e memb er BF, as obtained from th e free-body diagram of the pulley. The cable tension of 3.92 kN is also shown act ing on AD at its atta chment point . Next, the compone nts of all u nknown forces are shown on the diagrams. Here we observe that CE is a two-force memb er. Th e force compone nts on CE have equa l and opposite reactions, which are shown on BF at E and on AD at C. We may not recogn ize the actu al sense of th e components at B at first gla nce, so they may be arbitrarily but cons iste n tly assigned. The solution may proceed by use of a moment equa tion about B or E for member BF, followed by th e two force equations. Thus, l~fB ~

400 kg

3.92(5) - ! Er(3 )

0

Er = 13.08 kN

Ans.

I:!:Fy

~

OJ

By + 3.92 - 13.08/2

0

By = 2.62 kN

Ans.

I:!:Fr

~

OJ

Br + 3.92 - 13.08

0

Br = 9.15 kN

An s.

Positive numerical values of the u nknowns mean that we assumed their direction s correctly on the free-body diagrams. The value of = Ez = 13.08 kN obtained by inspection of th e free-bod y diagram of CE is now entered onto the diagram for AD , along with the values of B x and By just determined. The equation s of equilibrium may now be applied to member AD as a check, since all the force s acting on it hav e already been computed. The equations give

lut ion would be mu ch lon ger , because th e th ree equilibri u m equ ations for mem ber BFwould conta in four unknown s: B~ , By, Ex, and Ey. Note th at th e direction of th e line joining th e two points of force appli ca tion, and not the sha pe of t he memb er , determi nes th e direction ofthe forces acting on a two -force mem ber . A,

A--S <

y I

I I

L

x

0.419.81) = 3.92 kN D

ex

[ ~c =

OJ 4.32(3.5) + 4.32(1.5) - 3.92(2) - 9.15(1.5)

0

I ~Fr =

OJ

4.32 - 13.08 + 9.15 + 3.92 + 4.32

0

l~ Fy =

OJ

- 13.08/2 + 2.62 + 3.92

0

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3.92 kN

3.92 kN~ 3.92 ~tr 3.9'1 kN

3.92 kJ'i

t-

3.92 kN

198

Chapter 4

Structures

Sample Problem 4 /6 Neglect the weight of th e frame and compute th e forces act ing on all of its members .

CD

Solution. We note first th at the frame is not a rigid unit when removed from its suppor ts sinc e BDEF is a movable quadr ilateral and not a rigid triangle. Con sequently the exte r nal reactions cannot be comp letely determined unt il the individual mem bers ar e analyzed. However , we can determine t he vertical components of the reaction s at A and C from the Free-body diagr am of the fram e as a whole . Thus, [ ~c ~

f:!:Fy

~

OJ OJ

50(12 ) + 30(40) -

Gy

-

3DA,

50(4/5) - 60

o o

A, =

60 Ib

C; = 100lb

30"

--1

An s. Ans.

Next we dism emb er the frame and draw th e free-body diagram of each part. Since EF is a two-force memb er . the direction of th e force at E on ED and at F on AB is kn own. We assume th at t he 30-lb force is applied to the pin as a part of member BC. Th ere should be no difficulty in assigni ng the correct directi on s for forces E , F, D, and Br Th e dir ection of By, however , may not be ass igned by inspection an d t here fore is arbit rarily shown as downward on AB and upward on BG.

Helpful Hints Member ED. The two unknowns are eas ily obt ain ed by [kM D =

OJ

[kF = OJ

50(12) - 12E

~

0

D - 50 - 50 = 0

CD We see th at th is fram e cor res ponds to

E = 50 lb

Ans.

the catego ry illustrated in Fig. 4/14 b .

D = 100 1b

Ans.

@ Th e di rect ions of Ax and ex are not ob-

Member EF. Clearly F is equal and opposite to E with t he magnitude of 50 lb.

OJ

50(3/5)(20 ) - B x(40 )

0

Bx

~

15lb

Ans.

~

15lh

Ans.

[kFx

0]

Ax + 15 - 50(3/5)

0

Ax

[:!:Fy

OJ

50(4/5 ) - 60 - By

0

By = - 20 lb

@ Altern at ively t he ao-lb force could he applied to the pin consi dered a part of BA , wit h a res u lting change in the reaction Bx •

MemberAB. Since F is now known , we solve for Bx' Ax. and By fro m [ ~A =

vious in itially and can be assigned a rbitrarily to be cor rected later if necessary.

Ans.

Th e min us sign shows that we assign ed By in th e wrong direction.

Member Be. The resu lts for B x' By, and D are now transferred to BC, and the remaining unknown e x is found from [kFx =

OJ

30 + 100(3/5) - 15 - Gx = 0

C;

~

751b

A ns.

We may apply the remain ing two equilibrium equ ations as a check. Thus, [:!:Fy

~

[ ~c ~

OJ OJ

100 + (- 20) - 100(4/ 5) = 0

@ Alte r natively we could have ret urned

(30 - 15)(40) + (- 20)(30) = 0

to th e free -body diagram of the frame as 8 whole and foun d eX' Marwa n and W aseem AI-Iraqi

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Article 4 /6

Frames and Machines

Sample Problem 4 /7

60 ,

120

Th e machine shown is designed as an overload protection device which release s the load whe n it exceeds a predet ermined value T. A soft me tal shea r pin S is inserted in a hole in th e lower hal f and is acted on by the upper half. Wh en th e tota l for ce on th e pin exceeds its stre ngt h, it will brea k. T he two halves th en rota te about A under the ac tio n of the tensions in BD and CD, as shown in the second sketch , and rollers E an d F releas e the eye bolt. Determ ine the maximum allowable tension T if t he pin S will shear when the tot al force on it is 800 N . Also compute the correspo nding force on the hin ge pin A .

199

50 , ->,"",r T

D

Dimensions in millimeters

Solution. Because of symmetry we a na lyze only one of t he two hin ged memo

CD

bers. Th e upper part is chosen, an d its free-body diagram along with th at for th e connecti on at D is dra wn . Becau se of symmetry th e forces at S a nd A have no zcompone nts . Th e two-force members BD an d CD exert forces of eq ual magn itude B = C on the conn ect ion a t D. Equilibrium of the connec tion gives

B cos O + C cos 8 -T =0 B

~

)" ,

2B cos O ~T

I

From the free-body diagram of t he up per pa rt we express th e eq uilibriu m of moments about point A . Substituting S = 800 N and th e express ion for B gives

(cos 0)(50) + -2 T (sin 0)(36) - 36 (800) cos 0 cos 0

-2 T

f

0

T(25 + 5(36) _ 13) = 28 800 2(12)

T = 1.477 kN

Ans.

F inally, eq uilibriu m in th e y-direc ticn gives us

Ir Fy

~

OJ 800

S - BsinO - A

o

14 77 5 _ A 2(12/ 13) 13

o

Marwan and Wase em AI-Iraqi

A

492 N

•

A

T

2

S A

Helpful Hints (26)

Subst itu ti ng sin Oleos fJ = ta n () = 5/ 12 a nd solving for T give

or

~

B B .....

~c I

T / (2 cos 0)

T = 1477 N

Released positi on

An s.

CD It is alwa ys useful to recognize symmetry. Here it tells u s that t he forces acting on the two parts behave as mirror images of each other with r e· spect to the .r-axis . Thus, we cannot ha ve a n action on one member in t he plus z-direction and its reacti on on th e ot her member in the negative .rdirecti on. Consequently the forces a t S and A have no x-compo nents.

@ Be care ful not to forget th e moment of th e y -cornponent of B. Note tha t ou r uni ts here are newt onmillim et ers .

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200

Chapter 4

Structur e s

PROBLEMS (Unless otherwi se inst ructed, neglect th e ma ss of the various members a nd all fri ctio n in th e pr oblems which follow.I

Introductory Problems 4/65 Det ermine th e magnitudes of all pin reacti ons for th e fram e loaded as shown . A ns. A = 5 12 N , B = D = 10 13 N

301b

Problem 4/67

4/68 Deter mine the com ponents of a ll forc es act ing on ea ch membe r of th e load ed fram e.

SOON

p

p

c

R/

kA_ _

----'4:50e. .~~,.4;"-50---

kA

y

I I I

Problem 4/65

x

L

4/66 For a n BD-N squeeze on the handle s of th e pliers, determi ne th e force F applied to the round rod by each ja w. In addit ion , ca lculate th e force supported by t he pin at A. BON

-1- - - - - 95 mm - - -- -1

Problem 4/68

4/69 Det ermine t he compone nts of a ll force s act ing on each member of the loaded truss. Wha t is th e prima ry difference between th is pr oblem a nd P rob. 4/ 6B? Ans. Ax = e x = B, = 0 A y = 0.707P, B, - 0.707P, C, = 0.293P p

p

C A

BON

1.J,-----4"':. ~/ y I I

Problem 4/66

IL

4/67 Compu te th e force support ed by t he pin at A for the slip-join t plier s un der a gr ip of 3D lb. An s. A = 157.6 Ib

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Problem 4/69

Article 4 /6 4/70 A force P is a pplied to t he midpoint D of link BC. State t he value of th e couple AI which would render (a) t he hori zon tal force tran sm itted by pin B zero a nd tb) the vertical force transmitted by pin B zero.

Problem.

201

4/72 Determine th e react ion at t he ro ller F for t he fra me loaded as show n . !

a .2m

~======~===jJ C A B

I

~-.- - O,5 m - -r- O,3 m-.-j

250 N

Problem 4/ 12

c

4/73 Th e device shown in th e figur e is designed to drive brad s int o picture-framing mat erial. For a gr ipping force of 10 Ib on th e ha ndle s, determine th e force F exerted on th e brad . Ans. F = 25 1b

p

Problem 4/ 70 4 /71 The automobile bu mper jack is designed to su pport a 4000-N down ward load . Begin with a free -body diagram of BCD a nd determine t he force su pporte d by roller C. Note th at roller B does not contact t he ver tical colu mn. Ans. C = 6470 N 150mm

150 mm

~"GfO

o

400

"H+-

A

0.75"

, , "" 1 1 2 ,0 -,,

-

mml

C 4000 N

- t-C=-:'::"':'-rI 1' 00 mm I .

,

~-- 24 0 m m

10 Ib

Problem 4/73

D

' F

~ lOoo mm--1 Probl em 4/7 1

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202

Chapter 4

Struc tur es

4/74 Th e device shown is used to st ra ighten bowed decking boards j ust prior to fina l nailing to th e joists. Th er e is a lower bracket (not shown ) at 0 which fixes the part OA to a joist, so th at th e pivot A may be considered fixed. For a given for ce P exerted perpe ndicular to th e handle ABC as shown, determine th e corre sponding norm al force N applied to th e bent board nea r point B . Neglect friction.

4/76 T he wingnut B of th e collapsible buck saw is tig ht ened until th e ten sion in rod AB is 200 N. Determ ine th e force in th e sa w blad e EF and th e magnitude F of the force supported by pin C.

25°:\ I \ , \,

..,

..,-_-'.::'I=====\~. B

E

p

Problem 4/76

4 /77 Determin e the magn itude of t he pin reaction at A and t he magnitude and directi on of th e for ce reaction at th e roller s. Th e pull eys at C and D are small. AilS . A = 999 N. F ~ 314 N up

Problem 4/74

D

Representative Problems 4 /75 The "jaws-of-life" device is utilized by rescuers to pry

apart wrecka ge, th us helping to free accident vict ims. If a pressure of 500 Ib/ in.2 is developed behind th e pist on P of area 20 in.2 , det ermine th e vertical force R which is exerted by th e jaw tip s on th e wreckage for th e positi on shown. Note tha t link AB and its cou nterpart a re bot h horizont al in th e figu re for thi s posit ion. Ans. R ~ lllllb

T 1

0.5m

·c

60 kg

-r-~ " '~

0.4 m - --, --- 0.4 m

- +- 0.4 m

Problem 4/77

~.J+~~~ 18"~-Problem 4/75

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Art icl e 4/6 4 /78 The figure illustrates a common problem ass ociated with simple stru ctures. Un der the loadings L, the rafters can rotate, t he ridge bea m at A ca n lower, and the walls Be and DE can rota te ou tward, as shown in part b of th e figure. Th is phenomenon is sometimes clearl y observed in old wooden farm struc tu res as a centra l sa.gging of th e rid ge beam when viewed from th e side. A simple remedy is shown in part a of th e figur e. A cha in or cable is st retched between fast eners at Band E , and t he tu rn buckle F is tightened u ntil a proper tension is achieved, thereby prev enti ng the outward tilti ng of th e walls. For given values of the dimension d and t he point loads L (wh ich resu lt fro m th e distributed loads of th e rafter and roofing weights and any additio na l loads such as snow), ca lculate th e tension T required so th at there are no out wa rd for ces on the walls at B and E. Assu me th at the support of th e rafters at the ridge beam is pu rely ho rizontal and that all joints are free to rotate. d

d

6

31 2l

A

t~7

4/79 Compou nd-lever snips, shown in th e figure, are designed to rep lace regu lar ti nne rs' snips when large cutting forces are requ ired. For the gri pping force of 150 N, wha t is the cutti ng force P at a dist an ce of 30 mm along th e blade fro m the pin at A ? Ans. P = 1467 N

Problem 4/79

L

t

20 lb

12

B

E

F (a)

O.6
c

D

L

t

L A

t

20lb Problem 4/80

I b)

Problem 4/78

Marwa n and Waseem AI-Iraqi

203

4/80 A pair of 20-lb forces is applied to th e handles of the small eyelet squeeze r. Th e block at A slides with negligible fr iction in a slot machined in t he lower part of t he tool. Neglect th e small force of th e light return spr ing AE and determine the compressive force P applied to the eyelet .

d

L

P r o b le m s

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I

204

Chapter 4

Structures

4/81 A small bolt cutter ope ra ted by hand for cutting sma ll bolts a nd rods is shown in th e sketc h. For a hand grip P = 150 N, det ermine th e force Q developed by eac h jaw on th e rod to be cut. Ans. Q = 2.7 kN

4/83 For th e paper punch shown find the punching forc e Q corresponding to a hand grip P. b Ans. Q = P>

a

- -

I

P -

20

60

p

30 1 - - - -180 - - ---1

P

Dimensions in millimete rs

Proble m 4/83

p

Probl em 4/ 81

4/82 In th e spring clamp show n, an inte rnal spring is coiled a round t he pin at A a nd th e spring ends bear aga ins t t he inn er surfaces of the handle halves in or der to pro vide t he desired clamping force. In t he position shown, a force of magnitud e P = 6 lb is requi red to release the clam p. Determin e the compressive force at B if P = o.

4/ 84 Th e clamp is adjus ted so th at it exerts a pair of 200N compress ive forces on t he boards between its swive l gri ps. Determine th e force in th e threaded shaft BC and the magnitude of th e pin reaction at D.

~1010

~

p

F

30 A

B Dim ensions in millimeters

Problem 4/84 p

f-----

4.4"

Prob lem 4/82

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Articl e 4 /6

4/85 An IS-Ib force is a pplied to the handle OAB of th e cork puller. Det ermine th e ext raction for ce F exert ed on the cork. Ans. F = 54.2 Ib

r

1.3"

- t-- - 2.75"

Proble m s

205

4/87 Th e dual-grip clamp shown in th e figure is used to prov ide added clamping force wit h a positive action. If the vertical screw is tightened to produce a clamping force of 3 kN and th en th e hori zon tal screw is tightened until the force in th e screw at A is doubl ed. find the total react ion R on the pin at B. An s. R = 7.00 kN

A

Problem 4/85 4 /86 Th e unstretched lengt h of spr ing EF is 300 mm . Determine th e magnitude of th e pin reaction at O.

r- 600 mm

i=----;a""

,VV\'\'\'V 3.6kN/m

GOO mm

I

A

B

Problem 4/87

-1 F

4/88 The special box wrench wit h head B swiveled at C to the hand le A will accommodate a ra nge of sizes of hexagonal bolt head s. For the nomin al size shown where th e center a of the bolt and th e pin C are in line wit h the ha ndle, compute the magn itude of th e force supported by th e pin at C if P ::: 160 N. Assume th e su rface of the bolt head to be smoot h.

I

c

30 ..,.-- - - mm

120 mm A

500mm

I

.!

• 0 Problem 4/88

Problem 4/86

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- - - --I

p

206

Chapter 4

Structures

. -------

4 /89 Compute the force in link AB of the lifting tongs wh ich cross without touching. An s. FAB = 1650 lb ten sion

B

A

,,

)'

C D

30" ------- - - -------

.

L_ -

x

C

Prob lem 4/91 4/92 Determine th e .r- and y-com pone nts of all forces acting on each member of the loaded fra me for th e conditi ons (a) tJ = 0 and (b) 0 = 30°. Forc e P is applied to th e midpoint of memb er BC.

A

1 - - 30·

B

Problem 4/89 4/90 Det ermine th e vertical clamping force at E in terms of the force P a pplied to th e handle of th e toggle clamp.

E

,,

)'

+ --

- - 160 mm - - -- Ip i6 mm 12 mm

L--

' 5 mm 12 mm

Problem 4/92

Problem 4/ 90 4 /91 Det ermine th e .r- and y-components of a ll forces actin g on eac h me mber of the loaded fram e for th e conditions (a) 0 = 0 a nd (b) 0 = 30· . Force P is a pplied to th e midpoint of member BC. P A ns. (a) Ax Bx Cx = Dx = 2'

Cy

=

Dy = 0.289P

=

0.433P

A.,

By

Ex

Ey

0

tb) Ax A,

B,

C,

Cy

0.75P, By = 1.25P

o,

= 1.299P , Dy =

0.25P, Ex

4/93 T he figu re shows a wh eel pu ller wh ich is designed to remo ve a V-belt pulley P fro m its ti ght-fittin g shaft S by tightening of th e cent ral scre w. If t he pu lley st a rts to slide ofTth e shaft wh en th e com pression in th e screw has reached 1.2 kN, ca lculate th e ma gn it ude of th e force su pported by eac h jaw a t A . The adjus t ing screws D support hori zontal for ce a nd keep t he side a rms parallel with th e cent ra l scre w. Ans. A = 0.626 kN

0.866P

Ey = 1.5P

(force magnitudes only)

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C

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Article 4/ 6

Prob lems

207

4/95 Th e elements of a front-hinged automobile-hood asse mbly are shown in the figure. The light linkages Be a nd CD an d t he gas-pre ssu rized strut EF hold th e hood in th e ope n position shown . In th is position, t he hood is free to rotate clockwise about pin 0 ; pin A is locked u nt il t he hood has been lowered to a nearly closed horizon ta l position . For a hood weig ht of 80 Ib wit h cen te r of gravity at G, determine t he mini mu m compress ion force C in t he st rut wh ich will mai nt a in the open -hood position . Note that there a re two links OA spa ced across the fro nt of th e ca r, but on ly one se t of the remaining lin ks loca ted on the insid e of th e right-front fend er. An s . C = 77.2 Ib

Problem 4/9J 4/94 The figure shows a high-pressure ha nd pump used for boost ing oil pressu re in a hydraulic line. Wh en the handle is in equilibrium at fJ = 15° under t he action of a force P = 120 N, det er mine the oil pressu re p which act s on the 46-mm-d iameter piston. (P ressure on th e top of the piston is atmospher ic.)

200 111l1J ----------

-- - - 8

P

AO =4"

DE= 3"

BC = CD = 9"

Problem 4/95

Problem 4/94

Marwan and Waseem Al-lra qi

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208

Ch apter 4

Structures

4/96 In th e special position shown for th e log hoist , booms AF and EG a re at right angles to one a nothe r and A F is pe rpend icular to AB . If t he hoist is handlin g a log weighi ng 4800 lb, com put e th e forces sup ported by t he pins a t A a nd D in thi s one position du e to the weigh t of th e log.

100

I

,1---+0' I ;

I

85

Probl em 4/ 96

~---- 22.0 - - - -.,..·50 --;

4/97 A 250 -N force is applied to the foot -oper ated air pump. T he return spring S exerts a 3-N · m mome nt on mem be r a BA for t his position. Det ermine t he cor respondi ng compressio n force C in t he cylinder IJ/J. If the diamet er of th e piston in the cylinder is 45 mm, estima te th e air pr essu re generated for these conditions. St at e a ny ass umptions . A ns. C ~ 510 N, P ~ 321 kP a

Dimensions in millimeters Problem 4/97

4 /98 A lifting device for t ransport ing I35-kg steel dru ms is shown. Calcu la te th e magnitude of th e force exe rted on th e drum at E a nd F. L~ 34 0

I

mm

nnn-l

340

I

p

A

~

250mm j

Problem 4/98

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Arti cl e 4/ 6 4/99 T he toggle pliers are used for a va riety of clamping pu rposes. For th e ha nd le position given by a = 10° and for a handle gr ip P = 150 N, calculate th e clamping force C prod uced . Note that pins A an d D are sym met ric about th e horizontal centerline of th e tool. Ans. C ~ 1368 N

Pr obl em s

209

4 / 10 1 Det ermine t he compre ssion force G exerte d on th e ca n for a n applied force P = 50 N when th e ca n

cru sher is in t he position shown. Point B is centered on th e bot tom of th e can . Ans. C = 235 N

p

p

Problem 4/99 Prob lem 4/ 101 4 /100 Determ ine t he compr ession force C exerted on the ca n for a n a pplied force P = 50 N when th e ca n crushe r is in the posit ion shown. Note th at t he re are two link s AB and t wo links AOD , with on e pair

of link ages on each side of th e sta tiona ry por tion of t he crusher. Also, pin B is on the vert ical centerli ne of the can . Fina lly, note t hat sma ll squa re projections E of the moving jaw move in recessed slots of the fixed frame.

J-- - -

4 /101 Th e SO-kg ventilation door OD with mass cen ter at G is held in t he ope n position shown by means of a moment M ap plied at A to the open ing link age. Memb er AB is para llel to t he door for the 30° posi-

240 rum - - -..j

Il .fi m

Problem 4/ 102

Prob lem 4/ 100

Marwan and Waseem AI-Iraqi

tion shown. Determine M .

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Chapte r 4

Str u ctures

4/103 Th e car hoist allows the car to be driven onto the pla tfor m. after which the rea r wheels ar e ra ised . If the loading fro m bot h rear whee ls is 1500 lb. deter mine th e force in the hydraulic cylinder AB. Neglect th e weight of th e platform itself. Member BCD is a right -an gle bell crank pinned to th e ramp at C. Ans. AB ~ 3970 Ib C

4 /105 A ca rpenter bui lds the square fra me ABCD and th en bra ces it wit h mem ber EF as s hown in order to preven t racki ng (distor tion into a r hombic shape) under the app lied force P. Determine the ten sion T in th e brace as a funct ion of x. Take all con nect ions to be pin join ts. Member DC is firmly fastened to the 0001'. ,'2 PL Ans . T = .. x

p

A •

j

!~

L__t I

1- -

L

E

Problem 4/ 103 4/104 Th e aircra ft landing gear consi st s of a spring- and hydraulically-load ed pist on and cylind er D and th e two pivoted link s DB a nd CB. If th e gear is moving alon g the runway at a constant speed wit h the wheel support ing a stabilized constant load of 24 kN, calcu late t he tota l force which the pin at A supports.

D

' -- - - - L ----. ~ Problem 4/105 4/106 An adju sta ble tow bar connect ing t he t ractor u nit H with the landing gear J of a large aircraft is shown in the figure . Adjust ing th e height of the hook F at th e end of th e tow bar is accomplished by the hydrau lic cyl inder CD activated by a small hand pump {not shown l. For th e nomin al position shown of the triangul ar linkage A BC. calcu late the force P supplied by the cylinder to t he pin C to positio n th e tow bar. The rig has a total weight of 100 Ib and is support ed by the tractor hit ch at E .

Problem 4/104

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C

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Articl e 4/ 6

Problems

211

H

F

30"

' 24"

8"

Problem 4/106 4 /107 T he pruning mecha nism of a pole saw is shown as i t cuts a branch S . For t he particu lar posit ion

drawn, the actuating cord is parallel to the pole and carrie s a ten sion of 30 lb. Determine the shearing force P applied to t he branch by the cutt er an d the total force supported by the pin at E . Th e force exerted by the light return spr ing at C is small an d may be neglected. Ails. P = 338 lb, E ~ 75.1 lb

4/109 For the pruning shears shown , det erm ine the force Q applied to the circular bra nch of i 5-mm dia met er for a gr ipping force P = 200 N. (S uggestion: First dra w a free-body diagram of the isolated bra nch.) A ns. Q ~ 2.15 kN - --

-

190 -

-

-

-, I

p

p

Dimensions in millimeters Problem 4/109

Problem 4/107

4 /108 A double-axle sus pension for us e on small tru cks is

shown in the figure . The mass of the centra l fra me F is 40 kg , an d t he mass of each wheel an d at tached link is 35 kg wit h center of mass 680 mm from t he vertic al cent erline. For a load L = 12 kN tra nsm itted to the fra me F, compute t he tot al shear force suppor ted by the pin at A.

4/110 Th e design er s of lamp mecha n isms, such as t hat shown in the figure, usually rely on joint fr iction to aid in maintai n ing static equili briu m. For the present probl em, ass ume that sufficient frict ion exists at point C to preven t rotation t here , bu t ignore friction at all ot her joint s. If th e mass of th e lamp fixtu re is 0.6 kg wit h mass center at G, det ermine th e spri ng force FI; necessa ry for equilibrium in the position shown. A

~~..(.~~

' I;

r- 500 mm - l L

F

~/.

j ·k 65 111~ f _ 225 mm E! I~u ~r--

225 nun

750 mm --.~ Problem 4/108 Marwan and Waseem AI-Iraqi

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Problem 4/1 10

I':'m 111111 1 ,8

I

.

t"

~

I

212

Ch a pt er 4

Struc t u r e s

4 /111 In t he particular posit ion shown, the excavator ap-

plies a 20-kN force parallel to the gro u nd. Th ere are two hydraulic cylinders AC to control the arm DAB . Deter min e the force in th e hydraulic cylinders AC and t he pressure p agai nst th eir 9S-mm-dia meter pistons. Neglect the weight s of th e members compared. with th e 20-kN force. ATI S. F A C = 48.8 kN , P = 6.89 Ml' a

0.45 m

B

1.75 m

· 0.14 m 0.05m

0.09m' =F==~~~~

0.21 m

0.15 m :I=====.~';('

:£: .L----"'d~:!......-

20 kN

Problem 4/113

Problem 4/111

4 /114 Th e angle of elevat ion II of th e up per arm of a

4 /111 Det ermine th e force in hydraulic cylind er DE of th e

excavator of Prob. 4{111. Also dete rm ine the pressu re p aga inst th e l Of i-mm -diam eter pist on of th e single cylinder. Neglect th e weight s of the members compa red with th e other forces ac ting. 4 /113 Det ermine th e force in hydraulic cylinder GH of the

excavator of P rob. 4/111. Also det ermine th e pressu re p aga inst th e 9S-mm-diam et er piston of th e single cylind er . Use the additiona l dimensional detail supplied in the figur e. Neglect the weights of th e members compared with th e 20-kN force. A il s. GH ~ 45.2 kN, P ~ 6.38 MPa

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20kN

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"cherry picker " is controlled. by th e two hydraulic cylinders attached to the upper end of th e lower boom of th e rig. Eac h pisto n rod of t he cylinde r is con nected to the chai n which engages th e sp rocket A , as shown in th e enlarge d view. The sprocket is welded to the end of th e upp er arm. Det er mine the magn itude R of t he total force su pported by the hinge pin B and th e oil pressure p in th e upper cylind er to support th e arm in th e position (I = 30°. A constant pressure of 80 kPa is maintained in th e lower cylinder wit h the lower arm in the position (3 = 60°, The net area of the pisto ns su bjected to hydra ulic pressure is 7.85 ( W ·· :!) m 2 . Th e mass center of the 120-kg upp er boom is at midlength, and th e combined mass of the hinged bucket and man is n o kg.

Ar t icle 4 /6

Problems

213

E

- - - - 48" - - - -" Problem 4/114

Deta il of hoist ing mechanism

... 4/115 T he design of a hoist ing mechani sm for the dump truck is shown in t he enlarged view. Determine the com pression P in the hydrau lic cylinder BE an d t he magnit ude of the force supported by the pin at A for the part icu lar position s how n, w he re BA is perpen dicular to OAE an d link DC is perpendicu la r to AC. Th e dump a nd its load together weigh 20,000 Ib with center of mass at G. All dimensions for the indicated geom etry a re given on th e figure. Ails. P ~ 26,900 lh, A = 14,6001b

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Problem 4/115

214

Chapter 4

Structures

.... 4/116 T he shipboa rd cra ne is support ing a loa d of 4 to ns in th e posit ion shown whe re () = 30°. T he hoist ing drum B is operated by a high-torqu e elect ric motor . Ca lcu late th e added compress ion P in t he hyd rau lic cylinder and th e ma gn it ude R of th e additio na l force supported by th e pin at 0 , bot h du e to t he effect of t he 4-to n load. Ails. P = 43,100 Ib, R = 35,400 lb 1200 lb

Problem 4/117 4/118 Determine th e force acting on member ABC at conne ct ion A for the loaded space fra me shown . Each conn ect ion may be treated as a ba ll-a nd-socket joint.

.

[

l kN

z,

,, ,

2m

,j c

4'

r

Problem 4/116

4m

4/117 In th e sche ma ti c representation of an actua l structure, T represents a t urnbuckle, C an d D a re nonthrust -bearing h inges whose rows a re a long t he line CD, a nd B , E, a nd F are ba ll-a nd-socket joi nt s . Dete rmine t he te nsion T in t he t urn buc kle a nd t he force in mem ber EF . Ails. T = 1569 Ib, EF = 429 Ib

Problem 4/11 B

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Chap te r Revi ew

CHAPTER REVIEW In Cha pte r 4 we have applied th e prin ciples of equilibrium to two classes of pro blems: (a) simple trusses an d (b) frames and machin es. No new theory was needed, since we merely drew the necessary free-body diagram s and applied our familiar equations of equilibrium. Th e structures dealt with in Chapter 4, however, have given us t he opportunity to further develop our appreciation for a systematic approach to mechanics problems. The most essent ial features of th e analysis of these two classes of stru ctures are reviewed in the following statements .

(a) Simple Trusses 1. Simple trusses are composed of two-force members joined at their en ds and capable of supporting tension or compression. Each internal force, th erefore, is always in th e dir ection of its member.

2. Simpl e trusses a re built from th e basic rigid (noncollapsible) unit of th e t rian gle for plane trusses and the tetrahedron for space trusses. Additional units of a truss ar e formed by adding new members, two for plan e trusses a nd three for space trusses, attached to exist ing joints a nd joi ned at their ends to form a new joint . 3. The joints of simple trusses are assumed to be pin connections for plane trusses and ball-an d-socket connections for space trusses. Thus, the joints can transmit force but not momen t.

4. External loads are assume d to be applied only at the joints. 5. Trusses are statically dete rminate exte rn ally when the ext ernal constraints are not in excess of those required to maintain an equilibrium position. 6. Trusses are statically determinate internally when constructed in the manner described in item (2), where internal members are not in excess of t hose requ ired to prevent collapse. 7. The method ofjoints utili zes th e force equations of equ ilibrium for each joint. Analysis normally begins at a joint where at least one force is known and not more than two forces are unknown for plane trusses or not more than three forces are unknown for space trusses. 8. Th e method of sections utili zes a free body of an ent ire sect ion of a truss containing two or more joints. In general, the method involves the equilibrium of a nonconcurrent system of forces. The moment equation of equilibrium is especially useful when th e method of sections is used. In general, the forces acting on a section which cuts more than three unknown members of a plane truss cannot be solved for completely because th ere are only three independent equations of equilibrium. 9. The vector represent ing a force acting on a joint or a section is drawn on the same side of the joint or section as the member which trans mits the force. With this convention, tension is indicated when the force arrow is away from the joint or section, and compression is indicated when the arrow points toward the joint or section. Marwan and W aseem AI-Iraqi

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215

216

Chapter 4

St ructures

10. When the two diagonal members which brace a quadrilateral pan el are flexible member s inca pable of support ing compression, only t he one in tension is retai ned in the analysis, and the panel remains statically det erminate. 11. When two joined members under load are collin ear and a third memo ber with a different directio n is joined with thei r connect ion, the force in the third member must be zero unl ess an external force is

applied at th e joint with a component normal to th e collinear members. (b) Frames and Machines 1. Frames and machines are structures which contain one or more multifor ce members. A multiforce member is one which has acting on it three or more forces, or two or more forces and one or more couples.

2. Frames are st ru ctures designed to support loads, gene rally under static conditions . Machines are stru ctures which transfor m input forces and moments to output forces and moments and generally involve moving parts. Some structures may be classified as either a frame or a machine. 3. Only frames an d mach ines which are statically det erminate exte rnally and internally are considered here. 4. If a frame or machine as a whole is a rigid (noncollapsible) un it when its external supports are removed, the n we begin the analysis by computing the exte rnal reactio ns on the entire uni t. If a frame or machine as a whole is a nonri gid (collapsible) unit when its exte rnal supports are removed, then the analysis of the exte rnal reactions can not be completed until th e structure is dismembered . 5. Forces acting in the internal connectio ns of frames and machines are calculated by dismembering the structure and constructing a sepa ra te free-body diagram of each part. Th e principle of action and reaction mus t be strictly observed; oth erwise, error will resu lt.

6. Th e force a nd moment equ ati ons of equilibrium are applied to the members as needed to compute the desired unknowns.

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Review Problems

REVIEW PROBLEMS 4/1 19 Calculate th e forces in memb ers BH, HI, and Be for th e truss loaded by th e 40- and 60-kN forces. A ns. BH ~ 47 .1 kN e HI ~ 40 kN T Be ~ 6.67 kN e

4/121 Show that the force in the hor izonta l memb er BD is independent of its posit ion x with in the triangular truss. Each side of the over all truss su pports a centered vertical load L as shown by th e two dash ed vectors, and th eir loads are th en distributed to th e joints as shown by th e solid vecto rs.

I

4 panels at 5 m

lW tSf Jl Problem 4/119

4/120 Det ermine th e components of all forces act ing on each memb er of th e loaded frame.

e

d 2

x

L 2

d -x

60 kN

40 kN

%('-;)

.

.

d 2

~\k%( ' ol Problem 4/122

4/121 Th e basic stru ct u ra l shape and loading of Prob. 4/ 122 is now treated as th e loaded fra me shown in the figu re. Deter mine t he force in the hor izontal memb er BD as a function of its position x wit hin the fra me. 0.289Ld A ll ,'; , BD

v "

,,

L

217

x

Prob lem 4/120

4/121 Calculate th e force in member B O using a free-body diagra m of t he rigid member ABC. Ans. BG ~ 1800 Ib e 10'

G

10' 10'

Problem 4/12l

A

B

900lb

Problem 4/121

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x

218

Cha pt e r 4

St ru ctur e s

Arc of radi us 50 m _

4 /114 The nose-wheel assemb ly is ra ised by the a pplication of a to rque M to link BC th rough th e s ha ft at B . If the arm and wheel AD have a combined weight of 100 Ib with center of gravity at G, find the valu e of M necessary to lift th e wheel when D is direct ly und er B , at which position angl e fI is 300 •

H

2- -~-"'~:::::::--o:::::::---=------,G

I

z I ~

16"

Problem 4/126 4/117 A pneumatic cylinder pivoted at F operates the lever AB of th e qu ick-actin g toggle clamp, which holds the work in position while it is machined. For an air pressu re of 400 kP a above at mospheric pressure against th e 50-mm -diam et er piston, determine the clamping force at G for th e positi on a =: lOa, For t his positi on th e piston rod is perpendicular to AB. AilS . G = 2.30 kN

Problem 4/124 4 /125 Det ermine th e forces in membersAB, BI, and Clof the simple tru ss. Note th at all curved memb er s are two-force membe rs. An s. AB = 2.26L T, BI I. T , CI = 0.458L T

Arc of radius 50 III -

.... H ___~;i-----=-------.--

j

G

1- 100 - +Dimens ion s in millimeters

A

Pr------::?--- -:'C.- B

-,,\<...........----;:-;------Ad D

C

l-15m-.J-~ 15 mI

E

L

L

20

L

m---J.-15 m-+-15 m

I

Problem 4/127

4/118 Determine the force in each member of th e loaded. truss.

Problem 4/125 4 /126 Th e stru ct u re of Prob. 4/ 125 is modified in that the four curved. members are replaced. by th e two mem ber s AI H a nd HGF. In strumentation indicates the ten sion in memb ers CH and DH to be 0 .5L each . Deter mine the forces in member s AB, BI , and Cl. Is the problem solvab le wit hout th e infor mation about CH ?

C

E

Problem 4/128

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Rev iew P rob l e m s 4/129 The loading on the tru ss of Prob. 4/ 128 is mod ified as shown in th e figure . Deter mine the force in each member. Ails. AB ~ 1.471L C. Ai( = 0.1057L T BC ~ L C. BK ~ CD = 0.333L C CE = 0.471 L C. CK = 0.1381L T DE ~ EK = 0.255L C

1.13

L/3 D

45"

C

:\ B I . I

A

rec tly abo ve the center of rec tangle FGHI. Determin e the force in member CD . An s. CD = 2.4L T

1.13

1.13 L/3

219

F G

E

X

fir:

45"

J

22.5°- ----

r

L/3

• I I

~2 . ~ '___

0

At

FI I I I I I I

I.

1

Problem 4/129 4 /130 An antitorqu e wrench is designed for use by a crewman of a spacecra ft where he ha s no sta ble platform against wh ich to push as he tightens a bolt. Th e pin A fits into an adjace nt hole in th e structure which contai ns t he bolt to be t urned. Successive oscillations of t he gea r an d ha ndle unit tu rn th e socket in one directi on t hrough th e a ct ion of a ratch et mech anism . The reacti on aga inst the pin A pro vides the " a ntitorq ue" character ist ic of the tool. For a gr ippin g forc e P = 150 N, determ ine th e torque M tran sm itt ed to t he bolt an d the exte rna l reaction R against the pin A normal to t he lin e AB . (One side of th e tool is used for ti ghtening a nd the oppos ite side for loosening a bolt. I ,_17_.5

120mm _ _~ ~ 40 _ -,-~_--,==,--

mm

mm

P

1.25a

; I I I

I

I·

I

1.250

E~ ~ L

Problem 4/1J 1 4/132 Th e elements of a st u mp gr inder wit h a total ma ss (exclusive of th e hydraulic cylin de r DF and arm CE I of 300 kg wit h mass center a t G a re shown in the figu re . The mechan ism for a rti cu lati on a bout a ver tica l ax is is omitted, an d t he whe els at B a re free to turn . For the nomina l positi on show n, link CE is hor izont al a nd t he teet h of th e cutti ng whee l a re even wit h th e gro u nd. If the magn itude of th e force F exe rted by t he cut ter on t he st um p is 400 N, determ ine th e force P in the hydraulic cylind er a nd the magnitude of th e force supported by th e pin at C. The problem is to be treated as two-dim en sional. If -

E

150 D

~-- 1 300--~

I

600

450 1 - - - 900

Problem 4/130

Dimensions in millimeters

4 /131 The depicted structure is under consideration as the upper portion of a tran smi ssion -line tower a nd is supported at points F, G, H , and I . Point C is di-

Marwan and Wa seem AI-Iraqi

- - -!--,+-

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Problem 4/132

F ..-20"

220

Chapter 4

Stru ctures

4/133 T he st ructure shown is bein g consi de red as pa rt of a la rge cylindr ical vessel which m ust support exte rna l loads. Strai n-gage inst rument ati on ind ica tes that the compressive force in member BE is 0.8L. Determine the forces in membersAB and DE. Mak e use of sym met ry . An s. AB ~ 0.434 L T , DE ~ 1.166L C

4/135 T he tree feller cuts ofTla rge trees nea r gro und level a nd th en cont inues to grasp t he trunk. Determine th e force in hydrauli c cylinder AB for the posit ion shown if the tree weighs 6000 lb. Dete rm ine t he required pressure on the 4.72-in. -diameter pist on of the cylinder . Ans. F A R ~ 59,900 Ib, p = 3420 Ib{in. 2

I--- 7.5' -----,.- 6.8' ---'---1 2.4' Problem 4{135

L a = 22.5"

Problem 4/133 4/134 Det ermine th e punching force P in terms of the gripping force F for th e rivet squeezer shown.

... 4/136 Each of the landing struts for a planet exploration spacecra ft is designed as a space tru ss symmetrical about the vertical x -z plan e as shown . For a landing force F = 2.2 kN , ca lculate th e cor respo nding force in mem ber BE. Th e assu mptio n of st a tic equilibr iu m for the t russ is permissible if the mass of th e t russ is very sma ll. Assu me equa l loads in the sym me trically placed members. Ans. F il E ~ 1.620 kN

e_ j:U 1

Problem 4{134 F Problem 4{136

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Review Problems

~

4 /137 Th e bar bender for formi ng reinforcing steel consists of t he lever DAB hinged at 0 to the fixed base plat e. The forming rollers atA, 0 , and C a re a ll fre e 50 to rot at e. For th e 60° position shown, a force Ib is required on th e handle. Calcula te the x- a ndycomponents of th e force R exerted by the base plate on th e shaft of roller 0 for thi s position . (S uggestion: Anal yze se pa ra te free-b ody diagram s of th e bent bar a nd of the lever and attached rollers. Observe th at th e force exerted by th e bar on th e roller at a is in addition to the force R supplied by th e ba se plat e. ) Ans. Rx ~ 838 Ib, R, = 43.3 Ib

221

A 6'

or

100lb

Problem 4/IJS

·4/139 Det ermine a nd plot th e forc es in all members of th e loaded simple truss as fun cti ons of th e a ngle n over th e range 0 S n S 90°. St ate th e minimum va lues for the force s in members A C and CD a nd the values of a at which th ese mini ma occur.

An s. AB = 3 ,' 3 kN T , BC

/'

AC

~A ' / 1-'60"

>:;'

/' y

AD

"

10/

1

L-- x

/1"

2 (~ ' , , 0.5"

· 3 r-- j;f '~O

0 .5" ---'.

I

".

v C

1" 2

CD

3 ./3

~

15 + sin 4

4(

n -

n

2

2(, /3 cos a + sin n )

--

A - I = 1.25" 0 -2 = 0.75" A - O = 3" O- C = 2" 0 - 3 = 1.5"

fA C )min

2.60 kN T at "

(CD l min

6 kN C at

It

~

3D'

= 0

2m 30"

3 kN

'Computer-Oriented Problems

-4/138 Th e simple truss of Prob. 4/2 is modified as shown in th e figure, in th at the a ngle of th e support surface at C can be var ied from 0 (vertical ! to 90° thorizon ta ll. fa) Plot the force in mem ber BC as a function of (love r this ra nge . Note any unusu al condition s. (h I For wh at value of 0, if any, is th e force in member BC zero? (c ) If member BC is designed to fail at a load of 1000 Ib in eit her tension or compression. wha t is the a llowab le ra nge for th e a ngle If !

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C

D

kN T

a)

15 - + 9 si~ n - 3 "f;j cos n kN C

A

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3 , "3 cos + sin

,!:3 cos n

Problem 4/13 7

'!

6 kN C

3kN

Probl em 4/139

222

Cha p t er 4

Structu r e s

-4/140 Th e type of marine crane shown is uti lized for both

dockside a nd offshore operations . Determin e a nd plot the force in member Be as a fun ction of the boom angle 8 for 0 s (} :S 80°. Neglect t he radius of a ll pulleys an d t he weight of th e boom.

-4/142 A door -opening mechani sm is s hown in t he figure. The spri ng-loade d hin ges at 0 provide a mo ment K r (J which te nd s to close t he door, where 8 is th e

door-open ing a ngle and t he to rs ional spring constant K r = 500 lb-in ./rad . The mot or unit a t A provides a variable moment M so tha t t he slowly ope ning door is always in qu asi-static equ ilibri um. Det ermine th e mom ent At a nd th e pin force at B as fu nctions of (} for the ra nge 0 :S 8 :S 90°.

e

a

_.----

3"

I

6"

Problem 4/ 140 - 4/ 14 1 T he "jaws-of-life" dev ice of P rob. 4/75 is redrawn

here wit h its jaws ope n. The pr essure behind th e pisto n P of area 20 in .2 is ma in tained at 500 Ib/ in .2 Calculate and plot the force R as a functio n of 8 for 0 :s tJ :s 45°, where R is the vertical force acting on the wreckage as shown . Determine t he max imum val ue of R a nd t he corres ponding val ue of t he jaw a ngle. See the figu re of Prob. 4/ 75 for dim en sions and the geometry associated with th e condition 8 = O. Not e that link AB and its cou nterpart are both ho rizontal in th e figure for 8 = 0 but do not remain hor izont al as the jaws ope n. A ns. R mul( = 1314 Ib a t (J = 45°

11"

lb.-r:_

B -

_

14"_

-

"-i

Prob lem 4/142 - 4 / 14 3 The st ru ct u ral members support the 3-kN load whi ch may be applied at any angle tI from essen tial ly - 90° to + 90°. T he pin at A must be designed

to support t he maximum forc e tran sm itted to it . Plot th e force FA at A as a fu nction of (J a nd dete r mine its maxi mum value a nd t he corresponding angle 0. Ans. FA max = 6 kN at 8 = - 26 .6°

f-- mm 3oo --j r-, I

-,-

(<;)J"

I

VD

400

mm

I

A

Problem 4/ 14 1

i (~J r ·

c

~ (, \)

2B 600

mm

I

,

600

mm

Problem 4/143

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~3

kN

Revi ew Problem s

\

· 4 / 14 4 Th e uniform 30-kg ventilat ion door OAP is opened

by t he mechanism shown. Plot the required force in t he cylinder DE as a fu ncti on of the door opening a ngle 0 over the range 0 ~ e :S 0 mllX ' wher e (}max is the maximum opening. Determine th e minimum and maximum values of this force and th e angles at which th ese ext re mes occur. Note th at th e cylinder is not horizontal when (} = o.

\ \ \

\ \ \

-

. . ; - - - - 525 - - -':1 p,

AB = 300, Dimensions in millim ete rs Problem 4/144

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223

When forces are continuously distributed over a region of a structure. the cumulative effect of this distribution must be determined . The cables of this cable-stayed br idge support the weight of both the roadway and any vehicles distributed along its length. Marwan and Waseem AI-Iraqi

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Chapter

DISTRIBUTED FORCES

CHAPTER OUTLINE 5/1

Introduction

SECTION A. Centers of Mass and Centroids 5/2 Center of Mass 5/3 Centroids of Lines, Areas. and Volumes 5/4 Composite Bodies and Figures; Approximations 5/5 Theorems of Pappus SECTION 8. Special Topics 5/6 Beams-External Effects 5/7 Beams-Internal Effects 5/8 Flexible Cables 5/9 Fluid Statics

Chapter Review

5/1

INTRODUCTION

In the pr evious chapte rs we treated a ll forces as concentr ated along t heir lin es of action and at thei r points of applicat ion. This treatment provided a r eason able model for tho se for ces. Act ually, " concentrated" forces do not exist in the exact se nse , since eve ry exte r na l for ce applied mechanically to a body is distributed over a finite contact area , however sma ll. The force exerted by th e pavem ent on an automobile tire, for instance, is ap plied t o the tire over it s entire area of contact, Fig. 5/ 10, which may be appre ciable if th e tire is soft . When analyzing t he for ces act ing on t he car as a whole, if the dim ension b of the contact area is negligible compared with the other pert inent dim en sion s, such as the distan ce between wheels, then we may replace the actual distr ibuted contact for ces by t heir resu ltant R treated as a concentrated for ce. Even th e for ce of contact between a hardened steel ba ll a nd it s race in a loaded ball bea rin g, Fig. 5/ 16, is applied over a finit e tho ugh ext remely small contact area. The forces applied to a two-force memb er of a truss, Fig. Marwan and Waseem AI-Iraqi

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225

226

Chapter 5

Di st ributed For ces

OR o

(a)

Enla rged view orconta ct

I I

R

R Ib'

C I I

I liNII

c

~ I

I

C lei

Figure 5 '1

Marwan and Waseem AI-Iraqi

5/ 1c, are applied over a n actual area of contact of the pin against the hole and internally acro ss the cut section as shown. In these and othe r similar examples we may treat the forces as concentrated when analyzing their external effects on bodies as a whole . If, on th e other hand, we want to find the distribution of internal forces in the material of the body near the contact location, where the internal stresses and strains may be appreciable, then we must not treat the load as concentrated but must consider the actual distribut ion. This problem will not be discussed here because it requires a knowledge of the properties of the ma te rial a nd belongs in more adva nced treatmen ts of th e mecha nics of mat eri als a nd the theori es of elast icity and plast icity. When forces are applied over a region whose dimensions are not negligible compared with other pertinent dimensions, then we must account for the actual manner in which t he force is dist rib uted, We do this by summi ng the effect s of the distributed force over the entire region using mathematical integrat ion. This requires that we know the inten sity of the force at any locat ion . Th ere are three cat egories of such problems. (1) Line Distribution. When a force is dist ributed along a line, as in th e continuous vertical load supporte d by a sus pended cable, Fig. 5/2a , the intensi ty w of the loading is expressed as force per unit length of line, newtons per met er (N/ m) or pounds per foot (lb / ft). (2) Area Distribution. When a force is distribu ted over an area, as with the hydraulic pressure of water against the inner face of a section of da m, Fig. 5/ 2b, the intensity is expressed as force per unit area. Th is intensity is called pressure for the action of fluid forces and stress for the internal distribut ion of forces in solids. The basic unit for pressu re or stress in SI is the newto n per square meter (N/ m 2 ), which is also ca lled the pascal (Pa). Thi s unit , however, is too s mall for most applications (6895 Pa = 1 Ib/in.2 ). The kilopascal (kPa ), which equals 103 Pa, is more commonly used for fluid pressure, and the megapascal, which equals 106 Pa, is used for stress. In th e U.S. customary system of units, both fluid pressure and mechanical stress are commonly expressed in pounds per square inch Ob/ in?). (3) Volume Distribution. A force whi ch is distributed over th e volum e of a body is called a body force. Th e most commo n body force is the force of gravitational att raction, which acts on all elements of mass in a body. The det er mination of the forces on t he supports of the heavy cantilevered st ructure in Fig. 5/2c, for example, would require accounting for the distributi on of gra vitatio nal for ce t hrough out th e st ructure . Th e intensity of gravi ta tio nal force is th e specific weig ht pg, where (' is the density (mass per unit volume) and g is the acceleration due to gravity . The units for pg a re (kg/ m3)( m/ s 2 ) = N/m 3 in SI un its and Ib/ ft 3 or lb/in." in the U.S. customary sys te m. The body force du e to th e gravitational attraction of th e ea rth (weight ) is by far the most commonly encountered distributed force. Section A of thi s chapter treats the determination of the point in a body through which the resultant gravita tiona l force acts , a nd discusses th e associated geometric properties of lines , areas, and volumes. Section 8 WNW.gigapedia.com

Ar ticle 5 /2

~I I I I II

I I 1 IIII1 I

fal

227

Cen t er o f Mass

fbi

fel

Figure 5/2

treats distribu ted forces which act on an d in beam s and flexible cables and distributed forces which fluids exert on exposed surfaces.

SECTION A. CEN TE RS OF MASS AND CENTROIDS 5 /2

CENTER OF MASS

Consider a three-dimensional body of any size and shape, having a mass m . If we sus pend the body, as shown in Fig. 5/ 3, from a ny point such as A, the body will be in equilibri um un der t be actio n of the ten sion in the cord a nd th e resultant W of the gravit ationa l forces acting on a ll particles of t he body. This resul tant is clear ly collinea r with the cord. Assume that we mark its positi on by drill ing a hypo th etical hole of negligible size along its line of action. We repeat the experiment by suspending the body from ot her points such as Band C, a nd in each instance we mark the line of action of the result ant force. For all practical purposes these lines of action will be concurrent at a single point G, which is called th e center of gravi ty of th e body. An exact a nalysis, however , would account for th e slightly differing d irections of th e gr av ity forces for the var ious particles of th e body, because those forces converge toward the center of attraction of the earth. Also, because t he particl es are at different distan ces from the ear th, the inten sity of the force field of the earth is not exact ly const ant over the body. As a resu lt , th e lines of actio n of the gravity-force resu ltants in the experiments just described will not be quite concurrent, and therefore no unique cente r of gravity exists in the exact se nse. This is of no practical importance as long as we deal with bodies whose dimen sions are small compared with those of the earth. We therefore assume a uni for m and parallel force field due to the gr avitational att ract ion of the earth, and this assumption result s in the concept of a unique center of gravity . Determining the Center of Gravity

To determine mathematically th e location of the cent er of gravity of any body, Fig. 5/4 0 , we apply th e principle of moment s (see Ar t . 2/6 ) to the parallel syste m of gravitationa l forces. T he moment of the resulMarwan and Waseem AI-Iraqi

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T T. .J o r 'cr' . . , t A: B

,,

c•

A. ,8 "

"

e .,......

rc

A

IV

IV

IV

fal

fb i

fel

Figure 5/3

228

Chapter 5

Distributed Forces

z

z

, , ,

,, ,,

G dlV

IV

I I I I I

. 1_- ) ' Z _-- I

_ ......

,,--

"

--1 ,' .\'" x,

..---',,->

Z I

.~

-

\ ,.---

,, '.....---' ,

'x (b l

'a l

Figure 5/4

tant gravitational force lVabout any axis equals the sum of t he moments about th e sa me axis of th e gravita tiona l forces d lV acti ng on all particles treat ed as in finitesim al elements of th e body. Th e res ultant of the gra vitational forces acting on a ll eleme nts is th e weigh t of th e body a nd is given by the sum IV = f dW. If we apply th e moment principle about th e y-ax is. for example, t he moment abo ut this axis of the eleme nta l weight is x dW, and the sum of th ese mom ents for all eleme nts of th e body is f x dW This sum of moments mu st equa l IVx, th e mom ent of the sum. Thu s, XW = f x d W. With similar expression s for the other two components, we may express t he coordinates of the center of gravity G as

JXdW x

=

\V

_ JydW y

=

IV

J zdlV z =

lV

(5/ 1" )

To visualize the physical moments of the gravity forces appearing in the t hird equa tion, we may reori ent the body a nd attached axes so th at t he z-axis is horizonta l. It is essential to recognize that the numerator of each of these expressions represent s the Slim of the m om ent s , whereas the product of W a nd th e corres ponding coordina te of G re presents t he moment of the Slim . This moment principle finds repeated use throu ghout mechanics. Wit h the substitution of W = IIIg an d d W = g dill , t he express ions for the coordinates of the center of gravity become

y 111

(5/ 1bl

Equ at ions 5/ 1b may be expressed in vector form with t he aid of Fig. 5/ 4b, in which the eleme ntal mass and th e mass center G are located Ma rwan and W aseem AI- Iraqi

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Ar t icl e 5 / 2

Ce nte r of M a ss

by their respective position vectors r = xi + yj + zk and r = xi + yj + Ek. Thus, Eqs. 5/ 1b are th e components of t he single vector equation

(5/2 )

The density p of a body is its mass per unit volume . Thus, the mass of a differ ential element of volume dV becomes dm = p dV. If p is not const ant throughout th e body but can be expressed as a function of t he coordinates of the body, we must account for thi s variation when calculating th e num era tors and denominators of Eqs . 5/1b . We may then write these expressions as

x

JxpdV J pdV

Y

JYfl dV JpdV

Jzp dV JpdV

(5/ 3)

Center of Mass versus Center of Cravity

Equ ati ons 5/ 1b, 5/2, and 5/ 3 are independent of gravitatio nal effects since g no longer appears. They the refore define a unique point in the body which is a fun ction solely of the distribution of mass. Thi s point is called the center of mass , and clearly it coincides with the center of gravity as long as t he gravity field is treated as uniform and parallel. It is meaningless to speak of t he cen te r of gravity of a body which is removed from the gravitational field of the earth , since no gravitational forces would act on it . Th e body would , however , still have its un ique center of mass. We will usually refer henceforth to the center of mass rather than to the center of gravity. Also, the center of mass has a special sign ifica nce in calculat ing the dynamic respon se of a body to unbalanced forces. This class of problems is discussed at lengt h in Vol. 2 Dy namics . In most pr oblems th e calculation of the position of the cente r of mass may be simplified by an intelligent choice of refer ence axes. In gene ra l th e axes shou ld be placed so as to simplify the equations of t he boundaries as much as possible. Thus, polar coordinates will be useful for bodies with circular boundaries. Another important clue may be taken from considerations of symmetry. Whenever there exists a line or plane of symmetry in a homogeneous body, a coordinate axis or plane should be chosen to coincide with this line or plane. The center of mass will always lie on such a line or plane, since the moments due to symmetrically located elements will always cancel, and the body may be considered composed of pairs of t hese eleme nts. Thus, th e center of mass G of the homogeneous righ tcircular cone of Fig. 5 /5a will lie somewhere on its central axis, which is a line of symmetry. The center of mass of the half right -circular cone lies on its plan e of symmet ry, Fig. 5/5 b. Th e cente r of mass of the half rin g in Fig. 5/5 c lies in both of it s plan es of symmetry and th erefore is Marwan and Waseem AI-Iraqi

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(b l

(e l Figure 5 /5

22 9

230

Chapter 5

Distributed Forces

situate d on line AB. It is eas iest to find the locat ion of G by using symmet ry when it exists.

5 /3 CENTROIDS OF LINES , AREAS, AND VOLUMES When t he density p of a body is uniform throughout , it will be a cons tan t factor in both the numera tors and den ominators of Eqs . 5/3 and will therefore cancel. The remaining expressions define a purely geometrical pr oper ty of t he body, since any reference to its mass prop erties has disappeared. The term centroid is used whe n the calculation concerns a geomet rica l sha pe only. When speaking of an actual physica l body, we use the te rm center of mas s . If the den sity is uniform throughout th e body, the positions of the centroid a nd center of ma ss are identical, whereas if the density varies, these two point s will, in general, not coincide. Th e calculation of centroids falls with in three dist inct cat egories, depending on wheth er we can model th e shape of th e body involved as a line, an area , or a volume. (I) Lines. For a slender rod or wire of lengt h L , cross-sectional area A , and density p, Fig. 5/6, t he body approximates a line segme nt , and dm = pA dL. If P a nd A are constant over t he lengt h of the rod, t he

coordina te s of th e center of mass also become th e coordinat es of the cent roid C of the line segme nt , which, fro m Eqs. 5/ 1b, may be written

_= -f - -

xdL

x

L

y =

f

ydL

L

z =

f

zdL L

(5/4)

Figure 5/6

Note that , in general, the cen troid C will not lie on th e line. If the rod lies on a single plane, such as the x-y plane, only two coordinates need to be calculated.

A

(2) Areas. When a body of den sity p has a small but constant t hickness t, we can model it as a su rface a rea A, Fig. 5/ 7. Th e mass of a n element becomes dm = pi dA . Again, if p and t are consta nt over the ent ire area, t he coordinates of the center of mass of the body a lso become the coordinates of th e centro id C of th e surface a rea , an d from Eqs. 5/ 1b the coordinates may be written

_f

xdA x = - - A x

Figure 5 /7

y

f

ydA A

z =

f

zdA A

(5/ 5)

The nume ra tors in Eqs. 5/5 are called t he first moments of c rec ." If t he surface is curved, as illust ra ted in Fig. 5/7 wit h the she ll segme nt, all "Second momen ts of areas (moments of first moment s ) appear later in our discussion of second mom enta of urea, also called area moment s of inertia, in Appendi x A.

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Art icl e 5 /3

Centro ids of Lines , Areas , an d Vo l u m e s

three coordinates will be involved . Th e centroid C for the curved surface will in general not lie on the surface. If the area is a flat surface in, say , the x-y plane, only the coordina tes of C in th at plan e need to be calculated.

(3) Volumes. FOI' a general body of volume V and density 1', th e element has a mass dm = I' dV. Th e den sity I' cancels if it is cons ta nt over the entire volume, and the coordinates of the center of mass also become the coordinates of the centroid C of t he body. From Eqs. 5/3 or 5/ 1b th ey become

_f

xdV x = - - V

y

f

y dV V

f

zdV V

(5/ 6 )

Choice of Element for Integration Th e principal difficulty with a theory often lies not in its concepts but in t he procedures for applying it . With mass centers and cent roids th e concept of th e mom ent principle is simple enough; the difficult steps a re th e choice of the differentia l eleme nt and setti ng up th e integrals. Th e following five guid elin es will be useful.

(I) Order of Element Whenever possible, a first -order differen tial element should be selected in prefer ence to a higher-or der eleme nt so that only one int egration will be required to cover the entire figure. Thus , in Fig. 5/ Sa a first-order horizont al st rip of area dA = I dy will require only one integration with respect to y to cover the ent ire figure. The second-order element dx dy will require two integrations, first with respect to x and second with respect to y, to cover the figure. As a further example, for th e solid cone in Fig. 5/ Sb we choose a first -ord er eleme nt in the form of a circular slice of volume dV = r.r 2 dy . Th is choice requires only one integration. and thus is preferable to choosing a thirdorder eleme nt d V = dx dy dz, which would require three awkward integrations.

(2) Continuity. Whenever possibl e, we choose a n eleme nt which can be integrated in one continuous operation to cover the figure. Thu s, the horizontal st rip in Fig. 5/Sa would be pr eferable to the vertica l strip in Fig. 5/9, which, if used, would require two sepa rate integr als because of the discontinu ity in t he expression for the height of the strip at x = X l'

(a)

Iy I

IY 1

(3) Discarding Higher-Order Terms. Higher-order terms may always be dropped compared with lower-order term s (see Art. 1/7 ). Th us, the vertical st r ip of a rea under the curve in Fig. 5/10 is given by the first -ord er ter m dA = y dx , and the second-order t ria ngu lar area ~dx dy is discarded. In the limit , of course, there is no error. (4) Choice of Coordinates. As a general rul e, we choose th e coordin at e system whi ch best matches the bou ndar ies of the figur e. Thus, th e boundaries of th e a rea in Fig. 5/ 110 a re most easily describ ed in Marwan and Waseem AI-Iraqi

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(b)

Figure 5/8

231

232

Chap t e r 5

Distribu ted Force s

dy

.r T L _LLJ

x

L_.LL-'--_ ----' __ X

-1 f- dx

xI

Figure 5/9

Figure 5/10

rectangular coordinates, whereas the boundaries of the circular sector of Fig. 5/ 11b are best suited to polar coordinates.

(5) Centroidol Coordinate of Element When a first - or secondorder differential element is chosen, it is essential to use the coordinate of the centroid of the element for the moment arm in expressing the moment of the differential element. Thus, for the horizontal strip of area in Fig. 5/ 12a, the moment of ciA about the y-axis is Xc dA, where Xc is th e x-coordinate of the centroid C of the element . Note that Xc is not th e X which describes either boundary of the ar ea . In the y -direction for this element th e moment ar m Yc of the cent roid of t he element is t he same, in t he limit , as th e y-coordinates of the two boun daries. As a second exam ple, cons ider th e solid half-cone of Fig. 5/ 12b with th e semicircular slice of differential thickn ess as th e element of volume. The moment arm for the element in the x-direction is the distance X c to th e cent roid of th e face of th e element and not the x-distance to the bound a ry of the element. On th e other hand , in t he z-direction th e momen t arm Zc of th e cent roid of th e element is th e sa me as t he z-coordinat e of th e element . With these exampl es in mind , we rewrite Eqs. 5/5 and 5/6 in th e form

x =

y

f

yc dA

y I ~ I I .,:1 ~ 1 1I I

y

l

lLJ I

L _ _ -.J __ x

(b)

Figure 5/11 WNW.gigapedia.com

r

: 0

(a)

Marwan and Waseem AI-Iraqi

(5/5 a)

z =

A

--x

Article 5 /3

Centroids of Lines, Areas , and Vo lumes

/ /

Y

Figure 5/12

and

Y =

f

Ye d V

V

Z =

(5/ Gal

It is essential to recognize that the subscript c serves as a reminder that the moment arms appearing in the numerators of the integral expressions for moments are always the coordinates of the centroids of the particular elements chosen. At this point you shou ld be certain to under st and clearly the principle of moments, which was intr oduced in Art. 2/4. You should recognize t he physical mea nin g of t his pri nciple as it is applied to the syste m of parallel weight forces depicted in Fig. 5/40. Keep in mind th e equivalence between t he moment of the resu ltant weight W and t he sum (integrall of th e moments of the eleme nta l weights dW, to avoid mistakes in setting up the necessary mathem at ics. Recognition of the pr inc iple of moments will help in obtaining the correct expression for the moment arm XC' Ye' or ze of the cen troid of the chosen different ial element. Keepi ng in mind the physica l pictu re of the pr inciple of mome nts, we will recognize that Eqs. 5/4, 5/5, a nd 5/6, which a re geometric relationships, are descriptive also of homogeneous physical bodies, because the density p cancels. If the density of the body in qu estion is not constant but varies througho ut t he body as some func tion of t he coordinates , then it will not cancel from the numerator and denominator of the mass-center expressions . In this event , we must use Eqs. 5/3 as explained earlier.

Sa mple Problems 5/ 1 through 5/ 5 which follow have been carefully chose n to illustra te the applicati on of Eqs. 5/ 4, 5/5, an d 5/ 6 for ca lcula tin g the loca tion of t he centro id for line segme nts (slender rods), areas (t hin flat plates), and volumes (homogeneous solids). Th e five integration considerations listed above are illustrated in detail in these sample problems. Secti on C/ lD of Append ix C conta ins a tab le of integr als which in cludes those needed for the problems in thi s a nd subseque nt cha pte rs. A su mmary of the centroida l coord ina tes for some of the comm only used shapes is given in Tables D/ 3 a nd D/ 4, Appendix D. Marwa n and Waseem AI-Iraqi

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233

234

Chapter 5

Distr ibu t ed Fo rc es

Sample Problem 5 /1 Centroid of a circular arc. Locate th e centro id of a circu lar arc as shown in the figure.

CD

Solution. Choosing th e axis of sym metry as th e .r-a:xis makes y = O. A differ ential element of arc has the length dL = r d o expressed in polar coordina tes, and th e x-coordina te of t he element is r cos H. Applying th e first of Eqs. 5/4 and substituting L = 2ar give [LX =

J

.r dL I

~

(2nr li

f..

(r cos 0) r

de

2urx = 2r 2 sin a

x=

r sin

(~

- - a

Ans.

--

For a se micircular arc 2a = 1T, which gives x = 2r/ 1T. By symmetry we see immed iate ly tha t this res u lt also applies to the quarter-circular arc when the measu rement is made as shown.

! r

--

x

a

Helpful Hin t

CD It should be perfectl y evident that polar coordinates are preferable to rectangu lar coordinates to express th e lengt h of a circular arc.

Sample Problem 5/2 Centroid of a triangular area. Det ermine t he distance Ii from the base of a tria ngle of alt itude h to the cent roid of its area.

CD

Solution. The x-axis is taken to coincide with the base. A differential strip of area dA = x dy is chosen. By similar tri an gles x / (h - y ) = b/h . Applyin g th e second of Eqs. 5/00 gives lAy

and

~

J

Yc dAI

bh _ 2"Y = y =

r

2 b(h - Y) d = bh oY h Y 6

h

A ns.

3

H elpful Hint

This sa me result holds wit h respect to either of th e ot her two sides of th e trian gle consider ed a new base wit h cor res ponding new altitude. Thus, th e cent roid lies at t he intersect ion of th e medians, since the distance of this point from an y side is one-third th e altitude of th e t rian gle wit h that side considered th e bas e.

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CD We save one integr ation here by using th e first-order element of ar ea. Recogni ze that dA mu st be expressed in term s of th e integration varia ble y; hence. x = f ty ) is required.

Article 5/1

Centro ids of Lin es. Area s. and Volum es

235

Sample Problem 5/3 r

Centroid of the area of a circular sedor. Locate th e centroid of the area of a circular sector wit h respect to its vertex.

