I -
i
Understanding Structural Analysis DAVIDBROHN
PhD,CEng,MIStructE Princibal Lecturerin StructuralEngineering, BristolPo\techruc
Forewordby Sir OveAruP
GRANADA London Tbronto SYdneY NewYork
Contents
Preface Achnouledgements Part I The analysisof staticallydeterminatestructures 2 Staticalindeterminacy 3 The qualitativeanalysisof beams 4 The qualitativeanalysisof frames I
Part II 5 6 7 8 9 1O 11 12
The theoremsof virtualwork The flexibilitymethod The stiffnessmethod- frames The stiffnessmethod grids Momentdistribution Plasticanalysisof planeIrames The yieldlineanalysisofreinforcedconcreteslabs Influencelines
I
3 22 38 57 73
97 114 137 149 178 202
22r
Afpendix: Solutionstopracticebroblems
232
Index
282
Foreword
A force is not just a straight line with an arrow headat oneend. That is just a convenientabstractionor shorthandfor whatin reallife tums out to be abundle of particles under stress andstrain, alwayschangingandmovingunderthe slightest provocationfrom changingcircumstances.The theory of structures and, in fact, our whole scientificapparatusis foundedon suchabstractions. They have enabledus to imposesomeorder on tie chaoswith whichwe are faced when we look at the unendingandoverwhelrningwondersof nature which far exceedour powersof comprehension.Wehaveevenfoundtbat, if we assumethat this imaginaryworld of scienceis a true pictue ofreality andact accordingly,we caninfluenceandchangethe world we live in to suchanextent tlnt we canabolishwant anddrudgeryand,in fact, do almostanythingwe like, hcluding destroyingthe planetwe dependon, togetherwith its faunaandflora, in a few weeks - if we only couldagreewhere to start. This whole mechanisticworld-picture, the Cartesianor Newtonianworld of science,is under heaqystrain now; we 6ndin all disciplinesthat it doesnot work any more, nature simplydoesnot collaborate.I havenotime to elaborate on tlnt now andthere are, anyhow,thousandsofbooksandparnphletswritten about that theime.But how doesttris affectthe theory of structuresandthe n-holebusinessof structural engineering?Our structuresare all the time getting better, bigger, lighter andsafer, our machinesare moreefficient, smoother, usingless energy.We are all the time learningto domorewith less, so what is wrong with tllat? As David Brohn points out in hisbook, there is acriticallyimportantstageto be reachedbefore we caneven applyour numericalanalysisto a structure, namel),that we must havea structure to applyit to. Whenwe havethat our anzlysiswill tell us whetler the structure is capableof doingwhatit is supposed to do. Further, that the skills requiredto choosethis preliminarystructureare of an entirely different naturefrom thosewe maygraduallyattainwhenwe have masteredour structuraltechniques. Brohn confineshimselfto suggestingtlnt tie basisof theseskillsis the recognition of the relationshipbetweentlrc loadandthe resultingbehariourof the structure, in other words that we gainanintuitive understandingof ho'r' a structure will behaveunder load.This will. whenwe haveacsuiredOre
necessaryexperienc€,enabb us to choG€ the at leastapproximatelyright structure for a given task just by lookingat its shapeandproportionson a drawing. This is of coursemostimportant,for the increasinguseof computers hasto a great extent killedtlis understandingwhichis soessentia.l for rescuing the art of structura.ldesign. The computer hascometo stay, we must Livewith it, andthis bookteaches us how to do sowhilst remainingmastersofthe proceedings.Wemustbe able to check that the output from the computeris correct, andwhere andwhento use it, andwhat its limitationsare. If this bookdid nothingelseit wouldstill be the most important contributionto structual designwhichhasappearedfor a long time and shouldbe compulsivereadingfor anyoneinterestedin the subject. I havebeenextremely worried by the fact that giftedgraduatesfrom our universities enter the professionwith the ideathat it is belowtheir dignity to put pen to paper- the computerdoesit all. Here we havethe necessary remedy for suchconceit, andit is hightime. UnfortunatelyI haveneitherthe time nor the knowledgeto dojustice to the achievementsof Brohn'sbook, I have only a few hours left for the printers' deadline,whichis entirely my own fault. But he has seeminglygonethroughevery knownmethodof structural analysisand showsby clear diagramsandexplanationshow the structureis affected by the loading,thus at eachstagegivingthe readerthis essential understandingof structural behaviour.I hopethis bookwill give rise to a lively discussion. I cannotresist addinga comrnentof my own. Whilstrecognisingthe importance of what Brohn hasdone,I do not tlink he hasgonefar enough. Understandingof structural behaviouris very necessary,but are there not many more things that are equallyor evenmore necessary? Every structural designerof repute hasdeclaredthat structuraldesignis an art as well asjust an applicationof scienceandtechniqueto a givenproblem. You couldalsoput it the other wayroundandsaythat onlyif it is awork of art as well will it be adrniredandaddto the reputationof the designer.It is unfortunately impossibleto definewhat art implies,but it hasin anycase nothing to do with numericalanalysis.There are manyother mattersto consideraswell. The wholepurposeof a structuraldesignis to helpusto make the things we need, or fancywe need,orjust fancy.So we must makeit very clear to outselves what we want to achievewith our design,whichwill obviously affect its shape,the materialswe use andall kindsofother things. If we want to buildsomethingis this the right placefor it, couldnot our purpose be better achievedin a differentway altogether?Onlywhenwe havesortedout all these matters to ou.randour clients' satisfactionwill a structuralana.lysis become relevant. Obviously,wbat I wouldcalldzsigzis muchmoreimportant thar structural analysis,for that determineswhat we are goingto get for our efforts. And moreover raial we decideto do is muchmoreimportantthanhow to do it, andthat openstJresluke va.lvefor a wholefloodof questions,social, political, ethical which tlreaten us all with confusion,or worse, becausewe are not able to agreeon wbat to doHow to live in peacesitl or:r:neighboursonthis pJanetwitiout destroyingit is the ultimate andnow pressiugproblemandI wish I knew the answer. Ove Arun
Preface
This book is aimed at the identification oJ the fundamental princiPles of structural analysis together with the develoPment oI a sound understanding of structural behaviour. This combination leads to the ability to arrive at a numerical solution. Using a series of structural diagrams as a visual lanSuage ol structural behaviour that can be understood with the minimum oJ textual comments, the book aims to develop a qualitative understanding of the response of the structure to load. It is ideally suited to under8raduates studying indeterminate framed structures as Part of a core course in civil or structural engineerinS' but it is also suitable, because of its qualitative approach, for students of architecture and building technology. The book is in two parts. Part I' the first lour chapters, deals with the development ol qualitative skiils; that is' the ability to Produce a non-numerical solution to the loaded line-dia8ram ol a structure. It is considered that the ability to arrive at the qualitative solution to framed structures is a significantly imlortant component of the overall understanding of structural behaviour. Part II deals with current methods of structural analysis using the diagrammatic format to which the student has become accustomed. The need lor the developrrent of qualitative skills increases with the increasing use of the computer in design offices. In the near future, the computer will replace the majority ol analysis and structural desiSn calculations. Unfortunately, this will also have the elfect of eliminating much of the experience and consequent understanding gained by the student and trainee engineer. This work explains how that understanding is develoPed along with current analytical procedures, PreParing the student for the design olfi'e
design data' where the computer ls rne source of virtually all numerical Analgsis is an inteSrated approach to the Understandinq structural ol which teaching and learning of the PrinciPles ol structural analysis, are also this book in this textbook is a major part. The ideas embodied ol the same available in an audio/visual series of sel{-learning programmes name. The audio/visual programmes are backed by a suite ol micro-computer programs which have been used to produce the numerical and SraPhical solutions to the Practice problems, included in this text' software are The audio/visual programmes and comPuter-aided learning available from:
osE Ltd 197 Botley Road OXFORD
ox2 oHE, UK TeL 0865726625 Publishing' who should be contacted direct and not through Granada
Acknowledgements
The research project upon which this book is based has extended over a period ol ten years. In that time, many friends and colleagues have contributed to the development of my ideas of the way in which students can be encouraged to reach a better understanding of structural behaviour. Bristol Polytechnic has provided both time and resources and my Head of the Department, Dr Matthew Cusack, has been particularly supportive. These ideas would have been stillborn without the continuous slrpport, interest and encouragement of Peter Dunican, senior partner of the Ove Arup Partnership. Many other engineers in that remarkable organisation have helped me with their advice and constructive criticism. Perhaps the most successful period for the development and testing of the qualitative approach as a basis for the explanation 01 theories and methods ol analysis was the year I spent with the Department of Civil and Structural Engineering at Hong Kong Polytechnic. I owe much to discussions with Dr Kwan Lai and Dr Norris Hickerson. but most of all to the resDonseol the exceptional students. However, it has been the extensive and particularly fruitJul collaboration with Professor Peter Morice oJ the Department of CiviL Engineering at Southampton University which has led to many of the specific explanations and visual sequencesin the early part ol the book. I am indebted to all of them.
1 The Analvsisof Staticallv DeterminateStructures
The subject of this book is the behaviour and analysis of statically indeterminate structures.
However, this first chapter reviews the
behaviour of deterninate
structures, a thorough understanding of which
is essential before the topic of indeterminacy can be tackled.
The text
assumes a basic knowled8e of mechanics including an understandin8 of the principles of overall equilibrium, bending moments, shear and axial forces. It is possible to analyse determinate structures by consideration of equilibrium - in general terms, the application ol force and moment eouarions v 1 O. d = 0 and lt = 0. With most real structures, this is not possible as the presence ol redundant members (secondary load paths) makes it necessary to consider relative member delormation beJore a solution of the structure can be attained.
The number of unknowns which cannot be lound Jrom equilibrium
considerations is known as the degree oJ statical indeterminacy. The design oJ engineering structures usually starts from a need to sostain loads. Initially though, it requires an understanding ol the way in which a proposed system of members can provide the required support, and how it will deform. It is, however, clear that an understandin8 oi the behaviour of statically indeterninate cf deterrirrate
systems is based upon a thorou8h appreciation
systems.
This chapter develops the relationship between load and delormation for a range of structures which are amenable to solution by the application of equilibrium alone. Once we have analysed the behaviour of the proposed structures \re are then able to start an approPriate process of numerical analysis to llni
cJ:
UN'A,RSIA';]:;G SIRUCTURALENSIYSJS how much of each of the various parameters is involved, Jor example, the values ol the loads carried by each member and, as a consequence,the srze each will have to be to carry its load saJely. \ve can go on to find the values ol deformations which will result Jrom the loadins.
TI]E ANALYSIS AF STF.IICALLY DETERMINATE STRUCTURES
Thus we see that structural analysis must have two .omponents. The first is a qualitative understanding and ::e second the numerical procedures. It must be jderstood that qualitative analysis is not in any way a ,i:icsritute lor numerical analysis but should be regarded 3: a necessary complement to it so that the two allaoaches constitute a complete whole giving an -..erstanding of, and an ability to evaluate, the :::icrural perlormance.
\k
-, M A-
M QUALiTATIVE
ttu 5r eo- 3€
NLIA,1ER-tc.\L
:-e oasic principles of our structural analysis lie in the :a i --f sratics. Mosr srrucrures are required to be :-:---3 rn a static state. This is not to say that we :_g-::ci analyse dynamic behaviour such as may be caused :. --::rhquakes, wind gusts or moving loads but initially, a: -e3s:. we shall concern ourselves with statics. The ::aie: c.ane demonstrates the kinds of equilibrium ::r':-::cns which we have to satisJy. The vertical _..::::an ar the ground must balance the total downwards ::.:?:: counterbalance and load.
VELTIC'AL
EoUILIARIUM
:e:..c-). u'e can see that wind Jorces will tend to make _c-e r :-Je crane structure slide sideways and this too, 1-r--
by horizontal support lorces at the -.esistedj a.€- :1:s \r'e call ho.izontal equilibrium.
Fr't^4 _-.-.':' --------)
+
I1ARI: ONTAL @UILI 5P,JUM
---:_-1. -: :S Clear that the counterbalance weight cannot !c--i=-L= 1- .oldirions of loading on the jib so that any !rE-:i---::ce
$ill tend to topple the crane. If we add
E :rs :-e :.Citional toppling eilect oJ the wind Jorces r: lei =E::l:e base oJ the crane will have to provide _E\s=!-,= :c :hese out-ol-balance toppling moments. -r1s =* iioment equilibrium. -r /\^oMENT
Eeut!.tE;ui/At
UND' ?S' IlE DJ NC STRU CTURAL ANALYS I S
In our three-dimensional world we can express these equilibrium requirements in the following way.
Firstly'
we must ensure that in each of three directionst at an8les to each other' which we will label x'
Y and z'
the resultant oJ all forces acting on the structure must be zero.
In other words, reactions must balance loads.
Secondly, as we have seen in the case oI the toppling eifect, the tendency for the structure to rotate about any of these three axes must be resisted. We say that the resultant moment about each of the three axes x, Y and z must also be zero.
Thus the moment of
reactions must balance the moment oJ loads. This gives us all together six conditions of equilibrium.
6.
In much of the following explanation of behaviour, and in many real life engineering situations, we Jind it is possible to be sure that the lorces in the direction v are zero and that there are no moments of lorces about the x and z axes. If, indeed' this is the case, then our problem can be reduced to the consideration of three conditions of equilibrium only. Such simplification is described as a plane problem because all forces lie in one plane,
7.
we shaltr find it convenient to label the lorces in the .r direction with symbol d' to denote horizontal, and those in the z direction v, to denote vertical. Also we shall use the symbol l, for moments in the plane, about the Y axis.
8.
t
t {
Y =a VarttLt aluilihri.t
Vith these symbols we can write down the three
of equilibrium- all the horizontallorces must sum to zero, all the vertical forces must sum to zero and the sum oI the momentsmust also be zero.
H'o
Hon2o,{e q.fb/ilrf"1
M.o
iqon',t,f ?uilibrt-t
TEE ANALYSIS AF ::;'':'."LY
DETERMINATE STRUCTI]RES
\ow it will be remembered that a pure moment, =lled a couple, can be represented by two equal :^ d o pp osire p ar allel J or c es ar a dls r anc e apa r t . ,. lhis diagram we have the forces ar a olslance : grvlng a moment tv : F x d.
:-
-3: us now consider the moment of this couple ::cur two points, A and B in the plane. :::rsidering Iirst the point A we see that as ---,=Cownward force tr passes through A it wiil -Eie no moment about A and all that remains ::::e ::.ae r
anticlockwise moment ol the upward = at a distance d. The moment about
-: YA = a x d anticlockwise.
Considering
:_e:oinl
B the upward forces pass through
:*e:olnr
and the only moment is caused by
t
:*e :tr\r'n*ard force r at distance d. This :--:.. i ves an anticlockwise
moment
.s
f,"^( & " . r , o) 1-
-:: -. :o,* consider rhe moment about a point C a-=:_,ae: from the line oJ action oJ the upward E'e can also find the total moment of our = about this point. For the downward force
::r=:r:-
I
: =€-e -5 an arriclockwtse momenr due to rhe j plus e, and for the upward force F a ]:1:.:::i. :-\:cra.,= :roment due to the lever arm e. -as*-:-: ie
i,=
=omenr 4
is the force F multiplied by C j. :F_-:a:;oa still, of course, anticlockwise.
I_€': r:
:aie shovn is one oJ the most powerful
:: ,-:-..::-:. al ar aly s is . nam ely t hat . as a c o u p l e
],as :-€ =-e
:noment about any point in the plane,
].e ?:r*--:r:ri:. cf a structure will require that the 'r:ra -,:r-3::: cf all Jorces must be zero about any !t!r-: - ---e :iare. _':a :r:rsiCer an actual structure. The beam ABCD -r* J s 3,Trc:r i,5!-!-!ed at A and B and loaded at C. We shall E!'*r=:-E .::ec-!s of any self-weight and only study those :j.tc :: :-E ::a-ied
load
t/.
I
The
{'+:- : < - q)
'l
F
e.
.) rl-Fo
UNDERS'IANDI N G STRUCTURAL A.VAIY5I5
We can start the qualitative analysisof the structure by observingthat since there must be equilibrium of momentsabout the supportA the only lorce which can balance the clockwise moment of the load t1lat C will be an upward support reaction at B, giving the necessary anticlockwise moment. Furthermore, since the support is on rollers, only a vertical lorce yB can act at this point.
1 4 . As the lorce 1/Bhas to produce a moment about A balancing that of the load , we can observe that it will have to have a magnitude greater than t/ because its lever arm about A is less. Thus if there is to be equilibrium of vertical forces, the vertical support reaction yA at A will necessarily be a downward force
tr."
to balance the excess of yB over ff.
15.
Ve can also usethe requirementthat momentssum to zero about any point in spaceto confirm this result. Taking moments about B we observethat there is a clockwise moment due to the load ,1rat C. A balancing anticlockwisemomJnt is required. Thus yA must act downwards.
Mv46o;
16.
Lastly, we should observe that as neither the force kr nor the reaction yB has a horizontal componentthe requirementof horizontalequilibriumensuresthat the horizontalsuDportreaction at A is zeto.
TEE ANALYSJS C- S::::C?LLY
I7.
DETERMINATE STRUCTURES
The concept oJ equilibrium also applies to any part of the structure. If we consider an imaginary cut separating one part of the structure from the rest, it is clear that rnternal forces at the cut will be required to maintain
lw
"t'
equilibrium. Let us supposethat we separate part of the structure to the right oJ B by introducing a cut at K.
lE.
I
cu'f
If the portion o{ the beam KCD is to be in overall vertical equilibrium then there will have to be an
tw
lnternal vertical force S at the cut point K to balance the load rf.
It must act from below upwards on the part
'l '
--*--D
KCD and have a value equal to the load, This internal force is called rhe shear force-
{v="
The position at which we choose to place the cut K between the support point B and the load point C has no eflect upon the value ol the shear force, which will be constant from B to C. To the right of C there are no external forces and so the shear force reduces to zero in the portion oi the beam CD, We can plot the distribution of s as a shear Jorce diagram, drawing it underneath rhe base line, that being where we have shown the shear force arrow.
,.,
The diagrammatic convention for shear force is that the value of the shear iorce is plotted on one side of the base line, above or below; that is corresponding to the vrew an observer would have sitting behind the cut K. The load tr would cause the part oJ th-^ beam KCD to fall lrelor,, the beam, thus the dia8ram is plotted below the base line. The arrow on the base line is required to confirm this convention.
I S shear s= w fo"'
1..;!J}E?S:}:i )J EG STRUCTIJRAL ANAIYSIs
21.
For a cut K in the beam between A and B we see that the greater magnitude of vB compared with t{ will lead
ls
a----l--+-r /<
tW
to vertical equilibrium requiring a shearing lorce .s from above acting downwards at K, being the dilference
4\
between
th"^ W)
Cl4g;
YB and t'r',
'4.. zr=o
Again the choice oJ the position of K between A and B does not aflect the equilibrium equation and thus s will
lw
!_.i_-_o IA a
also have a constant value oJ shear force between these points.
'lva
Yua
Supposewe had considered the vertical equilibrium of leJt-hand portion of the beam from K to A. You will see that this will lead to an upward shear force 5 to balance
i-l'-
V
the only vertical force on this portion of the beam, the
ls
suPport reaction vA.
Vp, -
24.
This is a necessary result because, when we close the cut by putting the two sides of the beam together, the two
_Al_ :-:-
ls
shear Jorces must cancel each other leaving no external out-of-balance
Ir
ork
vertical force at the cut section.
THE ANALYSIS OF S:Ai:CALLY
DETERMINATE STRUCTURES
-: r1lv remains to be clear about the plorting convention ::. :re diagram oi shear Jorce distribution. We have a-:\rn a horizontal arrov/ pointing to the right to indicate --:: $e we re look ing at t he f or c e ac t ing on t h e r i 8 h t -:-.: side of the cut. We could equally well make the =::r*
point to the left and look at the force acting on
:_e -eir-hand side of the cut when clearly the shear force :-:i:am
in this case becomes a vertical reflection of the Fir her . iiFor am is . . ) r r c c ! but it is i 'n D o r t a n t
*+=f,"<:=q-
:: -...:rde the horizontal arrow to show which way we are
:5-
-:: -:-.now return to the moment equilibrium at the r-:a::rarv cut point K between B and C. We see -F_ -'e loFd l, Dro.i||ces,] .lockwise moment about . :-:\i a s ,^ t . ,(, y $ and lor equilibr ium . r hi s m u s r b e w r\, _:sj:3i bl an internal anticlockwise moment within :-E :e3:n ar the cut K itself. As the shear force ,S -;:+r _ -.ol|o h K ir . , r n m ^k e no c onr r ibut io n r o ------
ls r:
"
t[tr,
B4czl Ma(x)
{ u 6 c1 = o
cd ,,itihr ir , m . Thiq int er nal m om enr i s c a l l e d
:E:=:::-::r
ll-
1W
--- - - ---Alzr-o
m om ent .
.rioYe the cut point K towards the load at C
-:ie. 3rm oi the load Ia.Iwill reduce so that the -: =-,- -: ,nrcr nr l bFndi' 1pm om ent will dec r e a s e
=':
:r:c\..::..3iely. r,:'-€.:
If K coincldes with C the bending
1|i]l be zero. Thus we can plot, on the
i=r-:-:e
!s Da(e. a linear graph representinS the o f r hp bFn. iinc m om enr J r om z e r o a 1 C
-: : -:\-rtJm
v alue ar B. c lear ly , as t her e a r e
-r :t::::.a.1 fo r c es Dey ond C. t he bending m o m e n t r:E:c::ion
ol t he beam CD m us t be z er o .
E.M.
aBS/a1,1
UNDERST}J.ID!NC STRUCTURAL ANALYSI S
2E,
Suppose now that we place our imaginary cut polnt K within the span length AB.
In considering
equilibrium oJ the whole structure we have already observed that the upward support reaction yB,
{ n cri=o
whilst being larger than the load
, will only
produce equal moments about A.
Thus at an
internal point K, less far away than A, the clockwise moment of t/ will still exceed the anticlockwise moment of yB and the additional anticlockwise moment of resistance will have to be supplied by the internai bending moment.
29.
It should be clear that this internal bending moment will be equal to that caused by the load tt alone when we place the cut point at the support B. The effect of the larger support Jorce yB will reduce the required balancing internal moment as we move to the left from B until, in order to satisfy overall moment equilibrium, it will necessarily disappear altogether when we arrive at A. Ve can thus plot the full distribution of bending anoment along the beam.
30.
^
[T^
M
c*/ -{--n-.--"
i\B
Had we considered moment equilibrium of the left-hand portion of the beam with the cut K in the span AB, we should have found the requirement for a clockrljse internal bending moment to balance the moment effect
{
of the downward support reaction yA.
t* 617-o
We can also
note that the value of this bending moment decreases to zero as the cut point approaches the support point A.
31.
:)7:_g.th. 8.rua\ .--
lf again we consider the imaginary cut being closed up, the two bending moments on the two portions of beam, one clockr.ziseand one anticlockwise, will cancel and leave no er.ternal moment acting at the section.
THE ANALYSIS OF STATICALLY jt-
DETERMINATE STRUCTURES
l: s:o be noted that the bending moments we have on both sides of the cut will cause bending tensile
iEr:r
:=esses, and therelore tensile strains, on the upper :-trt.esof the bearr. \\/e have already adopted rhe ::rrenlion
that we draw the bending moment diaSram on
:-e:e:sile
side oJ the beam which we may indicate with
\le always use the axis of the structure as the
a :'. a<:::re
for the bending moment diagramr the ordinates
:i r::ch
are drawn ,or.aral to the baseline.
:=-s:ie srrains in the upper libres and corresponding ::r-.1:ess;ve strains in the lower libres of the beam cause it to bend into a curve which is called a
=:::&i
:cE=ag curvature. The centre oi curvature, in such a iies below the beam axis.
:e.
ro
i:-:.-:nB r:
:::
io our particular beam probleln we can see that also Sive a qualitative solution to the dellected
:-e.r.. as in this diagram. We know th:.t it will hog and :'E: =ere will be no deflection at the su??ort points A 1-'. l]=:
l,-::ri in the direction of the load. Note, in
a3_,:.i. :l-
s e lrill expect there to be a deflection at the
ihat as there is no bending moment in Portion
:::s qill not bend. However' this Portion will
ad.e::
a:]d rotate as it is attached to the rest of the
:--r-:ie
l.BC which does bend. You should particularly
'rcir,=:
coherence between the deflected shape and the :t10rnent dla8ram.
-_'L<
r@-(ffit
t3
l4
:,rl::--iil"-2-Ir:
ysf.s 13 STRUCTURAL Z]V?SZ
3J.
Ve will end consideration oJ this very sirnple problem by observing the various characteristics of the solution which show consistency and lead us to be satisfied that our qualitative solution is correct: (i)
The bending moment diagrarr shows that there are no moments at the end simple support.
(ii)
It also shows that the slope of the dlagram cnanges abruptly at each point where there is a concentrated external force and that it is linear between, i,e. where there are no other external forces.
(iii)
We have already observed that there rs a Seneral coherence between the shapes of the bendinB moment diagram and the deJlections, with hogging curvature wherever the bending moment dtagram is drawn above the beam axis.
(iv;
The shear force diagram is constanr wnere rnere are no external forces on the Deam anq it changes abruprly ar rhe load points.
We shall now look at a slightly more complicated problem in which our beam is extended by having another portion hinged to it and supported at the right-hand end. We note that the support system has been arranged so that there can be no horizontal support forces at B and E and the applied load |1' at D is in the vertical direction.
?EE ANALYSIS OF STATICALLY
G *
DETERI'IINATE
now proceed more rapidly to use the various
rrr
conditions to determine the directions oJ
1fl-rtfrt gat t
forces.
. .' rErt
Firstly,
as the hinge provides no
of resistance, equilibrium of moments ol CDE C require an upward force vE to balance the
.f.f.rie t
moment caused by ff.
ai.i
be less, Thus vertical
riX
*
However, because
arm of vF is greater than that of h/ its equilibrium oi the
CDE requires an upward shear Jorce sC to Fb ill af rlE hinge point C. An upward shear on CDE a
C rill
imply an equal downward shear at C on
tr.t
d rl€ beam CBA. Moments about B will a downward support force yA to produce an moment to balance the clockwise dF
to sC.
Lastly, vertical
equilibrium
of
rErlri€s an upward support reaction vR to dF two downward forces YA and 5C.
ir irst one possible sequenceol reasoning hEds r.6 to establish the full set of support sl|ovn.
in a position to start drawing the
Er -G
ol the internal bending moment, dE bending moment diagram,
-t
From A
reaction Ya must cause tension on dE beam, consequently the bending moment
-
b plotted above the beam axis. q
From
dE reaction vE causes bending tension
- th
|fdei'side
f!-tidr
of the beam axis.
of the beam BCD we know that at the
d rtre hinge C the internal moment must be zero B and at D it must reach the same values as that -lc+t$ouring spans. In addition to this, however' tlrat the moment developed from C to B and lun
C to D is caused by the same shear force sC.
ft
dope of the bendinS moment must be the same sile of the hinge. The bending moment diagram
belore,
be given by a straiSht line from B to c
ftrough zero, that is the beam axis, ar C.
UNDE RSTANDI N C S TRUCTURAL ATVAIY.SIS
40,
At this stage we can draw the shear force diagram. Remember that we must indicate whether we are looking at the force on the right- or left-hand side of a cut. Here we are looking at the right-hand side. Notice that a hinge carries a shear force, and if there is no external load at the hinge, there will be no change in the shear force value at the point C. This is consistent with the constant slope oJ the bending moment diagiam across the hinge. We shall not often need to draw out the shear Jorce distrlbution to obtain the p.imary inJormation of our qualitative solution. In future, thereJo.e, we will concentrate upon the other three diagrams, namely the support reactions, the bending moment diagram and the deflected shape.
4I.
In this case we see that the whole ol the beam is hogging from A to C and sagging from C to E. We have indicated this by placing a T on the tensile side of the beam,
42. The other lactors which we know about the deJlected
lHE
ANALYSIS OF STATICALLY
DETERMINATE STRUCTURES
at a somewhat more complex problem
lb(
a number of new ideas. The structure is a symmetrical portal frame with three .!L C
F as shown in this diasram. It is -rd a E ee'pinned portal. The portal is loaded forcf
f'e:Ere
it placed unsymmetrically at rhe *lat the corners oJ the frame at B and
!-Eti6 co,rnections between the horizontal and lEitre?E
ihus forces and moments are fully these connections.
lcoss
On the other hand.
tr C tsrEres. as before, transmitting lorces but rlulglriq
rE (?jr (lse the principle oJ equilibrrum r1 i:,Ie whole structure, dr@:E
about any point.
ooint F then it is clear that the r-ronent llt t/ caused by the load ,f must
a,'r egual and opposite moment /14 R l'!.{'l} available reacrion Ra ' At thii^
!E!d !| G
f!
c!:nc:
define the value or direction of the
-:.\ except to note thatr as tt has a diEi::!€ Doint F. it cannot lie on the line
riEcicr:
d @ fii@r 1ffi?
-d !e io*'look at the equilibrium of the s=--EiJre ABC we realise that the line of -=I :]]usr pass throuSh the hinSe point c aa: )e no resultant moment about this
rud::E:e u
TlE s:-S:Jte.
l!tr @].! fi
m
are no other external forces on this
:-{re :E:rrc
convenient to separate the reaction componentsi yA in the vertical
l7
UNDERSTANDINC STRUCTURAL .ANAI,Y.sI.s
47.
We are now in a position to complete the of the remaining support reactions; namely those we take moments about the support point A, equilibrium demands that there shall be an upward vertical reaction yF at F to provide an moment to balance the clockwise moment of the about the support hinge A. Notice that the line of horizontal force aF passesthrough A and produces moment about that point.
48.
We may now slart to draw our bending moment Clearly a horizontal force, such as tA acting at a distance from a point on a vertical member, such produces a bending moment in the same way as vertical Jorce on a horizontal member.
Ve start
drawing the bending moment diagram on the t of the vertical members. Since the horizontal and aF are numerically equal and heights from A and from F to E are equal then the bending at B and E will be of equal value.
49. We have noted that the joints at B and E form rigid connections between the members at these corner Doi As no external moments are applied here the bending moments in the horizontal members at the corners have the same values as the bending moments in the vertical members at these points to maintain moment equilibrium.
TEE ANALYSIS
S:A':CA;LY
DETERUINATE S"RUCTURSS
"
{Dr .rerr consideration is with the internal hinge at C. qe =!n !,oint. as at all unloaded hinges, the bending tu[c:: !
itas to be zero. Furthermore,
because there is
6re!-.)a1 loading within the length BC the bending
1r!!Ixe-: can only change linearly from its value at B to a r=ire ar rhe hinge C. Indeed, the slope of the
Go
!a:: rill continue at a constant value until we reach h
6:er-nal load point D. We saw a similar ef fect when
m *rred
afie case of a beam with an internal hinge.
!Dr! r:
ioie rhar this effect is consistent with our
brl:
r=Iues oJ the support reactions.
The bending
r-!nnEl:
ai B is clockwise balancing the anticlockwise
rr[rE::
causd by the horizontal support reaction
I{-
}L.rever, as we move Jrom B towards C, whilst
&
Thcar.err due to ,yA remains constant, the r.4[rorr reaction yA begins to have a
{!r'EG.!
giect
rErEl: d
in the reverse direction.
The eJJect
rr, rs io diminish the moment due to ,?A as we
rrer
!..5
lh
B towards C until it reduces to zero at
bir:ce Dolnr C.
'[lc sarE€ .easoning may be applied to the right-hand c: o(' frame from the corner at E to the load
t!'tE: pE
D- -1: D. of course, the bending moments from the
rirfr arc i:cn: the right must come to the same value. We cn arr-,:are fr
complete our bending moment diagram for
ft-a-.e b_r &awing a straight line from the value at E
!o dlE rd--E at D.
[b
tusr
d ldlE t
:1cr- derermine the lorm of the deflected shape g e start by indicating, with the letter T,
rcs-.r-,e. :: :E
members which will be in tension and
..............drre oc ere outer side of the curvature,
ffiil-?
t9
. :;.. F.STANDI NC STRUCTURAL AdE-LYS.I.S
We note that the support points A and F are r both horizontally and vertically but, as they are the slopes oJ the members may change at these The corners B and E may rotate and move hor but may not deJlect vertically.
Also we observe
hinge C may deflect vertically but it will have the horizontal displacement as the corners B and E due assumed axial rigidity of the horizontal member However, the members connected at a hinge can different slopes on either side.
54.
Let us first look separately at the two halves of the Jrame; the part ABC and the part CDEF. For the moment we will assume that the corners B and E
HfP1" ti -'
l"
-'\-
in their original positions vertically above A and F.
I
-ii:
bending on the left-hand side ABC takes place with
l-
tension on the outside, We can expect that pa.t oJ the frame to deflect as we see on the left-hand side
A
diaSram, B. on the right-hand portion CDEF there is equal bending on the vertical member but the outside tension changes rapidly to inside tension before we D and continues to C. This side of the frame would delorm as shown. \yhat we particularly. notice is downward deflection of C on the left-hand,side is than that oJ C on the ri8ht-hand side.
55.
However, our frame is, oJ course, in reality and the two points C have to be coincident since they shared by both the left and right parts oJ the frame. This can be achieved if we allow the points B and E to move horizontallg,
rotating the lelt part ABC in
anticlockwise direction about the support A causing C to rise and move to the left.
The amount ol rota
will be the same for each part oI the frame and must just sufficient to make the two points C coincident.
2 StaticalIndeterminacy
So far in your study of the analysis of structures' the range of structures has been limited to those which may be solved by the application of the Th. ee equar ions of qr at ic dl e q u r l i b r i u m : all horizontal lorces must balance' all vertical lorces mLlst balance, all moments must balance. Consequently for a solution to be found there can oniy be t1rree However, virtually all real structures have more than three unknowns, they are'statically
indeterminate'and cannot be solved by the three
equations of equilibrium alone. That condition is known as'indeterminacy'and
there are two ways of
identiJying the degree to which a structure is indeterminate: statical and kinematical indeterminacy. This chaPter deals with the former' Kinematical indeterminacy is specilically related to a particular method of analysis, Stiffness, which is discussedin Chapter 7. Although this chapter will ultimately show how to find the number oJ times a structure is statically indeterminate, this number or 'degree" is only of importance in the Jlexibility method of analysis. The degree is the number of equations of compatibility which must be identiJied to add to the three equations of equilibrium for a solution of all the unknowns' However, as you will discover, it is the identification that the structure is or is not indeterminate which is of primary irnportance, tn other words, it is the qualitative apPreciation of this condition which should be uppermosr in your mind. The calculation of the de8ree of I nder er m r aaL! c an onl! be s a f e l ) c a r r i e d o u l o n c e t h l \ q u a l i r a t l v e uidersranai:rg iras been soundly established.
STATTCAL I NDE?ERMINACY
23
]
r,E-:::::.: the state af statical indeterninacg fiF!@l[a : : real structures do not satisfy this criterion .L-ccin as staticallv
-Ih
indeterminate
structures.
-!a u-rllrar-r objective oJ this text is the development
d-
r€rranding
O'!r!rtr.1a:e
of the relationship between
and indeterminate
ftuL.n_
structures and the
Darricular, of the latter.
Tllbh$r
\r'e will deiine the axes.
r'-=
iffiE
=.d most real structural
fuar
:!:ived
, - i !!6'e_:
SAetc4
U
d6i,r ut^;i,? &nr<-'u',.c:
aro thosq' 6 whah styeJr Tgjaa*j w*.hM tl1i- t^c)y\bc.rt + bh., EtrucvL a^4 fu.< sucte as ba.aart3 r ^01^1.,tl-s ,akV 6^g 46 bo- a.t4rr,\t^eta A tho o4ta,Ein1 tl a$;llbvk;.
Most of our
Problems are
into two dimensions set in the
{ll
reactions and lorces will be
=s:
-rEf,
'lrar'-zar::al in the x direction r?---::1
::t rhe z direction
nrr:er:
about the v axis, set normal to
a nEi-lp lan e.
EQutLtERJt4M.
:_E: a]l these lorces rnust be in
hiE'
=. -.-':t a.jilt
Afernn
on the structure.
Att v.*ek+t fi r...s
I
f
Atthwaura $tt"as -N+ tut.
D[
eq-,a:ior]s ot equilibrium
Nl
determi nate structures'
t:a---Jll
D&
are suf f icient rno*1dlttr qb.u.ta pothr-
A^L/S7 ge
ft
d.
,oor ir::
3: a simply supported beam
-:aa t. :"e :-=a::ons
d
a ri-!s-;-
r*c"
'/A and Vg.
The
m!
rr a.e 1.a:zor!tal direction at B has T lr.:-in8 restraint B on rollers tiq'-zr-=l movement. The reactions rs llLh
:::1e -::::
hfu @mnFtr
5r
r!--a
-'..ir:.
on the rre.r?be-r
f,e equal to the hoaizontal
g :_E -aac-
F4U|LIB
There are only three :A,
|b|ilne
LIki
lH
,4
f-; {
4 UM
.;:-='-31'R',;
:T:EE:!r;r:
A'\-'.EiZr':
5 atrso tra'icall! This second example of a ca(ItiLF ciEracter determinate. It has the three reacdqE {loe'rls' theref a built-in suPport. There are three ttr ol the this structure may be solved by the apPlication of the wall laws ol equilibrium. The direction action of the reaction is negative, anticlockwise' the
)-
wall on the memDer.
6.
Pnnca
supS.tC
e-2H
T
+
the qualitative It is critical to an undetstanding oJ conditions are analysis of structures that the supPort known as a properly understood. This first is generally rotation and nas r|o pinned or simple support' lt allows pr6vide hotizontal moment restraint. The supPort will
iv
rt/
vertical lorce reactlons'
lnr_ Allo, S rotaEtoh. f?aot lo^ N6 '^owlqtt
-,
7. P..llt r
su.Pr*'
in real The roller release is a relatively rare occurrence are common and structures; however, partial releases will helP you to basic understanding ol this lull release support condition understand the partial release' This to the direction ol capable of only one reaction, normal suPport will noi movement. Note particularly that this dotnltard react lift off the support, it is capable ol a as well as an uPward reaction'
Ftxa+
S'f,?dYt
++
}J
and depend uPon the loadin8'
t
tvn
encastr5 or The fully lixed support, sometimes called ,built in', will provide three external restraints' It is are worth emPhasisinghere that these restrarnts
M
STATICAL INDETERMINACY -:o Dr ov ide an inler nal r eleas e t o t he _- J:_:-_:. Same internal releases are difficult to
httgo
Ittetut
introduced here to prepare you 10r their
t,!iiai-=:-a:ae
-pYr 1\ anFlv r ic Fl . lev t c es . -_ : : .- .1 6 r nal r none_L r eleas e. a hinge. I t i \ - - - - --.^ c'Fr - ino d\ ia and ( hc ar lor . es , T h e rf1r.............is always zero at a hinge. The ioint is 1::- -:-::t: ' . -L-'1rr^
"
:esrrained in position and both members
_>
\l No uo,ne*
fua$fl
slope independently of one another.
.: :::?: -
!ll@r"
*
r, -:-:-: '-- r _:-.:5:
_:,e:se, the hinge, is fairly common in :::'ar.
the shear release and th-'axial
L-
.ut,s!e,.:_: !- -::e. are only likely to aPPear as part of a
f,
'1"r,.r,:lr ._ ::- =-=-\:ical procedure. In each case two _:5-_1_:: ::e lransmitted and one released' i,_:"?ri
(?**\( snea fitaxi*
e l \r ,
--€
Na a x1.\l firco
ru b-t::-er: :r:
Uawfet
::i.rciures are almost always
a'!&,':fr.'---:-:-
l-:
e\ceDtion in its many forms is the
r- -. lere are f our external reactions, one .llirn ::-_ -.. _; ::.r'ec b! the three equations oJ iae hinge provides another equatlon €r,!r,:r- r-- ::;:1::. ti'tf.*.r"-: l:::
ll ylrL--,:r,| _ii' t-{-_:!-
:-: : ::ioinent equilibrium is considered at _:::::t.s HCand VC orovide the fourth
mt"ii -'r_:: ::-,
, u0tt-i -'"€li'llu dr',{:1-j-r
l:::--: :_i.
UA
la -:_.
:
::le q u a lita tive
:,: :.,i
behaviour
element : -:j:a-. _: -3cn =ar itbrium. frl!_lr:-irE "s: :E
lllG vi*|: !!r:
'+\.
\/"
u n d e r sta n di nS of is the concePt or member
ol
in a
re+j1w vB\
*"*\'
4.r-
26
, :::1 ' ::
A ND IN G S TR U C TUR A L A N A LY S! S
13.
It is convenient to separate the concept of indeterrt into external and internal indeterrninacy and we wil, first at external indeterminacy. A third support ha-. added to the simply supported beam. This means beam cannot be solved by the three equations of equilibrium, The structure is then said to be:
statical lg indexernjnate.
The subsequentstudy of the analysis of structures \ provide us with the means to determine the value o: fourth reaction, but for the time being in this
sJ-i^ /A
#n,,tv{
a r F . o n c e r n e o w i t l 't h e . o - r . e o t
/ w.
a_d later the degre:
indeterminacy.
---
/
Similarly, the cantilever may be made indeter by the introduction of a vertical reaction at B
' rP
i^+^
r
nr
^n ^a.l
-'n+ila\/ar
The three-hinped arch has been fixed at the crown. Although this diagram shows a load on the arch it si be emphaslsed that the degree of indeterminacy is 3 property of the structure
and not of the loadin8.
16. Because any ioading will cause the ribs of the arch i: to spread, inducing a horizontal reaction, there will always be four unknown aeactions in a riSid arch, the vertical and horizontal reactions at each support.
STATI CAL I NDETERMIN ACY Tlllh !rj.:c--ar
importance of the condition oi
der"u-eq
is that the structural
ririb..,t'i.:-e ri@6.,
2:
properties of an
srructure will affect the distribution oJ -r1s is not so with a statically determinate
@. -.-:;.
rcf
iirtcl![r-
:
rc
.+
"k thnu-!
s,rpporred beam, supportin8 a unilormly
ffirc
,cad will produce the same bendin8 moment :e€-Jdless oJ the varyinS dimensions of the
af!-L
T:€:rsiribution
b.. flwllll!