CD CV

Solution I. Th e .r-axis is chose n as th e axis of symmetry. an d y is th erefore automat ically zero. We may cover th e area by moving an elemen t in t he form of a' partial circular rin g, as shown in t he figure, from the center to the outer periph ery. Th e ra dius of the rin g is "o and its thi ckn ess is dro, so that its area is dA ~ 21'0" dl'o· Th e .r-cocrdina te to th e centroid of the element from Sample Probl em 5/ 1 is Xc = "0 sin nl u, whe re "o replaces r in th e formul a. Thus, th e first of Eqs. 5/00 gives

2" ( 1l'T2 )x-

-

21T

=

i' (1'0

-sin --,,) (2roa

0

_ X

a

dro )

2 r sin a

Ans.

= ----

3

"

Solution I

Solution II. Th e area may also be covered by swinging a t rian gle of differe ntial area about th e vertex and through th e total angle of t he sector. Thi s tr iangle, shown in th e illustrati on , has an area dA = (r I 2)( r d m, where higher-order term s are neglected. From Sample Problem 5/2 the centroid of the triangu lar element of area is two-thirds of its alt itude from its vertex, so th at the .r-coordina te to th e centro id of th e element is Xc = ~ r cos it, Applying th e first of Eq s. 5/00 gives

IAi =

Jx, dA ]

Helpful Hints

CD Note

carefu lly that we mu st distin gu ish betwe en t he vari able "o and t he consta nt r .

CV Be carefu l not

to use "o as th e centroi dal coordi na te for the eleme nt. .I'

I

2

I-

3

1 .t"(' =- r cos 8 and as before

_ x

2r sina

An s.

= ----

3

"

I I I

a

For a semicircular area 20 = 1T, which gtves F 4r13rr. By symmet ry we see immediately that thi s resu lt also app lies to the quart er-circular area where the measur ement is made as shown. It should be noted t hat, if we had chose n a second-or der element "o d r o dO, one int egr ation with res pect to (J wou ld yield the ring with which Solution I began . On th e other ha nd, int egration with respect to "o initially would give th e t riangu lar element with which Solution II began .

- x

Soluti on II

41'13H) ~ c

J \-\vt': ~ I--- r

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I

'_

_

1'---1

236

Chapt er 5

Dis tr ibut ed For ces

Sample Problem 5 /4 Locate th e centroid of t he area u nder the curve x x = a.

CD

hi' from

x

010

Y I I

x = ky 3

I I I -X

Solution I. A vertical element of area ciA = y dx is chosen as shown in the

I I I I

figur e. Th e x-cocr dinate of th e cent roid is foun d from th e first ofEqs. 5/00 . Thus,

'---

[AX =

J

Xc

; ; - _.L----'

X

a

dA ]

Y I

bL - - - - - - - I

Subst itut ing y = (x / k ) IJ3 and k = a /b 3 and integrating give 3ab _ 4

3a'b 7

-

-x = - -

4

Ans.

= 'i a

X

I x = ky 3 I I I I I I

In the solution for y from the second of Eqs. 5/00 , the coordina te to the cen troi d of th e rectangular element is Yr: = y /2, where y is the height of the strip govern ed by th e equa tion of th e curve x = kY. Thus, th e moment principle becomes lAy =

b

I-- x----1 r- dx

J dAl Yc

Substitutingy = b(x /a )1/3 and integr ating give 3ab _

3ab'

Ans.

""4 Y = W

Yh- x = '!..!2:

--1

b ~ _~ __2__

Solution II. The hori zont al element of area shown in the lower figure may be employed in place of t he vertical element . Th e x-eoordina te to th e centroid of th e rectan gu lar element is seen to be Xc = x + ~ (a - x ) = (a + x )/2, wh ich is simply the average of t he coordina tes a and x of th e ends of the strip. Hence, [AX

=

J

Xc

dAl

x

r

(a - x) dy

=

r

(a ;

x) (a

- .r) dy

Th e valu e of y is fou nd from [Ay

=

J

Yc dA ]

y

t o

(a - x )

dy =

Jr"o y (a

-

.r }

dy

where Yc = Y for th e hori zontal str ip. The evalua tion of t hese integrals will check t he previous results for i and y.

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t --

I x = ky 3

:I I

I

dy

---.L

x+--a-x--ll

L---~aJ-x Helpful Hint

CD Note

that element.

Xc

x for the vertical

Article 5/3

Cen troids of Lines. Are as. a n d Volumes

237

Sample Problem 5/5 Hemispherical volume. Locate the centroid of the volu me of a hem isph er e of radius r with respect to its base .

z = 0 by symmetry. Th e most convenient element is a circu lar slice of thickness dy parall el to the x -e plan e. Since th e hemisph ere int ersectj t he y- z plan e in the circle y2 + z2 = r 2 , th e radius of th e circula r slice is z = + r 2 - l . Th e volume of t he elemental slice becomes

Solution I. With the axes chosen as shown in th e figure , x =

CD

dV = mr' - y') dy The second of Eq s. 5/ 00 requ ires

[Vy

=

I

y, dV]

whe re Yc = y . Inte grating gives

~m-3y =

±m-

4

Solut ion I

y = ~r

Ans. z

Solution II. Altern ati vely we may use for our differential element a cylind rical shell of length y , radius z, and thickness dz, as shown in th e lower figure. By

I~ YJ'

y,=."/2 ,-

.

expanding th e ra dius of th e shell from zero to r, we cover the entire volume. By symmet ry th e centroid of the elemental shell lies at its center, so t hat y, = y / 2. Th e volume of the element is d V = dzHy ), Exp ressin g y in term s of z from th e equation of the circle gives y = + r 2 - z2. Uelng tb e velu e of'{er- compute d in Solution I for th e volum e of the hemisph er e and subst itut ing in th e second of Eqs. 5/00 give us

<3m

[Vy

=

I

y, dV]

(~ 1Tr")y =

,,Ir' - r (2m,!r' - r ) dz f. f.o mr'z - dz 0

2

rrr4

T

=

y =

Z3 )

4

gr

Solut ion II

Ans.

Solutions I and II are of comparabl e use since each involves an element of simple sha pe and requires int egration with respect to one variable only.

Solution ",. As an alternative, we could use the angle (} as our var iable with limits of 0 and rr/ 2. The rad ius of either element would become r sin (J, whereas th e thickness of the slice in Solution I would be ely = (r dm sin 0 and tha t of the shell in Solution II wou ld be dz =(r dm cos 0. Th e len gth of t he she ll would be y = r cos O.

z I I I I dO

r

r d8

I

- Y Solut ion III Helpful Hint


ident ify the higher-order element of volume which is omitted from the expression for dV?

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238

Chapte r 5

D i stributed Forc e s

PROBLEMS

x~ , y -, and z-coordina tes of th e mass cente r of t he homogen eous semicylinder. Ans. E = O,y = - 50.9 mm , Z = - 180 mm

5/3 Specify the

Introductory Problems 5/1 Wit h your pencil, make a dot on the position of your best visual estimate of the cen troid of the t ria ngu lar area . Check t he position of your est imate by refer rin g to th e results of Sample P roblem 5/2 and to Table D/ 3. 10

_ / mm -x

I

z

/I 1/ /

8

6

1/

1/

Problem 5 /3

/

5/ 4 Specify t he X - , y -, and z-coordina tes of t he mass cen ter of the qu arter -cylindri cal shell.

1/

4

2

~

1/

/ /

1/ 1/

120 mm 2

4

6

8

10

Problem 5/4

Problem 5 /1

5/5 Determine the y-coordinate of the cent roid of t he

5 /2 With your pencil, mak e a dot on the posit ion of your

best visual est imat e of th e centroid of th e area of the circ u lar sector. Check you r est imate by using the results of Sample Problem 5/3 .

shaded area. Check your result for the special case 0 =0. _ 2Ch 3 _ 0 3) Ans. y = h 2 2 31

y

y

I I

I

I I I

T

I I

h

I I

I

I

I I

_ L J~o~~/~o~ _

200 mm I

30 0

I

--- x

30 0

-------- x

Problem 5/5

Problem 5 /2

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- a)

Article 5 /3

5/6 Det erm ine the x- and y-coordi nates of the centroid of the shaded area.

Probl ems

239

5/9 Determine the coordina tes of the centroid of t he shaded area . Ans. x = ftb , "y = ~a

.Y I

.Y

I I I I I I

I I I

b

I a

I

l r- - I I

OL - - _L._ --I_ - - - x 0 12

---------- x

Problem 5/ 6

Problem 5 /9

5/7 Det er mine the .r- and y-coord ina tes of t he centroid of the tra pezoida l a rea . 2 2 h lo + 2b l _ 0 + ab + b A ns . x = 3 (0 + b ) , y 310 + b)

5/10 Determine the coordinates of th e cent roid of th e shaded area. .Y

I I I

.Y

I I I

x = Ily 2

I I I

a

I

b

I

...J _ - - x

L

a

L _ _--,--_ _ --.J

Ii Problem 5/7

Problem 5/10

5/8 Locate the cent roid of th e shaded area shown. .Y

I at: )- - ......

I

4'

...

5/11 Deter mine t he coordina tes of the cen troid of the shaded area. Ans. X ~ 1.443, Y ~ 0.361/1

.'2./5

y

~ )' = ;) - X

"-

x

b

,

\ xv = k

.

\

\

11 I

\

\ o-o - - - - - - - - - 5. . . . -

x

Problem 5/8 Problem Sill

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240

Chap ter 5

Dist ributed Forces

Representative Problems

5/15 Deter mine the .r-coordinate of th e mass center of the

z

ta pered steel rod of lengt h L where the diam eter at t he large end is twice th e diam eter at th e small end . A ns, x = ~ L

5/12 Find th e distan ce fro m the vertex of the rightcircu lar cone to the centroid of its volu me.

1

{..---------

L

.-------i

I I

('~-"

Dia . = D

Problem 5/15 Problem 5/12 5/13 Locate th e centroid of t he sha ded area. i = 2a/5, Y =

5/16 Let c - co and determine th e .r- and y -coordinates of the centroid of the shaded area .

3b/8

y I

y

I I

a

I

I

b

'--_

_

--::-_ _..:J

x

a

00L---

-

--=:= =

Problem 5/13

c

- - x

Problem 5/16 5/14 The mas s per un it len gth of the slender rod var ies with position accordi ng to p = Po(l - x/2>, wher e x is in feet. Determine th e location of the center of mass of the rod .

5/17 Determ ine the .r- and y-coordinates of t he centroid of th e shaded area. An s. :< = 0.762, Y = 0.533

{

[L

_

Iy = kx II.3

I

I I I

I

oo -- - ---- --- - !.--x 2 Problem 5/14

Marwan and W aseem AI-Iraqi

Problem 5/17

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Art icl e 5/ 3 5 /18 Deter mine t he y-coordinate of the centroid of t he

shade d area.

Pro b lems

2 41

5/10 Determine the y -coordinate of the centroid of t he shaded area. y I I

y I I I I

b

a

I I

I I

b

x= ky 2 l b I

a

2"

I _ __ _ _ _ _ _ _ _ _ LI

/ 45" -

Problem 5/18

-

-

-

-

-

-

- -- r

Problem 5/20

5 /19 Deter mine t he x- and y-coordi nates of t he centroid of

the shaded area.

/

/

r

a

/ /

_ 7a _ Ans . x = 6 (77" _ 1) ,y =

5 /11 Determine the x- and y-coordinat es of the centroid of

th e sha ded area.

a 1T -

A ns. i = _ a_ y ,, - 1 '

1

y

=

7b 6(,, - 0

y

I

1

1

1

1

1

T b

2"

t: QI

21

'----

r- ~--1

-

-

--' - - - -

1:

x

_

- - -x

I'

Problem 5/19 Problem 5/2 t

5/11 Determine th e coordina tes of t he centroid of th e shaded area. y

y = bsi n ~ -...... 2a

1

I

<,

1 1 1 1

b

1 1 1 1

/

" - - - - - ----' - - - x

a

/ / Problem 5/22

Marwan and Waseem AI-Iraqi

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242

Chapt er 5

Dist ribut ed Forces

5/23 Locate th e centroid of th e shaded area between th e 24 6 tw o cu rves . Ans. x = is, Y = :;

5/26 Det ermine the z-coordina te of the centroid of th e volum e ge ne ra ted by revolving th e shaded a rea through 1800 ab out th e a-ax is.

\'= ~

y I I I

z

. I 4 1

I _. _.~.-

)'2

2 1- - - - - - - '..... ......X=2 I

I

I

I

I I I

I I I

: <. I I I

I

00

----- t---

Problem 5/26

x

5/27 Determ ine the .r-coordi nn te of the centroid of the solid spher ica l segm e nt.

Problem 5/23

Ans. x

5 /24 Det ermine th e coordinates of th e centroid of th e sha ded area.

=

~R

)'

I R

R

2 .2 , / - ' -- '""' _-1- I 1

)'

I I .r =

I I I

'<.

I I

I I

.:

I

I

I I I Ib I I I

I

_1. \

\

- - - -a- - - -...l..--- x

_- x

.1

, -,

......

__

......

Prob lem 5/27

Problem 5/24 5 /25 Locate th e centroid of the area shown in the figure by direct integr at ion .
5 /28 Determi ne the x- an d y -coordina tes of the cent roid of th e volu me gene rated by revo lving the cross-hatc hed triangu lar area t hrough 90 0 about t he a-ax is.

)'

I I I

I I I" I I

->---

" Problem 5/28 _~:::""_~_ _...J

-

-

-

x

Problem 5/25

Marwa n and W aseem AI-Iraqi

WNW.gigapedia.com

_x

Articl e 5 /3 5/19 The th ickness of the t riangular plate varies linearly with y from a valu e to along its base y = 0 to 2/ 0 at y = II . Determi ne the y-coordinate of the ce nter of mass of the plate. All s. 5' ~ 3,./8

P roblems

243

5/31 Deter mine th e a-coordina te of the centroid of th e volume obtained by revo lving th e shaded a rea within the parabola abou t the z-axis through 180°.

z I

s

I

I I I I I I I I

b-

-

I I

.u b

Problem 5/29 5/30 Ca lcu la te th e dist ance h measured from th e base to th e centro id of the volume of th e fru stum of th e right-circular cone.

Problem 5/32 5/33 Th e homogeneous slender rod has a uniform cross sect ion an d is bent into the sha pe shown. Calculate the y -coordinate of the mas s center of th e rod. (Reminder: A differential arc lengt h is d L = ,f(dx )2 + (dy )2 ~ h + (dx / dy )2 dy.) AilS. Y = 57.4 mm y

I I I

I I I 100 mm :

I

Problem 5/30

I

5 /3 1 Deter mine the a-coordinate of the cent roid of the solid genera ted by revolvin g the qu arter-circu lar area

through 9W about the z-axis.

,_ /Vt S .

z

_ Z

=

lOOmm

1Iu 2(4 + 3 ,, )

Problem 5/ll 5/34 Determine the z-coord inate of th e centroid of th e volum e obta ined by revolving th e shaded t ria ngu lar area about th e z-axis through 360".

I I

z

y -, <,

"

I a ~---

it..

I

/ u

Probl em 5/31 Problem 5/34

Marwan and Waseem AI-Iraqi

-- x

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244

Chapter 5

Distributed Forces y

5 /35 Determ ine t he z-coordina te of the cent roid of the volu me generate d by revolving the sha ded a rea a round t he z-axis through 3600. Ans. = 00/8

I I

~

z

z 1 1

r: '\__ ---\~- x

a

Problem 5/37 a

"'--

--'

->1_ -

_

.... 5 /38 Determine t he coordinates of th e cen troid of th e volume obtai ned by revolving th e shaded area a bout th e z-ax is t hrough th e 90 0 angle.

r

Ans. x ::: y = (.; -

~)a, Z = a/4

Problem 5/35

z 5 /36 Locate th e ma ss center of t he hom ogeneou s solid body wh ose volume is determined by revolving the shaded a rea through 360 0 about th e a-ax is.

1

---1- - -

---3 / 1

r

I

r--

I I I I 1 1 I

a

30 0 m m

I

y-

I

1 1 1

X

/ /

200mm

r = kz 3

Problem 5/38

z

.... 5 /39 T he cylindr ica l shell of u niform small th ickness has a rad ius r a nd height z which varies fro m zero at o = 0 to h a t lJ = 11' according to z ::: klJ wh ere k is a constant. Dete rmine the X - , y o, a nd a-coordinates of t he mas s center of th e she ll. 2r _ 4r _ Ii A ns. x ::: - ,y 11'2 ' Z 1T 3

Problem 5/36 .... 5 /37 Deter mine t he y -coordina te of t he centroid of t he pla ne a rea shown. Set h = 0 in your result a nd com -

pare with t he resu lt

y :::

4a for a full semicircula r 3 1T

a rea (see Sa mple Problem 5/3 a nd Tab le D/ 3). Also eva lua te you r result for the conditions h ~

/

I

00

Ii

/

a ;;;.

~ a nd

4

Problem 5/39

MalWan and Waseem AI-Iraqi

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Art icle 5/3 .,. 5/40 Locat e th e center of mas s of the homogen eous bellshaped she ll of uniform but negligible thickne ss.

Ans. z = __ o_ n - 2

Problem 5/42

.... 5/41 Locate t he center of mass G of the steel half r ing. (H i nt: Choose an element of volume in th e form of a cylindrical sh ell whose intersection with th e pla ne of th e ends is shown.) a2 + 4R 2 Ans. = -:;--;;-2rrR

~ , ----- It.( ----- - i

_R--l

R~

Problem 5/41

Marwan and Waseem AI-Iraqi

245

... 5/41 Determi ne the x-coordina te of t he mass center ufthe homogeneo us hem isph ere with th e sma ller hemispherical portion remov ed. 45 Ans.i = R 112

Problem 5/40

r

Pr oblems

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246

Chapter 5

Distr ibuted Forces

5 /4 COMPOSITE BODIES AND FIGURES; ApPROXIMATIONS \Vhen a body or figure can be conveniently divided into several parts whose mass centers are easily determined, we use the principle of moments a nd treat each part as a finite element of the whole. Such a body is illustrat ed schematica lly in Fig. 5/ 13. Its par ts have masses "' 1' m2. ma with the respective mass-center coordinates XI. X2. X3 in the x-direction. The moment principle gives

where X is the x-coordinate of the center of mass of the whole. Sim ilar relations hold for the other two coordinate directions. We generalize, th en, for a body of any number of parts and express the sums in condensed form to obtain the mass-cent er coordinates

(x -

-

=-};mi }; ",

"_ my-

y =-

(5/ 7)

}; m

Analogous relation s hold for composite lines, areas, and volumes, where the nr's are replaced by L's, A's, an d V's, respecti vely. Note that if a hole or cavity is considered one of the component parts of a composite body or figure, the corresponding mass represent ed by the cavity or hole is treat ed as a negat ive quantity.

An Approximation Method In pract ice the boundaries of an area 0 1' volu me might not be express ible in terms of simple geometrical shapes or as shapes which can be represen ted mathematically. For such cases we must resort to a method of approx ima t ion. As an exa mple, consider th e pro blem of 10-

f---- x, --~

- x,---.j -;:j ' _.~-+---;-\J I

I \

• G2

•

I I GI

\

\

G3 l1/ :l

~--- X ---l

Figure 5/13

Marwan and Waseem AI-Iraqi

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Articl e 5 {4

Composite Bodies and Figures; Approximations

eati ng the centroid C of th e ir regu lar area shown in Fig. 5/ 14. Th e a rea is divided into stri ps of width .lx an d variable height h . The a rea A of eac h strip, such as the one show n in red, is h ..h and is mu lt iplied by the coordinates Xc and Y c of its centroid to obtain the momen ts of the eleme nt of area. The sum of the momen ts for all strips divided by th e tota l area of the strips will give the corresponding centroidal coordinate. A sys tematic tabulation of the result s will permit an orderly evaluation of the total area ~A , the sums ~A\"(" and ~Ayc' and the centroidal coordinates y

Figure 5/14

We can increase the accuracy of the approximation by decreasing th e widths of the st rips. In all cases th e average height of th e st rip should be estimated in approximating th e a reas . Alth ough it is usu ally advantageous to use elements of constant width, it is not necessary. In fact, we may use ele ments of any size and shape which approximate the given area to sat isfactory accuracy. Irregular Volumes

To locate the centroid of an irregular volume, we may reduce the problem to one of locating the centroid of an area. Consider the volume shown in Fig. 5/15, wher e the magn itudes A of th e cross-sectio na l area s normal to the x-direction are plotted against x as shown. A vertica l strip of area under the curve is A ~x, which equals the corresponding ele ment of volume .l V. Thus, the area under the plotted curve represents the volume of th e body, and th e x-coordinate of the centroid of the a rea under the curve is given by

whic h equa ls for the centroid of t he act ua l volume.

xA

I

.

C

Figure 5/15 Marwan and Wa see m AI- Iraqi

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247

248

Chapter 5

Distributed Forces

Sample Problem S/6

y I ~- 1 2"'------j

Locate the centroid of the shaded area.

, 4-

Solution . The composite area is divided into the four elementary shapes shown

41 2"

in the lower figure. The centroid locations of all these shapes may be obtained from Table D/ 3. Note that th e areas of the " holes" (parts 3 and 4) are taken as negative in the following table:

A

i

2

PART

in.

1 2 3 4

120 30 - 14. 14 -8

Y

];A

in.

in.

in.

6 14 6 12

5 10/ 3 1.273 4

3

720 420 - 84 .8 -96

c-t-t - ---': 1 -

I

3"

yA 3

in.

600 100 - 18 -32

2

L-_--"

L-_---,_-----'

TOTAl.";

127.9

959

- I

650

The area counte rparts to Eqs. 5/ 7 are now applied and yield

[-X _- -LAX] U\

X

7.50 in.

Am.

[y ~ ~]

650 Y ~ 127.9 ~ 5.08 in.

Am.

~

959 -127.9

~

Sample Problem S/7

6

Approximate the x-coordinate of the volume centroid of a body whose length is 1 m and whose cross-sec tional area varies with x as shown in the figure.

~s

.,; Solution. The body is divided into five sections . For each section, the average area, volume , and centroid location are determi ned and entered in the following ta ble: A. v m2

Volume V m'

i

Vi

INT ERVAL

m

m'

0-0.2 0.2-0.4 0.4-0.6 0.6-0.8 0.6-1.0

3 4.5 5.2 5.2 4.5

0.6 0.90 1.04 1.04 0.90

0.1 0.3 0.5 0.7 0.9

0.060 0.270 0.520 0.728 0.810

TOTAl.";

4.48

I

5

V

4

3

2

<,

/'

V

1

0.2

0.4

0.6 I,m

0.8

2.388

Helpful Hint

G)

[x ~ ~::]

x

= 2.388

4.48

MalWan and Waseem AI-Iraqi

~ 0.533 m

A ns.

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G) Note th at t he shape of the body as a function of y and z does not affect X.

1.0

Article 5{4

Composi te Bodies and Figures; Approx im ations

Sample Problem '5 / 8 Locate th e center of mass of the bracket- and-sh aft combina tion. The vertical face is mad e from sheet metal which has a mass of 25 kg /m 2 • T he material of the hori zon tal base has a mas s of 40 kgjm2 , a nd the stee l shaft has a density of 7.83 Mg/ m 3 .

Solution. The composite body may be conside red to be composed of the five elements shown in th e lower portion of the illu stra tion . T he t r iangu lar part will be taken as a negative ma ss. For t he re ference axe s indicated it is clea r by symmet ry t hat t he .r-coordina te of the center of mass is zero . The ma ss m of each part is eas ily calculated and should need no further expla nat ion. For Part 1 we have from Sample Problem 5/ 3 _ 4r 4(50) z = - = - - = 21.2 mm 3.. 311

Dimensions in mill imet ers

For Pa rt 3 we see from Sample Problem 5/ 2 that the cent roid of th e triangular mas s is one-third of its altitude above its base. Measurem en t from the coordinate axes becomes

z = - [150

- 25 - !(75l ] ~ - 100 mm

The y~ an d z-coordi na tes to th e mass cente rs of the remaining parts should be evide nt by inspection . The terms invo lved in a pplying Eqs. 5/7 ar e best handled in th e form of a tabl e as follows :

m kg

y

Z

my

mz

PART

mm

mm

kg ·mm

kg ·mm

1 2 3 4 5

0.098 0.562 - 0.094 0.600 1.476

0 0 0 50.0 75.0

21.2 - 75.0 - 100.0 - 150.0 0

0 0 0 30.0 1l0.7

2.08 - 42.19 9.38 - 90.00 0

140 .7

- 120.73

TOTALS

2.642

Eq uations 5/7 are now a pplied an d the results a re

[V~ ~:l:mmy]

V= 2.642 140.7 = 53.3 mm

[z

Z

=

~mz] ~m

- 120.73 2.642

Marwan and Waseem AI-Iraqi

Ans. Ans.

- 45.7 m m

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D ,

,

249

250

Chapter 5

Distribu ted Forces

PROBLEMS

5/46 Deter mi ne the y-coordina te of th e cent roid of the sha ded a rea.

Introductory Problems

y I

5/43 Determine the coordinates of the cent roid of th e t rapezoidal area shown . Ans. X = 233 mm, Y = 333 mm

I U

"

"

,

600

,

mm

1

i

h

600

mm

,,: \ a I ,

60'

/ /

~ ~o: I I

x

Problem 5/ 46 ....

5/47 Determine the coordina tes of t he cent roid of the shaded area . Ans. X = 244 mm , Y = 117.7 mm

x

~':"'""J

300

mm

y I

Problem 5/43 5 /44 Ca lcu late the y -coordina te of th e cent ro id of th e shade d a rea.

1

200 111m ~ 150 mm

1 - 200 ml11

1

125 mm

\'

1-+-----+-

60mm

125mm

L b=~~,,=-"-~~--=---' 74mm

x

Problem 5/ 47

~_\l

~

5/48 Determ ine t he x- a nd y -coordina tes of th e centroid of th e shade d a rea.

f- 32 --+- 32 --1 mm

y 1

mm

I I

Problem 5/44

1

40

5/45 Dete rmine th e y -coordi na te of the centroid of th e sha ded ar ea .

I

165

+[ [

40

v

! - 240

Dimensions in millimeters

<\=> "

"

Problem 5/48

/-r T / Ii / 45" .1._ J x /

"'" _ _ _ 45""' _"

~

Problem 5/45 MalWanand Wasee m AI-Iraqi

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Articl e 5 /4

Problems

251

y I I I

5/49 Det er mine th e y -coordinate of th e centroid of th e shaded area. Ans. Y = 102.5 mm

-"~~----' -

2a - -''

- - -

x

Problem 5/51

Problem 5/49

5 /51 Locate th e mass center of the slender rod bent into

th e sha pe shown.

Representative Problems 5 /50 Determine t he dist an ce

y I I

H from the bottom of the

base plat e to th e centroid of th e built-up stru ctu ra l section shown. 10

I

-j

r~ 1O

~ 1 L r 11 10

rI

~- BO~

~ 160

120

~

Problem 5/ 52 5 /53 Th e rigidly connected unit consists of a 2-kg circular disk, a Lfi-kg round sha ft , an d a I -kg square plate.

10

,

Det erm ine th e a-coord inate of th e mass center of the un it. A ils. Z = 70 mm

Dimensions in millimeters Problem 5/50 5/51 By th e method of thi s article, determine th e .r- and y-coordina tes of the centroid of th e shaded area of Prob . 5/ 19, repeat ed here. 7a a Ails. X = 6(,, - 1) ' y ~ ,, - I

- .'I: x

Problem 5 /53

Marwan and Waseem AI-Iraqi

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252

Chapter 5

Distrib uted Forces

5/54 Deter mine t he height above th e base of the centro id of the cross-sect ional ar ea of the beam . Neglect th e fillets.

6.24-

5/57 Calcu late t he X-, y -, and a-coordinates of th e mass center of the bracket form ed from t he steel plate of uniform thi ckn ess. Ans. X = 38.3 mm , Y = 64.6 mm , Z = 208 mm

l.fo-"

1

13r : .40_

~l.-12.48---1 Problem 5/54 5/55 Th e homogeneous hemisphere wit h the smaller hemi sph eri cal portion removed is repeated here from Prob. 5/ 42. By th e met hod of this article, det erm ine th e .r-coordina te of th e mas s center. 45 Ans. X = 112R y

Problem 5/57

I

I I

5/58 The 400-mm x 400-mm aluminum plate is 6 mm thick. A steel wire wh ich ha s a mas s of 0.5 kg per

meter of length is welded to t he plate as shown. Determine the required posit ion s of th e 50-mm· ra dius hole if t he mass cente r of th e entire unit is to be at the geometric center of th e plate. Neglect th e diam ete r of the wire relative to th e 400·mm dim en sion of th e plate, and check to see that the hole clears the wire . s

-----1

Problem 5/55 5/56 Determine th e positio n of t he mass cente r of the cylind rical shell with a closed semicircu lar end. The shell is mad e from sheet metal with a mass of 24 kg/m 2 , and th e end is made from metal plat e with a mass of 36 kg/m 2 .

Problem 5/58

Problem 5/56

MalWan and Wa seem AI-Iraqi

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Article 5/4 5/59 Determine the coordinates of th e mass cent er of the welded asse mbly of u niform slender rods made from t he sa me bar stoc k. 3a 211 tm An s. X = - -, Y = - - - , 2 6 +.. 6 + 71'" 6 +r.

Probl ems

253

5/61 Determine th e distanc e H from the bottom of th e base to th e mass cente r of th e bracket casting. Ans. H = 1.717 in.

z I I I

/I a

/

J,'f j"

a

a a

Problem 5/61 5/61 Th e assembly shown is formed of u niform rod. For what valu e of the len gth I will th e mass cen te r be

Problem 5/59 5/60 The welded asse mbly is made of a un iform rod weigh ing 0.370 Ib per foot of length and th e semicircu lar plate weighing 8 Ib per square foot . Calculate t he coord inates of the center of gravi ty of th e assembly.

located at a heigh t of

~

above the support ing

sur face?

Z

I

I

Problem 5/62

Problem 5/60

5/63 Determ ine the coordinates of the mass center of the bracket , which is constru ct ed fro m shee t metal of u niform thickn ess. A ns. X = 2.48 in., Y = 2.71 in., Z = - 0.882 in.

z

Problem 5/63 Marwan and Waseem AI-Iraqi

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254

Chap t er 5

Distr ibut e d Fo rc e s

5/ 64 Calc ulat e t he coordina tes of the mass center of th e meta l die casti ng shown .

5/66 Determ ine t he .r-coc rdi na te of the centroid of th e pOI'· t ion of th e homoge neous sphe re shown.

z

y

I I

I R

R

1"2 2 , ~-

~r1

I

_ - 1/

25 35

I

J L·

"

f I

....-.

_- x

_

-\ \

, -, ......

_--

Problem 5/66 Dimensions in millimeters

Problem 5/64 5/ 65 The figure shows th e underwat er cross-sect ional a rea A a dist anc e x aft of th e bow at the wa terline of a sa ilboat hull . T he variation of A with x is shown in th e graph for a particula r hu ll. Determine th e distan ce X aft of point P to the cent er of bu oyan cy of t he hull (cent roid of the dis placed volume of water ). The location of the cente r of bu oyan cy is a cri t ical param et er in th e design of the hull . Ails. X = 12.53 Ft

5/67 Deter mine the dimen sion h of t he rect an gul ar ope ning in t he squa re pla te wh ich will result in the mass center of t he rem a ining plate bein g as close to th e upper edge as possib le. Ails. h = 0.58Ba y

a 2

I I

a

2

!

a I I I I

- , -..La

,~--- "2

--"'i

Problem 5/67

x

/

5/68 Det ermine th e depth h of th e circ ular hole in th e cube for whi ch th e a-coordi nate of the mass cen ter will hav e th e max imum possible valu e. x

I I

16 12 / / 1

I 1 1/ I I

1

h -j

I

II r---.l.__ 1

<,

I I

I , I

z_ _

I

-y '\

4

'/

./ 4

8

12 x,ft

16

20

-, 24

Problem 5/65 Problem 5/68 MalWan and Waseem A1· lraqi

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Ar ticl e 5 / 4 .... 5/ 69 A cylindrica l containe r wit h a n exte nded rectangular back and semicircu la r ends is all fabricated from the sa me sheet-metal stock. Ca lculate th e angle a ma de by the back wit h the vertical when the container re st s in an equilibrium pos itio n on a horiz ontal su rfa ce. An s. a = 39.W

2 55

.. 5/70 Th e circu lar disk rotates about an axis t hrough its center 0 a nd has t hree holes of diameter d posi tio ned. as shown . A fourth hole is to be drilled in the disk at t he same radius r so that th e disk will be in balan ce (mass center a t 0 ). Determine th e required diam eter D of t he new hole and its angu lar position . A ns. D = 1.227d , 8 = 84.9"

Problem 5 /70

Problem 5/ 69

Marwan and Waseem A l-lraqi

Probl em s

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25 6

Chapte r 5

D i st r i b ut e d Forces

v

'1 1

:

5 /5

~/

I

, .-;~

1

V'I

1

I

I

I

cT -

IL_l.:_

y)'

_

I I

_ ...L

I 1

I

IL_

x

Figure 5/16

THEOREMS OF PAPPUS '

A very simple method exists for calculating the surface area ge nerate d by re volving a pla ne curve about a nonintersect in g axis in the plan e of the curve. In Fig. 5/ 16 th e line segme nt of length L in t he x·y pla ne generates a surface when revolved about the x-axis. An eleme nt of this surface is the ring ge nerated by dL . The area of this ring is its circumfere nce tim es its slant he ight or dA = 2 11)' dL. Th e total a rea is th en

Because yL =

Jy dL, t he area becomes (5/8)

wher e y is th e y-coordinate of th e centroid C for the line of lengt h L. Thu s, the generate d area is t he sa me as the lat eral a rea of a rightcircular cylinder of len gth L and rad ius y. In the case of a volume generated by revolving an area about a non inter secting line in its plane, an equally simple relation exist s for finding th e volum e. An eleme nt of th e volume genera ted by revolving th e area A about th e .r-axis, Fig. 5/1 7, is th e elementa l rin g of crosssection dA and radius y . The volume of the ele ment is its circumference times dA or dV = 21Ty dA, a nd the tot al volume is

y I I I I I I I I

L

_

Figure 5/17

"At t ributed to Pappus of Alexand ria, a Gree k geome ter who lived in the thi rd cen tury x .n . The theo rems orten bear the name of Guldinus (Paul Guldin, 1577 -1 643 ), who claimed original authorship, although the works of Pappus were apparently known to him.

Marwan and Waseem AI-Iraqi

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Arti cle 5/5

Because jzi

f

y dA, the volume becomes

(S/ 9) where y is the y-coordinate of the centroid C of the revolved area A. Th us, we obtain t he gene rated volu me by multiplying the generati ng ar ea by th e circumferen ce of th e circular path described by its centroid. The two theore ms of Pappus, expresse d by Eqs. 5/8 and 5/9, are useful for det ermining areas and volumes of revolution. Th ey are also used to find the cent roids of plan e curves an d plane areas when we kn ow the corresponding areas and volumes crea ted by revolving th ese figures about a nonin tersecting axis. Dividing the area or volum e by 2". times the corresponding line segment len gth or plane area gives the distance from th e centroid to t he axis. If a line or an area is revolved through an angle 8 less than 2"., we can determine the gene rated surface or volume by replacing 2". by 8 in Eqs . 5/8 and 5/9. Thus, the more general relations are (S/ Sa)

and (S/ 9a)

where 8 is expressed in radians.

Marwan and Wa see m A l-lraqi

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Theorems of Pappus

257

258

Chapter 5

Distr ibuted Force s

Sample Problem 5 /9 Determine t he volume V and surface area A of the complete torus of circu lar cross sect ion.

Solution. Th e torus can be generated. by revolving th e circu lar area of radiu s a through 360 0 about the a-axis. With the use of Eq. 5/9a , we have

CD

v = OrA = 2m R )(17l12) = 2n2Ra 2

Ans.

Similarly, usin g Eq. 5/ 80, gives A

~

OrL

2n{R )(217l1 )

Ans.

4 n2Ra

Helpful Hint

CD We

note th at th e a ngle 0 of revolution is 2 tr for the complete rin g. Thi s common bu t special-case res ult is given by Eq . 5'9.

Sample Problem 5 /10

z

Calculate the volume V of the solid gene rated by revolving the su-mm right. triangula r area through 1800 abo ut the a-axis. If this body were constructed of steel, what would be its mas s m?

I I

I

~ I

I

Solution. With t he angle of revolution 0

CD

~

I I .1- - -

180°, Eq . 5/ 9a gives

V = OrA = ".(30 + 3(60)][~(60)(60Jl ~ 2.83(10')

mm3

Ans.

I ::~

_60 --im m

~m m

Th e mass of the body is then

m = pV = [ 7830

I

z

~~] [2.83(10') mm3{1O~tmmr Ans.

= 2.21 kg

I

30

l"'"'""m m

Helpful Hint

CD Note M arwan and W aseem AI-Iraqi

W'NW.gigapedia.com

that iJ mu st he in radians.

Article 5 /5

PROBLEMS

Problem s

259

5/73 The qu arter-circular ar ea is rotated t hroug h 360 0 about the y-axis. Determine t he volume of the resu lt ing body, which is a portion of a sphere.

Introductory Problems

2 7T

/2a 3

Ans . V = - '3--

5 /71 Usin g t he methods of this article, determine the surface area A a nd volume V of the body form ed by revolving the rectan gu lar area through 360 0 about th e a-axis. An s. A = 10 300 mm '2 , V = 24 700 mm3

y 1 1 1 1

1 145 0 1

a

- - - - - - - - - - - .'1:

x-Problem 5/73

5/74 Compute t he volume V of the solid generated by revolving t he rig ht triangle abou t th e a-ax is t h rou gh

Problem 5/71

180°.

5/71 The circular ar c is rotated through 360 0 about th e yaxis. Deter mine the outer sur face ar ea S of th e result ing body, whi ch is a portion of a sphere. y 1 1

/ /

1 1 1

/

/

°/

145 I

y

/

.>

a

~-- - - - - -I ,

:45 0 '

1

1 1 1

-,

-- x

Problem 5/74

-, -,

" , '

Problem 5/72

5/75 Th e body shown in cross section is a complete circu la r ri ng formed by revolving the octagonal area about the a-axis. The entire su rface is to be cover ed with a specia l coating. Determine th is su rface a rea . An s. A = 177 100 mm 2

Dimensions in millimeters

Problem 5/75 Marwan and Waseem AI-Iraqi

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260

Chapter 5

Distributed Forc es

5/76 Det erm ine th e volu me V ge ne rated by revolving t he qua rt er-cir cular a rea a bout the z-ax is t hrough a n a ngle of 90°.

5/79 Det ermine t he total surface a rea of th e body ge ne rated by revolving t he shaded a rea t hro ugh 360 0 about the x-a xis.

z I

AilS . A

= m ·2 ( 1T

-

1)

y I I

="-' __ x

L __

Problem 5/79

Problem 5 /76

Representative Problems 5 /77 Ca lculate t he volu me V of the space r in the form of t he comp lete rin g of the semicircula r sect ion shown . Also compute th e total su rface a rea A of the space r. Ans. V = 6.56 in .a, A = 29.3 in.:!

z

"""""""

5/80 T he water storage tank is a shell of revolution a nd is to be spra yed with two coats of paint wh ich ha s a coverage of 500 ft2 per gal lon. The engi neer (who remem be rs mechanics ) cons ults a sca le drawing of t he ta nk a nd determ ines t hat the curved line AB C has a lengt h of 34 ft a nd t hat its cent ro id is 8.2 ft from th e ce nter line of th e tank. How man y ga llons of paint will be used for t he ta nk includi ng t he vert ica l cylindrical column ?

I

:.",..,~~ ::?'C"'l

A

I I I I I

B

I I

J

I - - 12"."- 1

I S'

Problem 5/77

S'

5/78 Compute th e volume V a nd tot a l surfa ce a rea A of the complete circu lar ri ng whose cross sect ion is shown.

Problem 5 /80

z I

I

I--J

so '

mm

Problem 5 /78

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Articl e 5 /5 5/81 The t wo circu lar a rcs AB a nd Be a re revolved about the vertical ax is to obtain the surface of re volution shown. Compute th e area A of t he outs ide of this sur face. Ans. A = 15 7.9 in. 2

,

261

5/84 A steel die, shown in secti on, has th e form of a solid ge ne ra te d by revolving the sha ded a rea a rou nd t he zaxis. Calcu late t he mass m of t he die.

z

I

f-- 2"-1 ..-_ ;.,'. .. IA - -2,71

1

I

I I,

Problems

IB

200 mm

c Problem 5/81

5/82 Ca lculate t he weig ht \V of th e a lu minu m cast ing shown. Th e solid is generat ed by revo lving the trapezoida l area shown about t he z-axis through 180".

z, I ~--I---

I I

=t

1 "

I I I

'": I

----- r I

Problem 5/ 84

5/85 Determi ne t he volume V and total sur face a rea A of th e solid gene rated by revolving t he a rea shown thro ugh 1800 about th e a-axis. An s. V ~ 1.775(0 6 ) m m'' , A = 105 800 mm 2

2"

»>:":""''---'~-J

r--

I

2" ~

3"

~

J-1" J30 mm I

I

I

- ,,-

Problem 5/82

»

5 /S3 Determine t he sur face a rea of on e side of th e bellshaped shell of P rob. 5/40, shown again here, using t he theorem of Pappu s. A ns. A = 1TQ 2( 'IT - 2) x

:\ I

I I I

a '\

Prob lem 5/81

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>:

_-- I

--1

~ 75 m m

Problem 5/ 8 5

262

Chapte r 5

Distributed Forces

5 /86 A hand-operated control wheel mad e of a lu minu m has th e proportions shown in th e cross-sectional view. T he area of the total sect ion shown is 15 200 mm 2 , a nd th e wheel has a mass of 10.0 kg. Calcu la te th e dist an ce to t he centroid of th e half-section. T he a luminu m has a den sity of 2.69 Mg/m 3 .

r

5/89 A sur face is gene ra te d by revolving th e circular a rc of 0.8-m rad ius a nd subte nded a ng le of 120" com pletely about t he a-ax is. T he diameter of th e neck is 0.6 m. Dete rmine t he outside ar ea A ge ne ra ted. A ns. A = 4.62 m 2

z

I , O.8 m \

/\

120" ) ~I

O.6m

I Problem 5/86 Problem 5/89 5/87 Det ermine th e volume genera ted by rotating th e se micircu la r area t hro ug h 180°. AlIS. V = 361 000 m m3

z I

5 /90 The shaded a rea is bounded by one hal f-cycle of a sine wav e a nd th e axis of th e sine wave. Det erm ine th e volume gene rated by com plete ly revolvin g t he area about t he x-ax is. y

I

r:

b I

II

I ' r- 75 mm ~ I

+, a

---;:---'1 c

( I -

-'-- 1-- -

Problem 5/87 5 /88 Calcu lat e the volume V of the large neo prene washer in th e form of th e complete r ing of sect ion shown . Also compute th e overa ll surface a rea A.

Problem 5/90

-;"--

2" - . j

Problem 5/88

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---1- -

X

Arti cl e 5/ 5 5/91 Calcu late th e mass m of concrete requ ired to const ruct the arched dam shown. Concrete has a density

"

Ans. m = L 126(lOB) Mg

of 2.40 Mg/ rn .

- -I I-- 10

III

: 70 01 I

/

/ /

10 m

/

Sect ion A -A

/' /' 'i\ 60 0

" J -,

<,

<, <,

200m

-, Problem 5/92

Problem 5/91

Marwan and Waseem AI-Iraqi

263

5/92 In order to provide sufficient su pport for t he stone masonry ar ch designed as shown, it is necessary to know its tota l weight W. Use th e resu lts of P rob. 5/ 7 and determine W. The density of stone mason ry is 2.40 Mg/m'-

~ - --

-1 ' T ~~"-.l

Proble ms

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264

Chapt e r 5

Di st ribu t ed Forc e s

SECTION B . SPECIAL TOPICS

5 /6

BEAMS-ExTERNAL EFFECTS

Beams are structu ral members which afTer res istance to bendin g due to applied loads. Most bea ms are long prismatic bars, a nd t he loads a re usually ap plied nor mal to t he axes of the ba rs. Beams are undoubtedly th e most importan t of a ll structural member s, so it is importan t to un derstand the basic theory underlying their design . To a na lyze the load-carrying capacit ies of a beam we mu st first esta blish th e equilibrium requiremen ts of th e beam as a whole a nd a ny port ion of it consider ed separate ly. Second , we must esta blish t he rela tions between t he resulti ng forces and th e accompa nying internal resista nce of the beam to support these forces. The first part of this analysis requires the application of the principles of stat ics. The second part involves the st rengt h characte rist ics of the material and is usually treated in st udies of th e mechanics of solids or th e mechan ics of materials. This article is concerned with the external loading and reactions act ing on a beam . In Art. 5/ 7 we calculate the distributio n alon g th e beam of the internal force and moment. Types of Beams

Beams supported so that their external support reactions can be calculated by t he met hods of statics alone are called statically determinate beam s . A beam which has more supports than needed to provide equilibrium is statically indeterm inate. To determine the support reactions for such a beam we must consider its load-deformation properties in addit ion to the equa tions of static equilibriu m. Figu re 5/18 shows

Simpl e Continuous

Cantilever End-su pported cantileve r

Combination

Fixed

Statically determina te beams

Stati cally ind etermina te beams

Figure 5/ 18

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Article 5 /6

exa mples of both types of beam s. In thi s article we will a na lyze stat ically det erminat e beam s only. Beam s may also be iden tified by th e type of exte rnal loading they support. Th e beam s in Fig. 5/18 are support ing concent ra te d loads, whereas th e beam in Fig. 5/ 19 is support ing a distributed load . Th e inten sity w of a distributed load may be expresse d as force per uni t length of beam. The intensity may be constant or variable, continuous or discontinu ous. Th e int ensity of th e loading in Fig. 5/ 19 is constant from C to D a nd vari able from A to C a nd from D to B. Th e inten sity is discontinuous at D, where it cha nges magn itude abr uptly . Alth ough the intensity itse lf is not discontinuous at C, the rate of change of intens ity d w/dx is discon tinuous. Distributed Loads

Loading inte nsities which are constant or which vary linearly are easily handled. Figure 5/ 20 illustrates t he three most common cases and the resul tan ts of th e dist ributed loads in each case . In cases a an d b of Fig. 5/ 20, we see that the resultant load R is rep resented by t he area for med by the inten sity w (force per unit length of beam ) and t he len gt h L over which the force is distributed. The resu lta nt passes through the centroid of this area. In part c of Fig. 5/20, the t rapezoida l area is br oken in to a rectangular and a triangular area, and the corresponding resultant s R I and R 2 of th ese suba reas a re determined sepa rately. Not e that a single resu ltant could be det ermined by usin g the composite techniqu e for findi ng centroids, which was discussed in Art. 5/4 . Usually, however, the determination of a single resultant is unnecessary . For a mor e gene ral load distribution, Fig. 5/ 21, we must sta rt wit h a differ ential incremen t of force dR = w dx . The tot al load R is t hen th e sum of the differential forces, or

Beams -E xt e rnal Effects

C

B Figure 5/19

~ U2~-1

I I R = wL

~ l - - i.--1 (a>

R =! wL 2

1 w

f-- i. --I (nl

~~:/,~ ,:'i 4(W,-w,IL

t .l2

~-l

--

R = f wdx As before, the resul tant R is locat ed at t he cent roid of th e area under consideration. The x-coordinate of this cent roid is found by the principl e of moment s Ri =

I I

w2

L k)

Jxw dx, or

Figure 5/20

f xwdx R For the distribution of Fig. 5/21, t he vertical coordinate of t he centroid nee d not be foun d. On ce the distr ibuted loads have been reduced to the ir equivalen t concentrated loads, the exte rnal reactions acting on the beam may be fou nd by a straight forward static ana lysis as developed in Cha pte r 3. MarNan and Waseem AI-Iraqi

D

A

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f----I.------- X I d R = wdx I I I I I I

r----+-

x-l

R -I

~

'f

~ dx

Figure 5/21

265

266

Chapter 5

Distributed Forces

Sample Problem 5 /11

-

Det erm ine the equi valent concentrated load(s) and extern al reacti ons for t he simply su pported beam wh ich is subjected to the distributed load shown.

Solution. Th e area associated with th e load distribution is divided into th e rec-

CD

tangular and tri angular areas shown. Th e concent ra ted-load values are det ermined by compu ting the areas, and these loads are located at the cent roids of th e re spectiv e areas. Once th e concentrated loads are det enn ined, they are placed on the freebody diagram of the beam along with the exte r nal reactions at A an d B . Using principles of equili brium, we have

l:':MA

= OJ

1200 (5) + 480 (8) - R n (10 ) Rn

[ ~Mn

= 01

RA

= 984lb = 696

6' ~

t,

1201b'~0 1 b'ft A B Helpful Hint

CD Note th at it is usually unnecessary to reduce a given distribut ed load to a single concentrated load. 8'

4(160 1(61= 4~0 Ib

~=--=---5'-~ t~ 160 Ib/ft J "---j j 20 lb/ft A T B

120 lb'ft I

0

An s.

RA(lO ) - 1200 (5) - 480(2)

4'

(1201(101= 1200 Ib

0

1200 1b 480 lb

Ans.

lb

5'

- t=-

3'

~-~ B

hn Sample Problem 5 /12

w(:d ,

w=wo+k0

2024

Determine th e reaction at the su pport A of the loaded cantilever beam .

N/m

8m

CD

Solution. Th e constants in the load distribution are foun d to be we

1000

A

N/m and k = 2 N/m 4 • The load R is th en

R

=

J

w dx

=

f

(1000 + 2r' )dx

=

( 1000>: +

~) r

Helpful Hints

CD Use

10 048 N

caution with the units of the constants w a an d k.

o

@ The x-coordinat e of the cent roid of the area is found by 1 x = -J xw-R dx- = 10 048

= 10 1048 (500x 2

f"

x(1 000

@ The student should recognize that the calculati on of R and its location sim ply an applicat ion of cen troids as treated in Art. 5/3.

x is

+ 2x3 ) dx

0

+ ~x")

I" = 4.49 m 0

,

From th e free -body diagr am of th e beam, we have Ax

MA

[:':Fy

=

-

(10048)(4.49 ) = 0

MA

45100 N 'm

Ans.

10 048 N

Ans.

OJ

Note that Ax

o by inspection .

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10 048 N

4.49 m -

i

~::;A=========::JI B y , , Ay L_- x

Articl e 5/6

PROBLEMS

267

P r o b le m s

5/96 Calcu late the reactions at A and B for th e beam loaded as shown.

Introductory Problems 5/93 Det ermine th e reactions at A and B for the beam subjected to th e u nifor m load dist r ibut ion. Ans. R A ~ 1.35 kN, R B ~ 0.45 kN 6kN/m

L

.

A

'-'--'- 300 mm

B

400lbln

--1-

-->.J...!!

300 mm

IA

12 ' -

--+,- 6,J

Probl em 5/96

--1

5/91 Det ermine th e reaction s a t A for th e ca ntileve r beam subjected to th e distributed a nd conce ntrated load s. An s. Ax = 0, A.v 8 kN , M A = 21 kN · m

Probl em 5/93

5/94 Det ermine th e reactions a t th e support s A a nd B for th e beam loaded as shown.

.~
I

3m

1.5 m

y I I

L_- x

~- 1.5 m

Problem 5/97 5/98 Calcu late th e support react ions at A a nd B for th e load ed beam.

Problem 5/9 4 5/95 Determine th e reactions a t A and B for the loaded beam . An s. Ax = 0, A, = 603 Ib, By = 757 1b

y I I

'o ~ ~---~ "

y

I I I L __ -

:~ ~:

isonvn

x

I s. r3

•

• B

-3'~+---5'---i--4'--!

1_ ,

Prob lem 5/95

MarNan and W aseem AI-Iraqi

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I I

21 _ _--j 3 Problem 5/98

268

Chap te r 5

Distr ibu ted Forces

5/99 Determi ne the reactions at A and B for the beam subjected to a combination of distributed and point loads. Ans. Ax = 750 N, A" 3.07 kN , By = 1.224 kN

5/101 Calculate the support reactions at A and B for the beam subjected to the two linearly distributed loads .

~0 8 m

2kN/m

1.5 kN

y

L _ _%

.~ 1.2 m '::

1lllllllll13kN/m

6kN/m ~

B

1.2m

I

1.8 m

I,

1.2 m

J:J

•

I

I

6kN/m

!--4 m=-i 2 4 m

I

1.6m

Problem 5/ 102

Problem 5/99

5/1 00 Calculate the support reactions at A and B for the beam subjected to the two linearly varying load distributions.

5/103 Determine the force and moment reactions at the support A of the built-in beam which is subjected to the sine-wave load distribution.

An s. R A

~ 2wol, MA 7f

2

= wol 7f

6kN/m

Sine wave 4 kN/m

A •

B

~4m ---l-

6m -

Problem 5/100

Representative Problems

Problem 5/1 03

5/101 Th e beam is subjected to an ellipt ical load distribution and the point loads shown. For what value of the force P will the reaction at B go to zero? Ans. P = 6.96 kN

5/104 The cantilever beam is subjected to a parabolic distribution of load symmetrical about the middle of the beam. Determine the supporting force RA and moment MA acting on the beam at A.

4kN/m

w I

I I

2m

3m

W

o

2m

Problem 5/101 A

f--- -

//2 - -4 - - - //2 - J

Problem 5/1 04 Marwan and W aseem AI- Iraqi

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Art icle 5 /6

5/105 Th e load per foot of beam len gth varies as shown. For x = 10 ft , th e unit load is w = 300 Ib/ ft. At x = 0, th e load is incr easi ng at t he rat e of 50 lb/ft per foot. Calculate the su pport reactions at A and B. Ans. RA ~ 667 lb, R n = 1167 1b

Probl em s

5/108 Determine the force an d moment react ions at t he support A of t he cant ilever beam su bjected to the load distribution shown.

r: r-.

w

e-

I

r-

IhJ

It'

j

x- -

~ f.-- -- - I0 "-

-

At

.

I

---

Problem 5/108

Problem 5/105

5/106 A ca ntilever beam supports the variable load shown. Calculat e t he support ing force RA an d moment M A at A.

5/109 Determine the react ions at A and B for the beam subjected to the distributed and con cent rated loads. An s. A, ~ 5.56 kN, Bx ~ 4 kN , B, ~ 1.111 kN w I I

w

k

26 9

I

2

I

~

90 lbllt

50 lblft

x- - -

]A

1.4

J

20'

Problem 5/ 106

5/107 A beam is subje cted to the variable load ing shown. Calcu late the support reactions at A and B. Ans. RA ~ 1900 Ib, R n ~ 1600 lb

0.4

III

m

Prob lem 5/109

5/110 Determ ine th e react ions at the supports of th e beam which is acted on by th e combination of un iform an d pa rab olic loadin g dist ribu tions.

w

I I I

6 kN/m ~

IV

= IVo- h x

3

200lb/~t

B

100 lb/ft ---x

-

-

-

- 3 m- - --

B

A •

Proble m 5/110

I

20'

I

Problem 5/107

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270

Chapt er 5

Dis tr ibu t ed Force s

.... 5/111 The transition between the load s of 10 kN /m and 37 kN / m is accomplished by mean s of a cubic funct ion of for m w = k o + klI + k2£ 2 + k:)X3, theslope

I

. . 5 /111

of which is zero at its end points x = 1 m a nd x = 4 m. Determine the reactions at A a nd B. An s. R A ~ 43.1 kN, R n = 74.4 kN Cu bic fu nct ion

The quarter-circular cantilever beam is subjected to a uniform pressure on it s uppe r surface as shown. Th e pressure is expressed in term s of t he force p per un it length of circu mfe re nt ial arc. Deter mine t he rea ct ions on t he beam at its support A in terms of t he compress ion CA , shea r V A , an d be nding moment MAo

--i

// w I I I I

10 kNlm A

/

/

I I I

((/

3 7kN/m

r

•

,tJ,

.L\

Problem 5/ 112

Problem 5/111

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Articl e 5/ 7

5 /7

Bea ms-Interna l Eff ect s

v

BEAMS-INTERNAL EFFECTS

The previous article treated the reduction of a distributed force to one or more equivalent concentrated forces and the subsequent determination of the external reactions acting on the beam. In this article we introduce internal beam effects and apply principles of statics to calculate the internal shear force and bending moment as functions of loca tion along t he beam . Shear, Bending, and Torsion

~ V

et=:::::::::;:: Shear :::::::::::~ M Bending

In addition to supporting tension or compress ion, a beam can resist shear, bending, an d torsion. These three effects are illustrated in Fig. 5(22 . Th e force V is called t he shear force, th e couple M is called t he bend ing moment, and the couple T is called a torsional moment . These effects represent the vector components of the result ant of the forces acting on a transverse section of the beam as shown in the lower part of the figure. Consider t he shear force Vand bendin g moment M caused by for ces applied to the beam in a single plane. The conventions for positive values of shea r V and bending momen t M shown in Fig. 5( 23 are th e ones generally used. From the principle of action and reaction we can see that the directions of V and M are reversed on the two sections. It is frequently impossible to tell without calculat ion whet her the shear and moment at a particular section are positive or negative. For this reason it is advisable to represent V and M in their positive directions on the fre e-body diagrams an d let t he algebra ic signs of the calculate d values indicate the proper directions. fu; a n aid to the physical interpretation of the bendin g coup le M , consider the bea m shown in Fig. 5(24 ben t by the two equal and opposite positive moments applied at the ends. The cross section of the beam is treated as an H-section with a very narrow center web and heavy top an d bottom fla nges. For th is beam we may neglect the load carried by t he small web compared with that ca rr ied by th e two flanges. The upp er flange of the beam clearly is shorte ned and is under compress ion, whereas the lower flange is lengthened and is under tension. The resul tant of the two forces, one tensile and the other compressive, acting on a ny section is a couple and has the valu e of t he ben ding moment on th e section. If a beam having some other cross-sectional shape were loaded

+M[ -J

-:1

.~

=

[ -J

~

E )

I

Figure 5/ 24 Marwan and Waseem Al-lraqi

271

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AI

-~~cTorsion

M V

Combined loading

Figure 5/ 22

+V

Figure 5/ 23

T

272

Chapter 5

Distr ibut ed Forces

in the same way, the distribution of force over the cross section would be different, but the resultant would be the sa me couple.

Shear-Force and Bending-Moment Diagrams The variation of shear force V and bending moment M over the length of a beam provides information necessary for the design analys is of the beam . In pa rt icular, the max imum magn itude of the ben ding moment is usua lly the primary cons ideration in the design or se lection of a beam, and its value and position should be determi ned. The variations in shear and moment are best shown graphically, and the expressions for V a nd M when plotte d aga ins t distan ce a long the bea m give the shear-force and bending-moment diag rams for the beam. Th e first ste p in the det ermi nati on of the shear an d moment relations is to establish the values of all external reactions on the beam by applying t he equations of equ ilibr ium to a free-body diagram of the beam as a whole. Next , we isolat e a por tion of the beam , either to the right or to the left of an arbitrary transverse section, with a free-body diagram, and apply th e equa t ions of equilibrium to th is isolated port ion of the beam . Th ese equations will yield expressions for th e shea r force V a nd ben ding momen t M acting at the cut section on the part of the beam isolat ed. Th e part of t he bea m which involves the sma ller nu mber of forces , either to th e right or to t he left of the arbitrary section, usu ally yields th e simpler solut ion. We should avoid using a transverse section which coincides with the location of a concentrated load or couple, as such a posit ion represents a point of discontin uity in the variation of shear or bending moment. Fin ally, it is import an t to not e t hat t he calculations for V and M on eac h section chosen should be consistent with the positive convention illustrated in Fig. 5/23.

General Loading. Shear. and Moment Relationships For a ny beam with dist ribu ted loads we can establish cer ta in general relationships which will aid greatly in t he determ ination of t he shear a nd moment distribu tions along t he beam. Figure 5/25 rep resen ts a portion of a load ed bea m, where a n element dx of th e beam is isolated. Th e loading w represent s th e force per un it len gt h of bea m. At the location x the shear V and moment M acting on the element are drawn in their positive directions . On the opposite side of the element where the coordinate is x + d x, these quantities are also shown in their posit ive

dx

(t~;'t) "'M M

Figure 5/25 M arwan and W aseem AI-Iraqi

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v + C/V

Article 5/7

dir ections. They must , however , be labeled V + dV and M + dM , since V and M cha nge wit h x. Th e applied loading w may be considered constant over the lengt h of the element, since thi s length is a differential quantity a nd th e effect of any change in w disa ppears in t he limit compared wit h t he effect of w itself. Equ ilibri um of the eleme nt req uires that the sum of t he vertical forces be zero. Thus , we have V - w dx - (V + d Vj

0

or

(5 /10)

We see from Eq. 5/ 10 that the slope of t he shear diagram mu st every where be equal to t he negative of th e value of t he app lied loading. Equat ion 5/ 10 holds on either side of a concentrated load but not at the concentrate d load because of the discont inuity produced by the abru pt change in shear. We may now express the shear force V in term s of the loading w by integrati ng Eq. 5/ 10. Th u s,

v

J dV Vo

=

-I

w dx

' 0

or V = Va

+ (t he negat ive of the area under the loading curve from Xa to .r)

In this expression Va is the shear force at Xo and V is the shear force at x. Sum ming the area under the loading curve is usually a simple way to construct the shear-force diagram . Equilibrium of t he eleme nt in Fig. 5/25 also req uire s that the moment sum be zero. Summing moments about the left side of the element gives M + w dxdx + (V + dV) dx - (M +dM) 2

0

The two M 's can cel, and t he terms W(dX)2/ 2 and d V dx may be dro pped , since t hey are differ entials of higher order t han those which remain. This leaves

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Beams-Internal Effects

273

27 4

Chapt e r 5

Di st ribu t ed Forc e s

which expresses the fact that t he shea r everywhere is equa l to t he slope of the momen t curve. Equation 5/ 11 holds on eithe r side of a concentrated coup le but not at th e concen trat ed coup le becau se of th e discontinuity cause d by th e abru pt cha nge in moment. We may now express t he momen t M in te rm s of the shea r V by integrati ng Eq . 5/11. Thus, M

f

dM

Mo

r Xo

Vdx

or M

=

M o + (a rea under the shear diagram from Xo to x l

In this expression M 0 is the bending moment at Xo and M is the bending moment at x. For beams where there is no exte rnally applied moment Mo at Xo = 0, the total momen t at any sect ion equals the area under t he shear diagram up to tha t section. Summi ng the area under the shea r diagram is usu ally th e simplest way to const ruct th e momen t diagram . \Vhen V passes throug h zero and is a continuous function of x with dV/dx 0, the bending moment M will be a maximum or a minimum, since dM /dx = 0 at such a point. Critical va lues of M also occur when V crosses the zero axis discontinuously, which occurs for beams under concentrated loads. We observe from Eqs. 5/10 a nd 5/ 11 th at the degr ee of V in x is one higher th a n t hat of w. Also M is of one higher degree in x th an is V. Consequen tly, M is two degrees hig her in x than w. Thu s for a beam loaded by w = kx , which is of th e first degree in x, the shea r V is of the second degr ee in x a nd the bending mom ent M is of the third degree in x. Equations 5/10 a nd 5/11 may be combined to yield

*

2M

( d 2 dx

= -wJ

(5 / 12)

Thus, if w is a known function of x, t he moment M ca n be obtained by two integr ations, provided that the limits of integra tion are properly evaluated each time. Th is method is usab le on ly if w is a continuous functio n of When bendin g in a beam occurs in more than a single plane, we may perform a se parate analysis in each plane and combine the result s vectorially.

x:

"wh en w is a discontinu ous funct ion of'x. it is poss ible to introduce a special se t of expr essio ns called singularity {unctions which permit writing- analytical expressions for shear V and moment M over an interval which includes discontinuitie s. The se functions are not discussed in this book.

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Ar ticle 5/7

Beams -In t e rnal Effects

Sample Problem 5 /1:S

4 kN

Determine th e shea r and moment distributions produced in th e simple beam by the 4-kN concentra ted load.

Jt

6m

Solution. From th e free-body diagram of t he entire beam we find the su pport reaction s, wh ich are

R , = 2.4 kN

!

4m

;,

4 kN

y I

R j = 1.6 kN

275

I - x

A section of th e beam of length x is next isolated with its free-body diagram on which we show the shear V and the bendi ng moment M in their positive directions. Equ ilibrium gives

l:£Fy

= 0]

l:£M R , = OJ

1.6 - V = 0

v

= 1.6 kN

M - 1.6x = 0

M

= 1.6x

R j = 1.6 kN

V + 2.4

o

V

= - 2.4 kN

- (2.4)(10 - x ) + M

o

M

~

[:£Fy = OJ l:£M", = 0]

V

y

These values of V and M apply to all sections of the beam to the left of the 4-kN load, A sect ion of the beam to the right of t he 4-kN load is next isolat ed with its free -body diagram on which V an d M are shown in t heir positi ve directions. Equi libr ium requires

':

~i

M

Ff 9

1.6kN

V

2.4 kN

2.4(10 - x)

Th ese results apply only to sections of th e bea m to th e right of t he 4-kN load. Th e values of V and M are plott ed as shown. Th e maximum bend ing mome nt occurs where th e shear cha nges dire ction . As we move in the posit ive x-direction starti ng wit h x = 0, we see that th e momen t M is merely th e accumulated area under t he shear diagram . M ,kN 'm I

I

96 ~ 0 1

o

-

6

Helpful Hint

ill We

must be careful not to take our section at a concent ra ted load (such as x = 6 m) since the shear and momen t relations involve discontinu ities at such positions.

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:I: III

10 "

276

Cha p t e r 5

Dis tr ib ut ed Forc e s

Sam ple Problem S/14 Th e cantilever beam is subjec ted t o the load inten sity (force per u nit length) which varies as w = Wo sin (rrx/ l ). Determ ine th e shear force Vand ben din g moment M as funct ions of the ra tio xl t.

Solution. Th e free -body diagram of the entire beam is drawn first so that th e shear force Vo and bending moment Mo which act at the support ed end at .r = o can be computed. By conven tion Vo an d Mo are shown in their posit ive mathema tical sense s. A summat ion of vertical forces for equilibrium gives

(~Fy


= 01

I'

Vo -

0 w

=0

dx

Vo

=

Ii

.

wo sm

0

2wl T1TX dx = ~

A summa tion of moments about the left end at x = 0 for equilibrium gives

f

I

[:;:M = 0]

- Mo -

x (w

Mo

dxl = 0

o

2

- w ol [ . 1TX ~ cos smT -

Mo = ~

=

_IIo WoX sin ~I dx

I

= _ woi 2

1TX

I

i

w

0

From a free-body diagram of an arbitra ry sect ion of length x, integration of Eq . 5/10 permits us to find th e shear force internal to th e beam. Thus,

@

[dV = -w dxl

f.

v

dV = -

IXwosin

Vo

0

7dx

0.637

1TX ]X wol [ - ". cos - I 0

wol ( cos T 1TX --;;

- 1)

or in dim ensionless form

~

woi

=

.!tr

(1+ cos :::=)

Ms .

l

The bending moment is obtained by integration of Eq. 5/ 11, which gives [dM = V dxl

(

dM =

J:

W;l ( 1 + cos ";) dx

2

. -m: - -w Oi- + -wol [ x + -I sm 11'"

11'"

Helpful Hints

11'"1

0]

or in dimensionle ss form 1 . 1TX ) 1 + -sm". I

Ans.

Th e variations of V/ wol an d M/ wol2 with x ll are shown in th e bottom figures. Th e negative values of M/ wol 2 indicate that physically the bending moment is in t he direction opposite to th at shown.

Marwan and W aseem AI-Iraqi

-:~:: ~


. -1TX]X M - M o = -wol [ x + -I sm 11'" 11'" 1 0 M

o l ----l._--'--_-'----="'-_ o 0.2 xII 0.6 0 .8 1.0

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this case of symmet ry it is clear th at the resultant R = V o = 2w ol/ rr of the load distribution act s at midspan, so th at t he moment requi rement is sim ply M o = - Rl/ 2 = - w ol'J. j 1r. Th e minus sign tells us tha t physically the ben din g mome nt at x = 0 is opposite to that represented on th e free-body diagram.