:: i;re slructure.
&Eugn:s
lb
of forces and reactions in a
aE:e:aninate structure are unafJected by the
ffi
W
Ea- i!6rr io see the importance of indeterminacy.
c
aeam shown has a greater structural
*,!m..c::
depth
cE'EE s{.T[,ort. Imagine the beam at the support
lrlfr q
!a_'a F?. !! re
oteM! mlllfl
so deep that virtually
no .load was
enc supports. The hogging moment at B
3!
hE
BC would be actins almost as cantilevers. -<
-7
fiansferred to the ends of the beam
U*g*.es:s
1E ].-aa s'ould be carried by the centre support.
[0 ffi
crcr--*ances
b S
h
rmel
the hog8ing bending moment at B
compared with the sagging moment in
lFI6icEBC. 1fre
Er:g:e e\amples illustrate the dilemma for the
re@i
::-:rer.
SrrE
:+:_:::
He must know the size oJ the rhe analysis is carried ou1. yet the
sd
a[G -r-e.-f,ers is the object ol the analysis. The
lnr
rs alE: :re designer must always make an
Lfiifr-
/,\
E g=:z:e. ihan the sagging moment in the spans,
&a-=!ive
tdlnrman=--re.
analysis of the structure before
numerical analysis can be carried out.
"
)0
I
iJRDERSTANDING STRUCTURAL A?VEIYSIS
The concept oJ a 'tree' as a statically determinate structure is a particularly powerJul one in the
Y^
qualitative analysis of structures because it is na and easily recoSnised in other structures.
34.
The effect of a load on a tree is always that of a cantilever, The efJect oJ the load goes straight to Sround. For this reason it is ojten preferable for us convert our indeterminate structures into an anal!.E Itree'. The resulting load effects are the simpler to identiJy and analyse.
We will now look at more complex frames, which represent the complexity oJ structures met by the structural designer. The two-bay, two-storey fixed at A and f and pinned at H. All other joints rigidly connected.
32. This two-bay frame has eight external reactions at H, and J,
Therelore the pxLprn-.z degree of
indeterminacy is: 8 - J (equations of equilibrium) = 5 times indeterminate. ,l 'l '
A
tv n
M7
r=
32
: r;:: i::ilrrlvc
ANALYSIS
The top Jrame BCDEF is reduced to a statically determinate Jorm by the removal of the moments and F. The lower frame by the removal of the restraint at H. The structure is therelore 3 times indeterminate.
This next exampie of a two-storey frame, has ao indeterminate sub-frame ABEFC, supporting an indeterminate sub-frame BCDE. We will now system of hinges to release the sub-structures to statically determinate f orms.
We will introduce hinges at B, C and E. Note thar introduction of a hinge between the tlree is, in fact, two separate hinge releases. Each of
f=t=l-
sub-frames is now a determinate three-hineed frame and the structure is 4 times indeterminate.
40. If there are less than three external restraints, no matter how many internal restraints, the structure fail. In this example, with horizontal ioller releasei both supports, the structure will ro1l, i.e. this is a
>^r
,/t ,fve Metlta^6^
^a
state of unstable equilibrium. Beware of the soDh: the notion that the structure could be stable under
,vs
tiarz.^iaL
!,ul,eti'|'
exactly vertical load. Indeterminacy or stabilitv is a property oJ the structure rot
the loading
STATICAL]NDE"':i.:: ;: : --
:arvert the whole structure into a 'Iree' we must . - -: F r^ F f . : m - RCnF A , , , r will . ele a 5 e r h r e e
- . -'\rn rn rer nal f or . es . , F. "F d1d Yf . in.la+arm
in:-w
Tqu( lhe
iq.
^f
do|e( utgko
tr.z
3 x
: : :le supports are pinned and the structure cannot be _::-:3i ro a tree because of the absence oJ a Jully fixed -.::
_':. :he .uppor l( m a) be r educ ed t o a 5t a t i . a l l y condition. This two-bay frame is pinned at
:::::.:,nate
_-:::jcture ::-
may be reduced to a statically determinate
:, lhe removal of the vertical and horizontal
-a:rr.is
at H and the horizontal restraint at C. The
:-:-:r3:-re ABCEDG is then statically determinate, with -::
:::a.minate
cantilever frarne EFH fixed to it. The
,:-
:,:e is J times indeterminate because it is necessary -: _: ::5e rhree reactions to reduce the structure to a =:
iitu
-.
:1-'. determinate f orm.
--::a.e !'Jr am es m ay be s olv ed in a s im il a r w a y b y .,- r ie\ nl \ , . ir i( allv dc c r m inale s u b - f r a m e s . --. - : _--_ -:ror ev lr am e has a hin8e at D and h a s p i n n e d '
FHI vH
STA?I CAL I NDETERH!N AC!
29
\s an alternative means of reducing the structure to a ieterminate form, a roller release has replaced the pin 'oint at D. This particular release should be noted since :i is frequently employed in analytical procedures. Thus .o horizontal reaction can be taken at support D. lecause the horizontal forces must be in equilibrium, ::e horizontal reaction at A will be equal to the :orrzontal component of the load.
:-is last example of reducing the structure to a :ererminate form is the axial release in the column. Ihe :€:rding moment capacity at this release is unaffected. :ich releases are rarely employed either in practice or in :-€1 r"t i cal proced ur es,
r- ::oment reaction at A has been introduced to the :,:rial frame turning it into a fully Jixed support, and :e
structure is now 2 times indeterminate,
2 x sGtt
l[: Tra].eCuce this structure to a statically determinate -r":* ::'. :eleasing the two reactions at D, yD and dD, T'!e .5 ::a:ricularly : 5 f,1€:-:est
important release system because
S',aEK4lLh e*2t,,^titt'.
to recognise gualitatively.
VO .^a ' Lrc*t
*
i^
ruc+
.- ' :;:.= :I' ID IN G
28
21,
S TR U C TU R A L3,V A IY S 15
Pin-jointed structures respond to the condition o{ 'n d e t e r m i n a c y I n r h e 5 a m e w a v . T h e p r e s e n c e o f additional members, here members PQ and PR, will the distribution of forces in the remaining members. those members have a large cross-sectional area the_ force in them will be correspondingly large,
22.
9xtra Jqf Prtr
To summarise then, indeterminacy may be the resul: l . additional external reactions,
2. internal rigidity, h3t4tv 3. the addition of members, ^Ltnb.g
The last two are classified as internal indeterminac\.
These three terms are used in other texts. They ha\'€ same meaning. The term indeterminacy will be usec rA
STAT I C'A LLY lNOLT L P-Ir1INATL ST R-u oT LtE Ejs Lho s am o es STLUoTqzE A RE0./^/DAV7 Is Eho sahto as 37 rul'fuREA, HlPE'.slATto
re\L and r.e dcgree oI inder.rminacy referred to as times indeterminate.
24.
lt is important to appreciate that structures may be converted back to a statically determinate condition lrom the staticaily indeterminate, by a variety oJ
sEatt
A
e^
DQ<-r VD
simpler form for the purposesof analysis. The portal
a I
S4
r,J
models the declsions made by the structural desiSner when he reduces the highly complex, real structure Ia =
s!,.aE tcallq
) | <-r-^
strategies. This technique is important because it
'L
rc.1L43Q.
frame is I times indeterminate because it has four external reactions. The frame has been released to a statically determinate structure by the introduction c: hinge in the beam BC.
THE ANALYSI S AF ST ATi CALLY DETERMINAT E . TRUC?UR'S
[e
_c-- :econnect the two parts of the frame and we
=:
lfi0uEf€ tu't'r
::..ecr
deJlected shape of our unsymmetricaliy
:_r:=-trtnned portal Jrame- The horizontal
fifi{F:c.:-::. =: :re top of the frame, which has been caused !!$ . ii[aa r-::: r.i{rr !!:€ it?irrrr:
\'ertica] component only, is calied the
:::me.
r -F_-:3ither the loading is appiied
rla,rrFi:nE:: ::-\ b .s D+:-r'. iGr/r-jlr: "r
i]
It always arises in a frame-type
or the structure itself is unsymmetrical.
:-- Je noted that the loaded point oJ this
:e:-:ats
rF-
both in the direction of the load and
1_E-:: :,. !r. that is, horizontally.
s $r-:* _::e:ali!g that in this example we have
["rio!rLur:p: : - _=)er of new ideas.
Firstly, that bending
inlt[Mfimcr]= :|: a_i'::. \'ertical members due to horizontal !0ll]Er jn : :.---=. \'3\'to that in which vertical forces @lG
:Fl[rig
-. ]::renls in horizontal members. Secondly,
fill]@ r_ T1L5: -:-e::3fie d iFF e-: _r: _:::.::ans dh tr-riln r---
analysis we are assuming that of the members in the axial
i..;lC place secondary restraints upon the
ro.r_x*JE [h rfr
:: ::.:3:. Doints in the structure. Thirdly, :!{ fe:-: :r _eei nol occur only in the direction of @]_g- Le :--i :rree-part solution of the three
,J_,,_"1-:_ * + ++
+,*
, -R
-a:
:.
: - -.i:io n .o n \tsts
1ic aJ:_--:_::
r@
\ ou -nay nor
fr o m o n e co m p o n e n t
- . t : ::- a il th r e e co m p o n e n t
3rd ::rs-.:3.: i,f,[r-*:r
.h a p e .
!:a.1 \\ith any one of the three
-a r n E 4 _ e , : : : : : i f ve 'is.iJf{:_€.
tn r e e p a r lr :
o: bending rnoments, the support
d - . : : - - : : i : - : : :e d
'ir, rc.=c.:
o
o f t he
p a r ts a r e
rArlh each olher. A typical
:::,::: a compuler analysis i5 given,
_Lll .tr='n*}
+
11{{
33 r j'r
ffirE
flire
ilE!-(:s
8li&A MG :fl!
=-'C:Je
__:
.frr:r.
aI A and D
--.e.e:c.e there is no
E 3 arc:-
1:,:
:E 3 :-e-_:::isn
CcllaPse
because
:rt.TTe-L_€. 3EF calid susrain a 3-F .re.e a cable-
:rE rre-Ee E I arE
:_,-
1c l e 1 = . --
{!as:t:
=L*-L-Jas.
r:ich
!tl|€:-sr:r_-s-
=_rca
:i€ basic rules of
-.:
is ihat
al]
-_:s: )e slr,all compared ther
do not affect
-::lal
d Ttc i=--r--r:. eirE:!r:
rtstrui^C'
C:a CUm SIa n Ce Sth e
--
-€= .Er.r :I: airs.:::oei:,:e
E fiE I:a:--€
It\a.jpr1$t'r ha sul!
a ,!-aC :a member DE
T:-ls. :n the context of
s=-l:-.r-.::i5
frame
is a
F sa,t=a J! linear analvsis. It is
Pscrao
Act teist 't
:_=r_-=-s :eccgnised as statically ![ :'E :-ree :ieanDers rn a trlansle. tffE :_r3:
fr:_n1r Y<
n arial ^f
forces
an',ilih.i,
which
can be
m
FAs
Triar,alo '. Ba -
I. -, :enain determinate no matter how ::.
adCed. The system can be extended
stat r
.
STATICAL I N DET'R!4;I;;'! The portal frame ABCD with pinned supports at A and D also has hinge releases at B and C. Therefore there is no tal
.estraint against sway and the structure will collaPse srdeways.
LL
This frame does not appear to be a mechanism because 'rnder certain conditions the member BEF could sustain a icad. IJ, for example, the member BEF were a cable,
l'^c.,lain
)roperly anchored at B and F, then a load in member DE
^o
$u1! frfiranE:
.ould be supported. However, in these circumstances the siructure would have to disobey one of the basic rules of :he behaviour of elastic structures' which is that all Ceflectjons due to the loading must be small compared $ ith the structural dimensions, so that they do not affect aie Seometry oJ the structure. Ihus, in the context of :ire analysis of elastic structures, this frame is a :nechanism and cannot be solved by linear analysis. It is
nL.t€4lsh1
: _:::,<tructures
The pin-jointed frame is recognised as statically
tB ,
ielerminate by the three members in a triangle' :esulting in the three unknown axial forces which can be
Triatglo
lclved by the three equations of equilibrium.
-
sta{tcelll adov" f^eto,
i:te structure will remain determinate no matter how i.any triangles are added. The system can be extended ::ldefiniteiy.
OJ
6ta|t.a lltl a.lz,/ ^;,4a-
- -. : t:S:.ii!'rl
NG STRUCTURAL 4,VeIvS15
45,
,:-\_::
We will look lirst at the Warren Sirder truss, which is statically determinate internally as it consists of a
i:
series of triangular pin-jointed frames. The truss has
-:s
\
statically determinate external reactions.
ao&rmr^ato
Lvllss
a ctzruisfz,
s uP ores, f
4 6 , If we now add an extra vertical reaction at the centre
5:-
:r=-
-^E
t l " e n ! h e . l r u . L u r e r s
:.E:
To emphasise the fact that this is an external
:
indeterminacy the actual frame may be ignored.
I x na.
47,
lf we now introduce additional members BE and EH in the truss the structure is 3 times indeterminate overall. It will always be simpler to separate internal and
:-"
:-r:5-f
external degrees of indeterminacy' lor pin-jointed
:-E
frames.
_iN::-:
=-€ :'--c.
:----r 4 8 . Although this text is written to emPhasisethe qualitative understanding of structural analysist the actual
Lr,./ts 2 x tM,s1aa,t4t^t-e' intuna(1
-
:_e.a
;ri-E--S.a
quantitive value is usefully examined in order to show how unreliable it is unless accompanied by an intuitive grasp of behaviour. For pin-jointed frames the relationship defininB determinacy is stated thus: 2 x j = n + 3'
2 J= t
-
^*z
+tutbzt
tr 1 = \L tfb q
nz
where j iovotr
v- v )l
belJ
-
number of joints
,?: number of members 3
t h r e e e q u a t 'o n \ o f e q u i l r b r i u m .
The three-member frame is thus calculated to be determinate.
:-E
?]J
'
S TATI CAL I NDETERMI Ii P.CY
ul.
: f \ _' is greater than m + 3 the structure is a .r.e:r^Eaism. :
Jf 2J > ,'t+z
: r 'is less than ,n + I then the structure is
1\
frE:eaminate.
n e..,'4/.' LJh1
X
<$s
[-
1.=-rna
the structure shown. The ca]culatior1
R
Jj 2j < vu+z thdetc,rr ^etz
shows
is statically determinate, but it should be clear _ris =€: structure will collaPse because the lirst panel :E::r
gr dRU, ' d,
4! -
uIdL",5.
J em ons r r at e \ t h a t r h e
I
r t 'r J
:r-En.iiat-Lvesolution must be judged qualitatively.
ii-
:a.ee-dimensional indeterminacy may be dealt with in :-e same way, although it should be appreciated that :::3s: buildin8 structures are reduced to a series of linked :-lo-dimensional Jrames Joa ease of analysis. T.ara
ira
ad,,,ri^n<
a^"il;hr",m
ih
thrFA
^f
:,:.,ensions: lorces and moments relative to the three l\e5.
.I-
:.:s cantilever ABCD is set in three drmensions. CD is la.rallel to the
y axis, Six reactions are required Jor
::e equilibrium of the frame at the'built-in'support
at
z,
V,
36
,,!,- a i:l;iiDr NG
RAL A,VE.LYS"TS STRUCTU Consequently, the release of a fully fixed support lrees six reactions. This bent frame, fully fixed at A
ard D
is 6 times statically indeterminate. The 'tree' method of idenriJyin8 the degree ol indeterminacy ;s even more efJective with three-dimensional structures.
,4.
The effect of the introduction of a pinned support at D, previously fully fixed, is to release three stress resultants, the moments about each axis. This frame is therefore 3 times indeterminate. The structure could be released to a statically determinatertreerby
the removal
ol the three lorce reactions at D.
If we now return to the frame Jully Jixed at A and D an internal
hinge would release three internal stress
resultants, the bending moments about each axis. This .
frame is now 3 times indeterminate.
,a t>
'Ezl-
Ywtl
h'^9t r^o'nAr
3rvr. vaaasa.
ca"A
axi
.
:4. i7. 5E.
:
-
STATICAL
& Determine the degree oJ indeterminacy oJ each of the structures shown distinguishing between rrte.na-7 and exter.ral
conditions.
r
3 The QualitativeAnalysisof Beams
ln the next two chapters, you will study an approach to the analysis of structures which is iikely to be unfamiliar to you' The traditional approach to the study of structural analysis has been based, almost hand' this is exclusively, on quantitative (i.e. numerical) methods' on one to Put entirely logical. The structural designer does, eventually, need of value values to loads and dimensions and determine the numerical reactions, bending moments and so on' There is, however, a critically imPortant stage before that numerical members' analysis; the preliminary analysis required to size the structural The nature of the analysis oJ statically indeterminate structures is such befor:e that the desiSner must know the size ol all the structural members an analysis is carried oL1t. Consequently, it should be clear that the detailed, numerical analysis of the strLlcture is a check on that Preliminary analysis. If This places great imPortance on that stage ol the desiSn procedure' be need to the structure is incorrectly sized' repeated analysis will carried out. This is true re8ardless of whether the analysis is being carried out bY hand or bY comPuter. It is generally assumed that the skills required for this preliminar.r analysis are the inevitable consequenceoI studying numerical techniques' The results of extensive research into the development oJ an understandlng oJ structural behaviour in undergraduates and trainee engineers suggest very stronSly that this is not the case. That preliminary analysis requires a quite dillerent set of skills which basrs are referred to here as 'qualitative'. Essentially non-numerical, the and of these skills is the recognition of the relationship between the load is the the resulting behaviour of the structure' In simple strLlctures' that
THE QUALITATIVI ;,"'Z:7JI.i
OF AEA&S
relationship between the load, deflected shaPe and the resulting reactions. These skills a.e distinctly different from quantitative skills; they rely on sets of coherent dia8rams rather than mathematical modelsr for example. However, the most obvious difJerence between the qualitative and the quantitative approach to the analysis of structures is that there is no obvious sequence for the steps in the qualitative approach. once a particular numerically based analytical technique has been learned, then the sequenceof the solution will almost always be the same. However, faced with a qualitative analysis it is not aPparent from which point the solution will emerge most elfectively. It may be best to start with the bending moment diagram in one case or the deflected shape in another. In many structural problems it is necessary to start with all parts of the full solution at the same time, which is potentially very confusing. Students studying this approach say that they understand each of the steps in the examples then' as soon as they try one of the practice problems they are stuck. This suESeststhat the teaching method is' in some way, deficient. This is not necessarily so. The exPlanation is that qualitative problems are inherentlq
diJficult.
This new aPproach requires a shift of
attitude and this takes time and practrce. However, you wilt discover as you begin to SrasPthis approach, that you understanding oJ find a significantly increased confidence in your overall structural behaviour which will contribute to your understanding of problems of structural desiSn and the methods of analysis studied in later chapters. Perhaps the Sreatest value of this qualitative approach will only be apparent to you once you start your oflice training. The use of computers is increasing rapidly. How can you check that the output is correct? The checking system must be independent and robust' You will discover that a sound understanding ol structural behaviour will play a siSnificant part in the overall checking procedures which must be constructed to ensure the correct use oJ the comPuter and the elimination of data errors.
)4 0
UNDERSTANDING STRUCTURAL EJVEiYSTS
l.
V/e will start by restatinS the assumptions \rhich underlie the elastic analysis of structures. The lirst is that, for all loading cases, the structure is within the elastic range of material behaviour, Srress is direc!ly proportional to strain and deflections directly proportional to load.
stvcin
',
The second assumption is that deformations due to the loadin8 do not create secondary bending moments. The cantilever bent ABC is subject to two point loads t{r
frFr
and ,{2.
Here the load ,{2 will cause bending in the
cantilever BC which will result in a horizontal deflection at B, ABh. The secondary moment t/1 x ABh is ignored. In the Engineer's Theory of Bending it is assumed that deformations due to loading do not.esult in signilicant changes in the structural geometry.
In the qualltative
analysis of structures and most
manual numerical analysis, axial loading, which would result in axial strain in the member, is ignored in the determination oJ bending moments. The lrame, ABC is loaded in the vertical direction at B.
4.
AlthouSh the load i/ would cause the column AB to shorten, it is assumed in the qualitative analysis that there is no secondary moment in beam BC. Generally, in the elastic analysis of structures, deJormations are considered to be small compared with the dimensions of the structure so that the equations of equilibririm ol the deformed structure are consistent with those ol the undeformed structure. It should be noted however, that this deformation Agu :s included in the computer analysis oj slrucrureS.
4l
THE QUALrTATil,il -:-','3:lsj.9 AF EEA|"LS It is important to restate the diagrammatic convention for
5,
the support conditions and the resulting reactions. The
-1
--
roller support has one reaction, normal to the Cirection of movement. The pinned support has two lorce reactions and the lully Jixed suppott t.,vo force and one moment
\
f'
reaction.
\
\*
The principle of superposition is assumed to aPPly as the
6.
structures are within the elastic range. We will illustrate this principle with this example. The load t/ on the two-sPan beam ABC Producesa hogging moment yC, to over support B with a downward reaction balance this bending moment at C.
.
If we remove the vertical reaction at C, the beam becomes a simply supported beam with cantilever' If we now aPPly the reaction at C, vC as a load, the combination oJ these two conditions wilt Produce the solution to the original two-sPan beam' This simple example illustrates one ol the most Power{ul desiSn tools in the armoury ol the designer, since it allows the reduction of comPlex structures into simpler lorms and the examination of the load ef f ects seParately.
t.
e-'rmrfllnr'-Vtr
ffir-i-t
The deflections may be treated in the same way. The vertical deflection at C due to the load n must be equal to the downward deJlection at C due to the lorce reaction yC since the real deflection at C is zero.
tL^./
Ao"
"n{--4*1
+
\.u-=
Y'-
.
-
FL#-
_(q_
-
N,
42
A N A LY S IS
UNDERST A N D IN G
9.
This is the first example of the qualitative analysis of structures, the simply supported beam ABC, with a
lw F--+=+-p
cantilever CD. The full solution of a qualitative analysis must always consist oJ: the deJlected shape, reactions, bending moment diagram, You must become accustomed to the idea that they are of equal importance and your qualitative analysis will not be correct until each of them has been determined and found to check against the other two, In the qualitative analysis oj beams, the best place to slart is likely to be the deflected shape. We will start the dellected shape by identifying the points throu8h which we are certajn
that the deflected beam
must pass, A and C and a downwards deflection beneath the load.
I0.
We can now draw the smooth elastic curve through these three polnts. Note that CD is straight. The curve must be smooth and with practice a very accufate deflected shape may be drawn. Try practising the identification of the deflected shape with a flexible plastic ruler,
I l.
This simple example of the bending moment diagram iilustrates the need to distinguish between the structure, which is drawn as the base line oJ the bending moment dragram, and the diagram itselJ. The point here is that the value of the bending moment over the portion of the span CD is zero. N -B -
The bendi ng
an ttle bendjng
nonenL
tersjon
di agran
j s a]w ags
sjde af the structure.
draw n
THE OUALITAT!Vi })iA'YSIS
OF BEAMS
There are certain simple rules which should be borne in mind when carryinB out the qualitative analysis, Experience suSgeststhar because thi. approach 's
tf .
*
pavb6f Ehe,
unfamiliar, students tend to become confused, reach a
sEruX,uro rutai3
solution containing obvious incompatibilities and are not
afi*
abie to spot them. Having found a solution consisting of
sEra!'iC
loaan3No EE-ND|N9MoMe^iT
the deflected shape, reactions and bending moment diagram you should take a cool look at each part and see how it relates to the other, Is there a bending moment at the support? Unlikely. Is the bending moment at an internal support ze(o? Very unlikely. Does the tension
|f e
identiJied by the deflected shape agree with the bending
d
moment diagram drawn on the tension side of the structure? We will call them 9olde,
rules.
Here is the Jirst.
\nd here is another.
l\ *
the. loaairp on a *ructuyo
BE'rDtNg M6 ME TT dAgR-AM ht4 alw44! ast\Sgt- 4sf zAt3ka L|NE-6,
Bending moments resulting from point loads are linear expressions,therefore the bending moment diagram consists ol straight lines. ln addition the dia8ram can only change direction at a load or reaction.
\nd another. That is, where the hinge occurs at the end of a structure, or at an internal hinge which we will study later in this chapter.
4t
z-a+a
s&+?r'lt,
:i.
budAj
Tho
/N nt otaatr
6 a
"1p"a Jl,ri,fl.2
-r
-fY
We wili now carry out a qualitative analysis on the propped cantilever AB, Ve will apply the principle of superposition to examine the effect of the fully lixed support at A, by the notional removal of the moment rea ctio n at A,
M A.
-N^ n
Jv
,o
.,o
,INAIYSJS UNDERSTANDING STRUCTURAL
44
15.
As a simply supported beam there will be a clockwise rotation at A. The propped cantilever however, will have a zero rotation at A and to return to our original
:sj+L+"
structure we must apply a
17.
antjcTockwise moment at A.
It is now apparent that there will be a change of curvature in the deflected shape, which is known as the point ol contraflextrre. This provides another point of reference between the deJlected shape and the bending
- F"trrf+^ffiun*"
M,c,H fve
moment diagram because the bending moment is zero at this point. The member is unstressedat the change of
rvc
curvature.
18.
We can now attempt to draw the bendinB moment diagram. We know that there will be a hogging bending moment at the fixed support at A. This is recognised by the anticlockwise reaction and the identification of tension on the top of the beam from the dellected shape. At the other end o{ the beam the upward reaction vB will cause a bending moment on the undersjde ol the beam. The bending moment diagram will be a straiSht line from B until it meets the line of action oJ the point load. The diagram is completed with a strai8ht line between A and the load. This line must pass through the point of contraflexure.
xU
botaina rtowe^,t.'
ai^3rois .r*
th^c, bato lmo f
"1; po&.1 csrfirqf lcxtlro.
t9.
Another golden rule.
THE QUALITATITa a-',':::.]-: 24.
::
ttat4s
The next example is the two-span beam ABC. The deflected shape must pass through the supports and the downward deflection under the load. Note that a structure will alrags
deJlect in the dr-rectro,
oJ the
load provided that it is the only load on the structure.
21,
The smooth curve shows the change of curvature, characteristic of all multi-span beams. The direction of the verticat reaction at C may be identified by imagining that the reaction is removed, The beam would deflect upwards and the direction of the reaction necessary to bring the beam back to the correct position would be oownwaros.
22,
Both vA and vB act upwards and the first lines of the bending moment diagram may be drawn from a consideration of the effect of the upward reaction at A producing tension on the underside oJ span AB and the downward reaction at C, producing tension on the toP of span BC.
The bending moment dia8ram is completed with a straight line between the load and point B. The coincidence ol the zero moment and the point of contrallexure may now be reco8nised and checked.
-ffi
46
UNDERS tA l l D tN G S l R U C TU R A , A rtA LfS IS
24.
25.
The analysis of a three-span beam would proceed in a similar way.
The Jirst step is to identify the points oi certainty on the deflected shape. The beam must pass over each oJ the supports and deflect below the load, A smooth curve may then be drawn between these polnts,
26.
The direction of the vertical reactions is identified by their notional removal. The reaction at D must act upwards and the reaction at C downwards.
27.
Ihe first, most obviolrs points on the bending moment dragram are usually those at each end of the beam. Here the upward reaction at A and D both produce bending tension on the underside of spans AB and CD
r*
respectively.
THE QUALITATIVE A:iF.:!s 3
3-
BEAMS 'F
:: complete the bending moment dia8ram you must be ::le to identify the fact that the beam is hogging over __e support at B. This beam has two points of contraflexure' in span BC =.a between point B and the Point load. Note the :e:-rtitication of areas ol bending tension on the :::lected shape and the visual check against the bending - r:nent dla8ram.
g.
-.^:s next example is a three-span beam with a lixed ;-:Jort at A loaded on the central span with a point load ;.
Note that the beam must enter the Jixed support
r-::
zero deformation compared with the unloaded Failure to draw this clearly is a common
i::icrure,
i:r:ce of error in drawin8 the deflected shape.
the bending moment diagram is drawn on the
K
-e:ause :=:s:rn side of the structure a check should alwaus be -:Ce between the deflected shape and the bending -,.:ient
diagram. To Jacilitate this check, identify the
of bending tension (T)' The reactions are -eas :-:ermined from the deflected shape by their notional i?::toval.
3r-
:-e most obvious lines of the bending moment diagram -:r
now be drawn: the sagging moment at A, the
-::ging moments at B and C and the sa8ging bending -.:nent
below the load. Note that it is known that the
, -e of the bending moment diagram from A will move to ::ass the baseline of the structure because the vertical --=:clion yA is causing anticlockwise moments which will ::aJce the clockwise moment MA that is moving alon8 --.e beam lrom A to B. The bending moment diagram -:\'rhen
be completed with straight lines as there are
-: orher loads between these identi{ied points on the :<1ding moment dtagram.
DI NG ST RUCTURAL AdAZYSIS UNDERSTATI
48
32.
Practice
problens
A-lr,r'aqrs complete the lull solution by drawing the three parts, however confident you feel about any one oJ them,
This is particularly true of the direction ol
the .eactiors
.
There is a temptation to miss this
step and attempt to construct the bending moment diagram from the deflected shape. Experience shows this to be most unreliable and the intermediate step of Producing the diagram of the reactions is usually critical in ensurinB coherence between each of the three diagrams: deflected shape, reactlons, bending moment diaSrams.
illl;i]llllilXl lTilllfi 33.
Externally applied moments are rare ln real structures.
This example is introduced here to
prepare the student for the use of external moments in the analytical procedures to follow. find them dilficult
.\a r--\r
You may
to visualise and this is probably
because they are unfamiliar.
1M
The moment aPplied to the end of a cantilever will cause a constant bending moment and is one of
tA
\7
'rM
the occasions when there is a bending r,noment at the end of a structure.
free
34.
The deflected shape is derived from the bending moment diagram which shows that the deflected shape must be of single curvature' and since it is the top fibres which are extending in tension, the deflection must be dowrttards. There is a balancing anticlockwise reaction moment, MA.
Note that there is no vertical
reaction as there are no loads to balance a reaction at A.
'fHE OUALITATIVE ,Ai.IAIYSI.SOF BEAMS Vhen the load moment is applied in the span of the simply supported beam AB, the Jirst point to recognise is that the beam nust rotate in the direction of the moment.
Note that the beam will almost always deflect
^ah |r
# -.4
lE
tla9, vertically or horizontally, from the undeflected position at the point oi application of the moment.
The
points oJ certainty on the deflected shape are the zero de{lection at the supports at A and B and the rotation of the beam at the point ol application of the load moment.
x.
Because the vertical equilibrium is only maintained by the reactions, the vertical reactions are equal and opposite and their value is the applied moment divided by the span. Taking moments at A, the applied moment M produces a clockwise moment on the structure. The position of the moment, relative to A, has no eJfect on the value oJ the external moment on the structure. This
-5_r +.. YA
clockwise moment must be balanced by the anticlockwise moment produced by YB.
F.
The bending moment diagram resulting from the application of an external moment is unusual in that there is a discontinuity at the point oJ application oJ the moment.
The moment is on both sides oJ the
structure at the sa.a?epoint,
The sum of the hogging
and sagging moments at the point of application of the external moment is equal to the value of the external moment.
The slopes of the lines ol the
bending moment diagram are equal, because they result from the same va.lue of the vertical reaction.
tt-
we will now apply the moment to the propped cantilever beam AB. Note the tension on the underside of the beam at the fixed support at A.
ffi r '
+., VE
Sf S UN DERS'IANDI NG ST RUCTURAL ANALY
39.
The resulting diaSram oi the reactions and bending moments. Note that the sloPe ol the bending moment diagram either side oJ the Point ol aPPlication of the moment is the same, resulting from the constant shear force across the beam,
40.
beam' We will now apPly the moment to a three-span will find You ABCD. Try to draw this dellected shape' rotation at B' it dilficult to recognise the anticlockwise throu8h Howeverr il you ensure tnar the curve is smooth appatent' the points oi certainty' that rotation will be
41.
The direction of reactions may be found Irom the part ol this dellected shape. Perhaps the most difficult ol solution is the recognition ol tension on the underside the beam at B. Try treating BC as a beam' 'built-in'at on the B and C. This will reveal more clearly the tension underside at B.
42,
The plotting of the bending moment diagram from a partial unilormly distributed load will now be explained' The reaction at A will be greater than that at B'
ffir*
TEE 2UALITATTVE .2,\t:1::_= sl
::
:a-t-.,:
The characte.istjc
paraboJa of the bending momenl :E8ram resulting from a uniformly distributed load aDL) is piotted on a new baseline between the :riotted values of the bending moment at the start and -:iish of the uniformly distributed load. Ihe :_raximum height of the parabola is wl2lg plotted -::!:a./ to the structure baseline where I is the _entth ol the uniformly distributed load. The
V=ru
curve may now be drawn to complete the -rabolic ae.dlng moment diagram,
A
I
::lre diagram is drawn to a large scale there are a -:jnber of additional geometrical properties which will ia
F
rhe accurate drawing of the parabola. I and 2. Note that the tangents to the start and
:,.ish of the parabola are the slopes of the bending ::oment diagram Jrom the reactions to the UDL. There a ill be no discontinuity in the diagram unless there is a :oint load at the end of the UDL, I and 4. Height at + span = f moment at mid-span :: L DL .
rt!
t
--::e tangent to the curve tor lhe naxinun
value oJ the
moment in the beam is parallel to the str:ucture
a
-trding it will not coincide with the maximum ordinate Jor the ':lL unless the load is symmetrically placed on the
t F
I
L}]le obvioirs limitation
ol the qualitative
approach is that
may be more than one qua.litative solution fof
--Ee srrtures
with more than one load, The simply
*-Dported beam AB is extended by a cantilever from B to C
The beam i9 only partially loaded wirh a uniformly J :srributed loddi the cantilever is fully loaded. There =:e lwo possible solutions. In this first case it is :.ssumedthat the load on the beam is sufficient to create a: upward reaction at A.
A
-1:
ffi
-:*
UNDERSTANDI NG S T R UC TUR A L A I/A i Y S IS
47.
ffr-
Consequently there will be a sa88ing bending moment from A to the start of the unilormly distributed load'
-: -::
Note that the plotting oi the parabolic distribution of bending moments due to the uniformly distributed load may be apPlied to the cantilever.
.
48.
:---J
Another galden rule. :
*
./ t\
Tha
r\, hertr
bu\dt^qw
4 ASrc^
4istrtbdt t.4 Io41
^ a lrvrSJ
--:T! _:{d
*:-
b\4?'r
f
'- ! --
is
cufvc4
49. If we now increase the sPan oI the cantilever to tnduce a
le". q
downward reaction at A, there will be a single curvature in the deflected shape, because the ef{ect of the load on
r "*- "
span AB is insufficient to countera't the effect of the
:
t"6![-!
,-{
hogging moment from the cantllever'
50. The bending moment diagram will be hoSSingover the span AB, desPite the sagging effect of the partial UDL'
_f,I
lt" .IF l
Jlltrfoofil,
l
THE QUALITATIVE ANAIYSIS OF BEAMS Jl.
This next example ol a propped cantilever beam has a hinge in the span at B. Remember that one of the most important features ol a hinge is that the bendlng momenr 15always zero. Because the support at A is simple, allowing fu.ll as the load causes the beam to deflect -aidr, &rrrards,
AB will curve and rotate.
The shear induced
ri-ll turn BC into a cantiiever. There will be -C &-'iauity in the deflected shape at the hinAe,
shows the reactions and bendins moment lJore ihat the resulting distribution of bending
Mc
is similar to that Jor a propped cantilever a hinge. The explanation is that in this example,
uo
tEs been Dositioned at the Doint of
i!+ect
the equilib.ium of each part ol the
fhe equilibrium of part AB is sustained by the Etical
shear Jorce sR,
This must be balanced
shear force on BC. It is this shear creates the hogging, clockwise moment -Jr ar C-
€rample is the three-span beam ABCDE, with a 5pan CE at D. The point load tv is to the leJt in span CE. As CD deflects downwards Ioad, DE will rotate without bending. , there is ngx6ear lorce across the hin8e. Till be undefo.med Jrom the load to the hinqe.
I i""'1"^ "j
DING STRUCTURAL A,^/A'Y.sT.' UNDERSTAN
55. This diagram shows the reactions and bending moment dlagram.
55. This next beam has two spans and is fully fixed at the in span support A. There is a hin8e at B and a point load CD. Ve can draw the points of certainty on the deflected shaPe' Note that the beam will have zero rotation at A.
57.
It is sometimes helpful' when there is a hin8e rn a structure, to imagine it replaced by a short member' helps The force in the member, in this case comPression' hin8e' to clarify the resuiting bending either side oi the a In this case, AB will act as a cantilever and BC as beam suPPortedat B.
J8.
This dia8ram shows the reactions and bendin8 moment diagram. Note that the linal solution for the bendin8 moment dia8ram is similar to that for a beam Fithout or a hinge. ln this example the hinge has been Placed at near to the natural point of contraflexure'
. i9.
THE QUALITATIVE A}JALYS'S CF BEAI4S
55
This next example is almost impossible to visualise. It is the application oJ equal and opposite external moments either side of the hinge, This is a device used later in one of the analytical methods. This deflected shape is difJicult to arrive at without an attempt at the reactions and bending moment diagram.
This diagram shows the reactions and bending moment diagram. In this example, because oJ the curious loading, we appear to have broken one of the golden rules. In fact the moment is still zero at the hinge, but has a value, the imposed moments, either side. The reaction at A may be determined by considering the equilibrium of AB. That downward reaction will cause a ho8ging moment from A to C. Taking moments about C of span AC, the applied moments cancel out.
tre will now study another example oJ alternative solutions. The cantilever ABC has two point loads, Here
!s
ihe downward load at C is greater than the upward load at B,
N ^+
N--"
67. This results in an anticlockwise moment at A and a single direction oJ curvature. The dotted lines on the
t
bending moment diagram indicate thernegative' effect
P.
of the upward load at B.
Mr^ 9L 4A + . vA t
' I
a
|
iiii,rr*
UNDE RSTANDI NG ST RUCTURAL ENAIYSIS
63. We will now examine the effect of making the load at B greater than the load at C. There is now a change in the
wl
curvature in the deflected shape.
64. This diagram shows the reactions and bending moment
M|c " +
diagram. There is a point of contraflexure between A and B and a clockwise moment reaction at A.
w^f
la:-1-,
All real structures have a complex loading arrangement. However, with a sound understanding ol the qualitative method, you will learn to understand how to model real loadings and the responseoJ real structures by reducing the complex structures to a series oi simpler sub-structures.
65. -f,
^ 'j---?'i----.-ff
lF=-'.t+/.;;4.-F-=iR
,/j-"ll!!Y .RRR+
^\
\
Practice
Problens
Produce the lull three-part solution for each of these structures. A typical computer analysis Jor each of these structures is given.
4 The QualitativeAnalysisof Frames
Ihe introduction 01 the second dimension in the plane, the vertical direction, required for two-dimensional structures, introduces the complication of horizontal loads or Jorces acting normal to vertical members. The effect ol loads or Jorces normal to members will always create internal bending moments. The fixity at the joints will allow the transfer of bending moments in vertical members into horizontal members. You will Jind it useJul to refer back to the last part of Chapter I which studied the distribution oi bending moments in statically determinate two-dimensional Jrames. Often the solution to a statically indeterminate frame is revealed if you try to see the effect of the load on the determinate frames within the indeterminate system, Thus the qLralitative approach to determinate Jrames may be seen to link with the analysis of the degree of indeterminacy, When you reach the end of this chapter and attempt to solve the practice problems you will find them dilficult at first.
There are a number of
reasons for this. The Iirst and most important aspect oJ this approach to understanding is that solving problems qualitatively js difficult in itselJ. Secondly, there is no immediate route into the solution. The problems seem open-ended. In fact they are internally coherent and that is the clue to the solution. You must check all the loads and forces against the eflects apply the golden rules and a?waqs produce the Jull three-part solution of bending moment, deilected shape and reactions. If your qualitative solution is correct then each oJ these parts to the solution will check against the olner two,
f
UNDERSTANDING STRUCTURAL E]VAI,Y.s15
J8
We will start by examining the simplest two-dimensional structure' the cantilever bent ABCDE' fully lixed at A'
l.
2.
The bent is loaded with two Point loads, ,t/r vertically at B and fl2 horizontally at D. Cantilever structures are always statically determinate and it is oJten simp to start the three-Part qualitative solution by drawing the bending moment dia8ram and using it to help to the deflected shape. Note that the bending moment is constant in thal part of the structure parallel to the load hr2, over the member BC.
We know that the beam ABC will deflect downward under load h/r and that load ry2 will create a tension in the top surface of BC. From this we are able to recognise that joint C will deflect downwards. Note when drawing the deilected shaPein two-dimensional structures the real shorteninB or extension of members under axial load must not be represented in the otherwise a misleading deflected shape could tesult. Such strains are insiSnilicantly small compared with bending dellections in the qualitative analysis. Consequently the deflected Position ol joint C is rhe line CE of the original structure.
X
ot afut!{una
sqP?ot? tho 1412'<.t44 / s\abQ bru&t &4jl.{) zo.t rftatri
at
thc
+ thc eqftrl€.
fa.
4.
de, ru le. Another 9'o-7 Virtually all structures with a fixed support are deJormed by the loading system just before the However, you must identify the zero rotation at the support as one of the points of certainty on the defl shape.
THE QUALITATIVE
ANALYSIS '?
59
'RANES
The final part of the solution is the diagram of the :eactrons, drawn, as usual, as the action ol the jojri ::: the nenber,
Thus MA is anticlockwise.
If you
:Eve difficulty visualising the support conditions, try .e-creating the support conditions and the structure aith a piece of flexible plastic.
This is a reminder that the full solution must always -icludel
.:--l-" l.
the deJlected shape,
2, bending moment diagram, l.
s
'r-
iDr>--'-'+-1
reactions.
+
r-
I
|
I
l+-
1-ou should start to identiJy the internal :elationship between each part of the full ;.rlution; bending tension and deflected shape, ":_-omentreactions and bending moment diagram {or "-xample, ::rere is a plotting routine for the bending moment at : joint or node which is particularly helpful. :om
We know
the deJlected shape that there will be a
:e.sion on the top of the member at A and from the :eaction diagram, that the reaction moment ItA is =lticlockwise.
Remember that we have adopted an
--;=ra-ll sign convention where clockwise moments are positive.
aollow this routine,very carelullyl Go from the joint under consideration along the member and rotate in the direction of the siE/?r of the moment.
The value oJ the bending moment
is plotted on the ordinate narmal to the member. Fortunately this is the same output convention used by (almostl) all computer programs lor the analysis of plane frames.
nnnr
ALll lll
A Ye,-
' ]\ ,
t't /-i.'
'L"
$ -''
UNDERS TANDIN C STRUCTURAL ENEIYSiS
60
9.
This next example is the same cantilever with an anticlockwise load moment at D.
.fm 10. The bending moment resultin8 from the application of a moment on a cantilever is a constant value thoughout structure regardless ol the shaPe ol the structure. In
lll ll lll lllllrillE l-
this example, the moment produces bending tension on the underside oJ the cantilever.
I
Once again, if in doubt'
try re-creating the structure with a piece ol plastic.
ll.
The deflected shapes which result from the application of moments are diflicult to Produce intuitively. Consequently it is the relationshiP between the deflected shape and the bending moment diagram and the convention of drawing the bending moment on the tensio,
side of the structure to which we must turn.
lmaSine the {ibres on the undersi(le of the frame from A to D extending and it will be apparent that point C will deflect upwards.
1)
There is only one reaction at A' the clockwise moment balancing the anticlockwise
load moment.
Note
particularly that there are no load reactions.
THE QUALTTATIVEANAI,YSIS O:' IR?.I!ES We will now reverse the two-member frame and extend it by a horizontal member CD, hinged at the junction C and supported on a horizontal roller supPort at D.
fl.
I e have generally ignored the siSnificance ol axial 5ress resultants in the members. However, it is helpJul :r€re to note that there is an internal compresslon In :nember BC.
To begin with we identily the points of certainty on :he dellected l.
shape:
the member must enter the lully fixed support at A without rotation from its original, unstrained position,
2. point B will be pushed downwards by the compression in BC'
3. because the right-angle will be Preserved at B, point C will deflect horizontally, 4. the horizontal deflection of C will roll support D to the right. le can now complete the deflected shape. Member BC r:ll remain straight because the structural conJisuration .liminates bending moments in that member and there is -.o horizontal reaction at D.
x
UNDE RSTANDI N G ST RUCTURAL ANALYSI S
17.
This figure shows the final bending moment diagram. The slope of the bending moment diagram from C to the line of the load will be the same as the slope from B to
Illilr'*f"
A. It is the compression in member BC which is creating these bending moments.
18.
Our original cantilever bent is now converted into a lrame with two supports, with the addition of a roller support at C. We will examine the effect of sway by subjecting the frame to a horizontal load at B.
To identify the direction of the reaction at C we will notionally remove it. Point C would deflect downwards as a.esult of the horizontal point load at B and the resultinB reaction at C would be upwards.
20.
We can now draw the deJlected shape, being careful to maintain the right-an8le at B. The reactions are derived as lollows: l.
the horizontal reaction dA will balance the applied load,
2. the vertical reaction, vA will be equal and opposite ro the reacrion vC.
HA
3,
There will be an anticlockwise moment at A, 14A, the sign of which is confirmed by the tension on the left-hand lace of the column AB, at A.
THE QTJALITA'IIVE
&-
A ELTSIS
OF FPI!,GS
The bending moment diagram may be derived from the r€actions and confirmed with the dellected shaPe. The rpward reaction at C will cause sagging moments in the bam
BC, IVe know that there is a tension on the side of the column at A, which completes ihe
Ht-hand HnB
moment diagram.
rdirmation
Ire
Note particularly
of the point ol contraflexure
the in member AB.
change the sign of the load by reversing the then all the stress resultants and delormations
-lction have opposite si8ns. -ll
As q,ith the dellected shape above so the reactions and moment diagram are reversed. Thus we have ft
effect of sway, either to the riSht or the left.
Ye will now remoYe the horizontal Ioad and apply a yertical load on member BC. Vhich way then' will the srpport at C deflect?
To the riSht or the lelt?
"Wp-it-
A1
64
I
UNDE RSTANDI N G STRUCTURAL ANALYSI S
25.
There are a number of ways in which we can examine the
,
horizontal deflection at C, i.e. sway. This particular approach, that ol introducing an artificial reaction at C to !\e\e\i\o.(\zo\ta\
mo\efi\e\\,
\s \\e same s\a\eal
\Ne
will employ later in the study of one of the rrajor analytical methods, Moment Distribution.
26.
In drawing the deflected shape, joint B will not move horizontally but will rotate, because of the artificial
u
+-
horizontal restraint at C.
vo
removal of the harizantal
If \re imagine a notional restraint at A (not shown)
then it is apparent that the restraining horizontal reaction ryA will be positive, to the right.
This must
be balanced by the artificial reaction a1 C, dC , i.e. to the left.
To remove the effect of the artilicial horizontal restraint at C, we apply an equal and opposite force to the original structure. This is an example of the principle of superposition. Thus the combination of the reactjon diagrams lor the artificially restrained structure and the application of the value of the artiJicial restraint at C as a load, will produce the real, original structural conditions of a roller support at C.
za-
$**f: -
This is the linal bending moment diaqram, the combination oJ restrained and ,sway'moment diaqrams. Note that the application oJ the principle of superposition, that is the combination of these diaqrams, cannot predict the constant bending moment in member
.$-
AB. It is only because we recognise the absence of the
H EI
horizantal
E \irl
reaction at A.
The slope in the bending
diagram in the columns oJ frames is alrlags due to the presence of a hortzo-ral reaction.
![
THE QUALITATIVE!];;::':
65
: :''RA'1AS
The frame will deJlect horizontally to the right, the eite cr ot the 'lo ad' H^. balan. r ns r he ar t if ic ' a l .eaction. Notice that there is single curvature in the :olumn AB. For there to be a change in curvature in the :rlumn oJ a lrame there must be a horizontal reaction to ::eate the zero moment and the consequent point of ::r:raflexure. The recognition of the effect of the - - To nta l re action is par r i. ular ly im por t ant as i r i s a :_:auent source of errors in the qualitative analysis of
I = n ill now solve the pin-supported portal frame ABCD. :: jetermine the direction of the horizontal reaction aDt .: \!ill notionally remove it. Support D will roll to the - :-r because the beam BC wili bend and as the _ : r-a ng le at B m us t be m ainr ained. B m us T d e f l e ( L t o :-: right. This alone would induce a horizontal :::lection at D. However, the riSht-an8le at C must also :e.naintained which will rotate CD, inc.easing the -:.-zontal deflection at D.
:: .eturn support D to its original position the -': /o nta l rea ction al D m us T be negat iv e, i. e. t o t h e .::r.
To satisfy horizontal equilibrium, the horizontal
-::ctio n a t A m us t be equal r o "O .
Bec aus e 'r A a n d
d re e qu al. d nd t he c olum ns AB and CD ar e t h e
.=me height, the bending moments at B and C will be :3ual.
-::cause of the later application to the method of 'lrment Distribution, and because it is a good example of :_e principle of superposition, we will examine the :.rblem of sway with the introduction oJ an artificial -:straint in the horizontal direction at C. This :jalitative
solution depends upon the relative vazues oJ
:-e bending moments induced at B and C for the :r:ificially
restrained Jrame. You must acceDt at thls
::age that the fixed-end moment at B is ereater than the -oment at C, for the case of the point load closer to B. -ris ls analogous to the distribution of moments in a :.\ed -en d b ea m .
rJ s
w
66
UNDE RSTANDI NG STRUCTURAL ANALYSI S
33,
This will result in a greater bending moment at B than at C for the restrained Jrame. Consequently the horizontal
:
a=
reaction at A is greater than the horizontal reaction at D, To balance the system in the horizontal direction the artificial reaction, aC, must be negative, to the left.
-_-:i _'
34.
To remove the artificial reaction flC we must apply an equal force in the opposite direction, We now have the two basic sets of diagrams for the application of the
mfr
;
principie of superposition to Jind the final distribution of bending moments,
r E- :a :
35.
#tr
The final bendingmoment diagram Jor the origrnal, unrestrained frame is the combination of the restrained
5f f r r
and sway solutions. As with the previous example, the
!f
::: -=^
combination of the qualitative bending moment diagrams will not identily the equality of the bending moments at B and C. This can only be recognised from the horizontal equilibrium of aA and ,?D.
_:
::
:,::
36.
We will now examine the qualitative solutlon of a structure with three mernbers at a joint. Firstly, we will identify the Doints of certaintv on the deflected shapel
twl
'.-,.-,.-L
tl
lD '\\
--Y -
l.
vertical restraint at A,
2, rotation of joint B with the Dreservation
)k
of the right-angles between the members at joint B,
3, must pass through the supports at C and D.
-1i:a
1-,: --a_e
-
S
67 : :Jmpleted deflected lorm reveals that there wilj be ..-s-cn on the top surface of AB and bending tension on :E -eft-hand side of the column BD. The direction of G
reactions shown in this figure is determined by the
E{bnal'.emoval
and replacementi technique described
bending moment diagram can now be plotted, --e --:rting
with the obvious pointsl hogging bendinS in AB resulting from the downward reaction at A, the upward reaction at C will produce sagginS in span BC,
1. The horizontal reaction at D, will produce tension on the left-hand side of the column BD. \ote the discontinuity of the bending moments in member ABC.
Let us examine the moments at joint B. These have of the joinL on Lhe -ction Thus the joint will exert a clockwise
been drawn here as Lhe
--e,nbe.r. reaction on member BA.
Note particularly the
relationship between the bending moment diagram and the direction ol the moments shown here. They are related via the plotting routine described in Thus the sum of bending moments UBA and MBD musr equal MBC since the joint must be Figure 8.
in overall moment equilibrium.
c
M&
Mtso
68
UN DERS?ANDI NG STRUCTURAL AA'A'TSiS
40.
,]
Remember that bending moments are Plotted l.
I
from the reference Point,
::E
&ecd
I
*1 *tq
2, along the member'
I
Lhd
gtg l
3. rotating in the direiiion ol the sitn of the
h;
bending moment,
z hoq
4. plottinB on the ordinate drawn normal to the members.
t
h€rr dE
41.
We will now study the qualitative analysis ol the frame ABCDE. The point load tr is located horizontally in column AB. Support A is fully fixed' E pinned and D is on a horizontal roller'
Ib 5r
b.rm{ rl|e rt
-EnbJ Ecrrt
42.
We may apply the Principle of suPerPositionto the point load by assuming that it is first applied to a simply supported'beam',which in turn transfers
i'i
!c can rD! This I Enent al
f8:
the suDDort'reactionsr to A and B.
43.
We will start the solution by identifying the points of certainty on the dellected shaPe. rvith Practice'
bte
and the study of the deflected shapesof two-dimensional structure!' the detlected shaPe
-agram atactioar al
shown may be drawn'
!
parti in
rel,ated to
THE QUALI'rATM
elVA:!S--j- :,: ::-:-t::
:-:re direction of the reactions is identified
from rhe
:eflected shape. We can srart ro draw the be;ding
69
!c>i
-:roment dia8ram with the most obvious points: l.
bending tension on the left of the Jixed support at A, reduced in value by the h^r
i?^n+21
ra2^ +i^n
''* " ""' ' '
.t , A,
hogging bending in member CD,
3 . bending tension on the right-hand face of the column EC.
The bendin8 moment diagram is comPleted and comPared rirh the reactions. Note that the bending moment at C :n member CB, is equal to the sum of the bending :noments in cD and cE at c.
Y9
H4
I e can now add the effect ol the Point load in column the -1,8. This 'simply supported' effect is 'added' to moment at A and B.
i:
\ote particularly that the slope in the bendinB moment diagram in column AB is related to the horizontal reaction at A. Similarly' the change in slope in BC is .elated to the downward reaction at A'
V
fot^
HE
f
70
UNDERSTANDI NG STRUCTU RAL AIVE'YST.S
This gozder ruie
{f
f,'," sloye i bf,ldlnl
\JJ
t^dtrrc*C
will remind you to carry out this
important check on the qualitative solution.
Lhc444/4u
is alwatt,r cssoqrtU wr'6 a forco NoR 4AL bo Uo sErwtwe..
Iil
u
t
lr ril
49.
'fya tvo
HinSesin frames may produce unusualsolutions. This orthogonal frame ABCD has a hinge at C. The structure swaysto the right to releasethe momentat B. The
dirirt
solutionis unusualin that there is a zero bendins
lErG5B{
fb
etpL
rays
abt
at the internal, rigidly jointed connection at B, The
M
horizontal reaction at D is zero because moments aDout C, considering member CD, must be zero. Thus the horizontal reaction at A is zero and the bending moment at B, due only to the horizontal reaction at A is zero.
ll l,u
Ve will now study the frame ABCDE, Ioaded with a
.-lcdly
load at D exactly over the reaction at E. There is a
dEe
hinge at B.
t,rl
iqect
d
is c d
l:ii
l;
tit! tjil l
) t-
This structure hasan unusualequilibriumin that the vertical reaction at A is zero. Taking moments about E, there is no out-of-balance moment due to the load since the line of action passesthrough E. AIso the horizontal reaction at A is zero becauseof the hinge at B. Since there is neither horizontalnor vertical reactionat A there are no bendingmomentsin either AB or BC. The bending moment due to the load r.,iin member CD, is resistedonly by memberCE.
The expli
in CE. n member.
QUALI'rATIVE
ANALYSI: "
aS
7I
"':1:S
:":r what happens to the deformation in BC? We can see --:ratjoint C must rotate, which appears to require a
i 3'a!,
:ownwards reaction at A, through a shear force at B, for eguilibrium.
lis
T'ne explanation is that the structure sways to the left
trture
and joint C sways and moves downwards as member CE sways about E. This allows CD and CE to bend, but
txhent
leaves BC and AB straight.
: bout
ment IL
poinr
L
Clearly there is a compressiveforce in memberCE. But there is only one vertjcal reaction at E. We need to
w
inspectthe force equillbriumof this support.
Nce -,,f\ l Yg
)).
The explanation is that there is an internal shear force
rt E,
in CE. It is shown here as the action of the joint on the
ince
member.
ftal F
he
72
UNDE RS?A.NDIN G S TRUCTURAL AMAI,YSI"5
56.
The resultant of the exterra.i
reaction to the shear and
axial Jorce is equal to the vertical reaction at A.
\ ) 5E
0*.:b-
+I
T 3.
+"
l+ I
I
-&
57, 58.
Practice
problens
Produce the full three-part
n
{*
"f'J'n
"{a* f]
these problems.
solution to each of
Computer solutions for
similar structural dimensions and loadings are Biven.
t h"
PartII
5 The Theoremsof Virtual Work
Tl9 k9l l. the soiutionof .1.,"19t9" M. "!gti911l_,!d.19!I]let.
abiliry to delermine structural dg[gllnations. In one meLhod of analysis, the deformations provide the basis of equations of compatibility which, in addition to the three equations of statical equilibrium, a.llow the solutron of the unknown effects and the lull solution to the distribution oJ internal forces and moments upon which the subsequentstructural design is based. In addition however, the deflection of a structure may well be a design criterion. All structures must satisJy two basic states of loadingl l.
Serviceability when the structure is subject to its working load. It is at this state that the deJlection of the structure is checked.
2.
Ultimate load, where the failure strength of a structure is compared with the serviceability load multiplied by a load factor.
There are two basic approaches to the analysis of structural deformations, .fteiq ClStgy gl9 u-tltual wotk- The latter has the advantage of being able to deal with congi1lo!: qlhgr ltat !b9!e lvltlrn !!9 glagtic range - a limitation o s!rai!'i elergy.
Only virtual work will be studied
here as it is generally agreed to be the more powerful of the two. The student should bear in mind that although virtual work calculations may be the key to the numerical solution of a problem in structural analysis, the mathematics content is relatively trivial.
The difficulty
encountered is the application of the concepts oI virtual work to the problem in hand, Once that can be understood qualitatively, the subsequent calculations although time-consuming, are straightforward. Without thrs qualitative grasp, methods like virtual work may quickly become an exercise in mathematical manipulation and the overall sense of the method and the behaviour of the structure, lost,
76
UNDERSTANDI NG ST PUCTURAL edetyS./ S
l.
In a real structure two conditions must always be
Similarly, we
satisfied:
_,
lw
s.--l-I
\A
\ and B. Thi l. --3-ft-
The sysrem of forces acting on the
slppor:t cond
structure and the internal forces must be
capable of sq
rn equilibriun.
ol two hinges
R.oal stru
Consequentlv
2. At the same time the structural -'
but not egrji
delormations must be compatible with the
Th e vj .:1
system of supports.
Because onlv Ou. first example is a propped cantilever ABC,
more
loaded with a point load at B.
rirtual states
easrl!-
and applicati,
)
Mra--IYu^
This ligure shows the load and reaction equilibrium. The built-in support at A will produce an
This chapter
anticlockwise moment ItA and a vertical reaction
lv.
upwards, vO.
:::ecre,'7,s oJ
There will be a vertical reaction
upwards at C.
':heorem disc
Since there are no horizontal
Vi r tu a l D
forces on the structure, the horizontal reaction at A is zero.
Virtual F
The deJlected shape oJ rhe strucrure must be compallb-lg yith
the ruploft
support A is built-in,
co!d!!!qlT.
u e will use
Since
'iheorems !h(
the deJlection aod r!!e!ie!_at
r h i ch i s su b
lhat point in the dellected
shape will !ql!_!::CI9, In addition, there will be a zero vertical deflection
at node D. i
at c.
4._l r In order to solve certain problems, particularly .t,.! ' those of deformations, we will introduce the concept of vj_rtual states. These virtual states need only satisfy ore of the two conditions: either equj.Zjbrr utn ar campatibilitg.
NJ
We can have a virtual
equilibrium of forces exists, as shown here. The load is supported by the moment and the vertical
.r-l-I
force state in which an
reactions at A only, but the deJlection at C is no
VIRTUALl.rt9l,{rs lea'\brDtF
longer compatible with the rigid support at that polnt.
l-
The efJect ( - a tD ,
Th
a deformed conditions-
77
AF VIR:':|'--;t'7 TEE THEOREMS Simi\arly, ve can create a state ol vir.-,jaaisplacenenL by the introduction of a hinge at .\ and B.
This arrangement is compatible with the
rs no ionger -:upport conditions, but the structure introduction The load. the capable of supporting o
two hinSes has creared a mechanism.
Consequently this system satisfies compatrbjlitq bLlt not equjljbr:rum' The virtual
NT ilRTUAL DISPLACEM\rg
state is an analytical device.
Because only one state needs to be g4tilfied ilis more easily evaluated. When we combine real and virtual states we can develop theorems of Sreat Power and application in the solution of structural problems'
,f,-
This chapter is concerned with the proof oJ the .aeorens oI virtual
watk.
The two forms of the
theorem discussed above are ol Particular interest:
6."
'.ri
THE)RE! oF V ls-TuNDISPL\CEJT,ENT6 lHEo PE&\oF VJRTUAL Foe.6
Virtual Displacements, and Virtual Forces.
\\e will use as a basis lor the development of the iheorems the deformable, Pinned structure ABCD ,r,hich is subject to two loads, h/r and t{r: acting at node D, which is a hinge'
f,
The effect of these twq loads i5 to c-qq99e-.deflection anq Ye ha.I9 , at D. T\e sllqq-t!!9jliq-equilibrium a de{ormedshapecqlBal-illg }'ith.the supPort conditions. This is a rea.l state.
rad a(le&tun
74.
UNDERSTANDI NG S TRUCTURA L ANALYS TS
9,
Ve will consider first the Theorem of yirtua_z Dispfacenents.
TH&REM
OF
n//n]D5Frna il U/ UIN U rt LUE=-
JEIIS/Pfl A N IEM1IEAI?E -u o u o u J
10.
The real structural virtual
system has imposed upon it a
displacement.
The direction of this virtual displacement is unrelated to the direction of the real deformation. This is a virtual or notional
th
displacement in the sense that, despite the introduction of the yirtual displacement, the internal forces, and the external loads ttt and t{2 are
-€ctidr 3 r IC G
unchanged by this virtual
N rc4 iWtnt
.pt
-
cxtern
-ti.n
deformation.
ll.
In order to make the diagrammatic presentation as clear as possible we will adhere to the convention shownhere.
12.
As we have assumed that the imposition of the virtual deflection has no effect upon the external
"s
q3_R3!
N rt&l a#iadr to.,.E A, raat a4tz*tn
d vfintuo( dlefiL
loads, equalJy it has no effect upon the balancing internal lorces, i/1, N2 and .V3 in members AD, BD and CD respectively.
Note that because all members are pinned at each end and the loading restricted to node loading the resistance of the structure
is limited to axial forces.
r!'""'st{ caEl
-€ctidlr lttual
to {
q
TI{E THEOREMS OF VIRTUAL
79
WARK
i lhe Theorem oJ Vllt-uAl,Di€pbrements the
qCNERAL
to--1 5-€qtl&l -.h: \-1\u_e, g!_\\e-Lea\-!9r_c9 -ir,-!hldine.q!i-o!,q!\h\e vi{+q-aL_dlgplacerogI _!:9_:-lhe corlrpone_q!-ol-tb€_llel
:rpression for the vit!]lal-\or.$ymhol
EQUATlo|'/
oF vtRTuAL WoR,( W = ,*Jrtal Forcz- ti'r 4ife/-tic 6F MERf UAU
force)_multiplig{ b+ -ths -value-ol the virtqg]
cts/4a^c.tr X [4uRtrUAL
displacement.
aBpl4@h\r4lt
Ti'e,9.x-t_el11r- lirt-uq!-y!41'q -d*e-Lelrn{9d*bv-E5or+i+g ihe externq!!_oj-cej I{ AJltg-..dllgc.i!9!,gll;hc-virJual displacetnent
The component of force
lr'r in the
l:
dlrection of the virtual displacement is equal to tL x cos 0r, where 0r is the angle between the line ol action oJ lorce tdI and the direction of the virtual displacement.
*Wu' .
W=
W r, ca?l .d)
For more than one external load actin8 at D we can determine the resultant Jorce R,,/ which is in the
rq
direction of the virtual displacement.-
w,rff1w, t<w ^ lvL
d tB".
bJ the- rea l ,1o-adsis wolk don-e_ equal to the resultant component ol th€ real loads !!l the The externa! yiltllal
{
Ni a1i
*+lc
direction oi the virtuaL displacernent multjplied by t!e virtual displacement.
w:5. e"
80
UNDERSTANDING
17.
Ie can nor
In a similar way, we may find the resultant of the internal axial forces RM in the direction of the
"v;
Rx
virtual displacement, The angle is that between the direction of the internal Jorce and the direction of the virtual displacement.
s t3*,, Eotxs e".{
I\NALYSIS
Nj. c.s..J.
1 8.
^v .r^ IL KN
R"
Since in the real state the structure is in equilibrium
I:
the sum of R,ir and RM, the resultants of applied loads
I I-.E-
a,
and internal forces in the direction of the virtual displacement, must be equal to zero.
,Rw+Rrv'o
19.
The internal vi
Ual work, rhe symbol for which is u
&f *
is equal!o l!9 rq:-u,!fert _-o.{1!_g 1ryelltq!rolces in the direction of the virtual displacement multiplied by the
ENTEPNAL VOR?VAU WOFI(
vrr Ludr
!
re' 'JPrdLcr
'
r, '
e! = e H .6
20.
-j\
However since Rrl/and RN are in equilibrium, the actual values ol the resultant of the external Jorce, and
egui[ibrium Rv = - R r y E x b VW = I 4 M W
W=(4
internal forces will be equal. Therefore, the internal virtual work must equal the external virtual work, since both Rrr'and RN are multiplied by the same virtual deformation,
!n
']hr]6
THE TIIEOREMS AF VIRTi]}.:
It
Ve c an no w s l a t e t h e T h eo r e m
o i Vir tu a l
i:'::
8l
Disp la ce m e n ts.
ThcbtaA fi
I
M|RT@AL Dis plaeeuet*1
L1 .1lJ svt4
@Ab
M|RTUAb wok 4 ttz- VtRTgA& 46pla*'^e& a.tis 13a(&- tte, r\w4w\
fdt<4 -
\Ie will examine the application of the Theorem of Iirtual
Displacements to the ultimate load analysis
:i the fixed-end beam ABC loaded at the centre rith the load H, such as to cause collapse. Note ::rat the ultimate resistance moment at B is twice
Wue''
:rat at A and C.
M* L
2Ll L^ Mr,
Ae will consider a virtual displacement system which .s compatible with the supports, by the introduction :i hinges at A, B and C.
Point B is assumed to
,:rdergo a virtual displacement oJ 6 and joints A, 3 and C undergo a virtual rotation which may be e\pressed in terms of the virtual displacement and :ie spaD dimension r,. Because the angular .ctations are small we can assume that the arc of : 1e .rrcle rad ius
"/2 is t he s am e as r he t angent . lngle 01 and 02 are expressed in radians.
IiIRTUAL sJs?" + [email protected]$vrJ
Tris is the bending moment diagram at ultimate -lad, due to the action oJ the ultjmate load ,r'. ;ris system is in for:ce equilib.ium.
That is the
-rrimate moments at A, B and C are in Jorce ::uilibrium with the applied, ultimate load ,/.
,/
/
A*u4
quihibrtua stlstart
+ tt'r'44^4 f4'>J
82
f
UNDER.sTANDINGSTRUCTURAL ANALYS IS
25.
To find the relationship between the ultimate load
Ve
tr and the moments of resistance M__,we determine
b##-*d
the internal and external virtual work.
Q)?
W " w.el W . z(0t.nu)+02.21,1u
time,
The internal
vrrtual work due to a moment is the mom6itvirttEl l 5a
In this example the direction of the force and virtual displacement are coincidental and therefore 2 6r
g
there is no need to determine the component of
w = p(28,,\u^
the forces in the direction of the virtual displacement.
(Y\rz ,'
+W67r)' znn . tzSMulu
26.
The .rea-Z ultimate ultimate
w8 = pSmqu
particular
load relationship between the
load I{ and moment of resistance lor this beam isi
W. t2 MrllL w = 12 x I'iulL-
Eirher it or Mu may be the unknown in this problem.
L
THbPeu ep
VflRT UAL
27.
We will now develop the equally powerful Theorem oI lirLual
Forces,
-
troR@85 -
d
28.
We will examine the same pin-jointed frame ABCD with the same real
B
P.ul i
stata
deflection A.
Jorces and the resulting real
THE THEOREMSOF VIRTUAL [:
-..]w impose a virtual
force
83
',1'5?]( p and, at the same
-rF.. disconnect the support at C, i.e, there will be -..=.naf virXuaf torces n, balancing the externally :Jplied virtual
force, p.
i:rtual force in member CD is zero.
Such a system
-!. a virtual force system because, although it is in €quilibrium,
it no longer satisfies the condition of
E o deflection at support C.
Reo ,-< , \6-\ -^l ^ 'x'4irccf,rott *
F
A' +".hsY=o)
It is not compatible
rrrh the real support systems. Note iE:lectio,
na-
Note that the internal
that the real
A is still identified because we are
torng to consider the virtual work condition in the drection
of that real deflection.
TIle external virtual
load p multiplied
rirtual :=-i
work, r/, is equal to the by the component of the
deformation A in the direction of the virtual
load p.
t.t.'.1-----
Pw,^ W= F x
The internal virtual
Ae o s0
work, u, is equal to the resultant
oI the internal forces, resulting Jrom the application of the virtual force, in the direction ol the virtual fo.ce multiplied
.a*( = - cag|
aon
by the component of the real
,'#:
deflection A in the direction of the virtual force. The angle 0 is between the line of action of the t-irtual force and the real deflection
A. Angle o
is between the resultant of the internal forces rn and the direction
of the real deflection
v^
Twcnd
ViIB.TUAL r"ork
V=-WAcn0
A.
Since this is a virtual force state, it is the condirion of force equilibrium which is satisfied by the virtual system. Therefore the virtual force p will be balanced by the resultant of the virtual iorces.
internal
The external virtual work r? is equal to the
:nternal virtual work u.
btfF^o gU LI = - lf"^ fg u " t/u
=-F.kA* 0 F A* 0 LlLU -
\!J
84
UNDERSTANDING STRUCTURAL Ai./AIYS.'.s
33.
At this
The Theorem of Virtual Forces.
The*cat + MIR.TCTAL Forc4 I^
a^1
*vu&\ra.l
and ttE
that du
SaF+a^^ V'L
exampl
L\tz,';at U/INT V6(1" wo'<- rfav, tt&$a^ ETSRT@A&. fo.Cli
ma]-. be
cvr', ac' REAJ--aiVlacoNr,ca,ta.hy is quat 6 U1o t^ta.|a^ VUR.TUAG. Wa/k-
dar\?,
alatNt-
MIR.TUAb i."*
(ha,
This rl
t^t-c,/^44
V th' e.j'-
ai+b
rat-
34.
Of the two theorems discussed above, il is the
_
Theorem ol Virtual Forces which is the most used in the development of classical methods of structural analtsis.
Ihe Theorem ol Virtual Forces will now be
ippliEd to the simply supported beam ABCD.
The
vertical deJlection at B is the object of analysis.
35.
We will apply a virtual force p, a unit load this time, to the same structure.
P=4
The structure is sufliciently
straightJorward Jor it to be unnecessary to change the support conditions, to ease the calculation of the eJfects of the virtual force.
The unit virtual lorce is
applied in the direction of the required deformation.
36.
We must now determine an expression for the jnternal virtual work due to bending. In the earlier example, the internal virtual work on the pin-jointed Jrame was related only to the resulting real axial displacement. Here the deflection at C is due to ,bendinq. 'We will examine a short length of the member dx,
:
a distance x, lrom the left-hand support. Under the real load o, dx has undergone a rotation Bx.
)
ffi
TI{E THEOREMSOF VIRT:UAL IIORX ft
dris section, the yalue of the real moment is ftx
d
the virtual
ffi
moment ,x.
due to the effect
craple
moment is
of the virtual action.
that is the unit virtual
be due to a yirtual -t
The virtual
load at B.
load or a virtual
F=iFo"
In this
i
That action
moment.
will be explained later.
-
b
mem!ga!1
Eal
rotation
--'-''-
is the virtual moment multiplied by the B .
t:-q-.Lt*-.
the action of the real loads, the stress in the fheq
39r,
distance y from the neutral axis is: rl---l [
-f
I
F+J u
I
q x-]/4.
I1-'r*ZJ
'lF* .4-
stl.as Mz.q T
Ihe mater-ial of which the beam is made is assumed ro satisfy Hooke's Law, where slress is directly Foportional
to strain.
Therefore the strain at level 9
G equal to the stress at level 9 divided by r, the modulus of elasticity.
-7\. t z,-Vet -[-
Substituting for stress from
above, we obtain an expression for the strain at level y in terms of the moment at x, the distance 9, ,E and r.
Strain
€,! = M.'!
86
UNDERSTANDING STRUCTURAL ANALYSi:
41.
The total strain over the leng1h dx at level 9 from the neutral axis, A6ro, is eqr.ralto the unit strain
TofAl ct4lTo u layU dx, al' lt4s!, t J 2g, etz' Oon, -
ovet
multiplied by the length dx.
Substituting for the
exprescion lor Jn.l slrain from above. the total strain over the length dx at level I from the neutral a y i - 's c o u a l L o t n e v a l u e o l L h e b e n d l n g m o r n e n t a t
M z - 4, . t P E7
point x, Mx multiplied by the distance 9 and dx all divided by the product of the modulus of elasticity and the second moment of area o{ the section.
42.
Since the actual rotation is small, the rotation may be assumed to be equal to the strain at level 9 divided by the distance to that Jibre, 9.
8x is expressed in
radrans.
B * - A a,q
p.=ast J
43.
Substituting for Adxg above we obtain an expression {or the rotation at x.
.' 1
$- =lu - . a o - 1 ! e1 I
Er
B" = (real moment at x)dx
t
Ma. az, oJ_
44.
We now substitute the expression for B* (Figure 43) in the equation for internal virtual work in the segmer.: length dx (Figure 38). Thus the irterral
M: ffix'
t \\X- , aa-
virtual
work is equal to the bending moment due to the virtual action, multiplied by the real moment by dx divided by 81.
bending
THE THEOREMS AF VIRTUAI ;":: -:e e.rpre'')on
.tor the e.rte,.i"ti
t .tua)
work )s
-ral to the unit virtual load multiplied by the reai -::lection A- .
--.
total internal virtual work in the beam is the
-:agration of the expression for the internal
v
:iual work u, for the len8th dx, for the Jull span
--e firll span contributes to the deJlection at the ::int of consideration, B.
/-L = /ffi. tu\az
4^
We now have both sides
:: the expression for virtual work.
\4\
\lLq -
:: lhe beam. The total internal bending strain in
6
-.u
The deflection
- rs the unknown.
.::tual farces : bar s i similar expression may be developed for the
4t
:rernal virtual work in pin-jointed Jrames with
t4'
Jading restricted to point loads at nodes. The three-member frame ABC is loaded
N
Ng.
riagonally at B, producing internal axial forces, NBA, '.96 and w6g . lf the vertical deflection at B is required w€ apply our unit virtual load in that direction.
lrt-
The internal virtual work in such systems is equal to :he value of the virtual iorce r for each member
,
multiplied by the real member displacement. The Cisplacement by simple consideration of stress and srrain is equal to the real axial force multiplied by the length of the member divided by the product of the area, A and the modulus of elasticity, -4.
- tt
r=
N/t
-+lak€=
A = NLt"/- n r {l .,1=
n
Lr .
FAI
Ll tl
At, tL
88
UNDE R'sTANDI NG S TRUCTURAL ANALYSI S
49.
(4= N!.Ul Ae
r'L
This could be expressedas an integral since the
':-
The subsequenti
length of rhe member r is equal to foi.dr. Ho*.u"r, since the axial force is almost alwaysconstantin
statically indete
such structures the expressionin the upper part of
d e Jo r m a ti o n s.'I
this diagram is the simplestto use.
oJ Virtual Force
degree on the at
- /N.W a*
ProDlems.
JT7 o
The first is th
't-
subject to two v units respective Tarsianaf
J0.
ratation i-
The vertical dell
tt.
The first step is
The bar oJ constant cross-section is loaded with the torsional moment, r.
The bar undergoes a total
rotation of O at the free end.
There is a balancinS
internal torsional reaction equal to the applied moment, 7.
The rotation 0 is the object of the
analysis.
51. Applying the Theoremof Virtual Forces,a unit virtual torsion is applied in the direction of the
real forces in thr
unknowntorsional rotation 0.
52.
It may be shown that the internal virtual work is equal to the inteSration of the real torsion ?, multiplied by the virtual torsion t, divided by the
W=4 " I n n - / L- , \1 -/ /i g
torsional constant J and the modulus of rigidity, c.
d z.
,YJ
J-
g.
eo.stt^4
.t ,tta C
r4tu l4uJal rgrarg
E.
We now apply a required delorm
THE THEAREI,ISOF VIRTUAL yIC:.
89
--nrchaplers on methods oI analysis oI _leterminate structures depend to a large :1e ability to determine structural :: ::rs. To extend the illustration of the Theorem ' : Forces we will now apply it to two typical
l"
126
Y
- : : .:: is the pin-jointed lrame ABCDEF which is -:: : : iwo vertical loads at C and E, 12 and 6load _::fectively. -: t:
*ie r 3.:icai deJlection at C is required.
-- first step is to determine the distribution of the -e .:-i forces in the members, N.
!t.3
t e now apply a unit virtual load in the direction of the -ecuired delormation, i,e. in the vertical direction at C.
^h. *{t' {
o
TI{E TI|EOREMSOF VIRTUAL WORK
89
The subsequent chapters on methods of analysis of sratically indeterminate structures depend to a large :egree on the ability to determine stiuctural :eJormations. To extend the illustration oJ the Theorem :i Virtual Forces we will now apply it to two typical lroblems. The lirst is the pin-jointed frame ABCDEF which is subject to two vertical loads at C and E, l2 and 6load $its respectively.
The vertical deflection at C is required.
The first step is to determine the distribution of the real lorces in the members, N.
r5.3
I e now apply a unit virtual load in the direction oJ the :equired delormation, i.e. in the vertical direction at C.
90
UNDERSTANDING STRUCTURAL A/VEI,YS.T.S
57.
The next step is the analysis oJ the distribution oJ
l:e
firsr srq
virtual Jorces. The characteristic of the application oJ the Theorem of Virtual Forces is that the slructure mus:
.::r
bending
be stable. Il the structure is statically deter,?jnate then releases such as were applied in the original explanation (Figure 29) cannot be made, otherwise the structure would become a mechanism. We cannot simplify this structure for this particular virtual load, i.e. by the omission of members AB, BD and BC Jor example without creating an unstable structure. Consequently it is necessary to determine the value oI the virtual force in each member.
,E-
Lsing the Th Jnit virtual I
Mau= H^".ho"
J8.
The internal virtual work done on member AB, is the
deformation.
value of the virtual axial force multiplied by the real
5ending mom
change in length. This calculation is to be carried out for each member in the Jrame.
royoyL, = (\ te.711/\ ,/A L
59.
The total
internal virtual work is the algebraic sum
oJ these terms Jor all members in the frame.
zu".rM={$. h
In the
{t-
The most reli
tabular meth(
expression Jor the equality of external and internal
are split into
virtual work, only the deflection ACv is unknown.
expressionsf(
= 6E/AE
f f i =f,. A" y
There is no I
the Jirst and
evaluation ol The second example is the simply supported beam ABCD
siSn of the x
loaded at B with a vertical point load of l8 load units.
the most con
The vertical deflection is required at C.
indicates the section ol th length along
application o structures. member will
global axis, s
THE TEEOREMSAF V Ihe first step is to determine the distriburi-; .eal bending moments, /y.
l
sing the Theoremof Virtual Forceswe apply a
lE l-
:
: \irtu al loa d in r he dir ec t ion of t he r equir ed
.€:..:nation.
We then determine the virtual load
:+ -: : nB mome nts, m .
--E most reliable approach to this numerical analysis is a -.::.lar
method. Firstly, the bending moment diagrams
:,-: split into convenient sections in anticipation of the
"_-f_",i 'tz
:tr!ressions for the integral.
'6
4l
ffi ts
ruP' /v\
ihere is no particular significance in the selection oJ :re first and last point for the integration since the 3\aluation of internal work is not related to the s:gn of the x direction along the beam axis and so :ie most convenient is chosen.
The diagram
ndicates the zero point lor the integration lor each section of the beam. The symbol s is chosen for the length along the member to anticipate the .pplication of the theorem to two-dimensional slrLrctures. The use of s for the length alorg
the
inember will avoid confusion with the horizontal global axis, symbol x.
l+/q
ha
UN DERSTANDI N G ST RUCTURAL A,VEIYSJS
92
65,
{il^,"=-f(+ix€/ , - . ! s= a
<{
AB A A+ B
W
M
zs
The inte8ral expression for the internal virtual work is
The Theorem of V
best created lrom a SraPhical summary. This is fot the portion oJ the beam AB. The total internal virtual work'
rotations. UsinB 1 is to lind the rota
u Jor part AB is 50 units.
3s
,3
-:3 I f 6.' sz)ds -l - €3-s' I l=q ' 127 I i
Lo
"
= eo ufr1
66.
/
("Mn", o, --f r?;v'-11 '-\lti"Y'/\ -J/
J
B a b b+ . .
3 6 -6 s
The diagram shows the expression Jor the real and
We apply a unit v
virtual bending moments in the portion BC. Note that
of the required d
sagging moments have been taken as positive.
bending moment
Thus the
negative sign in the expression lor the real moments in
E tr+ 5 t,S
BC recognises that the moment is reducin8 lrom B to C.
t%sz)4s J <6 "+ tas-
The completion of the integration expression for portion BC gives a value of internal virtual work lor this portion
AZ.7 twts
-
ol the beam of 55.1 units.
67.
portion CD is shown. Internal virtual work equals lll.l
{'^ * * * GX?) cD
D D>4
(t"+ !" ot =
/7)
6s
An expression for bending moments and integration for
units.
+/1s
-:.
The external vir multiPlied bY tt}
virtual work is d
product ol the e moments.
u't ux,fr
58.
The value ol the total internal virtual work is the sum ol the internal virtual work in the parts AB, BC and CD. Equating internal with the external virtual work, the rea! deJlection at C is equal to 23J.0 divided by Er' The
Aor,L= 235'o/EL
value of the dellection may be found by substituting lor R and r.
We will now stu
two-dimensiona
C, normal to mr
93
THE TEEOREMSOF VIRTUAL WORK t
--::.em
oJ Virtual Forces may be used to detern,t-:
:- . Using the same beam example, the object no\\ :-.r: rhe rotation of the beam at A, OA'
ce apply a unit virtual
nanent aI A in the direction
oJ the required deformation, The resulting virtual trendingmoment diagram, -,1is shown,
The external virtual work is equal to the real rotation multiplied by the unit virtual moment, The internal virtual work is determined as above, by integratinB the product of the expression Jor the real and virtual
= 4Jl \\'N1 \:./rJ
moments.
.
,q v
t4^
.L
M - /(wXv)st i
we will now study the application of virtual work to rwo-dimensional structures. The frame ABC is loaded at C, normal to member BC, with the point ioad t/'
'8
r
94
UNDERST AND I NC 51 P U C TUP A L A NA LV SI S
I
73.
74.
%K ;
The object of the analysis is the horizontal dellection ar C, Ar-. .