@ The free -body diagram serves to re mind us that the integration limits for V as well as for x must be accou nted for . We see t hat th e expression for V is positive, so th at t he shear for ce is as represe nted on th e free -body diagram .

Article 5/7

277

Beams-Interna l Effects

Sample Problem 5 /15 Draw the shear-force and bend ing-momen t diagram s for the loaded beam and determi ne the maximum mom en t M and its location x from the left end .

Solution. Th e su pport reactions are most eas ily obtain ed by considering th e resultan ts of the dist ributed load s as shown on the free- body diagr am of the beam as a whole. Th e first interval of th e beam is ana lyzed from t he free -body diagram of the section for 0 < x < 4 ft . A vertical summation of forces and a moment su mmation about the cut sect ion yield

v=

[}:F y = 0) [ ~M

= 0]

M

x

+ (I2.5x 2 ):3

-

I~M ~

0)

M

'-r--

M = 24 7x - 4.17x 3

247x = 0

4) + 200 - 247 = 0

x - 4

V = 447 - 100x 2

+ 100(x - 4) - 2- + 200[x - 3(4») - 247x

M ~ - 267

~

0

+ 447x - 50x 2

~

- 353 1b

an d

~

-

-

- - . ----.J

M

~

~p: ~~'" 653 1b 100(x I

4)

/C- ,r -L- x -

4

2

2930 - 353x

v

2471b

10

Ans.

732 Ib-ft

As before, note tha t t he moment M at any sectio n equals th e area u nder the she ar diagram up to t hat section. For inst ance, for x < 4 It, [aM

~

J

V dx)

and, as above,

M - 0

=

r o

(247 - 12.5x 2 ) dx

M ~ 247x -

4

4.17x3 -600

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- x

12.5x 2

2471b

Th ese values of Vand M ar e plotted on the shear and moment diagr ams for t he interval 8 < x < 10 ft. The last interval may be analyzed by in spect ion. Th e shear is consta nt at + 300 Ib, and th e momen t follows a straight -line relation begin ning wit h zero at the right end of the beam. The maximum moment occurs at x = 4.47 ft , where th e shear curve crosses the zero axis , and the magnitude of M is obtained for this valu e of x by substitution into the expression for M for the second interval. Th e maximum moment is M

-

R , = 653 Ib

Th ese values of V an d M are plotted on th e shear and moment diagrams for th e interval 4 < x < 8 ft . Th e ana lysis of the remainder of th e beam is continue d from the free-bo dy diagram of t he portion of th e bea m to the right of a section in t he next interval. It shou ld be noted that V and M are represented in the ir positive directions. A vertical force summation an d a momen t su mmat ion about the section yield

v

300 1b

12.5x 2

247

These values of V and M hold for 0 < x < 4 ft and are plotted for th at interv al in the shear and momen t diagr am s shown. From the free-body diagram of the section for which 4 < x < 8 ft, equi libr iu m in t he vertical direct ion an d a moment su m about th e cut section give

v + 100(x -

4001b I

278

Chapter 5

Distr ibuted Forces

PROBLEMS Introductory Problems

5/113 Determine th e shear-force and bending-moment dist ributions produced in th e bea m by the concentrated load . What are t he values of th e shear and momen t when .r = 1/2?

Ails. V

~

f------. x

I I I

P/ 3, M = PI/ 6

2P

I

I

I

4

2

4

A •

P

~ - -x

5/117 Dra w the shear an d momen t diagr ams for th e beam subjected to t he two point loads. Determine the maximum bending mom ent M max and its location. 5PI 31 Ails. M max = - a t x = 16 4

2/

I

• B

I

3- - -3 - 1 P B

A

Problem 5/ 117 5/118 Const ruct th e shea r an d mome nt diagr am s for the

Problem 5/ 113 5 /114 Dra w th e shear and moment diagrams for th e

bea m loaded by t he 2-kN force an d the 1.6-kN · m coup le.

loaded cantilever beam .

2kN

t

I

2

~§i==:::===t

p

P I

t

2

B

0.5 m

I

~

0.5 m

I

I

0.5m - ' 1

Problem 5/ 114 Problem 5/118 5/1 15 Draw th e shea r and moment diagr am s for the

loaded beam an d determi ne the distanc e d to the right of A wher e th e moment is zero. Ails. d = 2.67 m

5/119 Draw th e shear and moment diagr ams for th e uni-

4 kN

1

form ly loaded beam and find th e maxim um ben ding wp. moment M maxAilS. .M m a x = """"8"

I

2m

I

;:

2m

B

Problem 5/ 115 5 /116 Draw th e shea r and moment diagrams for th e

loaded beam . What are t he values of t he shea r and moment at midb eam ?

•

•

B

Problem 5/ 116

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Problem 5/119

Article 5 /7

Problems

279

5 /120 Dra w th e shea r and moment diagram s for the

5 /123 Draw the she ar and moment diagram s for the beam

loaded beam and determine the maximum valu e M max of the moment.

loaded as shown . Specify th e max imum moment All s. AImux

300lblll

•1•

A

1-"--'-. ~-- 4'- ---:-- - 4' - --;-- - 4'

~

Wo A

woL 2

= 24

Wo •

B

Problem 5/120

~-- ~2

5 /121 Deter mine th e shea r and moment diagrams for th e

loaded cantilever beam. Specify th e shear Va nd mement AI at the middle section of the beam . AilS, V ~ 4.8 - 0 .6x kN M = - 0.318 - x )2 kN 'm V ~ 2.4 kN, M ~ - 4.8 ktc -m

L _ _~ 2

Problem 5/1 n 5 /124 Draw the shear and momen t dia gra ms for the beam loaded by the force F applied to t he strut welded to

th e beam as shown. 600N/m

~

5,

F

1 "l

B

A •

~'---- 8 m --~

--'-

Problem 5/121

Problem 5/124

Representative Problems 5 /122 Draw th e shea r a nd moment diagrams for th e lin-

ea r ly loaded ca ntilever beam and specify th e bending moment M A at th e support A.

5/125 Th e l-beam su pports the 1000-lb for ce and th e 2000Ib-n couple, applied to th e 2-ft st ru t welded to th e end of th e beam . Calcu late the shear Vand moment M at th e sect ion midway bet ween A a nd B . Ans. V ~ 1467 lb , M ~ - 200 lb-ft

zooo n-n

1000lb

@ T2'

U'o

B

A

• I - - - - - ------,I ,Problem 5 /122

Marwa n and W aseem Al-lraqi

3, _

1 _

,

3, _

1 _

Problem 5/125

'NWW.gigapedia.ccm

.

3, -

280

Chapter 5

Dist ributed Forces

5/116 T he angle stru t is welded to th e end C of the l-beam a nd supports the 1.6-kN vertical force. Deter mine the bend ing moment at B and the dist ance x to the left of C at whic h the bending moment is zero. Also construct th e moment diagram for th e beam .

1.6 kN

5/119 Plot th e shear and moment diagrams for the beam loaded with both th e dist ributed and point load s. Wh at are t he values of the shea r and moment at x = 6 m? Determine the max imum bending moment M m ax • A ns. V = - 600 N, M = 4800 N· m at x = 6 m M max 5620 N - m at x = 4.25 m

1500 N

f--

IIIID--lt--, ;(B AS ~ 2 mL3 m 1 _I I

B A

- x

2m

•

•

5/nO Repeat Prob. 5/ 129, where th e 1500-N load has

400 m m - - 1--- 450 mm

been rep laced by th e 4.2-kN · m cou ple.

Problem 5/126

~

5/117 Deri ve expressions for th e shear for ce V and bending moment M as fu nct ions of x in th e can tilever beam load ed as shown. A ns. V = 2000 - 1Olb: - x 3 Ib .II = - 12,500 + 2000x - 5lb:' - ±x' lb-Ft IV

2m -

Problem 5/129

c

w

I

800 N/m

I

A~::;

l- 2 m -

= U'o + kx'l.

IIIID 800 N/m

- x

4.2 kN·m

-(0-,-· ;(B

I- 3 m

2m

I 2m ~

Problem 5/130 400 lb/ft

5/131 Draw the shear a nd moment diagrams for th e beam of Pr ob. 5/96 repeated her e and specify the shea r V and moment AI at a sect ion 6 ft to the left of the support at A. A ns. V = - 1800 Ib, M = - 6000 Ib-ft

Problem 5/ 12 7 B

400 Ib/ft

5/118 Th e adj usti ng scre w of th e special-pur pose offset clam p supports a comp ression of 500 N. Calculat e the shea r force V, t he te nsion T , and the bending mome nt M at sect ion A of the clamp bar for x = 250 mm . Wh ich of t hese three qu anti ties changes with x?

IA Problem 5/131

160 m m

L

50mm

I ;-- - -

400 mm - - --I Problem 5/128

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Article 5/ 7 5 /132 Draw the shea r a nd moment diagrams for the beam of P rob. 5/ 106 repea ted here a nd specify the shea r Vand mome nt M at t he mid lengt h of the beam .

k

IV

I

oo~

28 1

5 /134 Der ive expres sions for the shea r Vand momen t M in te rms of x for the ca ntilever beam of Prob. 5/ 104 shown agai n here.

IV 0

Problems

I I I

wa

50 lblll

x- - -

~A

I

20' Problem 5/ 132

A

5 /133 T he hea vy-duty paper clip is clamped to a 10-mm stack of pap ers a nd exer ts a clamp ing force of 40 N on each side of the stack at A with P = O. It requires forces of P ::::; 40 N to relieve t he compre ssion at A, an d th e l fl-rnm gap incre ases as P exceeds 40 N. Dete rmi ne a nd plot the force F and bend ing moment M su pport ed by the middle of th e spring at C as fu nctions of P from P = 0 to P = 80 N. Bear in mind that t he bending moment does not cha nge as long as the geometric con figu ration of t he clip does not chan ge. Ans. For P ~ 0: F ~ 40 N, M ~ 1.2 N ' m CCW For 0 < P < 40 N: F ~ 40 - 2P , M ~ 1.2 N · m For P > 40 N: F ~ - P, M ~ 0.030P

f..--

- 112 - ---...,- - - l/ 2 Problem 5/134

.... 5 /135 T he beam supports a u nifor m u nit load w. Deter mine t he loca tion x of the two suppor ts so as to minimize the maximu m ben ding mom ent M mux in the bea m. Specify M mux-

A ns. X = 0.207L, M ma x = 0.0214wL 2

IV

p

P

c

L-

-

Problem SI B S

fw1 mm Prob lem S /B3

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-

-

-

282

Cha pt er 5

Di stributed Forc e s

.. 5 /1l6 The u nifo rm Quarter-circu lar member of mass m lies in th e vertical plan e and is hin ged at A and su p-

ported again st the vertical wall by its small roller at B. For any section S , writ e express ions for the shear force V, comp ression C, and bending moment AI du e to the weight of th e member. 2mg ( OsmO . -cos ll) A ns. V = - orr

2mg

C = -orr

M

2mgr

- -

Marwan and W aseem AI-Iraqi

orr

(0

cos H + sin

fJ)

:\---I I I I

r

A •

0 cos fJ

Problem 5/136

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B

Articl e 5 /8

5 /8

FLEXIBLE C ABLES

On e important type of st ruc t ural memb er is tb e flexible cable which is used in sus pension bridges, transmission lines, messenger cables for suppor ting hea vy trolley or te lephone lines, and man y other applications . To design these structures we must know the relations involving the te nsion, spa n, sag, and length of th e cables. We determ ine these quantities by examining the cable as a body in equilibrium. In the analys is of flexible cables we ass ume that any resistanc e offered to bending is negligible . Thi s assumption means that th e force in the cable is always in the direction of t he cable. Flexible cable s may support a ser ies of distinct concent rated loads, as shown in Fig. 5/260, or t hey may support loads continuously distributed over the lengt h of the cable, as indicated by the variable-inte nsity loading w in 5/26b . In some ins ta nces the weight of the cable is negligible compared with th e loads it supports. In oth er cases t he weight of the cab le may be an appreciabl e load or th e sole load and cannot be neglected . Regardless of which of these conditions is pre sent, th e equ ilibrium requ irements of the cable may be form ulat ed in t he sa me manner .

General Relationships

If th e intensity of the var iable and continuous load applied to the cabl e of Fig. 5/ 26b is expressed as w units of force per unit of horizonta l lengt h x, th en the resultant R of the ver tical loading is R =

f

dR

=

f

w dx

y I I

I

- x~.:t -1

.T+dT

- ~;;..;.-~?r- - ) - .' T \e

e+de

wdx

w

Ie)

rR Ib) Figure 5/26

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Fle xibl e Cabl es

283

284

Chap t e r 5

Dis tr ibu t ed For ce s

where the int egration is taken over t he desired interval. We find the position of R from the moment principle, so that

Ri =

Jx d R

Jx d R

R

Th e eleme ntal load d R = w dx is re presented by a n eleme ntal st ri p of vert ica l length w and width dx of the sha ded area of th e loading diagram , a nd R is represented by t he total area. It follows from the foregoing express ions tha t R passes through th e centroid of the s haded area. The equilibr iu m condit ion of the cable is satisfied if each infinitesimal element of th e cable is in equilibrium. Th e free-body diagram of a differen tial eleme nt is shown in Fig. 5/ 26e. At the genera l position x the tension in th e cable is T , an d th e cable ma kes an angle 0 with the horizontal x-direction. At the sectio n x + dx t he te nsio n is T + dT, and th e angle is 0 + de. Note th at th e cha nges in both T a nd 0 are tak en to be positive with a positive change in x. The vertical load w dx completes th e free-body diagr am . Th e equilibr iu m of vertical an d horizon ta l forces requires , respectively, that (T + d'T) sin

(0

+ dO)

T sin 0 + w dx

(T + d'T ) cos (0 + dO) = T eas 0 The trigonometric expansion for the sine and cosine of the sum of two angles an d the substi tut ions sin d H = dH and cos dH = 1, which hold in the limit as d e approaches zero, yield

(T + d T )(sin 0 + cos Hd O) (T

+ dT)(cos

T sin 0

0 - sin 0 d O)

+ w dx

T eas 0

Dropping the second-order term s and simplifying give us T eas 0 d H + d T sin ()

= w dx - T sin HdO + d T cos H = 0 which we write as d t'T' sin 0)

=

w dx

and

d (T cos

0) =

0

Th e secon d relation expresses the fact th at t he hori zon tal compo nent of T remains unchan ged, which is clear from th e free-body diagra m. If we in troduce th e sym bol To ~ T eas 0 for this cons ta nt horizontal force, we may then substitute T = T o/ cos 0 into th e first of the two equa tio ns just derived an d obtain d (T o tan 0) = w dx . Because ta n 0 = dyf dx, the equilibrium equation may be written in the form

(5/13) Marwan and W aseem AI-Iraqi

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Article 5 /8

Equation 5/13 is th e differential equ ation for the flexible cable. The solution to t he equation is that functional relation y = {(x) which satisfies th e equation and a lso sat isfies the conditions at the fixed ends of the cabl e, called boun da ry condi tions. This relationship defines th e shape of t he cable, and we will use it to solve two important an d limit ing cases of cable loading.

Parabolic Cable When the intensity of vertical loading w is consta nt, the conditi on closely approximates that of a sus pension brid ge where the uni form weight of the roadway may be expressed by the constant w. Th e mass of t he cable itself is not dist ributed uniformly wit h the hor izontal but is relat ively small, and thu s we neglect its weight. For this limit ing case we will prove that the cable hangs in a pa rabolic arc. We start with a cable su spended from two points A an d B which are not on th e same hor izon tal line, Fig. 5/270 . We place the coord inate origin at the lowest point of the cable, where the tens ion is horizontal an d is T o. Integr at ion of Eq. 5/13 once wit h respect to x gives

wx + C To

dy dx

where C is a constant of int egration. For the coordinate axes chosen , dy f dx = 0 whe n x = 0, so t hat C = O. Thus,

dy ~ wx dx To which defines th e slope of th e curve as a function of x. One fur ther int egration yields

{dY

or

~

(5 /14)

~

Alternati vely, you should be able to obtain the iden tical results with the indefin ite integral toget her with t he evaluation of th e constant of

r tll--Ij-

---I

18

.~

-

-

- ,

~

1

y

'1

,

1'-"""'::"--1---=·1 IV =

A

il A - .. - - _ ..

Load per unit of horizont a l length

-xJ

T

i

X

)"

l

s

1

To

e

1

1 1

I- x/2 ~

R = wx Ib )

la) Figure 5/27 Marwan and W a seem AI-Iraqi

- -l

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x

Fle xibl e Cable s

285

286

Chapter 5

D istr ibuted Forces

integrat ion. Equation 5/ 14 gives the sh ape of the cable, which we see is a vertical parabola. The constant horizontal component of cable te nsion becomes t he cable tension at th e origi n. a nd y = h A in Eq. 5/ 14 Inserti ng th e corresponding values .r = gives

'A

so that Th e tension T is found from a free-body diagram of a finite portion of t he cable, shown in Fig. 5/ 27b. Fro m the Pyt hagorean theorem

Elimina tio n of To gives (5/ 15) The maximum tension occurs where x = fA and is (5 / 15a)

We obtain the length SA of the cable from t he origin to point A by integrating the expression for a di fferential len gth ds = J (dx )2 + (dy )2. Thus,

fo"" = fl"h ds

0

l"

+

(dy / dx )2 dx

=

f

0

h +

(wx / T o)2 dx

Altho ugh we can integrate this expression in closed form, for computational purposes it is more convenient to express the radical as a conver gent se ries and then integrate it term by term. For this purpose we use the binomial expansion (l

+ x )" = 1 +

IIX

+

n tn -

1)

2!

.,

x- +

nin -

l )(n - 2 )

3!

'j

.r '

+ ...

which converges for x 2 < 1. Replacing x in the series by (wx / T o)2 and sett ing 11 = give the express ion

4

fl"(1+ x: =, [1+ 3.3(h'A)2_ 3.5'(h A)<+ .. .J 2

SA

=

o

A

w

2T o

A

(5/ 16)

A

!,

This se ries is convergent for value s of h ~\/IA < which holds for most practical cases . The relationships which apply to t he cable sectio n from the origi n to poin t B can be eas ily obtain ed by replacin g hA , ' A ' a nd SA by h/J' '/J, and Sn. respectively. Marwan and W aseem AI-Iraqi

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Articl e 5 {8

, IA

L

B )'

I I I

Figure 5/28

For a suspension bridge where the supporting towers are on the sa me hori zontal line, Fig. 5/28, the total spa n is L = 21A , the sag is h = h A , and th e to tal length of the cable is S = 2sA- With these substitutions , the maximum ten sion and the total length become

~L h

T rna"

S = L [1 +

+ (L / 4h) 2

~

m 2 -

(5 / 15b)

3 52

(If

J

+ ...

(5 / 16a)

!.

This se ries converges for all values of h/ L < In most cases h is much sma ller than L/ 4, so th at th e three terms of Eq. 5/ 100 give a sufficiently accurate approximation .

Catenary Cable Consider now a uniform cable, Fig. 5/29a , sus pended from two points A and B an d han ging under t he action of its own weight on ly. We will show in thi s limiting case that the cable assumes a curved shape known as a catenary. The free-bo dy diagram of a finite por tion of the cable of length s measured from th e origin is shown in part b of the figu re. Thi s freebody diagram differs from the one in Fig. 5/2 7b in that the total vertical force su pported is equal to th e weight of th e cable sect ion of length s )'

'--IIJ--i----IA [ I

I

B

I

I I I

{ f---- x--j I I I I

R = ps (bl

Ic J Figure 5/29 Marwan and Wa see m A l-lraqi

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T

Fle xible Cable.

287

288

Chap ter 5

Distribut ed Forc es

rather than the load dist rihuted uniformly with respect to the hori zonta l. If the cable has a weight p: per unit of its lengt h. th e resultant R of the load is R = J1S . and the increm ental ver t ical load w dx of Fig. 5/2 6c is re placed by p. ds . Wit h this replacem ent the differen tial relati on. Eq. 5/ 13. for th e cable becomes d 2y dx 2 =

P.

ds

(5 / 17)

To d;

Because s = (C. x,y ), we must change this equation to one containing only the two variables. We may substitute the identity (ds )2 = (dx)2 + (dy)2 to obtai n

I

d2

_ u: - y2 1 + dx To \

(d- y )2

(5/18)

dx

Eq uat ion 5/ 18 is t he differ ential equation of th e cu rve (cate nary ) formed by th e cable. Thi s equa tion is easier to solve if we subst it ute p = dy Idx to obtain dp

J!:... dx To

In tegr ating this equation gives us In

+

(p

h

+

p 2)

Th e cons ta nt C is zero because dy / dx = p = 0 when x = O. Substituting p = dy ldx, cha nging to exponential form , an d clearing th e equat ion of th e radi cal give dy dx

= sinh

!J-X T

o

where th e hyperbolic function' is introd uced for convenience. The slope may be integrated to obtai n To !J-X y = - cosh - + K p. To The integration constant K is evaluated from the boun dary condition x = 0 when y = O. Thi s substi tut ion requ ires tha t K = - T o/ p., a nd hence, y

-T O ( cosh -!J-X - l ) p. To

(5/19 )

"See Arts. CIS and Cl IO, Appendix C, for t he definit ion and integral of hyp erbolic functio ns. Ma lWan and W aseem AI- Iraqi

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Art icle 5 /8

Equation 5/ 19 is the equa tio n of t he curve (catenary) formed by the cable han gin g under the act ion of its weight only. From the free-body dia gram in Fig. 5/29b we see tha t dy'[dx ta n e = JLS / T o. Thus, from the pr evious expression for the slope, s

T osm . h -ux =-

(5/ 20)

To

I'-

We obtain the ten sion T in the cable from the equilibr ium t riangle of the forces in Fig. 5/29b. Thus,

which, when combined with Eq . 5/ 20, becomes 2

To cosh

2 J- :

l'

" or T

=

To cosh

~

(5/21)

We may also express t he ten sion in terms of y with th e aid of Eq. 5/ 19, which, when substituted into Eq . 5/2 1, gives T

=

To

+ I'-Y

(5/22)

Equation 5/ 22 shows that the change in cable tension from th at at th e lowest position depends only on I'-Y. Most pr oblems dealin g with the catenary involve solutions of Eqs. 5/ 19 through 5/22, which can be handl ed by a graphical approximation or solved by computer. The procedu re for a graph ical or comp uter solution is illustrated in Sample P roblem 5/ 17 following this article. The solution of catenary problems where the sag- to-spa n ratio is small may be ap proximated by the relations developed for th e parabolic cable. A small sag-to-span rat io mean s a tig ht cable, an d the unifo rm distribu tion of weigh t along the cable is not very differ ent from th e sa me load inten sity dist rib uted uniformly along the horizontal. Man y problems dealing with both th e catenary and parabolic cabl es involve suspension point s that are not on the same level. In such cases we may apply th e relations just developed to t he pa rt of the cable on each side of th e lowest point.

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Fle xible Cables

289

290

Chapter 5

Distributed Forces

Sample Problem 5/16 The light cable supports a mass of 12 kg per meter of hori zontal length and is suspended between the two points on the same level 300 m apart. If the sag is 60 m, find the tension at midlength , the maximum tension, and the totaJ length of th e cable.

Solution. With a uniform horizontal distribution of load, the solution of part (b) of Art . 5/ 8 applies, and we have a parabolic shape for the cable. For h = 60 m, L ~ 300 m, and w ~ 12(9.81) (10- 3) kN/m th e relation following Eq . 5/14 with IA = L / 2 gives for the midlength tension

L2] Bit [To = W

T ~ 0.1177(300)2 = 22.1 kN o 8(60)

T max

~

1 + (4~.r] Ans .

Th e sag-to-span ratio is 60/ 300 = 1/ 5 < 1/4 . Therefo re , the series expression developed in Eq. 5/ 1& is convergent , and we may write for the total length

_3:W·

+

.. ]

300[1 + 0.1067 - 0.01024 + ...J

= 329 m

MalWan and W aseem AI- Iraqi

I 60 m

75 ;;;

/

l'7ifm

I I

.L - - x

R = 121150X9.81XIO-3) = 17.66 kN

CD Suggestion: Check the value of T max

3

~

_

Helpful Hint

12(9.81)(10 - )(300 ) / 1 + ( 300 )2 = 28.3 kN 2 \ 4(60 )

S =300[1 +~Gr

I

1-

Ans.

The maximum tension occurs at the supports and is given by Eq. 5/1 5b. Thus,

CD [Tm~ ~ W2L

)'

A ns .

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directly from the free-body diagram of the right-hand half of the cab le, from which a force polygon may be drawn.

Article 5/8

Sample Problem 5/17

1---

-

Fle xibl e Cabl es

-

-

300 m- - - - - .,

Replace the cable of Samp le Problem 5/ 16, which is loaded uniformly along t he horizo ntal, by a cable which has a mas s of 12 kg per meter of it s own lengt h and su pports its own weight only. The cable is suspended between two points on the same level 300 In apar t and has a sag of 60 m. Find the ten sion at midlength, the maxim um tension, and the total length of the cable.

Y I I I

Solution. With a load distributed unifor mly along the length of the cable, the solution of part (c) of Art. 5/8 app lies, and we have a catena ry shape of the cable. Eq uations 5/ 20 and 5/21 for t he cable length and tension both involve the minimu m ten sion T o at midlength, which must be found from Eq. 5/19. Thus, for x = 150 m, y = 60 m, and" = 12(9. 8 1 )(10 ~ 3) = 0.1177 kN / m, we have 6

or

o

=

---.IsL [ 0.1177

7.06 = cosh 17.66 _ 1 To To

~

0.32 P -..:---

-j--j--

-j--

-

-j

-j- 7.06 To

0.29

Thi s equ ation can be solved graphically. We compute the expression on each side of the equals sign and plot it as a function of To. The int ersection of the two curves est ablishes the equality an d determines th e correct value of To. Th is plot is shown in the figure accompanying this problem and yields the solut ion To

Solution To = 23.2 kN

0.31f--=---:k -+-

sh (0.1177)(150) - 1] co To

29 1

0.28,'-=------;2-;;,-----~-o-----,;-!

22.5

23.0

23.5

24.0

To, kN

23.2 kN

Alternatively, we may write t he equation as f (T ) = cosh 17.66 _ 7.06 - 1 = 0 o To To

and set up a computer program to calculate the value rs) of To which re nders f (T0) = O. See Art. C/ ll of Appendix C for an explanation of one app licable numerical met hod. The maximum tension occur s for maxim um y and from Eq. 5/22 is T rn ux = 23.2 + (0.1177 )(60)

= 30 .2 kN

Ans.

G) From Eq . 5/20 the total len gth of th e cable becomes 2s

=

23.2 . h (0.1177)(150) - 330 23.2 m 2 0.1177 SIn

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Ans.

Helpful Hint

CD Not e

that th e solution of Sampl e Problem 5/16 for t he parabolic cable gives a very close approximat ion to the values for the catenary even though we have a fairly large sag. The approximation is even better for smaller sag-to-span ratio s.

292

Chapter 5

Distr ibuted Forces

PROBLEMS (The problems marked with a n as te risk (. ) involve tran scendental equations which may be solved with a compu ter or by graph ical met hods.I

-5/140 A ligh t fixture is suspended from t he ceiling of an outs ide por tico. Four chains, two of wh ich are shown, prevent excessive mot ion of th e fixtu re during windy conditions. If t he cha ins weigh 15 Ib per foot of length, det ermine t he chain ten sion at C and th e lengt h L or chain BG.

r

Introdudory Problems 5/137 A mason st retche s a str ing between two points 50 ft apart on the same level with a tension of 10 lb at each end. If th e string weighs 0.1 lb, determine th e sag II at th e midd le of th e stri ng. A il s. )' = 0.750 in .

A

6S m ""-----------

1 G

~20, -j

Sm

1A

B

5'

-5 /118 A power tug is towing a barge with a cable which has a mass of 14 kg per meter of its lengt h. It is observed that the ta ngent to th e cable at point A is ho rizontal. Determ ine th e ten sions at A and B.

L

I

Problem 5/ 140

-

Problem 5/138

5/139 Th e Golden Gat e Bridge in Sa n Fran cisco has a ma in span of 4200 ft, a sag of 470 It , and a tota l static loading of 21,300 lb per lineal foot of horizontal mea sur ement. Th e weight of both of the ma in cables is included in thi s figure and is assume d to be uniforml y distributed along the hori zontal. Th e angle made by th e cable with th e hori zont al at the top of th e tower is th e same on each side of each tower. Calcu lat e t he midspan tension To in each of th e mai n cab les a nd th e compressive force C exerted by each cable on th e top of eac h tower . An s. To = 50.0 (06 ) Ib, G = 44.7(0 6 ) Ib

5/141 A cab le weighing 25 newton s per meter of length is suspended fro m point A and pas ses over th e small pu lley at B. Calculate th e mas s m of th e attached cylinder which will produ ce t he sag of 9 m. Also determ ine th e hori zontal distan ce fro m A to C. Because of the small sag-to-span ratio, use the approximatio n of a parabolic cab le. An s. m = 270 kg, AG = 79.1 m

1- ""~ mL --';· ;

B

A

~

~====--C1 9 m

rJ m

I Problem 5/141

470'

,

- - - - - -

4200' - - - - - --1

-5 /141 Repeat P rob. 5/ 141, but do not use t he ap proximation of a par ab olic cable. Compa re you r resu lts with th e prin ted answers for P rob. 5/ 141.

Problem 5/139

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Articl e 5/ 8 5/143 A cable supports a load of 50 kg/m uniformly dist ributed wit h re spect to t he horizontal and is suspended from the two fixed point s located as shown. Determine th e maximu m and mini mu m ten sions T max and To in the cab le. Ans. T lll llX = 35.6 kN, To 21.0(103 , N

.

I_oo_m--, - - - -'i

P ro b lem s

2 93

Representative Problems ·5/146 The glider A is being towed in level fligh t and is 400 ft behind an d 100 ft below the tow pla ne B. The ta ngent to the cable at t he glider is hor izon tal. The cable weighs o.s Ib per foot of length. Calculate th e horizontal tension To in the cable at the glider . Neglect air re sistance and compare your result with that obtained by approximati ng the cable shape by a parabola .

B

A

50 kglm

Problem 5/ 146

Problem 5/14 1 5/144 A hori zontal 350-mm-diameter water pipe is supported over a ra vine by th e cable shown. The pipe and the water wit h in it have a combined mass of 1400 kg per meter of its lengt h. Calculate t he compression C exerted by the cable on eac h support. Th e angles made by the cable with the horizontal ar e the same on bot h sides of each suppo rt .

5/147 Th e light cab le is suspended from two points a distance L apart and on t he same horizontal line. If th e load per u nit of hor izontal d ist an ce supported by the cable varies from Wo at the center to W I at the ends in accordance wit h th e relation w = a + bx 2 , deriv e t he equation for th e sag h of the cable in ter ms of the midspan te nsion To. Ans. h

~----- 40 m ------

L'

= 48T

( Sw o

+

W I)

o

2.5 m

1--- Ll2 --~:---- LI2-----I

Probl em 5/144

5/ 145 A cable of negligible mass is suspended from the fixed point s shown and has a zero slope at its lower end. If th e cable supports a unit load W which decreases uniformly with x from Wo to zero as indicated, det ermine th e equa tion of the curve assumed by the cable.

. _ 3hx 2 Ans. y - 2/2

(

1 -

ft

)

Problem 5/ 147 ' 5/ 148 Find the total lengt h L of ca ble which will ha ve th e configuration show n when suspended from points A and B.

~----

f<-- - - - -

I

Problem 5/1 48

Problem 5/ 145 Marwan and Waseem AI-Iraqi

40 m - - - - ---1

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294

Chapt er 5

Distr ibuted Forces

-5 /149 Find th e to tal lengt h L of cha in which will have a sag of 6 ft when suspended from two points on the sam e hor izonta l line 30 ft apart. Ans. L = 33.0 It

I

30' -

~

-

-

-

-

---1I

=7T

-5 /151 Calcu late th e ten sion T requ ired to stea dily pu ll the cable over a roller support on th e ut ility pole. Neglect the effects of frict ion at th e support. T he cable, which is horizontal at A , has a mass of 3 kg/m . Also det ermine the len gth of cable fro m A to B.

---r T

B

I

10m

__ ----------L

I

Problem 5 /149

- 5/ 150 A series of spherica l float s are equa lly spaced and secu rely faste ned to a flexible cab le of length 20 m. End s A and B are anchored 16 m apart to th e bottorn of a fresh-water lake at a dept h of 8 m . The floats a nd cable have a combined weight of 100 N per meter of cable length, and th e buoyancy of the wat er produces an up ward force of 560 N per metel' of cable lengt h. Calcu late the depth h below th e su rface to t he top of the line of floats. Also find t he angl e tJ mad e by th e line of floats with the horizonta l at A .

~~---- 30

01 - - - - -1

Problem 5/152

-5 /153 A rope 40 m in le ngth is suspended between two points which are separated by a hor izontal dist ance of 10 m . Compute t he distance II to the lowest part or th e loop. A il S. h = 18.53 m

' - 10 m-J

I

Ii

1-- - - - - 16 01

- - - - _J

Problem 5/153

Problem 5/1 SO ' 5/ 15 1 Numerous small flotat ion devices are attached to

th e cable, and t he differe nce between buoyan cy and weigh t resu lts in a net up ward force of 30 newtons per met er of cable lengt h. Det ermine th e force T which must be applied to cau se the ca ble conflgu ration shown. An s. T = 1210 N B

•

)

T

-5 /154 A cable which weighs 50 newt ons per meter of lengt h is secu red at point A a nd passes over th e small pu lley at B on t he same level un der a tension T. Determi ne the minimum value of T to support th e cable and th e corresponding deflect ion h . f--

-

-

-

-

lOO 01 -

-

-

A

Ii

Problem 5/154 25 m - - ---1

Problem 5/151

Marwan and Waseem Al-l raqi

-

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--1

81

--------------------------

m

A r-' ~ ..

-

-T

Art icl e 5/8

5/155 Th e blimp is moored to th e ground winch in a gentle wind with 100 m of 12·mm cabl e which ha s a ma ss of 0.51 kg/m . A torque of 400 N · m on th e drum is requ ire d to sta rt winding in t he cable. At this conditi on th e ca ble mak es a n a ng le of 30 with th e vertical as it a pproaches the win ch. Ca lcu late the heigh t H of t he blimp. The diam et er of th e drum is 0.5 m. A ns. H ~ 89.7 m Q

Problem s

295

-5 /157 The moving cab le for a ski lift has a mass of 10 kg/m and ca rries eq ua lly spaced cha irs a nd passe nge rs , whose a dded mass is 20 kg/m whe n av eraged over th e length of th e cable. The cab le lead s hori zontally from th e supporting gu ide wh eel at A. Ca lculate the te ns ions in t he cab le at A and B a nd th e length s of t he cable betwee n A and B. A ns. T A ~ 27.4 kN . T il ~ 33 .3 kN, s = 64.2 m GOm

B

r--------------- ~~~ . ~~-

I 20m I I rA

H

Problem 5/157

Problem 5/155

·5/156 A lengt h of ca ble which has a ma ss of 1.2 kg/m is to have a sag of 2.4 m when sus pende d from t he two points A and B on the same hori zontal line 10 m apa rt. For compa rison pu rposes, determine th e len gth L of cab le required and plot its configuration for th e tw o cases of (a) assu ming a parabolic sha pe a nd (b) using th e proper ca te nary model. In orde r to more clearly dist ingui sh between t he two cases, a lso plot th e di ITeren ce (yc - Yl' ) as a func tion of x, whe re C and P refer to catenary an d pa rab ola, res pect ively.

-- - - - - 10 m - - --

81

-5 /158 A sphe rica l buoy used to ma rk the course for a sa ilboat race is shown in th e figure . Th e re is a wate r current from left to right which ca uses a hori zontal drag on th e bu oy; the effect of the cu rrent on t he cable can be neglected . Th e length of t he cable betwee n poi nts A a nd H is 87 01 , a nd t he effect ive cab le mass is 2 kg/ Ill whe n the buoyancy of th e ca ble is accou nte d for. Det er mine the te nsions at both A and B .

BOm

---,

IA 2.4 m

~_..L_

_

ec

_

_

-- r

Problem 5/156

MarNan and Wasee m AI-Iraqi

Problem 5/158

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296

Chapt er 5

Di stribut ed Forc e s

-5 /159 The tree surgeon attempts to pull down t he pa rtially sa wn-throug h tree trunk. He creates a tension T A = 200 N in th e rope, whi ch has a mass of 0.6 kg per meter of its length. Deter mine the a ngle 0A at which he pu lls, the lengt h L of rope bet ween point s A a nd B, a nd the tension Tn at point B. Ans . IJA ~ 12.64°, L ~ 13.06 m, Til ~ 229 N m r--- - - -12 ---- -B

6m

I 5m l

I I

IJ.J.

I

A

Problem 5/ 159

Problem 5/161

'5/160 Recon sider Prob. 5/ 159. If tbe length of the rope between points A a nd B is 13.02 m, determine th e ten sion T A wh ich th e tree surgeo n must exe rt a t A. the a ngle 0A at wh ich he pull s, the tension Tn a t B, a nd the a ngle On wh ich t he rope mak es with the horizontal a t B. The rope has a mass of 0.6 kg per met er of it s len gth.

-5 /162 A 50·kg t raffi c signal is sus pe nde d by two 21-m cables which hav e a mass of 1.2 kg per meter of length. Determine th e vert ical deflecti on l) of th e ju nct ion ring A re lati ve to it s position before the signal is a dded.

'5/161 For ae st hetic reasons, cha ins are somet imes used instead of down spouts on sma ll buildings in order to dir ect roof runoff water fro m the gutter down to ground level. The a rc hitect of th e illustrat ed bu ilding specified a 6-m vert ical cha in from A to B , but t he builder decided to use a 6.1·m cha in from A to C as shown in order to place th e wate r farthe r from t he st ructure. By wha t perce ntage n did th e buil der increase th e magnitude of th e force exerted on th e gut ter at A over th at figured by th e a rchitect? Th e cha in weigh s 100 N per meter of it s length . Ans. Il = 29.0t;f,

Marwan and Waseem AI-Iraq i

i<

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+

~_. __

50 kg Problem 5/162

~~ "I

Articl e 5 / 9

5 /9

Flu i d Sta ti c s

297

FLUID STATICS

So far in this chapter we have treated the action of forces on and between solid bodie s. In this article we cons ider th e equilibr ium of bodies subjected to forces du e to fluid pressures. A fluid is any continuous substance which , when at rest, is unable to support shear force. A shear force is one tangent to the surface on which it acts and is developed when differential velocities exist between adjacent layers of fluids. Thus, a fluid at rest can exert only normal forces on a bounding surface. Fluids may be eit he r gas eous or liquid. Th e statics of fluids is gene ra lly called hydrostatics when the fluid is a liquid a nd aerostatics when th e fluid is a gas . Fluid Pressure

The pressure at any given point in a fluid is the same in all directio ns (Pasca l's law). We may prove this by considering the equilibrium of an infinitesimal triangular prism of fluid as shown in Fig. 5/ 30. Th e fluid pressures normal to the faces of the element are PI ' P2' P3' a nd P4 as shown. With force equal to pressure times area, the equilibrium of forces in the x - and y-directions gives PI dy dz

= P3 ds dz sin

H

y I

I I

Si nce ds sin I) = dy a nd ds cos H = dx, these equations require that

dx dy

= P2 = P3 = P

dp

o

= pg dh

(5 / 23)

This differ en tial relation shows us that the pressure in a fluid incr eases with depth or decr eases with increased elevat ion. Equation 5/23 holds MarNan and W aseem AI-Iraqi

Q<

e(J.~ dx

I

P 2 dx d z

Figure 5/30

By rotating the element thro ugh 90°, we see tha t P4 is also eq ual to th e other pressu res. Th us, the pressure at any point in a fluid at re st is the same in all directions. In this analysis we need not account for the weight of th e fluid element because, whe n the weight per un it volume (de ns ity P times g ) is mul ti plied by t he volum e of th e elem ent, a differenti a l qu antity of third order results which disappea rs in th e lim it compared with the second-order pressure-force terms. In all fluids at rest the pressure is a function of the vertical dimension. To determine this function, we consider the forces actin g on a differential element of a vertical column of fluid of cross-sectional area dA , as shown in Fig. 5/ 31. The positive direction of vertical measuremen t h is taken downward. The pressure on the upp er face is P, and th at on th e lower face is P plus th e change in P, or P + dp , The weight of t he eleme nt equa ls pg multiplied by its volume. Th e normal forces on th e lateral sur face, which are hori zontal and do not affect t he balance of forces in the vert ical direction, are not shown . Equilibrium of t he fluid element in the h· direction requires

P dA + pg dA dh - ( p + dp ) dA

,t )/

'I

P·l - 2-

P,

P' -2I / P3 d s dz

PI dy dz --.;. ~

dy

P2 dx dz = P3 ds dz cos Ii

dxdy

dz

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pdA

PlldAdh

t -----t i t t p +dp )dA

Figure 5/l1

I h

Tdh

--',,

,

298

Chap t er 5

Di5tribu ted p er e e s

for both liquids and gases, and agrees with our common obse rvations of air and water pressures. Fluids which are ess entially incompressible are called liquids, and for most practical purposes we may consider their density p constant for every part of the liqui d.' With p a consta nt , integration of Eq. 5/23 gives ( p = Po + pgh )

(5/ 24)

The pressure Po is the pressure on the sur face of t he liqu id where h = O. If Po is due to atmospheric pressure and the measuring instrumen t records only the increment above atmospheric pressu re," the measu rement gives what is called gage pressure. It is computed from p = pgh . The comm on unit for pressure in S I units is th e kilopascal (k .Pa}, whi ch is the sa me as a kilonewton per square me te r (103 N/m 2 ) . In computing pressure, if we use Mg/m 3 for p, m/ s 2 for g , and m for h , then the produ ct pgh gives us pressure in kPa directly . For exa mple, t he pressure at a depth of 10 m in fr esh wa ter is

p = "gh =

(1.0 ~;) (9.81 ~) (10

= 98.1 kN/m 2 =

m)

98.1 kPa

In th e U.S. custo ma ry sys te m, fluid pressure is gene rally expressed in pounds per square inch Ob/ in.2 ) or occasionally in pounds per square foot Ob/ft 2 j . Th us, at a depth of 10 ft in fresh wa ter the pressure is 3

1 in ft '" ) (120 in.) . · .2 p -_ "g!I -- ( 62.4 ftIb3 ) ( 1728 -_ 4.33 Ib/ In Hydrostatic Pressure on Submerged Rectangular Surfaces

A body subme rged in a liqu id, suc h as a gat e valve in a dam or the wall of a tank, is subjected to fluid pressure acting normal to its surface and distributed over its area. In problems where fluid forces are appreciable, we must det ermine th e res ultant force due to the distribution of pressure on the surface and the positio n at which this resultant acts. For sys tems open to the atmosphere, the atmospheric pressure Po acts over all surfaces and thus yields a zero resultant. In such cases , then, we need to cons ider only the gage pre ssure p = pgh , wh ich is the increment above atmospheric pressure. Consider the specia l bu t common case of t he action of hyd rostati c pressure on the surface of a rectangular plate submerged in a liquid. Figu re 5/3 2a shows such a plate 1-2-3-4 with its top edge hor izontal and with the plane of the plate mak ing a n arbit rary ang le 0 wit h the vert ical plane. The horizontal surface of th e liquid is represented by the x -y' plan e. Th e fluid pressure (gage) act ing nor ma l to th e plate at poin t 2 is

"See Table 0 / 1, Appendix 0 , for table of densitie s. tAtmospheric pressure at sea level may be taken to be 101.3 kPa or 14.7 Ib/in.2 Marwan and W aseem AI-Iraqi

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Art icle 5/9

represented by th e ar row 6-2 and equa ls pg times the vertica l dist an ce from th e liquid surface to point 2. Thi s same pressure acts at all points a long th e edge 2-3. At point 1 on th e lower edge, th e fluid pressure equals l>g tim es th e depth of point I, and thi s pr essure is th e sa me at all points along edge 1-4. Th e var iatio n of pr essure p over th e area of th e plat e is governed by t he linea r dept h relationship and th erefore it is represent ed by the arrow p, shown in Fig. 5/ 32b, which var ies linearl y from t he valu e 6-2 to t he value 5-1. The resultant force produ ced by thi s pressure distrib ution is re presented by R , which acts at some point P called the center of pressure. The cond itions which prevail at the vertical sect ion 1-2-6-5 in Fig. 5/320 a re identical to th ose at section 4-3-7-8 and at every othe r vertical sect ion norm al to the plate. Thus, we may a na lyze th e pr oblem from th e two-dimensional view of a vertical section as s hown in Fig. 5/32b for section 1-2-6-5. For this sect ion the pressure distribu tion is trapezoidal. If b is th e hori zontal widt h of the plate measu red norm al to the plan e of th e figure gil act s is dA = b dy, and an incremen t of th e

la )

y'- - -

---------;,~-

s' - - - - - - - - - - - - , , -

LI2 ciA'

5

Ie!

Ibl

Figure 5/3 2 MarNan and Waseem AI-Iraqi

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Flu id Static s

299

300

Chap t er 5

Dis t ribu t ed Forc e s

resu ltant force is dR = p dA = bp dy. But P dy is merely th e shaded increm ent of trapezoidal area dA ' , so th at dR = b dA ' . We may t herefore expres s t he res u lta nt force acti ng on th e entire plat e as the t rapezoida l area 1-2-6-5 times the widt h b of t he plate,

R = b

JdA '

= bA'

Be careful not to confuse the physical area A of the plate with the geometrical area A ' defined by the trapezoidal distribution of pr essure. The trapezoidal area representing the pressure distribution is easily expre ssed by using its average altitude. The resultant force R may ther efore be written in terms of the average pressure P a y = P 1 + P2) times the plate area A. The average pressure is also the pressu re which exists at the average depth, meas ured to the cent roid 0 of the plate. An alte rnative express ion for R is therefore

!<

R

\

\

\,

R

(a )

- - --

----

_

Y =

(b )

~

/I "

I

I'ghA

where Ii = Y cos 8. We obtain t he line of acti on of the resul tant force R from the pr in ciple of momen ts . Using the x-axis (point B in Fig. 5/ 32b) as the moment axis yields RY = J y ( pb dy ). Sub sti tuting p dy ~ dA' and R = bA' and canceling b give

B

y ' --

= PavA

P,

+-__-,

; 1 - ._ _

C

J Y dA' '--c---_

JdA '

which is simply the express ion for the cent roidal coordinate of the trapezo idal area A'. In the two-dimens ional view, therefore, the resu ltant R passes through th e cen troid C of the trapezoidal area defined by the pressure dist ribution in the vertical section. Clearly Y also locates the centroid C of the tru ncat ed pr ism 1-2-3-4-5-6-7-8 in Fig. 5/320 th rough which the resultant passes. For a trapezoidal distribution of pr essure, we may simplify the calculat ion by dividing th e trapezoid into a rectangle and a triangle, Fig. 5/32c, and separately consider ing the force represented by eac h part. Th e force re presented by th e rectangular portion acts at the cente r 0 of the plate and is R z = pzA , wher e A is th e area 1-2-3-4 of t he plate. The force R , rep resen ted by the triangul ar increm ent of pr essure distributio n is PI - pz)A and acts t hrough th e centro id of the triangular portion shown.

!(

Hydrostatic Pressure on Cylindrical Surfaces

(e )

Figure 5/33

M arwan and W aseem AI-Iraqi

Th e dete rmination of the resul tant R due to distributed pr essure on a submerged curved surface involves more calculation than for a flat surface. For example, consider the submerged cylindrical surface shown in Fig. 5/330 where the elements of the curved surface are parallel to www .gigapedia.com

Article 5 /9

th e hori zontal su rface x -y' of the liquid. Vert ical sections perpendicu lar to the surface all disclo se the same curve AB and the sa me pressure distribution. Thus, th e two-di mensiona l representation in Fig. 5/3 3b may be used. To find R by a direct integr ation , we need to integrate the .r- a nd y-compone nts of dR alon g the curve AB, since dR cont inuous ly changes direction . Thus,

n, =

b

f

(p

au, =

b

f

p dy

an d R , = b

f

( p dL l., = b

f

p dx

A moment equation would now be required if we wished to establish the position of R . A second method for finding R is usu ally much simpler. Consider the equilibr ium of the block of liqu id ABC directly above the surface , shown in Fig. 5/330 . Th e resultan t R the n appea rs as the equa l and opposite reaction of t he su rface on the block of liquid . Th e res ultants of the pressur es along AC and CB are P; and Px' respectively , and are eas ily obta ined. The weigh t W of the liquid block is calculated from the area ABC of its section multi plied by the const an t dimension b an d by pg. Th e weight W passes through t he centroid of area ABC. The equilibrant R is the n det ermined complete ly from the equilibr ium equa t ions which we apply to th e free-body diagra m of the fluid block. Hydrostatic Pressure on Flat Surfaces of Any Shape

Figure 5/340 shows a flat plat e of a ny sha pe submerged in a liquid. The hori zon tal surface of the liquid is t he plane x -y' , a nd th e plane of the plat e ma kes a n angle H with the ver tica l. The force act ing on a di fferentia l st rip of area dA parallel to the surface of the liquid is dR = p dA = pglt dA . Th e pressure p has the same magnitude th roughout th e lengt h of the st r ip, because th ere is no cha nge of depth along

(b l

Ca)

Figure 5 /34 MarNan and Waseem AI-Iraqi

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Flu id Sta tics

301

302

Chapter 5

Dis tr ibu t ed Forc e s

th e strip. We obtain the to tal force act ing on the exposed area A by integration, which gives

Su bsti tu ting the centroidal relat ion

IrA =

II! dA gives us (5 /25)

The qu antity pgli is th e pre ssure which exists at the depth of th e centroid 0 of the area and is the average pressure over the area . We may also represen t t he resul tant R geomet r ically by t he volu me V' of the figure shown in Fig. 5/ 34b. Here the fluid pressure p is re presented as a dimension normal to the plate regarded as a base. \Ve see th at th e resulting volume is a truncated r ight cylinder. Th e force dR act ing on the differential area dA = x dy is represented by the eleme ntal volume p dA shown by the shaded slice, a nd t he total force is represented by th e tot al volum e of the cylinder. We see from Eq . 5/ 25 th at the average alt itude of t he t ru nca ted cylinder is th e average pressu re pgli which exists at a dept h corresponding to the centroid a of the area exposed to pressure. For pr oblems where the cen troid a or the volum e V ' is not rea dily apparent, a direct integration may be perform ed to obta in R . Thu s,

R =

I dR = I p dA = I pgl!x dy

I!

where the depth a nd th e len gth x of t he horizontal st rip of differential area must be expressed in terms of)' to carry out the integration. After the resultant is obtained, we must determine its location. Using the principle of moments with t he x-axis of Fig. 5/ 34b as the moment axis , we obtain

RY =

I

y dR

or

_I I

y( px dy ) Y = '---;---

(5/26 )

pxdy

Thi s second relat ion satisfies th e definition of th e coordinate Y to th e centroid of the volum e V ' of th e pr essure-area t ru nca te d cylinder. We conclu de, th erefore, th at the resultant R passes through the cen troid C of th e volume descr ibed by th e plat e area as ba se and the linearly varying pressure as the perpendicular coordinate. The point P at which R is applied to the plate is t he center of pressu re. Note that the cente r of pressure P and the centroid a of th e plate area are not the same.

Buoyancy Archimedes is credited with discovering the princip le of buoyancy . Thi s principle is easily expla ined for an y fluid, gaseous or liquid, in equilibrium . Consider a portion of th e fluid defined by a n imagina ry closed Marw an and W aseem AI-Iraqi

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Article 5 /9

/- /'

I

\

......

,

I

, _- .... / I .....

(a>

(b)

(c )

Figure 5/35

su rface, as illustrated by the irregular dashed boundary in Fig. 5/300. If the body of the fluid could be sucked out fro m with in the closed cavity and replaced simulta neously by the forces which it exerted on the boundary of the cavity , Fig. 5/35b , th e equilibrium of the sur rounding fluid wou ld not be distu rbed. Furthe rmo re, a free-body diagram of th e fluid portion before remova l, Fig. 5/350, shows that th e res ultant of th e pr essure forces distributed over its surface must be equal and opposite to its weight mg and must pass through the cente r of mass of th e fluid element. If we replace th e fluid element by a body of th e same dimen sions, th e surface forces act ing on the body held in th is positi on will be ident ical to th ose acting on the fluid element. Th us, the resu ltan t force exerted on t he su rface of an object immersed in a fluid is equal and opposite to th e weight of fluid displaced a nd passes th rough th e center of mass of the displaced fluid. Thi s result ant force is called the force of buoyancy (5 / 2 7)

where p is th e density of the fluid, g is th e accelera tion due to gravity , and V is the volum e of th e fluid displaced. In th e case of a liquid whose density is constant, the center of mass of the displaced liqu id coincides with th e centroid of the displaced volume . Thus when the density of an object is less th an th e density of the fluid in which it is fully immersed, there is a n imbalance of force in th e vertical direction, a nd th e object rises. When th e immersing fluid is a liquid , th e object continues to rise u nt il it comes to the surface of the liquid and then comes to rest in an equilibrium position, ass uming that th e density of th e new fluid above th e su rface is less tha n the density of th e object . In th e case of th e su rface boundary between a liquid and a gas, such as wat er and air, th e effect of the gas pressu re on that portion of th e floating object above the liquid is ba lan ced by the added pressu re in the liquid due to th e action of th e gas on its surface. An important problem involving buoyan cy is the det ermination of th e stability of a floating object, such as a ship hull shown in cross section in a n upri ght position in Fig. 5/360 . Poin t B is t he centroid of the displaced volum e and is called th e center of buoyancy . The resultant of th e forces exerted on the hull by th e water pressu re is the bu oyancy force F which passes through B a nd is equa l and opposite to the weight Waf the ship. If the ship is caused to list through an a ngle c, Fig. 5/3 6b, MarNan and W aseem AI-Iraqi

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Fluid Statics

:50 :5

304

Chapter 5

Dis tributed Forc es

I

I I

I

I I (a ) F

(b)

F

(e ) F

Figure 5/16

th e shape of the displaced volume changes, a nd the center of buoyancy shifts to B'. Th e point of int ersection of t he vertical line t hrough B ' with the centerline of the ship is called t he metacenter M, a nd th e dist an ce h of M from the cente r of mass G is called th e me/acentric height. For most hull sha pes h remains practically constant for an gles of list up to about 20°. When M is above G, as in Fig. 5/ 36b, t here is a righting moment which tends to bring th e ship back to its upright position. If M is below G, as for th e hull of Fig. 5/360 , the moment accompa nying the list is in th e direction to increase the list . This is clea rly a condition of instability a nd must be avoided in th e design of any sh ip.

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Art icl e 5/9

Sample Problem 5/18

A

,- "

1m

A rectangular plate, shown in vertical section AB , is 4 m high and 6 m wide (normal to the plane of t he pap er ) and blocks the end of a fres h -water channel 3 m deep . Th e plat e is hinged about a horizontal axis along its upper edge through A and is re strained from openi ng by th e fixed ridge B which bears hor izonta lly agai ns t th e lower edge of t he plat e. Find th e force B exerted on the plate by the r idge.

I

3m

B

Solution. Th e free-body diagr am of the plate is shown in section and include s the vertical and horizo ntal componen ts of t he force at A, th e u nspecified weight W = mg of th e plate, th e unkn own horizon tal force B, an d the resultant R of the triangular distribution of pressure against th e vertical face. Th e den sity of fres h water is p = 1.000 Mgjm:J so th at the average pressure is

l

2

Pay ~ 1.000(9. 8 1 ) (~ ) = 14.72 kP a

R

(14.72)(3)(6)

~

i--. ~ B

= 265 kN

H elpful Hint

Ans.

B = 198.7 kN

CD Note that the units of pressure pgh are ( 10'

~;) (~) (m) = ( 10' k~~m)C~2) ~ kN/m 2 ~ kP a.

Sample Problem 5 /19

Al

The ai r space in the closed fresh-wat er tank is maintained at a pressure of 0.80 Ib/ in.2 (above atmosp her ic). Determine th e res u ltant force R exerted by th e air and water on the end of the tank.

Solution . Th e pressu re distribu tion on th e end surface is shown, where Po = 0.80 Th e specific weight of fresh wat er is I' ~ pg ~ 62.4/1728 = 0.0361 Ib/ in.3 so tha t th e incre ment of pressure up due to the wat er is

Ib/in .2

t:.p ~ I' ::'11 = 0.0361(30) = 1.083 Ib/in

CD

~

PoA,

~

R 2 = ::'P.,A2

Th e res ultant is t hen R

~

0.80(38)(25) ~

R 1 + R2

~

~

760 + 406

~

~

406 Ib 1166 lb.

Ans.

We locat e R by applying th e momen t pr inciple about A noti ng th at R 1 acts th rough the center of th e 38-in. depth an d t hat R 2 acts through t he cent roid of th e trian gu lar pressu re dist rib ution 20 in . below t he sur face of th e wat er and 20 + 8 =: 28 in . below A. Thu s, 116611 = 760(19) + 406(28) Marwan and Waseem AI-Iraqi

8"

Water

30"

h ::: 22.1 in. www.gigapedia.com

25"---c1

T

1

B

Side view Po

--

End view

A

R

R-

I~I-

~m I. 28" II

~ --L I[

R,

760 Ib

") -1.083 2- (30 )(20

Air

2

Th e resul tan t forces R 1 and R 2 due to the rectangul ar and trian gu lar dist ributions of pressure, respectively, are

R1

1

4m

-I-I R-

This force acts through the cen troid of t he triangular distribution of pressu re , which is 1 m above th e bottom of t he plate. A zero mom ent summation about A establishes the unknown force B. Thus, 3(265) - 4B = 0

1

mu

III

Im

The resultant R of the pressure forces aga inst the plate becomes [R ~ Pa,A1

305

Fluid Statics

An s.

Po B

Helpful H int

CD Dividing

the pressure distri bu tion into these two parts is decided ly th e simplest way in which to make the calculati on.

306

Chap ter 5

Distr ibu ted Fo rc e s

Sample Problem 5/10

a

Determine completely the resultant force R exerted on the cylindrical dam surface by the wate r. The density of fresh water is 1.000 Mg/ m3 , and the dam has a length b, normal to the paper, of 30 m.

Solution. The circular block of water BDO is isolated and its free-body diagram

D

~

I r =4m

Bi

is drawn. The force Px is ., pgr b (1.000)(9.81)(4) 30 () 23 kN Px -_ I'g -,~ = ""2 r = 2 ( )4 = 50

(j)

ar---,=-

J

The weight W of the water passes through the mass center G of the quartercircular sect ion and is

D - .,- - .r

I

L-_~~ ""'

Equilibrium of the section of water requires [ ~ F,

01

n;

P, = 2350 kN

[ ~F,

01

R,

mg = 3700 kN

c

- -- --\

mg = pgV = (1.000)(9.81) m: )2 (30) = 3700 kN

B"_ ' -4 t

A

/I

y

The resultant force R exerted by the fluid on the dam is equal and opposite to that shown acti ng on the fluid and is

[R

=

J R x 2 + Ry 2)

R

=

/( 2350)2 + (3700)2 - 4380 kN

Ans.

The x-coordinate of the point A through which R passes may be found from the principle of moment s. Using B as a moment ce nter gives

r

2350

4r

Px 3 + mg 3" - R,x = 0, x =

W

+ 3700 3700

(~) = 2.55 m Ans.

Helpful Hints

@ Altemative Solution. The force acting on the dam surface may be obtained

by a direct

integration of the component s dR;r

wher e p

=p

ciA cos H

= I'gh = I'gr sin

e,

=

Ry

=

R

Thus , = values gives

f'2

0 and dA

dRy

and

"2

0

pgrb sin 2 0 dO = pgr2b

JR/

if there is any question about the units for- pgli.

(1

= bir dOl. Thus,

pgr'b sin 0 cos 0 d o = _ pgr2b

f

P dA sin

=

[0

2-

Q) This approach by integra tion is fea-

[co~ 20]:,2 = ~pgr2b . 20] , ,2 0 4

S10

sible here mainly becau se or the simple geometry of the circular arc.

= ~ 1Tpgr2b

+ R/ ' = ~pgr2b,h + r?/4. Substituting the numerical

R = 4(1.000)(9.81)(42J(30)J! + . '/4 = 4380 kN

Ans.

Since dR always passes through point 0 , we see that R also passes through about 0 mus t cancel. So we write R;rYl = R y X l , which gives us

o and, therefore. the moments of R;r and Ry x, /y,

=

Rr/ R y = (~I'gr'b)/(~ "I'gr'b )

=

2/"

By similar triangles we see that

x/ r = x, /y, = 2/"

and

x = 2r/" = 2(4)/" = 2.55 m

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(j) See note (j) in Sample P roblem 5/ 18

Ans.

Article 5/9

Flu id Stati cs

307

Sample Problem 5/21 Determine t he resu ltant force R exer ted on th e semicircular end of the wat er tank shown in the figur e if the tank is filled to cap acity. Express the result in terms of th e radiu s r an d the water den sity p.

Solution I. We will obtain R first by a direct integr ation. Wit h a horizon tal stri p of a rea dA = 2x dy acte d on by the pressure p = pgy, th e increm ent of th e resu lt an t for ce is dR = p dA so that R

~

f

P dA

~

f

h--:

r

pgyl2x dy)

=:-1

An s.

In tegr ating gives

Th e locat ion of R is determ ined by usin g th e pri nciple of mom ents. Taking mome nts about the x-ax is gives

fRY

~

f

y dR]

In tegr atin g gives

-y

a nd

~

3nT

­

A ns.

16

SolutIon II. We may use Eq. 5/ 25 directly to find R, where th e average pressu re is pgh a nd h is the coordinate to t he centroid of th e area over which t he pre ssu re acts. For a se micirc ular a rea Ii = 4r / (31T). 4r 1Tr2 2 ~ -pgr' 31T 2 3

[R ~ pghA ]

CD

R ~ pg -

Helpful Hint

CD Be very careful not to

make the mis take of assuming that R passes through the centroid of the area over which the pressure acts .

A ns.

which is the volume of the pre ssure-ar ea figure. The resu ltan t R acts t hrough t he cent roi d C of the volu me defined by th e pressure-area figure. Ca lculation of the cent roida l distance Y invo lves th e sa me integra l obtained in Solution I.

Sample Problem 5/22 A buoy in th e form of a uniform 8-rn pole 0.2 m in diameter ha s a mass of 200 kg and is secured at its lower end to th e bot to m of a fres h-water lak e wit h 5 m of cable. If the depth of t he water is 10 m, ca lculate t he angle 0 mad e by th e pole wit h the hori zontal.

Solution. T he free-bo dy diagram of the bu oy shows it s weight acti ng th roug h G, th e vertical ten sion T in t he an chor cable, a nd the buoya ncy force B which passes through centroid C of the submerged porti on of the buoy . Let x be t he distan ce from G to th e waterline. Th e density of fresh wate r is p = 10 3 kg/m 3 , so th at the bu oyancy force is [B

~

pgV]

Moment equilibri um ,

B

+

~ 103(9.8 1),,(0 .1)2(4

~ MA

x) N

= 0, about A gives 4 +x

200(9.8 1)(4 cos OJ - [103 (9.8 1),,(0.1)2(4 + xJ] -2T hus ,

x = 3.14 m

an d

I L+5fi .L

tJ = sin - 1 (4

Marwan and Waseem AI-Iraqi

5

+ 3.14

cos 8 = 0

) = 44.50

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Ans.

T

308

Chapter 5

Distributed Forces

PROBLEMS Introductory Problems 5/ 163 The submersible diving cha mber has a total mass of 6.7 Mg inclu ding per sonnel, equipment , and ballast. When th e chamber is lowered to a depth of 1.2 km in the ocean, the cabl e ten sion is 8 kN . Compute the tot al volu me V displaced by the cha mber. A n s. V = 5.71 m :J

5/1 65 A recta ngu lar block of den sity 1'1 floats in a liquid of den sity P.2' Deter mine the ra t io r = h i e. wher e h is th e submerged depth of block. Evalu at e r for an oak block floating in fresh wat er and for steel floatin g in mercu ry. An s. r = !2 r = 0.8, 0.577 /}.2·

P2

Problem 5/165

5/ 166 The form s for a small concre te ret ainin g wall are shown in section. Th er e is a brace Be for every 1.5 m of wall lengt h. Assu ming that the joints at A . B, and C act as hinged conn ect ions, comp ut e th e compression in each bra ce BC. Wet concrete may be treat ed us a liquid with a densit y of 2400 kg/ m:J• Pro blem 5/163

r-r.,..,---~:--------,-

5/164 Specify th e magnitu de and locati on of the resulta nt force which acts on each side and t he bottom of the aqua riu m due to the fres h water inside it. 0.301r---

/ -----

3m

0.7 m

~..:..........

A

0.4 01

·- -

Problem 5/166

Problem 5/164

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2 01 - - - ,

Article 5 /9 5/167 A deep-submersible diving cham ber designed in the for m of a spherical she ll 1500 mm in diameter is ballasted wit h lead so that its weight slightly exceeds its buoyancy . Atmospheric pressur e is maintai ned within the sphere during an ocean dive to a depth of 3 km . The t hick ness of t he shell is 25 mm. For this depth calculate the compressive stress a whic h acts on a diametral section of the shell, as indicated in the rig ht- hand view. Ans. (J' = 463 MP a

(J

Proble ms

~09

5 /169 When t he sea-water level ins ide the hemisp herical

cha mbe r reac he s t he C.B-m level shown in the figure, the plunger is lifted , allowing a su rge of sea water to enter the vertical pipe. For this flu id level (a) det erm ine the average pressur e (T su pported by the seal area of the valve befor e for ce is app lied to lift th e plu nger and (b) determine the force P (in addition to th e force needed to support its weight) required to lift the plunger. Assume atmospheric pressu re in all airspaces and in the sea l area when cont act ceases under t he act ion of P. Ans. a ~ 10.74 kPa, P ~ 1.687 kN

Seawater supply -

-=::\

.

I

I ~

Problem 5/167

a .6m

5 /168 Fresh water in a channel is contained by the uniform 2.5-m plat e freely hi nged at A. If the gate is

designed to open when the dept h of the water reach es 0.8 m as shown in the figur e, what must be the weight w (in newtons per meter of horizontal lengt h into t he paper) of the gate?

-r =-0---==-==;:--;0;:;:::--

Problem 5/169

5/170 One end of a u niform pole of length L and density p' is secured at C to the bottom of a tank of liquid of density p and dept h h . For t he conditions p' < p and h < L , find the angle () assumed by the pole.

- -\'"

a .8m

-L

B 30'

Problem 5/16B

Problem 5/170

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310

Chapter 5

Distributed Forces

5/111 The cross section of a fresh-wate r tank with a slanted bottom is shown . A rectangular door 1.6 m by 0.8 m (normal to th e pla ne of th e figure ) in th e bottom of t he tank is hi nged at A a nd is ope ned against th e pr essu re of the water by the cabl e u nder a ten sion P as show n. Ca lcu late P. Ans. P = 12.57 kN

5/113 The solid concrete cylinde r 6 ft long a nd 4 ft in diameter is supported in a half- submerged position in fre sh water by a cab le which passes over a fixed pul ley at A. Compute th e te nsion T in the cab le. The cylinder is waterproofed by a plastic coating. (Consu lt T ab le D/ 1, Appe ndix D, as needed .) An s. T = 8960 lb

P

T

1.2 m

Problem 5 /171

Problem 51tH

Representative Problems 5/112 A block of wood in t he form of a waterproofed 16· in . cube is floa tin g in a tank of salt water with a a-In. layer of oil floating on the water. Assu me t hat the cube floats in t he a ttit ude shown, and calculate the height h of th e block above th e surface of th e oil. The specific weights of oil, salt water. a nd wood are 56 , 64, a nd 50 Ib/n 3 , respecti vely.