We will now examine the eJfect of both axial fo.ces d and bending moments tt, Because the real load ,1/is
...the axial
Ve will now
1
Virtual For(
normal to member BC there is no axia.l force in the member BC. However, there is a real compressive Jorce
the contrntx in span BD.
AB, equal to the horizontal component of ,/.
N
75 . We now apply a unit virtual load in the direction of the
\**
\
\
&n
Wl
76.
reactions at
diagrams are shown. Note that the virtual bending moment in AB is constant because the virtual load is
are convers
parallel to AB, also that the virtual axial Jorce in BC is the component of the unit virtual load in the direction oI
ol reactioa!
BC. Consequently this will be less than the value of the virtual axial force in member AB which is equal to l
solved beca
This ligure shows graphical summary of the integration for the bending efJects...to which must be added..,
L"t^ JL = /( hX /a
+ / (4\
Y4
r--'-. )ds PT lgg
This figure I
required deflection, that is horizontally at C. The resulting virtual bending moment and virtual axial load
This stru
chapter tre
only be apd
real load b(
to.
We will nor Virtual F(rt reducing q
This will a.f bending ftl
virtual loaa deformar;q
baaaig
TEE ?EEAREMS AF VIRI:J-Z:
77,
;":'
95
, . . t h e a x i a l l o a d e ffe cts.
-
/
mrnnV:\ /B 7 \
'/\
ds
)Te
*/ -'-t--- x\)ff' oZ,,o,t
;t.
We will now study the application of the Theorem of Virtual Forces to a statically indeterminate structure, the continuous beam ABCD loaded with a point load r/ ln span BD. The vertical deJlection at C is required.
This figure shows the qualitative distribution oJ _.eal reactions and bending moments. This structure is indeterminate, consequently until we are conversant with analytical methods which follow this chapter we cannot determine the numerical distribution of reactions and bending moments. Thus the theorem can only be applied to a structure which has already been solved because it is necessary to know the distribution of real load bending moments, lt.
r-
We will now be able to see the power oJ the Theorem of Virtual Forces. Ve can release the fixed support at A reducing the structure to a statically determinate Jorm. This will allow us to determine the distribution ol bending moments due to the application of the unit virtual load, applied in the direction of the required deformation.
^€=t
Jt JL
96
eMeLysr.s
UNDERSTANDING
81.
Note that because the values of virtual moments are zero in span AB, the product of the expression for the real and virtual moments will be zero. Consequently the
Ul =-/ M Kn as
integration will be limited to span BD only,
=f*wx-v,"'
Ya" *
*nJ 3.
.Xa-., I
Ya, {v
"r./ 1
-a
Practice
Problens l
These structures are to be analysed lor delormations at
*
4. I
l--\
-dl
82.
,rl--rc',
a+-I'r x<
X and v.
The subscript q indicates a rotation
(i.e. xo
rotation at x) and 6v vertical deflection. dn is a horizontal deflection, Draw the appropriate real ltj and virtual ,?6 or ,ns diagrams.
'/x"l'"' f.
Practice
prabiens 2
You should now practise the application of the Theorem of Virtual Forces to numeric problems. A full .ompurer solution to each of the pracrice problems Jrom Chapter I is given. Since each oJ these problems is statically determinate, they may be used as problems in the application oJ the Theorem oJ Virtual Forces. You should choose a particular deflection or rotation and find the value. Since the real moments are given in the computer solution it is only necessary to identify the direction of the appropriate virtual action and determine the dislribution of virtual moments, Remember, you should a-zh,a9scarry out a qualitative analysls of the effect o{ the virtual action before commencing any numerical analysis. Ihc col,rtinr ro rhF Se,]rn)roblems should check exactly with the computer solution. However, you must remember that the computer solution to two-dimensional structures includes the effects of axial displacement. , uu
,L ,d j
''' - "r ,,c -
P'o L L L i c
-- icatron oi thi> rheorem to
d PPL
statically indeterminate structures using the problems in Chapter 6.
6 The FlexibilitvMethod
The Flexibility Method is one of the two main approaches to the analysis oJ redundant structures, the other being the Stilfness Method. Both metnoos are based on identical theories of material behaviour and differ only in the mathematical treatment of the basic structural data. Neither method ts particularly suited to the hand analysis of real structures since the conclusion to both methods oI analysis is a set of simultaneous equations, dependent upon the degree of indeterminacy oJ the structure. In a real structural Jrame, the degree of indeterminacy is likely to be large and althou8h theoretically possible, this hand analysis is too lengthy and error-prone to be a feasible design office procedure, If such an analysis is.equired, a computer program based on the stiffness method is the mosr likely course of action with the ready availability of such proven software. Alternatively, in real structures the symrnetrical shape of the frame ! ay allow an approximate analysis of forces and moments or the use of an interative method such as moment distribution. This is based on the stifJness approach and will be discussedin Chapter 9. The analyst must always bear in mind, however, that he will, at best, be approximating to the behaviour of the real structu.e. He will be creatinR an analytical model which is a theoretical model of the complex, real structure, The advantage of the llexibllity method is that it contributes signiJicantly to an understanding of the real structure because the procedure of the method, that oJ reducing a complex structure to a statically determinate form, is closely related to the fundamental design decisions oJ rationalising the real, complex and highly redundant structure into a simpler form suitable for analysis. To some extent that simplifying procedure is likely to be adopted even when a computer is used. Three-dimensional structures are usually reduced to a series of linked two-dimensional frames, for example. These frames may be reduced to a
A'VA'YSTS STRUCTURAL IJNDERSTANDING I i1€
beams' The series of sub-frames or lurther simplified to continuous of these simplification of the real structure which allows the creation
re5_-5
behaviour of analytical forms mus, not yield a solution less safe than the that you the real structure and ultimately it is towards this understanding
ihe c
must see your study of the flexibility method directed' lor which the The flexibility method is based on structural deformations Theorem of Virtual Forces is used'
Th e :i
horr loacsg eti
equa
ihe stari the fl detern{ structrl deterni applied
bendin€
vert lcal
to a cai
w ]In a i
Now in
which I reactt(
positio
the val
beam i
THE FLEXIBILITY METHAD
99
Tlre two principles on which the method of flexibility rests wi.ll be illustrated with the propped cantilever
AB. The Jirst step in the analytical proceoure ts to determine the degree 01 indeterminacy. There is a potential horizontal reaction at A; however, for this particular loading arrangement it is zero. There being Jour notional externat reactions the propped cantilever is 4 _ 3 equations of equilibrium
-
I x statically indeterminate.
The second step is to release the srructure to a statically determinate form. The reason Jor this is that the flexibility method depends upon the ability to determine the deformations of the released determinate structure. The Theorem of Virtuat Forces is used to determine the deformalions and this method can only be applied to a statically determinate structure if the real bending moments and other load effects are known. The vertical reaction at B is removed reducing the structure
s--U-,' '<
\+*-'xa,"
to a cantilever beam. The structure deflects downwards with a deflection of AR at B. Now imagine that we know the value of the reaction which has been removed, yB, and we apply it at B. The reaction yB will return the structure to the original position at B, exactlg, because that is what defines the value of that reaction, the Jact that it holds the beam in that position with a zero vertical deflection at
+"-E{
f Va V6
E 10 0
UNDEP,STANDING STRUCTURAL ANALYSI S
$-+"To""
4.
However, we do not know the value of yB.
apply a unit load or action in the direction of the removed reaction the.ate
S
It is the
object of the analytical procedure. If, however, we
of deflection for the
application oJ a unit load applied at this point, i.e. dB B per unir load, may be lound. We also t-v determine the deJormation due to the applied load
fr.".'
in the direction of the removed restraint.
These
are the second and third stage in the analytical procedurel
2. Determine deformation due to the application oi the design load Aa in the direction of the removed restraint.
3. Determine deformation(s) resulting Irom the application of unit load(s) and/or reactions, 6BuBu.
Not" particularly that the unit loads
are applied in the direction of the removed restraint.
To complete the analytical procedure, we apply the principle of superposition. This principle allows us to add separately identified load effects together, the final solution being the sum of the parts.
Abr* Vr.fe,er=O
The
separate effects are the removal and replacement of vB.
One efJect is the release producing the
deflection at B, ABv. The other effect is the deflection at B due to the unit load at B.
If we
multiply this rate of deJlection by the unknown reaction yB it is clear that these two deJlections ABv and yB. 6gugu must have the same ya_tue,
6.
The last stage in the analytical procedure is the setting up and solving of the equation(s) oJ compatibility (there may be more than one degree of indeterminacy). The equation of compatibility is equated to zero because the true deflection at B of the original structure is zero.
t0l
THE ELEXIBTLITY METAC} --: iefo rriario ns whic h r es ult f r om t he anal) t ic al :':€edure are given a sign appropriate to the overail ::r'\'ention. In this case AR_. would be negative and ::-, Bv positive. The choice of direction for the Ltnit :.:a and thus the direction of the Lrnknownreaction yB : ::bitrary because in a more complex structure the true :.:ection will not be as apparent as it is in this simple :!=mple.
S--= j
i a
tve L+'
36
+--
For this propped cantilever however, it is
::i ious that
YB will act upwards.
:-? s olu rion of th e e quaTionof c or npaubilr t ) ic r hat v B : )ositive. This means that the choice of direction for ---e unit load (Figure 4), is correct confirming that the : :e. tro n o f th e rea c t ion / B is upwar ds . l' t he s t gn been negative then the direction of the reaction -:d : f,uid have been opposite to that chosen for the unit load.
Agu + Vg. f,"ner- o (-v.)*
VE(+vo)=o Yg r s +vo
"'
1e must remember that the evaluation of the unknown :eaction is not, normally, the only object of the :nalysis, Usually, the distribution of bending moments is :equired. The principle of superposition may be apPlied ',reretoo. The bending moment diagrams which result
I
ITiTfirfi-
:rom the application oI the point load and reaction at B 3re combined to produce the final bendinS moment iiagram.
\'iost real structures are highly redundant and we will now examine the application of the flexibility method to the beam ABC which is 2 x indeterminate. Step l.
Red uce t he s t r u. Lur e t o a s t at ic allv
deter mina te fo rmi rem ov e y B and v C.
Ilirr-*
rc
ta2
UNDERSTANDI NG STRUCTURAL E?r'AI,Y.S15
ll.
Step 2. Determlne delormations due to the application ol
To
the real load(s) in the direction of the removed restraints: ABv and ACv. Try to see the flexibility method as collecting delormations due to the real and unit loads at a particular point, in a particular direction.
12.
Step 3. Apply unit loads and determine deformations in the direction
of the .e,,noyed.eactiors.
Now we
can see the reason for the rather cumbersome system ol subscripts. It allows a clear delinition oJ a particular deJormation. Note the complication oJ
rne
more than one degree of indeterminacy. The
b ya
application of the unit ioad at one point contributes
Deattl
to the deformation at another.
13.
Interpretation ol subscriprs. The capital Creek symbol A
17.
For
18.
we
is used for deflections due to the real loads and the lower case 6 lor deflections .esulting lrom the application of the unit loads.
Deo L' tj1.. apph
14.
Step 4.
Set up and solve the equations oJ
compatibility.
A*: Vr.f,e,Bu" V,&r*=o
This step is confusing at first.
deJormations at ane point
in ore directior
make up one equation of compatibilty. |
l)crt
,,
F
vg. devBv* Vc.borcv-O
the
Remember that you are collecting all the to
t03
TEE FLEXIBILITY METi|CD To begin with, check each equation. The Jirsr subscript of the unit load deJormation musr agree *'irh the real load deformation,
The second unit load
deformation subscript must relate to the unknown reaction.
16.
Having lound the unknown reactions vB and vC, the beam may also be solved for the remaining reactions at A, ItA and vA, equations of equilibrium.
by aPplying the three
Note that we can identify
the direction of the unknown reactions vB and vC by a consideration ol the qualitative analysis of the beam. You should always carry out such a check. Numerical errors are easy to make in this method.
17.
For example, a minor error in the calculations oJ a deJormation could yield values of reactions which produce a net hogging moment at A. Only a sound understanding of the qualitative analysis of structures would help you to spot such a mistake.
It.
We will now examine an alternative method ol reducing the structure to a statically determinate form by the removal of a moment instead of a load reaction. Step l. The removal ol the moment reaction at A and the vertical teaction at B. The beam is now statically determinate, simply supported at A and C.
IinhJw
t
T W
'+'@-' LNooE'zeeT sel-q TtoN
104
UNDERS TANDI NG S TRUCTURAL ANALYSI S
19.
Step 2. DeJormations due to the application of the design load. Now instead of a deflection there is a rotation at A due to the removal of the moment reaction 11A. In addition we must determine the ve r t i c a l d e f l e c t i o n a r B , A B v -
20.
Step 3. The application oJ the unit loads, At A, instead ol a load, a unit moment is applied. Note the new subscript 'm'.
Used as a first subscript it refers
to the Jact that the deformation is a rotation.
As a
second subscript it identifies the fact that the particular deformation is due to a moment and not a force.
'6BvAmr is interpreted as a deformation due to a unit action (6) at B in the vertical direction (Bv), due to a unit moment at A (A.).
21.
Step 4. Equations of compatibility.
It is inevitable
that you will experience some dilliculty
+ VB,*A,hBv=o 94+ M4.oaa-A," t-
t-
Ae"" l{e, dg, Aa + \/E, Ag"c, - O
in
deciding which deformation is appropriate in which equation. You should keep reminding yourself that you are collecting together bits of the same defornation
in the sa,ne direction.
Having solved the equations of compatibility we can check the direction of the reactions. We know the direction of the unknowns, MA and yB Jrom a consideration of the qualitative analysis. lrA is clockwise and yB acts upwards. The unit moment at A was applied in the anticlockwise direction (figure 20), consequently the sign lor t4A in the solution of the equations of compatibility would be negative indicating that the djrection
originally chosen was incorrect. The
choice ol direction for the unit action is arbitrary. However, the sign of the resulting displacements must be in accord with the overall convention, i.e. upwards and to the right, and clockwise positive.
THE FLEXIBILITY METEOD 4
The next problem is the two-dimensional frame ABCD. fixed at A and pinned at D. A question often raised by students is the problem of sway. This is because the eJfect of sway is considered as a separate loading case in the method of moment distribution. In the ftexibility method however, it is automatically accounted lor in the analytical procedure.
?t.
We are now going to adopt the symbol convention for the deformations which is used in most current texts. It provides less information about the particular delormation, but it matches conventional matrix notation. The structure is twice times indeterminate and we will choose to release the structure to a statically determinate lorm by the removal of the two reactions at D, creating the determinatertree'discussed in the Indeterminacy chapter. These are deJined as x with a numerical subscript which does not deline the axis of application. The choice could equally well have been x2
25,
in the horizontal direction.
Step 2. Deformations due to the application of the design load. The subscripts now relate to the direction oJ the unknown lorce (or moment) as identified above.
26.
Step 3. Deformations due to the application of tbe unit Ioad in direction l. Note that the unit load in direction I, horizontal, produces a deflection in direction 2, 6r,.
JID' RST ANDI NG S TP U C IU R A L A N A LY 5JS
27.
clua Ea a unt totad21
This figure shows a clarification oJ the interpretation of the subscripts for the unit load deformations.
in tt'," avrrtti q A*l rr.$\NU
VC4rltgA, XZ
I
dto to tht- adtn S? q ttv* baa, u ailotivy
28.
The application of the second unit load, and the resulting deformations are shown.
The final step in the analytical procedure is the setting up oJ the equations of compatibilit,.
A,* X,.6t* Xz.iz =o Az* Xt.to*Xz,t""- o
30.
fd,1 l "u oJlx,l l l .l =-l l LA" E*llx"l i& I [- t^
t- J
The equations oJ compatibility may De rewrlrren in the form oJ a matrix of deformations due to the unit loads the unknown forces, x and the design load deformations, A,
t07
THE FLEXIBILITY I4ETHOD !,iote that the unit load deformation mattix &pendent only on the properties of the
is
nructure' and is known as the llexibility matrix' '[he values ol Ar and A2 are dePendent uPon the Frticular
loading arrangement.
P.actice
Probiens -1.
fn".,t;r;n,flx,] fo,l
L*^"jL-,1=L^:l
Eeams
degree of A number ol structures are showD' The choice the indeterminacy has been determined and
-l Bf
F--
R
of release made. Draw the following diagrams {or each problem: of the The lull three-part solution for the elJect of the unit application ol the desiSn loads and each Ioads: l.
dellected shaPe'
2. bending moment diagram,
3.
reactions.
the Indicate using the lirst of the subscriptsystems' delormationsaPProPriateto the solution'
tJ.
Practjce
PrabTens 2.
Ftanes
Adopt the same Procedure.
There is a potential conlusion between the unit load load applied for the flexibility solution and the unit applied as part ol the analysis of a partlcular We deJormation using the Theorem of Virtual Forces' deformation will examine the qualitative solution for the studied Previously' due to the applied load lor the lrame Reler back to diagram 5.25.
rL A.t
- T-
UNDERST ANDINGS TRUCTURAL E]VAZY.SI.S
35.
This is the bending moment diagram due to the application of the vertical point load n in the member BC. These are the trealtbending moments d of the virtual work solution.
36.
We will now treat the determination of the vertical deJlection A2 as a quite separate problem lor the Theorem of Virtual Forces. In order to determine the vertical deformation at D we apply a unit vjrtua_l load in the direction ol the required deformation.
37.
This diagram shows the resulting distribution of bending
41. c:
moments due to the aDDlication of the unit virtual load.
lr11
nm
38.
The true vertical deflection at D should include the effects of axial shortening in member AB. However, these are usually small compared with deflections due to bending and we are going to ignore them. This diagram shows a graphical summary oJ the integration of the internal virtual work,
u
required Jor the evaluation
of the vertical deJlection at D, due to the desi8n load tt.
w = a2,!
u=4/flvn)"
Note that the real bending moments lr,
beyond the
point of application oi the load tt are zero. Consequently it is unnecessaryto calculate the value ol the virtual load moments for this part oJ the structure, since when they are multiplied in the expression for the internal virtual work integration, the result will be zero.
to by
We load
t09
THE FLEXIBILTTY METEOD now study the Jull qualitative
Fe !i]] dk€r
analysis for the
load deformations Ar and Az usin8 the Theorem
Ni l iral
Forces.
-[nese two delormations Al and A2 require tnree Eding
moment diagrams to be evaluated. Real
4'crnents 14, due to the design load tf and the bend\n8 moment
d\re \o eac\ aL \\e
\\\\
virlual
actions.
\ra
ne
t\\
n
^[-1,@ 41.
Clerk-Maxwelt' s
f heor en ol Rec t p' ac at D - f l e c t ; o n s
The flexibility matrix is always square because the number of equations (rows) is always equal to the number of unknown forces or reactions (columns). Clerk-Maxwell's Theorem oJ Reciprocal Deflections allows us to reduce the number of elements we need to calculate those below and including the leading diagonal of the matrix, Deformations on either side oI the leading dia8onal, 6 rz/6 zr,6 r:/6:r ,
et c . ar e oI equal v alue ac c o r d i n 8
to Clerk-Maxwell's theorem.
We will now prove it
by stud yin g the t wo deJ lec t ions 6r z and 6z r .
12.
we will apply the real, horizontal, (llexibility) unit load and examine the resulting unit load deformation' 62r which is in the vertical direction. This is now the ^kia-+
^+
+ha.^.lvci<
arXotnL.,,
I l0
UNDERS TANDI N G S TRUC'IU RAL AIVAZY.S.IS
43,
The virtual Jorces solution would require the application
This fiSure I
of a unit
work soluti
vjrtua-7 load in the vertical direction, that is,
in the direction of the required deformation.
44.
^ffim
We can now draw the graphical summary of the virtual
We w i l l
work solution for 621. The real moment M diagram is
is appare
due to the application of the real (flexibility) unit
interch
load, in the horizontal direction at D.
result oi other a \
6,.L= / M, W as J
dragrams bending
E.Z.
45.
We will now look at the application of the real unit load at D in the vertical direction and the resulting
that onl)-'i
horizontal deJlection, 6r 2.
be calcul{
set up thel
46.
The appropriate unit virtual load for the deformation 612,
,0.
Statically
is applied in the direction of the required deJormation,
solved b1-'
horizontally at D.
frame is i
.lll
TEE FLEXIBILITY METHOD ---. iigure shows the graPhical summary of the virtual r,:r< solution lor deformation 6r 2.
#
t,.,.&= fM.m a t E
.le will now compare these two Eraphical solutions' It ,s apparent that the bending moment diagrams are :rrerchangeable' ln one equation the diagram is the .esult of the application of a real unit load and in the r:her a virtual unit load. Howeverr because tne llagrams are both for unjt loadsr the actual values ol
Er.[,,{(y,Xn)r. Err,,-/(#Xn)',
rending moments and consequently of the deformations :.:
:
a nd 6rz are the s am e.
This may be expressed in this general lorm and means to lhat only the leading diaSonal of matrix elements need to be calculated and those below this diaSorial, in order set up the equations of comPatibility'
I-
f,t1=fr"
Statically indeterminate pin-jointed frames may be solved by the flexibility method' In this example the Irame is internally and externally indeterminate'
ti = {ii
t1 2
J ]: DER5 TANDI NG S TRUCTURAL ANALYS I S
51.
We will release the structure by cutting the member AC, and identifying the axial force as xr and removing the vertical reaction at E, X2.
55, 56. Deterr axial t
subslir
ProPer \o
compu
solullo
deforIr
compir J2.
Having determined the degree of indeterminacy and an approprrare release system which is the first step in the Jlexibility procedure, the second step is to apply the real design load and determine the deJormation in the direction of the releases. That is the displacement of the'cutr in member AC, Ar and the vertical deflection at E, A2,
Third step. The application oj the unit loads and the identification of the unit load deformations. In direction I, the cut in member AC, an equal and opposite unit load is applied because the effect of the cut .eleases the bar on either side of the cut. This produces a deJormation 6 r r, which is the ex-tent to which the cut closes (or opens depending on the direction and application oJ loads) as a result of the unit axial load, This load also causes a deflection at E,6zr.
54.
+ l4
The application of the unit load in direction 2, This produces a vertical de{lection at 2, 6zz and a displacement of the cut in member AC, d I 2. Note that the Theorem of Virtual Forces would be used to determine the deformations in this pin-jointed frame, The procedure was illustrated in the previous chapter.
Jor rhr 6 .3 3 .
def or,-n
TEE ELEXIBlLITY METI|OD
fr"
113
Practice Prablens Determine the reactions and full distribution of axial forces in each of the beams and frames shown, substituting values lor dimensions, loads and section proPerties, You should have access to a'Plane Frames' computer program which will give you the check solution and may be used to generate the deformarions of the released structure.
txample
computer solurions are provided in the Appendix. You should practise full numerical solutions for the problems shown above in Figures 6.32 and 5.33. Use the computer to generate the delormations and produce the check solution.
"ff.*r1
7 The StiffnessMethodFrames
The Stiffness Method oJ analysis is similar in its aPproach to the flexibility method of analysis, the major difference being that the unknowns in the procedure in the stilfness method are the node disPlacements, whereas in the flexibility method they are the internal lorces, moments and reactions. The other major difference between the two methods is that the stiffness method requires no iudgement to be exercised by the analyst' whereas the flexibility method reguires a choice of releases. This makes the stilfness method particularly suited to automatic computation and it is the basis oi virtually all 'Plane Framesr and 'Space Frames'comPuter programs. The stren8th of the stiflness method, i.e. its suitability as a basis of automatic computation, is its weakness as a method of hand analysis. Because the method requires little or no judgement to be exercised, practice in stiffness problems promotes little understanding of structural behaviour. On the other hand' the flexibility method requires a sound understanding of structural behaviour and statics in the first steps of the analytical procedure. The reader should be aware that neither the flexibility nor stiffness method are praclical manual methods oJ analysis in the desiSn oflice because real structures are too hiShly redundant and must be checked for a number of dilferent load cases. The analytical Procedure of both methods, requiring the setting up and solving of large numbers of simultaneous equations is cumbersome, time_consuming and error-prone. It is also illusory in terms oJ accuracy.
t ica.l -19 accurate lt may be na tile.4?a
but no more accurate in terms oJ the relationship of the analytical model to the real structure than the method oI moment distribution. The latter is sometimes described as 'approximate' as if it were aPProximate to the behaviour of the real structure. It is no more or less close to the real
THE STIFFNESS METHOD- FRAUES structure than the so-called'exactr methods of stiffness and flexibility' which are based on idealised values of material and section properties.
tl 5
AIVAIYSIS UNDERSTANDING STRUCTURAL
I I5
l.
lr\
)a
M
The :.
BeJore starting to explain the stiflness approach to the analysis of statically indeterminate structuresr we need
the
to define certain relationships between deformations and
the
member properties. lt is assumed that the structural reac:rJl
members are straight and of the same cross-sectional
$__JA E"= rAa,
direc_-
area throu8hout. This is not a serious restriction as
!
r k\.-T
most structural members come into this cateSory. Ve will start by determining the value of the moment u which will cause a rotation of I radian at the pinned end of the propped cantilever beam AB.
To distinguish this moment from normal load moments it
6.
UsinF
is defined as 14Ab,the absolute moment, that moment ot
which produces a rotation ol I radian. The diagram of
N o te :
the deflected form above, reveals that there is a hoSSing
i si k\
moment at the fixed support at A. We refer to the 'near'
bend::.lg
end as that which has the absolute moment applied' consequently, the hog8ing moment is at the'far'end
3.
Dy leal
A'
In order to solve this structure, which is I x statically
7,
indeterminate, we will determine the reaction, yB. Consideration of the overall moment equilibrium at A will show that the vertical reaction at B is downwards, to be balanced by an upwards reaction at A, Note that there will be an anticlockwise moment reaction at A, which we can identify lrom the deflected shape.
4.
h/vtAb
V..43Aev
Using the llexibility method, we reduce the structure to a statically dete.minate form by the removal of the vertical reaction at B, vB,
The effect of the
8.
We will I
aSsumln The loai
absolute moment will be to produce an upward vertical
moment
deflection at B, ABv.
contain
fiillllTililJ'lJlllll1T||l nno"
Note that dAb, produces tension on the underside
oJ this statically determinate cantilever beam.
-IrF*-!l ,4
express
117
THE STIFFNESSY'E:-:: -
":,!:. The flexibility unit load is applied in the di.ec:ic: .l the removed reaction.
Here in order to illusrrate rhai
the choice of the direction oJ the unit ioad is arbilrar]. it has been applied in the opposite sense of the real
VE+d;.
reaction vB, that is in the positive z (vertical) direction,
The unit load produces a bending moment of
, kN.m a t A.
,|1]].T],'*
Using the Theorem of Virtual Forces to find the dellections we must apply a unit virtual load in the direction of the required deformations ABu und 6au.
rL ^ NA
Note that the value of the virtual bending moment at A is I kN.m. Remember that we distinguish virtual bending moment diagrams lrom the effect of real loads by leaving the virtual diagram without hatching lines'
l
/i
7o
This figure is a graphical summary ol the integratlon expressionsfor the internal virtual work for ABu .nd dBu' using the Theorem of Virtual Forces. Note that for 6Bv the iniegration is the effect of a unit load'squared'.
t.
We will now carry out the quantitative solution for ABv assuminS that saSSing bending moments ate posltlve' The load moment is constant for the '&' real bending moment in the expression and consequently does nor co nta in an 'x'te rm .
ELAq-ftMALX.)q j
The or igin is t ak en at B s o t h e
expression for the n moments is I x ,x.
[to* 'L'1' o L
.
)o
MAb,L' Z)
I
tl 8
UNDERSTANDING STRUCTURAL ANALYSI 5
9. .L
| (7l - J'/\'' o
* og ,
r
.lL-| 5t-
7-.\ aa' z-1 L
!l J. It
The figure shows the quantitative solution for 6Bv.
The
We could have lound this directly as this loading case on
plnnec
a cantilever is one of the standard cases. The dellection at the end of a cantilever loaded with a point load =
the
knorr--:
w13l3zr.
7
lO.
The final stage in the solution is the equation of compatibility Jor the deflection at B.
As,* Vs,$r-o
to r.l "+ V t . ! - - o
Note that the
neSative siSn of the solution for vB indicates that the choice o{ direction for the real unit load
Fo.
(Figure J) was incorrect.
srri moa_
vs-- lYt
ihe
..
ln _-:,e the
v' , rL-R
T"l\ns
Thus the value oJ the vertical reaction at B Jor the
Fro:-
propped cantilever, loaded with the anticlockwise
t.s_|
moment l, Ab is vB
(312\.M
be
Ablt , T\at is actinS
dce--
downwaros.
H--5^o. n'o ur=r!^*," l)
If we now consider the overall moment equilibrium at A:
lA6
;^\w,nn.
15. ,A - ,A b + r.(312).u OOl z= t) u^= -
7a
M
6612
We find that the value of the fixing moment at A is M
Ablz.
l _-.15
l 'l 9
THE STIIFNESS METllat - :a-:-:...4 Ihe relationship between the moment applied ar.:e pinned en d, th ern ea r endr of a pr is m at ic beam , and rhe moment created at therfar
end', end A, is
known as t,he carrV aver factor,
value ;.
For c sEmgluF
prun,tir'
bau
&to u,wu- avcy 44ov uq, \J
-
Ie
72t
now need to know the relationship between the
rotation at B and the applied moment MAb.
Wewill
Cetermine the value oJ 0B by the Theorem of Virtual Forces. This theorem may be applied to a redundant structure, as was discussed in Chapter 5. The bending
\a----.=#Ab t*S va
Ml+t
moment diagram is that for the redundant structure, ihe values oJ which have already been determined. ln the subsequent analysis this will be the '&' dia8ram,
\'\Ab
ahe effect of the real loads.
From the Theorem of Virtual Forces (see Figures 5.78 5.81) the structure sustaining the virtual load need only be able to satisfy equilibrium. The virtual structure does not have to be compatible with the original support
v"
conditions. The propped cantilever may be reduced to a statically determinate lorm by the removal of the support at B to make AB a cantilever. The unit virtLlal action is
ht
E*
then applied to the cantilever with the resulting constant bending moment diagram shown. Note that the unit virtual moment is applied to the beam in the direction of the required delormation 08. This diagram shows the Sraphical summary for the integration for the internal virtual work which results lrom the application oJ the unit virtual moment at B.
zx2fJ (\{o;*",
t20
UNDERSTANDI N G S TRUCTURAL AI/A'YsIS
17.
LrzE. ICM,$L+ ,"nw.z\!. )at /' L ' -L
The ligure shows the quantitative solution ol the internal virtual work expression for OB. The origin for the expression for the real and virtual bendinS moments has
=[tgr'" -zw!]:
been taken at A, clockwise bending moments, that is bending moments producing sagging in the beam, as positive. From this solution we find that the value oJ the moment to produce a unit rotation at the near, pinned end oJ the propped cantilever = 4Er/i.
18.
ry*i"
member properties is most important and is Jundamental
r. [lAs
r\A
This relationship between the absolute moment and the
to the application oJ both the stilfness and moment distribution methods of analysis. Note that this is an
I re4tqn
:
analytical device, No real structure could sustain a
tsc--S' lve {V6=a!1-
1a
rotation as large as I radian.
MAb=4eLr,
:n,
Note the units: F lL'z,LqlL = F.L, (E)
i9.
$lLl
Similarly, if the end of the beam is pinned, it may be shown that the moment required to produce a rotation of I radian at the near end is 3ErlL .
The carry-over
23. :dlil
factor
is clearly zero, since A is simply supported.
20.
Ve need to establish one more relationship, that of the effect of the settlement oJ one end of a fixed-ended
ia:-1|r
beam, and the resulting bending moment. Note that, in this case, both ends are fixed.
Sab€l
End B settles
downwards relative to end A, This implies a vettical Jorce (but ,ot
a moment) release at B.
c ::"| s ::, nl
THE STIFFNESS IIETHOD
Fi;,!::-
t2l
We will assume that the bending moment caused by the settlement vertical
of AB
is 1.r. There is a balancing
reaction at A, equal and opposite to the
force required to produce the settlement.
B
jlA
If we
consider overall moment equilibrium about A:
.'.
v^
^M t{vg=zur,
v-.I^M
1
-M-M+V^ , L= 0 = 2M lL
By applying the Theorem of Virtual Forces to the evaluation of the deflection ABv we can reduce the lixed-ended beam to a cantilever lor the application ol the unit virtual force at B. This is applied in the direction of the Lhknown deflection, ABv, that is in the vertical direction at B.
$:HN
#*
nql
This ligure shows the resulting virtual bending moment diagram, .a?. Note that the sign of the bending moment is negative at A, that is the action of the joint on the
of--.---.t---.-"
This diagram shows the graphical summary of the integration for the internal vi.tual work to determine the value of ABv.
Note again the particular
advantage
ol the Theorem oI Virtual Forces in that the original structure may be simplified to a statically determinate form, allowing the evaluation ol the expression for m, which is now based on a cantilever.
.L
ErAB,=_f (\)(:)*
.1 2 2
UNDE RSTANDI NG STRUCTURAL EI/A,LYSTS
The guartitatjve i
Er A Bv. $ +zA,* L+x f,- )at le F
,
.
2
!fit
expressi'bnfor the bending mdment due to the virtual unit load is the bending moment at A, - r plus the
rff}
llrl
of the reaction, I upwards at A.
.lMLx,- 3Wt + zlltt"- bI ,
Thus the
is shown. The origin has been taken at'4.
,.
Tk rl
solution for the internal virtudl work
3LL
MLZ 7
t9
26. Thus for a settleirent oJ A the resulting bending moment
30.
fl
is 6ErLlL'?,
#+A
#
Note that this negative (i.e. downwards) deflection will create a negative (anticlockwise)
"\w,
settlement
u= 64 L/LL
deflected shaDe.
A and B.
!d
{fl
bending moments
*
However, the siSn ol moments due to d*
should aTurags be checked against the
resulting qualitative
=ry
bendin8 moment diagram and the
14 dr fll fl
If one end ol the beam is pinned and there is a relative
rfrr'
vertical deflection or settlement then it may be shown
rdl
that...
ll.
frd :rd' dl (fi F6r rd 1!lI!
28.
t^lliillllnnn',,,* , M = 3LE A
Lz
...the relationship between the settlement deformation
dft ,d
and the bendins moment induced at the fixed end A is:
32. t-' fld di{
123
,fHE S?IFFNESS METHOD-
'?-!.U:: --e stiflness analysis Procedure will now be illustrated !-l the frame ABC' loaded vertically in the beam BC the point load tt which is central in the sPan' :1e height of the column AB is equal to the sPan BC' r-Jl
Ie will show that the stiJlness analysis method lor this 5iructure has only one unknown' the rotation at B, The €noring the ellect of the axial strains. s;gnilicance ol this assumption will be exPlained later in :Ie chapter. lJ this rotation is determined, the lull :;tribution
of bending moments may be found usin8 the
:elationships between bending moment and rotation' rendin8 moment and deflection developed above' This jnknown rotation is known as a'degree of freedom' rhich defines the number of unknowns to be found in the sriffness solution. This is known as'kinematic' rstatical' indeterminacy indeterminacy as compared with used in the flexibility method of analysis' of -qxial deformations will Sive lurther 'degrees freedom'. However, for the time being axial delormations are ignored, column AB.
We will start by studying the
Point B is fixed in sPace, but free to rotate'
t
l'116= 4.486 L .trl A L
For member AB, we know the relationshiP between the bending moments at A and B and their values in terms of the rotation Og. Bending moments are induced at each end, 4ErOB/i at thernear'end, 'carried over'to
end B' and half is
the far end' end A.
Now we can look at the effect of the rotation 0 on member BC. Simitar to the column' bending moments are induced at the'neartend
B and the lar end C'
29et Pd4
Mr=ofet
r
UNDERSTANDING STRUC?URAL A]VAI,YSJS
33.
To consider the efJect of the load, we will start by assumin8 that joint B is prevented Jrom rotating. There will, therefore, be a fixed-end moment at this artificial
I
The nex: suJficie.: Note
restraint due to the application of the applied load t/.
members
The bending moment at B is equal to this artificial
radian,
restraint, just as iJ end B were fully Jixed.
Se+ V cf+
34.
The linal bending moments in the beam BC will be the
,8.
sum ol the effect ol the rest.ained joint B and the effec: of the rotation 08.
tvO
r-d ==
\*/
This in the
This is another example of the
rolalroc
principle oJ superposition. The application of the real
bendinE
rotation 0B releases the artificial restraint at B.
membea-
This will produce the true condition of equilibrium at B,
m o m e n ::
i.e, no external restraint and a rotation oJ 0B.
joint or externdta distribL: Because tnducec
35.
This figure shows the distribution of the final bending moment. Note that the bending moment at B must be
To
the same Jor both members. The only unknown in this
are lne
analytical procedure is the rotation at B. This then is
DOIn rr
the object of the stiffness method ol analysis. joint a--r means I radia-'! 36.
We will now carry out a quantitative analysis of the frame ABC, previously studied qualitatively.
36
artificially
R,
n
We start b-r
restraining the rotation at joint B. The
momenr required to prevenl the roratjon at B, R, is
\ r-[
In fac: f,oint B
tr'z/8, assuming that the point load is in the centre of the beam BC.
Z---trL/A
40.
Because both ends of beam BC are fully
the ar:
lixed against rotation, the moments are known as i ' fixed-end moments. This distribution o{ bending
applLec
moments lor a fixed-end beam is a standard solution,
09.
\\hich may be proven by rhic and rhe subsequent
at the
"-"1-ri-.1
maih^,-1.
rolalroa
THE STIFFNESS METHOD_ FRAXES --e next step is to aPPly a moment tr'
t25
at Bt
, -::' cienr to cau se a ro t at ion ol I r adian aT joinL B' .,::e particularly that this means that ends B of botn -.::lbers AB and BC have underSone a rotation of I
lredqln
-- s diagram shows the distribution o{ bending moments ; :re frame due to the application of 'r' of I radian at B.
-::.rion
and the
You will note that the
:F--ding moments at B are on different sides of the They are in equilibrium with the aPPlied
-E:iber. T:.ient
r, that is they are shown as the action of the
:-rr on the member, i.e. in a clockwise direction.
The
:!.:ernally aPPlied moment r is balanced by the r.:ribution into the column AB and the beam BC' i:-aause the rotation at B is one r:adian the resultlng ::..:ced bJnding moment in each member is 441/l' -: simplify this calculation it is assumed thai the spans :-_erhe same, and that the Er value is constant for :'::1 memDerS. is known as the (absolute) stiffness of the :--: and is the sum of the stifJness (remember that : totaf
f = 4* bum LL
t 4ZL zolunn
the moment required to Produce a rotation of -::ns ::dian) of each of the members framing into the joint'
i:act
the artificial restraint at B does not exist.
:a:rt B is free to rotate to moment equilibrium ::.sistent with the stiflness of the members' Therefore artificial external restraint at B induced by the -e
load ,f is equal to the moments induced by the -)lied -.:ation of I radian multiPlied by the actual rotatron :.,.
This is the equation of equilibrium of the moments
:: rhe joint.
h,e,u4
Jot.JF' rLsit^rY,o
=O
R, + r. Dg = o US-- oy
r&cf,tcn.
DI N G ST RUCTURAL ENA LTSTS UN DERSTAN
lrl.
R-=- w?a
action of the ioint on the member)' the artilicial restraint R = -wLl8' that is in an anticlockwise
y . 8fr4 - wLyr+s,rEC OE:o 0z = NLz
'dezJ
lf we adoPt the normal si8n eonveiltion of clbckwise moments Positive' (for bending moments, that is the
direction. The total absolute (i.e, Producing a rotation of I radian)moment lor both membersBA and Bc = 8EJl'' Thus the rotation at B may y'e determined: /
Ydtlqf
wL2
This is a positive value' indicating that the direction assumed for the action of the absolute moment r at B correctly
42.
defines the direction
of the real rotation
0B'
The actual value ol the rotation at B may now be substituted to determine the final distribution of moments in the structure, first the column AB' The bending moment at B is equal to the rotation at B multiplied by 4Erl,
which equals
r/16. Half of
end this bending moment is carried over to therfarr end'
43.
The linal bendinS moments in the beam are the algebraic sum of the distribution ol bending moment resulting from joint B bein8 artificially
restrained
against rotation and...
44.
EI ..,the bending moments due.to the.rotation ol wL2164 at B.
9.^\
0r'- Nt2 - '6+tz
w[fu/wvt"
UETHOD
127
-:i:-:-Y::
:re final bending moments in the beam BC are the s,JrrrJ the restrained and ireleasedrconditions. This is another example of the principle of superposition.
-wL/s * n% N
trl *ta /ll .'qr'
=-"2""ff'.s*2"
!6.
This figure shows the linal lrame bending moments. Note that the sagging moment in sPan BC may be found by superimposing the 'lree' bending moment diagram over the restraint bending moments at B and C.
47.
Had this problem been solved by the Jlexibility method of analysis, the first step would have been to determine the degree of statical indeterminacy. The frame is I x ind ete rmina re. ! ( e c ould r educ e it r o a s t at ic a l l y determinate form by the removal of one support, three reactions, at C. Remember that the degree ol indeterminacy is the number of unknowns in the flexibility method oJ analysis.
48.
Applying the sti{fness method there is apparently only one unknown, the rotation at B. Howeverr we have ignored the axial deformation in both the horizontal and vertical directions,
There would be a large axial
lorce in members AB which would cause it to shorten. Thus point B would 'settle' relative to Point C' inducing the bending moments shown' In addition member BC is also in compression, which would cause it to shorten. These are, however, of a second order when compared with the primary bending moments and may usually be ignored in hand calculations.
T=*r
UNDERSTANDI NG STRUCTURAL ANAIY.SIS
tl9. Now we can see that there are .n.ee unknowns at B; one rolatlon and two deflections. Indeterninacg
in
the stiffness
"= :.nr!!ul a
nethod
This is identified by the degree of freedom at all unrestrained joints and is known as kjnenaxlcaj (movement) indeterminacy. As has been demonstrated aDove, lnere are always th.ee degrees of kinematical rnoerermrnacy at any joint.
Cenerally, in rjgid_jointed frames, however, the axial deformations are ignored rn the manual analysis of'plane frames, leaving only one, the rotation.
I h/
r
All commercially available computer programs, however, include the effects of the axial shortening.
A€lF-
f-J*
50.
The effects of novenent at a joint ( i.e. relative to a support) are always included. If we now introduce a horizontal and a moment release at C we create a further degree of kinematical inderermlnacy, the
54.
55.
horizontal deformation, which will change the distribution of bending moments, This unknown deformation ACh rnust be found to solve the frame. We would still ignore the axial shortening of AB, however the sway of the frame at C will produce bending moments oJ the same order as those oue to the rotation oJ joint B and must be included.
51.
In order to iliustrate the application of the stiJfness method to more complex structures, we will study this three-bay frame. The joints are numbered l_7,
52,
Step l.
The first part of the analysis is the identification of the degrees of freedom, the
degree of kinematical indeterminacy. Ignoring axial strains, they are the rotations at l, 2 and 3 and the horizontal deflection of member l-4, identified as the horizontal deflection at 4. Because we are ignoring axial strains, it is assumed that there is the same norlzontal deflection at j o i n t s l , 2 a n d J ,
56.
t29
TEE STIFFNESS METEOD- FA|-U.'S Step 2.
Each of the degrees of freedom is now notionally
.estrained. 'fixed'and
Thus the rotation at joint 1, 2 and 3 is the horizontal deflection of joint 4 prevented
by the introduction of an artificial horizontal reaction. Rr to Ra are the rclampr forces required to prevent rotation and deflection.
Rl to R3 are moments, Rq is
a force,
tr.
Because of the artificial restraints, there are unbalanced bending moments at each oJ the joints. The bending moments diaSram shows the characteristic discontinuity at joints l, 2 and 3. The difference between these moments must be balanced by the artificial restraining mome nts, Rl, R2 and R3.
))-
This out-of-balance or restraining moment (or Jorce) is taken by the clamp.
If, as shown in this example' l42r
is less than lr23 then the clamp moment to hold joint 2 in equilibrium would be in an anticlockwise direction' in order to balance the tendency to rotate this joint in a clockwise direcrion due to the effect of the larger bending moment M23.
t6.
Step 3. This next step in the analytical procedure is the release ol each of the restrained joints j]] turn and the application ol the moment required to induce the rotation of I rad ian . r,r i s t he m om ent r equir ed at join t l . Note that this rotation at joint I will introduce a horizontal reaction, r:qt, as well as a carry-over moment at 2. The subscripts are interpreted as rrr, moment induced at joint I due to a rotation at join t I, r2r, moment induced at joint 2 due to a rotation at joint l, rqr, force induced at joint 4 due to a rotation at ,oint I.
Rz Q <{zz
T
t3 0
UNDERSTA.IIDINGSTRUCTURAL ANALYSIS
57.
This dia8ram shows the distribution of bendinS moments due to the application ol a rotation ol I radian at joint
r4
'I
l.
r.{+
The horizontal reaction at 4, r:q1,is induced by the
rotation of I radian at joint I and is equal to the horizontal reaction at joint 5. The general expression
I
for the value ol these balancing reactions was
+k
determined above in Figure 18.
' *ri' J8,
Joint 2 is released and a rotation of I radian induced. Bending moments are carried over to joint l! .r 2 , joint 3: r:z and a reaction, -ra2 is created at joint 4.
59,
This fiSure shows the distribution of bending moments due to a rotation ol I radian at 2.
4E+t
Note that because the
far end of member 2 is pinned, the bendin8 moment at 2 in member 26 is 3Er2lLz.
Similarly, the horizontal
reaction.,,2, equal to the horizontal reaction at joint 5, is that lor a member with the far end pinned.
50.
This figure shows the release oI joint 3 and a rotation of I radian.
: t_ lllllr, I ![r
'rTIE STIFFNESS METHAD- FRI]!':
lll
Bending moments fuom the rotation of joint J and the force induced at 4, .,,3, are shown. Because the far end of joint 4 is pinned, the bending moment at 3 in memb er 34 is 3E r 5lL 6.
(q3
fl.", TR+ Le '2tL;
6a.
1qg..9,
The last release is the artiJicial horizontal restraint at ru u is that force which will induce a unit horrzon tal displa c em ent at 4. 4,
53.
This ligure shows the bending moments induced in columns 15, 26 and,37 which are similar to 'settlementl moments.
See Figures 20-28 above. Note again the
eflect of th e pin at t he' f ar ' end
64.
ol m em ber 2 6 .
We may now establish the eguations of equilibrium, Jor the displacement at joints l, 2, 3 and 4. ClerkMaxwell's Theorem of Reciprocal DeJlection applies as it did in the evaluation of matrix elements for th e lle xib ility met hod.
r - 3 = r lr and s o on,
P,+ r, 31+r1r02+ \z9a + A+=o ,q+ P, + ro0, + 6"$ + 6a07 + r24Nt- o 23 + q,9t + ru02 + l9r0r+ fn 4.
o
P*' qtq * 4z0z, rq1, . rn N=o
t32
UNDE RSTANDI N G S TRUCTURAL ANALY S I S
6J.
This may be expressed in a matrix form where (r) is the sfjffress matrix. It will be noted that the elements oJ the stiffness matrix are the forces o. r?o,?erts induced at the various artificial restraints (or degrees of lreedom) by each unit displacement
-i-l [,]Lul-
aPplied j,
tu-rn.
f,
These moments or forces aae
functions of the structural dimensions and material properties only.
66, Fora gvea
strqAwo
caa loql,
A .roi{ttrh tF \46r,n&l a^d
At this point in the explanation, it is usejul to summarise and compare the analytical procedure for both the stiflness and flexibility methods, The output oJ the flexibility method is the unknown forces, whilst the stifJness method produces
A srkrfrdr
defornations.
You should remember that the
analytical results lrom each are identical.
Neither method of analysis is preferable on theoretical grounds.
FLEXrBtLiTy
67.
We will now carry out a quantitative analysis of the prevlous example, and, for simplicity, we will assume that r values are constant throughout and the lengths ol beams and columns are the same.
68.
We may now draw the artificial restraining moments at joints l, 2 and 3, The Jixed-enq momenrs que 10 a uniformly distributed load are wt2l12.
Note that Jor this particular loading arrangement the restraining or ctamplng moment at joint 2 is zero since the fixed_ end moments from members either side of the
&= uL2- w
PZEO
Pa! tL Y
-3ty t9
joint have the same value.
-1. ]t ,l ?
t33
THE STIFFNESSMETHOD-
'-'2-U.'.' The artificial horizontal reaction at 4 is equal to half oJ the horizontally applied unilormly distributed load' At this stage in the analytical procedure ioint 2 is clamped, Consequently the distribution of reaction forces and bending moments is that for a lixed-end member.
FTTJ;
The next step in the analytical procedure is the application of the absolute moments in turn' The bending moment required to induce a delormation of I radian at joint I is the sum of moments requlreo to produce a rotation of I radian in members l5 and 12. There is a carry-over to joint 2, r2r' No bending = moments are induced at joint 3, consequently 131 Q' rqr, as above, is 6etlz2.
4,= 4LT7r+4*7, h.
zE+L
/3t o -
41. - cfr1Sz
This figure shows the bending moments and the horizontal restraint at 4 due to the aPPlication ol a rotation oJ I radian at joint 2. By applying the Theorem of Reciprocal Deflections, we can equate rt2 and 12r'
/L= r'zt ru. 4L* rO= -
t7; N1
*tl?u,. 'IJ
3'Jr,
eu
Bending moments and horizontal restraint at 4 due to the unit rotation at 3 are illustrated.
a" '.9 = n.t
+ 4EJ/.,,2 = n. .a '€J/ .L __
4z= o
r4^= -6=/,
l oE4
I
z
-t'"
t34
UNDERS TANDI NG ST RUCTURAL ANALYSI S
73.
73
Bending moments due to the unit deflection at joint 4 are shown. The value of r+r is that lorce required to produce a unit displacement at the end of members lr, 25 and 37. The expression for the value of that force was developed from Figures 2A-26, The reaction resulting
44=14 rn= 4z r+. r+l
from the settlement force equals 2talL (Figore 21)
'* ''T* E . t T
which substituting for ff from Figure 25 eguals l2EIlL3
74.
lo2
*r"
t|
-"ZtE
t-
7Lm
this ls a particularly crucral stage in the analysis,
r r
check each element carefully against Figure 54 above.
b
unit actions are the elements of the stiffness matrix. As
o
ttEE1L2a/L-3EElC
'a/'
These values oJ the bending moments and lorces due to
'1D
rifr/r' eztl*
'n/,
lI o
{D 0
-3f/. -6E | -cgr, .- 'L 'E''Lz L'
Z1EE,
* !
75.
f-r1 la1=
a
expression are the restraining moments and reaction due
| z1
to the applied loads, This set of four simultaneous
,)
equations would be solved for 01,02,0:
P -w1 =fI w E F I
-81
1S
and Aq.
These deJormations would then be substituted back into the expressions for the bending moments at each oJ the
lol
I - wtrrt I |t-l - uLt.
The elements in the column, the right-hand side of the
tl
points of kinematic indeterminacy in the structure.
I
-. llil
76.
lo 0
At this sta8e in the explanation of the stiffness method it will be simpler if we use a numeric solution. We will substitute loadinS and dimensions lor a three-bay lrame. The EI value is assumed to be J0 MN.m'z units.
EL = 5O , AN, I . ' L
THE
STIFFNESS
ME?HOD
135
.:.i.!,VJ'J
We are now able to set up the stiJJnessmatrix by reference to Figure 74 above. The units are MN.m.
7L
- lt J5
roo
2-5
t-5
t375 Z5
o
Ls
-ttt'
-13J5 -ttJS
0
-1t75
87.5 - ttts 2't a4+_
The solution of the four simultaneous equations gives us the values of rotations at nodes l, 2 and 3 and the horizontal delormation at node 4. Note particularly that the sign ol the solution to each of the displacements is a conlirmation or denial of the arbilrary choice of the direction oJ the unit rotations at 1,2 and 3 and the unit
Et. ' aooA4l( izc-,et'oo478r 0g- .a.ez4, Aa= 3,tJ6 na
displacement at 4. Thus the rotation at joint 2 is actually in the anticlockwise direction, consequently the solution is negative.
T'.
It is interesting to compare these results with those obtained from a computer analysis. The output of nodal (GLOB A L A X TS ) N OD A LD IS P LA C E IIE N TS
joint displacements shown is taken from a computer program based on the stiffness method which takes into
Z D D ILE C TION R OTA TION N OD E X D E F'LE C TION (nM) (R A D S ) (Mrl )
account the axial sttains. You will notice that there is
I 2 3 4 5000
a dilference between the hand and the computer analysis abo ut rh e se co nd o r r hir d s iSnif ic ant J igur e ot t he
3.181 3.\78 3.175 3.175
0.000844 -0.005 -0.014 -0.000048 0.000623 -0.016 0 -0.001318
solution for the delormations. This difference is due to 0 .001216 0
the fact that the axial strains have been iSnored in the hand calculations.
80.
We can virtually eliminate the difference between the hand and computer analysis by inputtinS an artificially large value for the area of all the members. This computer solution is based upon a member area of 150000 x 103 mm2 compared with those shown in Figure 79 above which is based upon an area of 150 x 103 mm2. As you will see, apart from the difJerence of I in the fourth signilicant fiSure for the horizontal deformation at computer node 4, the manual and computer solutions now a8ree.
8o (GLOB A L AXES) N OD A LD IS P LA C E ME N TS ROTATION (R A D S )
NODE x DEeLBCTIoN z DeFiEcTIoN (N t4) (MM) I 2 3 4 5000
t,175 3.175 3,175 3,rrs
0.000841 0 0 -0. 000050 0,000624 O O -0.001312
0. 001216 0
t36
UNDERSTANDING STRUCTURAL EA'E'YSiS
Sl.
"N
combining the ciamp moments with those from the
t]{\-rf 'it
The final distribLrtion of bending.moments is a result of
3'rfrt 64 '0.o4! | 25€7 14L=1z.oS ts.g&z
t7 t
application of the real deformations, _
The iinal distribution
of bendint
moments is the algebraic sum of:
8?.
.E
We will look at the
of bending moments in member lJ to clarify
the procedure.
tii;l
fr.' .6 .tt4 r uLz E .4/ --
distribution
l.
restrainedmoments,
2.
clockwise rotation at node I,
3.
horizontaldisplacement.
This figure showsthe final distribution of bending moments in member lJ with the effect of the free bending moments resulting from the application of the uniformly distributed load, superimposedon the 'restraint' bendingmoments. Ptactice
Problens
You should now practise the application oJ the stiffness method of analysisto numerical problems. As with the previous chapter, you should choosea structure from the range of full computer solutions.
8 TheStiffnessMethod- Grids
So far in this text, we have been almost exclusively concerned with the analysis of two-dimensional frames' set in the vertical xz Plane' All real structures ate, of courser three-dimensional. However, most structural frames, particularly those for multi-storey structures, are resolved into a series of sub-frames which are set in this two-dimensional xz plane. Consequently' this academic emphasis is a Senuine reflection of the day-to-day activities
of the structural
though, it is necessary to desiSn a structure nomaf
Occasionally
Plane. This is defined as a 8rid. The same approach is emPloyed' the stiffness method.
10 the structural
basic theoretical
engineer.
capable of carryinS a load
; 13
IJNDERS TANDI NG S TRUCTURAL ANALYS I S
l.
We have studied the analysis ol this cantilever bent ABC, loaded at C from a number oJ different aspects.
J.
'ti' "r_fl pr llr [i
I
2.
IJ we now introduce an additional member, CD, in the third dimension v, a new stress resultant is introduced, the torsion in member BC. That is the most signilicant
a
difference between plane frames and grids, the torsional
fi
stifiness oI the member is normally included in the
I
resistance oI the grid structure.
t
Consequently, there is
a rotation about the longitudinal ( x) axis ol the member due to the torsional effects.
3.
The stress resultants oI axial, shear force' bendin8 moment and torsion are identilied in relation to the ne.nber, rather than the global axes. The member axes will be identified with lower case symbols' There is the bending moment about the major axis 149. This is the same as the bending moment in the analysis oJ plane Jrames. There is a shear Jorce in the
Mehbcr 4x!!
z direction and
a new load elfect' i.e. the torsion about the x axis, lYx.
4.
We will now study the qualitative analysis of this cantilever bent frame, ABCD.
The vertical load at D
produces a bending moment at C, My, that is about the nenber g axis.
In addition, there must be an upward
shear force at C in the member z axis resisting the downward load at D.
?AE S?IFF]VESSMETHAD' ::e hat
139
"::: shear Jorce in CD at C is equal to the load 2t. shear force is transmitted to member BC as a poinr
.ad at C which in turn produces a bending moment, about ::e
i' axis of member BC.
-:]embers CD and BC. !-,
The shear lorce is constant in
The bendin8 moment in CD at C,
is resisted by the torsion in BC, which is constant
=roughout the member.
6
joint B, the moment about the 9 axis in member BC, t- is transmitted into the column about jts 9 axis. li
The shear force at B, is resisted by the axial force in .1B. The torsion in BC, &x, is resisted by the bending moment in the column AB about the z axis of that member.
M!
147
This figure shows the final shear force and axial load diagrams. There are no axial forces in either BC or CD since the loading is normal to these members.
.ffirt'k" ^ki)t ^Ki) &qr
!-
4z.tel
The moment reactions at A resist the l4qand Mz moments in column AB.
There is no torsion moment in
AB since the only load on the structure is parallel to the axis of AB.
For there to be a torsion in AB, there would
have to be a load at an angle to the global z axis.
Pzaettottl
Ea/fis^
UN DERSTANDING StRUCTU RAL ENAIYSTS
140
9.
The deflected shape may be built up by a consideration .of the resulting extension oi the fibres in bending tension. From a vertical start at A the bending tension will deflect the column AB in the positive direction ol both x and v axes. Since there is no bending about the z axis of BC or rotation of AB, point c will deflect directly downwards in a plane Parallel to the xz plane. Similarly, there is no horizontal dellection of point D' other than that induced by the horizontal dellection of point B.
PLacti ce
P robl ems
The example above and these Problemsare true structuresand not simplegrids which three-dimensional
N txia
are the subiect of this chaPter. However, it is this new torsional stress resultant which will present problems in developing an understandingof the behaviour of grids.
'.?
Practice in the qualitative analysisof those three-dimensional cantilevers will help to prePareyou
EA,
for the analytical procedure which follows. Dra'v the qualitative diagrams for each ol the stress resultants.
ll.
The stiflness method may be applied with equal facility to the analysis of Srids' although the aPpearanceof this type of structure on the desk of a design engineerwould be a very rare occurrence. ve will develoPthe stiflness approach with the frame 12345'set in the hotizontal x v plane. This grid is fully fixed at supports 2 and 3 and pinned at 5. Member 14 is a cantilever loaded vertically, that is parallel to the z axis at l.
The presentation will be clearer il the structure is reduced to a line dia8ram, however you should always bear in mind that the structure has a three-dimensional reality. This is particularly helPful when considering torsioneffects.
THE STIFFNESS I'|ETHAD- C..:: __€.e
aae three degrees oJ freedom, three degrees ol
xr-ematical indeterminacy. :e
l4l
The rotation oJ node 4 in
I and v direction and the vertical deflection of
-L.
If we look at member 245 in elevation there will
L
:e r verrical deflection at no.lc 4, A.z, and a roration :.: tu)de 4,oqy
about the y axis. This is likely to be
. ihe clockwise (i.e. positive) direction.
The axes
5lrown are the global axes and the deformation will be
t-.*
determined relative to these axes and not member axes.
Taking an elevation oJ member 143 the rotation at node 4 about the x axis, oqx, is certain to be in an aniiclockwise
direction,
The first step in the stiffness analysis oJ grids is the same as that for plane frames, the notional restraint and identification ol the degrees oJ freedom.
The kinematic
indeterminacies are given a numerical reference. is the rotation about the Y axis, is the rotation about the x axis, is the vertical displacement. The symbol rRr is the restraint required at the clamp to prevent rotation and deflection at node 4.
With node 4 clarnped against rotation and vertical displacement, the point load ,{ causes a clamp moment R2 and a clamp force R3. This particular loading arrangement does not induce a clamp moment in direction I about the Y axis.
Pt =a
t42
UNDERS?A.IIDING STRUCMRAL
17.
?,V,q'YSiS
The next stage in the analytical procedure is the application oi moments and forces in the direction of the degrees of kinematic indeierminacy to induce either a unit roration or unit deflection, as appropriate, This 'unlocks' the clamp in direction 1, but leaves the clamp in directions 2 and 3. rrr is the moment required to cause a rotation of I radian in direction l. rotation about the Y.axis.
18.
]"
For the first time in the application of this method of analysis we must take into account the torsional resistance of a member, here member 34. This is the diagram ior the bending moment and torsion combined. Note that although member -[4 rotates under the action of r11, it has no stiffness as it is a cantilever. Consequently,this member cannot sustain a torsional
<:
moment. The application of moment ilr induces a reaction in direction 3, r3r, but has no effect in direction 2, rotation about the X axis. Consequentlyr2r is zero, 19.
This figure shows !'22,the moment required to induce a rotation of I radian at node 4 in direction 2, that is about the x axis.
20,
Member 14 will rotate freely.and since the far end of member 45 is pinned (i,e, free to rotate), this member has no torsional stiffness.
r22 vill
be the sum of the
moments required to rotate member 24 about its longitudinal
axis, that is in torsion and rotate member
43 in bending. rr2 is zero.
143
,IEE STIFFNESS METHOD_ CRIDS ,- : :s lhe Jorce required to cause a unit displacenet-.a: :rode 4 in the z direction.
This produces bending
]rlainents in members 24, 45 and 34. Remember that s-j-r rhe vertical restraint has been removed leaving :re rotational restraints, the notional 'clamps' in c.-rections I and 2.
This upward movement does not
Geate torsion in any member since the members of :E
grid are at ri8ht-anSles to one another.
le
rill
"ffi=
now produce a quantitative solution by providing
.a r-Jre for the point load and dimensions for the E_.riuaal members in the 8rid. Er is constant for &a inembers at 40 MN.m'z units. The product oJ the Ersional constant J and the modulus ol riSidity G is t: lt\.m2
units. I
.' f a=
k h
lrN,r* Mv,tr.z
The lirst step in the analytical procedure is to clamP -*le lree node, node 4. The resulting clamP resistance
z
e ill be:
l/r' R,
Pt= o pr= - Lio
and R,
the vertical reaction - +j0 kN'
Note that the siSns are in relation to the overall globa1axes. Note also that the elfect of the clamp on member 143 at 4 is to isolate 43' as if l4 were a cantilever'
-\y
about the x axis = -250 kN.m.
lully fixed at 4.
* j. + s o
UNDE RSTANDI NG SZRUCTURAL ANALYSI S
t44
24.
\ve now apply the unit actions in the directions of the degree of kinematical indeterminacy, the de8rees ol
n=++,(z+) L
freedom,
+ 3EJ: L4E) L.
+29 C#)
L= 95.8
Remember that directions 2 and 3 are
clamped when rrr is applied. rrr is the moment required to induce a rotation of I radian about the v axis.
tr U.h.
Consequently, each
member at joint 4 must rotate I radian. rrr will now
74- e
include the torsional resistance of member 43. rrr is applied in the positive direction about the v axis. The torsional rotation
of a straight bar = JGIL where J is
the polar moment of inertia oJ the section and G is the modulus ol rigidity.
strength of materials for the proof oi this relationship.
(24)
A.= e4
r2r would be the result in direction 2 of the
L2-
application of rrr.
_ 3 4 tu a l
rrr has no effect in direction 2.
Consequently r2r is zero.
L: rq .z M N
-
Refer to any standard text on
25.
The effect of this rotalion of I radian in direction I has an effect on the vertical rclamp', r31. This reaction is positive for the bending moments induced in member 42 and negative for the bending moments in 45. Reler back to Figure 18, Chapter 7, Jor prool r hi .
^f
ti-
r F I.+i ^n
4ZE
aaal T \-/
-
26.
+ Jg (2+) Z'
r22 will produce a rotation of I radian in
direction 2. Member 4J has no torsional resistance as the far end 5
= 5a. G Mu.q
is pinned. There is, however, a torsional resistance in
o
fz-
The moment
24,
122 has no elfect in direction I therelore
.r 2
ts zero.
27,
This Jigure shows the effect of the unit rotation in direction 2 which is to produce an upward reaction in
rt€EE(+)
t: -
15M^l
direction 3, r: z.
TI{E STIFFNESS ILET;""
2E.
t45
:', J-:
The force 133 is that required to induce a veai:a1 deflection
of unity.
Remember that joint 4 is dan-.Jec
against rotation in directions I and 2. r:: 'settlement force'.
.rr= |ZEE8B)
is a
See Chapter 7, Figures 20-26 Jor
prooJ of the relationship
between the deflection,
I tzEE (+z l
bending
+ 3ZF GE)
moments induced and the value of the force required. For a member with both ends Jixed, the force lo induce
= 27 o6 Mrt
a relative settlement of unity is 12ErlL3. ll the lar end is pinned, the force is 3Er/L3. The value of rgr is 27.08 MN.
The associated elfects ol inducing an upward
deformation of unity in the two other directions do not need to be calculated.
Clerk-Maxwell's Theorem of
Reciprocal Dellections provides them lrom previously determined values.
We may now set up the grid stilfness matrix' unknown rotations and clamP lorces.
10.
\, ',lftl [' r,, va?llsLl= 'l
a rl
I
ezl -J
I
The values ol the elements of the stilfness matrix are now substituted lrom the calculations above. Note that the calculaled clamp forces are in kN units whereas the stiffness elements are in MN units, Therefore the clamp
qs.t
o
O
56 6
t1.z
t5.6
torces have been divided by 1000.
il.
-l
In, L', ",' .'.l L4l
"ll f' I "']
.-tI ',-l :."il e-l=-l-€".1 l 5a
The solutiqns to this set of simultaneous equatlons are the three unknown deformationsr 0rr 0z and A:. Note the siSnificance ol the sign of the de{ormations' The directions of apPlication of the actions 11I , rr 2 and .3 3 are in the positive direction in each case. The positive sign in directions I and 2 means that these are actual directions ol the deformations. Clearly the deflection at 4 is downwards, consequently the siSn is neSa!rve.
ot '
t2* to' ?
?2.
ts 6,o \ lo- recs
Lz' -
6'o2-
I
L 1;3J
-J
ral s
a*t
-
t46
UvDFDST4VDTNCST PU - U D A4\t4'w .- :
32.
Always draw the final solution for the deformed
36.
srruclure, Ir will be apparent if there ts a grosserror 'n t h e 'i p - o r v A l , r e . { r L e d c l o r m a t r o n s . Y o u w i l l I i n d when you practise the solution of similar problems that 6.oz,n6l
the solution of the simultaneous equations seems to be a .f arr^r
A s k e T c ho . t h e s o l u t i o n a t
this stage may help to avoid unnecessarycalculations Jor the distribution of bending moment and torsions.
33.
The solution is completed with the substitution of the deformations into the expression for the bending moments and torsions lor each member. We will look first at member 24J. The eJfect of rotation and
61 t0)
deformation have been separated. Remember that there is a carry-over factor of ; Jor the rotation and I for T h e d i f f e r e n t i a l d i s p l a c e m e nt .
r 6i N + N
!'g\.-ii
.*
5o2-tc1
(A)
34.
This figure shows the effect of rotation and settlement in member 45.
la)
3 '34 .lz .3a f_3 , 6.oz +2
=+5
/i\
WtrF
( e)
WEF
(^)
+
This is the final distribution of bending moments in member 245. The ditierence between rhc bending moments at 4 is accounted for by the torsion in member 43. The difference of I kN.m is due to errors in rounding up.
Tors!"h 43 = 5a, tz-
+
I5
?t
THE S'|IFFNESS METEAD,
GRIDS
t47
_' j ::i-re sho\rs deformations in member 143. Bending -:-_:'==
:n member l4 are unaffected by the
@rr-E::ons
of the grid, as it is a cantilever.
-iE ij.tre
shows the separate effects of rotation and
djs-*.-n
in member 43. The plotting oJ these diagrams
3i.,imr.!: :lq-ays be related back to the structural tEr_:;rions
to check that they are drawn on the
:E'Tsr:r: side of the structure,
4,4o. 6
ll\*
e =z+"_l,lil'lr. o a
r=rs':r: on the top oJ member 43 at node 4, Similarly,
\,r"
,
For example, the
::br:{-r:-:e rotation of 0.005 radians will produce bending
::nr ::.r:r*ard
rc)
T (A )
,n, q.
6.4a.6..2 ' = tlo
dellection at node 4 will produce bending
:r!ifs-!:r- on the underside of member 43 at node 4.
-ie:-iE
bending moments in member 143 and the
rar-Grc:g torsion in member 24.
Torsir
e1 . S 2r_ a | oO
-
:-.E:r.nplete E-c-:E
diagram of bending moments, that is
moments about the q member axis, of each
lAg
manber
axcs
UNDE RSTANDI N G S TRUCTURAL EAIAI,YSJS
40,
lul1,
Torslonmoments.
t,crssvt
I
41.
PracLice Ptoblens Carry out a Jull qualitative analysis for the diagrams
,P€4
which are relevant to each stage of the analylical procedure. Typical numerical solutions from a compu:analysis are included.
qqI
9 MomentDistribution
The method of Momeht Distribution
was developed by the famous. Amencan engineer Hardy Cross and was known Jor some time as the Hardy Cross Method. The impetus for the development of this analytical method was the reinforced concrete frames in the l93Os. Prior to that time' larger comprex structures had been built in structurai steelwork. The nature of the joints in steelwork and the ease with which effective and reliable hinges may be introduced meant that the real structure could be resolved into a series of statically deteminate sub-systems with relative confidence. In reinforced concrete, however, the continuity which creates the problem for the analyst is inherent in the system and to destroy it, as designers using precast systems tend to do. is to reduce substantially the real, uncalculated strength of the structure. Consequently, the development of modern structural analysis was born out of the practical need of the structura.l designer. The method of moment distribution is based on the relationship between deformation and moments which was developed in the earlier chapter on the stiffness method. Indeed the method of moment distribution is the iterative version of the stiffness method. They are quite different, though, in practical use, in that moment distribution is a practical design office method of ana.lysis whereas the structural designer uses the stiffness method almost exclusively via the medium of the comDuter. increasing use of multi-storey
UNDERST ANDI NG STRUCTU RAL AN ALYS I S
l.
In the stilfness method of analysis explained in the previous chapter the Theorems of Virtual Work were used to develop the relationship between the effects ol a unit rotation or displacement and the resulting
6.l/\.
krTiil"
o*' Arca
\.r'"
distribution of bending moments and reactions. In this chapter we will employ an alternatiye
approach using
Mohr's Moment-Area method. The proof of this method may be lound in any book on structural mechanics. There are two theorems - the first is that the chanSe of rotation between any two tangents to the curve of a deflected beam is the area of lhe
Nlar
diagram between the two points under consideration. ln this diagram the rotation
069
is the change in
rotation between the tangents at points A and B. The bending moment diagram is shown below and the
M6\r & IJF ,{rrtid,t - H 4g1r4*1 +
relevant portion of the M/aI diagram in the lower part ol this figure. We will distinguish between the bending moment and t't/a-r diagrams by using a chain line lor the latter.
The secondtheorem is that the deflection between one point, here point B and the tangent to another point, here the tangent passingthrough point A, is equal to the first moment of the area of the r4/r-r diagram betweenthesetwo points,taking point B as the point ol reference lor the lirst moment. It is convenient to reduce the bending moment diagram to simple Seometric shapes,here two triangles, and the lower diagram showsthe dimensionsrelevant to the first moment of area of the MIEI diagram.
MOMENT DI:-'::-
:.::.
Unless you particularly want to srud\ :::- ::,:-::-:_-: ol these relationships using the rnomei:-ejea :--e:_ra{lt ls Unnecessaryfor those students \rho::are s_-ja:€€ the stifJness method above to repeat rhe D.oo: aE3ji. The moment-area proof has been includeci here ic: those students studying structural analysis on ceriain courses who would not no.mally need to srud).rhe Theorems of Virtual Work, We are going to find rhe relationship between the moment applied at the pinned end oJ this propped cantilever AB, which produces a rotation of I radian at point B. This moment is defined as MAb, i.e. the ,absolute' moment, to distinguish it Jrom the normal load moments. It apparent from the deflected shape of the structure that the application oJ the moment at B results in a hogging momenr aT A. Note that the rotation of I radian is an analytical device. No real structure could sustain so large a rotatron,
s*.-Jf, Mfiilliflillil ^0,
f, e will use the flexibility method of analysis, sometimes known as the unit force method, to solve inrs structure and the first step in that method is to :elease the structure to reduce it to a statically ieterminate form. This we have done by removing the rertical reaction at B. The efJect of the application of 3e absolute moment is to cause an upward deflection ar AB. The bending moment diagram is shown below.
,|
. ede
'illllll]lr**.*
I e now apply a unit load in the same direction as the .emoved restraint, The resulting deJlection is 68. \ote the value of the bending moment at A is i kN.m. \oq/ if we return to our original problem of the :..pped
cantilever we can see that the application of
:ie absolute moment produces a downwaro reacTlon at 3 Ehich must be balanced by an upward reaction at A. :--e reaction at B, yB, is the object of this 1-Jvtical
procedure.
y-_4-\a
naAs \R
\nir
n'G"ot ivl.-
{vs
pl UNDERSTANDI N G STRUCTURAL E]VAIYs'S
r52
7.
We can set up an equation of compatibility displacement at B.
The dellection
lor the
at B due to the
applied absolute moment, AB + the real value ot the reaction, vB multiplied by the deflection due to the
A e* V c.f,s'o
unit load must be equal to zero because the actual deflection
t
is zero.
at B of the real propped cantilever
Mohrrs methods will now be used to find the values of the displacements in the equation of compatibility above. Usin8 Mohr II, the displacement between the
McNvE
tan8ents at A and the tangent at B oI the released cantilever AB is in lact the vertical dellection at B. This is a particular
MAb/?J
case for the application of this
method since the tangent at A is horizontal,
A a= M A A , t ,t-t -E-z z
-
i.e,
coincident with the original unloaded position of the
M t,.t -L€" -2-
cantilever.
Mohr II is the first moment of area ol the
u/EJ diagram about B.
The dellection at B is equal
to the absolute moment multiplied by L'l2Er.
Similarly, we can find the vertical deflection at B due to the application of a unit load at B. The result is 'lhe signs for the dellections are relative -L3l3Er.
$twE
%tnsffl
to the global convention previously established.
t B =- C ? e t ' l ' 1 ' 9 t t = -LL. aeL --
10.
Setting up and solving the equation of comoatibility gives us the value for the vertical reaction at B which is equal
A s + Vs.E
u6.!*vr1-!
Ez z
rc 3u ^,l2t. r.\o
-o
'3E tr/
w- z tllh
)-.
A Dositive siRnfor the solution of
this equation indicates that the choice of the direction of the unit load in Fieures J and 9 above was correct.
MOMENT DISTRIBUTTON
lI-
153
We now substitute the value lor the vertical reaction at B and determine the value of the bending moment at A which equals-M
Ab/z.. The negativesign indicates that it is acting in the anticlockwise direction, i.e. the action of the joint on the member.
i
\^
{:^tAb h,tlAb
!-ilA
v 4rMtu/c
f---
aAa.-(-rr,rar+ N ^4" ., ) l,t A?:.
I
-
lL
The result of this analysisis to prove the first of a series of important relationships ,tuhichis that the carry-over of moment to the far end of a propped cantilevei for a moment applied at the pinned (near) end
h4'^ot
is equal to a half.
^*b
l.a
This distribution
of moments on the propped cantilever is
now slightly redrawn Jor convenience lor solution by the moment - area method, by separating the hogging and sagging moments.
cc6
29EA
ab/tz,
The rotation at B, which we know to
be I radian, is equal to the area oJ the ,rr/Er diagram
e
area MTaA
between AB. This rotation a1 B is equal to M x Ll4Er. Ab
l+.
H*ig MAt.L/+etr
This provides the second and most important relationship which is that the value of the bendrng moment required to induce a rotation of I radian at the pinned end of a propped cantilever is equal to 4Er/-1.
!=
MAb.L/+zE
M4g= aLI7"
UN DERSTAND ! NG STRUCTU RAL ANA LYSI S
t5 2
7.
Ve can set up an equation of comPatibility fo. the displacement at B.
The dellection
at B due to the
applied absolute moment, AB + the real value of the reaction, vB multiplied by the deflection due to the
As* Vb.tg' o
unit load must be equal to zero because the actual deflection
8.
at B of the real propped cantilever
is zero.
Mohr's methods will now be used to lind the values ol the displacements in the equation of compatibility above.
Using Mohr tl, the displacement between the
tangents at A and the tangent at B of the released
$--Jia'":r
cantilever AB is in lact the vertical deflection at B. This is a particular case lor the application of this method since the tangent at A is horizontal, i.e.
AB= s .'. c-- #a.t
coincident with the original unloaded position ol the cantilever.
Mohr II is the first moment of area of the
M/Er diagram about B. The deflection at B is equal to the absolute moment multiplied by Lzl2Er. Similarly, we can find the vertical deflection at B due to the application of a unit load at B. The result is ' L313E1. 'fhe signs for the dellections are relative to the Blobal convention previously established. ?L
de=-(.EE -
=- t'a=-
.lJ
2/
t.2 4 t
3-
l
i 10.
Setting up and solving the equation ol compatibility Sives us the value for the vertical reaction at B which is equal
As + V s .f,.' o M Ab .t+v B (E EEo
.
\
Vz= 2 tlbt --L_
lF7j
A positive siSn for the sohition of Abl2L. this equation indicates that the choice of the direction
to 3M
)-o
of the unit load in FiSures 5 and 9 above was correct.
MOI4ENTDISTRIBATIOI' ll.
Ve now substitute the value Jor the vertical
t53 reaction at
B and determine the value of the bending moment at A which equals -M '
t\D ^./2.
The negative sign indicates
that it is acting in the anticlockwise
direction,
\L_.__a;^lAb 4--::'nns
i.e. the
action of the ioint on the member.
f 4rMturt
f
lvla. -(-iraa"+ | = _ nb/z_
12-
The result of this analysis is to prove the lirst of a series of important relationships ivhich is that the carry-over of moment to the lar end oi a propped
^+r.r-)
€,noo
cantilever for a moment applied at the pinned (near) end is equal to a half.
^*b
II
This distibution
ol moments on the propped cantilever is
now slightly redrawn for convenience for solution by the moment - area method, by separating the hogging and sa8ging moments.
c4r
2^LA
The rotation at B, which we know to
b€ I radian, is equal to the area of the MlEr
dia+ram
between AB. This rotation at B is equal to M x Ll|Er. Ab
06
area l\'t7at
=H;-"#
-
-
M/o.L/4eI
This provides the second and most important relationship rhich is that the value of the bending moment required to induce a rotation of I radian at the pinned end of a Fopped cantilever is equal io 4EIlu.
t=
t^Ab.L/+
M4g= aLJ7"
L
lEl UN DERSTANDING STRIJCTURAL AIVAIY'SI'S
154
15.
D,
lAs
Similarly, it may be shown that the absolute moment required to induce a rotation ol I radian in a simply supported beam is equal to 3Erli'
Because the far end
ol the beam is Pinned' clearly the carry-over lactor is
=qW
zero.
M4g= 3fr7, a ,o= o
L6.
That is
displacement deflection at one end relative to the other' From the dellected shape it is apParent that
^ Iiitt***^.
iil
needs to be determined'
the relationship between the bending moment induced in a beam, both ends of which are Jixed' and the
1_gj^
there will be arhogging'bending moment at A and a 'sa88ing'bendin8 moment at B.
Redrawing the dia8ram to Produce the M/Er diagram, the deflection at B is equal to the distance between
,, h +"t29-t i lF.*
-ez Lf.t\--n
One last relationship
the tangents of the two relative Points. Here the tangent at B would be horizontal to the tangent at
^, -TtJ ?-
Nevertheless, Mohr's theoiem still holds' and the deflection, which would be an aPplication of Mohr II,
H*
A.
is equal to the first moment of the lt/Er diagram, about B. The resultin8 relationship is that the deflection
A =4 . ! 2(-r* t.) i cz - deL EIa\.
is equal to r'l.l'/ 6r-r.
18.
\L++a
**
q,
M z 6EI L/Lz
This gives us the last and most imPortant fundamental relationshiP between section Properties and displacement which is that for a relative displacement of A at the end of the fixed beam' the momenl induced is equal to 5arA/r'?.
'fir|lflimilm
I
'
MOMENTDIS?RIBUTION 19-
155 If we now analyse the beam, knowing the value of the fixed-end moments, we find that the reactions are egual to l2ErA/ L3.
fu", tvA -
lvg
lJout I' + L.Vs no -M-M v1 zM/LtzEE A/L3
Ve will now deyelop the method ol moment distribution using the relationships between rotation, displacement and induced bending moments shownabove. We will consider first the four-member frame ABCDE with a clockwise moment applied at the centre node B.
This moment would cause node B, and the four members conni:cted to it, to rotate in a clockwise direction, OR. Note that, becausethe connectionbetweenthe four members is consideredto be rjgid, the rotation of eacl] oI the membersat B is Oa. va
The figure showsthe resulting qualitative bending moment diagram. From relationships developedabove we know that the carry-over of bending moment to the fully-fixed end of each memberis haTt that at rhe end B.
UNDERSTANDING STRUC'rURAL ZTVEtrYsiS
23.
We now assume that the moment at B is that which causes a rotation oJ I radian, that is the moment, MAbt as delined above.
24.
The value of
lrAb is the sum of the absolute
stiffnesses o{ each of the members lraming into joint B.
MAb ' s/l
The actual
,-+E-.!
moment
4 -g
cvE=
rotation 0B will be equal to the actua-l
uB, divided by the total absolute stiffness at
the joint.
Mg
<+EE \r25.
The proportion of the moment
tdB, induced into the
end of a particular member is the actual rotation 0B
I'A6 $r =
Z*u!4' MA
-
multiplied by the stiffness, i.e.4Erlt
ey1 ucr,rt*
1+ex
particular member, Substituting lor 0B in the
4d/Lh?abq x 1eE/,
t'
for that
exPresStonabove...
r,ict^bq
26.
ME for e1g Meubor = A?ttea r^ota?^tx n*abw stifttw11
...we find that the distribution of bending moment into any member is the total moment ItB multiplied by the absolute stiJiness of the particular member divided by the total absolute stillness of all members at that ioint,
Tota of .tL s+\(ncaa l-.
t4 rAbF+J aF th eJott"t-
MAMENT DISTR'Z:J..:);
27.
157
The lraction (4Er / L)/(l,4Er / L) member is kno*-n as rhe distribution factor. This may be simplijied br cancelling ,48, from the expression. Thus ir is onl! necessary to determine 4I for each member. This relationship is known as the trelative stiJfness'.
P-a-La.t'va' fiiffaa,1
Since, in the method o{ moment distribution it is only ever used to determine distribution factors, the
t-
l value too may be relative.
2t.
It is customary for theoretical problems to be set in terms of the relative r values or relative stiJfness. In real problems of course, the r value must be calculated.
We will extend our qualitative
approach by a numerical solution. The circled values are the relative I values.
5.
The lirst step in the moment distribution procedure is to determine the value of the relative stiffness of each member.
The symbol K is used here. The
reader shotrld note that there is no generally accepted symbol Jor stiffness. used.
Sometimes 's'
is
The total relative value of the stiffnesses
of all the members joining inro joint B is 6.63.
< = rzJ&ftva stlffhoJJ < . . -= kBu. Eso k,ls-
-
8= ^= F-_
|z n ,.n
! -r.., t.so v
25o
aJ.
3:-
qe will use the symbol ,a' to define the distribution iactor for any member.