5/114 A ma rk er bu oy cons isting of a cylinder a nd cone has t he dim en sions shown an d weighs 625 Ib whe n out of th e water. Det er mine th e protrusion h whe n t he buoy is floating in sa lt wa te r. T he buoy is weighted so that a low center of mas s ensures sta bility.

r 'l 2

I

6'

--r

Oil

Salt wate r

16"

~~3 _-----,l Problem 5/172 Problem 5/174

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Article 5 /9

5/175 A channel-marker buoy consists of an 8-ft hollow stee l cylinder 12 in. in diamete r weighing 180 lb and anchored to th e bott om with a cable as shown. If h = 2 ft at high t ide, calculate the ten sion T in th e cable. Also find t he value of h when the cable goes slack as th e t ide drops. Th e specific weight of sea water is 64 Ib/ ft 3 . Assume the buoy is weighted at its base so t hat it remains vert ical. Ans. T ~ 121.6 Ib, h ~ 4.4 2 ft

Probl ems

311

5/177 The hinged gate ABC closes an opening of widt h b (perpendicular to t he paper ) in a wate r cha n nel. The wat er has free access to the u nder side as well as the right side of th e gate. When t he wat er level rises above a certain value of h, th e gate will open. Determine t he critical value of h . Neglect the ma ss

Ans. h ~ a ./3

of t he gate. C

12"

-I I-

'~l 1~

h

8'

J

ru--l A

B

~ a-I~ r Prob lem 5/177

Problem 5/ 175

5/176 A fre sh-water channel 10 ft wide (normal to the plan e of the paper ) is blocked at its end by a rectangular barr ier, shown in section ABD. Su ppor ting struts Be are spaced every 2 ft along the lO-ft width. Det er mine th e compre ssion C in each st rut. Neglect the weight s of t he members.

5/178 Th e rectangular gate shown in section is 10 ft long (perpendicular to the paper ) and is hin ged about its upp er edge B. The gat e divides a chann el leading to a fresh-water lak e on th e left and a salt -water tidal basin .on the righ t , Calculate the torque M on the shaft of the gate at B requi red to pre vent the gate from opening when th e sa lt-water level drops to h ~3 ft.

Problem 5/178

Problem 5/176

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Cha pte r 5

Dis tr ibu ted Force s

5 /179 Th e hydraulic cylinder operates the toggle which

closes t he vertical gate agai nst t he pressure of fresh water on the opposite side . Th e gat e is rectangular with a horizon ta l widt h of 2 m perpendicular to t he pa per. For a depth h = 3 m of water, calculate t he requ ired oil pressure p which acts on the 150·mm· diam eter pist on of the hydraulic cylinder. Ans. p ~ 7.49 MPa

5 / 18 1 The barge crane of rectangular proportions has a 12·ft by 30·ft cross section over its ent ire lengt h of 80 ft. If the maximum permissible submerge nce and list in sea water ar e represented by th e position shown, determine the corresponding maximum sa fe load w whic h the bar ge can handle at the 20-ft exte nded position of th e boom . Also find t he tot al displacement W in lon g tons of th e un loaded barge (I long ton equ als 2240 lb). Th e distribution of ma ch inery and ballast places t he cente r of gravity G of th e barge, minu s the load w. at th e center of t he

hul l.

A ns. w = 100,800 lb, IV = 366 long tons

Probl em 5/179 5 /180 The triangu lar and rectangu lar sections are being

conside red for the design of a sma ll fresh-water concrete dam . Fr om the sta ndpoint of resistan ce to overtu rning abou t C, which section will require less concrete, and how much less per foot of dam length? Concrete weighs 150 Ib/ ft 3 .

Problem 5/181 5 / 182 Th e cast-iron plug seals th e drainpipe of an open fresh-w ater tank which is tilled to a dept h of 20 ft. Deter mine the tension T required to remove t he

plug from its ta pered hole. Atmospheric pressure exists in th e drainp ipe an d in t he seal area as t he plug is being removed. Neglect mechan ical frict ion between th e plug and its su pport ing sur face.

c

20'

f--a Problem 5/180

Prab lem 5/182

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A rticl e 5 / 9

5/183 The Quonset hut subjected to a hor izon tal wind a nd the pr essu re P agains t the circu la r roof is approximated by Po cos O. Th e pr essu re is positive on the wind ward side of the hut and is negat ive on the leeward side. Det ermine the tota l hori zon tal shea r Q on the foundat ion per un it length of roof measured 1 normal to the paper . All s. Q

=

Prob lem s

313

5/185 Th e up stream side of an a rch ed da m has the form of a verti ca l cylindrical surface of 500-ft radius an d subte nds an a ngle of 60°. If th e fres h wate r is 100 ft deep , det ermine th e total force R exer ted by th e w atel' on the dam face . An". R ~ 156.01l0 6 11b

FT1'PO

3-0~----~ -- --- - -------~-" 30"/ ~<JO

,,

,

, ,,

,/

Problem 5/183 5 /184 T he semicylindrica l stee l shell with closed ends has a mass of 26.6 kg. Determine th e mass m of th e lead ba llast which mu st be placed in the shell so that it floa ts in fres h water a t its hal f-rad ius depth of 150 mm.

Problem 5/185 5/186 Th e fresh-wat er side of a concrete da m has the sha pe of a vertical parabola with vertex at A. Determine t he positi on b of th e base point B throug h wh ich acts the resultant force of t he wate r against th e dam face C. 27 m

l

Problem 5/184

Problem 5/1 86

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Chapter 5

Distributed Forces

5/187 A st ru ct u re design ed for observat ion of sea life benea th th e ice in pola r wate rs consists of the cylindrical viewing cham ber connected to th e sur face by th e cylindrical s haft open a t the top for ingr ess a nd egress. Ballast is ca rr ied in t he rack below the cha mber. To e ns ure a sta ble condition for th e st ructure, it is necessa ry th at its legs bear on th e ice with a force that is at least 15 percent of th e tot al bu oya ncy force of t he subme rged st ructure . If th e st ructu re less ballast has a mass of 5.7 Mg, calcu lat e th e requi red mass m of lead ballast. Th e de nsity of lead is 11.37 Mg/m". Ans. m = 4 .24 Mg

, ((I )

r

( ,

x

:'iI O.375 I m j,.~,~.,.."."",,_~L ' O.55m

12 m

tb)

Problem 5/188

5/189 Th e 3-m plank shown in section has a den sit y of800 kg/ m 3 an d is h inged abou t a hor izont al axis through its upper edge O. Calcu late the a ngle H as su med by th e pla nk with t he hori zontal for t he level of fre sh wate r shown. A ns. /I = 48.2°

/~""t-T-'

.> / ·0-'1-

3m

1m I

!

Problem 5/187

Problem 5 /189

5/188 The small access hole A allows maintenance workers to enter the storage ta nk at gro un d level whe n it is e mpty . Two designs, (0) and (b), are shown for th e hole cover . If th e tank is full of fresh wa te r, est ima te t he ave rage pressure (7 in t he seal a rea of design (a) a nd th e ave rage increase u 'I' in th e initial ten sion in eac h of t he 16 bolts of design (b). You may take th e pre ssure over the hole a rea to be consta nt, an d the pressure in the sea l a rea of design (b) may be assu med to be at mospheric.

5/190 Th e deep-submer sible research vesse l ha s a passen ger compart ment in t he for m of a sp herical steel shell wit h a mea n ra dius of 1.000 m a nd a thickness of 35 mm . Calcul ate the mass of lead ballast wh ich th e vessel mu st ca rry so tha t th e combined weight of th e steel shell a nd lead ball ast exact ly ca ncels t he combin ed buoya ncy of th ese two part s alo ne. (Consu lt Tabl e D/l, Append ix D, as needed.I

Problem 5 /190 Marwan and W aseem AI-Iraqi

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Article 5 /9 5/191 The elem ents of a new method for constructing concr et e foundation walls for new hou ses ar e shown in th e figure. Onc e the footin g F is in place, polyst yren e forms A are erected and a thin con cr et e mixture B is poured between the form s. Ties T pr event th e form s from se pa ra ting. After the concr et e cures, th e forms a re left in place for insul ati on . As a design exercise, make a con servat ive es t ima te for the unifor m tie spacing d if th e ten sion in each ti e is not to exceed 6.5 kN. The hori zontal tie spac ing is th e sa rne as the vertica l spaci ng. St at e a ny assu mptions. Th e den sity of wet concre te is 2400 kg jm3 • A ns. d ~ 0.300 m

Water level --------------------------------,-----

c

Ramp

I

d

I I

-[

Problem 5/192

d

I

-J,1/2 r

T A

B

A

F

I

Problem 5/191

Marwan and Waseem AI-Iraqi

315

5 /192 T he trapezoidal viewi ng wind ow in a sea-life aquarium has the dimensions shown. Wit h th e aid of a ppropriate diagram s a nd coordin at es, describe two methods by which the resultant forc e R on th e glass due to water pr essure, a nd th e vertica l locat ion of R , could be found if numeri cal values were supplied.

I

:301

Problem s

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316

Chapt er 5

Distr i bu t ed Forces

CHAPTER REVIEW In Cha pter 5 we have studied variou s common examples of forces distributed throughout volumes, over a reas , an d along lines. In all these problems we ofte n need to determine the resu ltant of th e distributed forces and the locati on of th e resu ltant.

Finding Resultants of Distributed Forces To find the resultant and line of action of a dist ributed force : 1. Begin by multiplying the intensity of the force by the appropriate

element of volume, area, or length in terms of which the intensity is expressed. Then sum (integrate ) the incremental forces over the region involved to obtain their resultant.

2. To locat e t he line of action of th e resultant, use the principle of moments. Evaluate the sum of the moments, about a conve nient axis, of all of th e increments of force. Equate th is sum to th e moment of th e resultan t about t he sa me axis. Then solve for th e un known moment arm of the resultant.

Gravitational Forces When force is distribu ted th roughout a mass, as in the case of gravitational attraction, the intensity is the force of att raction pg per unit of volume, where p is th e den sity a nd g is th e gravitat ional accelera tion. For bodies whose density is constant, we saw in Section A that PC cancels when t he momen t pri nciple is applied. Thi s leaves us wit h a strict ly geometric prob lem of findin g th e centroid of the figure, which coincides with th e mass cente r of th e physical body whose boundary defines th e figure. 1. For fla t plates a nd shells which are homogeneous a nd have constant

thickness, th e prob lem becomes one of findin g th e properties of an area. 2. For slender rods and wires of uniform density and constant cross secti on, th e problem becomes one of findin g th e properties of a lin e segment.

Integration of Differential Relationships For problems which require th e integration of dilTerential relationships, keep in mind the following considerations. 1. Select a coordinate system which provides the simplest description

of th e boundaries of t he region of integration. 2. Elim inate higher-o rd er dilTerential quantities whenever lower-order different ial quantities will remain. 3. Choose a first-order differential eleme nt in preference to a secondorder element and a second-order element in preference to a thirdorder element. Marwan and W aseem AI-Iraqi

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Chapter Rev iew

4. Wherever possible, choose a differential element which avoids discontinuities within the region of integration.

Distributed Forces in Beams, Cables, and Fluids In Sectio n B we used th ese guidelines along with the principles of equ ilibrium to solve for the effects of distributed forces in beams , cables, a nd fluids. Rememb er th at: 1. For beam s and cables th e force intensity is expressed as force per unit len gth.

2. For fluids the force intensity is expressed as force per unit area, or pressure.

Alth ough beams, cabl es, and fluids are physically quite different appli cations, their problem formulations share the common elements cited above.

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318

Chapter 5

Distr ibu ted Forces

REVIEW PROBLEMS 5/193 Determine th e x-coordinate of the centroid of th e shaded area. Ans. X = 3.66 in .

{ o

To

7.5"

4.5"

+

Dimensions in millim et ers

4 .5"

~

Problem 5/1 95 5/196 Det ermine t he volu me V and total su rface a rea A of the complete rin g wh ich is shown in sect ion. All four corne r radii of t he cross section are 10 mm . z

es. 1

Problem 5/ 191

1

5/1 94 Determ ine th e x- a nd y-coc rdine tes of the cent roid of th e sha ded area. y I

IOmm , """"""",",,:-+,,,,-

L2 0 I

o o

mm

10

y = kx l/3

~Jmm

40--1

mm

o

Problem 5/ 196

I o o o o o

I

5/197 Th e assembly consists of four rods cut from th e same bar stock . The curved me mbe r is a circu lar a rc of radius b. Det ermine t he y- a nd a-coordina tes of t he mass center of th e assemb ly.

o o

Ans. Y ~ 0.461b, Z = 0.876b

I I

I

°0L------OJ.5-- - --~----x Problem 5/194 5{195 Calculate t he y -coordinate of the centroid of the sha ded area. Ans. Y ~ 99.7 mm

b .J--<,

<,

-,

.... -,

-, .... x

Problem 5 / 197 Marwan and W aseem AI-Iraqi

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y

Review P rob lem s 5/198 Determine the position of the center of mas s of the homogeneou s thin conical shell shown. z I

319

5/101 Determ ine th e maximum be ndi ng moment M m ax for th e loaded beam and specify the distance x to the right of end A where M m ax exists. Ans. Mm ~ = 186.4 N · m at x ~ 0.879 m

I

600Nlm

IB .

- - - - 2 m ---~~1 m

Problem 5/201 Problem 5/198 5/ 199 Draw the shea r and moment diagrams for the beam, which su pports t he un iform load of 50 lb per foot of beam length distribu ted over its midsection. Determin e th e maximu m bending momen t an d its location. Ans . Mm ~ ~ 500 Ib-ft at x ~ 6 ft

5/202 Th e triangu lar sign is attached to the post embe dded in the concrete base at B. Calcu late the shear force V, the bend ing moment M , and th e torsio nal moment T in the post at B du ring a sto rm where t he wind velocity nor mal to the sign reaches 100 km/ h. The ai r pressu re (cal led stagnation press ure) aga inst the vertical su rface corresponding to thi s velocity is 1.4 kP a .

Problem 5/199 5/100 Sketch th e shear and moment diagrams for each of t he fou r beam s loaded and supported as shown. p

p

~

r

2m At

...L

"5

dL

(c)

(0 )

p

c:: =

M

=

M

~

P (d )

(b )

Problem 5/200

Marwan and Waseem A l-lraqi

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Problem 5/202

320

Chapt er 5

Dist ributed Forc e s

5 /103 A sign board is supported by two posts e mbedded in concre te a t A a nd B. Det ermine t he moment M which th e concrete exerts on each post at A and B during a storm when the wind velocity is 100 km / h . Th e air pressu re (called stagnation pressure) against the vertical su rface correspondi ng to this wind velocity is 1.4 kP a. An s. M = 92.4 kN ' m

- - OA O.6m

Problem 5/204 5/105 A cable is suspended fro m points A and B on th e same horizontal line a nd supports a total load \V unifor mly distributed along t he hori zontal. Determin e the length S of th e cable. !Recall tha t th e convergence of the series ofEq. 5/ 160 requ ires that the sag-to-spa n ratio be less than 1/4.) A ns. S = 147.9 ft

A

Problem 5/203 5 /104 The fram e shown ena bles tran sfer of a 75-kg disa bled per son to a nd from a wheelc ha ir a nd a freshwate r swimming pool. A sma ll hand pump at B pressurizes the up per end of the cylinder to control the tension an d length of link A C. For th e position (J = 60°, link AC is un der a ten sion of 670 N. Calculate th e volume of the submerged portion of th e per son . Neg lect th e weight of the fram e ass embly. Recall that th e den sity of fresh water is 1000 kg/m 3 .

I

100' -

-

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-

-4

~------ --l - - - - --- -~ 50'

Probl em 5/ 205

Marwan and Waseem AI-Iraqi

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Review Problems

5/2 06 Determine t he dept h Ii of the squa re hole in th e solid circ ular cylinde r for which th e a-coordi nate of th e mass cen ter will have th e maxi mu m possible value .

321

5/209 A flat plat e sea ls a t rian gular open ing in th e vertical wall of a tank of liquid of dens ity '1. Th e plat e is hinged about th e upp er edge 0 of t he tri angle. Determine the for ce P requ ired to hold th e gate in a closed posit ion aga inst the pressur e of the liqui d.

"gab ( Ail S. P = 6-

,

"+-a) 2

\ I I I I

h

I

t

I / / /

o ---- \V--i="'=it

a

L~ p

Problem 5/206 Problem 5/209 5/207 Det erm ine the y-coord ina te of the cent roid of the volume obtained by revolvin g th e shaded area abo ut th e .r-axis through 180°. _ l Sa A fl S. y

5 / 2 10 A solid floatin g object is composed of a hemisph er e an d a cone of equa l ra dius r made from the same

= -

14rr

homogene ous material. If the object float s with the cente r of the hemi sp here a bove th e wat er surface, find th e maximum altitude h which th e cone may have before the object will no longer float in the u pright positio n illust rat ed .

- -- x

"

/

Problem 5/207 5/108 Locate the mass center of the body which is constru cted of sheet metal of uniform thickness .

Problem 5/21 0

Dim en sions in millimeters

Problem 5/208

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Chapter 5

Dis t r ib ut ed Fo r ces .r I

5/2 11 The figu re shows the cro ss sect ion of a rectangular

gate 4 m high and 6 m long (perpendicular to the paper ) which blocks a fresh-water chan nel. Th e gate has a mass of 8.5 Mg and is h inged about a hori zontal axis through C. Compute th e verti cal force P exerted by the fou ndation on t he lower edge A of th e gate. Neglect the mass of th e frame to which th e gate is att ac hed. Ans . P = 348 kN

: 121b1f\'

900'

I

p = k ·l i

-

,

bI

- - p

A

Problem 5 /213 Prob lem 5/ 2 11

5/212 The hor izontal steel shaft with two dia mete rs is welded to the fixed block at A . Construct the dia grams of shea r and mom ent in the shaft du e to its weight. Determ ine the valu es of th e shea r force and bending moment at th e discontinuity in diameter.

.... 5/214 Regard th e tall building of Prob. 5/213 as a uniform upright beam. Deter mine and plot the shear force an d ben ding momen t in t he st ruct u re as a functi on of the height x above th e gro und . Evalu at e you r expressions atx = 450 ft. A ll s .

.v I I I

A

I

I

- ---'-!1 -400 _ nun

--..;._

_

160 3

x3 ~ Ib

M = 7.780 0" ) - 1.44011O"lx + 64 x 5i ' lb-ft 3 0.931(106 ) Ib VJ.r ~ 450· 2.2 1(10" ) Ib-ft .II~ " 50'

48mmdia.

60 mm dia.

V = 1.440 (106 )

,==::.,

!!

J

_ 400 mm

Probl em 5/212

5/2 13 As part of a preliminary design study, th e effects of wind loads on a 900-ft building ar e investi gat ed. For th e parabolic distribution of wind pressure shown in th e figur e, compute the force and moment reactions at th e base A of the buil ding due to the wind load . The depth of t he building (perpendicular to th e paper) is 200 ft. An s. A = 1.4401106 1 1b, M = 7.78(08 ) Ib-ft

' Comp uter-Oriented Problems

·5/1 15 Const ruct th e shea r and moment diagr am s for th e loaded beam of Prob. 5/ 107, repeated here. Determin e the maximum values of the shea r and moment and t heir locations on t he beam . An s. V max = 1900 Ib at x = 0 .IImax = 9080 Ib-ft at .r = 9.63 f\ IV

I I

~ w = wO - kx3

I

~'~

100lblft -

B

A •

'---

-

-

-

- 20' - - - - --'

Problem 5/215 Marwan and Waseem AI-Ir aqi

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-

-

,\'

Revi ew Pr obl em s

-5 /216 Th e 30 0 cylindrica l sect or is made of copper a nd is attached as shown to t he semicylinder made of aluminum . Det erm ine th e angle H for the equ ilibriu m positi on of t he cylinder rest ing on a horizontal sur face.

32 3

-5 /2 18 Set up t he integral expression for t he y-coordinate of the centroid of the u niform slende r rod bent into t he sha pe shown. T hen , for the values a = 2 an d b = 8, numerically eva lua te your integr a l expression.

,,

y I

I a -------T------,

r

, I I I

I I I

y =k lx l3

:b I I I I I I

, I

Prob lem 5/218 Prob lem 5/2 16

-5 /217 A homoge neo us charge of solid propellant for a rocket is in the sha pe of the circular cylin de r formed with a con centric hole of depth .r. For th e dimensions show n , plot X, th e r-coordi nate of the ma ss center of th e propellant , as a fu ncti on of t he dep th x of t he hole fro m x = 0 to x = 600 mm . Deter mine the max imum value of X a nd show t hat it is equa l to t he correspond ing value of x. Ans. Xmax = 322 m m

'5/21 9 An underw ater de tectio n inst ru ment A is attached to t he midpoint of a 100-m cab le sus pen ded bet ween tw o ships 50 m a pa rt. Determi ne th e dep th h of the ins tr ument, wh ich has negligible mass. Does th e resu lt depend on t he mass of the cab le or on th e den sity of t he wa ter? Ans. h = 39.8 m

A

Problem 5 /2 19

Problem 5/217

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324

Chapter 5

Distributed Forces

· 5/ 110 As a preliminary step in the construction of a tramway across a scenic r iver gorge , a 505-m cable wit h a mass of 12 kg/ m is st ru ng bet ween points A a nd B. Determine th e hori zont al dist a nce x to th e right of point A to th e lowest point on th e cable and compute th e ten sions at points A and B.

·5/122 A right circular cylinder of density PI floats in a liquid of density 1"2. If d is the diameter of th e cylinde r and h is th e submerged depth, plot the . h _ PI rati o r = d as a fun cti on of - over the range

o :s;

o

P.2

...! :s; 1. Evalu at e r for a pin e cylinder floating 1'2

A

In

n_-_-_~_~_O_~_'-_-_-_-_-_---:'-"I'~ ;

in sea wate r.

50 m

Problem 5/220 · 5/ 11 1 A lengt h of pnwer cable is suspended from t he two towers as shown. Th e cable has a mass of 20 kg per met er of cable. If th e m aximum allowable cable ten sion is 75 kN , det ermine t he mass I' of ice per meter which ca n form on th e ca ble with out th e max imum allowable ten sion being exceeded . If additi on al stretc h in th e cable is neglected , does t he addition of the ice change th e cable configura tion? A ns. I' = 8.63 kg/ m A

~

200 m --- - ---------- -- , 130 m 10m t ~-

81

'!;

Problem 5 /221

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Problem 5/222

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When contacting surfaces tend to slip on each other, friction forces are produced and must be accounted for. This mountan climber depends on friction between his body and the rocks as well as friction between the rope and mechanical devices through which the rope can slip. Marwan and Wase em AI-Iraqi

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FRICTION CHAPTER OUTLINE

6/1

Introduction

SECTION A. Frictional Phenomena

6/2 6/3

Types of Friction Dry Friction

SECTION B. Applications of Friction in Machines

6/4 6/5 6/6 6/7 6/8 6/9

Wedges 5crews Journal Bearings Thrust Bearings; Disk Friction Flexible Belts Rolling Resistance

Chapter Review

6 /1

INTRODUCTION

In th e preceding chapters we have usually assumed that the forces of action a nd react ion between contact ing sur faces act normal to the sur faces. This as sum ption character izes t he interact ion between smooth sur faces and was illustrated in Examp le 2 of Fig. 3/ 1. Although this ideal assum ption ofte n invo lves only a relati vely sma ll error , ther e are many problem s in which we mu st con sid er the ability of contacting sur faces to su pport tangentia l as well as norm al for ces. Tangential forc es generated bet ween contact ing surfaces are called [riction forces and occur to some degr ee in th e interaction between all real sur faces. When ever a tendency exists for one contacting surface to slide al ong another sur face, th e frict ion for ces developed are always in a dir ect ion to oppose t hi s tenden cy. In some types of machines and processes we want to min imize the r etarding effect of fricti on for ces. Examples are bearings of all types, power scre ws, gears, the flow of fluids in pipes, and t he propu lsion of ai rcraft a nd missiles through the atmosph ere. In ot her situat ions we wish to max imize th e effects of fri cti on , as in brakes, clu tch es, belt Marwan and Waseem AI-Iraqi

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328

Cha pt e r 6

Frict io n

dr ives, and wedges. Wheeled vehicles depend on friction for both starting and sto pping, and ordina ry walking depend s on friction between the shoe and th e ground. Friction forces are present throughout nature and exist in all machines no matter how accurately constructed or carefully lubricated. A machine or process in which friction is small enough to be neglected is said to be ideal . When friction must be tak en into account , th e machine or process is termed real. In all real cases where there is sliding motion between parts, the friction forces result in a loss of energy which is dissipated in the form of heat. Wear is another effect of friction.

SECTION A.

FRICT IONAL PHENOMENA 6 /2

TYPES OF FRICTION

In th is article we bri efly discuss the types of frictio nal resist a nce encountered in mechanics. The next article contains a more detailed accoun t of the most common type of frict ion , dry frict ion. (a) Dry Friction. Dry friction occurs whe n the unlubricated surfaces of two solids are in cont act under a condit ion of sliding or a te ndency to slide. A friction force tan gent to th e surfaces of contact occurs both du rin g t he interval leading up to impending slippage and while slippage takes place. The direction of this friction force always opposes t he mot ion or impending motion . This type of fr iction is also called Coulomb friction. The principles of dry or Coulomb fricti on were developed largely from th e experiments of Coulomb in 1781 and from th e work of Morin from 1831 to 1834. Alth ough we do not yet have a comprehensive th eory of dry friction, in Art. 6/3 we describ e an ana lytica l model sufficient to handle th e vas t majority of problem s involving dry friction. Thi s model form s th e basis for most of this cha pte r. (b) Fluid Friction. Fluid friction occu rs whe n adjace nt layers in a fluid (liquid or gas ) are moving at different velocities. Th is motion causes frict ional forces betw een fluid elements, and th ese forces depend on the relative velocity between layers. When th ere is no relative velocity, the re is no fluid friction. Fluid friction depend s not only on the velocity gradients with in th e fluid but also on the viscosity of th e fluid , which is a measure of its resistance to shearing action between fluid layers. Fluid friction is treate d in th e study of fluid mechan ics a nd will not be discussed furt her in this book.

(c) Internal Friction. Internal friction occurs in all solid mat erials which are subjecte d to cyclical loading. For highly elastic materi als t he recovery from deformation occurs with very little loss of energy due to internal fricti on. For materials which have low limits of elasticity and which und ergo appreciable plast ic deform ation during loading, a consid era ble amount of intern al frict ion may accompa ny this deform ation. The mechanism of internal friction is associated with the action of shear deformation, which is discussed in references on materials science . BeMa rwan and W aseem AI-Iraqi

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Article 6 /3

cause th is book dea ls primarily with the exte rn al effects of forces, we will not discuss internal friction further.

6/3

DRY FRICTION

The remainder of this chapte r descr ibes the effects of dry fr ictio n acting on the exterior surfaces of rigid bodies. \Ve will now explain the mechan ism of dry friction with the aid of a very simple experiment.

Mechanism of Dry Fridion Consider a solid block of mass m resting on a horizontal surface, as shown in Fig. 6/ 1a. \Ve ass ume that the contacti ng surfaces have some roughness. The experiment involves the application of a horizontal force P which continuous ly increases from zero to a value sufficient to move th e block a nd give it an apprecia ble velocity . Th e free-body diagram of th e block for a ny va lue of P is shown in Fig. 6/l b, where the tangential fr iction force exerted by the plan e on th e block is labeled F. Th is fr iction force acting on the body will always be in a direction to oppose motion or the tendency toward motion of the body. There is also a normal force N which in thi s case equa ls mg, a nd the to ta l force R exerted by the support ing surface on the block is the resulta nt of N and F. A magnified view of t he ir regu lar ities of th e mat ing surfaces , Fig. 6/ l c, helps us to visualize th e mechanical action of friction. Support is necessarily intermitt ent and exists at the mating humps. The direction of eac h of the reactions on th e block, R I , R z, R o, etc. depends not on ly

( a)

(b'

F

Impending motion I I Stati c Kinet ic I friction friction (no motion ) I (motion)

"I I

~t:l\~~~ i

-

t

p

R,

IdJ

(c)

Figure 6/1

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Dry Friction

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330

Ch apt er 6

Fr iction

on th e geomet r ic profi le of the irregularities but a lso on th e extent of local deformation at each contact point. The total normal force N is th e sum of the Il-components of th e R' s, and the total fricti onal force F is th e sum of the I-components of the R 's. When the surfaces a re in relative motion , th e contacts are mor e nea Iy along the top s of th e humps, and the r-components of the R's are smaller than when the surfaces are at rest relative to one another. This observation helps to explain the wellknown fact that the force P necessary to maintain motion is generally less than that requi red to sta rt the block whe n th e irregul a riti es are more nearly in mesh. If we perform the experiment and record the friction force F as a function of P, we obta in th e relation sho wn in Fig. 6/l d . When P is zero, equilibrium requires that there be no friction force. As P is increased, the friction force mu st be equa l and opposit e to P as long as t he block does not slip. During t his period the block is in equilibrium, a nd a ll forces act ing on th e block mu st sa tisfy th e equilibr ium equations. Finally, we reach a value of P which causes th e block to slip a nd to move in th e di recti on of th e applied force. At this sa me t ime th e friction force decr eases slightly and abru ptly. It then remains essentia lly cons ta nt for a time but then decreases still more as the velocity increases.

Static Friction The region in Fig. 6/ ld up to the point of slippage or impending motion is called the range of static friction , and in this range the value of the friction force is determined by the equations of equil ibrium. This friction force may have any value from zero up to and including the maximum value. For a given pair of mating surfaces the experiment shows that this maximum value of static friction F mllx is proportional to the normal force N. Thus , we may write

(6/ 0 wher e Ji, is th e pr oportionality constant, ca lled th e coefficient of slo lic fric tion . Be aware that Eq. 6/1 describes only the limiting or maximum value of the static friction force and not any lesser value. Thus , the equation applies only to cases where motion is impen ding with the fr iction force at its peak value. For a condition of static equilibrium when motion is not impending, the sta tic friction force is F < Ji,N

Kinetic Friction After slippage occurs , a condition of kinetic frict ion accompanies the ensuing motion. Kinetic frictio n force is usually somewhat less than the maximum static friction force. The kineti c friction force F" is also proportional to th e norm al force. Thus,

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A rticle 6 /1

where I'-k is the coefficient of kinetic friction . It follows that I'-k is gen erally less than 1'-., . As the velocity of the block increases, the kinetic friction decreases somew hat, and at high velocities, this decrease may be signi ficant. Coefficients of fr ictio n depend greatly on t he exact condition of the surfaces, as well as on the relative velocity, and are subject to considerable uncertainty. Because of the variability of the conditions governing the action of friction, in engineering practice it is frequently difficult to distinguish between a static and a kineti c coefficient, es pecially in the region of transition between impending motion and motion . Well-greased screw th reads under mild loads , for example, often exhibit comparable frictional resistance whethe r they are on the verge of turni ng or whet her they are in motion. In the enginee ring literature we frequentl y find express ions for maximum static friction and for kinetic friction writte n simply as F = J.lN . It is unders tood from th e problem at hand whet her maximum stat ic frictio n or kin eti c friction is described. Alth ough we will freque ntly distingu ish between the static and kineti c coefficients, in ot her cases no distinct ion will be made, an d the friction coefficient will be written simply as 1'-. In those cases you must decide which of th e friction condit ions, maximum sta tic friction for impending motion or kinetic friction , is involved. \Ve emphasize again that many problems involve a static friction force which is less than the maximum value at impending motion , and th er efore un der th ese conditions th e friction relati on Eq . 6/ 1 cannot be used. Figure 6/ lc shows th at rough surfaces are more likely to have larger angles bet ween the reactions and the » -direction than do smoother surfaces. Thus , for a pair of matin g surfaces, a friction coefficient reflects th e roughness, which is a geometric prop erty of the surfaces. Wit h this geometric model of friction, we describe mating surfaces as "smooth" when th e frict ion forces th ey can support are negligibly sma ll. It is mean ingless to speak of a coefficient of friction for a single surface. Friction Angles T he direct ion of the resul tant R in Fig. 6/ 16 measu red from the direction of N is specified by ta n" = F/N. Wh en th e friction force reaches its limitin g static value F max - the angle a reaches a maximum value cbS" Thus, tan ePs =

J.l.~

When slippage is occurring, the angle a has a value the kinetic frictio n force. In like man ner,

tan
=

C ht corresponding to

I'-k

In practice we often see the express ion tan eb = 11. in which the coefficie nt of friction may refer to either the static or the kinetic case , depending on th e partic ular problem . Th e angle
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Dry Friction

331

332

Ch apt e r 6

Fric t ion

Cone of kinet ic fr iction

Figure 6/2

frict ion a ngle for eac h case clearl y defin es the limiting directi on of t he total reaction R bet ween two contacting su rfaces. If motion is impending, R mu st be one element of a right- circular con e of vertex a ngle 2(['s' as shown in Fig. 6/2. If mot ion is not impendi ng, R is within the cone. This cone of vertex angle 2
Fadors Affeding Fridion Further experiment shows that the friction force is essentially independ ent of th e apparen t or projected area of contac t. Th e t rue contac t area is much sma ller than th e projected value , since on ly th e peaks of the contacti ng surface ir regu lari ties support th e load . Even relativ ely small nor mal loads result in high stresses at the se contact points. As th e normal force increases, th e true contact area also increases as the mat eri al un dergoes yielding, crus hing, or tea r ing at th e points of conta ct . A compr eh ensive theory of dry fr iction mu st go beyond t he mecha nical explanat ion present ed her e. For exa mple, there is evidence th at molecu lar attract ion may be an importa nt cause of fricti on under condit ions where th e mating surfaces are in very close contact. Other factors wh ich influence dry friction are th e generat ion of high local temperatures an d adhesion at contact point s, relat ive hardness of mating surfaces, and the presence of thi n surface films of oxide, oil, di rt , or ot her substa nces. Some typical values of coefficients of fr ictio n are given in Tabl e D/l , Appendix D. Th ese values a re only approximate and are subject to considerable var iat ion, depending on the exact conditions pr eva iling. They may be used, however, as typica l examples of t he magni tudes of fr ict ion al effects. To make a reliab le calculation involving fr iction, the a ppropri at e frict ion coefficient should be deter mined by exper ime nts which duplicat e the surface cond it ions of the application as closely as possible. Marwan and Waseem AI-Iraqi

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Articl e 6{3

Types of Fridion Problems

We can now recognize the following th ree typ es of problems encoun te red in applicati ons involving dry friction. The first ste p in solving a frictio n problem is to identify its type. (1 ) In th e first type of problem, th e conditio n of impending motion is known to exist. Here a body which is in equilibrium is on the verge

of slipping, an d the friction force equals the limiti ng static frict ion F max = p., /I. The equa tions of equilibrium will, of course, also hold. (2) In the second type of problem, neither the condition of impending motion nor the conditio n of motion is known to exis t. To determ ine the actual friction conditions, we first ass ume static equilibrium and then solve for th e friction force F necessary for equilibrium. Three outcomes are possible: (0) F < (Fmax = p.,/I): Here the friction force necessary for equi librium can be supported, and th erefore the body is in static equilibrium as assumed. We emphas ize that th e actual friction force F is less tha n the limiti ng value F max given by Eq. 6/ 1 and th at F is determined solely by th e equations of equilibrium. (b) F = (Fmax = p., /I): Since th e friction force F is at its maximum value Fmax' motion impend s, as discussed in problem type (1 ). The ass umption of static equilibrium is valid. (c) F > (Fmax = p., /I): Clearly thi s condition is impossible, because the surfaces cannot support more force than the maximum JLsN. The ass umption of equilibrium is therefore invalid, and motion occurs . Th e friction force F is equal to p."N from Eq. 6/2. (3) In th e third type of problem, relativ e motion is known to exist bet ween th e contacting surfaces, and thus th e kinetic coefficient of friction clear ly applies. For thi s problem type, Eq. 6/2 always gives th e kinetic frict ion force directly.

The foregoing discussion applies to all dry contac t ing surfaces a nd, to a limited exte nt, to moving surfaces which are partially lubricated.

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Dry Fri ction

333

334

Chap ter 6

Friction

Sample Problem 6/1 Determine the maximum angle 8 which the adju stable incline may have with the ho rizontal before the block of mas s m begin s to slip. The coefficient of stat ic friction between the block and the inclined. su rface is ILB •

~ mg, the normal force N , an d the friction force F exerted by the incline on the block. Th e friction force acts in th e direction to oppose the slipping which would occur if no friction were prese nt. Equ ilib rium in the x- and y -directions requires

Solution. The free -body diagr am of the block shows its weight W

OJ

mgsin8 - F =0

F =mg sin 8

= OJ

- mg cos 8 + N = 0

N =mgcos8

[ ~F% = [~F,

Dividing t he first equation by t he second gives FIN = ta n 8. Since the maximum angle occurs whe n F = F max = J.LBN, for impending motion we have or

8Jn1llt = tan " ! J.L,

Ans.

y

\ W= mg

~~N Helpful Hints

CD We choose referen ce axes alon g a nd norm al to t he dir ect ion of F to avoid resolving both F an d N into compo nen ts.

(%) Thi s problem desc ribes a very simple way to det ermine a static coefficient of fr iction. Th e maxim um value of tI is known as the angle of repose.

Sample Problem 6 /2 Deter mine the range of values which the mass mo may have so that the lOO-kg block shown in the figure will ne it her start moving up the plane nor slip down t he plane. The coefficient of static friction for th e contact surfaces is 0.30 .

y

~ \98 1 N

~ T=moe

Solution_ The maximum value of m o will be given by the requirement for mo-

"",, :::J-':"

t ion impe nding up the plane. The friction force on th e block th er efore acts down t he plane, as shown in the free-body diagram of th e block for Case I in the figure. With th e weight mg = 100(9.81) = 981 N, the equations of equ ilibriu m give [~F, =

[Fm=

OJ

= I',NJ

N - 981 cos 200 = 0

N

»> x

20 0

Fmu.

N

Case I

= 922 N

F m ax = 0.30(922) = 277 N mo(9.81) - 277 - 981 sin 20" = 0 mo = 62.4 kg

Ans.

The min imum value of m o is determined when motion is impe ndi ng down the plane. The friction force on t he block will act up the plan e to oppose the tendency to move, as shown in the free -body diagram for Case II. Equilibrium in the xdirection req uires mo(9.81) + 277 - 981 sin 20" ~ 0 mo

~

6.01 kg

Ans.

Thus, mo may have any valu e from 6.01 to 62.4 kg, and the block will remai n at rest . In both cases equilibrium requi res that th e resulta nt of F max and N be concurrent with the 981·N weight an d the tension T.

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Helpful Hint

CD We

see from the resu lts of Sam ple Problem 6/1 th at th e block would slide down the incline with ou t th e restraint of attach ment to "tu since ta n 20" > 0.30. Th us, a value of mo will be required to mai ntain equi librium.

Article 6/3

Dry Friction

3 35

Sample Problem 6 /3 Determine the magnitude and direction of the friction force acting on t he 100-kg block sbown if, first, P = 500 N and, second, P = 100 N. Tbe coefficient of sta t ic friction is 0.20, and the coefficient of kinetic friction is 0.17. The forces are applied with the block initi ally at rest.

p

- - -20'-

Solution. Th er e is no way of telling from the state ment of th e problem whether the block will remain in equilibrium or whether it will begin to slip following the application of P. It is therefore necessary th at we make an assumption, 50 we will tak e t he fricti on force to be up the plan e, as shown by th e solid arrow. From th e Free- body diagram a balan ce of forces in both x- and y-directions gives 0]

P cos 20" + F - 981 sin 20"

01

N - P sin 200 - 981 cos 20" = 0

~

y \

\ 100\9.81) = 981 1' \

\

_ x

P

0 N

Case I. P = 500 N Sub stitution into th e first of th e two equations gives F

~

- 134.3 N

The negat ive sign te lls us that if the block is in equilibr iu m, t he friction force acting on it is in th e direction opposite to th at assu med and therefore is down the plane, as rep rese nted by th e dashed arrow. We cannot reac h a conclus ion on the magn itude of F, however, until we veri fy th at the sur faces are capable of suppor ting 134.3 N of friction force. Th is may be done by subst itu ting P 500 N into the second equation, which gives N = 1093N

The maxim um static friction force which the su rfaces can support is th en Fmax = 0.20(1093 )

~

219 N

Since thi s force is great er than that required for equilibr iu m, we conclude that th e ass u mption of equilibrium was correct . Th e answer is, th en , F = 134.3 N down t he plane

Ans.

Case II. P = 100 N Substitution into th e two equilibrium equ at ions gives N

= 956 N

But the maximum possible sta t ic frict ion force is Fmax

~

0.20(956)

~

Helpful Hint

191.2 N

It follows th at 242 N of friction cannot be su pported . Therefore, equilibrium (j) cannot exist, and we obtain th e correct valu e of th e friction force by using th e kinetic coefficient of frict ion accompanying th e mot ion down th e plan e. Hence, th e answer is

F

~

0.17(956)

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~

162.5 N up th e plan e

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Ans.

CD We

should note that even though

":i:.Fx is no longer equal to zero, equi-

libriu m does exist in t he y-direction , so that ":i:.F.v = O. Therefore, the normal force N is 956 N whether or not the block is in equil ibrium .

336

Chapter 6

Frict ion

Sample Problem 6/4

,

b-

- ',

-r-

T he homogeneou s rectangu lar block of mass In, width b, and height R is placed on th e hori zon ta l sur face and subjected to a horizontal force P which moves t he block along t he surface wit h a constant velocity . Th e coefficient of kinetic friction betwee n the block and t he su rface is J1.k' Determ ine (a) the gr eatest valu e which h may have so that the block will slide without tipping over and (b) th e location of a point C on the bottom face of the block through which the resu ltant of the friction and normal forces acts if h = H / 2.

1

P

H

I

m

h

j y I

I

Solution.


(a) With the block on the verge of tipping, we see th at t he entire

reaction betwee n the plan e an d t he block will necessaril y be at A. The free -body diagram of t he block shows thi s condi tion. Since slipping occu rs, the friction force is th e limiting valu e J1.kN, and th e angl e H becomes H = tan - I J1.k' The resultant of Fk and N passes through a point B through which P mu st also pass, since three coplanar forces in equilibrium are concurrent. Hence , from t he geometry of th e hlock

F, I

I

b/2

tanO =llk = -

h

h

=~ 21',

1/

Ails.

/~

__

l __ N

J!... 2

If h were gr eat er than thi s value, moment equilibrium about A would not be satis fied, and th e block would tip over . Alternatively, we may find h by combini ng the equilibrium requirements for the .r- and y -dir ections with th e moment-equilib rium equation about A. Thus, [ ~Fy

=

OJ

r:O:Fx = 01

N-mg = O F, - P

=0

N = mg P

C/

b 2

An s.

lbl With h = H/2 we see from the free-body diagram for case (b) th at the resultant of Fk and N pa sses through a point C which is a dista nce x to the left of th e vertical centerline th rough G. The angle 0 is still 0 = c/J = t an " ! Ilk as long as the block is slipping. Thu s, from the geometry of the figure we have x H/ 2

- - = tan

0

=

Il k

so

x = I', H/2

Ans.

If we were to replace Il k by the sta tic coefficient Il " then ou r solu tions would describe th e conditions under which the block is (a) on the verge of tippi ng and (b ) on the verge of slipping, both from a rest position.

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/

:~- - - -;; p mg

-

~

F", -il"'t-;:---L_-~ -'.r I

= F, = I',N = I', mg

Ph -mg - =O

I

-i

I /8 1/

l_

I

N

Helpful Hin t s

CD Recall

that the equilibrium equations apply to a body moving wit h a constant velocity (zero acceleration ) ju st as well as to a body at rest .

@ Alternatively, we could equate the moments about G to zero , which would give us FIR /2' - Nx = o. Thus, with F k = IlkN we get x = I'k H / 2.

Art icl e 6/3

Dry Fric t ion

337

Sample Problem 6 /5 The th ree flat blocks are positi oned on the 30° incline as shown, and a force P parallel to t he inclin e is applied to the middle block. The upper block is prevente d from moving by a wire which attaches it to t he fixed support. The coe f'. ficient of static friction for each of the three pairs of mating su rfaces is shown. Determine the maximum value which P may have before any slipping takes place.

CD

Solution. Th e free-body diagram of each block is drawn. Th e friction forces are assigned in t he direct ions to oppose t he relative mot ion which would occur if no frict ion were present. There are two possible conditions for impending motion. Either the 50-kg block slips and t he 40-kg block remains in place, or the 50- and 40·kg blocks move together with slipping occur ring betwe en the 40-kg block and th e incline. The normal forces, which are in t he y-direct ion, may be determ ined without refere nce to the friction forces, which are all in t he .r-direction. Thu s, r ~ Fy ~

01

(30· kg) N,

30(9.81) cos 30"

o

(50· kg) N 2

50(9.81) cos 30"

255

(40·kg) N a

40(9.81) cos 30"

680 = 0 N a

N,

255 N

0 N2

680 N 1019

'" 30<9.8 1I N \

300

F , = 0.30(255) = 76.5 N

F2

~

N 5019.811 N

0.40(680) = 272 N

The assumed equ ilibri um of forces at impending motion for t he 50-kg block gives [~ Fx ~

0]

P - 76.5 - 272

+ 50(9.81) sin 30"

= 0

p /

x/

/

.Y

P = 103.1 N

\

\

We now check on the validity of our initial assu mption. For th e 40·kg block with F 2 == 272 N th e fr iction force F3 would be given by [ ~ Fx

= 0]

272 + 40 (9.81) sin 30" - Fa

~

Fa

0

~

468 N

But th e maximum possible value of F3 is F3 == JLsN 3 == 0.45(1019) == 459 N. Thus , 468 N canno t be supported an d our initial assumpt ion was wrong. We conclude, the refore, that slipping occurs first between the 40-kg block and the incline. With th e cor rected value F 3 = 459 N, equ ilibrium of th e 40-kg block for its impen ding motion requi res

@

[ ~Fx ~ 0]

F 2 + 40(9.8 1) sin 30" - 459 = 0

263

N

Equi lib ri um of the 50. kg hlock gives, finally, [ ~ Fx

= 0]

P

+ 50(9.81) sin 30" - 263 - 76.5

P

~

o Ans.

93.8 N

Thus, with P = 93.8 N, motio n impend s for t he 50·kg and 40-kg blocks as a un it .

Marwan and Waseem AI-Iraqi

T

- - - - '\

We will assume arbitrarily t hat only the 50-kg block slips, so that th e 40-kg block re ma ins in place. Thus, for impend ing slippage at both surfaces of the 50-kg block, we have [Fm ~ ~ IJ.~ ]

.

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/

x/

/

N3

Hel pful Hi nts

CD In the absence of friction

the middle block, under the in fluence of P, would have a greater movement th an t.he 40.k g block. and the friction force F 2 will be in the direction to oppose this mot ion as shown.

mWe see IJ.,N,

~

now that F 2 is less than 272 N.

338

Chapter 6

Friction

PROBLEMS (Note that, fur checking purposes on ly, the computed results for all frictio n coefficients are reported to three significant figures, rega rdless of wheth er th e resu lt begin s wit h the digit I .>

6/4 The coefficien ts of static and kinetic friction between the lOO.kg block and the inclined plane are 0.30 and 0.20, respectively. Determine (0) the friction force F acting on th e block whe n P is applied with a magni tude of 200 N to th e block at rest, (b) the force P requ ired to init iat e motion up the incline from rest . and (c) the friction force F act ing on th e block if P 600 N.

Introductory Pr oblems P

6/1 Th e 85-lb force P is a pplied to th e 200-lb crate, which is stationary before the for ce is applied. Determine th e magnitude and dir ect ion of th e fricti on force F exert ed by th e hori zontal surface on th e crate . An s. F = 85 lb to th e left P ., = 0.30 P. = 0.20

200 lb P = 85 1b

Ilil ;I

is-

Jp = 0.50

Problem 6/4

/ )p; - 0.40 r

Problem 6 /1 6 /2 Th e 700-N force is a pplied to th e 100-kg block, which

is stationary befor e th e force is applied. Determ ine th e magnitude a nd directi on of th e fricti on force F exerted by the hor izontal su rface on th e block.

6/5 The 1.2. kg wooden block is used for level support of the 9-kg ca n of paint. Determine the magnitude and direction of (0 ) th e fr iction force exert ed by th e roof surface on t he wooden block and (b) th e tot al force exert ed by th e roof sur face on th e wooden block. Ans. (0) F = 3 1.6 N , Ib) P = 100.1 N up

P = 700 N

100 kg

l

.,/ 30'

I' = 0.80 = 0.60 ' \

II;

4

r--=----... 12 Problem 6/5

Problem 6/2

6/3 The design er of a ski re sort wishes to have a portion of a beginn er's slope on which th e skier 's speed will remain fairly consta nt. Tests indicate the average coefficients of friction between skis and snow to be 1', = 0.10 a nd 1'. = 0.08. What should be th e slope angle tJ of th e const an t-speed section? Ans. (I = 4.57 0

6/6 Th e magnitude of force P is slowly increased. Does th e homogeneou s box of mass III slip or t ip first? St at e th e value of P wh ich would cau se each occurrence. Neglect any effect of the size of the sma ll feet. P ----,

A ~~

9

c Problem 6 /3

r--.- - - -

2d - -

-

-

Problem 6/6 MalWan and Waseem AI-Iraqi

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- -I

Ar ticle 6 /3

Pr oblems

339 F

6/7 T he light bar is used to suppor t t he 50-kg block in its ver tical gu ides . If the coefficient of stat ic frict ion is 0.30 at the upp er end of t he bar a nd 0.40 at t he lower end of the bar, find th e frict ion force act ing a t each end for x = 75 mm. Also find th e maximu m value of x for wh ich the ba r will not slip. Ans. FA = F n = 126.6 N, Xm a .x = 86.2 m m

50 kg F B

Problem 6 /9

6/10 T he 300·lb cra te wit h mass center a t G is supported on the horizon ta l su rfaces by a skid at A a nd a roller a t B . If a force P 0[60 Ib is requ ired to in it ia te motion of the crate , determ ine the coefficient of sta t ic frictio n at A.

A

-t----- --r i -+ \/ i I 12" AI-- 24"--+- 24T

Problem 6/7

P

G,

6 /8 Th e 3D-kg homogeneous cylinder of 400·mm diam eter re st s aga inst the vertica l an d inclin ed sur faces as shown. If t he coefficient of sta t ic frict ion between the cylind er a nd the su rfaces is 0.30, calculate the a pplied clockwise couple M whi ch wou ld ca use the cylind er to slip.

12"

B

Problem 6/10

6/11 T he illust ration shows the design of a ca m-typ e locking device, which , in th e presence of su fficient friction , limit s the movem ent of body B to be to th e left only; rightwa rd movem e nt is pr evented. Th e su rface of the cam near point D is circula r with center a t C. Given the distan ce L , specify th e ca m offset d so t ha t th e device will work if t he coefficient of st atic fr ict ion fJ...~ is 0.20 or gr ea ter. Ans. d es 0.2L

30°

d

Problem 6/8 6/9 The to ngs are design ed to ha ndle hot ste el tu bes wh ich a re being heat-treated in a n oil bath . For a 20 0 jaw open ing, wha t is the min imum coefficient of sta tic friction between the jaws a nd the tube which will ena ble th e to ngs to gri p t he t ube wit hout slipping? A ns. fJ..:; m in ::::. 0.176

~.

;/,' 0. L

0

I I A

/ Y I Allowable No / <, I / motion motion ~ "B , ,;-- - - = -::-"'''-- - - - . D

---

Problem 6 /11

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340

Chapter 6

Friction

6/12 Th e uniform 14-fi pole weighs 150 lb a nd is support ed as shown . Calculate th e force P requ ired to move th e pole if t he coefficient of static friction for eac h cont act location is 0.40.

6/14 T he strut AB of negligible mass is hin ged to the horizonta l su rface at A and to th e uniform 25·kg wheel at B. Determ ine the minim um couple M applied to th e whee l which wiJI cause it to slip if t he coefficient of static friction between the wheel a nd th e surface is 0.40.

r

I

I ",,,~~

B

A

I

~ -

I"

-- --

I

6'

1_ _

. --

L

M

I I

~~=_ P

B

Problem 6 /14

1 ~-- 8, ---1

6 }15 The 180·l b man with center of gravity G supports the

75-lb drum as shown. Find the greatest distance x at which th e man can positio n himsel f wit hou t slipping if t he coefficient of static friction betwee n his shoes a nd t he gro u nd is 0.40. Ans. X = 10.52 ft

Problem 6/12

Representative Problems 6/13 Th e uniform pole of len gth I and mass /II is placed again st the su pporti ng surfaces shown. If the coeffldent of sta tic fricti on is fL.~ = 0.25 at both A and B , det ermine the max imum angle fJ at which th e pole ca n be placed before it begin s to slip. A ns. () = 59 .9° 15 '

75 1b

L . . - - - - - - x - - -- - - - ' Problem 6 /15 B

Problem 6/13

6/16 The force P is app lied to (a ) the 60-lb block and (hi the lOO-lb block. For each case, det erm ine th e magnitud e of P required to in itiate motion.

~, = 0.40 ~r-'~~~-_ P ia ) I'. = O. 1 2 ~

--~ P ( b )

Problem 6 /16

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Articl e 61l 6 /17 Th e syste m of two blocks, cable, and fixed pull ey is in itially at rest. Deter min e th e hor izontal force P nec essa ry to cause mot ion whe n (c ) P is applied to th e 5-kg block and (bl P is applied to th e In -kg block. Determine t he corresponding tens ion T in th e cable for eac h case . An B. (0) P = 137.3 N, T = 112.8 N Ib) P ~ 137.3 N, T ~ 24.5 N

Problems

341

6 /20 The lO-kg solid cylinder is rest ing in the incli ned v -block. If the coefficient of stat ic frictio n bet ween t he cylinde r and the block is 0.50, det ermine (0) th e frictio n force F acting on the cylind er at each side befor e force P is applied. a nd (b ) the value of P required to start sliding the cylinder up th e incline.

J..I" = 0.60 <, <,

Problem 6/17 6/18 Deter mine th e range of weights \V for which th e lOO·lb block is in equilibriu m. All whee ls and pulleys ha ve negligible friction.

II = 0.30

JI: = 0.25

, P

!

30 ' Horiz. - - _ 1-

_

Problem 6 /20

I

6/11 The homogeneous se micylinder rests on a horizontal sur face and is subjected to th e force P applied to a cord firmly attached to its per iphery. The force P is slowly increased and kept norm al to the flat su rface of th e semicylinder. If slipping is obser ved just as fJ reaches 40°, det ermine the coefficient of static frict ion J-Ls and th e value of P when slipping occur s. Ans .11.11 = 0.122. P = 0. 166 1m g

10°

Problem 6/18 6/19 The u niform rod wit h center of mass at G is suppor ted by the pegs A and B, which are fixed in th e wheel. If th e coefficient of frict ion betw een the rod a nd pegs is 11 . det ermine t he a ngle 8 t hrough which th e wheel may be slowly tu rn ed abou t its horizo ntal axis through 0 , sta rt ing from t he posit ion shown, before th e rod begins to slip. Neglect the diam eter of t he rod compared with th e ot her dimen sions. A ils.

(J

= tan - I

(

11. 0- +-

II ,

b)

0

-

Problem 6/21

8

Problem 6/19

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342

Cha pt er 6

Fricti on

6/22 The sliding glass door rolls on the two small lower whee ls A an d B . Unde r normal conditions th e upper wheels do not touch their horizontal guide . (a) Cornpute th e force P required to slide th e door at a stea dy speed if wheel A becomes "frozen" and does not tu rn in its bea ring.
6/15 The figure shows the design in sectio n of a loaded bracket which is supported on t he fixed shaft by the ro ller at B and by friction at the corn er A. Th e coefficient of sta tic friction is DAD. Neglect th e weight of th e brac ket and show t hat th e bracket as designed will remain in place. Find th e friction force F. A ns. F = 400 Ib

,

20" 4"

400lb

B p

Problem 6 /25

Problem 6/22 6/23 A clockwise couple Al is appli ed to th e circular cylind er as shown. Det ermine th e value of .\1 required to in itiate motion for th e condit ions m B = 3 kg, me = 6 kg, ( fLlI)n = 0.50, (lIsle = DAD, and r = 0.2 m. Friction betw een th e cylinder C and the block B is negligib le. An s. M ~ 2.94 N r m

6/26 Determine the magnitude P of the horizont al force required to initiat e moti on of the block of mass nlo for th e cases (a) P is appli ed to the r ight and
m

Problem 6 /26

Problem 6/n 6 /14 Repeat Prob. 6/23. except let 11l,lc = 0.20. All ot her conditio ns of Prob. 6/ 23 remain th e sa me.

MalWa n and Waseem AI-Iraqi

B

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Article 6 /l 6/27 Find the maximum distance x from th e hori zontal cente rline of th e drawer at which th e force P may be a pplied an d st ill allow the drawer to be opened wit hou t bin ding at the corne rs. Neglect friction on the bottom of the drawer and take t he coefficient of static frict ion at the cor ners to be J.l.~ . Alls. X =

Problems

343

6/19 The movable left-hand jaw of th e Cvclarnp can be slid along t he frame to increase the capa city of the clamp. To prevent slipping of t he jaw on the fra me when th e cla mp is under load , t he dimension x must exceed a certa in minimum value. For given values of a and b an d a static fr iction coefficient J1.", specify th is design minimu m value of x to prevent slipping of the jaw. Ans. x =

a - bJ1..~

b

-rr-

J

a

1

I

-L

x

p

, L r.--

Problem 6/27

-

----j

I

b

6/28 The two un iform slender bars const ruc ted from th e sa me stock materia l a re freely pin ned toget her a t B. Determin e t he minim um angle fJ at which slipping does not occur at eit her contact point A or C. T he coefficient of static frict ion at both A and C is J.l.~ = 0.50. Consider on ly moti on in th e vertical plan e shown.

Problem 6/29

6/30 Two auto mobiles, both of which have the mass cen ter located as shown midway betw een the fro nt and rea r axles, are identi cal except t hat one is fron t-wheeldrive and the ot her is rear-wheel-dri ve. Th e cars are driven at consta nt speed up ru mps of various inclina tion s. F rom a th eor etical design viewpoint, wh ich car could climb t he ramp of highe r inclination angle o? Justify you r an swer.

c

A

Problem 6[28

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•• •0

Problem 6/30

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344

Chapter 6

Fric tion

6 /31 Determi ne th e distance s to which the 90-kg paint er

ca n climb wit hou t caus ing th e 4-m ladd er to slip at its lower end A. The top of th e I 5-kg ladder has a small roller, and at th e gr ound the coefficient of static fr icti on is 0.25. The mass center of th e painter is directly abov e her feet. A n s. s = 2.55 m

6/34 Th e un ifor m slender rod of mass 111 and lengt h L is initially at rest in a cente red horizontal positi on on the fixed circular surface of radiu s R = 0.6L. If a force P norm al to the bar is gra dua lly applied to its en d u ntil th e bar begin s to slip at the angle (I = 20°, det er mine the coefficient of static friction J.lS '

B

~j

1 1

L/2

<:

IR 1 1 1 I I

+ Problem 6/34 1.5 m --{ Problem 6/31

6 /31 Th e two blocks are placed on the incline with th e cab le taut. (0) Determ ine the force P requi red to initia te moti on of th e I S-kg block if P is applied down th e inclin e. (b) If P is applied up the incline and slowly increase d from zero, det ermine the valu e of P which will cau se motion and describe th at motion.

6/35 The body is construc ted of an alumin um cylinder with an attached half-cylinder of steel. Deter mine th e ramp angle (J for which t he body will remain in equilibrium whe n release d in th e position shown where th e diametral sectio n of th e stee l half-cylinder is ver tical. Also calculate the necessary min imu m coefficient of sta tic friction Ji". A ns. II = 8. 98°, u, = 0.158 ( 40 mm

16 m~. r: 1

9

Problem 6/32 6/33 Repeat Pr ob. 6/ 32, but wit h Jis = 0.50 between th e blocks. All oth er condit ions re main the sa me. A n s . (a ) P = 71.4 N, (b) P = 16 2.0 N

Problem 6/35

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Article 6 /3

6/36 Th e uniform slender rod is slowly lower ed from the upri ght positio n ( Ii = 90°) by mean s of th e cord attached to its up per end and passin g und er t he small fixed pull ey. If th e rod is observed to slip at its lower end whe n 0 = 40°, det ermine th e coefficient of sta tic frict ion at the horizontal surface.

Problems

345

6/18 Th e uniform slender bar of length I is placed in th e opening of width d at the 3D'" a ngle shown. For what range of lld will th e bar remain in static equ ilibrium? Th e coefficient of sta t ic friction at A and B is 1111 0.40 .

T

-,

A

~-Problem 6/36

Problem 6/38

6/37 The three identical roller s are sta cked on a hor izontal su rface as shown. If th e coefficient of stat ic fricti on Il s is th e sa me for all pairs of conta cting surfaces, find th e minimu m value of Il s for which th e roller s will not slip. A ns. 11:; = 0.268

6 /19 Th e single-lever block brak e prevent s rotation of th e flywh eel under a cou nterclockwise to rque .M . Fi nd th e force P required to prevent rot ation if th e coef-

ficient of sta t ic friction is 1111• Explain what would happen if the geometry permitted b to equal Il:;e.

M (b- - e ) rI l l:;

Ails. P = -

Problem 6/37

Problem 6/39

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346

Chap t e r 6

Fric t ion

6/40 A block of mass rna is placed betw een th e vertical wall and the upper end A of the un iform slender bar of mass m. If th e coefficient of static friction is J-L s between the block and the wall and also between the block and the bar , determi ne a general express ion for the minimu m value 0mi n of the angle (J for which the block will rema in in equilibriu m. Evaluat e your expression for the condit ions J-Ls = 0.5 an d m fa ) = 0.1, 1110

(b l m

1, and

1110

(e)

m

10.

B

rna

For each case, state the minim um coefficient of static fr iction ( J-Ls)l1 necessary to preven t slippage at B.

Problem 6/41

6/41 A woma n peda ls her bicycle up a z -percent grade on a slippery roa d at a steady speed. The woman and bicycle have a combined mass of 82 kg with mass center at G. If th e rear wheel is on th e verge of slipping, determine t he coefficient of fricti on P- H between the rear tire and the road . If the coefficient of friction wer e doubled, what would be th e fricti on force F acting on the rear wheel? (Why may we neglect friction u nder th e front wheelv)

B

Prablem 6/40 6 /41 A block of mass rno is placed betw een t he vertical wall and th e small ideal roller at the uppe r end A of the uniform slender bar of mass m. Th e lower end B of the bar rest s on th e horizo ntal su rface. If th e coefficient of sta tic frictio n is MH at B an d also between t he block an d the wall, determ ine a genera l expression for the minimu m value 00lin of (} for which the block will remain in equilibrium. Evaluate your expression

for MH = 0.5 and .!!!.... = 10. For th ese cond ition s.

mo

check for possible slipping at B. A n s . Hmin -_ tan

- 1

Marwan and Waseem AI-Iraqi

(2mo) --, J-Lsm

_ •

0min -

21.8

- --

=;"",= = = 5 100

460

mm 1080 mm -

j

Probl em 6/ 42 6 /43 Th e industrial truck is used to move th e solid 1200~ kg roll of paper up th e 30 0 incline. If th e coefficients of static and kinetic fricti on bet ween the roll and th e vertical barrier of the truck and betw een the roll and th e incline are both 0.40, compute th e req uired tract ive force P between th e tires of t he truck and the horizont al sur face. An". P = 22 .1 kN

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Articl e 6/l

P ro b lem s

347

zont al force to the carr ier ha ndle. Assum e th at each work er supports hal f t he weight of th e pa nel. Ans. J1..~ = 0.126

348

Ch apt e r 6

Fr i ct i on

SECTION B. APPLICATIONS OF FRICTION IN MACHINES In Sectio n B we investigate the action of friction in various machine applications . Because the conditio ns in these applications are normally either limiting stat ic or kinetic friction, we will use the variable J1. (rather than Ils or Ilk ) in general. Depending on whet her motion is impendin g or actually occurring, Il can be interpreted as either the static or kinetic coefficient of friction.

6 /4

m

-

-

\

l-+- p

a

t (u )

\ \

/I"

t

mg

/I"

mg

fb)

(e)

Forces to raise load Figure 6 /3

MalWan and W aseem AI- Iraqi

WEDGES

A wedge is one of th e simplest a nd most u seful machin es. A wedge is used to pr oduce s mall adjust ments in th e position of a body or to a pply large forces. Wedges la rgely depend on fr iction to funct ion. When sliding of a wedge is impending, the resultant force on each sliding su rface of the wedge will be inclined from th e normal to the surface by an amou nt equa l to the fr iction angle. Th e component of the resultant alo ng th e surface is the friction force, which is always in the direction to oppose the motion of the wedge relative to the mating surfaces. Figure 6/ 3a shows a wedge used to position or lift a large mass m, whe re the vert ical load ing is mg. The coefficient of friction for each pair of surfaces is u: = ta n '" and ver ify th at th e wedge is self-locking as long as a < 2
Article 6/5

Range of R 2 20 - a

1-----1

9 -0~rf 9 \\ I

"'-[if-- -r~-I+:-__J

"'-----+-r---'- - x

,I\ 9

9

-O-fi',.\ 29 -

0

Range of H J

(b) Range of R , = R2

Slipping impending at upper surface

(0 )

(c )

for no s lip

Slipping impendi ng at lower surface

Figure 6/4

with the case of raisin g th e load . Th e free-body diagram s and vector polygons for thi s conditio n are sh own in Fig. 6/5. Wedge probl ems lend th emselves to graphical solutio ns as indicated in th e t hree figur es. Th e accuracy of a graphical solut ion is eas ily held withi n tolerances consiste nt with the uncertainty of friction coefficients. Algebraic solut ions may also be obtained from the trigonometry of the equilibr ium polygons.

mg

+ a p Forces to lowe r load

Figure 6/5

6 /5

SCREWS

Screws are used for fast ening and for transmitting power or motion. In each case the friction developed in the threads largely det ermines th e action of the screw. For transmitting power or motion the square thread is more efficie nt than the V-thread , and the analys is here is confined to t he square th read. Marwan and Waseem AI-Iraqi

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Screws

349

350

Cha pt e r 6

Fric t io n

Force Analysis

Consider th e square-t hrea ded j ack, Fig. 6/6, un der t he action of the ax ial load W a nd a moment M applied about the axi s of th e screw. The screw has a lead L (advancement per revolution ) and a mean radius r . The force R exerted by the thread of th e jack fram e on a small representative portion of the screw thread is shown on the free-body diagram of the screw. Similar reactions exist on all segments of the screw thread whe re con tact occurs with the thread of t he base. If M is just sufficient to t urn the screw, the thread of the scre w will s lide around a nd up on th e fixed thread of the frame. The a ngle e made by R with th e normal to the thread is th e angl e of friction, so th at tan '" = 1-'. T he mom en t of R about t he vertical ax is of the scre w is Rr sin (u + 4,), and t he total moment du e to all reac tio ns on the thread s is ":::.Rr s in (« +
= II' sin

(" +
~R

Equ ilibrium of forces in t he axia l directi on fu rther requi res t hat W=

~R

cos (u +
~R

Combining th e expressions for M and W gives M = WI' tan lu

+ 4,)

(6 /3)

To determine the helix ang le « , unwrap the th read of the screw for one compl et e turn and note that u = tan - 1 (/, /2 "1'). \Ve may use the unwrapped thread of the screw as an alternative model to simulate the action of the entire screw, as shown in Fig. 6/7 a . T he equivalent force requi re d to push th e movabl e threa d u p the fixed incline is P = M/ r, and the tri an gle of forc e vecto rs gives Eq . 6/3 immediately.