This factor is the
)roportion of any moment applied to the joint which will be taken by the particular member. It is the stiffness, or relative stifiness, of the particular member at a joint divided by the total oJ the srilfness oJ all members at that joint,
For
member AB, the distribution lactor is 0.18. Therefore the bending moment at end B of member -f,B is 0,18 x the applied moment t1.
a =
dLtv;b6lst^
4a^=
l20 6-.62
., 18
M BA- a'18x tA.
foo*
I'8 31,
MA
The analytical procedure is directly analogous to the stiJfness method in that the first step in the actual moment distribution procedure is the ,clampingrof the
Ma
joints. Therefore we need to calculate the fixed-end moments Jor each member. For the time being, these two standard cases will suJ{ice.
W*a
tlazh --:--
32.
We witi apply the niethod of moment distribution to the two-span beam ABC. The 1 value is the same for both spans. The method proceeds by examining the diflerence in bending moments at supported joints and distributing the difference in moments through the structure by a series of iterations,
33.
The joints which are Jree to rotate are notionally fixed with an analytical clamp, including the pinned joint at C. The real loads are applied and the resulting Jixed-end bendinS moments evaluated. The difference between the bending moments at joint B is maintained by the artilicial clamp.
We must now correct the fixed-end
moments to match the real support conditions of the beam.
34.
Node C is actually pinned, i.e. zero moment. We can release this joint by applying an equal and opposite
sA
&
:J.........Er.-Zac '\f(-ft
t6
h32
moment, here the anticlockwise moment ol l2 kN.m. This moment applied at C is'carried-over' to the'lar' end at B. Note the direction oJ the bending moment in the diagram. Both are d.awn as anticlockwise, that is Ithe action of the joint on the member'.
MAMENTDISTRIBUTICE The linal distribution
of bending moments in BC is the
3: N
ern of the 'clamped' condition and the 'release' at C.
A32
BN#.
This resuhs in a hogging moment ol 48 kN.m at B. At C,
.-+
ol course, the moment is zero.
;:N,,
W
This roundabout procedure avoids the necessity ol
I
..tnembering
the expression for lixed-end moments for a
Had we known this relationship, we cqlld have lound the rclampr moment at B for member
Fopped cantilever. BC directly.
f-
The routine above is particularly uselul lor members with rnsymmetricalpoint loading. The fixed-endor'clamped' moments are deteimined as shown.
*. oJW c --rc-J
*o
R LR wbzo
I
M&b
Then the release oi the pin-joint is made to car.y-over half of the momentto the far end,
_F-a -1:*
3r .€ft
lverb -r:
7' 160
UNDERSTANDING STRUCTURAL IXAiYSI.S
39.
The final Jixed-end moment at the ,clamped'support is the sum of the 'clamped, and 'released, sets of bending moments,
ry& , NG6 2-:Iz
40.
40
To return to the oritinal
problem, we now have the
distribution of bending moments assuming only joint B is clamped. The clamp moment at B required to maintain this condition is 29.2 kN.m anticlockwise.
R--27.L
41.
Mtr
Il the lar end of a memberis pinnedthe absolute stiflnessis 3Erlr.
Therefore,since'4E'is
cancelledout to determinethe,relative'stiffness, the
ffiR
relative stilfnessoi a member,the far end of which ii P rnneorl s i r/r.
? fllmnt'"",42.
The relative stiffness ol each member is calculated as Jollows:
Kl.s=tlt kAe= p".=
Eta *1. E/6
el.^
=
o64
og a
-_ o. 36
KBC=rl8x314 because the far end (C) is pinned. Since r is the same for each span in this particutar problem, the distribution factors are 0,64 and 0.36 for BA and BC respectively.
I MOMENT DISI'.:':-::L\
brt is
43.
The next step in the procedure is to elimirEre
rile ctai:?
moment by applying an equal and opposire momenr ar
!ding
B. This moment is distributed in accordance s,irh tlte distribution factors.
tr|e
44.
,-&
The iinal distribution of bending moments is the sums o{ the clamped and distributed conditions. The reader should note that:
- |tBis Eintain
l.
The approximate nature of ali
structural
calculations means that it is unnecessary to express the distribution factors to more than Lwo
2.
ucLr"rdr
PrdLcJ.
The distribution factors must always sum to 1.00 just as the distributed moments 10.5 and 18.7 must be numerically equal to the balancin8 moment.
ttE bis
45.
Moment distribution is a tabular method. The diagrams above would be summarised in this Jorm, Study this table carefully and relate it to the diagrams. Note that it is convenient to release the pinned support at C belore starting the full moment distribution procedure' since this'matches'the
fixed-end moments with the true
structural suPPort system.
hd as
46.
If we now compare the final bending moments with the deflected shape, the points of contraflexure on the deflected shape may be checked against zero bending moments and the overall shape of the tension zones checked visually.
tiune lftrrion hlr.
D'strb*ta {;dvs F.E.M s p.*s
Fiui
rqtqo.ts
ANALYS IS UNDERSTANDING STRUCTURAL
t62
47.
Generally, moment distribution is not completed with
51.
only ore distribution of the out-of-balance moments. This three-span beam will be used to illusttate the more usual procedure, Span CD has the same section proPerties as spans AB and BC.
The relative stiffnesses' K and
distribution Iactors, a are determined.
48.
The next step is to identiJy the clamp moments. Remember that these are the difference between the
t3.2
'L
fixed-end moments at each support. The correct identiJication of the direction is crucial. Always carry out a qualitative
check.
49. Each clamped joint is released, one at a time, with the
t3.z
S^4-'s-i r R + - rR
balancing moment.
5L
The joint is released by the
application oJ a moment of equal value and in the opposite direction to the clamp moment. Note particularly the carry-over to support C ol 2.8 kN.m, half of the value oJ the balancing momenr distributed into span BC at B. Both the distributed and the carry-over moments should be rounded off to one decimal place, Remember thou8h, that the sum of the distributed moments must equal the balancing moment.
This next figure shows the release oi joint C by the application of the balancing moment of 32 kN.m in an anticlockwise direction. Support D is pinned; consequently there is no carry-over moment to D. Note again, the carry-over from joint C of 8 kN.m to joint B.
5+.
MOIIENT DISTRIBUNON
vith lrrts.
51. The carry-overol momentsto B and C from the release ol these joints has unbalancedthe system again
le more
and requires external clamp moments to sustain
FoPerties and
equilibrium.
52. Itte
Here is the Drocedure so far. shown in the more usual moment distribution table. The clamp moments at B and C must now be released and distributed.
Ttfy
Because ioint A
is fully fixed in the original structure it'receives' moments but does not release or redistribute them.
r the
It is now apparent that the procedure is, theoretically, endless as there will always be a carry-over moment to unbalance joints B and C, Practically, it is usual to stop the distribution where the largest value being
m, hall trto wer oe.
distributed is about 5% of the original lixed-end moments,
values either side of the support reference Iines are the algebraic sum of all the moments to that side oJ the
ligures which are the balancing moments are not included.
l an
The final bending moment diagram is drawn, lirst by plotting the support bending moment and then superimposing the 'free'span bending moments. The next
Note irt B.
step in the design office would be to determine the shear forces. This is dealt with here by the method most often used by the practising engineer. The first step is the notional release oJ the moments at each joint, turning the continuous beam into a series ol simply supporled spans, with imposed end moments.
.,5J5 +a2: 32')
E .q L+ 8\
but there are no hard and last rules. Note that the
support, clamp and distributed moments. The bracketed
e
4L -D .A +8 8 -32 (!13 ,+ 7.7
tzt .t
+at 5
a-
t
E 164
UNDERSTANDING S'IRUCTURAL,qXAIYSTS
55.
Firstly, we treat the reactionsto the loadingas that
59.
resultingfrom the analysisoJ a seriesof simply suPportedspans.
t
t
I
4
E2'S
4
56.
ry'+
aL.€a 'ft' - 'e
,3ts
We will look at the span AB. The simply supported span has moments applied at each end which modily the
@.
distribution of shear forces and reactions. Taking
-^rflf,frfrfl
Aboq€ A - l2'+ + 31.5- aVB.o
moments about A there is an out-of-balance moment, the diflerence between 12.4 and 31.5, ol l9.l kN.m which is
Vge 3.2-
anticlockwise. This mus! be balanced, for moment equilibrium, by a reaction 'couple'. This couple is the shear due to the effect of the restraint moments and is
7r^ Ir,
known as the 'elastic shearr. The value of the elastic shear is the out-of-balance moment divided by the span.
57.
3t,5
EZ.+
T
t,z
We then determine the value of the elastic shears on
61.
each span. Note particularly that the direction of the elastic shears must be checked qualitatively.
32
b7 3,1
1,7
a' l
J8,
The final true shearsare the sum of the simply supported shearsand the elastic shears. The algebraic sum of these shearsat each support is the linal support reaction, The
,.f '"r4,',u!1" tu+ f5
f8
dragram of reactrbns .b comPleted with the moment reaction of 12.4kN.m. at A. Always checkthat the appliedloadsare exactlv balancedby the reactions'
6L
MOMEN?DISTRIBUTION
59.
The settlement distribution settlement
t65
of beams may be included in the moment
procedure.
Here there has been a vertical
at C downwards.
The gualrtatjve
distribution ol bending moments is
shown below. This is always a good idea in any analytical procedure as it serves as a check on the final numerical solution.
The bending moments caused by the settlement
are dealt with in the moment distribution
procedure in
exactly the same way as the bending moments resulting lrom normal vertical loads. EI
50.
This relationship was developed in the chapter on the stiffness method, using the Theorem of Virtual Work and
I the
earlier in this chapter using Mohrrs moment-area
iis
methods. The bending moment created at the supports is
5.+*"
M llihT*
equal to 6ttAlr'? .
e
-\/Yl
is
lA= €frL/Lz
!' lL
t
P
61.
This figure shows the qualitative
fixed-end moments due
Remember that the joints are prevented from rotating by the external rnotiona| clamps. If to settlement.
I
settlement
is combined with vertical loading, these
moments are added algebraically to the fixed-end moments resulting from the vertical loads prio.
to the
start of the distribution of out-ol-balance moments.
td
62,
We will now return to the Droblem of the four-member
F."
frame we started with and look at a more usual loading
lhe
arrangement. Span AB has a uniformly distributed load and BC an unsymmetrically
placed point load,
,t
UNDERSTANDING STRUCTURALANALYSIS
63,
67. r
The distribution lactors were determined previously.
t
The next step in the procedure is to determine the fixed-end moments and from them, the'clamp,force
b
at
B required to maintain equilibrium. This is 60.j kN.m in
11
an anticlockwise direction, the difference between the fixed-end moments in members AB and BC at B.
64,
The tabular layout is a little
more comp.licated for
6E.
two-dimensional structures. It is usual to set out the horizontal members on the left-hand part ol the table and other members on the right. Since there is only one clamped joint, the distribution oJ the balancing moment completes the moment distribution procedure. Always check that the algebraic sum of moments at a joint is
cta.k-
1ta6=a
zero.
55.
The figure shows the {inal distribution of bending ' moments. The sagging moments may be determined
69.
graphically by drawing this diagram to scale and superimposing therfree'span bending moments onto the support moments.
66.
z.o 4--- 3.-d A Frjrc{ ,.rss
za.+ s iR
About B -Vo +27.+ - 3. ::.S + E VA =o uA- a.7z kN. r ' ?2/ . = 115 2.5 fnn A=
B.M.saSig
= t'rl,ti.2 = 1.4 k,.t4
Alternatively, we can isolate a particular member and find the point of zero shear and calculate the value of the maximum sagging moment.
70.
r
167
MOMENTDISTRIBUTIOII We will now examine the application of the method to two-dimensional
w
structures which are lree to sway. To
t
begin with we will determine the qualitative solution.
h
Because the horizontal reactions must be in equilibrilrn, flA = aD'
68.
This fiSure shows the qualitative
solution for the bending
moment diagram and the deflected shape. The value ol the bending moments at B and C must be the same' because lor this particular structure flA = dD and the heights ol the columns are the same. Note that the structure will sway to the left. For a more detailed discussion of this phenomenon see Chapter 4, Fi8ure 30 onwaros.
69,
In the moment distribution Procedure' the sway condition must be examinedseparatelyand so we mist prevent that sway lrom occurring by the introduction of an artilicial restraint, here a horizontal reaction at C' The joints are prevented from rotating by notional clamps and the fixed-end moments are calculated'
J(aa' a,tF
70.
The columns are equivalent to supports and the moment distribution is conducted as if the lrame were a u cc_rP4r
I us4,,
'.
+*.1= o'+z; 8...= t-L a'tt '2'*'a "''' ) Ec'-
UNDERSTANDING STRUCTURAL EUAI,YSIS
7t.
1.7 7'2
It is particularly important to understand the physical
75.
No
responseof the structure to the application oJ the
dis
balancing moment at B and C, here applied separately.
Alt
Note the unusual bending moment diagram which results
oi
from the application oJ a load moment at a joint.
th(
onl
72.
This figure shows the final distribution of bending
76.
mt
moments and deflected shape for the restrained lrame,
73,
The value of the horizontal reactions at A and D is equal to the moments at B and C divided by the height of the columns. The difJerence between these reactions is the balancing horizontal artificial
74.
Tl
77.
Fl
re
te
c
restraint at C, here
0..1JkN, to the right. This artificial reaction is idenlified at eo .
c<
To eliminate this restraint we must apply an equal and
e(
opposite force, Q I, However, we cannot determine the
h(
value of the distribution ol fixed-end moments directly,
tt
c(
so we use our knowledge of what are, in efJect, settlement moments, These moments are calculated in L, B
iEr^/c ) s! 5^
terms of an arbitrary horizontal deflection A,
Ve
7E,
T
ls
have no other way of finding the set ol fixed-end moments with which to start the moment distribution procedure. Joints C and D a.e notionally clamped against rotation. An arbitrary value oJ the iixed-end moment is assumed of the sane arder as the otiginaf fjxedend nanents. This is expressed in terms of the arbitrary deflection A .
Il
t69
MOMEN?DISTETffiYIIX Note carelully distribution
the sign of the sway nromenE in $|e
table.
This is a lrequent source ol error.
Always check the sign very carefully against iie diagram ol the deflected shape of the clamped structure.
hs
the structure is symmetrical
Since
the moment distribution
for
only one column has been completed.
76,
The result of this analysisis a distributionof bending momentswhich are a lunction of the arbitrary dellection A.
+'2a
il
77.
From the bendingmomentsat B and C the horizontal reactions at A and D may be found and the value' still in
faolcv = Q.
terms of the arbitrary deJlectionA, oJ the lorce at C,Or, Iye know that the value oJ the lorce to correct the restrainedlrame is 0.lJ kN. The correcting iactor to be applied to the sway moments is the real value of the artilicial restraint divided by the equivalenthorizontalfoice resultingfrom the arbitrary horizontal displacement. This lactor is then aPPliedto rhe arbitrary sway moments,thus eliminating A. 78.
-
-n:"^ -
A
The final solutionto the distributionol bendingmoments is the sum of the restrainedand sway distributions. The
+
4.2 A by the correcting factor of 0.1614.
t
nq- JrrV
L
t.o6a
O , la
4.2410=o 53A
value ol 0,67kN.m lor the sway momentsat B and C is the result ol multiplyingthe arbitrary sway momentof lsr
O'15
tl s w4!
o ,6 7
UNDE RSTANDI N C STRUCTURAL ENEIYS-TS
79.
This Jigure shows the final distribution oi bending moments. Note the difference between the moments at B and C. These should be the same value. The difference is due to the rounding off in the moment distribution procedure.
Frn.t
bt*n,3
IAo+\eA'E!
80.
It is perhaps worth explaining the way in which the arDllrary sway moments are determined for an unsymmetrical frame. fully fixed at D.
Frame ABCD is pinned at A and
Under the imposition of a sway of A,
the moment induced at B is 3Er r\lLl properties of column AB.
a Junction of the
Similarly, column CD Jixed at
D has a distribution ol moments which are a Junction of
o=.
the properties of CD.
The fixed-end moments used in the
distribution table must be in these proportions.
Et,
Frames with sloping members present a further complication.
When the arbitrary sway is applied the
elfective 'settlement' oJ the sloping member is different from the value of the horizontal deformation, because of the frame geometry.
aa
In addition, the horizontal
deflection at C imposes
a bending moment in the beam BC. The eflective deflection for member AB is ABA, i,e. normal to member AB at B, a dellection which is greater than the horizontal deflection A at C.
83.
We will study the analysis of this frame by examinin8 the analysis of the sl'ag moterts
It
moments are distributed supported at B. rotation
since the restrained
as if the beam BC were
t'aA
rF iF_
The first steP is to prevent the
of the joints for the calculation
of the
fixed-end moments due to the sway which will be a function of the arbitrary delormation A. This ligure shows the deflected shape for sway to the right.
Remember that we are assuming that any
shortening of member BC is insiSnilicantly small and so we can assume that the horizontal deflection at C is the same as the horizontal deflection at B,
A rd
84.
It is clear now that the detlection AAB is a function
6,
of the geometry of the frame.
!e
Ieg AB rotates about A and that the deflection is sufficiently small for the overall Seometry to be
la t ro f
Xot -t3
Ve assume that the
Z,1B
trlx^"
unafJected. The triangle of delormations A, ABC' A o is s im ilar t o t he J r am e di m e n s i o n s , A B ' 'A B ^ and rO", where the latter is the actual length of
I rne
member AB.
85. Flr
The fixed-end moments in member AB are based on AAB and the length of the member, rAB.
rol
dAg
M Ae-
86,
. Ap,g
2
The other major dillerence in the analysis of lrames with sloPing members is that hotlzontaz sway induces vertical
deflection'
The horizontal
member BC' unaflected by sway in the right-angle frame studied earlier, now has fixed-end sway moments' induced by the horizontal deflection at B'
M 2., =
$EEga,
Aga.
( Le c ) '
UNDERSTANDI NC STRUCTU RAL AIVA'Y.91S
87,
:ior
I
The fixed-end moments in the vertical member CD are straightforward and are simply related to the
91.
3l
arbitrary horizontal deflection A.
D
88.
We will now carry out a quantitative analysis ol the slightly more complicated structure shown here. The only load on the frame is the horizontal load of 4 kN at
89.
There is no loading normal to any member, consequently there are no fixed-end moments for the horizontally restained condition. The applied horizontal load oj 4 kN at B is taken through member BD to the artificial restraint at D.
We now clamp joints B and D againstrotation and impose the a.bitrary horizontaldeflection A. The settlement deflectionof memberDE ls /2L, memberDE beingset at 45".
T
94.
I D
MOMENTDISTRIBI]ITE
91.
The fixed-end moments for the membei BC are equal to 3ErL/L2
since the far end c is pinned.
"fu
t\. lEE A 6z
S a44-tn s1- 8 34
The diagram shows lixed-end moments lor column AB'
e he
tt4= r, lc( , A IzL 1
N at
= +'za
+.2A
enfly
93.
The diagram shows fixed-end moments for beam BD"'
f T4 K N
A'
M- 6.100.6
WA lnt
...and the sloping member DE. Note Particularly that the rr' in the expression for the moment induced by
iset
'settlement' is the true len8th of the member.
lpose
T
-
s'+L
M= 6.roo\fzA
(2r4,
, 2, 7 A
174
UNDE RSTANDI NG ST RUCTURAL AN ALYSI S
95,
Zehttio sfiffner (4"= loofl"=9.3
We now need to determine the relative stiJlness for each
99.
Th
member, in preparation Jor calculating the distribution
ITIO
factors.
dei
Keo . *+,100/6=t2'5 R a D - t o EE .tz's ks1 = loo/1zc.5'i
96.
Dbtviburtrl, = 8'3 e-^ o^
fah4t
8.3+ 12..8+2 5
aEo 4BD
ooa' 4 De =
-
o.z+
Distribution factors. Remember the advice about two decimal places for the distribution factors and the check that they must always sum to unity.
100.
c
be
5h
o'38 = o''E
tz2t='s*s.q o'8 o'32-
It is always advisable to draw the diagram ol the
t 01.
Jixed-end moments as a check for the sign of these moments ln the moment distribution table. Bending moments are always plotted as the action oJ the joint on the member, and it is this sign that appears in the
+ 2A'
moment distribution table. This was discussedin the earlier chapters on the qualitative approach. So, for example, the sign of the moment in BC at C is positive, that is clockwise, since it is plotted by going along the member Jrom B and rotating clockwise to plot the value oJ 8.3 A on the ordinate normal to member BC at B, tt
A
B
D
In
ro2.
EE
98.
Only the first line of the distribution table has been shown to give you the opportunity to practise the remainder of the operation. The final moments are shown on the bottom line.
MOMENTDISTRIH'|I(' The distribution
of final sway moments- The betding
moments are, of course, in terms of the arbitrary dellection
f00.
A.
In order to lind the horizontal external reactions at A,
o.€aA
C and E, from which the artificial restraint at B may
l<-
be found, each member must be analysedfor shear. Shearin memberBc.
+!
l !=" ^
o,62L
l0l.
Shearin memberBD.
1ot
I o.aca
'f
roz.
Shear in member AB.
s.4,
UNDERSTANDI N G STRUCTURAL 4N,1IYSIS
103,
The key to this particular solution is the axial force in member BD which balances the shears at B in members AB and BC. IJ we consider member DE in isolation this is balanced by the horizontal reaction at E.
"i"ft\r, 104.
At this point in the procedure a check may be made on this value for the horizontal reaction at E. We may consider the equilibrium of DE as a lree bcdy diagram. Taking moments about D, there is a small error due to the approximate procedures of moment distribution.
74* 36
105.
The arbitrary sway lorce at B, 3.36A is equal to the sum ol the reactionsat A, C and E.
6EA
cotrcaitg
f".t""
- fi
o-
l:Eo
log.
This lactor is then applied to the distribution of arbitrary moments, as shown in Figure 99 above, to produce the Jinal distribution of bending moments.
tM
m h lr' t
) t77
MOMEN? DISTRIBUTIOE
to7.
It is usaful to check any manual solutions n ith that from a computer 'plane frames'program. As you see, there are minor diiferences between moment distribution solution and the computer analysis,due in part to the approximate nature of the former and due
+'t Va.7
to the inclusion of axial strains in the latter. 6,t D, 2- + i^a
h r ^ }! 1 6 m .
A*lPttCY
Carry out the full moment distribution procedure usinS the computer solutions included: Do not.leave your hand calculations until they agree closely with the computer analysis!
l
10 PlasticAnalvsisof Plane Frames
The development ol structural theories so far in this book has been based upon an important assumption, i.e. that the responseoJ the structure to load is such as to keep stresses within the elastic range where stress is proportional to strain, Thus the eJfect of the load, such as stress and deformation, is directly proportional to the loading. Such structures are described as being within their elastic limit.
There is, however, another
approach to the desi8n of structures developed from research work at the University oJ Cambridge known as the Plastic Method of Analysis. Instead ol considering the responseof the structure at working load and comparing the elfects of that loading with a stress limit normally derived by applying a factor of safety to the elastic limit of that material, structures designed by the plastic method oJ analysis are considered at ultimate toad. Ihis is a substantial shilt in philosophy lor structural designers and one which is under some criticism at this time. However. it is Dlanned that ali structural codes will eventually be transferred to a consideration of strength at what is known as the ultimate limit state, that of the collapse load of the structure. The structural material Jor which this method of analysis was lirst developed was structural steelwork and this led to the development of an analytical procedure which is quite difierent lrom the elastic analysis we have considered so far. That is the subject of this and the Jollowing chapter on the design of reinforced concrete slabs using the Yield Line method. It should be noted that the plastic method analysis does not take into account the real deformations of the structure and that these, in any event, have to be checked at workinS load, known in the modern codes of practice as the'serviceability'state
since it is the deflection at working load
which is a design criterion insofar as the prevention of damage to linishes ts concerned.
179
PLASTIC ANALYSIS O'
-::-Y:i ':;-": ---::-:alytical procedures considered so far have been :r :he assumption that the structures are within
!,.:
: { ::Ei-::c limit oJ behaviour for all conditions of Lr,b!iE- [ e idealise the stress-strain curve of the ricrs,r
from which the structures are made to this
ynrr--
srraight-line relationship; the elastic and
plis;f,:
fJnes.
{*i :r: .o$ Boing to examine the relationship between ulac:1_€stress resultants when the loading causes the io exceed the yield strain.
d-.r-,re Er!:+r:ae
This analytical
is known as plastic collapse, or ultimate load
6tvaln
-,i !*::r
with, we will examine a simply suPPortedbeam.
al€:.ss-section
is rectan8ular. In this first staSe of
arr -':edin8, the strain at the extreme fibres is within rre:=sric limit and the stress is linearly proportional iE :e
r;:
strain.
w
aertain load, the strain in the extreme fibres
r=r:.es
that approPriate to the yield stress. The bending
fifrent
at which this yielding {irst takes place is
c_ ::jied
as r he ut ' J d
nanenL, Yy .
'4f'- [ {"
DI NG STRUCTURAL ANAL1:, : UNDERSTAN
The section is still capable of taking load. However, as
*J+{ F!
the maximum sir.€ss, that is the yield stress fy, cannot be exceeded, more oJ the section yields as the strain across the section increases, bringing fibres which are
W tE[A
closer to the neutral axis, to yield stress,
{9
25 '
Elentually the whole section is fully stressed at the
f-+d
'yield v'
roW
a
t---7
v._4
?4'
stress f,., except for the fibres very close to
the neutral axis. At this stage in the loading, the beam cannot sustain any more load and a plas.jhingre is formed,
This loading condition is known as
the u-Ztimate or callapse E=
subscript c.
#
load and is given the
The bending moment, identiJied on the
diagram 14^ is shown with a bar. P
7.
The p-lastic
hirqe
has certain properties which need
to be clearly understood:
#{
l.
The plastic hinge created by t4p has an effect on the structure similar to that of a true hinge, as discussed in the chapter on Indeterminacy.
2.
However, the value of the rnoment at the hin8e is not zero, it is the fully ptastrc moment MP.
I
The deflections at the ultimate load are of a considerably larger order than those due to elastic bending so that although the parts of the beam between the supports and the plastic hinge, AC and CB, are curved with elastic bending, they are normally drawn as straight lines,
::-:,Y85
PLASTIC ANALYSJS:.- .:]:,!;: le
t8 l
will now define two important relationships: l.
The Yield Moment.
7
rha
Sh:ha
Yie la
Fi-r^r
M o r ".o n r ;
Sh"pa fa.tzr.
-tr-
:-.e vield momenr t4 . is that which causes the stress -: ihe extreme libre just !r
to reach the yield stress.
serass
may be determined by substituting the value oJ
I
=e vield stress into the elastic formula:
:fu
+,8'
vv IY
Mg,Yiota n.^"t= f7.t
[]L :-:e shape facto_r. At the fully plastic moment 14^, P
r1n
tension and compression laces are fully stressed
a: ihe yield stress, because the increase in strain rEs not increase the stress beyond the value of the l-e:d stress. Taking a rectangular section, the of the fully yielded stress blocks are at a
=rrroids
:-s_.anced/4 from the neutral axis, The lever arm, is d/2.
:erefore, r:eflal
The fully plastic moment, the
moment of resistance, for a rectangular of homogeDeousmaterial is:
---.1on :
1 M
ov--f
-
htl'
.t
4
Shapo JiL l-e shape lactor is equal to the fllly plastic moment drded
by the yield moment.
::_-:n8ular
)
section is 1.5.
The shape factor
Jor a
M!
Fqatar
Fg lz - t,s f-a aaz
(ar a ra4al3uhr
s*.ttot
r82
AtvAtvs-as
UNDERSTANDINC
13, Generally sections are 'I'shaped and not rectangular, structural steelwork for example. The shape lactor for 'I ' s e c t i o n s i s a r o u n d I . 1 5 .
Shapc {",-tor arouaA l.15
14.
The qualitative approach to the analysis of structures is particularly appropriate to plastic analysis because the final so[rtion may only be determined by the study of alternative collapse mechanisms, that is the arrangement oJ plastic hinges at collapse oJ the structural system.
\_+
To begin with we will see how the collapse mechanism
-o R
and the relationship between the collapse moment is arrived at for straightforward structures. The first step, as always, is to determine the qualitative solution; our first structure is a propped cantilever; structure and deflected form.
t5.
A quantitative analysis of this beam will show that the value of the elastjc
bending moment at A is greater
than that below the point load p.
16. Let us now study the eflect of increasing the load p until the structure collapses at the load Pc.
rg-'
"'il**
As the
load increases, the yield stress is reached in the outer fibres at A. It will occur at A because bending moments within the elastic Iimit are in direct proportion to the loading and MA is greater than that below the load. In other words the relationship between the moment at A,
{r
EX
3pL/15 and under the loadatC,5pLl32will remain constant until the yield stress is reached at A, That moment at A is then defined as My, the yield moment,
)
PLASTIC
ANALYSIS
:i
!::i':
183
.:.:;]YE-5
As the load is increased, more bending moment will be I-
absorbed at A until a fully plastic state is reached. However, this is not the condition at C. The span AB can still sustain an increase in load. Point C may still be within the elastic limit depending on the relative values
$J4
N.*
of the bending moments at A and C. However' the bending moment at C, beneath the point load, is shown here in what is known as the relasto-plastic' state. A
=P
EA
proportion of the section is at the yield stress but the section is not yet full plastic.
ts
t&
i
./
l,lore load is applied until a plastic hinge forms at the next largest bending moment after ItA' that is under
I
the load. We now have the collapse load Pc. The structure cannot sustain a.try'further load. We are able
F
to identify three particularly important features of structures at the collapse load: -, l.
The structure is a mechanism,that is the
The value of the fully plastic moments will be the same at A and C - if the section of the beam is the same at these points. The elastic relationship of 3PLl15 at A and 5PLl32 at C no longer holds.
!l
The value of cqEideration
I
!e
may be determined lrom a ol statical equilibrium.
may separate the free and restrained moments.
The
free moment is that due to the aPplication of a central Fint
load to a simply supported beam' BN.4= wLl4, in
ifiis case P.t/4.
)
wx ru* EE
condition of statica.l determinacy + I release.
?
lP.
UNDERSTANDING STRUCTURAL AfiAITSi.s
18 4
21.
.-.A -.-
The restrained moment is the fully Plastic moment at A. l't . -p
P
22.
If we now add these together by takin8 the base line Jor the free moment as the hogging line ol MP it is clear that there is an easily identifiable relationship between the net moment at C and the restraining moment at A.
a (*,
aP-
PL
c -]-
cp
+ -Y P P
Mp
2
PL
c
we now see why the plastic method of a\alysis may ' PI 6L
eh&ti
produce (theoretical!) economies, because smaller structural effectively
Pla*ie
sections may be employed'
We have
reduced the maximum moment in the
beam lrom 3PLl16, 0.1875PL, to PLl6,0.166PL, by 'redistributing' mdments to the mid-span.
hea ' tookN | wtktP = E'' s?*, L F+S-2 ,lg,-4 c.'t/ bhL s4o s€rats - 23o ^t/''t'^z 5.4
-
ojesz.
to6tn3
24. This calculation will illustrate the potential economy ol the plastic method ol analysis. We will assumea working load of 100 kN, a span of 8 m and a lactor ol salety on the yield stress ol 2 to give the sale working stress: The elastic section modulusrequired = 0'652 x 106 mm3.
)
185
PLASTIC ANALYSIS O' :,..],';: .=.i.T-Y:-: -
analysis at ultimate load, assuming a shape
:€
Se-
l.l5 the section modulus required is
a:a-.f
.(
€a.t./
-
= L jg2 x 10 6 m m 3. "lbs r,c{-ild give a substantial saving.
correct in detail because ol the partial
nrE recisely
t
'e-::rs
of safety ac tually employed in practical
[{F
"4" -
-
Mq -
+9
r-.E--Jal design.
lh
analysis ol this two-span beam ABCDE will
i-stic
the judgement required by the analyst in the
llrs:3ie
of the plastic method of analysis.
miErion E I;i'€ct rqes I
to a point load P, at mid-span.
Each span
Clearly
$ill most likely occur at D, B and E, the points
Tr-rimum
bending moments.
E r,e:row increase both loads to collapser Pc, the hinges will coincide with the lar8est values of
dhd.
bending moments; i.e. 14B and lrE, and they rrilll:ccur in that order, because with equal central
dEi:<
loads and unequal sPans the elastic value of
Frlt:
f,q srtl be greater than ME. \cte lidi.€
that the bendin8 moments at collaPse in span AB are still within the elastic or
dli6lD-plastic
limit.
Alc value ol Mp is determined' as before, by a simple drdderation
of statical equilibtium.
l,5 M
p
= P L"
c' 4
?o Lz
+
)
Io44 44o
+
-
,i a" =
\cre that this simple design is an illustration and
s*dJ!
2-.& (N .h.
f00.2.8
MF:
c^
r41.cr
b.tit^ !ie1
2
266
-{<.tor T+
2zt a.<
z3l
s.,xtoe
lE6
UNDE RSTAND1 NG S TRUCTURAL ANAI,Y.'JS
29. Note that the degree of indeterminacy is l, thereJore a minimum of two hinges are required to c r e a t e d c o l l a p s e m e c h a ni s m .
/i
t
4r
I x i,\aetttui^&o
30,
So far it has been assumed that the sizes ol the structural members in each span have the same value. IJ they differ, then the collapse moment will differ. Here you must be able to identily that at the change of size, the support B, the collapse moment will be based on the weaker ol the two sections.
3t.
The Theorem oJ Virtual Displacements is used to determine the numerical relationship between the collapse load ,Pc and the plastic moment of resistance
MP e
Mp for more complex structures.
We will illustrate
the application of this theorem to the beams studied so far.
At the same time we will introduce the
concept of 'upper' and'lower
bound' solutions.
\Ye will impose a virtual displacement state such that there is a hinge at C which has undergone a virtual, vertical displacement of unity, a)
The external virtual work is the collapse load x virrual deJlection of uniry,
f h e i n t e r n a i v i r t u a l w or k
is the moment rip x rotarion at C, 20, We i;;;;a
the
bending energy absorbed in the deJormation of the beam from A to C and from C to B. The angle O is expressed in terms oJ the span and virtual deflection. Since 0 is small we may assume that
0r
,4
' ?
L. v . N-
P1. 4-
I. v.w-
MF.2e
M P 2 L/t-.
LLu Pc. -- 4^^p /L
Equating external and inte.nal virtual work, the solution Jor the collapse load is
c
___L L
)
t87
PLASTIC ANALYSIS AF PLANE FRA}1;S is, of courser the same solution we would have -eis :drieved
by studying the collapse bending moment
:=8rami
__3
'
W*r*
L
:-e: us now assume that w-e understood so little about =[.Lapse mechanisms that we assumed that the hinge
t"
rG-]d be at I span for a collaPse load at the centre'
a+'oi D
:1e result of this analysis is that the ultimate load is jx.!'d to be 8up/1, twice as Sreat as the solution lt is imPortant that you appteciate the
aEre.
r€rjficance
of this result,
Firstly,
the mechanism is
that in more +-:-.-.:i-le, consequently you may imagine a.lilrDlex structuresr you may have collaPse mechanisms However, it is clearly not the .rs obviously incorrect. one :I-rect mechanism because another mechanism, the yields a rfiae with the Plastic hinge at the centre, load less
o.iaree :, r +an. :rad'e
r
clearly
than the mechanism with a hin8e at the beam would collapse at 4 ti plL
A.V.W o Po. 2/z Z,.V.W - M?Cet+ 02)
' ur(t*"- I*r) P.,+,. E lP P..
t-
it could attain 8l'tp/i.
is yalue of Pc for the hinge at I span.8[P/t rE|E-fn as an 'uPPer bound' value. lf we Plot the value d rE collapse load against the Position ol the hinge Le
! r.;
:!e sPan, the concePt of 'upPerr and 'lower' bounds srsrt to emer8e. \n upper bound solution is de{ined thus:
Bte?L
q?' L
_Tne prescribed loads are equal to or Sreater :En the collaPse loadin8, il calculated from a:: assumed collapse mechanism.'
)
lMP
-
:\,./ ll
pir;.,.. "1- pitnl hiago
188
UNDERS?MDING STRUCTURAL ?qr!'a:t: -.
37.
There are two orher relevant definitions.
The lower
bound may be siated !hus: 'The prescribed external loads are equal to or less than the ccllapse loading provided that the bending moments throughout the structure are equal to or
v4t4a!
+ - \' /" ' 3 v &
-L_
+ r@b
bot 6.l
v414.4 ,
less than the ultimate moments and satisJy the equations ol static equilibrium,'
38.
We may illustrate this in another way. In the virtual work solution above, with a hinge incorrect.ly placed l/4 span, Mp the plastic moment is evaluated at Pctl8.
There is however an ultimate bending moment
at the centre which is greater than pcI/g,
VZr-#L
Mp= Pa a
+
These two concepts oJ the 'upperr and rlowerrbound are combined and result in the uniquenessrneorem:
=WK,", +
rThe prescribed
external loads are equal to the collapse loading when a collapse mechanism is produced, provided that the bending moments throughout the structure do not exceed the ultimate moment and satisfy the equations of equilibrium.r In the simple example above oJ the centrally loaded beam, the unique solution is that with the plastic hinge beneath the load.
40,
We will now return to the two-span beam and carry out an analysis using the theorem of virtual displacements. We can identify two possible collapse mechanisms: Mechanism I where plastic hinges occur at B and E. Mechanism 2 where plastic hinges occur at D and B.
.l
PLASTIC ANALYSIS :'
::.:.":
:-:::al rlark calculatians -_e exPressionsfor the internal and external work are :-"---ar for each mechanism. Clearly, if lr is less than ::ren Pc in mechanism I is less than Pc in -e:.,anism 2r since the bending moments would be :
i:ss :n span AB because of the shorter span'
This
.-e:.s that span BC would collapse before the collapse :E: :
,{1C
189
::':-Y::
Mara. L x. v, t^,/-- M pe + 2'.Pe = S,t^Pa _ .Lz lz L. V.w = 6M?/, 'e2
g.v.W =
P < ,.L
Po-- e!? L2
in mechanism 2 is reached.
c
li>-aranism I would be chosen as the lower ol these -rpper bound solutions. If we then comPare r.TF:_.anismI with the equivalent ultimate load =oment d)agram jt is apparent that it satisJies *-E.-€ :I.:
:!
r
--ieness
theorem.
le:]ding moments do not exceed the ultimate -3ment5. :quiiibrium is satisfied because any value of -::oment at B will satisfy equilibrium'
r&
; a: := : :ec tjon a!:-c-3" i ia-:.:
re f at Tanshrp
.he deflection ol ultimate load systems is not the solution' collaPse being essentially a
relationshiP, the load/deflection curve offers er,E -:a:8ht into the nature of the collapse of a the :ajE:::i-. in the case oJ the simPly suPported beam r-s-r--
until :Gas::-.:r is directly proPortional to the load a:E '.:-: stress is reached at l'y'
rt"
ls :iF,?-c ::lis load, the second moment of area relationship erl reduced as the stress/strain .65-; :. be a constant across the section' As the increases laE:_-r,- :l value reduces so the deflection :e !'hole section is at yield stress' Beyond this :rmr!-L _-_3 structure cannot sllstain any increase in load' lErr'! :a
!!:
B-
aontinue to deflect.
-+*
--'l
190
UNDERST AN D IN G STRUCTURAL A/V'ZYS.I.S
45.
The col l apse Although frames,
there
of
frates
is a graphical
method
based on a manipulation
lor the solution
of the ultimate
of
bending
moment diagram, it is cumbersome and conceals the first, and most important decision in the collapse analysis, the choice of mechanism. alternative
Let us examine
mechanisms for the pinned-base portal frame
ABCDE.
45.
B-Effi
4f1o
IJ
We will now separate the eJfect of each load on the portal, assumed to be within the elastic limit. These are the quaiitative solutions for the distribution of bending moments, resulting Jrom the application of the horizontal and vertlcal loads.
FI
47.
Dependingon the relative value of the horizontal and vertical loads, there are two possibleelastic solutidns. ln both, it is obvious that the combination of the effect of the horizontal and vertical load will produce a maximum value of the bending moment at D, since the bending moments in both load casesat D are on the same
n a''h, L
side ol the structure. The alternativearrangementof the secondhinge required for the creation of the collapse
M 4't\ z
mechanismis then identified by the position of the maximumbendingmomentsresultingfrom a combination of these two load cases. Mechanisml. Hingesat C'and D. Mechanism2. Hingesat B and D. 48.
Mo.h
L
Ma*
2
At this point in the procedure, you should be able to draw a collapse mechanism,i,e. the deflected shape,which satisfies these conclusions,from a study of the elastic distributionof bendingmoments.
'
F
1II-1FU
l9 l
PLASTIC ANALYSIS OF PLANE FRAMES now solve each ol these collapse mechanisms using
m. 1il
1!i" i:Eorem
of Virtual Displacements.
We will look lirst
If we assume a v.irtual displacement of
e rE.dlanism l.
l]rry a.i B in the horizontal direction, we may determine r-Ejal displacement at C below the load of 3Pc.
fr
of rotation of AB is 0' i.e. l/t.
aqie
D
of member BC is also l/r,
nift
The
hence the vertical
deflection aI C = ll2.
rEo
Morhaarsm
L
virtual work is the value of the load x in
th e
lrlernal
t
d ir e ctio n
of
th e
lo a d .
E.V.W - P"'4- + 3P.,l. -- tL. p-
virtual work is the value ol the
rEroanentol tesistance x virtual rotation.
Both
C .v.H = 2e Mp- zqa?
hd€es at C and D underSo a virtual rotatioD
= +O t!'i.7 resrdtingexternal virtual work is 2.5P. and
t
-
virtual work 4lt p/r.
t
EI.E
oJ the collapse load Pc
'
n?. o- it-
for this
oJ the hinges isl
Far
-G€rnent P
.
hrs
a'
__3
?--Eiv'p
now study the second mechanism, with
at B and D. Note that this mechanism does Fodrce
d*.
.L
zL e. = +up
= 8M
5L
riII
tw*aarsi
a virtual
displacement below the load
r.
1l E -rl
F.
fe
Mea,'a^!ttr
Z
-tE]'' 192
UNDE RSTANDI NG STRUCTU RAL ]qTVE-',IS-iS
53. Equatingexternal and internal virtual work:
E.v.v t.v.w -
P".L P
Mp.0+
Mp,O
= 2M_ CP
;
Zr^pe g !/t.' . P.. -- 2'^^?/L-
54.