1\'

";

C

lli:J L L

r

Figure 6/6 MalWan and W aseem A1~lraqi

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.\1

I ~

Article 6/5

p = .\1

I'

R ; --

-

-

2al'

- - --1

fa) To raise load

(b)

(cJ

To lower load fcx < ¢)

To lowe r load (a > 9)

Figure 6/7

Conditions for Unwinding

If the moment hI is removed, the friction force change s direction so that .p is measured to the other side of t he norm al to t he thread . Th e scre w will re ma in in place and be self-locking provided that " < ,P. and will be on the verge of u nwind ing if" = .p. To lower the load by unwinding the screw, we must reverse the direction of M as long as (l < cb. This condition is illustrated in Fig. 6/7 b for our simulated thread on the fixed incline. An equivalent force P = M/I' mu st be appli ed to the thread to pull it down the incline. From the triangle of vectors we therefore obtain the moment required to lower the screw, which is

M

= WI' tan (.p - " l

(6 /30)

If " > .p. the scre w will unwind by itself. a nd Fig. 6/7c shows that the moment required to prevent unwind ing is M = WI' tan (" - .pl

Marwan and Wa see m A l-lraqi

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(6/3 b)

Screws

351

352

Chap t er 6

Frict ion

Sample Problem 6 /6

p

The horizont al positi on of th e 500. kg rectangular block of concr ete is adju st ed by the 5° wedge u nder the action of the force P. If th e coefficient of st atic friction for both pairs of wedge sur faces is 0.30 a nd if th e coefficien t of stat ic frict ion between th e block and the horizontal surface is 0.60, determine the least force P required to move t he block.


Solution. Th e free-body diagrams of th e wedge and the block are dr awn with t he reactions R t - Rb and R:) inclined with respect to th eir normals by th e amo un t of th e frict ion angles for impending motion. Th e friction an gle for lim iting sta tic friction is given by eP = tan I p.. Each of th e two frict ion ang les is computed and shown on the diagram. We start our vector diagram expre ssing the equilibrium of th e block at a conveni ent point A an d draw th e only known vector, the weight W of the block. Next we add R 3 • whose 3 1.0° inclination from th e vert ical is now kn own. The vecto r R.z, whose 16.70° inclinat ion from the hori zont al is also known, must close th e polygon for equilibrium . Thu s, point B on th e lower polygon is determined by th e intersection of the known direc tion s of R 3 and R,l and the ir magn itudes become known. For the wedge we draw R.z, which is now known , and add R I , whose direction is kn own. Th e directions of R I and P int er sect at C, thus giving us th e solutio n for the magnitude of P .

. .. .. "

.

5"

'

.

.

'

.

, .'. . 500 kg··., :

............ ..

. ..

'

..... ' .

p

w = 50019.81) N

----1

a

b

0.30 = 16.70'

¢I = t an " !

Q2=

tan-I 0.60

= 3 1.0°

p

AlgebrDic solution. The simplest choice of reference axes for calculation pu r-

@

poses is, for the block, in t he dir ectio n a-a norm al to R 3 an d, for t he wedge, in th e direction b-b normal to R I • Th e an gle between R 2 an d the a-di rection is 16.70" + 3 1.0" ~ 47.7". Thus, for th e block

B

A

50019.8 1) sin 3 1.0" - R 2 cos 47.7" = 0

R2

~

IV = 4905 N

3750 N

For the wedge th e an gle between R.! and t he b-direct ion is 90° - (2tb1 + 5") = 51.6°, and th e angle between P and the b-direction is cPI + 5° = 21.7°. Thus, [~ Fb

=

0/

3750 cos 5 1.6" - P cos 21.7" = 0

Ans .

Helpful Hints

CD Be certain to note GraphicDI solution. Th e accu racy of a graphical solution is well with in the un cer ta inty of th e friction coefficients and provides a simple and direct resul t. By laying ofT th e vectors to a reasonabl e sca le following the sequence described, we obtain th e magn itudes of P and th e R 's easily by scaling them directly from th e diagr am s.

that the reactions are inclined from their normals in the direction to oppose the motion. Also. we note the equal and opposite reactions R.!.

Q) It shou ld be evident t ha t we avoid simu ltaneous equations by eliminating reference to R j for the block and R I for the we-dge.

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Articl e 6/5

Screws

353

Sample Problem 6 /7 The single-threaded scre w of th e vise has a mean diamet er of 1 in. a nd has 5 square threads per inc h. The coefficient of stat ic frict ion in th e threads is 0.20 . A 60-lb pull a pplied normal to th e handle a t A pr odu ces a clamping force of 1000 lb between th e jaws of th e vise. (a) Det ermine th e fric tiona l moment M B' developed at B , du e to th e thrust of th e scre w aga inst th e body of t he jaw. (b) Det ermine the force Q a pplied normal to the handle at A required to loosen th e vise.

I

I

6" I

""""TTI

8" 10"

A

c

Solution. From t he fre e-body diagr am of the jaw we first obtain t he te nsion T in the screw. I~Mc =

OJ

1000(16 ) - lOT = 0

T

~

1600 lb

1000 lb

The helix a ng le a a nd th e friction an gle c/J for th e thread are given by T

CD

a

=

tan

- 1

L

2w

= tan

- 1

1/ 5

27l'i0.5)

=

3.64

e

c wh er e the mean rad ius of th e thread is r = 0.5 in .

R

(0) To tighten. The isolated screw is simulated by th e free -body diagram show n whe re all of th e forces acting on the th read s of the scr ew are represe nted by a single force R incli ned at the friction angl e cP from the normal to the thread. The moment applied about the screw ax is is 60(8) = 480 lb-in. in the clockwise direction as seen from th e fron t of the vise. The frictional momen t M B due to th e friction force s act ing on th e collar at B is in the cou nterclockwise direction to oppose the imp ending motion . From Eq. 6/ 3 wit h T subst it ute d for W the net mom ent acting on th e scre w is AI

-

r

T

J 60(8 ) = 480 Ib-in.

~:f/~~ a M u

(a) T o

tight en

Tr tan (a +
( b)

Ans.

To loosen

H el p ful H ints

CD Be careful (b) To loosen. Th e free -body diagram of th e screw on the verge of bei ng loose ned is shown with R acting a t th e fri ction a ng le from the normal in th e dir ection to counteract the impending motion. Also shown is t he fri ctional moment M B = 266 lb-in . acting in th e clockwise direction to oppose the motion. T he an gle between R a nd th e scr ew ax is is now eb - 0', a nd we use Eq . 6/3a wit h the net momen t equal to the appli ed moment M ' minu s M n . Thu s M

M ' - 266 M'

Tr tan (
Thus, the force on t he ha nd le req uired to loosen th e vise is

Q

~

AI ' [d

Marwan and Was eem Al-lra qi

~

374/8 = 48.81b

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An s.

to calcu late the helix angle correctly . Its ta ngen t is the lead L (adva ncement pe r revo lution ) divided by the mean circ u mference 2 ttr a nd not by the diam et er 2r.

@ Note t ha t R swings to th e opposit e side of t he nor mal as t he impe nding moti on reverses direction .

354

Chap t er 6

Friction

PROBLEMS (Unless ot herwise instructe d, neglect t he weights of the wedges and screws in the problems which Iollow.t

Introductory Problem s 6/47 If th e coefficient of frict ion between the steel wedge and the moist fiber s of the newly cut st ump is 0.20, determine th e maximum angle a which the wedge may have and not pop out of th e wood aft er being driven by th e sledge. Ans. C.l' = 22.6°

6/49 The au-rom-diamet er scre w has a doubl e square thread wit h a pitch of 12 mm and a lead of 24 mm. Th e screw and its mating t hreads in the fixed block are graphite-lu bricated and have a friction coefficient of 0.15. If a torq ue M = 60 N · m is applied to th e right -hand portion of th e shaft. dete rm ine (0 ) th e force P required to advance the shaft to th e right a nd (b) t he force P which would allow th e shaft to move to the left at a constant speed, A ns. (aJ P ~ 75.3 kN , Ib ) P ~ 8.55 kN

I

I

40 mm

I I

I

I I

Problem 6/49 6/ 50 Th e device shown is used for coarse adjustment of t he heigh t of an exper ime nta l apparatus wit hout a cha nge in its hor izonta l posit ion. Because of the slip. joi nt at A . tur ning th e scre w does not rotate the cylindrical leg above A. Th e mea n diam eter of t he thread is ~ in, and the coefficient of friction is 0.15. For a conservative design which neglects friction at the slipjoint, what should be th e min imum number N of t hr eads per inch to ensure th at t he singleth readed screw does not turn by itse lf u nder t he weight of th e appara tus?

Problem 6/47 6/48 Th e 7° wedge is driven u nder the spring-loaded wheel whose support ing strut C is fixed, Determ ine th e mini mum coefficient of sta tic frictio n J1. s for which th e wedge wiII remain in place. Neglect all friction associa ted with th e whee l.

P

c

Problem 6 /50 Problem 6 /48

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Article 6/5 6/51 T he 10° door stop is inse rte d wit h a rightward horizontal force of 30 lb. If th e coefficient of sta tic friction for a ll su rfaces is J1..~ = 0.20, deter mine the values N u and N L of the normal forces on the upper and lower faces of the doorsto p. With the given information, ca n you deter mine th e force P require d to extract th e door sto p? Ails. N u ~ 53.5 lb, N 1• ~ 50.8 lb

12 m 5 0

T

-

__ Door

-

355

6/53 Determine th e force P req uired to force t he 10° wed ge u nder th e gO-kg u niform crate which rests a gain st the small stop at A . The coefficien t of fri ction for all surfaces is 0.40. Ans. P = 449 N

9\0

~

0.7 m

I \g

100

30 lb

Problems

A

I 100

P

\ Problem 6/53

I I

10: ' - '

Representative Problems Problem 6/51 6 /52 T he bar clam p is bein g used to clamp two boa rd s together whi le the glue between th em cures. Wha t to rq ue M mu st be applied to the ha ndle of the screw in order to pro duce an SO-Ib comp ression between t he boa rds? The ~- i n. -diameter single-thread screw has 12 squa re t hreads per inc h, a nd the effective coefficient of friction is 0.2 . Negle ct any frict ion in th e pivot contact at C. Wha t torque M ' is re qu ired to loosen the clamp?

6/54 Th e coefficient of static fr icti on J.l s bet ween t he 100Ib body a nd t he 15° wedge is 0.20. Determ ine th e magnitude of t he force P req uired to raise the 100·lb body if ta) rollers of negligible friction a re pre sent un der the wedge, as illustrated, a nd (b) t he rollers are removed a nd t he coefficient of stat ic frict ion J1.s = 0.20 applies at thi s sur face as well.

lOO lb

P

150

Problem 6/54 Problem 6/52

Marwan and Waseem AI-Iraqi

6/55 For bot h conditions (a ) and (b) as sta ted in Prob. 6/54, determine th e magnit ude a nd direction of the force p i required to lower t he 100-lb body. A ns. (a) P' ~ 6.45 lb, (b) P' ~ 13.551b

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356

Chapter 6

Frict ion

6 /56 Th e two 50 wedges shown a re used to adjust th e position of the colum n under a vertical load of 5 kN. Determi ne the magnitude of the forces P req uired to ra ise the colu m n if the coefficient of frict ion for all su rfaces is 0.40 . 5 kN

t

-

-

A

i

(

p

6/59 Ca lcu lat e the hor izontal force P on th e light 100 wedge necessa ry to initiate movement of the 40.k g cylinder . Th e coefficie nt of sta tic fricti on for bot h pairs of con tact ing su rfaces is 0.25. Also determine the frictio n force FH a t point B . (Caution: Che ck carefully your ass umpt ion of wh ere slipping occurs.) Ans. P = 98.6 N, Fn = 24.6 N

p

5

I

B 10'

\ Problem 6/56

6 /57 If t he loaded column of P rob . 6/56 is to be lower ed, ca lcu la te th e horizontal forc es P ' required to with d raw th e wedges . Ans. P' = 3.5 1 kN 6/58 The large t u rn buc kle supports a cable tension of 10,000 lb. Th e 1 ~.in . scre ws ha ve a mean diamet er of 1.150 in. an d have five squa re threads per inch . T he coefficient of fr icti on for t he greased t hreads does not exceed 0.2.5. Determi ne the mome nt M applied to the body of the turnbuckle (a) to ti ghten it a nd (b) to loosen it. Both screws have single thread s a nd a re pr evented from t urni ng. T

Problem 6/59 6/60 The threaded collar is used to connect two shafts, both with right-hand thread s on th eir ends . Th e sha fts a re under a ten sion T = 8 kN. If th e threads ha ve a mean diameter of 16 mm a nd a lead of 4 mm , calcu la te th e torqu e M required to turn th e collar in either direction wit h the shafts pr evented from turning. Th e coefficient of friction is 0.24.

T

T

T

M Problem 6/60

Problem 6/58

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Art icl e 6/ 5

6/6\ The colla r A has a force fit on shaft B and is to be removed from the shaft by th e whee l-puller mechanism shown. T he screw has a single squa re thread wit h a mean diam eter of 20 mm and a lead L of 6 mm . If a torque of 24 N· m is required to tu rn wheel C to slip th e collar off th e shaft, determine the average pr essure p (compressive st ress) betwee n the collar and the shaft. Th e coefficient of friction for the screw at E is 0.25, and t ha t for the shaft an d collar is 0.30. Friction a t the ball end D of the shaft is negligible. Ans. p ~ 2400 kPa

P rob lem s

357

6/63 T he vertica l position of t he 100-kg block is adju sted by th e screw-activated wedge. Ca lcu late the moment M wh ich must be applied to the handle of the screw to ra ise th e block. Th e single-threaded screw has squa re threa ds with a mean diam eter of 30 mm a nd adva nces 10 mm for eac h complete turn . Th e coefficient of friction for the screw th reads is 0.25, a nd t he coefficient of fricti on for a ll mating su rfaces of th e block a nd wedge is 0.40 . Neglect frict ion a t t he ba ll joint A . Ans. M ~ 7.30 N vm

Probl em 6/63

B Problem 6/6\ 6/62 The two 4° wedges a re used to posit ion th e vertical colu m n u nde r a load L. What is the least value of t he coefficient of friction J.Lz for th e bott om pair of su rfaces for which the colu mn may be ra ised by applying a single horizont al force P to the upp er wedge? L

t

6/64 Th e bench hold-down clamp is bei ng used to clamp tw o boar ds togethe r while t hey a re bei ng glued. Wha t torque M must be applied to t he screw in ord er to produce a 200-lb compression bet ween th e boards? T he ~-in .-diameter single-t hread screw has 12 square threads pe r inch , and t he coefficient of friction in the thread s may be ta ken to be 0.20. Neg lect an y friction in the small ball contact a t A and ass ume that t he contact force at A is dir ect ed a long th e axi s of t he screw. What torque M ' is req uired to loosen th e cla mp?

p

~

Problem 6/62 Problem 6/64

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"'

t

\\

3"

f

358

Chapt er 6

Fric t ion

.... 6/65 Replace the square thread of the screw jack in Fig. 6/ 6 by a v -thread as indicate d in the figu re accompan ying thi s prob lem and determine the moment AI on th e screw required to rai se t he load W. Th e for ce R acti ng on a represe ntative small section of t he t hrea d is shown with its relevant projections. Th e vector R I is th e projecti on of R in the plan e of t he figu re containing the axis of the scr ew. Th e ana lysis is begu n with an axial force and a moment summation a nd includ es substitut ions for the an gles )' and f3 in te rms of iJ, 0' , and the friction angle eb = ta n ~ 1 p.. The helix a ngle of th e single th read is exagge rated for clarity. tan AilS. M

H

+ J.L

IVr p.

where

H

tan a

\

2 tJ 2 1 + tan - ens 2

Ii

1 + ta n- - cos 2 \

.... 6/66 A scissors-type jack with a single square th rea d which engages t he th readed collar G an d turns in a ba ll th rust bearin g at D is being designed . Th e thread is to have a mean diamet er of 10 mm and a lead (advancement per revolution ) of 3 mm . With a coefficient of frict ion of 0.20 for th e greased threads, (a) calcu late the torque M on t he screw requ ired to raise a load of 1000 kg from the position shown an d (b ) calculate the torque M required to lower the load from th e same positio n. Assume that platfor m AB and line DO will rema in hori zon tal und er load. Neglect frict ion in the bearing at D. Ans. (a) At = 35.7 N -m ,
H

2

a

tan - I ~ 2",. I

L

f

ill = ill = 160 mm , fll = Ci = 80 mm Problem 6/66

Problem 6/65

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Ar t i cl e 6 /6

6 /6

JOURNAL BEARINGS

A jo urnal bearin g is one which gives lat eral support to a shaft in contra st to axial or thrust support. For dry bearings and for many partially lubricat ed bearings we may apply the principles of dry friction. These principles provide a sati sfact ory approximation for design purposes. A dry or partially lubricated journal bearing with contact or near conta ct between th e sha ft and th e bearing is shown in Fig. 6/8, where t he clearance between th e shaft and bearing is greatly exaggerated to clarify t he act ion. As the shaft begins to turn in the directi on shown, it will roll up th e inn er surface of the bearing until it slips. Here it will remain in a more or less fixed position during rotation. The torque ..\1 required to maintain rotation and the radial load L on the sha ft will

Sha ft

Bearing

Figure 6/ 8

cause a reaction R at the contact point A. For vertical equilibrium R must equa l L but will not be collinear with it. Thus, R will be tangent to a sma ll circle of radius r called th e friction circle. Th e a ngle between R and its normal compone nt N is th e fricti on angle sb. Equ ating t he sum of the moment s about A to zero gives

M = Lrf = Lr sin
(6/4)

For a s mall coefficient of fr iction, th e a ngle l' is small, and th e sine a nd tan gent may be int erchanged with only s mall erro r. Sin ce f-I. = tan
(6 /4a)

This relation gives the amount of torque or moment which must be appli ed to th e shaft to overco me friction for a dry or partially lubricated journal bearing. MarNan and W aseem AI-Iraqi

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Journal Bea ri ngs

359

360

Chapte r 6

Fri ction

6 /7

THRUST BEARINGS; DISK FRICTION

Friction betw een circular surfaces under dist ributed normal pressure occurs in pivot bearings, clutch plates, and disk brakes. To examine these applications, we consider the two flat circular disks shown in Fig. 6/ 9. Th eir sha fts are mounted in bearings (not shown) so th at t hey ca n be brought into contac t u nder the axial force P. The maximum torq ue which this clutch can t ransmit is equa l to the to rq ue M required to slip one disk aga ins t the other. If p is the norm al pressure at any location between the plates, the frictional force acting on an elemental area is !1P dA , where !1 is the friction coefficient a nd dA is th e area r dr dt) of t he eleme nt. The moment of th is eleme ntal fricti on force about th e shaft axis is !1P" dA , a nd th e total moment becomes

f

M =

!1P" dA

where we evaluate the integr al over the area of th e disk . To ca rry out this integration, we must know the variation of J-L and p with r, In the following examples we will assume t ha t !1 is cons ta nt. Fu rthermore, if the surfaces are new, flat, and well supported, it is reasonable to ass ume that the pressure p is uniform over the ent ire sur face so th at rrR 2p = P. Su bst itu ting the cons tan t value of p in th e exp ress ion for M gives

M

J1P

=-

1TR

f2. fR

,. 2

2

0

dr dO

=

0

~J1PR

(6/ 5)

We may interpret this result as equivalent to the moment due to a friction force J,LP acting at a distance ~R from the shaft cent er. If the friction disks a re r ings , as in th e collar bearing sho wn in Fig. 6/ lD, the limits of in tegration are the ins ide an d outsi de radi i R, a nd Roo respectively, and the frictional torque becomes M =

(6 /50 )

After the initial wearing-in period is over, the su rfaces retai n their new relative shape and further wear is therefore cons tan t over the surface. This wear depends on both the circumferential distance traveled __ dO IL _

-

-- -

-- ---

--:.-;--.. I

'-1

-*J __ __ li/,. yc:::. R _ IlpdA \ l1 , d,.

_

_

_

_

_

_

_

Figure 6/9 M arwan and W aseem AI-Iraqi

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_

_

_

O'---_~

Article 6 /7

Figure 6 /10

and the pressure p . Si nce the distance traveled is proportional to r, the express ion rp = K may be written, where K is a constant. The value of K is determined from the equilibrium condition for the axial forces, which gives

P =

With P"

K

~

f

2.

P dA = K

J 0

Jil dr d (i 0

= 2 rrKR

P/ (2rrR) , we may write the expression for M as

M

f

lIP" dA

= -tLP2 rrR

J2. JR,. dr d O 0

0

which becomes M

= ~tLPR

(6/ 6)

Th e fricti onal mom ent for worn -in plat es is, the refore, only (~) /(3), or ~ as much as for new surfaces. If the friction disks are rings of inside radi us R, a nd outside radius R o ' subst itution of th ese limits gives for the frictional torque for worn-in surfaces (6 /60)

You should be pr epared to deal wit h other disk-fr iction problems where the pressure p is some other function of r,

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Th ru st Be arings ; Dis k Fric tio n

36 1

362

Chapter 6

Fr ic t ion

Sample Problem 6 /8

T ~--c==;~

Th e bell crank fits over a 100-mm-diameter shaft which is fixed and can not rotat e. The horizontal force T is applied to maintai n equ ilibr ium of th e crank under t he action of the vertical force P = 100 N. Determine the maximum and minimum values which T may have with out causing the crank to rotate in eit her dir ect ion. The coefficient of sta tic friction 1-L between the sha ft and the bearing su rface of th e cra nk is 0.20.

1

120mm

1 t-- -

Solution. Impending rota tion occurs when the reaction R of the fixed shaft on t he bell cra n k makes an angle eJ> = tan - 1 1-L wit h t he norm al to the bearing surface and is, th erefore, ta ngent to the friction circle. Also , equ ilibriu m requ ires t hat th e three forces acting on the crank be concurrent at point C. Th ese facts are shown in th e free-body diagram s for the two cases of impendin g motion. Th e following calcu lat ions are needed : F riction angle

eJ>

=

t an " ! Jl

=

tan " ! 0.20

-

180 mm - - -1 P = 100 N

= 11.3P

Radius of friction circle rf = r sin 4J = 50 sin 11.3 P = 9.8 1 mm I 120 = 33.7" Angle 0 = tan- 180

sin"! rf = sin"! 9.81 OC ) (120)2 + (180)2

Angle fl

~P

2.60"

T,

(0) Impending counterciockwise motion. Th e equil ibrium triangle of forces

(al Counterclockwise motion impends

is drawn an d gives T,

P cot (0 -

T,

T max

=

fl)

= 100 cot (33.7" - 2.60°)

Ans.

165.8 N

(6) Impending clockwise motion. The equilibrium triangle of forces for this case gives

+ fl) = 100 cot (33.7" + 2.60°)

T2

P cot (IJ

T2

T min = 136.2 N

An s.

-

R2~ pP= 100 N

4L-J T,

(bl Clockwise motion impends

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Articl e 6 /7

PROBLEMS Introductory Problems 6 /67 Circu lar disk A is placed on top of disk B a nd is subjec ted to a compressive force of 80 lb. The diameters of A and B are 9 in . a nd 12 in ., respect ively, and t he pressure under eac h disk is constant over its surface. If the coefficient of friction between A and B is 0.40, determine the coup le M whic h will cau se A to slip on B. Also, what is the minimum coefficient of friction Il bet ween B and the supporting surface C which will prevent B from rotating? Ans. 1.\ [ 96.0 Ib-in. I' = 0.300

Pr oblems

363

6 /69 A torqueM of15 10 N · m must be applied tothe 50-mmdiam et er shaft of t he hoisting drum to ra ise t he 500- kg load a t cons ta nt speed . Th e drum a nd shaft toget he r hav e a mass of 100 kg. Ca lculat e the coeffi cient of fric tio n IL for t he bear ing . A ns. IL = 0.271

80 lb

~M

500 kg

Problem 6/69 6/70 Det erm ine t he tension T in th e cable to raise t he BOO-kg load if t he coefficient of friction for the 30-m m bearing is 0.25. Also find th e ten sion To in th e stati onary sect ion of the cab le. Th e mass of t he cab le a nd pull ey is sma ll a nd may be neglect ed .

T

300 mOl

Problem 6/67 6/68 Th e two flywh eels a re mounted on a com mon sha ft which is su pported by a jou rnal beari ng between t he m. Each flywheel has a mas s of 40 kg, a nd the diameter of the shaft is 40 m m. If a 3-N · m cou ple At on the shaft is required to mainta in rota tion of th e flywhee ls and sha ft at a constant low speed , comp ute (0) th e coefficient of frict ion in the bearing a nd (b> th e radius rf of th e friction circl e. 800 kg

Problem 6/70 6 /71 Calculate the tension T required to lower th e BOO-kg load described in Prob. 6/ 70. Also find To. Ans. T = 3830 N. To ~ 4020 N

Problem 6 /68

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3 64

Ch a p t e r 6

Fr ic tion

6/72 Th e 20-kg stee l ring A with inside a nd outs ide radi i of 50 mm and 60 mm , respect ively, rest s on a fixed hor izont al shaft of 40-mm ra dius . If a downwar d force P = 150 N a pplied to the periphery of th e ring is just sufficient to ca use th e rin g to slip, calculate t he coefficient of friction Ii and th e angle fl.

6/75 Th e axial section of th e two mating circu lar disks is shown . Der ive th e expressio n for the torque M requi red to turn the up per disk on th e fixed lower one if th e pressure p betwee n th e dis ks follows the relat ion p = k /r 2 , wher e k is a consta nt to be determin ed. Th e coefficient of frictio n Ji is consta nt over t he entire surface.

L

elM

~

=1lt1

Problem 6/72

Fixed

Representative Problems

I

6/73 Th e weight of th e dru m D a nd its cabl e is 100 lb, and the coefficient of frictio n Ji for th e beari ng is 0.20. Det ermi ne the for ce P required to ra ise the BO-Ib cylinder if the bearing frict ion is (0) neglected and Ib) included in the analysis. The weight of the shaft is negligible. 50 Ib, Ibl P ~ 52.9 1b Ails. (a ) P

Problem 6/75 6/76 The fro nt wheels of a n experimental rear- drive vehicle ha ve a rad ius of 300 10m and are designed with disk-type brakes consist ing of a ring A with ou tside and inside radii of 150 mm and 75 mm . respecti vely. Th e ring, which does not turn wit h the whee l, is forced aga inst the wheel disk with a force P . If the pressure between th e ring and th e wheel disk is u niform over th e mat ing surfaces . compu te the frict ion force F betwee n each fron t tire and th e horizontal roa d for an axial force P = 1 kN whe n the vehicle is powered at constant speed with the wheels turning. The coefficient of friction between th e dis k and ring is 0.35.

80 1b Problem 6/7:J 6 /74 Det ermine the force P required to lower th e BO-l b cylinder of P rob. 6/73. Compare your answer ,·vit h t he stated res ults of that problem . Is the no-friction value of P equal to the average of the forces requ ired to raise and lower the cylinder?

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Problem 6 /76

Articl e 6 / 7

6/77 Th e telephone-cable ree l has a mass of 250 kg and is moun ted on an SO-mm-diam eter shaft. If the coefficient of frictio n betwe en the shaft and its bea ring is 0.30, calcu lat e th e horizontal tension T requ ired to turn th e reel. Ans . T = 56.4 N

Probl em s

365

6/80 An automobi le disk brake consist s of a flat-faced rotor and caliper which contains a disk pad on each side of the rotor. For equal forces P behind the two pads wit h the pressure p uniform over th e pad . show that the moment appl ied to th e hub is independent of th e angular span fJ of th e pads. Would pressure variat ion with (I chan ge th e moment?

,. -;::::P~::---- _ T

/ ,

'

I/ soomml

,

BOm m

I

/

I \ \ \

,, Problem 6 /80

Problem 6/77

6/78 Th e pulley system shown is use d to hoist th e 200-kg block. Th e diameter of the bear ing for the upp er pul ley is 20 mm, and tha t for th e lower pulley is 12 mm. For a coefficient of frict ion p. = 0.25 for both bearings, calcu late th e te nsions T, T I , an d T'2 in the th ree cables if the block is being raised slowly.

6{81 In a design test on friction , shaft A is fitt ed loosely in the wrist -pin bea rin g of the connect ing rod with center of gravity at G as shown. Wit h th e rod initiall y in the vertical position, th e shaft is rotated slowly unti l the rod slips at the angl e «. Write an exact expression for th e coefficient of frictio n u , Ans .

\

II

,I

Problem 6 /81

6/79 Calcu late th e ten sions T . T I • and T2 for Prob. 6/78 if the block is bein g lowered slowly. Ails. T = 899 N. T , = 949 N, T, = 10 13 N MarNan and W aseem AI-Iraqi

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=

,====;;== \I(r

d /2 ) ' - 1 sin tr

Vertical

Problem 6/78

J.l

366

Chapter 6

Friction

6/81 For th e nat sa nding disk of radius a , th e pr essure P developed between th e disk a nd th e sa nded sur face decr eases lin early with r from a value Po at th e cente r to Po/2 a t r = a . If th e coefficient of fri ction is u , derive th e express ion for the torque M req uired to turn the shaft under a n ax ia l force L .

,

I----a

P $b-' I

I

6/84 The Iu-Mg cra te is lowered int o an undergr ou nd stora ge facility on a t wo-scr ew e levato r design ed as shown . Each screw has a mass of 0.9 Mg, is 120 mm in mean diam eter, an d has a sing le squa re th read with a lead of 11 mm . The scre ws a re t u rn ed in synchron ism by a moto r unit in the base of t he facility. The entire mass of t he crate, scre ws, a nd 3·Mg e leva tor pla tform is su pported equ ally by na t collar bearings at A. eac h of which has a n outsi de dia meter of 250 mm a nd an ins ide diam ete r of 125 mm . The pressure on the bea rings is assumed to be un ifor m over the bearing surface. If the coefficient of fr ictio n for the collar bea ri ng a nd th e scre ws a t B is 0.15, calcu late the torque M which mu st be a pplied to eac h scre w (a) to raise the elevato r a nd
""

-1 P

2 0

r

A

A

"

l

Problem 6 /82

6/81 Ea ch of the four wheels of th e vehicle weigh s 40 Ib and is mou nted on a 4-in .-diameter jou rnal (shaft ). The tota l weight of th e vehicle is 960 lb, includin g whee ls , and is distributed equa lly on a ll four whe els. If a force P = 16 Ib is required to keep the vehicle rolling at a cons tant low speed on a hori zonta l su rface, calcu late th e coefficient of friction whic h exist s in the whe e l bearin gs. (Hint: Dra w a com plete Freebody diagram of one wheel.) A ns. J1. = 0.204

j 250 mm IOMg

1·.·1 Ji B

B

I Detail of collar bea ring atA

Problem 6/84

Problem 6 /83

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Article 6/7 6/85 Th e hemispher ical thrust beari ng on t he end of the shaft supports an axial force P. Der ive the express ion for th e momen t M required to rot ate th e shaft at constant speed if the pressu re p is proportional to sin a and the coefficient of frictio n is u , rA ns. M = ppr

367

6/86 Determine the expression for the torque M requ ired to tur n the sha ft whose t hrust L is su pporte d by a conical pivot bearing. Th e coefficient of frictio n is Il, an d t he bearing pre ssure is consta nt.

Problem 6/86 p

Problem 6/85

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Probl ems

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3 68

Chapte r 6

Fri ction

6 /8

(u)

n

r i ar

I I

_ ~\ I -- 1, T-1

C ' dN ) de dN

2

/'

I I

BELTS

The impending slippage of flexible cables, belts, and rop es over sheaves and drums is importa nt in th e design of belt dri ves of all types , ba nd brakes , a nd hoisting r igs. Figure 6/ 11a shows a d rum subjected to the two belt ten sions T 1 and T 2 • the torque M necessary to prevent rotation, and a bearing react ion R. Wit h M in the direction shown, Tz is greater than T I ' The free-body diagram of an eleme nt of the belt of length r de is shown in pa rt b of th e figure. We analyze t he forces acting on this differential element by establishing the equilibri um of the element, in a manner similar to that used for other variable-force problems. The tension increases from T at the angle II to T + dT at t he angle e + d e. The normal force is a differential dN , since it acts on a different ial element of area. Likewise the friction force, which must act on the belt in a directi on to oppose slipping, is a differential and is p. dN for impe nding motion . Equilibr ium in the I-direc tion gives

de

de

FLEXIBLE

T cos

2

do

'2

+ u: dN p.dN

or

(T + d'I' ) cos

=

do

'2

= dT

Ib )

Figure 6/11

since the cosine of a differentia l quantity is unity in t he limit. Equilibrium in the n-direction requires that

' do . do dN = ( T + (i T ) sm '2 + T s m '2 dN =Tdll

or

where we hav e used t he facts t hat the sine of a differential angle in th e limit equals the ang le a nd th at the produ ct of two differen tials mu st be neglected in the lim it compa red wit h the firs t-o rder differentials remaining. Combining the two equilibrium relations gives

dT

-

T

= p.d e

Int egr ating between correspon ding limit s yields T2

f

T)

or

dT T

-

In

=

fPu: do

T: =

0

T

p.{3

where t he In (Tz/ T 1 ) is a natural logari thm (base e ), Solving for T z gives

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Articl e 6 /9

Ro lling Resis tance

Not e that {3 is t he tota l an gle of belt contact an d mu st be expressed in radians. If a rope were wrapped around a drum n times, the angle /3 would be 27Tn rad ian s. Equat ion 6/7 holds equa lly well for a no ncircu lar section where the total angle of contact is f3. This conclusion is evident from th e fact t hat the radius I" of the circular drum in Fig. 6/ 11 does not ente r into the equa t ions for the equilibri um of th e differential element of the belt . The relat ion expressed by Eq. 6/7 a lso applies to belt drives where both the belt a nd t he pu lley are rotat ing at constant speed. In this case the equation describ es th e ratio of belt tensions for slippage or impending slippage. When t he speed of rota tion becomes la rge, the belt te nds to leave the rim, so Eq. 6/ 7 involves some error in this case.

6 /9

ROLLING RESISTANCE

Deform ation at t he point of contact bet ween a rolling wheel an d its supporting surface introduces a resistance to rolling, which we mention only briefly. This resistance is not due to tangential friction forces and ther efore is a n entirely different phenomenon from that of dry friction . To describe rolling resistance, we consider the wheel shown in Fig. 6/ 12 u nder the action of a load L on the ax le and a force P applied at its cente r to pr oduce rolling. Th e deform ation of th e wheel a nd support ing surfaces as shown is greatly exaggera te d. Th e distribution of pressure p over the area of contact is similar to the distribution shown. Th e resultant R of t his distribution acts at some point A an d mu st pass through the wheel center for the wheel to be in equilibrium. We find the force P necessary to maintain rolling at consta nt speed by equating the moments of all forces about A to zero. This gives us

r

Figure 6/12

where the moment a rm of P is taken to be 1". Th e rat io p.,. = a / I" is called the coefficient of rolling resistance . This coefficient is the ratio of resisting force to normal force and thus is analogous to the coefficient of static or kinetic friction. On the other hand, there is no slippage or impending slippage in t he inter pr etat ion of P.r' Because the dimension a depends on many factors which are difficult to quantify, a comprehensive theory of rollin g resistance is not available. Th e distan ce a is a function of t he elast ic a nd plasti c pr oper ties of th e mating mater ials, the ra dius of t he whee l, th e speed of travel, and the roughness of the surfaces. Some tests indicate that a varies only slightly with wheel ra dius, and t hus a is often taken to be independen t of the rolling radius. Unfortu na te ly, the qua ntity a has also been called the coefficient of rolling friction in some references. However, a has the dimension of length and therefore is not a dimensionless coefficient in the usual sense .

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369

370

Chap ter 6

Fr ict ion

Sample Problem 6 /9 A flexible cab le which supports t he lOa-kg load is passed over a fixed circular dru m and subject ed to a force P to main tain equ ilibrium. The coefficient of sta tic friction JL between the cable and the fixed drum is 0.30. (a) For a = 0, dete rmin e th e max imum and min imu m values which P may have in order not to raise or lower t he load . (b ) For P = 500 N, deter mine th e minimum value which th e angle a may have before th e load begins to slip.

"r I

p

' 0

--~

100 kg

Solution . Im pending slipping of th e cable over th e fixed drum is given by Eq. 6/7, which is T 2 / T , ~ er" .

CD

(a) With a ~ 0 the angle of contact is (3 ~ . /2 rad. For impending u pwar d motion of th e load, T z = Pmwr.' T 1 = 981 N, and we have P max/ 981 =

eO.30 h r/2 )

P m,u = 981(1.602) = 1572 N

For impending downward motion of the load, Tz = 981 Nand T 1 Thu s,

P min

981 /1.602

612 N

{3

~

eO.30 ,

2.25 rad

Ans.

a

981 N (bl P = 500 N

0.30{3 = In(981/ 500 ) = 0.674

or

{3

~

2.25

128.7" - 90°

~

(~6:) ~

H elpful Hin ts

128.7°

38.7°

(a ) a = O

P m in ·

(b) Wit h T 2 = 98 1 Nand T , = P ~ 500 N, Eq. 6/7 gives us 98 1/500 =

981 N

Ans.

CD We are careful to Ans.

note that {3 must be expressed in radians.

@ In our derivation

of Eq. 6/7 be cer tain t.c note that T'l. > T I •

@ As was noted in the derivat ion of Eq. 6/7, the radius of the drum does nut enter into the calculations. It is only the angle of contact and the coefficient of friction which determine the limiting conditions for impending motion of the flexible cable over the cu rved surface.

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Art icl e 6{ 9

PROBLEMS Introductory Problems 6 /87 What is th e minimum coefficient of friction IJ. between t he rope and the fixed shan which will prevent th e un balanced cylinders from moving? A il S. 11 = 0.221

P r obl em s

371

6/89 It is observed th at the two cylinder s will remain in slow steady mot ion as indicat ed in th e drawing. Dete r mine the coefficien t of friction 11 between th e cor d and th e fixed shaft. Ans. J.l = 0.244

.. t

m

10

100Ib

m

Probl em 6 /89 Problem 6 /87

6/88 Deter mine th e force P required to (a ) raise and (b) lower th e 40·k g cylinder at a slow steady speed. The coefficient of fr iction between th e cord and its supporting su rface is 0.30.

6/90 A force P = mg j6 is required to lower t he cylinder at a constant slow speed wit h th e cord making l ~ turns around the fixed shaft. Calcu lat e th e coefficient of frictio n Il between the curd and th e shaft.

Problem 6/90 40 kg

Problem 6/88

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Chap te r 6

Fric tion

6/91 T he 180-1b rock climber is lowered over t he edge of t he cliff by his two compan ions , who together exert a horizo ntal pull T of 75 Ib on the rope . Compute th e coefficient of fricti on JJ. between th e rope a nd th e rock. A ns. I' = 0.699

6/93 For a certain coefficient of frictio n JJ. a nd a certai n a ngle tr, th e force P required to rai se m is 4 kN a nd that required to lower 111 at a constant slow speed is 1.6 kN . Calculate the mass 111 . A ils. m = 258 kg

T

m

Problem 6/91

Problem 6/93

6 /92 In western movies, cowboys are frequ en tly observed

hitchin g th eir hor ses by cas ua lly winding a few turns of th e rei ns a round a hori zontal pole and letting th e end han g free as shown-no knot s! If th e freely hanging len gth of rein weigh s 2 oz and the number of turns is as shown, what tension T does th e hor se ha ve to produce in the direct ion shown in orde r to ga in freedom? The coefficient of frict ion between the re ins a nd wooden pole is 0.70.

Representative Problems 6 /94 A 50-kg package is attached to a rope whi ch passes

over an irregularly sha ped boulder with uniform su rface tex t ure. If a downw ard force P = 70 N is requ ire d to lower t he packa ge at a consta nt ra te, (a ) determine th e coefficient of frict ion JJ. between th e rope and th e boulder. lb> Wha t force P ' would be required to ra ise the package at a cons tant rate?

t p

50 kg Problem 6 /94

Problem 6 /92

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Article 6/9 6/95 T he 180-lb tree su rgeo n lowe rs himself wit h the rope ove r a hori zontal limb of th e tree . If th e coefficient of frictio n between the rope a nd t he limb is 0.60, com pute the force wh ich the man mu st exert on t he rope to let himself down slowly. Ans. P = 23.7 1b

Problems

373

6/97 A ga rden hose wit h a mass of 1.2 kg/ m is in full conta ct wit h the gro u nd fro m B to C. Wh at is the horl zc ntal component P:r of th e force which th e gardener mu st exert in ord er to pull t he hose around the s ma ll cylindrical gu ard a t B ! The coefficient of fr iction bet wee n th e hose a nd the gro u nd is 0.50, a nd th a t betwee n th e hose a nd th e cylinder is 0.40 . Assume t hat t he hose does not to uch th e gro un d between A an d B. Ans . P, = 55.2 N

Problem 6/95

6 /96 Magn etic ta pe passes around th e light idler pu lleys B a nd over t he fixed circu lar reco rd ing head A with a consta nt speed. Th e ta pe ten sion is u ncha nged as it passes a round th e idle r pulleys. Ca lcu la te the minimum spacing a in the design of thi s u nit for which th e ra t io of the tensions T I a nd T 2 will nut exceed 1.15. T he coefficient of fr iction bet ween t he tape a nd the head is 0.10.

Problem 6 /97 6 /98 Ca lcula te th e hor izonta l force P req ui red to raise th e

l OO·kg load. T he coefficient of fr ict ion bet ween th e rope a nd the fixed bar s is 0.40 .

Problem 6/96

100 kg

Problem 6/98

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Chapter 6

Fri ction

6 /99 Deter mine t he range of cylinder weights lV for which

th e syste m is in equilibrium. The coefficient of frictio n between the 100-lb block and the incline is 0.20 and tha t between the cord and cylindr ical support sur face is 0.30. A il s . 8.66

s IV s 94.3 Ib

tion to th e corn er turns, are required for a 75-kg man to lower himself at a constant rat e wit hout exerting a pull on the free end of th e rope, calcula te th e coefficient of friction ts bet ween the rope and th e contact sur faces of the device. Neglect th e small helix angle of th e rope arou nd the shaft . Ans. J-l = 0.195 L

L

t

20'

IV

Problem 6 /99 6/100 The u nifor m 3-m bea m is suspended by the cable which passes over th e lar ge pulley. A lockin g pin at A prevents rotation of the pulley. If t he coefficient of frict ion betwee n th e ca ble and th e pull ey is 0.25, deter min e t he min imu m value of x for which th e cable will not slip on the pulley.

L

Problem 6/101

1m

6/101 The cylinder of mass 111 is att ached to th e rin g A. which is suspended by t he cable th at passes over th e pulley, as shown in part a of th e figure. A coupleJf applied to th e pulley turns it u nt il slipping of th e cable on th e pulley occu rs at th e position tI = 20", shown in part b of t he figure. Calculate the coefficient of frict ion J-l betwe en the cable and t he pulley.

.-.\'

3m

Problem 6/ 100 6/101 A device designed for lowering a person in a sling down a ro pe at a constant contro lled rat e is shown in the figur e. Th e rope passes around a central shaft fixed to th e fram e and leads freely ou t of the lower collar. The nu mber of turn s is adju sted by tu rn ing t he lower collar. wh ich winds or unwinds th e rope arou nd th e shaft. Ent ra nce of t he ro pe int o the upper colla r at A is equ ivalen t to of a tu rn, and passage arou nd t he corner at B is also equivale nt to ~ of a turn . Fricti on of th e rope through th e st ra ight portions of the collars avera ges 10 N for each collar. If three complete turns around th e shaft, in addi-

A

m

m

1

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Problem 6/1 02

Art icl e 6 /9

6/103 For th e design of th e ba nd br a ke shown, find t he couple M requ ired to turn the pipe in t he V· block against th e action of the flexible ba nd. A force P = 25 Ib is ap plied to th e lever , whi ch is pivoted about O. The coe fficie nt of Fricti on between th e band a nd the pipe is 0.30, a nd that between th e pipe and th e block is DA D. T he we ights of t he pa rts are negligible. Ail s. M ~ 1834 lb-in .

Problems

375

6 /105 Shown in th e figu re is th e design of a band-type oilfilter wrench . If the coefficient of frict ion betw een

the band a nd t he fixed filte r is 0.25, det erm ine t he mini mum value of h whi ch ensu res tha t th e wren ch will not slip on t he filter , regardless of th e magn itude of the force P. Neglect the mass of the wren ch a nd assume that th e effect of the sma ll pa rt a t A is equ ivale nt to tha t of a ba nd wrap which begins a t the th ree-o'clock position and ru ns clockwise. Ans. h = 27 .8 mm p

~1• ~.~!!!'!!~ + '-~ --~~'+i - --kJ; A 8

1 ~ ~I ! ~ 7 mrn -

I

---J

1

-

26

-

-

1.

mm Problem 6{103

Problem 6 /105

6/104 Replace the flat belt a nd pu lley of Fig. 6{1l by a Vbelt and mat ching grooved pulley as indicated by t he cro ss-sect iona l view accompanying t h is probl em . Derive t he re lat ion a mo ng t he bel t t ensio ns, the angle of co ntact , an d t he coefficient of fri ction for t he V-helt wh en slip ping impe nds . A Vvbelt design with a = 35° wou ld be equ ivale nt to increasing the coefficient of frict ion for a flat belt of t he sa me material by what factor n?

.. 6 /106 T he chain has a mas s I ) per un it length. Determine the overhang h below the fixed cylindrical guid e for

which the chain will be on the verge of slipping. T he coefficient of fricti on is J.L. (H i n t: T he result ing differ ential equation involving the va r iable chain ten sion T at th e cor respon ding a ngle (J is of t he form d T /dO + K T = ( Ull, a first -order , linea r, nonhomogeneou s eq uation with constant coefficient. Th e solut ion is

T = Ce- K II + e - K II

J~'I{{(I)

where C a nd K a re cons ta nts. )

-~~~::-c71

V-belt cross section

AilS .

h =

~ (l 1 + J.L2

l

Problem 6 /104

Problem 6/106

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+

e~")

376

Chapter 6

Frict ion

CHAPTER REVIEW In our st udy of friction we have concentrated on dry or Coulomb friction where a simple mechanical model of surface irregularities between th e contact ing bodies, Fig. 6/1, explains th e ph enom enon adequat ely for most engineering purposes. Thi s model helps to visualize the three types of dry-friction problems which are encountered in practice. Th ese problem types are: 1. Static friction of less than th e maximum possible value and det er-

mined by th e equations of equilibrium. (Th is usually req uires a check to see that F < p-,N.l 2. Limiting static friction with impen ding motion (F = p-,N ). 3. Kinetic friction where sliding motion occurs between contacting surfaces (F = P-kN) . Keep in mind th e following when solving dry-friction problems: 1. A coefficient of friction applies to a given pair of mating surfaces. It

is meaningless to speak of a coefficient of frict ion for a single surface. 2. The coefficient of static friction Il s for a given pair of surfaces is usuall y slightly greater than the kinetic coefficient P-k.

3. Th e friction force which acts on a body is always in the direction to oppose the slipping of the body which takes place or th e slipping which would take place in th e absence of friction . 4. When friction forces are distributed over a surface or along a line, we select a representati ve element of the surface or line and evaluate th e force and momen t effects of the elemental fri ction force acti ng on th e element. We th en integrate th ese effects over the entire surface or line. 5. Friction coefficients vary considerably, depending on the exact conditi on of th e mating surfaces. Computing coefficients of friction to three significant figures represents an accuracy which cannot easily be dup licated by experiment. When cited, such values a re included for purposes of computational check only. For design computations in engineering practice, any handbook value for a coefficient of static or kinetic friction must be viewed as an approximation.

Other form s of friction mentioned in t he introductory article of the chapte r ar e important in engineering. Problems which involve fluid friction , for exa mple, ar e am ong the most important of th e frictio n problems encou nte red in engineeri ng, an d are studied in the subject of fluid mechanics.

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Review Problems

REVIEW PROBLEMS

p

6/107 A 100-lb block is placed on a 30° incline a nd released from rest. Th e coefficient of static frict ion between th e block a nd th e inclin e is 0.30. (0) Determine the maximum and minimum values of the initial tension T in the spring for which th e block will not slip when relea sed . (b) Calcu late the frictio n force F on the block of T = 40 lb. Ans . (0 ) T max = 76.0 lb, T illin = 24.0 lb

A }J"

IbJ F = 10 Ib

}J.

377

= 0.25

Prob lem 6/109

6/110 Th e figu re shows a jam cleat design ed for a sailboat whe re t he large fr iction forces develo ped by t he cam s prevent th e rope from slipping. Determine th e force R supported by eac h cam beari ng for the 900N rope tension. The coefficient of static friction between the rope and th e cam s is 0.80.

= 0.30

e

30'

•

Problem 6/107

6/108 Three boxes are placed on the incli ne in contact with each other and released fro m rest. Th e coefficients of static friction u nder boxes A , B, and Care 0.30, 0.20. and 0.35, respectively. Descri be what happens.

"\

80 mm

~-1

r~ 20 mm

Problem 6 /1 10

Problem 6 /108 6 /109 Th e hom ogen eous disk of mass m is resting on the right-angled supporting sur faces shown. The ten sion P in the cord is very gr adually increased from zero . If the frict ion at bot h A an d B is cha racterized by J-ls = 0.25, what happens first-does t he homogeneous disk slip in place or does it begin to roll up the incline'! Determine the valu e of P at wh ich thi s first moveme nt occur s. Ans. P = 0.232mg

6 /111 A frictional locking device allows bar A to move to th e left but prevent s movement to the right. If th e coefficient of friction between th e shoe B and th e bar A is 0.40, specify th e maximum length b of th e link which will permit the device to work as descr ibed . Ail S. b = 96 .9 mm

Allowable mo tion r , - - - --'-'''-''''''--'----'---,

--- A

Problem

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11

378

Chapter 6

Fric tion

6/112 The circula r cylinder weighs 50 Ib an d is held by a cord fixed to its per iphery at B and to th e gr ou nd at A . If the coefficient of stat ic friction is 0.60, cal culat e th e force P required to cau se the cylinder to slip. B

p. ~--

Problem 6/114 6 /115 A compressive force of 600 N is to be app lied to th e two boa rds in the grip of th e Ccclamp. Th e th readed screw has a mean diam eter of 10 mm a nd adva nces 2.5 mm per tu rn . Th e coefficient of stat ic frict ion is

v, = 0 .60 Problem 6 /112

0.20. Determine t he force F which must be ap plied normal to the handle at C in orde r to (0) tight en and (b) loosen th e cla mp. Neglect frict ion at point A. An s. (0) F = 8.52 N, (bJ F ~ 3.56 N

6/113 Calcu late the torque M which th e engine of th e pickup truck mu st supply to th e rear axle to roll th e fron t wheels over th e curbing from a rest position if th e rear wheels do not slip. Deter mine th e minimum effect ive coefficient of friction at th e rear wheels to prevent slipping. Th e mass of th e loaded t ruck wit h mass cente r at G is 1900 kg. Ans . AI = 3.00 kN . m, J.Lmi n = 0.787

100 mm

c

I 130 m m

1600 mm

1200mm

Problem 6/115

Problem 6 /113

6 /116 A 500.kg log is bei ng steadily pull ed up the incline

6 /114 Th e detent mechanism consis ts of th e spr ing-loaded

plun ger wit h a spher ical end, which is designed to position th e horizontal bar by engaging the spaced notches. If the spring exerts a force of 40 N on th e plunger in the posit ion shown and a force P = 60 N is required to move th e detent bar against the plu nger , calculate th e coefficient of friction between th e plu nger an d the deten t. It is known from ea rlier tests that the coefficient of fr iction between the light bar and th e hori zontal sur face is 0.30. Assume that th e plunger is well lubricated and accurate ly fitt ed so that th e friction between it and its guide is negligible.

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by mean s of the cable attached to th e winch on t he truck. If th e coefficient of kinet ic friction is 0.80 between th e log and the incli ne and 0.50 betwee n the cable and rock, det ermine th e ten sion T which mu st be developed by th e winch .

Problem 6 /116

Review Probl ems

6 /117 Th e 4600-lb lathe with mass center at G is posi-

tioned with the aid of th e 5° stee l wedge. Determine t he hor izontal force P requ ired to remove th e wedge if the coefficient of fricti on for all contacting surfaces is 0.30. Also show t ha t no horizon tal movement of th e lat he tak es place. Ans. P = 913 Ib

379

6 /119 Th e small roller of th e un iform slender rod res ts aga inst t he vertic al sur face at A while th e roun ded end at B rests on t he platform which is slowly piv-

oted downward beginning from th e hori zon tal position shown. If th e bar begin s to slip when tJ = 25°, det ermine the coefficient of stat ic fr iction 1411 between th e bar and t he platform . Neglect friction in th e roller and th e s mall thickness of th e platfor m. A ns. 1', = 0.767

A r

5° 2/

- - - - 92" ' - - - - - --1

Problem 6 /117

B

6/118 The movable head of a un iver sa l tes ti ng machine

has a mass of 2.2 Mg and is elevated int o testing position by two 78-mm -diameter lead screws, each with a single thread and a lead of 13 mm . If th e coefficient of frict ion in the t hreads is 0.25, how much torq ue At must be supplied to each screw (a) to raise th e head and
Problem 6 /119

6/120 Determine the ran ge of cylinder mass m for which th e system is in equ ilibrium . Th e coefficient of frict ion betwe en the 50-kg block and th e incline is 0.15 and that between the cord a nd cylindr ical support is 0.25.

A

Problem 6/118

m

Problem 6 /120

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380

Chapter 6

Friction

6 /111 Under t he act ion of the ap plied coup le AI the 25-kg cylinder bears aga ins t the roller A , which is free to tu rn. If th e coefficients of stat ic an d kinet ic frict ion between t he cylinder an d the horizontal surface are 0.50 a nd 0.40, res pect ively, determine th e frict ion force F acting on the cylinder if (0 ) AI = 20 N . m a nd Ibl .II ~ 40 N ·rn. An s. la l F ~ 133 .3 N, Ibl F ~ 127.6 N

B

150 mm

~1

.1/

D

Problem 6 /121

Prablem 6{121 6 /111 Th e cylinder weighs 80 Ib and th e attache d uniform slender ba r has a n un known weight W. T he u nit remai ns in sta t ic equ ilibri u m for values of the angle n ranging up to 45° but slips if Hexceeds 45°. If t he coefficient of stat ic fricti on is kn own to be 0.30 , determine w,

6 /114 Th e element s of a rollin g mill are shown here. In th e design of t he rolle r spacing, determine the maximu m slab thic kne ss b so that the slab will enter the rollers by mean s of friction alone if the coefficient of kinetic fr ict ion is J1.k ' Assu me th a t (b - a) is s mall com pared wit h d.

Problem 6 /122

Problem 6/124

6 /113 Th e jack shown is designed to lin small unit-body ca rs. Th e scre w is threaded int o th e collar pivoted at B , and th e shaft turns in a ball thru st bearing at A. The thread has a mean diameter of 10 mm a nd a lead (adva nce ment per revolution ) of 2 mm . T he coefficient of friction for the threa ds is 0.20. Deter mine the force P normal to the ha ndle at D requi red ( 0 I to rai se a mas s of 500 kg from the positi on shown and (bl to lower t he load from th e sa me positio n. Neglect frict ion in th e pivot a nd bearing at A . An s. (a ) P = 78.6 N, Ibl P = 39.6 N

6 /115 Th e 8-kg block is resting on th e 200 inclined plane with a coefficient of sta tic friction J1. s = 0.50. Dete rmine th e minimum hor izonta l force P which will cause the block to slip. A ns. P ~ 25.3 N

20'

Problem 6 /125 Marwan and W aseem AI-Iraqi

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Rev iew Problems

.... 6 /126 The device shown is designed to prevent clockwise rotation in t he hor izontal plane of th e cent ral wheel by means of fricti onal lockin g of th e two s ma ll rollers . For given values of Rand r and for a common coefficient of friction 1.1. at all con tact surfaces, determine th e ran ge of values ofd for which th e device will opera te as described.

Ans.

2r + (1 - p.2JR < d < (R + 2r ) 1 + p.2

381

'6/128 The 100-kg load is e levated by t he ca ble which slides over the fixed drum with a coefficient of Ir iction of 0.50. The cable is secu red to t he slider A which is pulled slowly a long it s smoot h hori zontal guide bar by th e force P. Plot P as a fun cti on of (/ from (J = 90" to (J = 10" and determine its ma ximum value a long with th e corresponding an gle 0. Check your plotted value of P max a nal yti cally . p = 0.5

- ~I I

I

r

B

p A

Problem 6/126 Problem 6/128

e

'Computer-Oriented Problems

-6 /127 Plot th e force P requ ire d to move th e lOO-kg cr ate

u p t he 10" incli ne sta rting from rest at va rious values of .r from zero to 10 m. Det e rmin e t he least possibl e value of P and the correspo nding val ue of .r. A ns. Pm in = 517 N at x = 7.5 m

-6/129 T he sma ll roller on th e upper end of th e uniform rod rests agai nst the vertical su rface at A whi le t he roun ded end B rests on th e pla tform which is slowly pivoted down wa rd begi nn ing at the horizon tal positio n shown. For a coefficient of sta tic fr iction 1.1.8 = 0.40 at B. deter mine t he angle 0 of the platform at which slipping will occur. Neglect the size a nd friction of th e roller an d the sma ll th ickness of t he platform. An s. tJ = 5.80"

, \

p

3~ \ \

10" Problem 6/127

B

B

Problem 6 /129

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3 82

Chapter 6

Fri ction

*6 / 13 0 The u niform slende r pole rest agai nst a smal l roller a t B. End A will not slip on the horizontal su rface if th e coefficient of stat ic frict ion J-Ls is su fficiently la rge. (a) Determ ine the requi red minimum va lue of J-Ls to prevent slipping for any va lue of 0 fro m o = 0 to (J = 60° an d plot J-Ls versu s e. From these resul ts find th e range of tJ for which t he pole will be u nstable if J-L s = 0.4. (b) At wh at angle (1 is t he pole most un stable, a nd what is th e least coe fficient of stat ic friction J-Ls wh ich wou ld be requ ired. to prevent slipping for this angle?

*6 / 131 A heavy cab le with a ma ss of 12 kg per meter of lengt h passes over the two fixed pipes 300 m a part on t he sa me level. One end supports a 1600· kg cy lind er. By experime nt it is fou nd that a downw ard force P of 60 kN is requi red to indu ce slipping of t he ca ble over bot h pipes at a constant rat e. De-te rmi ne the coefficie nt of kine tic fric tio n J.lk be· tween t he cable an d th e pipes , t he maxi mum tension T in the ca ble between the pipe s, a nd th e sag h in th e cab le.

I

300 mm

I

~

\ 600 kg

Problem 6/132

A

Problem 6/130 ' 6 / 13 1 The semicy lindrica l shell of mass m a nd radiu s r is roll ed through a n a ngle () by th e force P which re ma ins ta nge nt to its per iphery a t A as shown . If P is slowly inc reased, plot the til t an gle 0 as a function of P u p to th e point of slippi ng. Det ermine the tilt angle 0max and the correspondi ng value P max for which slipping occu rs . The coefficient of static friction is 0.30. 0

*6 / 133 The u niform slende r rod of mass m and lengt h L is in itially at rest in a centered hori zontal positio n on t he fixed circu lar surface of radius R = 0.6L. If a force P norm al to t he ba r is gr adu ally a pplied to its end, det ermine th e max imu m equ ilibri u m an gle 0 which the rod ca n reach before slipping ta kes place. T he coefficient of sta tic frictio n bet ween t he rod a nd its suppor t is 0.15. A n s. 0 = 11.04°

59.9 0.295mg

B I 1 1

I I IR I I I I I

A B

+

11,= 0.30

e>

Problem 6/133

Problem 6/131

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L/2

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Review Prob lems -6 / 13 4 Th e device is designed to per mit an adjustment to the horizontal tension T in t he cable passing

around t he two fixed whee ls in order to lower the mass m. If th e coefficien t of friction betw een t he cable and the wheel sur faces is DAD, determi ne and plot th e rati o T/ mg as a function of 0 in the range o < 0 < 90". Also find the value of th e shea r force V in th e adjusting pin at D terms of mg for 0 60". DO= 3r BC = 2r

m

Problem 6/134

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383

The analysis of multi-link structures which change configuration is generally bes t handled by a virtual-work approa ch . This construction platform is a typ ical example. Marwan and W aseem AI-Iraqi

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Chapter

VIRTUAL WORK CHAPTER OUTLINE 7/1 7/2

7/3 7/4

Introduction Work Equilibrium Potential Energy and Stability

Chapter Review

7 /1

INTRODUCTION

In the previous chapters we have analyzed the equilibrium of a body by isolati ng it with a free-body diagram a nd writing th e zeroforce and zero-moment summation equations. This approach is usually employed for a body whose equilibrium posit ion is known or specified and where one or more of the exte rnal forces is an unkn own to be determined. Th ere is a separate class of problems in which bodies are composed of interconnected members which can move relative to each other. Thus various equilibrium configurations are possible and must be examined. For prob lems of t his typ e, th e force- a nd moment-equ ilibrium equations, alt hough valid and adequate, are often not th e most direct and convenient approach. A method ba sed on th e concept of the work don e by a force is mor e dir ect. Also, the method provides a deeper insight into t he beh avior of mechanical systems and enables us to examin e the stability of sys tems in equilibrium. This met hod is called th e method of virtual work.

7/2

WORK

\Ve must first define the term work in its quantitative se nse, in contrast to its common nontechni cal usage. Ma rwan and Waseem AI-Iraqi

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385

386

Chapter 7

F

Vi rtual W ork

A

- - A-' -

~_....L_~_l-_

Work of a FOKe

Consi der the cons ta nt force F acting on the body shown in Fig, 7/ 10 , whose movem en t along the plane from A to A ' is represen ted by the vector cls, called t he dis placemen t of the body. By defin ition th e work U don e by the force F on th e body du ring this displacement is the component of th e force in th e direction of the displacement t imes th e displacement, or

Cal

U

=

(F cos

" j

cls

Ibl

From Fig. 7l Ib we see that the same resu lt is obtained if we multiply th e magnitude of th e force by th e compo nent of th e disp lacement in th e direction of th e force. Thi s gives

Figure 7/1

U = F(cls cos a )

Becau se we obtain th e sa me result regardless of th e direction in which we resolve the vectors, we conclude that work U is a scala r quanti ty. Work is posit ive when the working component of the force is in the same directi on as the displacement. When the working component is in the direction opposite to th e displacement, Fig. 7/2, the work done is negative. Thus,

Figure 7/2

F A' __ e _ ~ -- A2 ...-"'- --

- !!

o ~

, A,

U r + dr

r

o Cal

F

~ \

\

dr

(t.

\ f co;' ~, "'" /

»>

_0,1,

a /

- - -;;..,)

A d.~ COs

""'/ /

Q""' !dt-t

COs Q

=

(F

cos "I cls

=

-

(F cos

Ii)

cls

We now genera lize the definition of work to account for conditi ons und er which the direct ion of th e displacem ent a nd th e magnitude and direct ion of the force are variable. Figu re 7130 shows a force F act ing on a body at a point A whi ch moves along th e path shown from A , to A 2 . Point A is located by its position vector r measured from some arbitrary but convenien t origin O. Th e infinite sima l displacem en t in t he mot ion from A to A ' is given by th e differential cha nge d r of th e position vecto r. Th e work don e by th e force F du ring the displacement d r is defined as

°A ,

( d U = F' dr )

Ibl

(7/0

Figure 7/3

If F denotes the magnitud e of the force F and ds denotes the magnitude of th e differential disp lacem ent d r , we use the defin it ion of t he dot prod uct to obtain dU

=

Fds cos a

\Ve may again interpret this express ion as the force component F cos a in th e directi on of the displacem ent times the displaceme nt, or as th e displacement component ds cos a in the direction of the force times the Marwan and Waseem AI-Iraqi

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Articl e 7/ 2

force, as repr esented in Fig. 7/ 3b. If we express F and d r in terms of their rectangular compone nts, we have dU = (iFx = Fx dx

+

jFy

+

kFz) · (j dx

+j

dy

+

k dz )

+ Fy dy + F, dz

To obtain the total wor k U done by F during a finite moveme nt of point A from A l to A 2 , Fig. 7/ 3a , we integrate dU between th ese posit ions. Thu s,

U

=

f

f

F · dr

sr, dx

+

r; dy

+

r, dz )

or

U ~ f Fcosads To carry out this integration, we must know the relat ion between the force compone nts and their respectiv e coordinat es, or the relations betwe en F and s and betw een cos a and s. In the case of concurrent forces which are applied at any particular point on a body, t he work done by the ir resu ltan t equa ls t he total wor k done by the several forces. This is because the component of the resulta nt in the direction of the displacement equa ls the sum of th e components of the several forces in the same directi on. Work of a Couple

In addition to the wor k done by forces, couples also can do work . In Fig. 7/40 the coup le M acts on the body and cha nges it s angu lar position by an am ount dO. Th e work done by the couple is easily determined from the combined work of the two forces which cons titute the coup le. In part b of the figu re we represen t the couple by two equal and opposite forces F a nd - F acti ng at two arbitrary points A a nd B such that F ~ M/ b. Duri ng the infinitesim al movement in t he pla ne of t he figu re , line AB moves to A"B'. We now take the displacement of A in two ste ps, first , a displacement d rB equa l to that of B and, second, a displacement d rAIB (read as the displacem ent of A wit h respect to B) du e to the rota tio n abo ut B. Thus t he wor k don e by F during t he displacem ent from A to A ' is equal and opposite in sign to t hat due to - F acting through the equa l disp lacem ent from B to B' . We t herefore conclude tha t no work is done by a couple durin g a trans lat ion (movement wit hout rotation ). During the rotation , however, F does work equal to F ·drAIB = Fb di), whe re drA IB ~ b de and whe re dO is the infinitesimal angle of rotation in radians. Since M = Fb, we have

(al

-F

(bl

(7/ 2) Marwan and Waseem AI-Iraqi

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Figure 7/ 4

Wor k

387

388

Ch apt e r 7

Virtual Wo rk

The work of the coupl e is positi ve if M has th e sa me sense as do (clockwise in this illustration), and negative if M has a sense opposite to that of the rotation. The total work of a couple during a finite rotation in its plane becomes u =

f

M do

DimensiDns Df WDrk

Wor k has th e dimen sions of (force) x (distance). In SI un its the un it of work is the joule (J ), whi ch is th e work don e by a force of on e newton moving through a distance of one meter in the direction of the force (J = N · m ). In t he U.S. cus to mary syste m the unit of wor k is the foot-pound (ft-lb), which is t he work done by a one-pound force moving th rough a distan ce of one foot in the direction of the force. Th e dimen sions of the work of a force a nd the momen t of a force are th e sa me although they are enti rely differ ent physical quantities. Note t hat work is a scalar given by the dot product a nd thus involves t he product of a force and a distance, bot h measured along th e sa me line. Moment, on th e other hand, is a vector given by th e cross product and involves the product of force and distance measured at right angles to the force. To distingui sh between these two quantiti es when we write their units, in SI units we use t he jo ule (J) for work an d reserve the combined units newt on -meter (N · rn ) for moment. In the U.S. custo mary sys tem we normally use th e seque nce foot -pou nd Ift-lb) for work and pound-foot (lb-Ft ) for mome nt. Virtual WDrk

\Ve consider now a particle whose static equilibrium position is dete r mined by t he forces which act on it. Any ass umed and arbitrary small displacement Dr away from this natural position and consiste nt with the system constraints is called a virt ual displa cement . The term virtual is used to indicate t hat t he displacemen t does not rea lly exist but on ly is assumed to exist so that we may compare various possible equilibrium positions to determine the correct one. Th e work done by any force F acti ng on the pa rt icle dur ing the virtual displacemen t ~ r is called virlual work and is ~u

=

F ' ~r

or

~u =

F & cos cr

wher e a is th e angle betw een F an d ~r, and & is the magnitude of ~ r. Th e differ enc e between d r and ~r is that dr refer s to a n act ua l infini tesimal change in position and can be integrated, whereas Dr refers to an infinitesimal virtual or assumed movemen t and cannot be integrated. Mathematically both qu antit ies are first-o rder differ entials. A virtua l displacemen t may also be a rotation ~o of a body. Accord ing to Eq. 7/2 th e virtual work done by a coup le M du ring a virtua l angu lar displacem ent ~ H is ~ u = M ~ (J. We may regard the force F or couple M as remaining const ant during any infinitesimal virtual displacement. If we account for any M arwan and W aseem AI-Iraqi

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Articl e 7 /3

Equ ilibrium

change in F or M during th e infinitesimal motion, higher -order terms will resu lt which disappear in the limit. This consideratio n is the sa me math em ati cally as that which permits us to neglect the product dx dy when writing dA = y dx for t he eleme nt of area u nd er the curve y

=

[ix),

7 /3

EQUIL IBRIUM

We now express the equilibr ium conditions in terms of virtual work, firs t for a particle, then for a sing le rigid body, and th en for a sys te m of conn ect ed rigid bodies.