We now have two'upper bound' solutions. There may be another solution which could yield a lower value of Pc. In this lairly straightforward structure we can be satisfied that the lower of these two solutions, mechanism I, is the unique solution since the distribution of bending moments at collapse does not exceed the ultimate moment as defined by the hinges and the
Collapso
equilibrium is satisJied by an inspection of the qualitative
solutions for the eflect ol each load, shown
in Figure 45 above.
55.
Clearly, conJidence in the choice ol the mechanism is related to the elastic distribution ol bending moments since it is this distribution which is the equilibrium condition. A typical numerical distribution is shown for an arbitrary set ol loads in the correct value relationship,
,,,.,,
56.
't . l I
It is clear from the distribution oJ elastic bending moments that the first hinge will occur at D. Il we
Plqt+p a.aquis = z|zo .at 1 3 = + e o N /" a L Ap. IzqT ka."n.
P.- 2 yJ = + l5k N.
assume a value ol 2820 cm3 for the plastic modulusfihe lully plastic moment is equal to 1297 kN.m. From the lower bound solution above, Pc = 8,4pl5L therefore the collapse load P. is equal to 415 kN.
""""'''1tr.::]:". -lltfl
1 93
PLASTIC ANALYSIS OF PLA}]E FRAI,IES h more complex structures, there will be a number of collapse mechanisms and the risk for the
iiiernative
dlalyst is that the unique solution will simply be missed. :{orhing in the calculations subsequent to the decision you oill take on alternative
collapse mechanisms will reveal
jrai the unique solution is there, but that you have Only your perception and practice will ensure
m6s€d it.
a',ar the correct solution has been lound. I e will now study the collaPse mechanisms for this :ro-bay portal frame, loaded with three point loads' one at B, and centrally in spans BCD and DFC.
iErzontally
fe can separate the efJects of the horizontal and vertical i:arb.
ttorzotdaA
l
tie::4:isn
TIE structure is 3 x indeterminater Jor a mechanism.
llqrired
therefore 4 hin8es are
The combination ol the two
moments could produce a mechanism with hin8es
Hng
rc C. D io the beam and G. The Beometry of deJormation will require a further b€e at D in the column DE. This is conf irmed by of the bending moment diagrams above.
qecrion
;Et5aanisn
tAc* '!-
2
I de bending moments due to the horizontal ]oadinB are than those due to the vertical loading, a possible
eEater
D.dranism
is a simple sway. Note that the hinge at D is
b lhe column and that there are only three hin8es in this ldEnism. rE*dless r@jn
Nevertheless the structure has collapsed, of the fact that beams BD and DG could still
the load.
Me.h Z
t94
UNDERS? AJ]DI NG STRUCTU RAL AJVAZ,YS-.-.S
61.
Mechanisn 3
65, Mechanl
ll/e can identify another mechanism in the collapse of the beam BCD. only three hinges are required lor this partial
Hinges E
mechanism, nevertheless it is a collapse ol the frame. However, such mechanisms are usually one ol the upper
G.
bound solutions.
Mut^ 3
52. We will now determine the quantitative solution for each of these three mechanisms. The relative values of collapseloads are shownin this figure. Ve will assume that the plastic moment oi resistance for the beams is 2l.tp and the columns, r'4p.
$cat+tc colt
2 u1 t p
Nechanisn
f
Assuminga virtual horizontal dellection oi unity at B, the vertical virtual deflection at C = Il2. The plastic hingesat D in the beam and column and G each undergoa virtual rotation of O.
The hinge at C undertoes a
rotation of 20. IAIJ)1A rs|,'.
L
64.
a4
l
e.V.W- Pa*L+ 3?e,lz ' .2LP",< c I.V.N . taF(V.el+2M, tzg) ,
.,ir1 @1 ,
. . tMpP
t . 1-
?o-- 3,zWL
Equating the internal and external virtual collapse load pc is equal to 3.2 t|p/L.
work, the
beam is
-qq PLASTTC ANALY$S Of. P:'iri:
,=:kanisn 2 jinges are introduced at B and D in the column DE' but the Ern
195
-=-:-u'r5-'
is continuous across joint D. The last hinge is at
4l +l l--
3P e
a
'c dr : F s e
load Pc
Mnl,ll.quil^Jtn
Z
F-.V .h/=
P e., L
is e q u a l to 3 M p lL -
Lv.w t.
at B. C and D in the beam BCD.
= i^P 3e
?o =
3ua L
Clearly the
?a
yirtual work will be Sreater in the beam BCD @mDared with DFG because the central point load tb span are Sreater. Note that the plastic hinge
a kms
in the column' since the plastic moment of ot the columns is half that of the beams.
3
Mt4'v^teq
ifE equation of virtual work'
Pc = 4.7MplL.
is an upPer bound solution.
mchanism
The
ism which produced the lowest value for Pc
dsnism
2.
E. v.V./t SPot L C.V.W . Mp.e+ 2tvr?.24
__.,
'
e e=r4.L ^AF Po-- t!!P 3,
* ).u7 &
, rrl. lit Ftsl[ll-Muwi|iMmlm$lli 196
.,
UNDERS?ANDI NG ST RUCTU RAL A];3.:: J',-J
69.
The elastic anal)'sis is a valuable asset in taining co.fidence in rhe iinal choice of'unique'solution.
If
we determine The elastic analysis it appears that s_ince hrnges must occur in descending order ol the elastic bending moments, if rr' + I hinges may be identified, a
tq+ eta*L
'unique' solution has been found in mechanism I since the three greatest elastic bending moments are at D in the
a^.9sls
co.lLrmnand beam, C and C. However, mechanism 2 gives the lowest value of pc because ol the partial sway Jailure resulting from the lower resistance of the columns.
70.
We will now study a slightly more complicated structure, that of the pitched portal frame ABCDE loaded horizontally at B and vertically at C. It is assumed that the portal is oJ constant plastic moment of resistance tnroughout.
7t .
tr
tr
It is always uselul to separate the elfect ol the vertical and horizontal loads and to study the resulting collapse
ffi
mechanisms. With the horizontal Ioad at B, the frame will collapse with hinges at B and D. The collapse mechanism due to the vertical load at C will produce a change in Beometry as hinges form at B, C and D, It is assumed that
fl*f,
both the columns have been rotated outwards. Since the structure is assumed to have the same plastic momenl of resistance throughout, hinges at B and D will occur simultaneously. However, the mechanism requires only 2 hinges, i.e. a degree of statical indeterminacy of I plus l release,
72,
Mechanisn l One possible mechanism is a combination oJ the sway and vertical load. The problem now is to relate the two virtual deformations required lor the solution, i.e. the horizontal deflection at B and the vertical deflection at
c.
l PLASIIC
197
ANALYSIS OF PLAIIE FPJ,,f,S
rlnstantaneous Centre' can use a device known as the l97l). This !b.ne, P-lastjc Theotg of structures' fe
Fint
is the junction ol the lines taken lrom the points which the collapsed parts of the structure rotate' the hinges.
-o.rt tqgh
bntaneous
.r4,=
We will use the symbol I lor the
centre.
now consider this collapse mechanism and in member CD, the angle of rotation of C l-tiqiar' r C' and D to Dt about I must be the same because
J G
distance between C and D is fixed'
le
ih{|tify
A I
Ve will
.4E l,
.'7
this angle as o.
H
ll we assumethat the horizontal virtual deformation The at B is unity' the angle I is I divided by '' an8le between the line. AC and Ac' is also 0 because rfie bent ABC has rotated about A. Since the value of the dimension AC is the same as CI (this is a coincident of the symmetrical Seometry of this particular portal frame) O has the same value as
o.
-
Now imagine a line from c meeting the baseline AE at F. As ABC rotates' the Point F would drop to F', having rotated through the angte d' The vertical (that is, virtual) deformation of Point c is llL x 6 since the horizontal dimension AF is tir" Therelore the vertical virtual delormation at C is ll unirs'
a
l\
t\
-^tIotl l a 4Av .44
,i, l\
f
te ab
fltuag El
7" 198
UNDE RSTAND f NG ST RUCTURAL AA'AZYSJJ
77.
a-z 4l
T
displacements.
/t t'ie \"_ l
<11
-1 1l*
We now need ro be able to describe the rotations in terms of ihe irame dimensions and the virtual 0 = l/t
lzL z .ll\ \l IT
and ID
The dimension of DD' = 2, since -
2t, rherelore the a gle \ = 2lL.
:l s t- 1l w'r.7 iliI ll
i rF-
1- 9
Tbn'1,
78.
The total virtual rotation at C is 2 O and at D, o +Y .
.e
1o * Y
The total internal virtual work is equal to N, x 20 + &p (O + y). Substitutingappropriate
'fct4 Z.v.w. M?ze+ M?(0+l I 0- , DD' . z '-1 .
va.lueslor the angles, the total internal virtual work 5M^l L.
*,
t . V . N' z . t l^ F t. ,,L + M,| (L J Z I \t- t_J
a sr^ftu
80.
The virtual disDlacements in the direction of the loads at B and C are I and lj
units respectively.
PLAS?IC ANALYSIS OF PLZIIE FREUES
199
l-ie total external yirtual
work is equal to x I + 3pc x lj. Thus the relationship between -=c jF collapse load pc and rhe plastic moment resistance of the structure
Pc = 10.
is:
'sLe' sle.. su112 ?c-tZgJ
Mp
ll
L
I
J
E.V.W* P",4"+ 3?o,tl
aanisn 2
E
fhe pitched portal has a type of mechanism which 6rld
not occur in a right-angle
portal frame.
If
:free
hinges occur at B, C and D, members AB and DE rotate outwards because of the change in geometry d rl|e pitched part of the portal, BCD.
Mc<-hq6nt
TlEe
is a relationship between the vertical
2
deJlection
d C, CC', and the horizontal deflection at D, DD,, rlri<J| may be proved by using Pythagoras.
Provided
frrat these displacements are considered to be small with the overall frame dimensions, the Glationship of x (CCr) to 9 (DDi) is equal to the
=c>__.-_:.-
@npa.ed
rdationship oJ sides,a'the
pitch of the portal, r/2
half the span 3rl2. -d'b',
lf re assume that the virtual horizontal displacement a
B and D is I unit, the vertical virtual disDlacement
& C is J units.
The hinge at C has undergone a total trsIation of 20 and the hinges at B and C (0 + y) each.
b
8s Pglhgont
l l r.i
-riF ?
t
200
UNDERSTANDING STRUC?URAL AtlE TSIS
85.
76t&
E,Y,h/. <'1, + O)M? + 2,e M?
0-2.t.2 FZ
l,9L
t-
-{^L z . v . w . z ( 2 Jtrrt* L/
To determine the intcrnal
virtual work we need to know the value oI rl|e hinge rotations, 0 and 1 in terms of the virtua.l displacements and the frame dimensions.
O is equal to the vertical
displacement
at C divided by the appropriate horizontal dimension
! ut
FE.
L
= rogP L
E6. The calculation for the external virtual work is unusual in that there is negative work done against the horizontal load at B. J
87.
E,v,w -- 3 P.,'l6Pa -
The result for mechanisrn2 is that pc = l.2j[p/L.
P1.4-
'. 8p.. to3rp 't.
. ' . p c . t.i sMF t-
a,' 88.
Mechanisn 3 This last mechanism is a simple sway with hintes at the top of the columns at B and D. We assumea horizontal virtual deformation of unity at B.
Me:ac^te^ 3
PLASTIC
F*
201
ANALYSIS OV PI'AEE FB"-EES
The solution to the equation of the internal and = 2uplL' external virtual work gives a value of Pa
2,7
x. v.w-
P 2 , ! ^^ uP
L.v,w.
?c, L
r
:, e.. zfP
ridge Thus mechanism 1, a combination ol sway and c' gives beam collaPse, with Plastic hinges at B and the lowest value of Pc in terms ol the lully Plastic will moment Mp. Ve can check that the hinge occur at these Points by carryinS out an elasrlc
L
analYsis. is This distribution of elastic bending moments vertical load based on a horizontal load of l0O kN' a hinges the o{ 3OOkN and a sPanoi 3 m' Ctearly' will occur at D and C, in that order' t-
since only two hinges are required for collapse' runique' mechanism I may be assumedto be the solution. This subject ol the plastic theory ol structures' or are ultimate load theory, is one to which lull texts devoted. The coverage in this and the following chaPter on the yield line method of analysis for reinforced concrete slabs is' therefore, necessarily here limited. The reader should see the exPlanations as an attemPt to show how a qualitative understanding load relates to the concePts ol analysis at ultimate and and turn to other texts lor a lull exPlanation Practice Problems' Futther readjng Horne, M R (l97Il Plastic
Theorg of structures'
Nelson, London. steeL and Moy, 5 S J (1981) P.zastjc Methads for conc rete StEucturesr Macmillan, London'
L. 3q Pr ls!|1{
,rr"-
11 The Yield Line Analysisof ReinforcedConcreteSlabs
The method of yield line analysis lor the design of reinforced concrete slabs was developed by the Danish engineer Johansen. The method began to appear in design oflices in the late l96Os, u".on..pt,
of ultimate load
analysis were applied to reinforced concrete structurei. Before the advent of ultimate load analysii, reinforced concrete slabs were designed on the assumption that they were elastic plates. The analysis oJ such elements, taking load normal to the member, and spanning in two directions, is complex.
i ;
difficulty
The structural designer normally circumvented this
by the use of standard tables which gave the distribution
of
bending moments for orthogonal plates with varying support conditions. The current code of practice for reinforced concrete, CP ll0 uses tables ior the design of slabs spanning in two dimensions, based on the yield line method. However, these slab tables are limited to rectangular panels,
I'
lr i t
t I
I f
I !
I
Admittedly,
the majority of slabs come into this category.
However, one of
,the inajor advantages of reinforced concrete over other structural materrals is the complete flexibility
of form.
This may produce a structure wlth
non-orthogonal slabs, or orthogonal slabs with openings. In these circumstances, the code tables cannot be applied. One of the hand methods open to the designer is the yield line method. To begin with however, the yield line method will be developed for rectangular slabs before demonstrating its application to non-rectangular slabs.
THE YIEI.D
LINE
AIJZ:!s:3
:'
._'3:::ICRCED
CANCRETE SLMS
We will develop the yield line method by studying tirst a rectangular slab' simply supported along each edge'
u u ,c,(,
carrying a uniformly distributed load throughout' The slab is assumed to be of constant depth and reinforced in each direction parallel to a suPport' The moment of resistance in each direction may vary.
L
If that slab is loaded to iailure, it will eventually fail yield alon8 a series of nearly straiSht lines, known as The tesultin8 Pattern of yield lines and Panels is known as the col.lapse nechanisn' Note that the It yield lines and panels have a Eeometric compatibility' is assumed that each of the panels formed by the supPorts ljres.
and yield lines is a flat plate.
The bending deformation
within the panels is iSnored because the overall deformation due to the yielding of the slab is ol an order
w_
the rhich reduces the deformation of the Panels between s,pports and yield lines to a second-order effect'
l|
at is a three-dim ensional view of the same slab at the frfl,l.-e. There will be a maximum deflection f,ke
Because the panels are assumed to be dellection fE: plates, and the slab is rectangular, the will be point d dis centre yield line is constant' This cantre yield line.
d oflelrtoh
of the yielded clErrfied later as we study the geometry riaIt.
L
These C.sr€rally, however' we draw slabs in Plan' the flag.ammatic conventions are used to identify dtterent
suPPort conditions'
D=
st;^Plo
tr3
fxa
f-*-
suiTort
|
€ro.
204
UNDERSTANDI N G STRUCTURAL A]VA'TS/S
We will call the area bounded by the yield lines and suPports a paret.
6.
A ssq.e.c
between all parts on the panel and its boundary of yield
L Pc/ne,ltqro
2.Yi4d
The signilicance of the assumption that panels are plane surfaces is that it defines the geometrical relationships
PLANe surfaoel
lines and supports. If we know the angle oJ rotation oJ the panel and its plane dimensions, we can determine the slope of any line within the panel. Similarly, the
lMoJ ero fird;gW
Junction ol any two flat plates must be a straight line.
7.
We are going to apply a concept studied earlier, the Theorem of Virtual Displacements. This allows us to impose a virtual
pattern of deformations
on the
structure and compare the equilibrium of the.real loads and the reat resistance of the structure.
Note
the symbol for the ultimate moment of resistance oJ the slab in a particular direction based on the reinforcement, 'r?r. The numerical suffix will be used to identify diflerent values of moments oJ resistance, Since the slabs are ol constant depth throughout, this dilference will be almost exclusively due to the amount of reinforcement provided. There is a minor dilference due to the fact that reinforcement
(o
in difJerent directions must be at diflerent levels, one
'i{or.iy
tn
mat lying on the other; consequently the effective
bar
depth wiu be difJerent in each direction.
8.
T h i s 'b a r 's y m b o l i s t o b e i n t e r p r e t e d t h a t t h e reinforcement is set at right angles to the bar. Bear that in mind. When you see the symbol imagine the reinforcement at right angles to it.
I
TAE YlELD
I
LINE
ANALYSIS :5
.t_:.I:].:':-.,
te will inspect the equilibrium of one Panel to begin riti\
panel A, bounded by cefd, and look at the
i"+iilihrium
of that panel in isolation.
lf we assume that the Llniformly distributed load 't is frat which causes a lull pattern of yield lines to be we have an Iroduced, i.e. a collapse mechanism' laad condition. At ultimate load' the i.:1iate work, based on the reinforcement, will b. equal to the external virtual work based on the
draernal virtual
liEflection ol the ultimate
load.
will assume that the virtual deJlection of the peld line ef is I unit. Due to this deflection the panel c€:d has rotated gA from its oriSinal horizontal Ie
I.sition. the panel T!€ Iigure shows a three-dimensional view of and f is e cffd at failure. Note that the dellection at panel' crit-v because of the plane Seometry ol the
Ie will now imagine a reinlorcing bar in a typical poition across yield line ef . The symbol -'? does' of cdrse,
represent
reinforcement
width of the slab' -*:
distributed across the
CONCRETE SLABS
205
206 13,
m = ul?n&A Juomatv? of rcsisfanco
-.jt ii
u
=
t\.. ^a JE
reinforcemenr, rs :ie area oJ tensile reinforcement x yield stress x le\er arm divided by the partial factor of
. ]11 , z,-
safety on the steel! l,lj,
Iis pct
ttetYo
lc^4lh
t4.
!rf.eta4
Vtrt,acl
From the ultirna:e .a: :.e.-:\ oJ reinforced concrete we know that the ul.i:::::i. =,rmenr, based on the
Wolk
= trn, 9A prr r|et r.
From the chapter on virtual work we know that the internal virtual work is equal to the moment of resistance multiplied by the virtual rotation. Consequently the internal virtual work lor yield line ef is equal to the moment oJ resistance per metre multiplied by the angle oJ rotation.
The total internal virtual work for the whole length of the yield line ef is Jound by mulriplyinS the internal virtual work per unit length by the total
lobat, J, V,N1 ovov of
=
3 .C
"J
lenarh of the ) reld line,
* n17 ?rq
In rhis expressior
length of yield line ef, utllmate moment of resistance of reinforcement crossing this yield line aI right-angles, rotation of the yield line eJ with respect to panel A oriy.
t5.
We will now look at the internal virtual work of the reinforcement which crosses yield lines ce and fd. A section is taken parallel to support ac. Remember that we are still only concerned with the internal virtual work with respect to the rotation of panel A, bounded by yield lines cefd, the equilibrium oJ which we are studying in isolation.
?EE YIELD LINE
2 47
ANALYSIS OF REINFORCED CONCRETE SLABS
the yield line ce is one of the boundaries of
i*use
alE Fnel cefd, the rotation oi the yield line paral?el
'l -
[E,E
i--.ldy the equilibrium of the bar crossed by
acd d
The forces in comPression --.re at an an8le. E :esolved nornaf and paralle.l to the yield
[fr*,.
f&-serer,
this leads to a complicated analytical yield moments are referred to the
{'dr
r5ere
d0!c!dr
.j the yield line.
&rerer.
I
the external and internal virtual
liotd Ilro
work
I
i3ar8htforward
aPproach is revealed.
We may
::E ultimate resistance of the slab by ihe yield line as a series of steps' parallel E-ilIEi
to lhe direcLion
af
reinfotcenent.
d!.'5-rep' of the yield line normal to the is assumed to contribute towards the sEength.
&
The 'strength' of the step parallel
reinJorcement is ignored. (Jones, L.L.' a E : . a !i
An a lg sis
of
Re in fo r ce d
anal
a Conc r et e St t uc t ur es , 1968 . )
b
re nay assume that the effective
length of
lecH line is that dimension Parallel to the ol the reinlorcement as indicated by the !-
rr' *rnbol.
Jiotd
m -t
lu,lth
UNDERSTANDING STRUC'IURAL EI]'J?S:S
21.
The total internal lirnEl
qork of the yield line at
an angle to the reinJorcement
is equal to the moment
of resistance oi the reiniorcement
per metre x the
effective length of the yield line x the angle of rotation oJ the reinforcement.
Therefore the total
internal virtual work for panel A, the work done by the reinJorcement, is equal to the eflective length of
Total EVN e,ofa, panA A = L1 x trtl- O/ ^
yield lines ce and fd plus ei which is the length Zr x the ultimate moment of resistance per metre length , , n, x 0 ^ .
22.
Ve will now look at panel B. The reinforcement in direction mr is not bent by plate B. Therelore no internal virtual work results lrom the rotation of panel B Jor reinJorcement in direction providing the resistance moment ,?r.
c\--s\\sss\a
We will now introduce a new.layer ol reinforcement, parallel to the long side oJ the slab which provides a resistance moment m2. We will examine the internal virtual work developed in this reinforcement, by rotation of the triangular panel B. We will study a section taken just above yield line ef.
Since this reinforcement is parallel to the line of rotation of panel A, that is the simple support cd, the reinlorcement is not'bent'by the rotation
of panel A.
Consequently ,r2 ooes no1 contribute to the equilibrium of panel A. The position of the junction of yield lines at e and f is a guess. The dimension which will give us the rotation of this panel is identiJied asr.l'.
24, IJ we now study section 2-2 we can see that because panel B is a ftat plate, the angle of rotation oJ the
vB =
,e
reinforcement in .n2 will be the unit deJleclion at point E divided by the dimension at right-angles from the line of rotation ac to e. the radius ol rotation,l,.
We will deJine this as
THE trELD llrNE e,fltJTs:s c: ;:--sa:{5t SirEi;tying G-s ]16
209
coNcRErE SIABS
the behaviour ol yield [nes ae and ce inlo a
ol 'stepsr, the ettectiYe length ol these t\,o -vield is equal to , 2.
LL
Itz It
Tll€ ioral internal virtual
rJf eottio IWth doo
-
L-
work, created by the rotation
d parlel B, relative to reinforcement t' iesistance t2 x the elfective
,t?2is the moment
lenSth of yield lines ae
arxj ec, the dimension i2 x rotation of panel B.
Tot"* EVN Fr lt;P, q a.JtolA = L2, ttt2 x 99
F important point in the procedure above needs to be -:]|-. !.yphasised. We are relating the length of the yield line ifd
rhe rotation ol the panel to the line of the
:-elniorcement. :€inJorcement
Because the slab is rectangular and the placed parallel to the sides, the rotation
ri a panel, always taken as that within the Plane lyin8 !o rhe line oJ rotation, is the same as the angle -.i] 3l roration ol the reinforcement. -!-
In the Seneral case' studied later, this does not :€cessarily occur.
\ow we will turn our attention to the extertal
virtual
r-ork which is created by the virtual deflection of the c€rtroid of the total load acting on a panel. This would, oI course, include the self-weight of the slab. In panel B. we may assume that the total load on the panel acts at rhe centroid. The virtual delormation of this Point is l/3.
The external virtual work created by the
deformation ol the panel is equal to area oJ panel B x total ultimate load per unit area' w x virtual deflection ol the centroid.
Yof-aato
Ponot
ol
voC^EtoA+ r en{or"'.r'^a^'e
210
UNDERS'Ir'dJJ DI NG STRUCTUPAL ANALYS : S
29.
/,i A i-l
The external virtual work oJ the trapezoidal panel celd is best dealt with by dividing it into easily calculated shapes, triangles and rectangles. This allows the
\l e r
identification ol the position of the centroid Irom which the value of the virtual deflection of the centroid may be found.
free ,
t.Ll iei
the s
slab i
are t
and -
sr_.rpp
righn pand
30,
rvw
=j/Cth)
The object of the yield line analysts rs to dete.mine the relationship between the ultimate load t,/ and the ultimate moment of resistance z oJ the band(s) oJ reinfoacement. The internal virtual work is a function of /r, the external virtual work oJ ra. These are equated:
Evw-JC, ) TVN = E V W Sotvo for m th tcyns * u
internal virtual work
-
external virtual work
We then solve Jor the ultimate moment of resistance ,? rn terms of the ultimate load !r. 1l
For a givo,i valAo + W
The position oJ e and f and consequen y the value of the dimension I is the analystrs choice. The value of r, for a given value of ra will vary as a function of the angle of rotation of the triangular panel.
For a given loading the designer is rrying to find the worst case, that is the greatest value of n. This is theruniquervalue, as discussed in the previous chapter. It is possible to set up a general expression with the variable I and lind the maximum value of ,? which can be satisiied by the reinforcement and the chosen set of yield .lines by using calculus. However, this approach ls cumbersome when applied to non_orthogonal slabs. Iye will develop the more generally useful trial and error approach.
THE .IELD
lilvi
;li-:-:::':i
2tl
CONCRETE SLABS
:' "::;-tt-'ED
s:uc) we will now examine a slab which allo*s us 'o triantular the solution in a more general form' The
mz /
ab and ad and slab abcd is simply supPorted along sides ac' There tree along bcd. The assumed yield line is
i1
by -i: are two bands ol reinforcementr represented the to and -mu r neither ol which is Parallel al supported edges of the panel, nor are they the two right-angles to each other' We will identify by abc' panels as A, bounded by acd and B' bounded of Point c is lr is assumed that the virtual dellection unrty. lor To determine the internal virtual work of the rotation reinforcement ,nr which relates to the two vield lines bounding panel A, we need to identify
}.
siBnificant dimensions: I.
in The 'radius of rotalion' of reinforcement band mr, dimension llrr;.
in the 2, The effective length of the yield line direction normal to the line ol the actual reinforcement'
ti
rotation \ow we can see the distinction between the of the panel' and the rotation of the reinforcement' to the line The angle d is in the Plane at riSht angles ol rota tion ' ad. Angle O is in t he plane c o n r a i n i n g :he direction ol the reinlorcement'
x-
r o&€icre c1 , ornftra,^o',E ;
the Taking a section parallel to the reinforcement' A' panel lor 3ngle of rotation oI the rejnfo?:cenent I divided by the aand mr, is the virtual dellection of
fttaCto 7an\*
+
lll!,V
q
radius ol rotation l(r?r).
la
o
lt\4 3-5
l-
\^n)
UNDERS TANDI NG STRUC?URAL ANALTS :S
37.
We may now fiM
tl|e relevant dimensions for
reinforcement band r2.
Note that the radius of
rotation l(r?2)is the liDe ce, where e is an extension of the line of rotation ol panel A, ad.
lye will take
a section parallel to the reinforcement in band .n2.
38.
The angle of rotation of the reinlorcement in n2 it 0A(rr).
This is equal to the virtual delormation of
I at C divided by the radius of rotation
+_4 4... vACMz)
=
1r?2).
I/ /LL-r)
39.
Now for the significant dimensions for the internal virtual work for panel B. Firstly, Jor reinforcement ,n r. The radius of rotation ol panel B is the dimension cf, where I the coincidence of the extended line of rotation ab and a line drawn parallel to the direction of the reinlorcement in ,n, which passes through c. is taken parallel to the reinlorcemenr.
40.
To understand this section you must be able to recognise the rotational
-;t'--
+
A section
panel B, abc. area bcf.
continuation ol the plane of
The panel is notionally extended to the
Thus the angle of rotation of reinforcement
in band nt, crossing yield line ac is equal to the unit deflection at c divlded by the radius of rotation Z(n])
.5-5
for this panel, that is the dlmenslon c{.
TIIE YIEI.D LIEE
rlL
}.Ii}:75:S:?
The figure shows panel B and the signilicant ior the reinforcement
?::!iICRCED
dinrensions
in band ,n2. The radius o{
rotation is the dimension f(m2). The analysis would be completed with a calc1ilation ol the external work, from which the relationship between ,,, and w can be Jound.
ael,vsis of simply supported triangular slabs
b
impo.tant geometricai relationships. We iErrne
rhat the yield line is at an anSle q to
ryorr
ad and that there are two bands ol ,nr and m2. The angle between the
r+ports ab and ad is 6.
@lculate
the relationship between the ultimate
r- lnd ultimate
moments of resistance ,n, we find
is a position for the yield line which gives a ralue of v in relation to ,i,. The result oJ arElysis would show that the yield line bisects
r
dab. ftr
This is; ol course, the maximum value
rhe slab is capable o{ supporring a! ultimate
L- L. ib idl
r| rhis study of the behaviour of yielded slabs, reinforced yield lines have been sagging. IJ the b restrained a.long a support, there will be an
hogging yield line, usually shown dashed, with seDarate band of reinforcement and contribution rDft
irternal
virtual work.
CONCRETE SLABS
2t1
UNDERSTANDING 5 T PU 'T U 4AL
2 t4
45. b
AN AIJ .:
I
.'
To bring all these ideas ioteiher'
t
we will now look at
the full analysis of the slab abcd. -47.2-/+.r)
The slab is simpl) supported along sides ad and bc and lully restrained along ab. The edge cd is free.
The
co-ordinates oI each of these Points, in metrest are given. You will find it useJul to draw the slab to scale
d(1.aa)
x
*1
so that you will be able to check certain critical
-tl -.rE
dimensions,
1
I
46.
There are rhree bands oJ reinlorcement: Band l,
I
I
ta
Thcl Aed
Over the lixed support ab, in the top layer
?.irr
of the slab, shown dashed. The reinJorcement is normal to ab.
:: e!E
Band 2, Reinlorcement parallel to side bc, bottom oJ the slab. Band 3. Reinforcement parallel to side cd, bottom of the dab.
47.
We will assume that each o{ these bands has a diJferent
t-
value for the ultimate moment of resistance. The main reinforcement, i5 n, and will be given the greatest
t l ta1
I rn!-rr1 )
value 1.3 units as this will support the majority of the
-m3l
load. Reinforced concrete slabs will almost always resist loading by taking rhe shortes! parh. In a rectangular slab, for example, the majority of the bending resistance
,hZ= O.7tn q3= l'5h
is taken by the short span. The remainin6 values are n r =n a n d n r =4 . 7 n ,
48.
In a real desi8n problem we would have to choose a range
I 5Z
-rlrcl
of patterns of yield lines. This is our Jirst choice.
red
There is a hogging yield line at ab and we will assume
inrd
point e to lorm a triangle oJ yield lines abe. We will
lir€i
assume a notional virtual deflection of unity for point J.
letU
The panels are given references, A lor the triangular panel, B and C for the trapezoidal paneis. Note the co-ordinates of points e and f.
THE YIELD LINE
215
ANALYSIS AF REINFORCED CONCRETE SLABS
lhe panels surrounded by the yield lines are to be flat Plates, consistent geometry requlres ol yield line ef to pass through the fitrr dslsion ol the continuation of the lines ol rotation bc tE!6t
@r@d
This is identified as point o. If both supPorts --fid:d
are simPly suPported, the point f will be the
m.!
between c and d because the pattern oJ yield
b.
ffi.
_.....-__r_];
luNx -s similar to that for the triangular slab discussed &c-
rrual
& le:se
of Point f is assumed to be unity' of the Seometry of the slab, the dellection ol deflection
e is less than unity.
lFiE
The angle of rotation of Panel
the ?aiel B, is 112.4' i.e. in the plane normal to .: rolation bc. From this the virtual deflection at
bih
]"
)e found to be 0'79.
|t let
(
= "
r{l
hrg
line ab. The approPriate rotation is in -vield lira1e parallel to the reinforcement ,trr, which in case, is the plane normal to support ab.
*;enicular L:e.evant d
-- o 7q (vvtuaa :
srart our yield line analysis by looking at the
*
tb
L3*
radius of rotation is 1.7 m.
The rotation
ilE .einf orcement in .a?1is equal to the virtual
dhc::on
]b
0.79 divided by the radius of rotation' 1.7 m,
4g.opriate
length of the yield line ab is namal
to
,r l that is side length ab. The total Gcement, virtual work is equal to the length of the yield G:EI lh qtr-
r salue of ultimate moment of resistance Per untt r angle.
4o5
// 4a thL
=
3 9th
. t.v. u/ ( hqguj at n, = 3.s , 77.14-- t.ezn
216
UNDERSTANDIN3 STRUjTURAL eNel'ys.-:
l. m 5 3'Yi e Id l i n es aeandbe, pa r€ l -\,a re s a tg i n 8 y i e l d Ii n e 5 . , j:1 9. .
I
I
I
f--f
,l\,f
-
l^"'
|
Tlte IOI
from rhe reintorcemenr in band :r _. The radius of
|
-
t .
We will study the internal virwal work which derives
I 1
resl
,.ot"rion ror thisbandor reinrorcement, ror panelA is
|
il|fT'Ithedimen5ion8e'l.6m.TheeffectiVe|en8thofthe |lIff-|,|yieldline,norma|toreinforcemenla.is3.4m.
J4.
Thus we may determine the internal virtual work for yield line aeb, reinforcement band, nz.
A . V . lc , /ao b
h2
,f< rtL
vlrl
A
reir
= e ' + r o .7 q, ,o ry.e .
l'tA Ln
5J, Sincethe reinforcementin m3 is not parallel to the line of rotation ol supportab, yield lines ae and be
JlL
Ie bo!
deform the reinforcementin this band,
i si tie. )ie coll
reir pal 5in of rot
in band r? 3 rotates is equal to the deflection at e, 0.79, divided by
The angle through which reinforcement
the radius of rotation dimension he, 7.0 m.
6.
No
cr€ eiI
sid
?HE YIELD LINE
illl|Geffective
lengths of yield lines be and ae, normal
ls 4\ ba
re info r c em ent in n. t ar e I . 2 m and 2 . 0 m
!jE
2t 7
ANALYSIS OF REIN?ARCED CONCRETE SL,aS
nryctlvely.
tn3
ltrtgth
ffiun lhis data we may determine the total internal !r'ork due to yield lines ae and eb for panel A, rEal tltlltio(cement
co
I.V .W eob n3 | f^aL A
,a 3.
= C t.2+ z.o)' t3n,
ozl
= o.47r4
le
rill
now turn our attention
to panel B and the
yield lines be and eJ. Reinlorcement band nl
hrdlt
b nr J|e top of the slab and since these are sa88in8 virtual work in these !!5 -Lines,creates no internal [nes. ReinJorcement band ,n2 also makes no ld ar@rD(ltion towards the internal virtual work' The dirc€ment
musr be ben! by the rotation oJ a
yield line' Panel, with respect to a particular fr'qrdar .!lle reinforcement in ,?2 is parallel to the line lhE i rc-ition of panel B, j1 will not be bent by the rrr-rn
of this panel. e.f -- 6.aq1
rc the evaluation of the internal virtual work h oreted by yield line bef, reinforcement band n 3' @clive d
length is that normal to the reinforcement,
bc and the radius ol rotation is cf, 2.4 m'
The
r],n2 "7 ltr*f @l
UNDERSTANDI N G ST RUCTURAL E]VAIY5I5
2tE
61. t . V.Vl bol .
6 .4,
,
5.a7 q
Thus we may determine the internal virtual work, yield line bef, lor the reinforcement in band ,'?3 lor panel
I/42
1. 3, 1,
?4n*
ts
!
2,+
62.
The reinforcement in band ,n,
is almost parallel to
yield line ef. Conseguently the rotation ol panel C about the line of rotation ad causes the reinforcement in m 2 to undergo only a small rotation. The diagram shows the radius of rotation, line fk, where k is the junction of the line passing through f, parallel to the reinforcement and the extension ol the line of rotation of panel C, ad. Note the eJfective lengths of yield lines ae and ef.
63.
From this data we may determine the internal virtual work for yield line aef, for the reinforcement in band.n 2 for panel C.
A,V.ll aof q2 lq.t o (1.5+o .) a'7,r' ' 13
64,
The reinforcement in ,n. will take the major part of the load and will thereiore contribute the most internal virtual work lor this panel. The radius of rotation is the line df and the eflective length of yield line ael is
I
--Treid 61vl$cfuirr L' - 24t 1
the dimension normal to the reinforcement.
Tt]E YIELD
LTNE ANALYS]S CF P!!!!'D.'!D
219
CONCRETE SLABS
s-,:rfi, ihese dimensions we may determine d€ inret-:ral rcrlfilgl *ork for yield line aef for panel C' rGfficement
r. v.N aof m3 f'\4t .. 'r,z / l'36' I 2,+
/n 3.
ol the io:al internal virtual work is the sum The total panel' ol each yield line lor each G6tance &
kN'm' dlis choice of yietd lines is 10'76m
f
t. I w
Tfr'4 ?aqal
/'6t 148 o17
A
tnth2 ' tn3 =
6
,rt- a hz' a h3' 5 +7 B -,o.tl
-
.'^. z 7/to 7e''
the identification of external virtual work requires areas for the eifective ft centroids of easily calculated will ol the unit deflection at 1' Calculation Fqdrion r I l'0 to is equal that the external virtual work ior ultimate Ioad on the [!SJn, vhere Iil is the total ol per unit area' inclusive of the self-weiSht srEture,
&
th
load (l'4 x dead I r-e assume that the total ultimate then the had + 1.6 x imposed load) is l4 kN'm2'
Frtern
ot
T&.L e.V.W= n ' ou
slab.
ffmate
,8a
moment of resistance, n lot this of yield lines is l4'31 kN'm/m run'
Partlcular
Fa/ ,kk frn.ds l'n4 t Jttu l o Taan
ftei
I ol a)
Sg +tT.t' utf,ri&z' ha' t^
'
l1'U
EN 4'
&lu,z
UN DERSTAND] NG STRUCTU RAL AN ALYS TS
69, 70. :'
Prac.tice
PFobTe',s
Problems shown below, the yietd line Pattern has been chosen' You should consider the equilibrium of Panel A'
ln each ol the fotrr Factice
Then identify,
Ior each band oJ reinforcementt
the radius of rotation and the effective length of the yield line as shown in this example'
LZ InfluenceLmes A. q h
The subject of Influence Lines has been left until last because the basic approach to the analysis of stdtically indeterminate structures and inflLrencelines is completely dillerent and it is important for the student to separate the two approaches quite distinctly. Normally, the loads on a structure are static. The subsequentanalysls determines the distribution of load elfects; bending moments, shear torces' etc. at points along Ihe structural members. The influence line is the result ol a procedure which is almost exactly opposite. The lnJluence line is a diagram, ustng the structure as rhe base' and is a measure of the load eflect at one point ior a nov ing load' Clearly it is particularly useJul in the analysis of structrires which have to be designed for moving loads; crane Santries and brid8es being the most obvious examples. The technique of influence lines allows the designer to examine the effect on one member in a bridget for example, under the effect of a load or a combination oJ loads moving across the bridge' This may reveal rhat a Particular member has to acr in borh tensjon and compression for a certain combination oI loads.
222
I]NDERSTANDINC STRUCTURAI A','.i:] -' -'
l.
We will start b) e\a.: .-::
::e qualitative behaviour of
this continuous bea!:. ABCD. loaded in the span AB with a point load. The resri:rirg dellected shape is shown.
2.
This diaSram shows the resulting reaction and shear for.e dr.pram.. The .rza nf the reactton arrows has been drawn to show the approximate relative value of the reactions. The reaction at B will have the greatest value. Note that we are'lookinS' alon8 the beam Jrom left to right.
3.
We now construct the qualitative bending moment diagram. We are goin8 to inspect the effect of moving a point load along the beam, from A to D, at a particular point and eJfecU the bending moment at B. The bending moment at B for a point load in span AB is hogginS.
We will now look at the effect of moving the load into span BC. Note that the direction of all but one ol the reactions, vB is reversed from that resulting from the application of the load in span AB.
INFLUENCE iI\.: 'of
This diagram shows the resulting qualitatire si-ar {orce
rith
and bending moment diagrams. IJ the spans of ihe beam were equal the bending moment at B for the point load in span BC would be less than the bending moment which results Jrom the application of the point load in span AB, AB is similar to a propped cantilever whilst BC is more like a fully fixed beam.
5.
We will now move the load into span CD. The diagrams are the mirror image oJ those due to the load in sPan AB.
been
7.
ga
Reactions and shear force diagrams for the load in span
cD.
ir tn8
8.
The distribution of bending moments lor the load in span CD. Notice that the bending moment at B resulting from the application of the load in span CD is causin8 tension on the urdelside
of the member.
T 224
! I
UNDERSTANDING STRTJCTURALelvAi 1i_- i_
9.
t
,;-
We will start ro plor rhe riiiluence line for the bending
n
moment at B, .'lB. The ordinates drawn are the values of t4Bwhen rhe load is a. xhat point
along the
bean.
MB rs zero \r.hen the point load is over the supports at A, B, C and D. you should confirm these
'I
first points on the influence line diagram by referring
t . L, M 6
ba.k ro the bending mornenr diagrams above,
"'h"^---. 10.
We can now complete the influence line with a smooth
:!l
curve, although we will go on to prove that it is a smooth
j
curve, and not a series of straight lines.
a
a
tl.
Now let us look at the influence line lor the vertical reaction at B, YB.
!L
I
When the load is over the support
at B, the reaction is equal to the load.
!
x t2.