Equilibrium of a Particle Cons ide r the particle or small body in Fig. 7/5 whic h attains an equ ilibri um position as a result of th e forces in t he attach ed spri ngs. If t he mass of th e particl e were sign ificant , th en the weight mg would also be included as one of the for ces. For an ass umed virtual disp lacement 8r of the particle away from it s equi libr ium position, the total virtual work done on the particle is

We now express ~ F in terms of its scalar sums and 8r in terms of its component virtual displaceme nts in the coordinate direction s, as follows: 8U =

+ k ':iF) ·( j Bx + j 8y + k 8zl = ':::.Fx 8x + "i.Fy 8y + ss, liz = 0

~ F '8r =

(i ':iFx

+j

~ Fy

The sum is zero, since ~ F = 0 , which gives ":5:. Fx 0, ~ Fy = 0, and ':i Fz = O. T he equa t ion 8U = a is t he re fore an alte rnative statement of the equilibrium condition s for a particle. This condition of zero virtua l work for equilibrium is both necessary and sufficient, since we may apply it to virtual displacements taken one at a time in each of the three mutuall y perpendicular directions, in which case it becomes equivalent to the three known scalar requirement s for equilibr ium. The principle of zero virtual work for the equilibrium of a single particle usu ally does not simplify this alread y simple problem becau se 8U = 0 and ~ F = 0 provide the same information. However, we introduce t he conce pt of virtual work for a particle so that we can later apply it to sys tems of particles.

Equilibrium of a Rigid Body We can easily extend the pri nciple of vir tua l work for a sing le particl e to a rigid body t re ated as a syste m of small eleme nts or particles rigid ly attached to one anot he r. Becau se the virtual work don e on each particl e of t he body in equilibriu m is zero, it follows t hat the virtual wor k done on the entire r igid body is zero. Only t he virtual work don e by external forces appears in the evaluation of aU = 0 for the entire Marwan and Waseem AI-Iraqi

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Figure 7/5

38 9

390

Ch apter 7

Virtual Work

80

I - --I

R

Figure 7/6

body, s ince all internal forces occur in pairs of equal, opposite , and collinear forces, and the net work done by these forces during any movement is zero. As in the case of a part icle, we aga in find t ha t t he principle of virtual work offers no particu lar advantage to the solution for a single rigid body in equilibr ium . Any ass umed virt ual displ acem ent defined by a lin ear or angular movement will appear in each term in 5U = 0 and when canceled will leave us with the same express ion we would have obtained by using one of the force or moment equations of equilibrium directly. T his cond ition is illust rated in Fig. 7/6, whe re we wa nt to determ ine th e react ion R under t he roll er for th e hinged plate of negligib le weigh t under the actio n of a given force P. A sma ll assumed rotatio n 8/1 of th e plate about 0 is consistent with the hinge constraint at 0 and is taken as the virt ual disp lacement. The wor k done by P is - Pa 88, a nd th e work done by R is +Rb 811. Therefore, th e principle 8U = 0 gives

- Pa 88 + Rb 88 = 0 Canceli ng 80 leaves

Pa - Rb = 0

A

o•

Equilibrium of Ideal Systems of Rigid Bodies

(a ) Active

forces

B

A

o (b )

wh ich is simply th e equation of moment equilibr ium about O. Therefor e, nothing is gained by using the vir tual-wor k principle for a sing le rigid body. T he pr inciple is, however , decided ly advantageous for interconnected bodies , as discussed next.

Reactive forces

(c)

Internal forces

Figure 7/7 Ma rwan and W aseem AI- Iraqi

We now extend th e principle of virtual work to the equilibrium of an interconnected system of rigid bodies. Our treatment here will be limited to so-called ideal systems. These a re systems compose d of tw o or more rigid members linked togeth er by mechanical connections wh ich are incapable of absorbing ene rgy through elongat ion or compress ion, and in which friction is sma ll enough to be neglected . Figu re 7/7a sho ws a simple exa mple of a n ideal sys tem where rel ative motion betw een its two parts is possible and where the equilibrium position is determined by th e applied exte rnal for ces P and F . We ca n ident ify three types of forces which act in such an interconnected syste m. Th ey are as follows: (I ) Active forces are external forces capa ble of doing vir tual work during possible virtua l disp lacem ents. In Fig. 7f7a force s P a nd F a re active forces because they would do work as the links move.

(2) Reactive forces are forces which act at fixed support positions where no virtua l displacement takes place in the direct ion of the force. Reactive forces do no work during a virtual displ acemen t. In Fig. 7/7b the hori zontal force F B exe rted on th e ro ller end of the member by the vertical guide can do no work because there can be no horizontal displacem ent of th e roller. T he reac tive for ce F 0 exerted on the system by the fixed support at 0 also does no work because 0 cannot move. WNW.gigapedia.com

Articl e 7 /1

(3) Internal forces are forces in the connections between members. During any possible movement of the sys tem or its parts, the net ioork done by the inte rnal [orces at the connections is zero. This is so because the interna l forces always exist in pairs of equal and opposite forces, as indicat ed for the int ernal forces FA and - FA at joint A in Fig. 7/ 7c. Th e work of one force there fore necessarily cancels the work of the other force during their identical displacements.

Principle of Virtual Work

Not ing th at only the extern al activ e forces do wor k durin g a ny possible movement of the system, we may now state the principle of virtual work as follows:

The virtual work done by external active forces on an ideal mechanical system in equilibrium is zero for any and all virtual displacements consistent with the constraints. By constraint we mean restriction of the mot ion by the supports. \Ve state the pr inciple math ematically by th e equat ion

(7/ 3) where au sta nds for th e total vir tua l work done on the sys tem by a ll active forces during a virtua l displacement. Only now ca n we see th e real ad vantages of the method of virtua l work. Ther e are essentially two. First , it is not necessary for us to dismember ideal sys tems in order to es tablish the relati ons between the act ive forces, as is genera lly the case with th e equilibr ium meth od based on force and moment summations. Second, we may determine the relations between the active forces directly without reference to the reactive forces . Th ese advantages make the method of virt ua l wor k particularly useful in deter mini ng the positi on of equilibrium of a sys tem under known loads. Thi s type of problem is in cont ras t to th e problem of determining the forces acting on a body whose equilibrium positi on is known . Th e method of virtual work is especially useful for th e purposes mentioned but requires tha t th e int ernal friction forces do negligible work during any virtual displacement. Consequently, if internal friction in a mecha nical sys tem is appreciable, the met hod of virtual work can not be use d for th e sys te m as a whole un less the work don e by intern al friction is included. When usin g th e method of virtua l work , you should dr aw a diagram which isolat es the system under considera tion. Unlike the free -body diagram, where all forces are shown, th e diagram for the method of virtual work need show on ly the active forces, s ince the reactive forces do not en ter into th e a pplication of au = o. Such a drawing will be termed a n actiue-force d iagram. Figu re 7/7 0 is an active- force dia gram for th e sys· tern show n. Marwa n and Waseem AI- Iraqi

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Equilibr ium

391

392

Ch a pt e r 7

Virtual W ork

P,

P,

P,

P,

mg Exampl es of one-degree-of-freedom syste ms

( a)

(h I Exampl es of t wo-degr ee-of-freed om sy s te ms

Figure 7/8

mg

Degrees of Freedom P

N lal

mg

Th e number of degrees of freedom of a mechanical syste m is the number of indep end ent coordina tes nee ded to specify completely th e configu ratio n of the syst em. Figure 7/Sa shows three examples of onedegree-of-freedom sys te ms . Only one coordinat e is needed to establish th e position of every part of t he system. The coordina te ca n be a distan ce or an a ngle. Figu re 7/8b shows th ree examples of two -degre e-of-freedom systems where two independent coordinat es are needed to determin e the configu ration of t he system. By the additio n of more lin ks to the mechanism in the right-hand figure, there is no limit to the number of degrees of freedom which ca n be introduced . The prin ciple of virtual wor k fiU = 0 may be applied as many times as th ere a re degrees of freedom . With each an plicat ion, we allow only one independent coord inat e to change at a tim e while holding the others constant. In our treatment of virtual work in this chapter, we consider only one-degree -of-freedom syste ms:

N tbl

Systems with Friction

When sliding frict ion is present to any a ppreciable degr ee in a mechanical system, the system is said to be "real." In real systems some of th e positive work done on th e system by exte rnal active forces (inpu t work) is dissipated in the form of heat gen erated by t he kinetic frict ion forces during movement of the syste m. When there is sliding between contact ing surfaces, the friction force does negative work because its direction is always opposite to the movement of the body on which it acts. This negativ e work cannot be regained. Th us, th e kineti c friction force J1.kN acting on th e sliding block in Fig. 7/90 does work on th e block du r ing the displacem ent x in t he a mount of - J1.kNx. During a virtual displacement fix , t he friction force does work equal to - J1.kN fix. The sta tic fricti on force act ing on the lei

Figure 7/9 Marwa n and Waseem AI- Iraqi

" For examples or solutions to problems or two or more degrees or freedo m. see Chapter 7 of t he first aut hor's Statics , 2nd Ed ition, 1971 , or S I Version, 1975. WNW.gigapedia.com

Art icle 7 /3

Equilibrium

rolling wheel in Fig. 7/9b , on the ot her hand, does no work if the wheel does not slip as it rolls. In Fig. 7/9c th e moment M f about the center of th e pinn ed joint du e to th e fricti on forces which act at the contacting surfaces does negative work during any relative angul ar movem ent betw een the two parts. Thus, for a virtual displacement lJO betw een t he two parts, which have th e separate virtual disp lacements 801 a nd 802 as shown, the negative work done is - M f 801 - M f 882 = - M / 801 + 882 ) , or simply - M f 8/!. For each part , Mf is in the sense to oppose th e relative motion of rotation. It was noted ea rlier in the article th at a major advantage of th e method of virtual work is in the analysis of an enti re system ofconnected members without taking them apart. If ther e is appreciable kin eti c friction internal to the system, it becomes necessary to dismember th e systern to determine the friction forces. In such case s the method of virtual work finds only limited use.

Mechanical Efficiency Because of energy loss due to friction , the output work of a machine is a lways less than th e input work. The ratio of the two amounts of work is the mechan ical efficiency e. Thus , e =

ou tpu t work input work

The mechanical efficiency of simple machines whi ch have a single degree of freedom and which operate in a un iform mann er may be determined by the method of work by evalua ting the numera tor and den omi nat or of t he express ion for e during a virtual displacement. As a n exa mple, consider the block being moved up the inclined plane in Fig. 7/10. For the virtual displacement 8s shown, the output work is that necessary to elevate the block, or mg 8s sin O. The input work is T 8s = (m g sin /! + IJ.km g cos 0) 8s. The efficiency of the inclined plane is, therefore,

sin 0 8 + IJ.k cos 8l 8s

1

mg se m g (sin

As a second example, cons ider th e scre w jack described in Art. 6/ 5 and shown in Fig. 6/6. Equ ation 6/ 3 gives the moment M required to raise t he load IV, where th e screw has a mean radius r a nd a helix an gle I a, a nd wher e th e fr ictio n angl e is = tan - IJ.k' During a small rotation 80 of the scre w, the input work is M 8/! = IVr 80 tan (a +
e

= ""W;-;r--=-80:-C-ta-n---"'(a-+'---eb77)

tan a

tan

(a

+
As friction is decreased, eb becomes smaller, and the efficiency approaches unity . MarNan and Waseem AI-Iraqi

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Figure 7/10

393

394

Chapt er 7

Vi rtu al Work

Sample Problem 7/1

p

Each of th e two uniform hinged bars has a mass m a nd a length l, and is support ed a nd loaded as shown. For a given force P determine t he a ngle (J for equ ilibr iu m.

Solution. Th e active-force diagram for th e sys te m compo sed of th e two members is shown separa tely and includes th e weight mg of each bar in addition to th e force P. All othe r forces acting externally on th e syste m are reactiv e forces whi ch do no work during a virtual movem ent ax a nd a re therefore not shown. Th e principle of virtual work requires that th e total work of aU exte r na l act ive forces be zero for a ny virtual displacement consiste nt with th e cons trai nts. Thus, for a movem ent at th e virtual wor k become s

Q) [W

= 01

P Ii:<

+

+x -

-

_

2mg Oh = 0

We now express each of these virtual displacements in terms of th e vari able th e required quantity . Hence, . 0 x =21 sIn'2

- --

a nd

(J,

Ii:<

Helpful Hints

@ Similarly,

CD Note carefully that with x positive to I 0 h = -cos 2 2

and

I . (I 5h = - - sm -50 4 2

Substitution into th e equation of virtual wor k gives us I . 0

2mg "4 sm

i! 00

= 0

from which we get

o

2P

2

mg

tan -

or

2P

8 = 2 tan - 1 -

mg

Ans.

To obtain thi s result by th e principles of force an d moment su mma tion. it wou ld be necessary to dismember the fram e and take into accou nt all forces acting on each me mber . Solution by the method of virtual work involves a simpler operat ion.

the right fu: is also positive to t he right in the direction of P, so th at the virtu al work is p ( + lix I. \Vit h " positive down M is also mat hematically positive down in the dir ect ion of mg , so that th e correct mathemat ical expression for t he work is mg( +Mll. Wh en we express fill in te rms of fill in th e next ste p, f~h will have a negative sign. thus br inging ou r mat hemat ical express ion into agreeme nt with th e physical observation th at the weight mg does negat ive work as eac h cen te r of mass moves u pward with an increase in x and o.

@ We obtain M a nd


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Article 7/3

Equilibr ium

395

Sample Problem 7/2 The mass m is brough t to an equilibrium position by the application of the couple M to th e end of one of the two parallel links which are hinged as shown. Th e links have negligible mass, and all frict ion is ass umed. to be absent. Determine the express ion for the equilibrium angle 0 ass u med by the link s with th e vertical for a given value of M . Consid er th e alte rnative of a solution by force and moment equilibriu m.

Solution. The active-force diagr am shows the weight mg acting thr ough the center of mass G and th e coup le M applied to the end of the link. There are no other exter nal active forces or moments which do work on the system during a cha nge in th e angle 9. Th e vert ical position of the center of mass G is designat ed by th e distance h below the fixed hori zontal referen ce line and is h = b cos 0 + c. Th e work done by mg du r ing a movement M in th e directi on of mg is +mg M

=

= mgtr-b sin

l!!!==~ I=""==~

The minu s sign shows th at the work is negative for a positive value of 80. Th e constant c drops ou t since its variat ion is zero. With 0 measured positiv e in the clockwise sense, {j(J is also positi ve clockwise. Thus, the work done by t he clockwise couple M is +M 80. Substitution into the virtua l-work equation gives us [SU = 0]

which yields ,M MJ

= mgb

(I =

_

sin 0 1

l}(I

At

sin " - mgb

Ans.

Inasmuch as sin 8 cannot exceed unity, we see that for equ ilibri um, M is limited to values that do not exceed mgb. The advantage of th e virtual-work solution for t his problem is read ily seen when we obse rve wha t would be involved with a solut ion by force and moment equilibriu m. For the latter approa ch, it would be necessary for us to draw separa te free-body diagr am s of all of t he three moving parts and account for all of th e internal reactions at th e pin connections. To ca rry out these ste ps, it would be necessary for us to includ e in the analysis the horizontal positio n of G wit h respect to the attachment points of t he two links, even thoug h referen ce to this position would fina lly drop out of the equations when they were solved. We conclud e, th en , that th e virtua l-work method in this problem deals directly with cause and effect and avoids reference to irrelevant qu antities.

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Helpful Hin t

CD Again. us in Sa mple P roblem 7/ 1, we are consis te nt mathematically wit h our definition of work. and we see t hat th e a lgebraic sign of t he res ulting expression agrees wit h the physical change.

o

M80 + mg 8h

+h

__ _ .-...,.~ _---'.I. _._

0 SO + 0)

= - mg b sin 088


I

mg 5(b cos 0 + c)

396

Chapter 7

Virtu al Work

Sample Problem 7 /l For link OA in the horizontal position shown, determine the force P on the sliding collar which will prevent OA from rotating under the action of the couple M . Neglect the mass of the moving parts.

1 ---

P

- -- jr-.y~"""I'l

h

I

I y 0$.~:=:,:==:
b

~a

x--1

Solution. The given sketch serves as the active -force diagram for th e system. All ot he r forces are eit her internal or nonwor kin g reactive forces du e to the constraints. We will give the crank OA a small clockwise angular movement 80 as our virtual displacemen t and det ermine the res ulting virtua l work done by M and P. Fr om th e horizontal position of the crank, the angu lar movement gives a downward displaceme nt of A equal to

where MJ is, of course, exp ressed in rad ian s. From the right triangle for which link AB is the constant hypotenuse we may write

-Ox

,--/ B'

I

se o;-,=",=~=-,,----:::=A~I.-;/ /

B

/.

He lpful Hints

CD Note

t ha t the displacement a bf1 of point A \v-ould no longer equal oy if the crank OA were not in a horizonta l positio n .

We now take the differen tial of the equation and get 0 = 2x1ix +2y1!Y

or

@ T he length

b is constant so that = O. Notice th e negative sign , which merely te lls us t ha t if one ()b

Th us,

1ix ~ -~aSO x

cha nge is posit ive, t he other must be nega ti ve.

and the virtual-work equat ion becomes

@

[W ~ 0]

M SO +P Iix ~O

Q) We cou ld just as well use a counterclockwise virtual disp lacement for the crank , which wou ld merely re vers e the sib'TIS of all terms.

o

p = AlI = Mx ya ha

Ans.

Again, we observe that th e virtual-work method produces a direct relationship between the act ive force P and the couple M withou t involving other forces which are irrelevant to thi s relationship. Solution by the force and moment equations of equi librium, although fairly simple in thi s problem, would require accounting for all forces initially and then eliminating the ir relevant ones .

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Art icle 1 /3

PROBLEMS (Assume th at th e negative work of frict ion is negligible in t he following pro blems u nless otherwise indicated .I

Problems

397

7/3 Th e foot-operated lift is used to raise a platform of mass m . Determine th e necessary force P applied at th e 10° angle to support th e 80-kg load . Ans. P = 458 N

Introductory Problems

m

7/1 Th e mas s of the uniform bar of lengt h I is m while tha t of th e un iform bar of lengt h 21 is 2m . For a given force P. det ermine th e an gle 0 for equ ilibrium .

• 75

Ans. H~ 2 lan -1 (4P) mg

mm

1--, -.. , ,, '' ,, '' , ,'

p 100 mm

Problem 7/ 3

Problem 7/1

7/1 Det erm ine the couple AI required to maintain equilibrium at an a ngle 0. Each of the two un iform bar s has ma ss m a nd lengt h t.

7/4 Th e spr ing of constant k is u nstretc hed when tJ = O. Derive an expression for the force P requ ired to deflect t he system to an angle O. Th e ma ss of the bars is negligible.

)0-

B

Problem 1/2

Problem 1/4

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-+- p

398

Chapt e r 7

Virtu al Wor k

7 (5 By means of a ra ck-and-pin ion mechan ism, large

forces can be developed by the cork pull er show n. If t he mean ra dius of the pinion gea rs is 12 mm, determi ne the force R which is exerted on th e cork for given for ces P on the handles. Ans. R ~ 11.67P

7(7 For each unit of moveme n t of t he free end of t he rop e in the direction of the app lied force P, the 250-1b load moves one-fou rth of a uni t. I f the mec han ical efficiency e of the hoist is 0.75, ca lcu late th e force P req uired to rai se the load and t he force p i req uired to lower t he load. Ans. P ~ 83.3 Ib, P' ~ 46 .9 1b

(

f~

70 mm

~ P

\

Problem 7(5

250 lb

7/6 The up per jaw D of the toggle pre ss slides wit h negligible frictional resista nce along the fixed vert ical colum n . Determ ine th e req uired force F on the handle to produ ce a com pr ession R on the roller for a ny given value of H.

Problem 7(7 7/8 Determine t he torq ue M on t he activating lever of t he dump t ruck necessa ry to bala nce t he load of mass m with center of mass a t G when the dump a ngle is H. The polygon ABDe is a para llelogr am .

F

, 1

1 .1 1

I

Problem 7 /8

Problem 7/6

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P

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Articl e 7/3

39 9

F

7/9 T he por table ca r hoist is opera ted by the hydrauli c cylind er which cont rols the horizontal moveme nt of end A of th e lin k in the horizontal slot. Determi ne th e corn pression C in th e piston rod of the cylinder to support th e load P at a height h.

An s. C

P rob lems

~ P J(~~)2 - 1

p

r

300 mm dia.

Problem 7/9 7/10 T he hydraulic cylind er OA a nd link OB a re a rra nged to cont rol t he tilt of t he load which ha s a mass 111 a nd a center of ma ss at G. T he lower corner C is free to roll horizontally as the cylinder link age elonga tes. Determi ne the force P in the cylinder necessa ry to mai nta in equilibr iu m at a give n a ngle 0.

Problem 7/11

Representative Problems 7/12 Th e speed reducer shown is design ed wit h a gea r ratio of 40 :1. Wit h a n input torqu e M 1 = 30 N . m, th e measu red outpu t torque is M 2 = 1180 N om. Determine t he mechan ical efficiency e of the un it.

A

// B /

b

b

.

-~- - - - - - - - - - - . 0

Problem 7/10 7/11 The ha nd -operated hoist is designed to lift a 100-kg load whe re 25 t u rns of the ha ndle on the worm sha ft prod uce one revolution of t he drum . Assu min g a 40perc ent loss of ene rgy du e to frict ion in the mechanism, ca lcu la te t he force F norm al to the ha ndle a r m requ ired to lift t he load . Ans. F ~ 61.3 N

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Problem 7/12

400

Cha pte r 7

Virtual Work

7/11 Th e folding link age is composed of n identical sect ions , eac h of which consists of two identical bars of mass m each. Determine the horizontal force P necessa ry to maintain equ ilibrium in an arbitrary position characte rized by the an gle fl. Does P depend on the number " of sect ions present? A il S. P 2

mg ta n

/I

2' no

7/16 Det erm ine t he force F which t he perso n mu st apply

tangent to th e rim of th e handwheel of a wheelc hai r in order to roll up th e incline of angle 0. Th e cornbined mass of the chair and person is m. Of s is th e displaceme nt of the center of the wheel mea sured along t he inclin e a nd {3 the cor respond ing a ng le in ra dians th roug h whic h th e wheel turns, it is eas ily show n that s = R{3 if the wheel rolls without slipping. )

n

A

B--- p

B

c

Problem 7/13 7/14 Replace t he force P act ing on the linkage of P rob. 7/ 13 by a couple 1.\1. Determine th e momen t AI of the

couple necessa ry to maintain equ ilibrium in an arbitrary positi on cha ra cte rized by the angle fl. Does AI depend on the number '1 of sect ions presen t?

9

Problem 7/ 16 7 /15 Determ ine the couple M which must be applied at 0

in orde r to support the mecha nism in the positi on o = 30°. Th e masses of the dis k at C, bar OA , and bar Be are "to- m, an d 2m, res pectively. 5 Ans. M = ( ~m + m o lgl ,/3

B •

7/17 Specify the hor izontal force F necessary to maintain

equ ilibriu m of the 80-kg platform in term s of th e angle tI mad e by the supporti ng links with t he hor izontal. Each of th e t hree un ifor m link s has a mass of 10 kg. (Compare th e solut ion by virtual wor k with a solutio n by force and moment equi librium.) An". F = 932 cot /I N F

'"

80 kg

"'0

Problem 7/15

Problem 7/17

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Art icl e 7/ 3

P ro bl em s

4 01

7/18 The cargo box of the food-delivery truck for aircraft se rvic ing has a loaded mass m and is elevated by t he application of a to rque M on th e lower en d of th e link whic h is hi nged to the truck frame. T he hor izontal slots allow th e linkage to unfold as the cargo box is eleva ted. Expre ss M as a funct ion of h .

b

b

Problem 7 /20 Problem 7/18

7/19 Each of t he foul' uniform movable ba rs has a ma ss m , a nd their equilibriu m positi on in t he verti cal plane is controlled by the force P applied to th e end of the lower ba r . For a given va lue of P, det ermine the equ ilibri um a ngle tJ. Is i t possible for the equilibri um position shown to be maintained by replacing the force P by a coup le M applied to the end of the lower horizontal ba r'? mg An s. () == tan " ! P' no

7/21 The existing design of a n aircraft cargo loader is under review. The plat form AD of th e loader is elevated to t he pro per height by t he mechanism shown. There a re two se ts of link ages a nd hydrau lic lift s, one set on each side. Cable s F which lift t he platform are controlled by th e hydraulic cylinders E whose piston rod s elevat e t he pulleys G. If t he tot al weigh t of t he pla tform a nd containers is W, determine t he compre ssive force P in each of th e two piston rod s. Does P depend on th e height h? What force Q is support ed by each link a t its cente r joint when W is cente red between A a nd D? Ans . P

~

W, no ; Q

~ ~2( 1 +

2e

/ /' - h' G

t

--€::X = = -::::J;f

t

F

p __

_

Problem 7/19 7/20 Th e portable work plat form is elevated by mea ns of the two hydrauli c cylinders a rt iculated at points C. Each cylinder is under a hydraulic pressu rep a nd ha s a piston area A. Determ ine the pressure p requ ired to su pport th e platform and show t ha t it is indepen dent of o. Th e pla tform, wor ke r, a nd supplies ha ve a combin ed mass m , a nd the ma sses of th e links may be neglected.

A'

I

.-

D

//2 //2

h

,

//2

//2

.

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B

1'1

.

, '

Problem 7/21

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,F

)

402

Chapt er 7

Virtual Work mg

1 /12 A device for cou nt ing t he body radiation of a pat ient is shown. The rad iation coun te r A has a mass m a nd is positioned by turning th e screw of lead L (advancement per revolution ) with a torque AI which cont rols th e dist anc e BC. Relate the torque M to the load mg for given value s of band 0. Neglect all fricti on a nd th e mass of th e linkage comp ared with m .

Problem 7 /23

Problem 7/22 7/13 T he postal scale consist s of a sector of mass "'0 hinged at 0 and with cen ter of mass at G. The pan and vertical link AB have a mass m 1 and are hinged to the sector at B . En d A is hinged to the u niform lin k AG of mass "' 2' which in tu rn is hinged to th e fixed fra me. Th e figu re DRA G forms a para llelogra m, and the angle GOB is a right angle. Determine th e relati on betw een the mass m to be measured and the an gle 0, assuming t hat H = 00 whe n m = o. a Ans. m = [;mO (tan 0 - tan /10 )

7/14 Th e elevation of th e platfor m of mass", supported by th e four ident ical link s is contro lled by the hydr aul ic cylinders AB and AG wh ich are pivoted at point A . Determine the compression P in each of t he cylinde rs required to suppor t th e platform for a specified a ngle o.

Problem 7/24

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Article 7 /3 7/25 The elevation of the load of mass m is controlled by t he adjus ting screw which connects joi nt s A and B. The cha nge in th e distance between A and B for one revolution of th e screw equals t he lead L of the screw (advancement per revolution ). If a moment AIf is required to overcom e frict ion in the thread s and thrust bea ring of th e screw, det ermine th e expression for th e tot al moment M. applied to the adju sting screw, necessary to ra ise th e load.

Ans. M

Problems

403

1/21 Express t he compression C in t he hydraulic cylind er of th e ca r hoist in te rm s of th e an gle o. T he mass of th e hoist is neg ligible comp ared with th e mass 111 of th e vehicle. b An s. C 2T. cos (}

mgL

M f + - - cot tI tr

m

- L-~ Problem 7 /27

7/28 Determine th e force P developed at th e jaws of th e r ivet squeezer of P rob. 4/134 re peated here.

e- -r

Problem 7/25 1 /26 In the design of the claw for th e remote-acti on actuator, a clamping for ce C is developed as a result of the ten sion P in the control rod. Expre ss C in ter ms of P for the configuratio n shown, where the jaw s are parallel.

""

C ".---

•

b $18; 0 '~~:: _;:_=====:Jji.lii~ c

Problem 7(28

Problem 7/26

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404

Chapte r 7

Virtual Work

7/29 Deter mine the force N exerted on t he log by each jaw of t he fireplace tongs shown. Ails. N ~ 1.61' p

7/11 A power- operated loading platform designed for the

back of a truck is shown in th e figu re. T he posit ion of the platform is controlled by t he hydraulic cylinder , which applies force at C. Th e links are pivoted to the truck fram e at A, n, and F . Deter mine the force P supplied by the cylinder in order to su pport the platfor m in th e position shown. T he mass of the platform and link s may be neglected comp ared with that of the 250-kg crat e wit h cente r of mass at G. Ans. P = 3. 5 kN 2fiO 300

mm mm

'-H+il~-l

C

N

l-5>- ,

7>

_1_ ,

7>-

,

- - - 8>-

"

'1/

D

E

Problem 7 /29 7 /10 Th e antitorque wre nch is design ed for use by a crew

member of a spacecraft where no sta ble platform exists agai nst which to push as a bolt is turn ed. The pin A fits into an adjacent hole in th e space st ruc tu re which conta ins th e bolt to be turned. Successive oscillations of th e gear and handle unit turn the socket in one direct ion through th e action of a ra tchet mechani sm. Th e reac tion against th e pin A provides the " a nt itorque " characteris t ic of the tool. For a gr ipping for ce P = 150 N, det erm ine t he torq ue M transmitted to th e bolt.
Problem 7/31

.... 7/32 Determine the force Q at the jaw of the shear for th e 400-N force a pplied with fJ = 30", (H i nt: Replace the 400-N force by a force an d a couple at the center of the small gea r. T he absolute angu lar displacement of th e gea r must be carefully determined.)

~ 7 .5, _ 40 ~--- 1 20 m m ----l P Imml

mm

A ns. Q

I

Problem 7/30

Proble m 7/32

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~

13.18 kN

Article 7 /4

7 /4

Potential Energy and Stab ility

405

POTENTIAL ENERGY AND STABILITY

The previous article treated the equilibrium configuration of mechanical sys tems composed of individual members which we assumed to be per fectly rigid. We now exte nd our meth od to account for mechanical sys tems which include elastic elements in the form of springs. We introdu ce the concept of potenti al energy, whi ch is useful for determini ng the stability of equilibr ium.

Elastic Potential Energy The work done on an elastic member is stored in the member in the form of elastic potential energy Ve. Th is ene rgy is potentially ava ilable to do work on some other body during the relief of its compression or extens ion. Consider a spring, Fig. 7/ 11, whicl is being compressed by a force F. We assume th at the spring is elastic and linear, which mean s that the force F is directly proportiona l to the deflection x. We write this relation as F = IlX, where k is the spring constant or stiffness of the spr ing. Th e work don e on th e spring by F duri ng a movem ent dx is dU = F dx, so tha t th e elast ic potential energy of th e spr ing for a compression x is the tot al work done on the spring

Ve

=

Thus, the poten t ial energy of th e spr ing equals t he trian gu lar area in th e diagram of F versus x from 0 to x. During an increase in the compres sion of the spring from X l to X:b the work done on the spring equals its cluinge in elastic potentia l energy or

=

x,

Ix,

hx dx

=

~iI(xl - x / )

which equals the trapezoidal area from X l to X 2' During a virtua l displaceme nt Iix of the spr ing, the virtua l work done on the spring is the virtual change in elastic potent ial energy bVe = F lit = kx lit

During a decrease in the compress ion of the spring as it is relaxed from x = X t to x = x,, the change (final minu s initia l) in th e pot en tial ene rgy of th e spr ing is nega tiv e. Consequ en tly, if Iix is negative. bVe is also negative. \Vhen we have a spring in tension rather than compression, the work and energy relations are the same as those for compress ion, where x now represen ts the elongation of the spring rath er than its compression. While the spring is bein g stre tched, t he force again acts in th e direction of the displacement , doing positi ve work on the spring and increasing its potenti al energy. Marwan and W aseem AI-I raqi

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l--;,--~umumuu

F- - - - -

- o~ =

h ex

F) ": /

x,

O \,· _-:"-_~_c:'-_--_

o

II

",

Figure 7/11

(7/ 4)

.l Ve

l ~~~\\~l\\\\\\\\\\\

I--- .r ---I I- Ox

J: F dx = J: hx dx

or

~

.r

406

Cha p ter 7

Virt ua l Wor k

Becau se the force act ing on the movabl e end of a spring is th e negat ive of the force exerted by the spr ing on the body to which its movabl e end is attached, t he ioork don e 0 11 the body is the negatioe of the potential energy cha nge of the spri ng. A torsional spring, which resists the rotation of a shaft or another element, can also store and release poten tial energy . If the tor sional stiffness . expressed as torque per radian of tw ist, is a constant K , and if II is t he angle of twist in radian s, then the resisting torque is !vi = K II. Th e potential energy becomes V, = J~ K II d II or

(7 /4al which is ana logous to the expression for the linear extension spring. The units of elastic potential energy are the same as those of work a nd are expressed in joules (J ) in 51 units an d in foot-pounds (ft-lb) in U.S. cust omary units.

Gravitational Potential Energy

T ~"

\\'

-

T

oU = - II' 0" or oV. = +1V0"

+h

Datu m Plan o+ - v" = 0 -

-

I

+h aln.-. m at ive I

\i =- Wh II' Figure 7/ 12

In the previous article we treated the work of a gravita tional force or weight acting on a body in the same way as the work of any other active force. Thus, for an upward displacement flh of the body in Fig. 7/ 12 th e weight IV = mg does nega tive wor k flU = - mg flil . If, on the other han d, th e body has a downw ard disp lacem en t Ilh , wit h h measured positive downward, the weight does positive work SU = + mg Sh. An alternative to the foregoing treatment expresses the work done by gravity in ter ms of a cha nge in potential ene rgy of th e body . Thi s alter native treat ment is a useful representation when we describe a mecha nical system in term s of its total energy. Th e gravitational potential energy Vg of a body is defined as the work done on the body by a force equa l and opposite to th e weight in br inging th e body to the position under consideration from some arbitrary datum plane where the poten tial energy is defined to be zero . The potent ial energy, then , is th e negative of t he work done by the weight. When th e body is raised, for example, the work done is convert ed into energy which is poten tially available , s ince the body can do work on some other body as it returns to its or iginal lower positi on . If we tak e Vg to be zero at iI = 0, Fig. 7/ 12, the n at a height h above th e datu m plane , t he gravitational potential ene rgy of the body is

(7/5) II + ~h Datum - 1- - - - -

-- , m

I

"I

I I

Figure 7/13 M arwan and W aseem AI-Iraqi

If the body is a dist a nce h below the datum pla ne, its gravitational potential energy is - mgh. Note that t he datum pla ne for zero poten tial energy is a rbit rary because only the cha nge in potent ial energy matters, and this change is th e same no mat ter where we place the datum plane. Note also tha t the gravitationa l potentia l energy is in dependent of t he pa th followed in ar riving at a particular level h . Thus, th e body of mass m in Fig. 7/ 13 www .gigapedia.com

Articl e 7/4

has the sa me potential-energy chan ge no matter which path it follows in going from datum plane 1 to datum plan e 2 becau se :1h is the sa me for all three pat hs. Th e virtu al change in gravitat iona l potenti al energy is sim ply

where 8h is the upward virtual displacement of the mass cen ter of the body. If the mass cente r has a downward virt ua l displacement, th en ~V" is negati ve. The units of gravitational potenti al energy are the same as those for work a nd elas tic poten tial ene rgy, jo ules (J ) in SI uni ts a nd foot-pounds (ft-Ib) in U.S. customa ry units. Energy Equation

We saw that the work don e by a linear spri ng on th e body to which its movable end is attached is the negat ive of the change in the elastic poten tial energy of th e spr ing. Also, t he work done by the gravitationa l force or weight mg is the negati ve of the change in gravitational potential energy . Therefore, when we apply the virtual-work equation to a sys tem with springs and with changes in the vertical position of its members, we may replace the work of the springs and the work of the weights by the negat ive of th e respective potential energy cha nges. \Ve can use these substitutions to write the total virtual work au in Eq. 7/3 as the su m of th e work ~U' done by all act ive forces, other than spr ing forces and wcight forces, and th e work - ( ~V, + ~Vg) done by t he spring and weight forces. Equat ion 7/ 3 t hen becomes or

(7/6)

where V = Ve + V" stands for the total potential en ergy of th e syste m. Wit h this for mulat ion a spr ing becomes internal to t he system, and t he work of spring and gravitational forces is accounted for in the 5V term. Active-Force Diagrams

With th e method of virtual work it is useful to construc t the activeforce diagram of the syste m you are analyzing. Th e bou nda ry of th e syste m must clearl y dist ingu ish tho se memb er s which a re part of th e sys te m from ot her bodies which are not part of th e syste m. When we include an elast ic member wit hin the boundary of our system, the forces of interaction between it and the movable members to which it is attached are internal to the system. Thus these forces need not be shown because their effects are accounted for in the Ve term . Simi larly, weight forces are not shown because their work is accounted for in the Vg term. Figure 7/1 4 illustrat es th e ditTerence between th e use of Eqs. 7/3 and 7/6 . We consi der th e body in part a of th e figu re to be a pa rticle for s implicity, and we assume that the virtual displacement is along the fixed path. The particle is in equilibr ium under th e acti on of t he applied forces F 1 and F 2 • the gravitational force mg, the spring force kx , and a Marwan and Wa seem AI-Iraqi

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Pot ent ial Ener gy and Stabil it y

407

408

Cha p t er 7

Vir tual Wo rk

Smooth path

m

(0 )

mg

Eq. 7/3, OU = 0

Eq.7/6 ' OU' =0 1; +oy,; =OV

Ih)

(e)

Figure 7/14

normal reac tion force. In Fig. 7/ 14b, where t he particle a lone is isolated, f>U includes the vir tual work of all forces shown on th e acti ve-force di-

agram of t he pa rticle. (The no rma l reaction exerte d on th e particle by th e smoot h gu ide does no work a nd is ornitted .) In Fig. 7/1 4c th e spr ing is included in the sys te m, a nd f>U' is the virtua l work of on ly F } and F 2 , which are the only external forces whose work is not accounted for in t he potential-energy terms. The work of th e weight mg is accounted for in th e f>Vg term , and th e work of th e spr ing force is inclu ded in the f>Vc term.

Principle of Virtual Work Thus, for a mechanical system with elastic members and members which undergo changes in position, we may restate the principle of virtual work as follows:

The virtual work done by aU external active forces (other than the gravitational and spring forces accounted for in the potential energy terms) on a mechanical system in equilibrium equals the corresponding change in the total elastic and gravitational potential energy of the system for any and all virtual displacements consistent with the constraints. Ma rwan and W aseem AI-I raqi

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Article 7 /4

Stability of Equilibrium Consider now the case of a mechanical system where movement is accompanied by cha nges in gravitational and elas tic potential energies and where no work is done on the system by non poten tial forces. The mecha nism treated in Sample P roblem 7/ 6 is an example of such a syste m. With 8U ' = 0 the virtual-wor k re lat ion, Eq. 7/6, becomes

(7/ 7)

or

Eq uation 7/7 expresses the requiremen t that t he equilib riu m configu ration of a mechanical system is one for which the total potent ial energy V of the sys te m has a stationary value. For a system of one degree of freedom where the poten tial energy and its derivatives are continuous functio ns of th e single variable, say, x, which describes the configuration, the equilibr iu m condition 8V = 0 is equivalent ma t hematically to t he requirement (7/8 )

Equ ati on 7/8 sta tes that a mecha nical syste m is in equilibrium whe n the derivat ive of its total potential energy is zero. For sys tems with se veral degrees of freedo m the part ial derivative of V wit h resp ect to eac h coordinate in turn must be zero for equilibrium." Th er e a re t hree cond it ions under which Eq. 7/8 applies, na mely, when the total potential energy is a minimum (sta ble equilibrium ), a maximum (unstable equilibri um), or a const ant (neutra l equilibri um). Figure 7/ 15 shows a simple exa mple of th ese three conditions. Th e potential energy of the roller is clearly a minimum in the stable posit ion, a maximum in the unstab le position, and a consta nt in the neutral pos ition. We may also characterize the stability of a mechanica l system by noti ng that a small displaceme nt away from the stab le position result s in an increase in potential ene rgy and a tendency to return to the position of lower en ergy . On the other hand, a small displacement away from the unstable position resu lts in a decrease in potential energy and

Stable

Unsta ble

Neutral

Figure 7/15

" For exa mples of two-degr ee-of-freed om system s. Sloe Art . -13, Chapte r 7, of th e first aut hor's Sta tics, 2nd Edition, 5 1 Version, 1975 .

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Potential Ener g y and Stabilit y

409

410

Chapt er 7

Virtu al Work

a tendency to move farther away from the equilibrium position to one of still lower energy. For the neutra l posit ion a small displacement one way or the other result s in no change in potenti al energy and no tendency to move either way. When a function and its derivatives are continuous, the second derivative is positive at a point of minimum value of the function and negativ e at a point of maximum value of the function. Thus, the mathematical conditions for equilibrium and stability of a sys tem with a single degree of free dom x are:

Equilibrium

dV

dx

0

Stabl e

d 2V - 2 >0

Unstable

d 2V - 2 < 0

dx

(7/ 9)

dx

The second derivat ive of V may also be zero at the equilibrium positi on, in which case we must examine the sign of a higher derivative to ascertain th e ty pe of equilibri um . When th e order of th e lowest remain ing nonzero derivative is even, the equilibrium will be stable or unstable according to whether the sign of this derivative is positive or negative. If the ord er of the der ivative is odd, th e equilibri u m is classified as un stable, and the plot of V versus x for this case appea rs as an in flection point in the curve with zero slope at the equilibrium value. Stability crite r ia for multiple degr ees of freedom require more advanced treatment. For two degrees of freedom, for example, we use a Taylor-ser ies expansion for two variables.

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Article 7 /4

Potential Energy an d Stability

411

Sample Problem 7 /4 The lO-kg cylinde r is suspended by th e spr ing, which has a stiffness of 2 kN /m . P lot th e potentia l ene rgy V of th e sys te m a nd show th at it is minimum at th e equ ilibrium position.

Solution . (Although the equilibrium position in this simple problem is clearly

CD

whe re t he force in t he spr ing equa ls the weight mg, we will pr oceed as t houg h t his fact were u nknown in order to illustra te the en ergy rela tionships in th e simplest way .) We choose th e dat um plane for zero poten ti al ene rgy at t he position whe re the spring is u nextended . The elastic potential energy for an arbitrary position x is Ve = ~kx2 and the gr avitational potentia l energy is - mgx, so that the total potentia l energy is

8l t T T / i

IV = V, + V, I

8 ~;I+-- ~

Equilibrium occurs where

4

dV dx

m = 10 kg

I

I

I

2'

x = mg/k

Although we kn ow in this sim ple case th at th e equilibrium is stable, we prove it by evalua t ing th e sign of th e seco nd derivative of V at th e equilibrium position . Thus, d 2V/ dx'2 = k , which is positive, proving that th e equ ilibrium is sta ble. Substituting nu merica l value s gives V = !( 2000Ix 2

-

1Ol9.8 1)x

or

i -I

o

I

I

I

~~OJ.?_4"",,9=c-~cc-_/. x , m

0.02

O.W I 0.06 .r = mglk I

0.08

.10

I

v =~ + ~I -4 f--f---',,--f--t-___ -6 f--1--1--~-t-___1

expressed in joules, and th e equilibriu m value of x is x ~ 10(9.81)/2000 ~ 0.0490 m

V: '!.,k\-2 < .

I

V.J

kx -mg =O

I

-8

49 .0 mm

Ans.

We calcul at e V for va r ious va lues of x a nd plot V versus x as shown . The = 0.0490 m wh ere dV/dx = 0 and d 2 V/ dx 2 is pos itive.

@ minimum valu e of V occurs a t x

Helpful Hints

CD The choice is arbitrary but simplifies th e algebra.

@ We could have chosen different datum planes for Ve and Vg without affecting our conclusions . Such a cha nge would me re ly sh ift the separate curves for Ve an d Vg up or down but wou ld not affect the position of t he minimum value of V.

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412

Chapter 7

Virtua l Work

Sample Problem 7 /S The two uniform links, each of mass m, are in the vertic al plane and are connected a nd cons trai ned as shown. As th e a ngle 8 betwee n the link s increases wit h the ap plica tion of the horizontal force P, the light rod, which is con nect ed at A and passes th rough a pivoted collar at B, comp resses the spr ing of stiffnes s k . If the spring is uncompressed in the position where 8 = 0, det erm ine t he force P which will produce equilibrium at the angle 9.

Solution. T he given ske tch serves as t he active-force diagram of the syste m. Th e compression x of t he spring is the distan ce which A has moved away from B , which is x 2b sin ()/ 2. T hus, t he elast ic potenti al ene rgy of th e spring is

Wit h the datum for zero gravitationa l potential energy taken through the support a t 0 for convenience, the express ion for Vg become s

lVg = mgh l

cos ~)

Vg = 2mg ( - b

Th e dista nce between 0 and Cis 4b sin 0/ 2, so that t he virtual work done by P is 8

ZPb cos 208 The virtual-work equ ation now gives

loU' = oV, + OVgJ ZPb cos

~ se = 0 ( 2kb' sin'~) = 2kb z sin

+ 0 ( - 2mgb cos~)

~ cos ~ so +

mgb sin

~ SO

Simplifying gives finally 8

I

8

P = kb sin - + - mg tan 2 2 2

Ans.

If we had been as ked to express the eq uilibrium value of 0 corresponding to a given force P, we would have difficulty solving explicit ly for (J in this pa rti cul ar case . But for a numer ical problem we could resort to a compu te r solution a nd gra phical plot of numeri cal values of the su m of th e two fu nctions of (J to determin e th e value of fJ for wh ich the su m equa ls P.

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Article 7 /4

Potential Energy and Stability

413

Sample Problem 7 /6 The ends of the u niform ba r of mass m slide freely in the horizontal and vertical guides. Examine t he stability conditions for th e positions of equilibri um . Th e spri ng of sti ffness Ii is undeformed when x == o.

CD

Solution. The system consists of the spr ing and th e ba r. Since t here are no external active forces, the given sketch serves as th e act ive-force diagram. We will take the x-axis as the datum for zero gravitat iona l potential ene rgy. In the displaced position the elastic and gravitational potential energies are and

Vg == m g

()

CD With

no exte r nal acti ve forces th er e is no (jU ' term, and iiV = 0 is equ ivalent to dV/dO ~ 0,

+ !mgb cos ()

Equilibrium occur s for d V/dO == 0 so that ! mgb sin

(1 -

(J

== (kb 2 cos (} - !mgb ) sin 0 = 0

Th e two solut ions to th is equation are given by sin

(J =

I-x--I "

b

H elpful Hints

V == Ve + Vg = ! kb 2 sin 2

== Ilb2 sin () cos

1111111111111111111111

x - - -

'2 cos o

T he total potential ene rgy is then

~~

o

0

an d

cos

(J

@ Be carefu l not to overlook the solution (J = 0 given by sin () = o.

mg = 2kb

We now determine the stability by examini ng th e sign of the second derivative of V for each of t he two equili brium positions. Th e second der ivati ve is d 2V

-

d(P

= kb 2 (cos2

= kb2 (2 cos2 Solution I. sin 0

~ 0,

0

sin2 H) - }.mgb cos

II -

2

(J -

1) - ~mgb cos

d . ~ hb' (2 dO'

1) - l mg b = hb' 2

= posit ive (stable)

@

(J

°

~

2V

=

(J

negative (unstable)

(1_

g) m 2hb

if h > mg/2b if h < mg/ 2b

A ns.

Thus, if the spr ing is sufficiently stiff, the bar will re tur n to the vertical positi on even though there is no force in t he spri ng at that position.

@ We might not have anticipated thi s re sult without the mathematical analysis of th e sta bility.

Solution II. cos 0 = ~~, 0 ~ cos - 1 ~~

~:~ = hb' [2(;;,~ )' - 1]- ~mgb (;~~) ~ lib' [(;~~)' @

1] Ans.

Since t he cosine mu st be less than u nity , we see th at th is solution is limited to th e case where k > mg /2b, which mak es the second deriv at ive of V negative. Thus, equ ilibrium for Solution II is never stable. If Ii < mg / 2b, we no longer hav e Solution II since the spring will be too wea k to maint ain equ ilibrium at a value of (1 between 0 and 90 Q

•

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@ Again , without the benefit of the mathematical analysis of th e stability we might have supposed er roneously that the bar could come to rest in a st able equ ilibrium positio n for some value of (J between 0 and 90°.

414

Chapter 7

Vir tual Work

PROBLEMS (Assume th at the nega ti ve work of fri ction is negligib le in the follow ing prob lems. I

7/36 The sma ll cylinder of ma ss In a nd radius r is confined to roll on the circ u lar surface of radius R. By t he met hod s of th is a rt icle, prov e tha t t he cylinder is unsta ble in case (a ) a nd sta ble in case (bl.

Introdudory Problems

f

7/33 The potential e nergy of a mech anical sys te m is given by V = 6x" - 3x 2 + 5, wh ere .r is the position coordi nate whic h defines t he configuration of t he single-degree-of-freedom system. Determine the equ ilibrium value s of x and th e sta bility condition of each . A il s. x = 0, un stable; x = sta ble; x = stable

4,

rei R

If I

1

Ie

/..-

I

I1

'-

'I

(a l

(bl

Problem 7/36 7/37 For the mechanism shown the spring is uncompressed whe n (J = O. Determ ine th e a ngle 0 for th e equilibrium position and specify th e min imu m spri ng stiffness k whic h will limit (} to 30°. T he rod DE passes freely through the pivoted colla r C, a nd th e cylinde r of ma ss m slides freely on th e fixed vertica l shaft. - 1 mg Ans. fJ = cos 2kb' k mm

L

Probl em 7/34 7/35 Th e bar of mass m with cente r of ma ss at G is pivoted about a horizontal axi s through O. Prove the stability conditions for the tw o positions of equilibr ium. Ail S. II = 0, u nstable; (J = 1800, stable

A

Probl em 7/37

Problem 7/35 Marwan and W aseem AI-Iraqi

R

,.

~

-4,

7/34 The uniform bar of mass m a nd length L is supported in the vert ical plane by two identical springs each of . stiffness h a nd compressed a distance f, in the vertical position /} = O. Determine the minimum stiffness k which will ens ure a sta ble equi libr ium position with tJ = O. T he spr ings may be assumed to act in t he horizontal direction during sma ll a ngu lar motion of th e bar.

I

I

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Article 7/ 4 7/38 The figu re shows th e cross sect ion of a un iform 60-kg venti lat or door hinged along its upp er hor izontal edge at O. Th e door is contro lled by the spring-loaded cable which passes over th e small pulley at A. The spring has a st iffness of 160 N per met er of stre tch an d is undeform ed when (J = O. Deter mine the an gle fJ for equilibri u m.

I II

Pr o b le m s

415

7/40 Deter min e t he equ ilibriu m value of x for the springsupported bar. Th e spring has a st ifTness h and is unst re tched when x = O. Th e force F acts in the directi on of t he bar , and the mass of t he ba r is negligible.

k

11I1 1

,, A I

I I

I

, ,,

I

I

I

I I I I I I I

,,

,

Problem 7/40 7 /41 One of the critical requirements in th e design of an

i-f Problem 7/38 7/39 For the device shown the spr ing would be u nstretc hed in the positio n (J = O. Specify the st iffness h of the spring which w ill establish an equilibrium posit ion II in the vertical plane . The mass of the links is negligible compa red with m. mg cot fJ Ans. h = -26 1 - cos (J

artificial leg for an amput ee is to prevent th e knee joint fro m buckling u nder load when t he leg is st ra ight. As a first app roximation, simulate the artiflcial leg by t he two light links wit h a torsion spr ing at their comm on joint. The spr ing develops a torqu e M = K{3 , which is proport ional to th e angle of bend {3 at the joinl. Deter mine the min imum value of K wh ich will ens ure stability of the kn ee joint for

f3

~

o.

A n s. K min = ~mgl

.

.\\\\\\

~

m

II

~.

~j ~ I

Problem 7/39

.JfJ. Prob lem 7/41

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416

Chapter 7

Virtual Work

Representative Problems 7/42 Th e cylinder of mass M and radius R rolls with out slipping on th e circu lar surface of radius 3R. Attac hed to th e cylinder is a small body of mass 11l . Determine th e required relationship betwee n M and m if th e body is to be stable in th e equili brium position s hown.

7/44 Determine the max imum height Ii of th e mass I1l for which the inve rted pendulum will be stable in th e vertical position shown. Each of th e springs has a stiffness k, and th ey have equal precompressions in this positi on. Neglect the mas s of th e remaind er of the mechani sm .

/~I I

R /4 3R /4 ---- /~,--- -

Problem 7/42

Problem 7/44

7/43 Each of the two gears ca rries an eccentric mass m and is free to rot at e in the vertical plan e about its bearing. Determine th e values of fJ for equi librium and identify th e type of equilibrium for eac h positi on. Ans. (J = 0, un stable (} = 120°, stable o = 180°, unsta ble (J = 240°, stable

7/45 One en d of th e torsion spr ing is secured to t he gro u nd at A , and th e othe r end is faste ned to th e shaft at B. Th e torsional st iffness K of th e elasti c spr ing is the torq ue required to twi st the spr ing through an angle of one radian . The spring resist s th e moment about the shaft axis caused by th e tension mg in the cable wrapped aroun d the dru m of ra dius r, Determ ine the equ ilibriu m value of h measured from the dashed positio n, where the spring is untwist ed . mg ,:J. An s. li = K

Problem 7/43

Problem 7/45

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Articl e 7/4 7/46 For th e mechani sm shown, th e sprin g of stiffness k ha s an un stretch ed length of esse ntially zero, and th e larger link has a mass m with mass center at B. Th e mas s of t he smaller lin k is negligible. Determ ine th e equ ilibriu m angle (} for a given downward force P.

Problem s

417

7/48 Th e 3-tb pendulum swings about axis 0-0 and has a mass center at G. When (} = 0, each sprin g ha s an initial stretch of 4 in. Calculate th e maximum stifTness k of eac h of the parall el spr ings which will allow t he pend ulum to be in sta ble equilibrium at the bottom positio n 8 = O.

p

k

·~--\ \ \ \ \ \ \ \ \ \ \ \ \ -----~1 b

b

•

8

B

b • A

Problem 7/46

7/47 The cross sect ion of a tra p door hinged at A and having a mass m and a center of mass at G is shown in th e figur e. Th e spring is compressed by th e rod wh ich is pinn ed to th e lower end of t he door and which passes through th e swivel block at B. When II = 0, th e spring is undeformed. Show that with th e proper stiffness k of the spring, the door will be in equilibrium for any angle 8. mgb A lI S.ll = - 2a

Problem 7/48 7/49 Th e solid he misph er e of diameter 2b and concen tric cylindrical knob of diam eter b are resting on a hori zontal su rface. Determine the maximum heigh t h whic h the kn ob may have wit hout caus ing th e un it to be u nstable in the upright positio n shown. Both parts are made fro m the same material. An s. h < b ,,''2

Problem 7J47

Problem 7/49

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418

Chapt er 7

Virtual Work

mine th e angle of tilt (I for equ ilibrium. (Hi nt: Th e defor mati on of th e spring may be visualized by a llow. ing th e base to tilt th rou gh th e required angle fl about a while th e seat is held in a fixed posi tio n. I

7 /50 Pr edict thro ugh calculatio n wheth er t he homogeneo us semicylinder an d the half-cylindrical shell will remain in th e positions shown or wheth er t hey will roll ofTt he lower cylinders .

~8 ----j \...--:-- -,.,:3 00

r

III III

FO'~ ~8 ~~

\

Problem 7/50 7 / 5 1 Th e uni form lin k AB has a mass m , and its left end A travels freely in t he fixed hor izont al slot. End B is

attached to the vertical plunger , wh ich compresses the spring as B falls. The spring would be uncompressed at th e position (I = O. Determine th e angle 11 for equilibriu m (other th an the impossible position corr espo nding to /J = 90°) and designat e the condition wh ich will ensu re sta bility.

Problem 7/52 7/53 A proposed parallelogr am link age for an adjus ta ble-

position lamp is shown. If th e un stretched length of the spring is b/2 , determine t he necessa ry sp ring stifTness k for equilibrium at a given an gle fl with th e vert ica l. Th e mass of th e lamp and triangu la r fixture is m . Check the stability withi n th e working range from fI = 2 sin " ! ~ es 29° to (J = 180°. mgl 1

Aus. sin II = mg /". > mg 2i1l' 21

A TIS.

= l7" -1-...,...'----,-

~ esc 0/ 2 sta ble wit hin specified range

1

b

I

8

V,

Probl em 7/51

7/51 The figur e shows a tiltin g desk chair toget her with t he design detail of the spr ing-loaded til ti ng mecha nism . Th e fra me of the seat is pivoted about th e fixed point 0 on the base. The incre ase in distance between A and B as th e chair tilt s back about 0 is th e increase in compress ion of th e spri ng. Th e spring, which has a stifTness of 96 kN/ m, is un compressed when H = O. For sma ll angles of ti lt it may be assu med with negligible erro r that the axis of the spri ng remains parallel to the seat. Th e center of mass of an 80. kg perso n who sits in the chair is at G on a line through 0 perpe ndicu lar to t he seat. DeterWNW.gigapedia.com

~\

I111 k 11 I11

~-Hl~ ./

Marwan and W aseem AI-Iraqi

I;

b/

Problem 7/53

Art icle 7 / 4

.. 7/54 Th e fro nt -end sus pe ns ion of P rob . 4/108 is repeated here. In a test of th e designed action, th e fram e F mu st be jacked up so th at h = 350 mm in orde r to relieve the compress ion in th e coil springs. Determine t he va lue of Jz whe n th e jack is re moved. Ea ch spr ing has a stiffness of 120 kN /m . Th e load L is 12 kN , and t he central frame F has a mass of 40 kg. Each wheel and a ttached link has a ma ss of 35 kg with a center of mass 680 mm from th e vert ical cen te rline . A ns. h = 265 mm

Prob lem s

4 19

.. 7/56 The un iform garage door AB shown in section has a mass m and is equipped with two of th e spring-loade d mech anism s shown, on e on each side of t he door . The a rm DB has negligible ma ss, and t he upper corner A of the door is free to move horizontally on a roller. Th e u nstretched len gt h of th e spri ng is r - a , so that in the to p posit ion wit h fJ = 1T th e spr ing force is zero. To en sure smoot h action of the door as it reach es the vertica l closed position 0 = 0, it is desirabl e that the door be insen sitive to movement in thi s position. Deter mine th e spri ng stiffness k req uired for t his design . mg tr + a J An s. k = &z2

A

" -, .... \

\ \ \

, I

Probl em 7 /54

I I

.. 7/55 The portable roller stand for supporting board s ejected from a wood plan er is designed wit h a microfine height adjustment produced by turning th e kn url ed knob of the adjus ting scr ew wit h a torque M . T he single-th rea d screw wit h square threads has a pit ch p (adva nce ment per revolution ) a nd is threaded into th e collar a t B to control th e distance between A a nd B (a nd hen ce C a nd D ). The roller E a nd supporting box ha ve a mass m i. a nd th e four uniform lin ks (two on eac h side) have a combined mass "' 2 a nd a le ngth 2h for eac h. Neglect a ll frict ion a nd find the to rque AI necessa ry to raise t he roller for a given va lue of fJ. 2")p,,,g -(:.:2::.m::..!.1_+;-:n:':'''cot fJ Ail S. AI

4"

E

EEi 0

b A

•

C Partial end view

8 b

-«.

B

• Probl em 7 /55

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I /' /' /'

Problem 7 /56

B

420

Chapte r 7

Vir t ual Wo rk

C H APT E R REVIEW In this chapte r we have developed the principle of virtual work. This principle is use ful for determining th e possible equilibriu m configurations of a body or a syst em of inte rconnected bodies where the external forces are known. To app ly the met hod successfully, you must understa nd th e concepts of virtual displacement, degr ees of freedom, and pote ntial energy .

Method of Virtuol Work When various configuration s are possible for a body or a syste m of interconnect ed bodies und er th e act ion of applied forces, we ca n find t he equilibrium position by applying the principle of virtua l work. When usin g this method, keep the following in mind . 1. Th e only forces which need to be cons idered when determining the equilibrium position a re those which do work (active forces) during th e assumed differential movement of t he body or system away from its equilibriu m posit ion.

2. Those exte rnal forces which do no work (reactive forces ) need not be involved. 3. For thi s reason th e acti ve-force diagram of the body or system (rather than th e free-body diagram) is useful to focus attention on only th ose external forces which do work during th e virtual displacemen ts.

Virtual Displacements A virtual displacement is a first-order differential change in a linear or angular position. This change is fictitious in that it is an assumed movement which need not take place in reality. Mat hematically, a virtual displacement is treated th e same as a differential cha nge in an actual movemen t . We use th e symbol S for t he differen tial virtual change and t he usual symbol d for the different ial change in a rea l moveme nt . Relating th e linear an d angular virtual displacements of the parts of a mechanical sys tem during a virtual movement cons istent with the constraints is often t he most difficult part of th e analysis. To do th is, 1. Write th e geometric relationships which describe th e configuration of the system.

2. Esta blish th e differential cha nges in the positions of parts of the syste m by differentiating th e geometric relationship to obtain expressions for the differen tial virtual movement s.

Degrees of Freedom In Chapter 7 we have restricted our atte ntion to mechanical sys tems for which th e positions of the memb ers can be specified by a single vari ab le (single-degree-of-freedom syste ms). For two or more degr ees of freeMa rwa n and W as eem AI- Iraqi

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Chapter Revie w

dom, we wou ld apply the virtual-wo rk equation as many times as th ere ar e degr ees of freedom, allowing one variable to change at a time while holding the remaini ng ones constant.

Potential Energy Method The concept of potential energy, both gravitational (Vg ) and elastic (Ve), is useful in solving equilibrium problems where virtual displacements caus e cha nges in the vertical positions of th e mass cente rs of th e bodies and changes in the len gths of elastic members (springs). To apply this met hod, 1. Obtain an expressio n for the total potential energy V of the system

in te rms of the variable which specifies the possible position of the system. 2. Exa mine the first and second derivat ives of V to establish, respect ively, th e positi on of equilibriu m an d th e corresponding stability condition.

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421

422

Chapter 7

Virtual Work

REVIEW PROBLEMS

Plane of each figure is vertical. Size and mass of each member and applied force are known.

7/57 A control mechanism consist s of a n input shaft at A which is turned by a pplying a couple M and an out put slide r B which moves in the x-direction again st th e action of force P. Th e mechanism is designed so that th e linear movement of B is proportional to the angu lar movement of A, with .r increasing 60 mm for every complete turn of A. If M = 10 N · m, det ermine P for equ ilibrium. Neglect int ernal fricti on and assu me all mechan ical componen ts are ideally connect ed rigid bodies. Ail s. P

~

p

Find B for equilibrium

p

Find reactions at A and B

fal

fb i

1047 N

f - - x--'1 Find r for equilibrium

Find forces at A. B. and C fdl

fe)

Problem 7/57

(n

7/58 Iden ti fy which of the problems (0 ) t hrough are best solved (A ) by the force and momen t equilibr iu m equations an d (!3 J by virt ua l work. Outline br iefly the procedu re for each solution.

f---- x----I Find x for equilibrium. fel

Determine maximum k for stable equilibrium at B = 0

ff!

Problem 7 j 58

7/59 The semicylindr ical shell of radiu s r is pivoted about a shaft through points 0 as shown. Th e mass of th e two support tab s is small compared with the mass of th e shell. Determine the maximum value of Ii for which equilibriu m in th e position shown is sta ble. Ans. h mllJf. = O.363r

Problem 7/59 MarNan and Waseem AI-Iraqi

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Re view P robl em s

4 23

7 /60 Neglect the mass of th e crossed links an d det ermi ne

the a ng le II for th e equilibrium position of th e symmetrical mechanism in the vertical plan e. Ea ch of th e identical rectangu lar blocks of mass m is homogeneous with mass cente r a t G. Evaluate 1/ for equilibr iu m whe n b = a.

Problem 7/62 7/63 Determine th e eq uilibri um values of

(J and th e stability of equ ilibrium a t eac h posit ion for t he un bala nced wheel on th e 10" incline. Stat ic friction is sufficient to preven t slipping. Th e mas s center is at G. Ans. II = - 6.82", sta ble; fI = 207", un stable

Problem 7/60

7/61 Th e sketch shows th e approxima te design configura t ion of one of th e four toggle-act ion hold-down as· semblies whi ch clamp the base flange of th e Saturn V rocket vehicl e to th e pedestal of it s platform prior to lau nching. Calcu late the pre set clamp ing force F at A if th e link CE is under ten sion produ ced by a flu id pressure of 2000 Ib/in .2 act ing on t he left side of t he pist on in the hydra ulic cylinder. T he piston has a net area of 16 in. 2 The weight of th e assembly is considerable, but it is small compa red wit h th e clampi ng force produced a nd is therefore neglected here. A ns. F = 960 ,000 Ib

='7'--

, Saturn V ~_ L __ ..:\_ -----~_..!.

I

Prob le m 7/63 7/64 Two semicylindr ical shells with equ a l pr ojecting rec-

I I

\:;:.4-- - ""'t'!f' B -

--

- T

= 60 mm A

60" - ~

6"

basc tlan ge

,. = 100 mm

r

I

tangles are form ed from sheet metal, one with configurat ion (a) an d th e oth e r wit h configurat ion (bl. Both shells rest on a horizontal sur face. For case (0) det ermine the maximum valu e of h for wh ich th e shell will remain sta ble in th e position shown. For case (b> prove th at sta bility in th e positi on shown is not alTecte d by the dimen sion II.

40" E

c• 2"

IL

40" .L

Problem 7/61 7/61 Th e figure shows the cross section of a contai ner

composed of a hem ispher ical she ll of radiu s r a nd a cylindrical shell of heigh t h, both made from th e sa me ma te rial. Speci fy th e lim itat ion of h for stab ility in t he upri ght posit ion when th e containe r is placed on th e horizontal sur face.

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(al

(bl

Problem 7/64

424

Chapter 7

Vir tual Work

7/65 An exploration device, which unfo lds from the body A of an unmanned space vehicle resting on th e moon 's surface, consist s of a spring-loaded pantograph with detector head B . It is desired to select a spri ng that will limit th e vertical contact force P to 100 N in the position for which 0 = 120°. If th e mass of the arms and head is negligible, specify the neeessary spring st ifTness k. Th e spring is uncompressed when 0 = 30°. A ns. k = 1.664 kN/m

I.. 1/61

The platform of mas s m is support ed by equa l legs and braced by the two springs as shown. If th e masses of the legs and springs are negligible , design th e springs by det ermining th e minimum stiffness k of each spring which will en sure stability of th e plat . form in the position shown. Each spri ng ha s a ten sile preset deflection equal to .l. Ans. k min = mg 2b ( 1

2

b ) + (i

A

;--

- -Problem 7/67

B

b = 300 mm

p

Problem 7/65

7/66 Th e uniform alu min um disk of ra dius R an d mass m rolls without slipping on the fixed circu lar surface of radi us ZR. Fastened to the disk is a lead cylinder also of mass m with its center located a distance b from th e center 0 of th e disk . Determine the minimum value of b for which th e disk will remain in sta ble equilibrium on th e cylindrica l sur face in the top posit ion shown.

... 1/68 In t he mechanism shown the spring of stiffness k is un compressed when 0 = 60°. Also the masses of th e parts are small compared wit h t he su m m of th e masses of th e two cylinders. The mechan ism is constr ucted so that the arms may swing past the vertical, as seen in the right-ha nd side view. Determine th e values of 0 for equi libri um a nd invest igate the stability of the mechanism in eac h position. Neglect friction . Ans. tI = 0, stable if k < mgf a un stable if k > mg/a

l(

g O = cos - I - 1 +m- ) 2 ka only if k > mg /a, sta ble

a

/

a

-u

/

III

~ A

/ ~ Ik

B k

Prablem 7/66

m

m

2

2

Prablem 7/68

Marwa n and W aseem AI-Iraqi

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Re view Probl ems

e

=,

' Computer-Oriented Problems

-7/ 69 Th e bar OA , wh ich weighs 50 Ib with center of gravity at G, is pivoted about its end 0 and swings in th e vertical pla ne under th e constra int of th e 20-lb cou n terweight. Write th e express ion for th e total potential energy of the system, taking Vg = 0 when II = 0, and compute Vg as a func tion of U from II = 0 to (} = 360 0 • Fr om your plot of the resul ts, determine the positio n or positions of equilibrium and the stability of equilibri u m at each position. A ns. 0 = 78.0 0 , stable; 0 = 2600 , u nstable

425

"7/71 Determine th e equi libri um value of th e coordina te x for the mechani sm und er the action of th e 60-N force applied. normal to the ligh t bar. Th e spring has a st ifTness of 1.6 kN /m and is un st retched when x = O. (Hint: Replace t he applied force by a forcecouple system at poin t B.> Ans. X = 130.3 mm

k = 1.6 Iu'l/m

B

i j "-! 3'

.~

Problem 7/71

20 lb Problem 7 / 6 9

- 7/7 0 The toggle mechani sm is use d to lift the 80· kg mass to a locked position whe n OB moves to OB ' in the 30 posit ion. To evaluate the design act ion of the toggle, plot the valu e of P required to oper ate t he toggle as a functi on of H from (J = 20 to (J = _30.

-7 / 71 The u niform link OA ha s a mass of 20 kg and is support ed. in th e vertical pla ne by the spring AB whose unst retched lengt h is 400 mm . Plot the tot al pote ntial ener gy V and its derivative dVl d l1as fun ctions of () from 0 = 0 to (} = 1200. Fr om t he plot s identify t he equilibrium values of (J a nd the corr esponding stability of equilibrium. Tak e Vg = 0 on a level throug h O.

0

k = 550 N /m

A

Problem 7 /72

Problem 7/70

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426

Chapt e r 7

Vir t u a l Wo r k

·7 /73 Deter mine the equ ilibr iu m angle

f} for the mechanism shown. Th e spr ing of stiffness I~ = 12 Ib/ in. has an unst retched length of 8 in. Each of the u nifor m links AB and CD ha s a weight of 10 lb, and member BD wit h its load weighs 100 lb. Motion is in t he vertical plane. A il s. f) = 71.7°

·7/74 Th e unifor m 25-kg trap door is freely h inged along

D

•

T 1

its bottom edge 0 -0 and is att ached to the two spr ings eac h of stiffness k = 800 N/ m. Th e spr ings are u nst retched when (1 = 90°. Take Vg = 0 on th e hor izontal plane through 0 -0 and plot th e pot ential energy V = Vg + V(:' as a fun ction of 0 from (I = 0 to fJ = 90°. Also deter mine the an gle fJ for equ ilibriu m and determine the stability of this position . D

l 00 lb

16"

B

.

-

Iii

Problem 7/74 Problem 7/73

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Appendix

AREA MOMENTS OF INERTIA

ApPENDIX OUTLINE

All A/2 A/3 A/4

Introduction Definitions Composite Areas Products of Inertia and Rotation of Axes

All

INTRODUCTION

When forces are distributed continuously over an area on which they act, it is often necessa ry to calculate t he moment of these forces about some axis either in or perpendicular to t he plane of th e area . Frequ ently t he inten sit y of th e force (pressure or stress ) is proportional to the d istance of the line of action of t he force from the moment axis. The element al for ce act ing on an eleme nt of area, then, is proport ional to distan ce tim es differential area, a nd t he eleme ntal momen t is proport iona l to distan ce squa red times differential area. \Ve see , t he refo re, t hat t he to ta l moment involves a n int egr al of form J (dista nce)2 d (area ). This int egral is called th e moment of inertia or th e second moment of t he a rea . Th e integral is a fun cti on of the geo metry of th e a rea a nd occurs frequ ent ly in the applicati ons of mechan ics. Thu s it is usefu l to develop its proper ties in some detail and to have t hese properties ava ilabl e for ready use when t he integral ar ises. Figure Al l illustrates the physical origin of these int egrals. In part a of th e figur e, t he surface area ABe D is subjec ted to a distributed pr essure p whose inten sity is prop ortional to th e dist an ce y from th e axis AB. Thi s situation was treated in Art. 5/9 of Chapter 5, whe re we descri bed t he actio n of liqu id pressure on a plan e surface . T he moment abou t AB du e to the pressure on the element of area dA is py dA = ky 2 dA. Th us, the integral in question appears when the total moment M = k J y 2 dA is evaluated. Marwan and W aseem AI-Iraqi

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427

428

Appendi x A

Are a Moments of Inertia

A

In Fig. A/ lb we show the distribution of stress acting on a t ransverse section of a simple elastic beam bent by equal and opposite couples applied to its ends. At any section of the beam, a linear distr ibution of force intensity or stress CT, give n by a = ky , is present. The stress is positive (tensile) below the axis 0 - 0 and negative (compressive) above th e axis. We see that the eleme ntal moment about t he axis 0-0 is dM = y (tr dA ) = hy 2 dA. Thus, the same integral appears when the tot a l momen t M = II Jy2 dA is evaluated. A third exa mple is given in Fig. A/Ie, which s hows a cir cular shaft subjected to a twist or torsional moment. With in the elast ic limit of the material, thi s moment is resisted at each cross section of the shaft by a distribution of tangential or shear st ress T, which is proportional to the radial distance r from the center. Thus , T = kr, and the total moment abou t the central axis is M = J r ( T dA ) = h J r 2 dA . Here t he integral differs from tha t in the preceding two examples in that the area is normal instead of parallel to t he momen t axis and in t hat r is a radial coordinate instead of a rectangular one. Although the integral illustrated in the pr ecedi ng examples is generally called the moment of inertia of t he area about the axis in question, a more fitt ing term is the second moment ofarea , since the first moment y dA is mult iplied by t he moment ar m y to obtain t he second moment for th e element dA . Th e word inertia appears in the ter minology by reason of the similar ity bet ween the mathema tical form of th e integra ls for second moments of areas and those for the resultant moments of the so-called inertia forces in the case of rotating bodies. The moment of inertia of an area is a purely mathematical property of the area and in itself has no physica l sign ificance.

A /2

DEFINITIONS

Th e following definit ions form t he basis for the a nalys is of area moments of inertia.

Rectangular and Polar Moments of Inertia ,

(c)

Consider th e area A in the x -y plane, Fig. A/2. The moments of inert ia of th e element dA about t he x- and y-axes a re, by definition, -u, = y" dA and dl y = x 2 dA , respectively. Th e momen ts of inertia of A about the same axes are therefore

Figure All

y I I 1

A

~f-- x

- ' f . ] dA

I 1

o

1/

r

(A/l)

/1

y

l

x

Figure A/2 Marwa n and Waseem AI- Iraqi

where we carry out the integration over the entire area. WNW.gigapedia.com

Article A/2

Th e moment of inertia of dA about the pole a (a-axis) is, by similar definition, dl z = 1''2 dA . Th e moment of inertia of th e entire area about a is (A/ 2)

Th e expressions defined by Eqs . A/I are called rectangular moments of inertia, whereas the expression of Eq . A/2 is called th e polar moment of inert ia: Becau se x 2 + y2 = ,-2, it is clear tha t (A/ 3l For an a rea whose boundaries are more simply described in rectangular coordina tes th an in polar coordinates, its polar moment of inertia is eas ily calculat ed with th e a id of Eq . A/3. Th e moment of inertia of an elem ent involves t he squa re of the dist anc e from the in ertia axis to th e eleme nt. Thus an eleme nt whose coordinat e is negative contributes as muc h to the momen t of inertia as does a n equa l eleme nt with a positive coordinate of the same magn itu de. Conseq uently th e area moment of inertia about any axis is always a positive qua ntity. In con trast , the first moment of t he area, whi ch was involved in t he comp ut atio ns of centroids, could be either positive, negative, or zero. Th e dimensions of momen ts of inertia of a reas a re clea rly L4 , whe re L sta nds for the dimension of lengt h . Thus, th e SI uni ts for a rea moments of inertia are exp res sed as qu artic met er s (m'') or qu artic rnillimeters (m rn"). The U.S. customary units for area mome nts of iner tia a re quartic feet (ft4 ) or quartic inches (in." ). T he choice of th e coord inates to use for th e calcu lation of moments of inertia is important. Rectangular coordinates should be used for shapes whose bounda ries are most easily expressed in these coordinates. Polar coordina tes will usually simplify problems involving boundaries which are eas ily described in I' and H. Th e choice of an eleme nt of area which simplifies th e int egration as much as possibl e is also impo rt ant. Th ese consideratio ns are quite a nalogous to tho se we discussed and illustrat ed in Cha pte r 5 for the calculat ion of cent roids.

Radius of Gyration Consider a n area A , Fig. A/ 3a , which has rectangular moments of inertia Ix and I and a polar moment of inertia Iz about O. \Ve now " as concentrated into a long narrow st rip of area A a visua lize thi s area dist an ce !Ix from th e .r-axis, Fig. A/ 3b. By definition th e moment of iner tia of th e st rip about th e r -axis will be th e sa me as that of the original area if !I/ A = Ix. The distance !Ix is called th e radiu s af gyration of the "T he pola r mome nt of inerti a of an area is so met imes den oted in mecha nics literatu re by the sy mbol.' .

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Definitions

429

430

App endi x A

Ar e a M om en t s of I n er ti a

y

.Y

1

1

l-r1

y I

Il l '

A

---"-j

.

1

A

A

1

- - ;I:

1

o

-- x

L __

(a )

--x

(ell

(c)

Figure A/3

a rea about the a-axis. A simila r relation for th e y-ax is is wr itten by consider ing the area as concen trated int o a narrow st rip parall el to the y-axis as shown in Fig. A/3c. Also, if we visua lize t he area as concentrat ed into a narrow r ing of ra dius I~z as shown in Fig. A / 3d , we may express the polar moment of iner tia as h z2A = lz. In sum ma ry we wr ite

t,

=

k/A

Iy

=

ky 2A

k x = J Ix/A

or

k y = ) Iy/A

I , = k;A

k, =

(A /4 )

jIJA

Th e radius of gyrat ion, then, is a measure of t he distribution of t he area from the axis in questi on. A rectangul ar or polar momen t of inertia may be expressed by specifying t he ra dius of gyra t ion and the a rea. When we substitute Eq s, A/ 4 into Eq . A/3, we have

(k;

y I f-o-

)'u

- d,. -~If.- Xo

1

1 1

A

1

I dA OI ~~ I Yo

Cl

1

_

- - xu

=

kx 2

+

k/ )

(A/5)

Thus, th e square of t he radius of gyration about a polar axis equals the sum of the squ ares of the radii of gyration abo ut t he two correspond ing rectan gu lar axes. Do not con fuse the coordinate to t he centroid C of a n area with the radius of gyration. In Fig. A / 3a the square of th e centro ida l dist ance from t he .r-axis, for exa mple, is y 2 , which is the squa re of the mean value of th e dist an ces from t he element s of the a rea to th e x-axis. T he qu an ti ty hx 2 , on t he other hand, is the mean of t he squa res of t hese distan ces. Th e moment of iner tia is not equal to Ay2 , since t he square of t he mean is less than th e mean of the squa res.

1

:

1/

(1~

d/

ok - - - - - - - - - - - - -

Transfer of Axes - - ;I:

Figure A/4

Marwan and Waseem Al-lraqi

The mome nt of ine rtia of a n area about a no ncentroidal axis may be eas ily express ed in te rms of th e moment of inertia about a parallel cent roida l axis. In Fig. A/4 the Xo-Yo axes pass t hrough the centroid C of t he a rea. Let us now determin e t he mom ents of iner tia of the area www.gigapedia.com

Article A /2

about th e parallel x-y axes. By definiti on, th e moment of in er tia of th e element dA about t he x-axis is

Expanding and integrating give us

t,

=

f

y02

dA + 2d x

f

Yo dA + d/

f

dA

We see that th e first integral is by definition the mom ent of inertia Ix about the centroidal xo-axis. The second int egral is zero, since J Yo dA ~ Ayo and Yo is au tom atically zero with the cen troid on the xoaxis. Th e third term is simply Ad/ . Thus, t he expression for Ix a nd the similar express ion for I y become I; = I, Iy

_ - Iy

+ Adx 2 + Ady

(A / 6 )

2

By Eq . A / 3 th e sum of these two equations gives (A/ 6a)

Eq uations A/ B and A/Ba are the so-called parallel-axis theorems . Two points in particular should be noted. Fir st , the axes betw een which th e transfer is made must be pa rall el , and second, one of the axes must pass through the centroid of th e area . If a tra nsfer is desired between two parallel axes neither of which passes through the centroid, it is first necessary to transfer from one axis to the parallel cent roida l axis and the n to transfer from t he centroidal axis to the second axis. The parallel-axis theo re ms also hold for rad ii of gyrat ion . With substitution of the definition of " in to Eqs. A/B , th e transfer relation becomes (A / 6b)

whe re Ii is the rad ius of gyrat ion about a cent roida l axis parallel to th e axis about which" applies and d is th e dist anc e betw een the two axes. Th e axe s may be eit her in th e plan e or norm al to t he plan e of the area. A summary of the moment -of-inertia relations for some common plan e figures is given in Table D/ 3, Appendix D.

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Defi ni t i on s

431

432

Appendi x A

Area Moments of Inertia

Sample Problem All

,,

y

Determine the moments of inertia of the rectangu lar area about the centroidal xo- and Yo-axes, th e centroidal polar axis Zo th rough C, th e x-axis, and t he polar axis z t hrough O.

Th

Yo ,

, ,

,,

2

Solution . For t he calcu lation of the moment of inertia Ix about the xo-axis, a

Q) hor izontal stri p of area b dy is chosen so that all elements of th e strip have th e sa me y-coordina te . Thus,

+

Ans. Th e centroidal polar moment of ine rtia is

An s. By the parallel-axis theorem the moment of inertia about the z -axis is I, = f-"bh 3 + bh

1, + Ad/l

J

c,

_L

- - xo

2

J.- !",o_-,--_~ __ x

By int erc hange of symbo ls, t he mome nt of inertia abou t th e centroida l Yo-axis is

~

I

h

r - b ------1

An s.

II,

dy

(~)

2

= l bh 3 = ¥h2

Ans.

Helpful Hin t

Q) If we had started with the second or der element dA = dx dy, integration with respect to x holding y constant amounts simply to multi plication by b and gives us the expression y 'J.b dy, wh ich we chose at t he outset.

We also obtain the polar moment of inertia about 0 by t he parallel-axis theorem, which gives us [I, =

I , = f-,A(b 2

1, + Ad 2 ]

+ h 2) + A [ ( b2 ) 2 + ( _h 2] 2)

I, = lA(b 2 + h 2 )

Ans.

Sample Problem A ll Determine the moments of inertia of the triangular area about its ba se and abou t parallel axes through its centroid and vertex.

CD

Solution. A strip of area parallel to the base is selected as shown in th e figure,

(1) and it has the area dA

II,

~

fy 2dAJ

I = ,

~

x dy

~

fh0 y 2 h

[(h - y )b/ hJ dy . By defin ition - Y b dy = b h

[i _y4]h 4h

3

0

= bh

3

12

Ans. Helpful Hin ts

By th e parallel-axis theor em th e momen t of iner tia 1 about an axis through t he

[1

~I

- A d 2j

1=

b~3

An s.

An s.

@ Expressing x in terms of y should

2 _ (bnm

= b;:

A tran sfer from the centroidal axis to the x ' -axis thro ugh the vertex gives [I ~

1 + Ad2 J

3 3 I . = bh + ( bh ) ( 2h)2 = bh x 36 2 3 4

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CD Here

agai n we choose the simp lest possib le clement. If we had chose n dA = dx dy, we would have to integrate J ''J. dx ely with respect to .r first. This gives usy~x dy . which is the expression we chose at the outset.

centroid, a distance hl3 above the x-ax ie, is

cause no difficulty if we observe the proportional relationship between the similar triangles.

Articl e A/ 2

Sample Problem A/3

Defin itio ns

433

y I

Calcu late th e moments of inertia of the area of a circle abou t a diametral axis and about the polar axis through the center. Specify the radii of gyr ation.

CD

Solution. A differential element of area in th e form of a circular rin g may be used for t he calculation of the moment of inertia about th e polar a-axis thro ugh since al l elements of the rin g are equ idistant from O. T he elementa l a rea is dA = 21Tro dro, and thu s,

o

[I ,

~

Jr

2

dA )

I,

~

r r0

J

2(

o

2rrro dro )

~

4

";

~

0 ,-2

Ans.

Th e polar radiu s of gyration is

= .!:-

k By sym met ry i , [lz

= t, +

Iyl

~

Ans.

,/2

Z

I y • so that from Eq . A/3 rrr 4 2 I x ~ 2'"z lr = _ 4 = l4 A r

Ans.

Th e radiu s of gyrat ion about th e diam etral axis is

k

x

= !:.2

Ans.

T he foregoin g det ermination of Ix is the simplest possible . T he resu lt may also be obta ined by direct integr ation, usin g the eleme nt of are a dA = "o dro d fJ shown in th e lower figur e. By defin ition

Helpfu l H ints

CD Polar coordinates are certainly indi cated her e. Also, as be fore, we choose th e simplest and lowest-order element possible, which is th e differ ential rin g. It should be evident immediately from the definition t ha t the polar moment of inertia of th e r ing is its area 27Tro dro times r02 .

@ Thi s int egration is st ra ightfor wa rd, Ans.

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but th e use of Eq . A/3 along with th e resul t for I z is certainly simpler.

434

App endix A

Ar ea Moments of Inerti a

Sample Problem A /4

,

)'

3 ,L - - - - - - -

Det ermine th e moment of inertia of the area und er th e parabola about th e x· axis. Solve b.... using (a ) a horizontal strip of area and (b ) a verti cal strip of area.

Solution. Th e constant k = ~ is obtained firs t by substituting x )' = 3 into th e equation for the parabola.

4

~ Jy2 dAl

t,

=

f

I

an d

O ~--------' -

o

(a) Horizontal strip. Since all parts of the ho rizontal stri p are the same distance from the .r-axis, th e moment of ine rtia of th e st rip about t he .r-axis is)'2 dA where dA ~ (4 - x ) dy ~ 4(1 - y2/9 ) dy . Integrating with respect to y gives us

IIx

, , ,I

4

,

)'

f- x

-)1':::::= :::::;:=1

52

Solution (0)

y L-

4y 2 ( 1 - y:) dy = 7 = 14.40 (u nits)'

-- x

-'-___'

X

A ns .

,

)'

(b) Vertical strip. Her e all parts of th e element are at different distances from th e .r-axis, so we mu st use th e correct expression for th e moment of ine rti a of th e elementa l rectangle about its base, which, from Sample Problem AyLvis bIl 3 / 3. For the width dx and th e height y the express ion becomes

To int ewate with respect to .r, we must express y in terms of x, which gives y = 3Jx / 2, and the integral becomes

Ix ~ 1 i'

CD

3

0

2

~

72 -5

~

Solution (bl

, I

y

I

1-- x- -++,---'- - - x .... six ~

dl , ~ ~(dr)y3

(3 Jx)3 dx

I I

. , 14 .40 (u nits)

Ans.

Sample Problem A I5

Helpful Hint

CD Ther e is litt le prefer ence bet ween Soluticns (a ) and (h) . Solut ion (b ) requi res knowing the moment of inertia for a rectangular area about its base.

6~-l\---x:,

Find t he moment of inertia about the x-axis of th e semicircular area.

- -- x

I

Solution. Th e moment of inertia of the semicircu lar area about th e x' -axis is one- half of t hat for a complete circle abou t th e sa me axis. Thus, from the results of Sa mple P roblem A/3 2 4

CD

15 0101

l

----- ---- -~----- - x

8

We obtai n t he moment of ine rtia I about th e parall el cen troidal axis Xo next . Transfer is made th rough the distance r ~ 4r/ (3,,) ~ (4)(20)/( 3 ,,) ~ 80/ (3 ,,) mm by th e parallel-axis theorem. Hen ce,

Helpful Hint

[1 = I - Ad2)

CD This problem

Finally, we transfer from th e centroi dal xo-axis to the x-axis. Thus, [l

~ 1 + Ad 2

J

I,

~ 1.75500' ) + ( 20;,,)( 15 + :~)2 = 1.75500' ) + 34.7(10' ) ~ 36.4(10') mOl'

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An s.

illustrates the caution we should observe in using a double transfer of axes since neither the x ' nor the .r-axis passes through the centroid C of the area. If the circle were complete with the centroid on the x ' -axis. only one transfer would be needed.

Article A / 2

Defini tion s

4 35

Sample Problem A/6 Calculate the mome nt of inertia about t he x-axis of the area enclosed between t he y-ax is and t he circu lar arc s of ra dius a whose centers are at 0 and A .

Solution. The choice of a vertica l differential str ip of area permits one integr a-

a

t ion to cover th e entire area . A hori zontal str ip would require two integrat ions wit h respect to y by virtue of the discontinuity. The moment of inertia of the strip about th e x-axis is t hat of a strip of height Y2 min us that of a str ip of heig ht )' 1' Thu s, from t he results of Sample P roblem A/ I we write d lx

CD

= ~( )'2 dX)Yl - ! (Yt dX)Yt 2 = ~(y~?

o

- Y 13) dx

Th e values of Y2 and Y l are obtained from the equ ations of the two curves, which are x 2 + yl = a 2 and (x - a )2 + YI 2 = a 2 , and which give Y 2 J a 2 - x 2 and Yl = J a 2 - (x - a )2. Thus, Ix

l

=:\ o

{ (a 2 - x 2) Ja 2 - x 2 - [a 2 - (x - aj2 ] J a 2 - (x - a )2}

dx

ra dicals here since both Yl and)'2 lie above the x-axis.

Simultan eou s solution of th e two equ ations which define the two circles gives the .r-coordinate of th e intersection of the two curves, which, by inspect ion, is a/ 2. Evaluation of the integrals gives

_( t2a2/ a2 [ 12

(r

_

a )2J a 2

Collect ion of t he integrals with the factor of ~ gives a4 Ix ~ 96 (9}3 - 2,,) ~ 0.09690'

An s.

If we had sta rted from a second-or der element dA = dx dy , we wou ld write y2 dx dy for the mom en t of iner tia of th e element about th e .r-axis. Integr atin g from Y I to)'2 holding x constant produces for the ver tical strip

whic h is th e expression we st arted with by having the moment-of-inertia result for a rectan gle in mind.

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Helpful Hint

CD We choose the posit ive signs for t he

ar2

436

Appendi x A

Area Moment s of In ertia

PROBLEMS

A/4 Th e t hin qu arter-c ircular ring has a n area of 1600 mm 2 • Determine th e moment of ine rt ia of the ri ng abou t the .r-axis to a close a pproximation.

Introductory Problems

j'

All If the mome nt of inertia of th e t hin strip of area about th e x-axis is 2.56(06 ) mm''. determine th e a rea A of th e stri p to wit hin a close a pprox ima tio n. Ails. A = 1600 mm 2

I I

- - x

o Problem A /4 Problem A/ I

All Determine by direct integratio n th e momen t of inerti a of the triangu lar area about t he y-axis.

A/5 Det erm ine the polar moments of inert ia of the sem icircula r area about points A and B. 3 , • III An s. IA = 4"rrr

Problem A /2

>_-

-

- ;::"--

-

r' (34" - :!3 )

- -'- - - x

A

All The moments of inerti a of the area A about the parallel p- a nd p '·axes differ by 50 in." Compute th e a rea A, which has its centroid a t C. Ans. A = 10 in.2

Problem A j5

A/6 Calculate th e mom ent of inert ia of th e sha ded area about th e y -axis.

p p'

c.

-: A

Problem A/6

Prob lem A/3

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Art icl e A / 2

A/7 Show that the moment of inertia of the rectan gular area a bout t he x-ax is t hrough one end may be used for its polar momen t of inertia about point 0 when b is small compared wit h a . What is the percen ta ge er ror n wh en b [a ~ 1/1O? Ans. n = - 0.249% 1

o!- - ----''------- - Ib a

1

Problems

4~ 7

Representative Problems A/lO T he area of th e narrow st rip of u niform widt h is 750 mm 2 . Using t he fact th at th e widt h is small compared wit h t he len gth of the st rip, approximate its moment of inertia about t he x-axi s. Compare your answer wit h t he erroneous result uf multiplying the ar ea by t he square of its distance from its centroid to th e x-axis.

-1-

1

1 .r

\00 ",,,,

- T-

Problem A/7

l OO m m

~r:

A/8 Deter mine the moments of iner tia Ix and l y of th e ar ea of the t hin semicircular ring about t he x- and y-axes. Also find t he polar moment of inertia I e of th e r ing about its centroid C.

l_x Problem All 0

y

Alll Determine the polar radii of gyration of th e t riangu lar area about points 0 a nd A. o

1 1

A ns. k o

1

~ t«

~ 0,

kA

~

jij

1

Ct

""'r r""" I:

A

____ ___ .'::>L

_

°

0

I I I I I I

---x o

Problem AlB

I 1

A/9 Ca lculate th e moment of inertia of th e rectan gu lar area about t he x-axis and find t he polar momen t of iner tia abo ut point O. A ns. Ix = 7.2(10 6 ) mm", 10 = 15.95(0 6) mm" y 1

o

1

25mm

60mm

1

--- - ----- . J O Problem AliI

Alll Determine the radius of gyration about a polar axis thro ugh the midpoin t A of the hypotenuse of the righ t-t rian gular area . (H int: Simplify your calculation by observi ng the results for a 30 x 40- mm rectangu lar area.)

lOOm m

A

Problem A/9

40 mm

30 mm Problem A/12

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438

App en d i x A

Ar ea Mo ments of Inert ia

A/1J Determine th e rectangu la r and polar ra dii of gyra-

A/16 Th e moment of inertia about the .r-axis of th e rec-

tion of the shaded area abo ut the axes shown. Ans. hx ~ 0.754, h.y ~ 1.673, h, ~ 1.835

ta ngle of area A is approxi mate ly equal to Ad'l. if II is small compa red with d. Deter min e an d plot th e percentage error n of th e ap prox ima te value for hId ratios fro m 0.1 to 1. What is the percent age er ror for h ~ d /4?

.v I I

p~l- -

b

- xu

d

1

;-

o --~'::- L-_---'--_ x o 2

--1-_

Problem A/16

Problem A/ll A/14 In two different ways show that t he mome nts of iner tia of th e square area about the x- and x ' -axes are

A/17 Calcu late the moment of inertia of th e shaded area about the .r-axis .

.v

the same.

;-

;-

-::...---

a I I

afl

I I

I 30 nun

I I

«n

I I

---- -40mm ---- -- ------ x Proble m A/17

Problem A/14 A/15 Determine the polar rad ius of gyration of the area of th e equilateral triangle about the midpoi nt M of its

base. An s. k,\1

=

A/18 Determin e t he momen t of inert ia of the quartercircular area about th e tan gen t .r ' -ax is.

a

./6

.v I

I

- - - - - -x·

a ""-- - - - - ' - - - - x

Problem All 8 afl

a fl

Problem A/15

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Ar ti cle A /2 A/19 Determine the momen t of inertia of t he shaded area ab out t he x-axis usin g (a ) a hor izontal strip of area an d (b ) a vert ical strip of area . 16ab3 An s. Ix = 105

Pr o bl ems

439

A/l1 Calculate by direct integration th e momen t of iner t ia of the sha ded area about t he x-axis . Solve, first , by usin g a hor izontal str ip having differen tia l area and, second, by usin g a vert ical st rip of differential area. y

.v

I I

4"

x2

y = b (1 - a 2

)

4"

b

--------- x l-

-x

-----'

Problem A/22

a Problem A/19

A/20 Determine th e moment of iner tia abou t th e .r-axis

and the polar rad ius of gyration about point 0 for the semicircular area shown. y I I I I

All1 The plane figu re is symmet rical with respect to the 45° line a nd ha s an ar ea of 1600 mm 2. It s polar moment of inertia about its cent roid C is 40( 04 ) rnm''. Compute (a ) the polar rad ius of gyr at ion ab out 0 and (b ) the radiu s of gyration ab out the xo·axis. Ans. (a) I~ o = 45 .3 m m, (b ) I~ xo = 11.18 mm y

r

I

I

Yo

I

I I

I

1---- 30 mm -r--l I I I

I I

I I

AI- - - - -

i

B

I I I

L

_

o

r

I I

1, I

A/11 Determine the mom ents of inertia of the shaded area about the x- and y-axes. Use the same differe nt ial eleme nt for both calculati ons. An s. t, = a 4 /28 , ly = a4 / 20

-- - x

I I

/

I

,,

I

I 0'

x2

a

8 -8 --------

R .r

a

Problem A/24

Problem A/2! Marwan and Waseem AI-Iraqi

1

y

1/

, : (1 ,, ,, _________ L

/

A/14 Deter mine the momen ts of iner tia of th e circu lar sector about the x- an d y-axes .

,

;-.v = ,

30mm

Problem A/l3

I /

,I

'

o~~~~ - - - - - - -

Problem A/20

.v

/

,?'-----r 'o

i

- - - - --x

./

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- -x

440

App endi x A

Ar e a Momen ts o f Inertia

AilS Determ ine t he rad ius of gyra tion about the y-axis of the shaded area shown. Ans . k)' = 53.1 mm

A/lS Calculate t he momen ts of ine rtia of th e shaded area about th e x- and y-axe s, and find the polar moment of ine rtia about point O.

y

y I

I I I

1

- _ _

'!::

I I I

I I I

_

,

I I I I I I I I I

,,, I I

' 4"

I I I I

y =k,r' , I

0' ------ -- - - Problem A/2S

y

I I I

r

I I I I I I II

A/l9 Determine th e rectan gular moments of ine rtia of the sha ded area about th e .r- and y-axes a nd the pola r radius of gyration abou t point O. b b3 Ans. Ix = ~h 4 - a4 ) , Iy ~ - -3(h 4 - a 4 ) 4h 4810 ko

-' -0

r- ~

Problem A/26

All1 Determine t he moment of inertia of t he shaded area about the y -ax is. y. mm I I

40 r - - - - - - - - - - - - -· I ' I 'f.~ : I 1I I '$: I ' I : I :

o o

JH

1 +

1~2)(h2 + a

: I

2

)

~ --j

iI

1 h

------~~l~~~ l__, o

Problem A /29

I

- - - - .J.. - · x, mm 20 40 Problem A /27

Marwan and Waseem AI-Iraqi

~

y

- - - x

0

x

Problem A/2S

A/l6 From considerations of symmetry show tha t Ix' l.v· == Ix = 1)' for the semicircular area regardless of the angle 0' .

y;

I _---l _ _

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Art icle A /2 A/10 By the methods of this art icle, det ermine the rectan-

gula r and polar radii of gyra tion of t he shaded area abou t the axes shown. .l' 1 1

a

L -_ _

~

..

:

~ ~I+

L -_ _

~

__ x

--- x

Problem A/30

.... A/ll Calculate the moment of inertia of the shaded area of th e two over lapping circles about the .r-axis. Ans. Ix = 0.19881'4 .l' 1

,,_--- __ 1,-----_' /

/""

r

:

I

---:---- --+-\

1

\

I I

"

,

,

/"\

\

- - --:--- x I

/ /

"

',,-----~I-- -----_/ 1

Problem Al31

Marwan and Waseem AI-Iraqi

441

.... Ajll A narrow st rip of area of constant width b ha s the form ora spiral r = k O. Afte r one comp lete tu rn fro m o = 0 to 0 = 27T, t he end radius to the spira l is R . Determine t he polar moment of ine rt ia and the ra dius of gyration of th e area about O. A ns. / 0 ~ 1.609R 3b, li o ~ O.690R

o

/

Problem,

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\ Problem A/ 32

442

Append i x A

Are a Moment s of I ne rt ia

A/3

COMPOSITE AREAS

It is frequently necessary to calculat e the moment of inertia of an a rea composed of a number of distinct parts of simple and calculable geometric shape. Because a moment of inertia is the integral or sum of th e products of dist an ce squared times eleme nt of a rea, it follows t ha t the moment of inertia of a positi ve area is always a positi ve quantity. The moment of inertia of a composite area about a particular axis is therefore simply the sum of the moment s of inertia of its component parts about the same axis . It is often convenient to regard a composite area as being composed of positive and negative parts. \Ve may then treat the moment of inertia of a negative area as a negative quantity. \Vhen a composite area is composed of a large number of parts, it is conveni ent to tabulate the results for each of the parts in terms of its area A, its centroidal moment of inertia 1, the distance d from its cen troidal axis to the axis about which the moment of inerti a of the entire section is being computed, a nd the product Ad 2 . For an y one of th e parts t he moment of iner tia about the desired axis by th e tran sfer -of-axis theorem is 1 + Ad 2 . Thus, for the entire section the desired moment of inertia becomes I = ~1 + ~Ad2. For such an area in the x-y plane, for example. and wit h the notation of Fig. A/4 , wher e Ix is the same as Ixo an d I,. is the sa me as I,. 0 the ta bu lat ion would include

Part

Sum s

Area, A

dx

dy

~A

Ad}

Ad, 2

Ix

I,

~Adx 2

~Ady 2

~I

~I .

From the sums of the four columns, t hen , the moments of inert ia for th e composit e area about the x - a nd y-axes become

t,

si,

I.v

~1J' + ~AdJ' 2

+ 2.Adx 2

Although we may add th e mom ents of in ertia of th e individua l parts of a composite area about a given axis, we may not add their radii of gyration. Th e radius of gyration for t he composit e area about th e axis in questi on is given by k = , II/ A , where I is the tot al moment of iner tia a nd A is th e tot al area of the composite figur e. Similarly, th e radiu s of gyration k about a polar axis through some point equals Hz/A, wher e I , = Ix + I , for x -y axes through that point.

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Article A I]

Composite Ar eas

443

Sample Problem A /7 Calcu late th e moment of ine rtia and radius of gyration about the s -axis for th e sha ded area shown . 30mm

~- 40 mm

Solution. Th e composite area is composed of th e positive a rea of the rectangle and the negative areas of th e qu arter circle (2) and triangle (3). For the rectangle the momen t of inertia about the .r-axis, from Sample P roblem All (or Table D/ 3), is

(1 )

1% = 0h 2 = ! (80)(60)(60J" = 5.76(106 ) mm'

I

X'~0f'J.[J LV

Xo

From Sa mple P roblem A/3 (or Table D/ 3), th e moment of inertia of t he negative quarter-circu lar area about its base axis x ' is

III ---

I . = - -I (""') = - -16 u-) " mm' ". (30), = - 0 1590Il " % 4 4 . We now tra nsfe r thi s result through the distance F = 4r / (37T) = 4(30)/ <37T) = 12.73 mm by th e tran sfer-or-axis th eor em to get th e centro ida l moment of ine rtia of part (2) (or use Tab le D/ 3 directly).

CD

2

1%= - 0.1590(106 )

(1 = I - Ad 2 ]

-

,,-(30 ] [ - - 4- ) 02.73)2

- 0.0445( 06 ) mm" Th e moment of inerti a of t he quarter-circular part about the .r-axis is now

@ (I = I +

Ad 2 )

1% = -

0.04451106 )

+ [-

"-(~0)2]

(60 _ 12.73)2

-

= - ,\;(40)(30)' = - 0.09(06 ) mm'

1.6240 0") - 0.09(06 ) = 4.05(06 ) mm'

Th e net area of the figure is A = 60(80) - !,,-(30 )2 - ! (40)(30) so t hat the radius of gyration about t he x-axis is

= 3490

k% = J I%/A = J 4.050 0")/ 3490 = 34.0 mm

MarNan and Waseem AI-Iraqi

CD Note

that we mus t transfer th e moment of inertia for th e quartercircular area to its centroidal axis Xo befor e we can transfer it to th e .raxis, as was done in Sample P roblem A/ 5.

® If

The total moment of inertia about the z-axi s of t he composite area is, conseq uently,

1% = 5.76(06 )

Helpful Hints

A carry neg ative signs .

Fin ally, th e momen t of inertia of the negative trian gu lar area (3) abou t its base, from Sample Problem A/2 (or Table D/ 3), is

- bbh'

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~-X

@ We wat ch ou r signs car efully here. Since th e area is negative, both I a nd

- 1.624( 06 ) mm"

1% =

/1

Ans. mm2

Ans .

ther e had been more than th e three part s to the composite area , we would have ar ranged a tabu lation of th e I term s and th e Ad 2 terms so as to keep a systematic account of th e term s and obtain J = ~ l + ~Ad2.

444

Appendix A

Area Moments of Inertia

PROBLEMS

y I I

Introductory Problems A/33 Det ermine t he momen t of ine rt ia about th e .r-axis of t he square area with ou t a nd with th e central circ ular hole. A ns. Ix ~ 21.3R ' , Ix ~ 20.6R'

2R

y I I

'----

--' --x

2R Problem AilS

A/36 Calcu late the polar ra dius of gyr at ion of the area of th e a ng le section about point A. Note t ha t the width of the legs is sma ll compared with th e lengt h of ea ch leg.

r

Problem Alll

r- 1.5 "

15"

A/34 Determine th e pola r moment of inertia of th e circu lar area wit hout a nd wit h th e cent ral square hole.

~~i======1~"

y

- - 20 " -

I

-..,

I

Problem A/36 A/37 Th e cross-sect ional ar ea of a wide-flange I-bea m has the dimen sions shown. Obtai n a close app roximation to th e ha ndbook value of Ix = 657 in ." by treating th e sect ion as bein g compos ed of three rectangl es. A ns . Ix = 649 in .4

R R

~+

R ---

--x

y

I I

-r

I

Problem A/34 A/35 By the method of thi s a rticle, determine th e rectangular an d pola r ra dii of gyra tion of the shaded a rea, repeat ed here from Prob. A /30 , about the axes shown. ./5 jiO A ils. i; = k y = -:ta, k z = 4 a

. -

x

16.25"

~J

0.380" -

f.- 7.073" -~ 0.628" Problem A/l7

Marwan and W aseem AI-Iraqi

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Art icle A /3

A/3S Determ ine the moment of iner tia of th e shaded area about t he s -axis in two ways. Th e wall thi ckn ess is 20 mm on all four sides of t he rectangle.

of the bea m about its centroida l xo·axis. An s. Ix = 10.76( 0 6 ) mm"

r--'2°~1 , ' !.

I

I I

20 mm_l~L=====l=_=--=-=-~I~:~-, 1-- - - -

360 mm

---~J

445

A/41 Calculate th e momen t of inertia of th e cross section

y

:-

Problems

20 mm

~-

_t-

20 mm

Problem A/38

r

I

-r

r--

---1

Problem A/4 1

Aj39 Determine t he moments of inertia of the shaded area about the x - and y -axes. Ans. l, = 4.94a4 , I y = 3.37a 4 y

Representative Problems A/42 Determine the momen ts of ine rtia of th e Z-secti on abou t its cen troidal xo- and yo·axes.

I I I

Li

, oo mm

-1

r : L

20 mm L l -_ _,

2a

:0

a

+

140 mm

--x

~ - xo

--Ie

! - 20 mm

Problem Al39

.-

~1 20 mm

'-

A/40 Calcu late the moment of inerti a of the sha ded area about the .r-axis.

-

100 mm ----1

-r

Problem A/42

Aj43 Deter mine th e moment of inert ia of the shaded area abou t th e .r-axis in two differ en t ways. Ans. Ix = 5f a 4 90 mm

1

40

a , 30 I mm

--- +'- - -

a

I

-------,-

--x

50 mm -1---50 mm --1

I I

a

-x a a

Problem A140

Problem A 14 3

Marwan and Waseem AI-Iraqi

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446

Append ix A

Area Moments of Inert i a

A/44 Determine th e polar radius of gyration about point A for th e sha ded area shown. 80mm

Aj47 Calculat e th e polar radius of gyration abou t point 0 of th e ar ea shown. Note th at the widths of th e elemen ts are small compared wit h t hei r length s. A ns. k o = 7.92 in. )'

80 mm

I I I

I I I I 60mm 1

- - - - -"'-I - ·A

3" 4

Problem A/44 A/45 Derive th e express ion for th e moment of inertia of th e trapezoidal area about the s -axis through its base.

====:=:::J-I

- - Ie . 0 1 1

-x

3"

4

Problem A/41

A/48 Th e rect an gular ar ea shown in part a of th e figure is split into three equal areas wh ich are t he n arra nged as shown in part b of th e figur e. Determine an expression for the mom ent of inertia of the area in part b about the centroidal .r- axis. Wh at per cent increase n over th e mome nt of iner tia for area a does th is repr esent if II 200 mm and b = 60 mm?

Problem A/45

A/46 A floor joist whic h measures a full 2 in. by 8 in. has a l -in. hole dri lled thro ugh it for a wat er-pipe installat ion. Determine th e percent redu ction II in the moment of inertia of th e cross-sectional area a bout th e .r-axis (compared with th at of the undrilled joist> for hole loca tion s in th e ran ge 0 S J' S 3.5 in. Evaluat e you r expression for y 2 in.

4"

1

-,

1"

,

T

)'

.L

- x

L :'::.

4"

"

Section A

Problem A /46

Marwa n and W aseem AI-Iraqi

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Art icl e A / 3

A/49 Develop a form u la for the momen t of inertia of the regu lar hexagonal a rea of side a about its central

5

x-axis.

13

An s. I = - ' .T 16

P robl em s

447

A/52 Determine the momen t of ine rtia of the shaded area about the x-axis . y

04

I I I

a

I

I

a

I 10

- -- j - - -

1 1

--x

I

I

Problem A/ 52 Problem A/49

A/50 Calcu lat e t he polar ra dius of gyration of the shaded a rea about it s cent roid C. 1001 11001

r--

1

--------- r--

--r

A/53 Calcu late the moment of ine rtia of t he standard 12 x 4-in. cha nnel sect ion about the centroida l xo· axis. Neglect the fillets a nd ra dii a nd compare wit h the handbook val ue of Ix = 16.0 in." Ans. Ix = 16.00 in .4

"1 r- 0.65" 070"

J

I

600

-
Xo

160

i...-

~L

I

Yo

~500

I

Dimensions in millimeters Prob lem A/50

A/51 Calcula te the polar moment of iner tia of the shaded ar ea about point O. An s. I z = 0.552(0 6 ) mm''

I 15 ~m I

_ ___ .i_ l

1--- - - -

0.75" __ _ t_ _~ ~L I

Problem A/53

.. A/54 For the 2 in. by 8 in . floor joist wit h t he circu lar hole of Prob. A/ 46, deter mine a n expression for the per cent n by which the momen t of inertia of the shaded area about its cent roida l x ' -axis (pa ra llel to x ) is reduced from t he moment of ine rtia of the complete u ndrille d sect ion about th e x-ax is. Expr ess n in terms of y for the range 0 =:; y :s; 3.5 in . Eval uate your expression for y = 2 in. Ans. n ~ 0.1953 + 2.68 y 2, n ~ 10.91%

_

o Problem A/51

Marwan and Waseem AI-Iraqi

12"

- - - xo

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448

App end ix A

Are a Moments o f In er tia

Ill- A/55 For the H-beam section, determine the flange width b wh ich will mak e the moments of inertia about th e central .r- and y -axes equal. (H i n t: The solution of a cubic equa tion is req uired here . Refer to Art . C/ 4 or C/ I I of Appendix C for solving a cubic equation.) A ns. b = 16 1.1 mm

r l

Ill-A/56 Calculate the value of h for which Ix = l y for th e sha ded area shown. (H int: Read the hint for the previous problern.) A ns. h = 20.0 mm y 1 1

10 110 1 1 1 1

y

,

b

+

-

h

I

%

30

l Omm

I I

30

Dimens ions in millimeters Problem A/56

-1

t.

L

IOmm -J · ~ l OO m m ~ -

lO mm

Problem A /55

MarNan and Waseem AI-Iraqi

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Products of I nertia and Rot at ion of A x es

Art icle A/4

A/4

449

PROOUCTS OF INERTIA AND ROTATION OF AXES

In this article, we define the pr oduct of inertia with respect to rectangular axes and develop the pa rallel-axis th eorem for centroidal and noncentroidal axes. In addition, we discuss the effects of rotat ion of ax es on moments and products of inertia. Definition

In certain problems involving unsymmetrical cross sections and in the calculation of moments of inertia about rotated axes. an expression d Ixy = xy dA occurs, which has the integra ted form

(A /7l

wher e x and yare the coordinates of the element of area dA = dx dy , The quantity I xy is called the product of inertia of the area A wit h respect to the x-y axes. Un like moments of inertia. which are always positive for positive areas, the product of inertia may be positive, negative. or zero. The product of inertia is zero whenever either of the reference axes is an axis of symmetry , such as t he x-axis for t he area in Fig. A/ 5. Here we see that the sum of the ter ms x( - y) dA an d x (+y ) dA du e to symmetrically placed elements vanishes. Because the entire area may be considered as composed of pair s of such elements, it follows th at the product of inertia I xy for the entire area is zero.

By defin ition the product of inertia of th e area A in Fig. A/4 wit h respect to the x- and y-axes in terms of the coordinates xo. Yo to the centroidal axes is

= =

f f

I

I ...X I

I

0T

1

+Y

----t------t---Y

1

1 1 k- X

O -.l

I

Figure A/ 5

Transfer of Axes

Ixy

-r

(x o + dyHYo + d x ) dA xoYo dA + d x

y I f-o-

f

Xo dA

+ d;

f

Yo dA + dxdy

f

Yo

I

- d,

I

xo

I

I

A

Yo

1 1 1 1

I

- - Xo

~

d

1/

o~ - - -- - - - - - - - - -

1

-- x

Figure A/4, Repeated Marwan and W aseem A l-lraqi

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dA

-- x

450

Appendi x A

Ar e a Moments o f In ert ia

The first integral is by defin iti on th e product of inertia about t he centroidal axes, which we write as IX)'. T he middle tw o integrals are both zero because the first moment of the area about its own centroid is necessa r ily zero. The fou rt h term is merely drdyA . Thu s, th e tra nsfer of-axis theorem for products of inertia becomes (A /8)

Rotation of Axes The product of in ertia is usefu l when we need to calculate the moment of inertia of an area about inclined axes. This consideration leads d irectly to th e important problem of determin ing th e axes about which the moment of inertia is a maximum and a minimum . In Fig. A/ 6 th e mom ents of inertia of t he are a ab out the x ' - a nd y'·axes are

Ir · Iy '

f =f

y'2 dA x '2

dA

f f

(y cos 0 - x sin 0)2 dA (Y

sin 0 + x cos

0)2

dA

where x ' and s' have been replaced by their equivalent express ions as see n fro m the geom etry of the figure.

Figure A i 6

Expanding and substituting th e trigonometric identi ti es . 2

sm 0

=

1 - cos 20 2

2

cos 0 =

1 + cos 20 2

and th e defining relations for Ir , Iy , Iry give us

_ Ix + Iy I - I l x, - - 2- + cos 20 - IX)' sin 20

Y

t, + I y

1.= y

Marwan and W aseem AI-I raqi

2-

t,

(A / 9)

2

Iy

cos 20

+ IX)' sin 20

Article A/4

Produ cts of Inert ia and Rotatio n of A xes

In a si milar manner we write the product of inertia about the inclined axes as I x 'y '

=

f

x 'y ' dA

~

f

+ x cos lI)( y cos 0 - x sin

( y sin 0

8) dA

Expanding and substituting the trigonometric identities sin () cos H = ~ sin 26

a nd th e defin ing relati ons for I x

cos'' fJ - sin2 6 = cos 28

t; I y , I xy give us

- I 2 y sin 20 + IXY cos 26

(A / 9a )

Adding Eqs. A/9 gives Ix' + I y' = Ix + Iy = I" th e polar moment of in er tia about 0 , which checks t he results of Eq. A/3. The angle which makes L and 1\, either maximum or minimum may be det ermined by sett ing th e deri vative of either Ix' or I y' with respect . to II equa l to zero . Thu s, dl x'

dO

~

U - Ixl sin 20 - 2I x.v cos 20 Y

=a

Den oting t his critical angle by c gives

(A/ lO)

tan 2"

Equ at ion A/l0 gives two values for 2" which differ by", since tan 2" = tan (2" + ,,). Conseque ntly the two solutions for" will differ by ,,/2. One value defines the axis of maximum moment of inertia, and the other value defines the axis of minimum moment of inertia . These two rectangular axes are called the principal axes of inertia . When we substitute Eq . A/ la for the critical value of 20 in Eq . A/9a , we see that the product of inertia is zero for the principal axes of inertia. Sub st itution of sin 2" and cos 2", obtained from Eq. A/ la, for sin 211 and cos 20 in Eqs. A/9 gives the expressions for the principa l mom ents of inertia as

[ max

l mi n

=

t, + Iy 2

= Ix + Iy 2

I

+ 2 Jux

l) y2

+ 4Ixy2

(A/ m I /

2 -, u,

- 1y )2

+

41xy2

Mohr's Circle of Inertia

We may represent the relatio ns in Eqs. A/9, A/9a, A/la , and A/l! gr apllm lf il)ljlJVd~"mw. Il'allled Moh r 's circit;wlil .~ tili>Ies of Iy ,

t;

451

452

App endi x A

Ar ea Moments of In ertia

lX)o I

1

1-- - - - - 1,- - - - --'

1

r- - - -

I ,. -

-

-

---1

1

~ lmin

1 1 1 1 1

>--- -

I

1 1 1 1

r-

I,.

D

I.' I 1- - 1.',. - -, ll max - - - - - -.J Figure A/7

and Ixv the corresponding values of Ix·, I,.., and Ix'v· may be determin ed from th e diagram for a ny desired a ngle rioA hori zontal axis for th e measurement of moments of inertia and a vertical axis for the measurement of prod ucts of inertia are first selected, Fig. A/7. Next, poin t A , whic h has th e coordina tes (1" Ixv )' a nd point H, which has the coordinates (1, ,, - Ix,Y )' are located. . We now draw a circle with these two points as the extremities of a diameter. The angle from th e radius OA to t he horizontal axis is 2a or twice the angle from the x-axis of the area in quest ion to the axis of maximum moment of inertia. Th e angle on th e diagram and th e a ngle on the area are both measured in the same sense as shown. The coordinates of a ny point Care (1x" Ix'y ' )' and those of the corresponding point D are (1,Y., - Ixy l. Also the a ngle between OA a nd OC is 28 or twice t he angle from the .r-axis to the x' -ax is. Again we measure both angles in the same sense as shown. \Ve may verify from the trigonometry of the circle that Eqs. A/9, A/90, and A/l0 agree with th e statements made.

Ma rwan and W aseem AI- Iraqi

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Article A/4

Products of Inertia and Rotation of Axes

453

Sample Problem A/B Det ermine th e product of inerti a of the rectangu lar area with centr oid at C with respect to th e x -y axes parallel to its sides.

Solution. Since the product of inertia IX)' about th e axes :Co-Yo is zero by symmetry, th e t rans fer -of-axis theor em gives us II", = / ", + d.,d,AJ

An s.

In this exam ple both d x and dy ar e shown positive. We mu st be carefu l to be consiste nt with th e positive direct ions of d, and dy as defined . so that th eir proper signs are observed .

Sample Problem A/9

Y I

Determine the product of inertia about the x-y axes for the area under t he parabola.

br - - - - - - - - - I

I I I

dx x --!jJdy

r--o

Solution. With t he substitu tion of x = a wheny = b, the equat ion of t he curve becomes x = ay2/ b2.

=

dx dy , we have

xy dx dy . Th e int egral over the enti re are a is

Ix, =

fbfa, o

xy dx dy =

ay 'ljb 2

(a 2_ab2~4) y dy Jo

02

0

20

:c

6a

b

I

I

:cal

0

Helpful Hin t

CD If

Yo

- - x

I I I

r" -21

sa ve one int egration by using t he resul ts of Sam ple Problem A/ S. Ta king a vertical st rip dA = y dx gives dIn = 0 + (h )(x )(y dxs, where the distan ces to the centroidal axes of the elemental rectangle ar e d, = y / 2 and dy = .r. Now we have o 2 2 Ixy = dx = [:Cb dx = b = ~a2b2 Ans .

[i:c

y I I

-

I I

Solution II. Alternatively we can st art wit h a first-order elementa l strip and

CD

y

I

'---------'--~ a-

Solution I. If we sta rt wit h the second-order element dA d Ixy

I

,'---- - - ----'--'---'--!.a- - - - x dx Y

a +x

Yo

1-- - - - - . ; I 2

r ---j/:::::~i:=::~~ x

I I I

we had chosen a horizontal strip. our expression would have become + x J[(a - xl dyJ. which when integrated. of course, gives us the same result as before .

dIn =

.dfa

Sample ProblemA /l 0

Y I

Determine t he product of inertia of th e semicircular area with respect to th e .r-y axes.

CD

Solution. We use the tran sfer-of-axis theor em, Eq . A/ S, to write

r =- -2 ( 311'4r") (m-') 2 3

:

Yo

I

I

4r

0-9:::· ( -

4

I

xy

= 0

+ --

(r)

-

Ans.

whe re the .r- and y-coordinates of the cent roid Care d y = + r and d x = - 4r/<3 11'"). Becau se Yo is an axis of sym metry, Ixy = o.

Helpful Hint

CD Proper

use of the tra nsfer-or-axis theorem saves a great dea l of la bor

in computing products of iner tia. Marwan and Waseem AI-Iraqi

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454

App endi x A

Ar ea Moments of Inert ia

Sample Problem A lII

y

Det ermine the orie nta tio n of th e principal axes of iner tia th rough th e centroid of th e angle sect ion a nd determine th e corresponding maximum and min imum momen ts of ine rtia.

Solution . Th e loca tion of th e centroid C is easi ly calcu lated, and its positi on is shown on the diagram.

r

10 ' mm II-~ 2.5 mm

5 0mm

-,+

L -_ _

troi da l axes parall el to the x-y axes is zero by symme try. Th us, t he product of inertia abo ut th e x-y axes for part I is

1.." + d,d,AJ I•.,

[ Hl mm ...l...- ~

d. = -( 7.5 + 5) = - 12.5 mm

and

d,

y'

=

+ (20 - 10 - 2.5)

=

7.5 mm

o

,

\ \

Likewise for part II,

\

II" . = 1.." + dA ,AI where

~ -'-

40mm ~

= 0 + (- 12.5)( + 7.5)(400) = - 3.750 0') mm"

where

Ie

· - - - - - - - 1 7.5~m

Products of inerti«: Th e product of inert ia for each rectan gle about its cen-

[l.." =

I I I I I I I

I., = 0 + (12.5)(- 7.5)(400) = - 3.750 0') mm'

d. = + (20 - 7.5) = 12.5 mm ,

d, = -(5 + 2.5) = - 7.5 mm

For th e complete angle

Ix., = - 3.750 0' ) - 3.750 0' ) = - 7.500 0') mm''

Moments of Inertia. Th e moments of inertia about the x- and y -axes for part I are

Ix = 1,,(40)(10)3 + (400)(12.5)2

6.5800' ) mm"

=

He lpful Hint

I y = 1,,(10)(40)3 + (400)(7.5)2 = 7.5800' ) rum"

Mo h r 's circle. Alte r natively we could use Eqs. A/I! to obtain th e resul ts for l mll.,(an d I min, or we cou ld const ruct th e Moh r circle from t he calcu lat ed values of l x, l;y, and Ix)". Th ese values are spotted on t he diagram to locat e poin ts A and B. wh ich are t he extremities of t he diameter of the circle. The angle 2(1 a nd I Olllx and I min are obtai ned fro m the figu re, as shown.

and the moments of inertia for part II about these same axes are

t,

= 1,,(0)(40)3 + (400)0 2.5)2 = 11.580 0' ) mm'

I,

= 1,,(40)(1013 +

(400)(7.5)2

= 2.58IlO" ) mm''

T hus, for th e entire sectio n we have I.

6.58(0)' + 11.58(10' ) = 18.1700") mm'

Iy

7.58(10' ) + 2.58(10' )

=

10.17(10' ) mm'

Principal axes. Th e inclination of t he prin cipal axes of inertia is given by Eq. A/l O, so we ha ve 21" '-] [ tan 2a = l;y - t,

1-- tan 2a

2(-7.50) 10.17 _ 18.17 = 1.875

2" = 61.9" for

a = 31.00

I max = 22.7 - ---j

Iy = 1O.17

Ans.

We now compute th e pri ncipal moments of inertia fro m Eqs. A/9 using a and get l max from t, and l m in from [;y" Thus,

(I

Im~

[18.17 ; 10. 17 + 18.17 ; 10.17 (0.471) + (7.50)(0.882)] 00' ) An s.

22.700' ) mm' l min

[ 18.17 ; 10.17

18.17 - 10.17 0 2 ( .471)

(7.50)(0.882)] 00' )

5.67(104 ) mm 4 Ma rN an and W aseem AI-Iraqi

Ans. W'WW .gigapedia.c om

A

Ar ticl e A/ 4

PROBLEMS Introductory Problems A/57 Determine the product of inertia of eac h of the four areas about th e .r-y axes. Ail s. (0) and (c) : IX). = 360(104 ) mm"

1 50 60

rn i[

:

501

1

y I

100

[
60

<el

I


40

100 I

-L

400

Dimen sions in millimeter s Problem A /59

10

4

A/60 Determine the prod uct of inertia of eac h of th e fou r areas about t he x-y axes.

30

y I

I

Dimen sions in millimeters

Probl em Ai57

A/58 Determ ine t he produ ct of inertia about the .r-y axes of the circu lar area with three equ al square holes. y I

. 3"_ 2" 1-

I I L

80

I I I I

D :

50 10"

-f- -- + +~ ---- -±,,-x D:D l " I

I I

Problem A/58

Marwa n and W aseem AI-Iraqi

I

x

30[ B --- - --t-- 60 - -L 60

I I

I

L.

30 -~ ----1- - - - -I- - 1

I

:f - ~-t-~ +-, "-v'

,

I~_' i 30!-- 1-+ T I

455

A/59 Determ ine Ix. 1•., and l xv for the recta ngu la r plate wit h t hree equal circular holes. Ails. I. ~ 2.44<10") mm" l , = 9.80110"1 mm" Ixy ~ - 14.14<1 06 , mm"


Probl em s

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Dimen sion s in millimet ers

Problem A I60

456

A p pendi x A

Are a Moments o f I nertia

A/61 Det ermine the product of inert ia of the shaded area abou t the x-y axes.

A ns. I

xy

= 18.40(106) rnm?

Y

A/64 Obtain the product of inertia of the quarter -circular area with respect to th e x-y axes and use t his result to obtain th e product of inertia wit h respect to the parallel centroidal axes.

I I

Y

L r-'------~---, 20

r:

60

I I

Yo

I

I

I

[2}-_,.

I

I

LL------------I I

120

r

-- - x

20

t

Prob lem A/64

A/55 Solve for the product of inert ia of the semicircular ar ea about t he x-y axes in two different ways.

Dimensions in millimeters

Ans.lxy = ;2V 4

Problem A/6 1 Y I

A/51 Determine th e product of inertia of th e shaded area with re spect to th e assign ed axes. (Hint: Locate the centroid of the symmetr ical area. )

I I I I I

y

[---- :

5°150 ~ I

I

I

'--- - - -'--- - ----'- - - x

t

Problem A/65

40

:--+

A/66 Determine th e product of inertia of the rectan gu lar st rip number 1 about th e x -y axes . Assu me that the width b is sma ll compared with t he lengt h L . From t his first re su lt , det erm ine the produ ct of inert ia for each of th e strips 2, 3, and 4. Verify th at t he su m of th e four results is zero.

40

__ _______ ___L x Dimensions in millimeters Problem A/ 62

A/53 Determi ne the product of inertia of t he area of t he qu arter-circul ar ring about th e x -y axes. Tr eat the case where b is small compared wit h r, br 3 An s. I xy = 2

Y 2

L

I I I I

L

I b

I I

I

o

~ - - -x

Prob lem A/66

B

Probl em A/63

Marwan and Waseem AI-Iraqi

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I

~b

Art icle A/4 A/67 Deter mine the produ ct of inertia of th e shaded area about the x -y axes. I 2 2 Ans. Ix)' = 120 b )'

Problem.

457

Aj70 Deter mine the product of inertia of th e rhomb ic area abo ut th e x -y axes . (Hint: Rega rd th e area as a combination of a rectangle and tri an gles and use th e results of P rob. A/69.)

1

v

1

-I

1 1

1

1

1

I

1

1 1

1 1

s

I

1

"--_--,-_ s

1

_

J. 60°

- x

1

1

Problem A /70

' -=--

-

-

---' - - - - -

x

b

A/71 Calculat e the product of inertia of th e shaded. a rea about th e r -y axes. (H i n t: Tak e advan tage of the transfer-of-axes relations. ) A ns.. Jxy ~ - 1968 In · .4

Problem A /67

)'

Representative Problems

1 1 1

A/68 Derive th e express ions for the product of inertia of the r ight-triangu lar area about th e x -y axes and abo ut the centroidal Xo·Yo axes.

6"

4"

)' 1 1

4" )'0

1

.~~~-,

6" Prob lem A/71

A /72 The product s of inertia of the sha ded area with respect to the .r-y and x ' -y' axes are 8(106 ) mm " and - 42( 0 6 ) mm'', respectiv ely. Compute the area of t he

b

Problem A/68 A/69 Derive th e express ion for the product of inertia or th e right-t riangu lar area about th e x -y axes. Solve, first , by doubl e int egr ati on and, second, by single int egration start ing wit h a vert ical st rip as th e element. b 2h 2

Ans. Ixy = -8-

)~ .

y'

y

I 50mm

r 1 1

---

_1

_

1

"T --- - - - - -

1

I

I

h

C

20mm

- - x b Problem A /72

Problem A/69

MarNan and Waseem AI-Iraqi

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x'

~

1 mm l

1

:

figure, whose centroid is C.

60 mm x

458

A p p e nd i x A

Ar ea M o m en t s of In erti a

A/73 Det ermine th e moments a nd product of ine rtia ofthe quarter-circu lar area with re spect to th e x ' -y' axes. ,.4

_

,. 4

Ans. Ix· ::::: - (7r - , /3 ) I . ::::: - ('rr 16 " y 16

+ ,131

,.'

16 y

I I

y' \

A/77 Deter mine th e maxim um a nd minimum momen ts of inertia with respect to axes t hrough C for t he composite of t he tw o areas shown. Find th e a ngle a measured cou nterclockwise from the x-ax is to t he ax is of max imum moment of iner tia . Ma ke use of the res u lts of P rob. A/ 68. A ns. I max ::::: 3.79a 4 I min = 0.373a 4 0 H = 111.5

\

{

\ \ \

""

\ \

I I

x'

\

a

\ \

2a

c 2a

r

a

Problem AlB

A/74 Determine t he mome nt s a nd pro duct ofi ner tia of the a rea of the squa re with res pect to t he x ' -y' axes . y

A/7S Sketch t he Mohr circle of inert ia for each of th e four rect a ngu lar areas wit h the propo rtions a nd position s shown. Ind icate on each diagram point A which has coord ina tes (/ x' Ix ~') a nd the a ngle 2a, wh ere a is the an gle from the .r-axis to t he ax is of maximu m moment of ine rtia.

I

,

Y\

I [ I J

\ \

b \

\ ""-30

"

"" 0

-

"

Problem A/77

x'

--x

b

Y I

Problem A/ 74

I \1

AIl S Determine t he maxi mum a nd minimum mom ents of inertia with re spect to centrnida l axes through C for the compo site of t he four squa re areas shown . Find the an gle a measured from t he x-a xis to the axis of max imum moment of inertia. A ns. I m ll x = 5.57a 4

_-

_~=~_x

I mi n = 1.097a' a ::::: 103.30

I I (,, )

Y

y I

"

a

I

I

I I

............

-~=~~

"

a

c

I

a

a

a

I I

a

Idl

Ie)

Problem A /7S

A/76 Prove that the ma gn it ude of t he product of inertia can be compute d from t he rela ti on Ix.v ::::: ,/ I., Iy - Imnx./min' Marwan and Waseem AI-Iraqi

II

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Problem A/18

Article A/4

A/79 Find l~ and I y for th e shaded area and show that th e x -y axes are principal axes of inertia. Ans. I, ~ 0.446 b' . I, = O.280b'

Problems

459

A/81 The moments and product of inertia of an area with respect to th e x-y axes are Ix = 14 in .", Iv = 24 in.' , and I xv = 12 in .4 Construct the Mohr circle of ine rt ia and use it to det erm ine th e principa l mom ent s of inert ia and the an gle a from th e s -axis to th e axis of maximum moment of inertia. A/B3 Determine the maxim um momen t of inertia abou t an axis thro ugh 0 a nd the angle a to t his axis for the tr iangu lar area shown. Also construct th e Moh r circle of inertia. . 4 An s. I m ax = 183.6 in . ,u = - 16.85

0;>

Problem Aj79

y

A/BO Determine the minim um a nd max imu m moments of inertia wit h res pect to centroidal axes t hrough C for the composite of th e two rectangular areas shown. Fin d the an gle a measured from the x-ax is to the axi s of maximum momen t of inertia.

I

I

y I

8"

I I

2a

a

c a

Problem A/83 2a

Problem A/BO

A/8 l T he maximum a nd mini mum mo ments of inertia of the shaded area are 12(106 ) mm '' and 2(06 ) mm 4 , respectively, about axes passing through th e centroid C, and th e product of inert ia with respect to th e x -y axes has a magnitude of 4(106} mm 4 • Use th e pr oper sign for th e product of inertia and calculate Ix and th e angle a measured counterclockwise fro m th e xaxis to th e axis of maximum mom ent of inertia . Ans. Ix = 10(106 ) mm '', tr = 26.6°

A/84 Calculate the maxi mum and minim um momen ts of inert ia of t he structural angle about axes th rough its corne r A an d find th e angle a measured counterclockwise from th e .r-a xis to the axis of maximum moment of ine rtia . Neglect the s mall ra dii and fillet. )'

- " i-'- 10 mm I

80 mm l Om m _

'----------'--.1 r

L -_ _---'_

A I- 60 m m

- x

Probl e m A/ 84

Pro ble m A/8l

Marwan and Waseem A l-lraqi

-I

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460

'§

= .'

App endi x A

Ar ea Mom e nt s of In e rt i a

Computer-Oriented Problems

·A /85 Plot th e moment of inertia of th e shade d area a bout th e x ' -ax is as a fu nction of Hfrom H = 0 to H = 90° a nd determine t he minimum value of Ix' an d the corresponding value of H. A n s. I X' min = 2.09 in .", H = 67.5°

· A/ 87 Plot t he mom en t of inerti a of th e sha de d a rea about t he x '-ax is as a function of H from H = 0 to (J = 180°. Determine t he maxim um a nd min imum value s of Ix" a nd t he cor responding values of (J from the gr aph. Check you r resu lts by ap plying Eqs. A/ IO Ans. I max = 0 .655b4 at (J = 45° I m in = 0.4 05b 4 at () = 135°

a nd A/ H .

y I

y I I I

Y I

I

I 1"

I I

I

I ~

3"

x' I I

I

~

I

~

I

b

~ ~

_ x'

I I I

~O

--

\I

2"

b

Problem A/8S

· A/ 86 P lot the momen t of inertia about t he x '-axis as a fun cti on of () from () = 0 to l} = 90° and determine the minim um value of I x ' a nd t he corres ponding value of 0. y I I I I I I I

-0

---- x

b

Problem A /B7

· A/ 88 P lot t he momen t of ine r tia of the Z-section area a bout t he x ' -axis as a functi on of (J from l) = 0 to 8 = 90°. Determine th e maxim um va lue of Ix· a nd t he correspon ding value of (J from your plot , t hen verify t hese results by usin g Eqs. A/ l0 and A/ H.

U- 50 mm - j

1Omm .IL _ _ ~

r

50mm ~

x

__ _ _ _ -k -'::='8

-- x

I

I I

70 mm

30 mm

I 30 mm 130 mm

I

Yo

x

-.-- - +:I v - i-x 0

-

~f.-

Problem A / B6

f.- 10 mm

Marwan and Waseem AI-Iraqi

~l l0 mm

I-- 50 mm -

Problem AI BB

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-.!

·f!

Probl em s *A/ 89 Plot the mom en t of ine rti a of th e sha ded a rea a bout the x ' -axi s as 8 fun cti on of {J from 8 = 0 to tJ = 180°. Deter mine the max imum a nd minimum values of Ix' and th e corresponding values of O. A ns. l msu = 0.286b 4 at (J = 13 1.1°

-A/ gO Determine th e moment of inertia of t he sha ded area about the x ' -axis through 0 in terms of H and plot it for th e range 8 = 0 to {} = 180°. Find t he maximum and minimum values an d t heir correspondi ng a ngles 8.

I m in = O.0547b4 at tJ = 4 1.1 0

y I I I 30

y I

I

I 90 120

L-_ -..I _ - - x 240

Problem A/89

Dimensions in mill imeters Problem AlgO

MarNan and Waseem AI-Iraqi

461

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App.endix, -_

MASS MOMENTS OF INERTIA See Vol. 2 Dynamics for Appendix B, whic h fully treats th e conc ept a nd ca lcu lati on of mass mom ent of inertia . Becau se th is quantity is a n important element in t he study of rigid -body dyn amics a nd is not a factor in statics, we presen t on ly a brief defi nition in th is St atics volume so that the st ude nt can ap pr eciate th e basic differen ces between a rea and mass moments of inertia. Consider a three-dimensional body of ma ss m as shown in Fig. B/ l. The ma ss moment of inertia I about the ax is 0-0 is defined as I

=

f

,.2

dm

where r is the perpen d icula r distance of th e ma ss element dm from the ax is 0 -0 and whe re th e integr at ion is over the entire body. For a given rigid body th e ma ss mom en t of inertia is a measure of t he distribution of its mass relati ve to th e ax is in qu estion , and for th at axis is a constant property of th e body. Note th at t he dimen sions are (rnassn length j", which are kg'm 2 in SI units a nd lb-ft-sec/ in U.S. cu st om ary units. Contrast th ese dim en sions wit h t hose of area momen t of ine rtia, whi ch a re (lengt h r' , m4 in SI unit s and ft 4 in U.S. cus to mary uni ts.

o

I

Figure 8 /1

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463

Marwan and Waseem AI-Iraqi

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Appendix_ _

SELECTED TOPICS OF MATHEMATICS ell

INTRODUCTION

Appendix C contains an abbreviated summary and reminder of selected topics in basic mat hemati cs which find frequent use in mechani cs. T he relati onshi ps a re cited without proof. T he stude nt of mech anics will have frequent occasion to use many of these relation s , and he or she will be handicap ped if t hey are not well in ha nd . Oth er topics not listed will also be need ed fro m ti me to tim e. As th e reader rev iews and applies mathem atics, he or she should bear in mind that mechanics is an applied science descriptive of real bodies and actual motions . Therefore, the geometric and physical interpretation of the applicable ma t hematics should be kept clearly in min d duri ng the developm ent of theory an d the form ulation a nd solution of problem s.

e /2

PLANE GEOMETRY

4. Circle

1. When two intersecting lin es are, respec-

Circu mfere nce = 2 ttr Area = ;rr2 Arc length s = r 8 Sector area = ~O

tively, perpendicular to two othe r lin es, the angles formed by the two pai rs are equal.

2. Similar triangles

5. Every tr iangle in scribed within a semicircle is a righ t triangle.

~ =h - y b h

3. Any triangle Area = !bh

LIl

6. Angles of a triangle 81 + 82 + 83 = 180 ' 84 81 + 82

b

Marwan and W aseem AI-Iraqi

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465

466

App endi x C

C/3

Sel ected Topics of Mathematics

SOLID GEOMETRY

3. Right-circu la r cone

1. Sp here

Volume ~ rrr3 Surface area =

41iT

2

I

Volume = A r;r2h Lateral area = n rl,

L = Jr2 +

II

jz2

.

~ r-~

2. Spherical wedge 3

Volum e = 3gr

C/4

I

----'--

4. Any pyramid or cone

n

Volume = ! Bh where B = area of base

ALGEBRA

1. Quad.-atic equat ion

ax 2 +bx + c =0

-b

:!:

/ b2

4. Cubic equat ion

x 3 =Ax +B Let p = A / 3, q = B/2 .

2a

Case I: q2 - p3 nega t ive (t hree root s real an d distinct>

2. Logarith ms

bX = y , x = log, Y

cos u = q / (p / p ), 0 <

Natural loga rithms

x,

b = e = 2.718 282 eo' = y, .r = log, y = In y log (ab ) = log a + log b log (a / b) = log a - log b log (l /ll ) = - log II log a" = 1l log a log 1 = 0 10g lO X = 0.43 43 In x

x2 x3 Case II: XI

at

a3

(1/ /3 + 240' )

= 0 (t hree roots real, two

=

X

+ a t b 2C3 + a2 b 3 Ct + a 3b l c2 - a3 b 2c l - G2 b t C 3 - a l b3 c2

MarNan and Waseem AI-Iraqi

2q 1l3, X 2

=

X3

=

_ q 1l3

For ge neral cubic equat ion

2

b l Cl b2 c2 b3 c3

(1/ / 3 + 120' )

= tq + / q2 _ p3)1/3 + (q _ / q2 _ p3)1/3

Xl

3rd orde r 02

(1/ /3 1

roots equa l)

a 1 b l l = a,b 2 - a2bl a2

< 180'

p 3 positive (one root real, two roots imaginary)

Case III : q2 - p3

3. Determin ant s 2nd orde r

l

q

2

2,1p cos 2,1p cos 2,1p cos

1I

3

+ ax 2 + bx + c = O

Substitute x ~ Xo - a/ 3 and get x0 3 = Axo + B. Th en proceed as above to find values of Xo from which .r = Xo - a/ 3.

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Art icle C/S

C/5

Anal yt ic Ge om e t ry

ANALYTIC GEOMETRY

L St rai ght line

3. Parabola

.v

y 1

I

a

y I- - ~ -

E-,

~m

~ 1

1

1

Ib

L------- x

1

__ ..L __ x

ye a w mx

,,2

2

x ,,= b=-a2

.r =

.

2. Circle

(I f,2

4 . Ellip se ."

v

1

1 1

--, lE) --/ I bl I

r-b __ ,

~

I I

L

..l

x

a (x _ a )2 + (y _ b J2

=,.2 5. Hyperbola y

.ty = a 2

C/6

467

TRIGONOMETRY

L Definitions

sin Ii ~ al e cos Ii = b]«: tan Ii = alb

esc Ii sec Ii cot Ii

~

ci a b

2. Signs in the four qu adrants

H * ~I+) 1Il

Marwan and Waseem AI-Iraqi

II

III

IV

(J

+

+

-

-

eos ()

+

-

-

+

tan

+

-

+

-

esc fJ

+

+

-

-

sec

(J

+

-

-

+

cot

(J

+

-

+

-

sin

= clb = bfa

T~ ~~ ~ I + ) H~

I

I_I

I_ I IV

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(J

468

App end ix C

Sele c t e d Topics of Math ematics

4 . Law of si nes

3. Miscellaneou s relat ions sin 2 8 + cos 2 8 = 1 1 + tan 2 8 = sec 2 8 1 + cot 2 8 = csc 2 8 8 sin - = 2

An

o

sin A si n B

b

- cos (})

cos - = J~( l + cos (}) 2 s in cos sin cos

b

28 = 2 sin 8 cos 0 28 = cos 2 fJ - sin2 (J (a ± b) sin a cos b ± cos a sin b (a ± b) = cos a cos b :;:: sin a sin b

C/7

5 . Law of cosines

c2 c2

0

=

0

2 2

+ b2 2ab cos C + b 2 + 20b cos D

VECTOR OPERATIONS

1. N uta t ion . Vector quan ti ties are printed in boldface type, and sca lar qu antities appea r in ligh tface italic type, T hu s, t he vector qu antity V has a sca lar magnitude V. In longha nd work vector qu antities sho u ld always be cons iste ntly indicated by a sy mbol such as!, or V to distinguish t he m fro m scal ar qu an titi es.

2. Ad d ition T ria ngle ad dition

+ Q =R

P

P + Q

Parallelogr am add ition Commutative la w Associative law

P

+Q

= Q

=R + P

P + (Q + R ) = (P + Q ) + R

3. S u btr act io n

P- Q 4. Un it vectors

where

P + ( - Q)

i, j, k

V

= V) + V,j + V, k

IVI =

V = J V/

+ V/ + V, 2

5 . Di r ection eosine s l , m, 1l are the cosines of the angles between V and t he X -, Y-, z-axes. Thus,

/ = Vxl V so t ha t and

-= D

L..!!..

(}

V

m

= /2

~

VylV

n

= V, IV

W i + mj + nk ) + m 2 + n2 = 1

Ma rwan and W aseem AI- Iraqi

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p

Articl e C/7

Vect o r Op er a t ions

469

6. Dot or scalar product p .Q

~

P Q cos 8

T his pro duct may be viewed as the magnitu de of P mult iplied by t he compon ent Q cos II of Q in the direction of P , or as t he magn itude of Q multiplied by the component P cos 8 of P in the direction of Q. P ' Q = Q 'P

Commutative law

From the definition of the dot pro du ct

=

i ·j =j ·j

k ·k

i -j = j . j

j ·k = k ·i

p .Q

~

~

~~ Qe , ,pco, e

/1 <-:

i f Qcos e

p

1 j ·k

~

~

k ·j

0

~


p .p ~ p 2 + p 2 + p 2 x y z

It follows from th e definiti on of t he dot product th at two vectors P and Q are perpend icular when their dot product vanish es, P' Q = O. The angle Ii bet ween two vect ors PI and P 2 may be found from t heir dot product express ion PI ' P 2 = P IP2 cos II, wh ich gives

where I, In, n stand for the respecti ve directio n cosines of the vectors. It is also observed tha t two vectors are perpendic ular to each othe r when their direction cos ines obey the relation 11/ 2 + m lm 2

+

lt lft 2

= o. Distributive law

p . (Q + R)

=

p .Q + P ·R

7. Cross or vecto r product. T he cro ss prod uct P x Q of th e two vectors P and Q is defined as a vector with a magnitude

~P Q

PxQ

IP x

Q[ = PQ sin 8

a nd a dir ectio n specified by th e right-hand rule as shown . Reversin g th e vecto r order a nd usin g t he right-hand rul e give Q x P ~ - P x Q. Distributive law

P x (Q + R )

=

P x Q + P x R

From th e definitio n of the cro ss product, using a right-handed coord inate system, we get j x j = k

j x k = i

j x i = - k

k x j

~

- j

k x i = j i x k

~

-j

i x i = j x j = k x k = 0 Marwan and Waseem AI-Iraqi

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1, C~ Q

\

\ \

""

Q x P =-P xQ

-,

-,

470

Appendi x C

Selected Topic s of Math ematic s

With th e aid of these identities and the distributive law, t he vecto r product may be writte n P x Q

= (P) + Pyj + Pzk ) x (Q.,i + Q) + Qzk l = (P, Qz - PzQyli + (PzQx - PxQz)j + (PxQy - PyQxJk

Th e cross produ ct may also be expresse d by the deter minant i

P x Q

k

j

r, r, r,

o,

Qy

o;

8. Ad d it ion a l relation s Tripl e scalar prod uct (P x Q I' R = R · (P x Q ). Th e dot a nd cross may be interchanged as long as the order of the vectors is maintained. Parenth eses are unn ecessary since P x (Q . R ) is meaningless because a vector P canno t be crosse d into a scalar Q . R. Thus, the express ion may be written P x Q ·R

= P 'Q x R

The triple scalar product has t he det erm inant expa nsion

e; P x Q 'R

a,

n,

Py Q,. Ry

r,

o.

n,

T riple vector produ ct (P x Q ) x R = - R x (P x Q ) = R x (Q x P l. Her e we note tha t the parentheses mu st be used since a n express ion P x Q x R would be ambiguous because it would not identify th e vector to be crossed. It may be shown t hat th e triple vector product is equivalent to Q) x R

R ·PQ

R· QP

P x (Q x R )

P ·RQ

P ' QR

(P

or

X

Th e first term in th e first express ion, for example, is t he dot prod uct R · P , a scalar, mul tipli ed by t he vecto r Q. 9 . Derivatives of vectors obey the same rules as they do for scalars. dP

dt d(PllJ

dt d (P ' Q I dt d(P x Q )

dt Marwan and Waseem AI-Iraqi

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p. Q + p . Q P XQ +p xQ

Article C/8

10. Integration of vecto rs. If V is a fu nction of x, y , and z and an element of volu me is d r = dx dy dz , t he in tegral of V over the volume may be wri tten as the vector sum of the three integral s of its components. T hus,

I C/8

V dr

=

I v,

i

dr

+j

I v,

+k

tl-r

I v,

dr

SERIES

(Expression in brackets followi ng series indicates range of convergence.)

=

x)"

(l :!:

sin x

=x

1 ::t nx +

1) x 2 :!: Il lll -

Il lll -

2!

x3 x 5 x7 - - +- - 3! 5! 7!

1 )(1l -

3!

2) x 3

+ ... Ix2 < 11

+ ...

x 2 x4 x 6 cosx =l - - + - - - + ' " . 2! 4! 6! •

e-x

(!'\' -

sm h x = cosh x =

! (x ) =

e' + e- x x2 x4 x6 = 1 + - + - + - + .. . 2 2! 4! 6! ~

I1

whe re a " =

1l 7TX

,;:21

I'

00

J

[x 2

<

00

J

[x"

< cc]

x3 x 5 x 7 = x+ - + -+ - + ' " 3! 5! 7!

2

a 2° +

Ix2 <

an COS - / -

_I

~ b

.

1l 7TX

+ '~1 "sin - /-

17n;x

fix ) cos - /- dx ,

b.;

I1

=

II . _I fix)

17m

sin - /- dx

[Four ier expa nsion for - I < x < IJ

C/9 dx" _ dx

DERIVATIVES

= 17X,, - l

lim sin j,x

d tuu t dv du - - = u - +v dx dx ' dx

'

sin dx

tan dx

cos d x

1

du dv v- - udx

dx

dx

v

2

dx

.lx -·O

lim cos j,x .lx-O

d sin x dx d sin h x dx

cos x,

=

cosh x,

d cos x dx

d cosh x dx

Marwan and Waseem AI-Iraqi

- sin .r,

sinhx,

d tan x -dx

sec2 x

d tanh x = sech'' x dx www.gigapedia.com

Se ries

471

472

Append ix C

C/l0

J~

Selected Topics of Mathe matics

INTEGRALS

= In x

2 I 3 3b ' (a + bx )

J J a + bx dx

2 (3bx Jx ,la + bx dx = ~ 15b

Jx2j a + bx dx =

J

bx

J J(l+X dx Jb -

Ja J

12abx + 15b2x 2)j (a + bx )3

-

2 j a + bx b

dx

../0 +

_ 2_3 (Sa 2 105b

2a )j (a + bx )3

- Ja+x ~

+ (a +

b)

sin "

!

x

x dx 1 + bx = b2 [c + bx - a In (a + bx)J

(a + bX)l- " (a + bx _ _ a_

xdx

(a

+ bx )"

b2

_-=d::,x,...." = Ja + bx 2

1

- - t an

Jar;

- I

2 -"

xJar;

1 - " 1 j- ab

or

--

a

)

- - t an

h- I x j -ab

---

a

x dx 1 --~2 = 2b In (a + bx 2)

Ja + bx

f

,'/x 2

::!: 0 2

Jj a 2 -

f

x 3 ,10 2

dx =

x 2 dx =

-

~fx./x2 ::!: 0 2

~(x,la2

::!: 0

- x2

x 2 dx = - !(x 2 +

+

2

In (x + ,lx 2

a sin-I;) 2

302 ) ../ (0 2 -

MarNan and W aseem AI-Iraqi

::!: 0 2 )]

x 2 )3 www.gigapedia.com

Articl e

f r====:;; ~ k..(c dx

+ bx+ cx

2

=

In ( Jo + bx + cx2 + x ..(c +

In (x + J x 2 :t

fJ

~) 2..(c

0 2)

xdx

0 2 :t

x2

04

- -8 In (r .

f f

sin x dx

- cos x

=

sin x

cos x dx

1 1 fs ecxdx = - In 2 1

f sin

2

x sin 2x =2 + 4

f

cos

f

sin x cos x dx

2xdx

sin x

sin 2x 4

x 2

x dx

+ smx

sin 2 x 2

f sinh x dx cosh x f x dx = x =

cosh

f f

sinh

ta nh x dx

~

In x dx

x In x - x

=

In cosh x

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or

e /l 0

Inte gr al s

- 1 - - sin- 1 ( b + 2CX ) Jb2 - 4ac Fe

473

474

Ap pendi x C

Sel ect ed Topic s of Math em atics

e'"

f eUX dX = a

Jxe'" dx =

l!'X

~

(ax

-

1)

Je"-' sin px dx

e"-'(a sin px - p eas px j a 2 + p2

Je'" cos px dx

eUX(a cos px + p sin px ) a2 + p 2

Je"-' s in" x dx

e -9 --

ax

4 + a-

(

a sin2x - sin 2x +

(a

eO.'

Jfflx sin x cos x dx = 4 + a

---2

f f

sin3 x dx

cos"

Jcos

5

f f f

2

sin 2x - cas 2x)

- -cos -x (2 + sin" x) 3

sin x (2 + cos 2

X

dx

X

dx = sin x

Jx sin x dx

-

~)

3

x)

= sin x - x cos x

x cos x dx = cos x + x s in x x 2 sin x dx = 2x sin x - (x 2 x 2 cos x dx

-

2)

cos x

2:r cos x + (x2 - 2) sin x

fix)'

Radius of curvat ure

,." + 2

(~~r

MarNan and Waseem AI-Iraqi

- r

d",.

r-r-x

dIP

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Article

e /l1

Newton 's Method for So lvi ng I n t ra ct a b l e Equati ons

Cj 11

NEWTON'S METHOD FOR SOLVING INTRACTABLE EQUATIONS

Fr equ ently, t he application of the fund amental pr inciples of mechanics leads to an algebraic or transcendental equat ion which is not solvable (or easily solvable ) in closed form. In such cases, an iterative techni que, such as Newt on's meth od, can be a powerful tool for obtaining a good est imate to the root or roots of the equation. Let us place t he equa tion to be solved in th e form f (x ) ~ O. Part a of the accompanyin g figure depicts an arbit rary fun ction f (x ) for values of x in the vicinity of the desired root x,.. Note that X ,. is merely the value

(lx l (lxl

'"- 'I

~\

1

1

Tangent to {l x ) at x=x l

I I

x,

(I.d

(lx l

L

I

i Xl

~

~.

X'_' __ x

.1:2

"-----::;c'--- x

.\'2

I

of X at which t he function crosses the x -axis. Su ppose tha t we have available (per haps via a hand-drawn plot ) a rou gh estimate of this root. Provided that Xl does not closely correspond to a maximum or minimum value of the fun ction f (x ), we may obtain a better estimate of the root x, by exte nding t he tangent to f (x ) at x, so that it intersects t he x -axi s at X 2' From t he geometry of the figure, we may write

x,

where j'(x , ) denotes the deri vati ve of f (x ) with respect to x evalua te d at X = X l ' Solving the above equat ion for X2 results in

Th e term - f (x , )/ j' (x , ) is t he correction to th e initi al root esti mat e x , . Once x2 is calculated, we may repeat the process to obtain X3 ' and so forth. Thus, we generalize the above equation to f (x,,) j'(Xk ) Marwan and Waseem AI-Iraqi

y

(c)

Ibl

(a )

~X l

1

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475

476

App end ix C

Sel ec ted Topics of Math em at ics

where

the

(k

+ l )th est imate of the desired root x,

th e kth estimate of the desired root

Xr

the functio n f<x) evaluate d at x = x. the funct ion derivative evaluated at x =

Xk

This equation is repeatedly applied un til f(x. +l) is sufficiently close to zero and x. + I es xh- The st udent should verify t hat th e equatio n is valid for all possible sign combination s of xi, f (x . ), a nd r (x. ). Several cautionary notes are in order: 1. Clea rly, r (x. ) must not be zero or close to zero. Thi s would mean, as restricted above, t hat exactly or approxi mately corresponds to a minimum or maximum of [ix ). If the slope r (x. ) is zero, then th e ta ngent to th e curve never intersects the x-axis. If t he slope r (x. ) is small, th en th e correction to x. may be so large that x• • I is a worse root estimate than X k ' For this reason, experienced engineers usually limit the size of the cor rectio n te r m; that is, if the absolute va lue of f (x. )/r (x. ) is larger than a prese lecte d maximum value, that maximum value is used.

x.

2. If there are severa l roots of t he equatio n f (x ) = 0, we mu st be in t he vicinity of th e desired root X r in order that th e algorit hm actua lly converges to that root. Part b of the figure depicts the condit ion when the initial es timate X l will result in convergence to x r 2 rather than x r l " 3. Oscillatio n from one side of th e root to t he ot he r can occur if, for example, the function is antisymmetric about a root which is an inflection point . The use of one-half of the correction will usually prevent this behavior, which is depicted in part c of the accompanying figure . Example: Beginni ng with an initial esti mate of xI = 5, estimate the single root of t he equation e' - 10 cos x - 100 = O. Th e table below summarizes th e application of Newton's meth od to the given equation. The iterative process was terminated when the absolute value of the correctio n - f (x. )/ r (x. ) became less th a n 10- 6 •

"

x.

f (x. )

j'(x. )

I

5.000 000 4.671 695 4.596498 4.593 215 4.593209

45.576537 7.285610 0.292 886 0.000 527 - 2(10- 8 )

138.823 916 96.887065 89.203 650 88.882 536 88.881 956

2 3 4 5

Ma rwan and W aseem A l-lraqi

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Xk +l -

Xk =

_ f(x. ) j'(x.)

- 0.328 305 - 0.075 197 -0.003283 - 0.000 006 2.25(10 - 1° )

Arti cl e CI t 2

Sele cte d Tech niq ues for Nu m e r ical In t e g ra ti o n

C/12

SELECTED TECHNIQUES FOR NUMERICAL INTEGRATION

I , Area determinatian. Consider th e problem of determining t he s haded area under t he curve y = ! (xlfrom x = a to x = b, as depicted in part a of t he figure , and suppose that ana lytical integration is not feas ible. The function may be known in tabular form from experimental measurements or it may be known in ana lytical form. The function is taken to be contin uous within the interval a < x < b. We may divide th e area into n vertical str ips, each of width Ar = (b - a l/ n, an d then add the area s of all st r ips to obtain A = f y dx. A representative strip of area Ai is shown wit h darker shading in the figure. Three useful numerical approximat ions are cited . In each case the greater the number of st rips, the more accurate becomes the approximation geometrically. As a ge neral rule, one can begin with a relatively small number of strips and increase the number until the resulting changes in the area approximation no longer improve the desired accuracy.

J'

y = { (:r )

-,

J'" Yn -l

J'i

v., 1

•••• Yo (a)

YI

)'2

• •

Y3

'----~-~-~-~---+---c~--.,,-'--+-- x Xo

=

Xl

X2

X3

{/

Rectangular Ai

A j= Ymtl x

J

A = Y d x ::= r Ymtl X

1. Rectang ular [Figure (b)J Th e areas of th e st rips are taken to be rectangles. as shown by the representative strip whose heigh t YIIl is chose n visually so that the small cross-hatched areas are as nearly equal as possible. Thus, we form the sum ~)'m of the effective height s and multiply by .lx . For a function known in ana lytical form, a value for Ym equal to th at of th e function at the midpoint Xi + ..I.x / 2 may be calculated and used in the summation. II. Trap ezoidal [F igu re (cl J The areas of th e str ips are ta ken to be trapezoids, as shown by the represent ati ve strip. The area Ai is the avMarwan and Waseem AI-Iraqi

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477

478

Appendix C

Sel ected Topics of Mathematics

erage height ( y , + Y' +I )/ 2 times .'n. Adding th e areas gives the a rea approximation as tabulated. For the example with the curvature shown, clearl y th e approximation will be on t he low side . For the reve rse curvat u re, the approxima tio n will be on th e high side.

Trapezoidal

A

J'i

(e )

Ai

i-

J'i + Yj •

2

A = J y dx Yi -+

1 fi

x

2:::(~ +)' 1 + Y:.! +

... + Yn - 1 +

~)LlX

1

Llx

Parabolic

..

v.

(
.1A =

dA Yi . I

J'j • 2

A=

1

3

(,)' j +

J), dx =:

4,)'j . 1 + J'j -+ 2 l.l x

~ (Yo + 4 ) '1 + 2)'2 + 4 )'3 + 2Y4 + . .. + 2y,. _:.! + 4Yn - I + y"l.1x

dx

III. Parabolic [Figure (dJ ] The area between t he chord and the cu rve (neglected in the trapezoidal solution ) may be accounted for by approximat ing the fun ctio n by a parabola passi ng t hroug h th e point s defined by three successive va lues of y . Thi s area may be calcu lated from th e geometry of th e parabola a nd added to th e trapezoidal area of the pa ir of st rips to give th e a rea ~ A of the pair as cited. Addin g all of t he .lA's produces the tabulation shown, which is known as Si mpson's rule. To use Simpson's rule, the number II of strips must be eve n. Examp le: Determin e the area under the curve y = x .J l + x 2 from x = 0 to x = 2. IAn inte gra ble fun ction is chose n her e so that th e three approximations can be compared with the exact value, which is A x h + x 2 dx = ~(J + x 2 )3 2 1~ = ~( 5,J5 - 1) = 3.393 447,)

f6

AREA AP P ROXIMATIONS NUMBE H OF SU BIN T E RVALS 4

10 50 100 1000 2500

Ma rwan and W aseem AI- Iraqi

RECTANGULAR

TRA PEZOIDAL

PARABOLIC

3.36 1 704 3.388399 3.393 245 3.393 396 3.393 446 3.393447

3.45673 1 3.403536 3.393850 3.393547 3.393448 3.393447

3.3922 14 3.393420 3.393447 3.393447 3.393447 3.393447

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Articl e e/12

Selected Tec hn iques for Numerical Int e gration

Note t hat t he worst approxima tion er ror is less th an 2 percen t , even with on ly four stri ps. 2. Integratian af first-order ordinary differential equations. The applicat ion of th e fu ndamental pr inciples of mechan ics frequ ently results in differen tial relationships. Let us cons ider th e first -order form dy jdl = f it) , where t he func tion f it) may not be readily integrable or may be known only in tabular form . We may nu merically integrat e by mea ns of a simple slope-projection technique , known as Eu ler integration , which is illustrated in the figu re. /

dv

Slope

dl = f lt)

yltJ

/

y( tJ ",, / ~

Slope = f(t ,1 ) <, /

Slope = f (l 1.) <,

~

-

I I I Y2

I I I I I Ya 1

I I

I I

Slo pe = f !t, J"

-----l.. \

/

I Accumulated I al gorit h mic I er ror I I I Y-t I I I -

-c

etc .

I I

Beginning at l}o at which the va lue y, is kn own, we project the slope over a hori zont al subinterva l or ste p (t 2 - ' I) and see tha t Y2 = Y l + f ll l lll 2 - I ,) . At 12, t he process may be repeated beginning at Y2' an d so for th unti l the desired value of I is rea ched. Hence, t he genera l expression is

If y versu s t were linea r, i.e., if f( t) wer e constant , the meth od would be exact , and t here would be no need for a num er ical approach in that case. Changes in th e slope over t he subinterval introduce er ror. For the case shown in the figu re, the estimate 12 is clearly less than the t ru e va lue of t he funct ion y (t) at 12, More accura te integra tion te chniques Isuch as Runge-Kutta methods) take into account changes in the slope over t he su binterva l and thu s provide better result s. As with the area -dete r mina tion te chniques , exper ience is helpfu l in t he selection of a subinterva l or step size when dealing with a nalytical funct ions. As a rough rule, one begins wit h a relatively large ste p size a nd the n steadily decreases th e ste p size u nt il the corresponding changes in the int egrated result are mu ch smaller th a n the desir ed accuracy. A step size which is too sma ll, however, can result in increased error due to a very large num ber of compute r operations . This type of er ror is generally known as " roun d-off er ror," while th e er ror which resu lts from a large step size is known as a lgorit hm error. Marwan and Waseem AI-Iraqi

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479

480

Appendi x C

Se lec ted Top ics of Ma th ema t ic s

Example: For the differential equa tion dyf dt = 51 with t he initial condit ion y ~ 2 when I = 0, determine the value of y for I = 4. Application of th e Eul er integrat ion te chnique yields th e followin g results: NUMBER OF SUBI NT ERVALS

STEP SIZE

10 100 500 1000

0.4 0.04 0.008 0.004

y at t

~

38 41.6 41.92 41.96

4

PERCENT ERROR 9.5

0.95 0.19 0.10

Th is simple exa mple may be int egrated an alytically . Th e resul t is y 42 (exactly ).

Ma rwan and W aseem AI-I raqi

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App,endix_ _

USEFUL TABLES TABLE 0 /1

PHYS ICAL PROPERTIES

Density (kg/ rn3 ) a nd specific weight Ob/ ft 3 )

Air· Alum inum Concrete (av .) Copper Ea rt h (wet , av.)

(dry, av .) Glass Gold Ice Iron (cast)

kg/rn 3

Ib/ ft 3

1.2062 2 690 2 400 8910 1 760 1 280 2590 19 300 900 7 210

0.07530 168 150 556 110 80 162 1205 56 450

Lead Mercury

Oil (av.) St eel T itani um Water (fresh) (salt ) Wood (soft pine) (ha rd oak )

kg/rn 3

Ib/ ft 3

11 370 13 570 900 7 830 3 080 1 000 1 030 480 800

710 847 489 192 62.4 64 30 50

•At 20°C
Coefficients of friction

(The coefficient s in the following table represent typical values under normal working conditions . Actual coefficients for a given sit uation will depend on the exact nature of the contact ing surfaces. A variation of 25 to 100 percent or more from these values could be expected in an actual applicat ion, depending on preva iling condition s of clea nliness, surface finish , pressure , lubrication, and velocity.)

TYPICAL VALUES OF COEF FICIENT OF FRICTION

Marwan and Waseem AI-Iraqi

CONTACTING SURFACE

STATIC, 1'-,

KINETI C, 1'-.

Steel on steel (dry) Stee l on stee l (greasy) Teflon on stee l Steel on babb itt (dry) Steel on babb itt (gr easy) Brass on steel (d ry) Brake lining on cast iron Rubber tires on smooth pavem ent (dry) Wire rope on iron pulley (dry) Hemp rope on metal Metal on ice

0.6 0.1 0.04 0.4 0.1 0.5 0.4 0.9 0.2 0.3

0.4 0.05 0.04 0.3 0.07 0.4 0.3 0.8 0.15 0.2 0.02

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481

482

App endi x D

TABLE D/2

Use fu l r a b ie s

SOLAR SYSTEM CON STANTS

G ~ 6.673110- 11 ) m 3 / Ikg · s2 ) ~ 3.439 110· ' ) ft' / Obf·s' ) m e = 5.976(024 ) kg ~ 4.09511023 ) Ibf-s' /ft

Unive rsal gravitational constant Mass of Eart h

Period of Earth's rotat ion (1 sidereal day )

= 23 h 56

min 4 s 23.9344 h w ~ 0.7292110-') rad/ s w' = 0.1991(10 . 6 ) rad/ s ~ 107 200 km /h = 66,610 mi/h ~

Angular velocity of Earth M ean angu lar velocity of Earth-Sun line

Mean velocity of Earth 's cent er about Sun

BODY Sun Moon

MEAN DISTANCE TO SUN km (mil -

384 398 ' (238 854)' 57.3 X 106 Mercury 135.6 X 106 ) 108 X 106 V enus 167.2 X 106 ) 149.6 X 106 Eart h 192.96 X 106 ) 227.9 X 106 Mars 114 1.6 X 10 6 )

SURFACE GRAVITATIONA L ESCAPE ECCENT RICITY PEIUOn MEAN MASS OF ORBIT OF ORBIT DIAMET ER RELATI VE ACCELE RATION VELOCITY solar days km (mil TO EARTH m/ s2 Ift/ s2 ) k m/ s Imi/ s) e

-

-

0.055

27.32

0.206

87.97

0.0068

224.70

0.0167

365.26

0.093

686.98

I 392 000 1865 0001 3 476 (2 160) 5 000 13 100) 12400 17 7001 12 742t (7 918)t 6788 (4 218)

333000 0.0123 0.054 0.815 1.000 0. 107

274 18981 1.62 15.32) 3.47 111.4) 8.44 127.71 9.82 1* 132.221* 3.73 112.3)

616 (383) 2.37 11.47) 4.17 12.59) 10.24 16.361 11.18 16.951 5.03

13.131

• Mean distance to Earth (center-to-center) t Diameter of sphere of equal volum e, based on a spheroidal Earth with a polar diamet er of 12 714 km (7900 mil and an equatorial diameter of 12 756 km (7926 mil * For nonrotating spherical Earth , equivalent to absolute value at sea level and latitude 37.5°

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Us eful Ta bl e s

TABLE D/3

PROPERTIES OF PLANE FIGURES FIGURE

Arc Segm en t

AREA MOMENTS OF INERTlA

CE NTROID

r=

a C 4 _,a r~.

r§ »-

<,

r sina

-

2r K

-

a

,

Qu a rter a nd Se micircu la r Arcs

~~

)' = -

I V r L ___ ' ___ _ ~ _ _

,

)'

Ci rcula r Area

I :c = I.v = -If4,. "'

0'

If ,.4

/z = -

2

l[

.

I

Semicircular Area

,.4

Ix = I,. = -

)'

GtJ r if

4r 3K

- (K 8) 8 - 91l'

t, =

)' = -

--x

8

,.4

Kr'

1,. = - 4

lOA

Ix = I y = . 16

)'

I I

Qu ar t er -Cir cul ar Area

{Sh

-

-

4r

x =Y = -

3K

~

I

aXe

"-

Marwan and Waseem AI-Iraqi

[

= ]

Iz = -

Ix =

--x

=

_

_ _

,.4

x

n ,.4

- - x

)'

Area of Circ ula r Sector

- -.v (K 4) 16 9rr

2 r sina a

x = ----

3

8

r'

-(a

4

r'

ly = 7((1 + '21 sin 2a )

/z = ~ r4a

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- '12 sin 2u )

483

484

Appendi x D

TABLE

n/a

Useful Tables

PROPERTIES OF PLANE FIGURES Con t in ued FIGURE

CENTROID

AREA ~IOMENTS OF INE RTlA

bh'

Recta ngul ar Area

1 =• 3

Yo

rf- ;c -1

• --x

!--b--.j

~Xl x C h

Y I

T ri angu lar Area

Y

I

I--b-J

L

-

bh'

I: =

bh Cb' +h' )

1 = • 12

-

x

12

bh'

1 = • 12

a wb x = -3

-

-

bh'

1= • 36

h

Y =3

I

bh'

=

"

Area of Ellipt ical Qu adr ant

-

4a x = -

Y I I

3 1 = -Jrab Ix = x 16 '

(-,,- i..-)ab' 16 9 ~

:I< Jra 3 b

- 4b )' = -

.

bE 5 D __ .,

I.v = 16' ly =

I = !rob (0 2 + b2) z 16

Subpa rabolic Area y =

Y I

kx 2

t,

b = _x 2

=

-

3a x =4 ly =

Area A = ab I

3 I

- 3b

x>tb /"

I

a

jy

Y =10

- x

I

b

y=kx 2 = _ x 2

·'P (

MarNan and Waseem AI-Iraqi

x- = 3a -

8

- 3b

y =-

5

ab' 21

a'b 5

(a' + -b' ) 5 21

I =ab z

Pa ra bolic Area

Y

(-,,- i..-)a'b 16 9Jr

:I<

a

AreaA = 2ab 3

4

2ab' t, = -73b 1 = -2a ' 15

(a'15 +b') 7

I = 2ab z

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Useful rabi es

TABLED/4 PROPERTIES OF HOMOGENEOUS SOLIDS (m = mass of body shown ) MASS CENTER

BODY

Circular Cylindrical

Z

MASS MOMENTS OF INERTIA

~mr2

+

l X lx l

~m r2

+ ~m12

Shell

x,

mr 2

I"

.r

In

I.o

~

n

= ~mr2 + Hal[ Cylin dr ical

x

2r

i -: =

tr

x,

Circ ula r Cylinder

Z

!m1 2

mr 2

lu

(I

In

:1"mr2 +

~) mr2

-

n

ml2

tmr2 + 3m l2

l X lx !

!mr

2

x ~

In

[ X IX I

Z r

'.v, I

-,

Se micy linde r

.v

X

~

4r

I

=

t'"

=

IY 1J'1

mr2

t

= m r2

3"

x

+ ~m12

!mr2

lu

~

G- 91~)

,

I", I" I Y\.Y I

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ntr

= t'2m (a 2 + [2)

f.lm (b 2 + [2 ) f2 m (a 2 + b 2 ) rzmb2 + !m1 2

I y z.v 2 = !m (b2

Marwan and Waseem AI-Iraqi

-bml 2

=

L

Rectan gu lar P arallelep iped

+

I"

I

Z

IY 1J' 1

I"

I"

x,

ml 2

= !mr 2 +

= -

Shell

x,

-bmi 2

In

+

(2 )

2

485

486

Ap pe n dix D

Usef ul Ta bles

TABLE 0 / 4 PROPEHTl ES OF HOMO GENEO US SOLIDS Con t in ued (m

=

mass of body shown)

MASS MOMENTS OF INERTIA

MASS CENTER

BODY

Spherical

Shell

Hemi spherical

Shell

~mr2

In

=

I yy

=

Iu

2

IJ'Y

=

Iu

=

f2mr 2

_ 3r x = 8

Ixx

=

I.'YY

=

Iu

_

r

x = -

I

=

x

Sphere

Hemi sphere )'

I x

I'

Unifor m Slender Rod G

Ma rwan and W aseem AI-Iraqi

~mr2

=

l.ry = lu = ~~m r2

I.'YY 1\. 'V . 1.1

WNW.gigapedia.com

1

= 12ml = ~m rl

2

Us e ful Tab les

TABLE D/4 PROPERTIES OF HOMOGENEOUS SOLIDS Continued tm = mass of body shown ) jl,\ASS MOME NTS OF INERTIA

Jl,IASS CENT ER

BODY

QuarterCircular Rod

x=

y

1.Y)'

I"

mr 'l.

2r tt

=

lu

!mr2

z

Elliptical Cylinder

lu

tma2 + -bmi'J.

1'Y

4mb + Eml

I"

~m (a 2 + b'l.J

[ ."'LVI

I

2

1

= tm b'J. +

2

km{l

.r

I I

y

Conical Shell

Z

2" 3

Hal f

,-

'r

.v

Conical Shell

4r

x = -

3"

Z

2" 3

x

I I

-

I"

~m r2

1",

1mr2

Cone

Marwan and W aseem AI- Iraqi

=

, =

3" 4

!mh2

+ -mmh'l.

I yy

= 1mr :'! I x 1x 1

+

+ ~mh2

= IY IYI

= 1m ? + !mh 2 I"

~m r2

izz

(!_ ~)mr2

l yy

:fum,.'l. +

;}m h'l.

1.'1 1_'1' \

'1. :wm r +

tom h2

2

9'"

:I

RightCircu lar

2

1mr2 + f"mh2

].'1 1.'11

lu

x\

tmr

J.\y

I"

-fumr2

i.~ 'Y

:fum? +

WNW.gigapedia.com

5

£mh

2

487

488

Appendix 0

Useful Tabl es

PROPERTIES OF HOMOGENEOUS SOLIDS Con tin ued

TABLE D/4

(m = mass of body shown)

MASS CENTER

BODY

MASS MOME NTS OF INE RTIA

In

x z

r

= tr

Halr Cone

r XI

YI

I yy

=

:fumr2 +

I"

x

~m h 2

= I.Y\YI = ~mr2

-3" 4

"

X

[ X IX I

=

=

!zz

~

In

=

Tamh 2

+

-&mr2

Uo- ~) m r tm (h

2

+ c2 )

IyY = ~m (a2 + _ z

3c 8

IyY = t. m Ca2 + ~C2 )

y

!mc2 = lima 2 + !mc2

In = y

IyY

I._

Elliptic

~ Paraboloid

z ,-_ _ c

_

z

2c

I"

=-

3

!m (a'2 + 4c'2)

i;

=

rom

(b'2

+ c2 )

x = -

Iry

=

fumea2

+ c2 )

_ y

I" =

a

_ Y

~

b -

Y

Marwan arid Waseem AI-Iraqi

= ifum(b2 + c2 )

In

c

IyY = fom(a 2 +

4

Half Toru s

romCa2 + b 2 )

4

z =-

/

!c'2 )

=

4

x_

i m (a 2 + h2 )

Iry

z

Rectangu lar Tetrahedron

=

~mb2 +

t; = ~m (b2 +

x

I

c2 )

2 2 I" = im (a + b ) I xx = ~m (b2 + file'l )

= -

Se miellipsoid

C

2

l zz

=

tom Ca2 + b

=

I yy

Izz

=

mR 2 + ~ma2

=

2)

!mR 2 + ~ma2

[xx

www.gigapedia.com

c2 )

PHOTO CREDI T S C hap ter Ch a p ter Ch a p ter Ch a p ter Ch a p ter Ch a p ter Ch a p ter C h a p ter

Openers 1: Andrea Pist olesilTh e Image Bank. 2: Key Sanders/Sto ne. 3: Ron Behrmannllnternational Stock Ph oto. 4: J oh n Edwards/Stone. 5: Ala in ChoisnetlThe Image Bank. 6: Ted Wood/Stone. 7: Courtesy of UpRight , Inc.

Marwan and W aseem AI- Iraqi

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INDEX

Absolute sys te m of units, 10 Acceleration , of a body, 8, 115 du e to gr avi ty . 11 Accu ra cy, 13 Act ion a nd rea ct ion. princ iple of, 8, 25, 105, 195. 216. 271 Act ive forc e, 390 Act ive-force dia gra m, 39 1 Additi on of vecto rs, 6, 25, 28, 468 Aerostatics, 297 Angle, of frict ion , 33 1 o[ repose,334 Appro xima tions, 14. 246 Archimed es, 3 Area, first mom ent of, 230 second mom ent of, 230, 427 Area mom ents of inertia, see Moments of ine rti a of a reas Atm osph eric pressure, 298 Axes, choice of, 27, 65, 108, 171, 229, 231 rotat ion of, 450 Axis, mom ent, 37, 74 Bea ms , concentrated loads on, 265 definitio n of, 264 distribu ted load s on, 265 external effects, 265 inte rnal e ffects, 271 loadi ng-shear relat ion for, 272, 273 res ulta nt of forces on cross sec tion of, 271 shear-mo me nt re la tion for, 272, 273, 274 sta t ically det erminat e a nd indeterminate, 264 types of, 264 Bea ring frictio n, 359 Belt fric tio n , 368 Be nding moment, 271 Bend ing-moment diagr am , 272 Bodies, inte rco nnected, 195, 390 Body, de formable, 5 rigid , -1 Body force, 24, 226 Bounda ry conditio ns, 285 Br it ish sys te m of units, 9 Buoya ncy, center of, 303 force of, 303 principle of, 302 Ca bles, ca te nary, 287 flexible, 283

490

MarNan and Waseem AI-Iraqi

len gth of, 286, 289 pa raboli c, 285 tensio n in, 286 , 289 Cajori , F., 8 Center , of buoyancy, 303 of gravity, 25, 227 of mass, 227, 229 of pr essure, 299 Centroids, 230 of composite figur es, 246 by integration, 230 of irregu lar volu mes, 247 tabl e of, 483 by t heo rems of Pappus, 256 Coefficient, of fricti on , 330 , 331, 481 of rollin g resistance, 369 Collinear force s, equi libriu m of, 115 Compone nt s, of a force , 26, 27, 28 rect a ngul ar, 6, 27, 28, 64 sca la r, 27 of a vecto r, 6, 26, 27, 64 Composi te a reas, mome nt of inertia of, 442 Composite bodie s, cen te r of mass of, 246 Composite figures, centroid of, 246 Compressio n in t russ memb e rs, 167, 169 Com pute r-or iented problems, 18, 100, 161,221, 322 , 381, 425,460 Concent rated forces, 25. 225 on bea ms, 265 Con current forces, equili br ium of, 115, 139 resu ltant of, 25, 57, 86 Con e of fr iction, 332 Cons ta nt of gravitation, 12,482 Cons t ra int , 118, 141 adeq uacy of, 119, 142 partial. 142 proper a nd imprope r, 119 redundant, 120, 142 Coord ina tes , choice of, 27, 65 , 108, 23 1,316, 429 Copla na r forces, eq uilibriu m of, 115, 116 resulta nt of, 28, 56 Cou lomb, 328 Couple, 48, 75 equivalent, 48 mome nt of, 48 , 75 resolution of, 49, 86 resultan t , 56, 85, 86

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Index vect or representation of, 48, 75 work of, 387 Cross or vector product , 38, 73. 469 D'Alembert, J" 4 da Vinci, 4 Deformable body, 5 Degr ees of freedom, 392, 410, 420 Den sity, 229 , 481 Deri vative of vector, 470 Derivatives, table of, 471 Dia gr am, act ive- forc e, 39 1 ben ding-mome nt, 2 72 free-body, 16, 104, 108, 139 shea r-force, 272 Differen ti al element , choice of, 23 1 Differe nt ials, ord er of, 14,231 ,3 16 Dime nsions, hom ogen eity of, 17 Direc tion cosin es, 7, 64 Disk fri ction. 360 Displacem ent, 386 virtua l, 388 Distributed forces, 25, 225, 226, 3 16, 3 17 on beams. 265 Distributive law , 39, 469 Dot or sca lar product, 65, 386, 469 Dyna mics, 4, 8 Efficie ncy , mecha nical , 393 Elastic potent ial ene rgy, 405 Energy, cri te rio n for equ ilibr iu m , 410 cr ite r ion for stability, 4 10 elas t ic, 405 potent ial, 405, 407 , 409 Equilibrium. alternative equations of, 117 ca tego r ies of, 115. 139 of collinea r for ces, 115 of concur re nt forc es, 115 , 139 condit ion of, 56, 115, 138, 389, 391 of copla nar forces, 115, 116 e ne rgy criterion for, 409, 410 eq uat ions of, 115, 138 of int erconnected r igid bodies, 195, 390 of mac h ines, 195 necessary and su fficient condit ions for, 115, 138 neutral ,409 of pa ra llel force s, 115, 141 of a pa rticle, 389 of a r igid body, 389 stability of, 119,409 wit h t wo degr ees of freedom , 392 by virtua l work, 388, 389, 39 1 Euler, 4 External effect s of force, 24

Fir st mom ent of area, 230 Fixed vecto r, 5, 24 Flexible cables, 283 difTeren t ial equatio n for, 284 Flu ids, 297 frict ion in, 328 Marwan and W aseem AI-Iraqi

incompressible, 298 pressure in , 297 Foot , 9 Force, action of, 23. 105. 106, 140 act ive, 390 body, 24, 226 bu oyancy, 302 compone nts of, 26, 27, 64 concent ra te d, 25, 225 concept of, 4 contact, 24 copla na r syste m of, 56 distri buted, 25, 225, 226, 3 16, 3 17 effects of, 23 fricti on, 107, 327 gravitational, 12, 25, 107, 226 inertia, 428 int en sity of, 226 int ernal, 24 , 226, 271 , 39 1 kind s of, 24 magn eti c, 25, 107 measu rem en t of, 25 mech an ical action of, 105, 106, 140 mo me nt of, 37, 73 polygon , 56, 117 reactiv e, 24, 390 remote action of, 107 resolution of, 26, 2 7, 64, 65 resulta nt, 56, 85, 86, 22 7, 3 16 shea r, 271, 297 specificat ions of, 24 un it of, 9 wor k of, 386 For ce-cou ple system, 49, 56, 76 Forc e syste m, concurrent , 57, 74, 86, 115, 139 coplan ar , 56 gene ra l, 23, 85 pa rall el, 26, 57, 86 For mu lat ion of problem s, 15 Frames, defined, 195, 216 eq uilibri u m of, 195 Fram es a nd ma chines. rigidity of, 195 Free-body diagram , 16, 104, 108, 139 Freedom, de grees of, 392 , 410 , 420 Free vector, 5, 6, 48, 75 Friction, angle of, 331 bea rin g', 359 , 360 belt, 368 circle of, 359 coefficients of, 330 , 33 1, 48 1 cone of, 332 disk , 360 dry or Cou lomb, 328, 329 f1u id, 328 interna l, 328 jo u rn a l bearing. 359 kineti c, 330 lim it ing, 330 in machi nes , 348 mech an ism of, 329 pivot , 360 WNW.gigapedia.com

4 91

492

In de x

Fricti on (Conti nued) probl em s in dry friction, 333 , 376 ro lling , 369 scre w t hread , 349 sta tic, 330 types of, 328 wedge, 348 work of, 392 Gage pr essu re, 298 Ga lileo,3 Gas, 297 Graphical repr esentati on , 15, 25 , 26, 56 Gra vit ation , cons tant of, 12, 482 law of, 12 Gravitationa l force, 12, 25, 107, 226 Gravitation a l potential energy, 406 Gravitational sys te m of units, 10 Gravity, acce lerat ion du e to, 11 cente r of, 25, 227 Gu ldin , Paul , 256 Gyration, rad iu s of, 429 Hom ogeneity, dim en sion al , 17 Hydrost at ic pr essure, 298, 300, 30 1 Hyd rost at ics, 297 Hyperbolic functions , 288 Ideal sys te ms, 390 Impend in g mo ti on, 330, 332, 333 In clined axes, a rea mom en ts of inertia about, 450 Inertia, 4, 428 ar ea mom en ts of, see Moments of ine rtia of a rea s pr in cipal axe s of, 45 1 produ cts of, 449 Inertia force, 428 Integr al s, table of selected, 472 In tegration , choice of eleme nt for , 231, 3 16 nu merical te chniques for , 477 , 479 of vectors, 471 Intercon nect ed bodies, 195, 390 Intern a l effect s of force, 24, 226, 27 1, 39 1 Intern al fr icti on, 328 Intern a tional Syste m of un its, 9 J oints, met hod of, 168, 189, 215 J oul e, 388 J ournal bea ri ngs, friction in , 359 Kilogram , 9, 10, 13 Kilopou nd , 10 Kin et ic frict ion , 330 coefficien t of, 331, 481 Lagr a nge, 4 Lap lace, 4 Law, associa tive, 468 com mutative, 468 of cosines, 468 distributive, 39, 469 Marwan and Waseem AI-Iraqi

of gravitation, 12 parallelogram, 6, 25, 56 of sines, 468 Pascal 's, 297 triangle, 6, 26 Laws of motion, Newton's, 8 Length, standard unit of, 10 Limit, ma thematical, 14 Line of act ion , 24 Liqnids, 298 Load ing-sh ear relation for beams, 272 , 273 Mach, Ernst , 39 Mach ines, defined, 195, 216 equilibr iu m of, 195 fric tion in , 348 ideal or real , 328 Mass, 4, 10 center of, 227, 229 unit of, 9, 10 Mathematical limit, 14 Mathematical model , 15 Mathematics, selected topics in, 465 Mechanical efficien cy, 393 Mechan ical sys tem , 104 Mech a nics, 3 Metacenter, 304 Met a centric height, 304 Met er , 10 Method, of jo ints , 168, 189,2 15 of problem solution , 16,95, 120, 156,215,316,376,420 of sections, 179, 189, 215 of virtua l work, 385 Metric units, 9 Minimum ene rgy , pri ncipl e of, 409 Mohr's circle, 451 Mom ent, bending, 271 components of, 74 of a coup le, 48, 75 of a forc e, 37, 73 tors ional, 271, 428 units of, 37 vecto r representation of, 38 , 73 Mome nt ar m, 37 Mom ent axis, 37, 74 Mom en ts, principle of, 57, 85, 227 , 233, 316 Mom ents of inertia of area s, 427 for composite areas, 442 dimen sion s an d units of, 429 about inclined axe s, 450 by integration, 428 maximum and minimum, 450,451 Mohr's circle representation of, 451 polar , 429 principal axe s of, 451 radi u s of gyration for , 429 rectangu lar, 428 table of, 483 tabula r computat ion of, 442 tr an sfer of axes for, 430 , 449 www.gigapedia.com

Index Repose, angle of, 334 Resolut ion, force, 26 , 27, 64 force and couple , 49, 56 , 76 Resu ltant , of concurrent forces, 25 , 5 7, 86 of coplanar forces, 28 , 56

Morin, 328 Motion, impending, 330, 332, 333 Multi-force members, 195 /

Ne utral equ ilibri um, 409 Newton, Isaac, 4 Newton's laws, 8 Ne wton {unit ), 9 Ne wton 's method, 475 Numerical integrat ion, 4 77, 479

coup le, 56, 85, 86 of fluid pressu re , 298 , 300

force,56,85,86,227,316 of forces on beam cross section, 27 1 of ge ne ral force system, 85

of parallel forces, 26, 57, 86 Right-hand rule, 37, 65, 73, 469

Orde r of differentials, 14, 231, 316

Rigid bodies, interconnected, 195, 390

Rigid body, 4 Pappus, 256

equ ilibri um of, 389 Rolli ng resistance, coeffic ient of, 369

th eo rem s of, 256 Parallel-axis theorems, for area moments of inertia, 4H Parall el forces , equilibrium of, 115, 141

Scalar, 5

resul tant of, 26, 57, 86 Par allelogram law, 6, 25, 56 Particle, 4

Sca lar components, 27

Scalar or dot product , 65, 386, 469

Part icles, equ ilibrium of, 389

Screw, friction in, 349 Second moment of area , 23 0, 42 7

Pascal (unit), 226 Pascal 's law, 297

Sections, met hod of, 179, 189, 215

Pivot frict ion, 360 Polar mom ent of inertia, 428

Shear force, 271, 297

Se ries, se lected expansio ns, 47 1 Shear-force diagram, 2 72 S hea r-mome nt relat ion for beams, 2 72, 2 73, 274 She ar stress, 428 Si ngu larity functions, 274 S I un its, 9

Polygon, of forces, 56, 117 Potential energy, 405, 407, 409 dat um for, 406 uni ts of, 406, 407 Pound , sta ndard, 10 Pound force , 9 Pound mass, 10 Pressure , 226, 297 atmospher ic, 298 ce nter of, 299

Sliding vector, 5, 24, 37, 75 Slug, 10 Space, 4 Space trusses, 188, 215 Speci fic weight , 226 Spring, linear and non linear, 107 potent ial energy of, 405 sti ffness of, 405

f1uid,297 gage, 298 hydrostatic, 298, 300, 301

Stability, of equilibriu m, 119, 409 of float ing bodies, 303

on subme rge d surface s, 298 , 300, 30 1 Principal axes of inertia, 45 1

for single degree-of-freed om syste m, 40 9

Principia, 8

of trusses, 170, 188

Principle, of action and react ion , 8 , 25 , 105 , 165 , 19l, 216,

271 of buoyancy, 302 of concurrency of forces, 116 of mi nimum ene rgy, 409

of moments, 57, 85, 227, 233, 316 of tran smissibility , 5 , 24 , 56

of virtual work, 389, 391, 408 Products of inertia, 449 about incl ined axe s, 450

Products of vectors, 38, 65, 74, 386, 469 Radius of gyration , 429 Reactiv e forces, 24, 390

Rectangular components , 6, 27, 28, 64 Rectangu lar mo me nts of inertia , 428 Redund an cy, exte rna l and int ern al, 170, 188

Redundant supports, 119, 142 M arwan and W aseem AI-Iraqi

Statically dete rmi nate st ructures, 119, 142 , 165, 170 Stat ically indetermin ate st ructu res , 119 , 142 , 170,

189, 195 Static friction, 330 coefficient of, 330, 481 Statics, 4 Stevinus,3 Stiffness of spring, 405

Stress, 226 shear, 428 Structures, statical determinacy of, 119, 142, 165, 170, 189, 195 types of, 165 Sub me rged surfaces, pressure on, 298, 300, 301 Subtract ion of vectors, 6, 468 Sym metry, consi derat ion s of, 229, 44 9 System , with elastic memb ers, 405 force-couple , 49 , 56, 76

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493

494

Inde x

System (Continued) of forces, concurrent, 25, 57, 74, 86, 115, 139 coplanar, 56 ge neral, 23, 85 paral lel, 57, 86, 115, 141 ideal , 390 of interconnected bodies, 195, 390 mechanical, 104 re al , 392 of u nits, 9 Table, of area momen ts of inertia, 48 3 of centroids, 483 of coefficients of friction, 481 of densities, 481 of derivatives, 471 of math ematical relations, 46 5 of solar system con stants, 482 Te nsion in tru ss members, 167. 168 Theor em, of Pappu s, 256 of Varign on , 38, 57, 74 T hree-force member, 116 Th rust bearing , frict ion in , 360 T ime,4, 11 Ton, 10

Torque , see Moment, of force Torsional moment, 271, 428 Tran sfer of ax es, for momen ts of inerti a, 430 for products of inert ia, 449 Transmi ssibility, principle of, 5, 24, 56 Triangle law, 6, 26 Triple scalar product , 74, 470 Triple vector product, 470 Trusses, de finition, 167 plane, 16 7 simple, 167, 188 space, 188, 215 stability of, 170, 188 statical determinacy of, 170, 188, 215 types of, 166 T wo-force members, 116, 167

MarNan and Waseem AI-Iraqi

u.s. cus tomary units, 9 Un its, 9, 37, 388 Unit vectors, 7, 27, 64 , 66 , 74 Un stable equili bri u m, 409

Varignon,4 Varignon's theo rem, 38,57,74 Vector equatio n, 8 Vectors, 5, 23 a dditio n of, 6, 25, 28, 468 components of, 6 , 26, 27, 64 couple, 48 , 75 cross or vector product of, 38 , 73, 469 derivative of, 470 dot or scal ar product of, 65, 386 , 469 fixed,5, 24 free, 5, 6, 48, 75 moment, 38, 73 notation for, 5 resolution of, 26, 27 , 64, 65 sliding, 5, 24, 37, 75 subtraction of, 6, 468 unit, 7, 27. 64 , 66, 74 Vector sum, of couples, 76 , 85 of forces, 25, 28, 56, 85 Virtual displacement, 388 Virtual work, 385 , 388 for elastic systems, 40 8 for ideal sys tems, 390, 39 1 for a particle, 389 for a rigid body, 389 Viscosity, 328 Wear in bearings, 360 Wedges, friction in , 348 Weight , 13, 25, 107, 226 Work , of a coup le, 387 of a force, 386 units of, 388 virt ual, 385, 388 Wrench, 86

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Conversion Factors

u.s. Customary Units to Sl Units To convert from

To

Multiply b y

meterseecond e (mls 2 ) metervsecondf (mls 2 )

3.048 X 10-1. 2.54 X 10- 2 •

meter2 (m 2 ) mete r2 (m 2 )

9.2903 X 10- 2 6.4516 x 10- '"

kilogramlmeter3 (kgfm 3 ) kilogramlm eter3 (kglm 3 )

2.7680 x 10"

newton newton

4.4462 4.4482

(Accelera tion )

foot/second' (ftioe<') inch/s econ d 2 (in ./sec 2 ) (Area ) foot 2 (ft 2)

inch2 Un.2 ) (De ns ity)

pou nd masslinch 3 Obm/in .3 ) pound mass/foot 3 (1bm/ft 3 )

1.6016 x 10

(Force)

kip (1000 Ib) pound force Ub)

( N) (N)

X

ui'

(u ng/h )

foot (ft) inch {in.)

mile (mil, (U.S. statute) mile (mi}, (inte rn at ional nautical)

meter Im) meter (m) meter (m) meter Im ]

3.048 X 10- 1. 2.54 X 10- 2 • 1.6093 X 103 1.852 X 103 •

kilogram (kg) kilogram (kg) kil ogram (kg)

4.535 9 X 10- 1

(Mass)

pound mass {lbm} slug lIb-oe<'/ft ) ton (2000 Ibm ) (Mo ment of force) pound-foot (lb-It) pou nd-inch (Ib-in .I

1.4594 x 10 9.0718 X 10'

newton-mete r

(N - m)

1.3558 0.11298

newton-m ete r (N - m )

(Moment of inertia. area ) inch" !.Moment of inert ia. area ) pound-foot- secon d'[ Ob-ft -se( 2 )

41.623 kilogram-meter- (kg m 2 )

X

10- 8

X

10- 2

1.3558

o

(Mome ntum , linear ) pound-second (lb-eec)

kilogram -mete r/second (kg mls )

4.4482

(Momentum, angula r) poun d-foot-second (lb-ft-see)

newto n-m eter-seco nd (kg m 2 / s)

1.3558

watt watt

2.2597

0

0

(Pow er)

foot -poun d/minute (ft·l blmin) horsepower (550 ft-lblsec)

(W) (W)

7.4570 x

ui'

(Pressure, stress)

atmosph ere Istd)( 14.7 1b1in .2 ) poundlfoot 2 Oblft2 } pou ndli nc h 2 Uh/in .2 or psi)

newton/rneter2 (N/m2 or Pa l newton/rneter2 (N/rn2 or Pa l newton/meter2 (N/m2 or Pal

1.0133 X 105 4.7880 x 10 6.8948 X 103

newton/meter (N/m)

1.7513

mete r/second meter/second meter/second

3.048 X 10- 1• 5.1444 X 10- 1 4.4704 X 10- 1. 1.6093

(Spring constant) pou nd/ inc h Ob/in .)

(Velocity) foot/seco nd (ft/ sec) knot (nautical mi/hr ) mil elhour (mi/hr ) milelhour (milh r)

(mls) ( mls ) (m/s)

kilometer/hou r

(kmIh)

X

102

(Volu me ) X 10- 2 X 10- 5

mete~ (m 3 )

2.83 17

meter (rn3 )

1.6387

British thermal unit (BTU)

joule

foot -pound for ce (ft -lb)

joul e (J) jou le (J )

1.0551 X 10' 1.3558 3.60 x 106 •

foot 3 (n 3 ) inch 3 Hn.3 )

(Work, Energy )

kilowatt-hour (kw-h)

(J)

"Exact value

Marwan and Waseem Al-lr aqi

www.gigapedia.com

SI Units Used in Mechanics Quantity

Unit

SI Symbol

(Base Unit,> Length Mas, Time

mete r " kilogram second

m

meter/ second'[ radian /secondmeter2

rn/ s 2 rad / 52

(Derived Units) Acceler at ion, linear Acceleration, angular Area Density Force Frequency Impulse, linear Impulse, angular Momen t of force Momen t of ine rtia. area Momen t of ine rtia, mass Momen tum, linear Momentum, angular Power Pressu re, stress Product of inertia, area Product of inertia. mass Spring constant Velocity, linear Velocity, angular Volume Work, energy

kilogram/mete~

newton hertz newton-second newton-meter-second newton-meter mete r'[ kilogram-mete r kilogram-meter/ second kilogram-mete r / second watt pascal mete r" kilogram-meter newton/meter meter/second radian/second meter3 joule

(S upplementary and Other Acceptable Units ) Distance (navigation) nautical mile Mass ton (metric) Plane angl e degrees (decimal ) Plane an gle radian Speed kn ot Time day TUne hou r T ime minute

kg s

m'

kg jm3 N (~ kg-m/ , ' )

H, (= 1/ . ) N -. N omos N -m

m'

kg 'm 2

kg- m/ s ( = N · s) kg· m 2 / s ( = N vm -e)

W ( = J/. = N -m /.) Pa ( = N'm1m 2 )

m'

kg' m 2

N/m ·

m/. fadls

m'

J

(=

Nv m)

1.852 km ) t ( = 1000 kg)

(=

(1.852 km/h ) d h min

•Also spelled metre.

Selected Rules for Writing Metric Quantities SI Unit Prefixes Multiplication Factor 1000 000 000 000 "" 10 12 1 000 000 000 "" 109 1 000 ()()() = 106 1000 = 103 100 = 10' 10 = 10 0.1 = 10 - 1 0.01 = 10- ' 0.001 = 10- ' 0.000001 = 10- 6 0.000000001 = 10-' 0.000000 000 00 1 = 10- 12

Prefix tera giga mega kilo hecto deka deci centi milli micr o n an o pico

Marwa n and W aseem Al -lraqi

1. (a) Use prefixes to keep numeri cal valu es generally between 0.1 an d 1000.

Symbol T G M k h d. d c m

"n P

(b) Use of the prefixes hecto , deka, deci, an d centi shou ld gene rally be avoided except for certain areas or volume s where the numbers would be awkward ot herwise. (e) Use pre fixes only in the numerator of un it combinations. The one exception is the base unit kilogram. (Example: write kN /m not N/mm ; J/kg not mJ/g) (d} Avoid doubl e prefixes. (Example: write GN not kMN ) 2. Unit designa tions (a ) Use a dot for multiplication of units. (Example: write N · m not Nm ) (b ) Avoid am biguous double solidus. (Example : write N/m 2 not Nlmlm ) (e) Expon ents refer to enti re unit. (Example: mm 2 mean s (mm )2) 3. Number gr ouping Use a space ra the r than a com ma to separate numbers in gro ups of three, cou nti ng from the decimal point in both directions. (Example: 4 607 32 1.048 72) Space may be omitted for nu mbers of four digits. (Example: 4296 or 0.0476) www. glgapedla.c om

Conversion Charts Between 51 and mm SOO _

in. --{=--2O

m

It

3O_ _

~- 100

u.s. Customa

Units

km

200 --=r -- I20

4OO- =t::-+-

90

-:::}-

80

--4- -100

- 15

+ - - 90

70

20- --=_ JOO -

-11--- 80

eo

-{: -

-=1-- - 7 0

--=1-- 10

---=1=-- 50

100 -

30

:::J---,5 100- ---+-

kg

Ibm

9O,--':=:::j--

N

Ib

--4-

-20

-:::}-

- IO

9OO- = E ,-I30

- t=---11O 700--:::::J:=:--I00 150

000----=:::1-

::::::f=--- 80

::=:j--- 12O

SOO---==E~70

SOO- ---l=_ lOO

400---::::::1--

400 ---==€~OO =;E -50

3OO- ---l=_

3OO~E_4O

2OO---=~~:50

200--':==1'= --30

lOO--!=--

lOO -=i~ 10

- E-

1O--:::::f=-- :2O

o

0

Mass

Marwan and W aseem Al-l raqi

140

8OO---=~~12O

800---::-:-1--

=r--

lb/ tn."

kPa

1000 -3~

1000--+-

:::J1--14O

4O---==t::-- + -- :80

30

Length

700- -=f==-

--+- =:}--IOO

----:::1--

o-~ '---o

9OO--::::::f=---2OO

50-

40

Length

-2OO

~§==I80

+ --

0 --=>=----0

Length 100- =..--220

- 50

20

10 ----''--- 0

---=1f-- oo --{-

40

0-

- uo

o

0

Force

www .gigapedia .com

o

2O

0

Pressure or Stress

Conversion Charts Between 51 and u.s. Customa lb-ft

N -m

1000

Mg/m 3 10

Ibm/h 3

600

260

ISO

9

800

bp

600

700 900

kW 200

500

8

240 220

160

200

700

500

7

140

ISO

400 120

8

600

160

400

5

500

300

140

100

120 300

400 300

200

4

SO 200

3 2

200

100

SO

60

60

40 100

100

40 20

100 0

0

0

0

m/s

ft/ sec

100

320

km/b 500

0

0

Densi ty

Moment or Torque

20

Power mi!hr

rn/ s2 SO

ft/ sec 2

160

300

300

90

140

280

SO

260

250

400

40

240 70

120

220

60

200

SO

160

200

100

30

300

ISO 140

40

SO

ISO

200

20 60

120 100

100

30

SO 20

60

40 100

10

SO

40

10

20

20 0

0

Velocity

Marwan and Waseem AI-Iraqi

0

0

Veloci ty

www.gigapedia.com

0

0

Acce leration

';