We may now draw the Jirst points of reference on the influence line for yB. ReJer back to the qualitative diagrams above which indicate the.relative values of the reactions. The ordinate in the mid-span of AB is shown in Figure 2; the ordinate at B is equal to the applied
L. L. V g
load; ordinates in spans BC and CD are shown in Figures 4 and 7 respectively. When the load is positioned directly over the support at C, the vertical reaction at B is zero.
lar
j
INFLUENCEL:n-i: The lull influence line lor the reaction ;'B is
nding
completed with a smooth curve. g the
Dese
!.L.
t4.
oth mooth
V6
Ve will now 8o on to Prove that the influence line for the particular efJect on a structure is the deflected shape which would result Jrom the aPPlication ol the object
of the influence line' as a released action.
Here the support at B has been released and the reaction applied as a load. This is identical in shape to the diagram above in FiSure 13.
lr.
Ll
VB
We will now prove this relationshiP by studying the analysis o{ the two-span beam ABC.
rt
L,L,
Note that the
vertical reaction at B, vB, is a function of the distance of the Point of application of the load from point A, distance x.
t 16, [e ne { the o|wn I lres 4 rtly tero.
We will adopt a llexibility
aPproach to the solution
of this two-sPan beam, which is 1 x statically indeterminate.
vB is removed and the deJlection due
The to the real load t{ is determined at Br A6"' deflection under the load a distance x from A is Ar."' Ve now apPly a unit load at B and the resulting vertical deflection is 6gr.gu at B. That is the dellection due to a unit load, at B in the vertrcal direction, resulting from the aPPlication of the load at B in the vertical direction.
,-
-\
o,J"*t"
lt
{to
UNDTRSTANDING ST RU'TU RAL el\iE:T-:--J
t7.
The actual deflecdon
at B is zero.
Therefore we can
set up the equation oi compatibility at that point. the solution of the equation the vertical
Ag- A e v * A e w - - o &uan.Ve*&w- o
vB is equal ro the deflection
reaction at B,
at B ol the released
structure divided by the deflection at B, resulting from the application of a unit load at B, Note the negative
Vp ,=-A e w /. -
From
sign. If upward deflections are assumed to be positive, 46, would be negative. Thus yB would be positive
/ 6BvBv
indicatinS that the choice of direction, upwards, is correc!.
18.
We will now use a slightly diJferent set of symbols. The deJlection at B due to the load r{ at x is defined as ABx. Il we place a unit load at x from A. the deflection at B is identified as 6Bx, rhar is rhe deflection due to the unit load (6) measured at point B (6") due to a load at x (6g*).
19.
We can now set up a slightly different equationol compatibility. ABx is equal in value to ABv for the real structure since the actual vertical dellection at
Ls'Az,r* Aro- o
point B is zero. We can substitute ry68" fo. A"* since ABx is equal to r/ x 6Bx from the above figure.
Grzr.Ve*W'ta;-o V"*- ts- . W Ee"B,
20.
We now apply a theorem developed earlier in the chapter on the flexibility method: Clerk-Maxwell's Theorem of Reciprocal Deflections.
A unit load
applied at point x will produce a deflection at B, 6Bx. A unit load applied at B will produce a deflection at x, 6xB. value.
These deflections are of equal
lNFLUENCE :l-1:-j C 1:
Substituting the deflection 6xB for iie ce:;ec:i":l
p.f--
:g* in the equation of compatibilit\
shour i;r Figure l9
\'e find that the reaction vB is equal 1o tie ieilection tr: r /!e to a u.'E t load at B multtpl'1ed bJ' .he Joad t'
l!{:r
ir
l\
c.
rhe dellection at B resulting Jrom rhe
t a"-e'
,4 I 'ilr
e-)
tz-B
i
-.'- :-e
.t- r e !E.r e ..e 11-
deflection
i- e
lin e fo r
at ang point
on the structure
o I a u n ;t lo dd at B . l herel ore
:> p .,ca lio n
vB is th e d e fl ected
a p p /i.a tio n
?IL.
eE,/ov
€
due
rc a
'ko, tr NFLjEN.e UNifot Vg k tho' a{lattert 4t- to Vd= L lelo
*!r-i,e
scale.
l:::a-v
summarise this procedure thus: To find the
r:-jence
a
rhe
shape resul ti ng
o f a o n it l oad at B , draw n
r.
I
of a unit load at B.
,mL':-:::Jn
ve- - f,' ,e . N
line for a reaction, we release the structure at
=e: ?oint and apply a unit point load. The resulting shape is the influence line, drawn to a suitable -:,ected <=e. The ordinate at point B is l.
.4_
A!- =- - r c
+V"
+
:ie influence line for the vertical reaction at A, Note =ar the influence line curve
is negative in span BC.
:::is is as you would expect since a load applied in span 3C would produce a hogging moment at B, inducing a :c\r'nward reaction at A.
I
t**
^\-.-,"
lL
*.
I I urrits
228
UNDERS TANDI NG ST RUCTURAL AXEI lJ. J'
25.
We w i l l n o w r e r u 'n '- - 'r r i : . - - : g . 1 a l
problem of the
three-span beam ABCD. The influence line for yB
is
the deflected shape oi r.e struclure resulting from the removal of the Yerrical reaction at B and the application of a unit load.
To determine the actual value of t/- , the ordinate dt +ha
27. !, P-ohov ? with
roa?it
Fonfr-
fha
l^rA
i<
n,tllin1
iad
-f
This is an appropriate point in the development of these
inlluence line.
and te?142
d4lcqea
ENIL\ENE
drewr\ to
)-at;-n
ar TA,i/.,ENT
2. Tho. rosulttry is tho
i
a^^1 -f
ideas ro summarise rhe procedure lor crearing an
ar1 3-NF'I/ENCE LINL:
To 4raw
^^;nr
q Sait.bl I
.s:g'fo,
uve/ S(,l a
28.
It may be shown that the same principles apply to the inflt-renceline Jor a bending moment at any point in a structure. The two-span beam ABC is fully lixed at A.
S-a--------!.RR
Il we wish to draw the inJluence line for the reaction moment at A, li4lf we must release the fixed part ol the reaction at A and appiy a unit moment.
229
I N F I ULNCELI : ' - : --::esulting
deflected shape is the influerce Lr.e. dra\\n
: : :ritable scale.
- : -..3r procedure is adopted for a shear force ': ---ae line. The structure is released at the object :: -: :: rhe influence line, here a point in the sPan AB.
rhe moment connection is retained at this -_3i ,-- :, .elease between A and B. The influence line is -. ::: .caed shape resulting from this release' The ":-:
loads to -- - trsi be replaced by equal and oPposite -:::: :ie correct deflected shaPe. Note that the the curve of the influence line is horizontal
--.::--:o
--=_ .-ae of the 'cu t ' .
l{ } i -- :cr' study the application of this method to Care -:_:e lines for two-dimensional structures' ',r:
:€:aken in two-dimensional structures not to
r: i-
:.e influence line deflection with the normal
ln beams the problem does not arise' We will -r-a :_: influence lines lor this two-dimensional frame -3:. i .s-: though, we need to determine the full :--:- _-:: ie solution for the dellected shaPe' reactions !i : :,a..
dragram. --: :e-:rarg moment
FF
UNDERST ANDINC5 TRUCTIJ RAL AI/Ai :.:, -:
Let us assume thar \r e .i :s: :c iind rhe influence line for the horizontal reacr:c. :: n. :f. used, implicitl\.
Tc crea.ie rire influence line can only be
for loads in eirher rite .:::zr:.ai
I
The loading which is
or verLical
direction.
The relea\ed ceflecred lrape t. the place Losrarr to
"1*
create the influence iine but it has to be carefully interpreted. I'ie must remember that the value oi the horizontal reaction A is the design load multiplied by the influence line ordinate jn xhe directjon of the design l aad.
34.
This diagram shows the influence line for horizontal loads acting at any point on the frame ABC. We are ignoring axial strain, consequently the horizontal
I
deflection of point B is zero. A horizontal load at B would be transferred directly through the axial resistance of BC to the reaction of C. Consequently no deformation
T,L
HA hortz6't'z.-r b4a3
would occur at B. AB would remain undeformed and the horizontal reaction at A resulting from the application oJ a horizontal load at B would be zero. This would be true tor any horjzantaT
load applied argr,/lere along BC.
The influence line for yertjcat
loads may be similarly
interpreted. A vertical load applied at B or anywhere along AB would not produce a horizontal reaction at A. I
I
^l t,L
UA vovttit^ to 4 d S
36.
Statically determinate structures present a special problem because the structure will become a mechanism if a release is applied. The resulting deflected shape of
C ::EI
the mechanism is in the form of a notional rigid body displacement and the key uDit ordinate must be determined by inspection. We will study the simply
I^ lr
supported beam ABC. The influence line Jor yC
required. Removing the vertical reaction at C and applying a unit load, the beam will rotate as a straight
L,L.
Ya,
member about A. Consequently, the influence line is straight Jrom A to a value of unity at C.
C:'o
is
Pral
Cilo
inflr it tl corl
23\
INFLIJENCE LINL: if the object of the analysis is the in:luence
: ---:rly,
1. ,
:_::ci the bending moment at B, we release rhe at this point by the introduction of a hinge and --,::jre ii- i equal and oPPosite unit moments, either side of
=...{W',.r. t,
-'
: :io-span beam ABCD is statically determinate
r!'-.::-.3 of the hinge release at B. We want to find the :: -i_ce line for the bending moment at C. Ve r:.:-:::re introduce a hinge at C and apply the equal and ::a,:,.,:3 unit moments at C. At first it would aPPear would be bendinS in CD, due to the apPlication
1-.::-.:e
::ght-hand moment' but this is not the case. These
:: ::
moments react against each other. The
-_:-a-
-.,'-r1-"iTl.]--
::-:e to the left of C is now a mechanlsm. -_, : r-!:e.:iently it has no bending stiffness. The two :G!:-:-:.3 moments react against the stable portion of CD, and the displacement takes place in ABC.
'-,::E:::
Plt
,.,:-:
a.e ck the s ens e of y our inf luenc e line .
:::::,aa d
llrr
If we
on t he or iSinal s t r uc t ur e in s Pan C D ' t h e
:r:. t-:
::aoment at C i5 zero.
fi: *:i
confirm this influence line by considering the
r::..1 of a Point load in span AB which will :F, : :. : nosgin8 moment at C, which will diminish as
ri=
:rproaches either Point A or point C. it is correct that the inlluence line is a l:r-:,::rly
:_.r :.:
---
\alu e at B,
--r ---
:r:,ii:
.
--.h1en<
:.e of the structures in the solutions Jor the
r.3:- ::::oblems :-j':i::
included in the earlier chapters.
rotnt and load eflect to be the object of the
:t -e-':: ,rne. Determine the deflected shape, and from - fn -:-rence line. Check that the influence line is ::r"--:- ::: the particular loading in the soiution.
*,
to Practice Problems Appendix: Solutions The practice problems shown in fig. J8 have been gjven point load values, J0 kN, and dimensions in metres, as shown be1ow. These structures have been analysed on the computer and the Braphical and numerical output are given in the lollowing pa8es, Structure 5, as shown in fig.58,
is not soluble by the theorems oJ
statical equilibrium as it is I x statically indeterminate, a concept discussed in the Jollowing chapter.
The solution is included here in case
you we.e able to produce a qualitative solution. The section properties are constant throughout, area = 175 x l0rmm', mom ent of iner t ia = 7150 x 106 m m a , t = 1 4 k N / m m '.
The hinges are
created by the introduction of a short member,0.002 m long of small ine r t ia 7l. J m m ' .
1A _
@z@@
@o@1 3@
+
e. sFU-.-@ lz <* +k-
ol
-{"
21J
APPENDIX: SOLUTIONS TO PRACTICE PROBLEI4S F .lt r:! l:t r:-, I*}']CIBE
I
r ::i
Er I
:= F
I'I FEFLEI_]T I !I{ ' .f1 H :1
L
lA f:
f
E tr'1Ei
t't -l- -=
{ GLIBflL
F:frTrlTI nh.l itRFrBS:i
frEFLEr-:T1rl.l ,::lln:1 B t1 .--E:" ::r!4
E1 E q
t:l" BE11:.r:r3 er. E1BS::31 -il, EEr! 1B-1 -8. AtJl l lr'3
li1
5
r:l
a
r.1t= r"l H l- l+.1 F: r:r li;: lt-: E *: I'lEIIFEF:
At:rfll. :,t FnE[l: ':'l':t{}
1.
lal E1
I
{1 ff
3 ':I tlr
LI
5
E
:
,;t-l]rinl.
SHEFIF] F[frr::E t l'i!.1:
RliEf .1
F!tE5:r
I
BEhINIHG tl T t.1crl.1E ':l{ r'1} E 104
n
-1A U r:r
EI El
B B
:ilit
EI l:1
fi ,B
Solution l.l
l
234
UNDERSTANDING STRUCTURAL A.VEiISis f-{ Cr l:t r::l l.*
iltrCIE
1
r .il ti 6
; *
I:r
]t :a; ta L- FI l]
}I fIEFLETT I OH r.l']l'l)
f'"1E
t-l -T. =r
t1
-i l . -:,
IJ rl
::t51 815
-.-1 . F3i rjl II
L]I
RI'1FIL i! FLIF:T:E 'i h:l.l:r !:l
F: fJF{r::
l:: *;
]
ifiEFtFl F[F][:F ,::Ht.ll -9,:l
f:l
Et" ::l
:
E
t-*i
:1
tlt
-1
ttl IJ
!ll;
4
ljr 6
:t:i
5 5 tl
!1 fl
.;:
6
{1
i Li
tl [i
Solution 1.2
Fr:'TFITI {ll.l !.F:Fr[r5]' -8. rjr*6l f 5 EI. UEFJ:iU B. r.iBBir;'5 fl,, {i{iE} 1{i ,".8" rtSn{3er3 -u, e1B1 I,:i:; -.ri1,EE66ijg Er.0{rfr::r:ril
t L.r:rL:RL frrE-i:!
:
{ir
': r_rLtrEtiL F:.lE::,l,
nEFLF[:TI r:r { t1t1:,
'i' lt tl {-1 F
l*"1[: l'"1E* El Fi. I']E:I.]F E F:
f
[:
uIf ltrI llr] T t,tr_rt.lHf.t t:l i l l
l 'l :r {i
[1 EI
rl a tl
a 6 g
:15 Lt,5
:in '-' li E
236
UNDE RSTANDI NG STRUCTURAL ANA'T5I.S I--l []
|::r T: ::i F: L--. F:l C: [:1. t"-l |:; h J -a =:
[:1rj::t t-.-.-
t.: uEt:rLIL:l I t]tl { r'lf'l:r
I I iJfrt:
L
;:: LIL-Ft_[r::l I r_!i] i i,I,lt:
.
i,'.:Eti:r --*=-i
F,nTFT I r:rt..l FiFf'5 :r '. E 8.81F1531.+ L-1 . ';1&*:4!
i:l
;: :,r
I L :-r**
l*$ E:: tr"I[:! [= tr;i tlHflEt-.fr:
F l:' :1 ftL ,
ti FtFr.::t: ,: filt i I
r.L r]Il:RL Fi::15::i:r
f:: r:r[:;:lL::tr4:5
ar:H[rIt]r:j f1i:'tlHlil
':il'lFtlFl
;: r-{tHL[ ,: i.t.l,l:r
'i !':l .l l l r
|:]
5ii1
",1:1rl l. fli:r
'tl il rIl
i:ill
-.1:i |:J l:l
lll
Solution 1.3
f-ll r:n i::r i:ri L__ |,llltrl:
L
f!
:t :;:ii F) L- tst ||.-: E: l'-il E: t'11-l* *::
l( nt{F L-Li:::Tllr!.1 { t.fi.r
3 frfl F L-E:[-l' :1. rJI i ,:t,[,t:,
r3 ,,I llr, ,55{ - 1.[i " ;,:].:l
r;i Ir
F,-[[l: tr-,l|83Ei Fr;: I,]EHEIE:Fi
1 ;t
F :,:,IF IL .. :'.i FflB[:!: ': f:if'l ]
F- r::!lF::j:r::: I::l :::;
lifl!:tt&l :I FIlE:I:E ,::t:il,t]
0
Solution 1.4
ra1 -11" r:ieri:i5ll,:i -{J,, el{:riJ'ir!Ei
ErEllrrlflrr t.lt:rHEt.t1 t:l l tl
f4l tljt$
a - 5 fi
I IlJf'l F1r:r-f1:l ,l lilFlt|:i.l
,:.l*[|:FtL t:t!::E.i:::)
EI
; ]-:l
'1Lit.-.lrgr::rL-nl:E::i r
tl
L1 I
APPENDIX:
SALUTIANS
2)7
Ta PRACTICE PRABLEMS
Solution 1.3
Solution 1.4
2 )8
UNDE RS?ANDI NG STRUCTU RAL A'AI
!:::r T ::1!.;lf:: l--- F::]lr::r ft::: ii'-ll E: t I -t
* 'll r::r it:.i !:lJ [-
:,1 flf:r:t, IIi-1 I [rl.l i:l'll'11
tlrlnH
1.':.' :=.
i
::::
'. - t r E : F 't - E : - t I : r . iai;
]l
f'11 [:: [-'11tlirr f:-i: {ii;:
t,1|Ill[JEr.j li
I
Fri |:.t t[:j:!r:],Er-::::: : L,r,l_ri:tl_r:r:;i:[::'"1::!
itrll,lI i:11.[: r:jl'::{:t] ,ali:l' l:j ) ) :;J
Ft'::'l i :i T Lr:.rl .l '; i;:r:r[:i:.i]l
ii iar,, i:]l]t:!:l:6ii; .-.r,:r ,.i::rI:ri:r "+ijill
!41 1 3 ";' ;:t:.:l tit,, ,rt:t::l
1 ::l
+:'',-rri :'
,i
:tlll F.i:ll; F[rFi{:E: 'l l :.l .l J
,li
" ::::
.;:i: . :::l ;l:,: , l"
t:r
ri [;l ][:r.l l i ::i l l r.:rl l l l i {l :l ': l.ii,l rl .-.r:l ,, :i ::l " :i: .-.:::r, ;.::
Solution 1.5
F.l ri:.lr[::[ lt-'r!--!-.r[.rE
lt.{ -f {j:;i j:r.jl.,{:rtajf1l. l:lr':t]:.::i.r !::]:rll 1:i lF:"ll .,.F:j$t-:::[: f'! lE::::
| ] L.:ltl rj l"rr:F:1.-t:l-: ,il'lI llr
1 i
!:j -..fi, {:r4,,+
.j
-,!:r,. t;:1..1.I
.1
t
,+ Solution 1.6
[![ F ]-tlr:::lI rlr:l t:f'llllr rir B --r;r,,l:l{:!:r
r;ir
t:1
r-|ltt:: !--l l!:r [i [:4t
tlr-|lL:r-F:
;
fi:, :l i:11,. t,,: Fl]FJ:tH { t i t..jr
il r:l
l.ljr --!, t:i!:it::r:: iir,,ri:r!J!:1"+jli:::r .'- lir ,, lll !:l lii {:li;l I
I -{r, ril[itjrr;it::
Liilll::::i:i F]: $::!I$;: {::::x:.]::::; ,;Lr::rr::t:]L-
i
ii !-ttr.t1t;: F{]Fr[l : 'i l,:.l'.1:' -.;;l:i:i --i::11
.--1l , ;r t ; ": r
.-.1r ,,;'
-. t::ir,
tll
:]::;
i; r:r1l::llIr'rf'l ,:.iil:ii-l!:l
1 ;1. ;'
|:l
FiL:llnIllr-i I lr:rllEllI ,i t{t.t tt r
rl
u []
APPENDfX:
SOLUTIONS TO PRACT:C'
P,RCBiA&S
219
Solution 1.5
Solution 1.6
240
UNDERSTAIIDI N G STRUCTURAL ElvAIY.'.-.; llntI
1
:,i
tlFrL[r,:-l-] |:|f.l ,il'11 11
;::
IlItrt-E[:T ] [ri] ,:l .l l l
'lFiRt151
lii
a:
E " t!li:;:'rlil Er.Ef.rl.:r;"18 !:'. E1!:rBilB:t ir. OOrJ:!l'j' '-i:', !-J{:: E1 f1;'1"i
[:l
L1
--t.r, tiLl;ti I ti
::: ::l
,+
u t!. t'tEt"tHrE::Fl
L
r :li' ;Ifll!: Ftrll{iE. f. !:f.l.r
';iFtrifiFl
,i I:uFti[: ,.|,: ::
5[:l
J:i ;:t
TiL:I
.-.ti* ; i::i
:;
-::jo
ir
...lti:l
ii
5[i
SolutionI.7
riHllfrl ftli f.tilHE:Hl' { l::Fl l.l.r tll J.ULl - L Er{:l
;1::i
;:::, :1.
Fir:!TFrlI f.ri.l
{:J {i
U .- t.ri10 1{1!:1 i:'
APPENDIX: SOLATIONS IO PRACTICE PROBLEMS
I,Jtr|lE
1 I
: 4 5 r'1Ft'lSEF:
]i frEFL.ff:T I trt.l ,l l'lt,i:l B ]. *ilrll 1' FFl l t . r3l3 lj'
HTiIFt, ],1 Fr:rftt:L. { ti tl }
1 t
!s . {1 -fe r.. q
t
1 !, : ,
:ij
f
DEFLEI:TIIFI { r.1t'1.) t:t
* f . i+Eig * 1 , "?r30 -r.i. Blirf r.J :;HEFIF: T FI:IRI::E { t;t l:' L] fJ :l E :. i .-16. i
:l
.+ 5
241 Itfi-rfiI1 t.l rFI.lf]S:, iJ,,atlrer,+la rir" {rL:$:t4.1:? *tr, litrt{]44r, :..F" e'$er19lJ I " rlrrjd*!j:r1
EEt'lJ:rIf .lLi
l{ntlgirT i l.:FlHJ
er fil
! B
r1 t:. 5 a1
Solution1.8
UI'IDERSTANDI NG STRUC'IURAL EN,q:]S-'-' chapter 2: Statical
indeterninacq
Craphical solutions to practice problems shown in fiSs' i5 to J8'
rl ,A EC D
-+iT
Solution 2.1
5 axtornql,
rg-st ratwts
- 3 e4^s
ay eryilibrilu = 2 la&s
httyo
l/a(@so
IrKwruL =
t^4etzt/' i^''+o
I \
{ d etz.v ,t|.t^a*p,
.
Solution 2.2
I
bre.z.
-
2 x tnd,.
ie 3.
Solution 2.3
2 x rr,*!..
L
Solution 2.4
.E
t 6x
ina.
APPENDIX:
241
SOLUTIANS -a
5.
Solution2.5
'rhQ.
lowar
is a
D Porta,l ABc- 6HJ
M oc^ent
s r+1.
SuTyorbs A ana
cr,
Solution 2.6
H aro afl suffie*" -tt^2' for st4biltY . ttuzo.ntaJ, rzacttoA a h g b t -c s a t * it . / x rna . intdvnatt! frauo
c-oz4
Solution 2.7
lta.t!\bg,
fat Ao
not ne<nssavj
AL
st"tlig
-r
O:a fmme,.
APPENDIX:SOLUTIONSi'
t:-:-:-:':
245 ":t:':::
a'a i '- =: : : : ': : - : frl' 32' n'jlnters I 10 6' and fi8' 55' 7 to in The practice problems shown as shown below' The point 12, have been Siven load values anrl dimens:cns Chapt er 3: The quallt at iv e
loads are 50 kN' pages' Sectlon The computer solutions are shown on the {ollowin8 shown as a mechanism properties as Chapter I' Problem 6 was oriSinally ,n ttg. 32.
'l
2
3
+ 5
6
?s?ff, 6 S? GfN
rffi
Moh^ 4.t 5oA^,/,lat u.D .L-
lo U4=9 LD+
& 5R
,t v-+Er^ui+&{ r + (9) -{ + €/@ )
'2ffi
25 kNlt4
lar$
246
UNDERSTA]IDI NE STRUC?URAL EdtrITSI.S l'"{ t:l [:] dq L--.
fr
d 5: F" l*
i.: ItFFL-E':TIdH ,:t,tft:1 1
t-l [:
E f.t E ]-{ -T- =
;I TIEFLEL-:TIAH iHtlr
rl
r,1 E; r-'I F3 E: F?.1 F-- r tr;: r*:il::= IIEI.lFEF:
Ft!:IttL :"i Ftri?tE rt!:l.t:i
t
EE1r3:14 BBB4I:T oB 1i 113 $63968
{ L':rt:FjL tl:tE:!.r
5dEfiq FliFl'::E ,i t{:t,].1
:
tll E1
14. r'1 ,14. 1i
::l
13 trJ
-.48. : 4-q' I
:r 4.
tl
I
a. _8, rt. E,
B Er _ 1:. =:r4
E &
4
Fl:':Es) FlrTgT I flN {R H 0 6 l ,
rJ
a :l
r-GLfl8FL
a
FEf{rrlftG f.1'li'l E f, T { HH t1:}
-4:.
el _t
+t . 3 158
5ff
-154 EI
a
Solution l.l
F"l 13 t:r r",t t_
)
I
IlooE
l:r I =iFL*Ft|:
X LTEFLE':T I Ni] r.t{r'11 _
1 f :J
f
Fli':I FIL t: FnhtrE 'i t"lHl
Ei
trEFLEf:T J trf.l 'i f'1t"1:l
lt ft E F
r'l E: l"n B [: Ft I'lE11BEE
Ef.IEt"{-f
T|fL$BFILRXES} -
ff
E & -8,317
F-fl [q |:: E -g f
F]IJTRTIO}.I tR fl nsl E , S B E1{3 59 -0. 'nErBt E. FFBB75 F. AEiEETE
{ Ln[:nL fitiE$]
S H E FN FOE C E { tr.f.ti,
EEHEII I,IG hnf.lEHI { HH T1:
1 Z
a
I
[J E
:J *il
* 11.9 B
E
s
n
E
:l
! 4.
Solution1.2
s
UI
.F
1
::, 5g
fa--3-il4
\\\i\l
,/
\
w-w
,/
\
24E
UNDE RSTANDING ST RUCTURAL ANA: ! : : : h-t r:::r t[::I rq L-
i.lfrfrE:
I::|l I
tj: F
)'l IJEFi-[{:TI [rt] ,:],lit.r
L-_ H i::] E: f.-l E t.{
I
a-+E .R L
frIFLFr::T I r-'li ,:t1lt r
a a
I
T':=:
E-r-l-11-l"I|:|f.l
r:F:Frn$l
a -Er. rlErlll46 Il " Llritril|llijl
i1
,41
f-: r:r F+ r::F:::--
i"1iar.-[t?Eilq
t'rEnBEt;:
t1 l:Il:r L :,: FIFl_:{:'i k.f'l.l
1 J
;
{1 r]l
Lr-,i.HL |li.rr;.
::rHE11R Ftti[:F: ,i l::t,t1
ErEtll-rItlr:i HL--r t 1Et.l-l ,it,at1H]
::tfi. :3
r.4, I
u
J
.;
r-!l E !r:,
-14. 1 0
t1r
Solution 3.3
:E:h-lE::t".l-T-:=;
It-l!:r[lFiLI]IUE
1 :l :I
t,; [.rEFLFi:T IUt] ttJ tl :' i:1 r1r
o a
r"1 E: l,.tE$ E: Ft IIEi'lFEF
1
rll{IFrL I'i F0FIL:E. r:la|.{ ) ra:l
:
'::rjillrEi]1l Fl:,lE!_i:] FJ:rIFt-l l: 1.-tFl ,i liFJ[r::i:r
NEFLEIT I TI}.I iI' ,l)
E -,t" oolr::,t. tr, tl{ifiFr4 u '. E A 1 , *1 3
ljl
a & ,-3. {:rlr:r
F. r:rFq {: E: ::;,iLr_rl:l]L SllEFFl 3 FnFir:E {:t:t.1.r F.4
t1:.1E5::, LrEf.lrrIflr:i !10tlr T ,i f,:t"l ll:' *1 ! " : i
a ;l 'lr
rl
il
*
'1
tl
Solution 3.4
|l|
.-+1"r '{..t. i' :i E -5|l|
1t:1{i - t Bt:] E
APPENDIX: SOLUTIONS ?A PRACTICE PRABLEI'IS N FTIN U N D E FLE C TION A N N
-.-<-.7-\./
MF).INUN
'G
DEFLEETIT]N
3 NN
UNDERSTANDING STRUCTURAL AIVA-Li'S:S
250 h-{ l:rff
r:--t L-
[r
I
F
L- F, |l:: E: f'l E t'l -f =i
fi:'.:E::i:l
:',:31-
-:i
tlr_rfrt:
:,,: nEFl-Er:-fI nH
:
|:IE FLE [:T I i !I]
F):rTF'f 1r:rl.l i.RFn5 )
!:l
E1.SEFt:rEt Lr. t1EF1{:i -E1. rlri16176 Lr. tiEr$EF*
I.l lll .) tIl
I .i ::i
- u, tr4Ll
i1
-l:1 , r:1; :ll - 8 . A7 ;l
E1 rjr
f"lE: f'l Fr E: Fi
F" tlr I;: |.; E:=
f ilt I nL I: FLIF:I:E tll.ltl!
I.]EI'lFEF:
:
'. L|- ::RL fr:,:E::i:r EEf.{nI r} nrrHEt.ll':.l::ti H:
:iHEFIF.: FirF]::E ti:: :, '
:f t:r
L
E 1::, 4
;: ;1i3' '+
:l{l
il ::i
fl
:i i1
E {ir
l 5,
1El. tl
E
-14,S E1
Solution 3.J
t.i t:|| [:r t-t t-* NI:IE
1 : :r .1
fr
I -'= F: l_ F: :: L: r,11= t { -l- :=
:..JDEFL.EC:III]I'] ,iflf'l:t
1 J f
u r:r
Solution 1,6
I'EFLEL|TI NH 'i Fll.l] r:r S ,1 1..153
S -ri.943 -,(1. t!.;'.: - Er , Bf g
FtEr."tE:rER llEHFFr-i
:
F|:rlq|::l=5r
Fr!:IrlL ll t:lrF:f:E r:l<.f.1.:
I
t] E
R':ITIITIL]I'I { PFE$] _*. SFE1BB fJ. Eofrlss _4, 0Brl.JFB -rl.oOA3EB
,iLLTCRLFr,iE5.r 5H E fi N FI:IR!::E: ,i fif..tl
BEt.lE I l.16 f.t0tlFIt'I I:P:FIH:,
s
::ttJ - 38 :::tr:l r:l
r.[iLtrErFlL FlfiES ]
15
:3" 7 EI tjr lll
IJ
I
APPEN DI X : SOI//ZaG'
m EIICE
PrcBL4t'lS NRXXSUN DEFLECTIEN A NN
,"1i'
nl'1 =", n 2r' 7 zi,s
''t''
Solution 3.5
nFrlnun
4F
3, 7
,ffi 27, 5
1 . 6 .3
I]EFLECTION .1 nfl
252
UNDERSTANDI N C STRUCTURAL ANALYS J S {--., r:r tf
l.l{rUE
F:! [*r
[:r
:'; F: t-
I
Fr r.:: E I,-t F: f_{ ]-
:.: tIHF L..E(:TI LIH i:1 .1 t1 .1
1 J
f
r.1#: r.ll El E: E f'18l.1l!E El
r-ail'_-]ERL F:.:E:i l
frEFt_EL:TI L-rtl ,ttlH.r
fl {1 el
:::
=.
RtrTfiTlr:rf.l ,lFflu!i!
B A ,IJB F:;88 E ,8014!5
E1 - 1 . !:tt:l
F
t:rF;:f::E =:
RI.iIFiL ].] FI]R[E: r . lif lI
r Lr: ::Fl L rl l tE S 1 ::iHEFIF:
BEf{n I f.rr_.i H|:|t{Ef.iI {l':tl f'l }
T F']E[:E {:!.:t.l l
I
ltl
;a:
E
1tl. r:l
5E
I 3
rll {1
* Er
5S
'- l fl,
l:l
Solution3.7
f-l r:r |:r tq LU|:ltrE
I
fj
I *
ll trEFLEf:T 1L'lf.l { ['1f'1)
:
E1 EI
i-J
a
}.| I3 T..I Ef F:: FE t.lEtlSFF
1
:
11I:I fiL l.: l--rlECE I l; Nl
I
{ rjL.frE{HL Ft,!iE51
IIEFLEITII]I..I t HFtl
RI:ITFITITI,I {F t F n S }
n
& -rl. qHEetir E
-8. l.&5 rl
F?|::rli;:rl:E::=
f
(Lcr[:flL
F[rRf:E 'i HIJ.)
FlilE:i], BEHNIIII:i l.tfltlEHT { l.::f,l Hj
{J
a
:
tIl
l
LI
Solution3.8
Fi t-_ F| |] t:- Fl E t-{ -I. g:
-.:5.7 -l:r.
3
.-.s.!
253
APPENDIX: SOLUTIONS TO PRACTICE PROBLEMS NflXINUN
TR
LA,. 7
\\-_-'____T-t
z5v
2 ffI
'1ss
$
ry{
DEFLEETMN
-..\ \
-
//-\58
r'a, E
NNXINUN
DEFLECTION
C NN
---...-_--_-,2: -t-----,,
Solution3.8
UNDE RSTANDTN G STRUC'IURAL'N'E'TS-'S t- { [IEiF tt*.
HONE
I:r I
:= F l*t- tt::E :f'tE Fl -r:=
:1 NETLETT I L':II.] {:t.lH}
T I-IE I.LE [:TII
.1
r'1E I.1EI'IBEE
f'l E-:!E }?
LI A. FF1:J6 -8. rllEr1333 EI
U
F: l:r Fl c:::F_ :=
l
r.LOt:tiL
fr):Er'.,
sHEFF: FrlrFt::E. { l.;H1
a fr
1B
t-J
'^ 1 a j 1r:l
:-1 'J r3
+
Fr|:|T11T]r-ll..l ,:RRn5i:r
-4.14+
: if
I.I
r:1
F{l{ I FlL. ,'{ FORita ( l.:ll)
L i
F:.:E::::,
rH t1.,
El E B t1
I
: :r
'rliliEISL
FE
fJ l ll[i
I'l0t'tEHT ,:l.il.l ll I - LJrl E
a fl
- l tll
!]
:l:l
1r,6,,5
Solution3.9
r-ifl[]rnL_
l{ooE
I:r I *i F} L r:1CIn:r'lEl"j
I: TIEFLE|]TIFI.I I H]
nr:r"fR-f1|]t{ {Riiu$l
a B
F - {3 . r i E B 1 6 ; !
rl
'n F |f F4 r:l:E g
r.'tEr.lFEE AHI RL !i FIEC:E t l";f.t:'
:
5HE11R FI]F:I::E {:F;H.r
ul F
+:1. t3
7
E
5.6
Solution 3.10
El
s, sr!08s3
riLnI.RL ntiEli]
1 I
tr
,:.r3l.nBr1L tll:E5l
NEFLEIIIOI'I t.l"lfll
l:1 tjr
1 T 3
}'1EI'JEEE
I
-r *]i
EJEf']N lHF HtrtiEi4I tF:t.t H:, *4 1 . t 16. r
6
APPENDIX:
SaLUTIA\S
;t
a?.ac:!:a
PRABLEMS
MNt1NUN DEFLEETIEN 4 NN
Solution1.9
NFTINUN
DEFLEITION
=._---z:
A NO
UNDERSTANDT N G S TRI]CTU RAL ANA,Y5J5
t{rtLtE
r.: LIEFLEI]-I|l1.1 ,:'1t1:l
t :
a 11 rl
'+
-i. -i.
Itl
s
5
r'rEt{EIn
J neFLE(:TI L-rH r.I,ltli
F} ' : 1RL l": f:r:rF.:[:E ,::tj.li',
1
:
E ql
: :
SHEFlFl FI:IR.E Ll ::l l i
ABli743 EE145r EBr355:+ DB1 1t1
FEilfr I l{r-.i llfrf"lEt{T
,:Ht.tu: 14,I
*7 4 " 5
.-31,5
{1 {l 4 5
s B. a, -8. -s,,
Jer. 1
|3 13
:jl
E ri +! 4 =:E1 B
R l 'T F T I n H
a
3 ; ', S
Lr F
1J
*
F
:J7. 5
Lil
Solution 3.1I
llflfrE
1 I
].i t'EFLE(:TInf.J ,;ntJ:,
:
il t:t L'1 kr
5
r:r
TlET]FEF:
F]FJIIIL }: FflFICE r:Flt,t:'
I
nEFLEi::TI Ot.l ,:r,lt,tl EI
* E. 6EE; tt, 6alLi r:1
I
::lHTFE F F]I::E: { tit.l:'
I t
Lj
r
E
l:I
rj
E
,J 3
5
Solution 3,12
a
ur
Er:rIllTI|:ltl r:F.:Ftr::] tr.t
-8. -8, E. B,
E{|FiSr3 OB03lt: BABT?I Ed8e:t
EENN I |.l[t tlfit1Ef.lr {1".H i.l1
t-;J. 5 F
o
a E
B
F
a
lll
F tll
E
E
a
APPENDIX: SOLU?IANS TO PR/-C::::
PROBLEMS
. 25
ffi
i-+
37. sv
R(R
sJ. s
=zr.s
Solution3.12
25E
UNDE RSTAND I NG STRUCTU RAL'XAiYS--J Chapter 4: The qualitative
analgsis of franes
The practice problems shown in figs.57 and J8 have been giyen load values and dimensions as shown below. The point loads are JO kN. The computer solutions are shown on the following pages. Section properties as Chapter .1.
APPENDIX: It.l r:l ffla IJNOI:
1
: :I
SOLUTIONS 7A PFl-CTa:s [*'
fr
I
:=F.Lr-r=EttEb-{-I-
!: [IEF L EI::T I L lI] r:t,ltl:,
I
:
259 €;
{FLI:IARL Fi.:F51
f'EFLET::TI r:rt-l ,:_tlt,]:r
rl t1 , til1 4 !:r. [11]6 1l
r,-!l=: r.t EI En*e f.rEtlEEti
?RABLE\S
Itr:'f tlI1t]l.l { RFlt|i :r
13 _8. B5E1 -8. F:]E !l
F r:r F-::r:: E =:
F IF ]IIiL :'i f:r"!F;t|:F ,l fj: ]
ll [1" F8gtsg -tt. F!:i 61S E j --.!:r.& rl l B rj l 4
.:L t:ta:Fll H!:Eg r
:iHEFIFI J Fr:lnr::E 'i lr.f'l:r
i4 "; *;t4 , ;r
EEllnIf{'-1 l,tlrflEtl"f ';tifl i.];,
-t ;:
;'
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t4.:
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--':rd.
,-5d,, 5
?
;:
:iE, s
*i
a
i :5, i l
NFXIN|JN
DEFLECIION
1 NN
Solution4,1
260 I
I
1 J :3
+ I.JETlEE F.:
NEFLEI:I
I I:II'I
,::r{n l
J
T]E FLE C T I L-'IH i t]t1:l E1
lll
tr"[1!1 s. E3r IJ
FtiiItlL l.: r:tFitt:
_a, 1B l -' 1r1":3! _.1rf. 5:f6 S H E FIE
:
FnFlr::E ':.i al.l.j
1 I ;1 r'l
50 --tE1
-.+,+" i:j ,{.4. i
4. + , $ *44. l]
5[r .-lj'i'
:::l
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F:r-rTRT 1Uf.l ,.F;rin51 IJ F, EEll:1{r1 B " S841.'+fl t!. F[14.t4u EllfltlIl.lrli l,lrHHHT ,:!,i11 t'r:r
1.5El - 15Li rj
{l
EI
t1
*
Solution 4.2
l{rlfrE
I '.nf'i
l
i I
i
:': nIFLft:Tl[rt]
I
:;t [:rEFLE:r:::T', I fr].1 { t.]t1:r [r --rlr.1r.l;l -.,+" tilrj .' I " 4,+::r .-.. ; . ,il.4s
tlt ti"Itl4].
iJ ::i
r:i
E, Er;:'l t:', !J1$ 13" ErlEi
n
Il
Ht_rTflt-Iitl'{ r':l;tF[r::i:l
-.r;1.AE{ttr;4 E , l : i [ l 1 t 'l i l ril,,Ftl:1,;t;i -rlr. EEJ;:::'r:; -.fi, Ar:rl l;16
l
: I
r'iE FrFl
I ;l
: ::l
Ft:,'iIt-lt_.. !: Fr:'Fr{:E r.t..}.{:,
-. i:ri"l :B
-ro
Slrl * IiF
4. :l ::i
Solution 4.1
.iil-lEIiSl Ft-rFlL::f: { l.::}.1:i
5fi *Efl
lEl
*,
;
;ltl '-lr;1
aE:H|:]I tlrj flttlEl.lT I. F:.1.1l,l l t:l 1A l J * 1 tlfi
rl
r:t
fi
t,I
0
lil
&
EI
[1
E1 Li
l-:l t1
APPENDIX:
SOLUTIONS
TO pRACarrCE ?ROBLEMS
261
Solirtion 4.2
Solution4.3
26 2
UNDERSTANDI NC ST RUCTURAL ANALY' -.
l.lr_rf'F:
!: EIEFLEITI I]}'1 ,ifitl:r
1
i
TEFLEIT ] L]IJ ,::l1l.l-, !l
[1
;
B. i:lrr ri1 , :11,
1
-{1. ,_n
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ii
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f]t,::tFll-. ]i Fr:rR{::!: ,::lill:,
IlEI'1HIEF:
-
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f
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FEI]N I I]G l'llrllEt.ll r:1,:.1'l 1.1:1
I
1:: " !.i I i:i , It
I
:r
"ilI
-;:1 I " 1
:jl
l
4
41 i;tl. ,,I
.+":: tll
5
Solution l+.4
t{-rf{E:
I
nEF LET]TIr-rt"l , it lt il
:l
5
I'lE l.lfi E Fl I
1
,:ftH.1
rj 1, : J l. E; 1, : r 1' : : 1": f 1d; +. hrl 1
1 :1
2
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RL 'I:':I I- t_1Ri::[ ,i t*t l
1l ii -.-!1" Il1,+ *1 . 1 l 1 '+ -,11"!:r:i7
a
I
filiEtlli: Ftrllr:rF r:l::ll tl el
!;l " I --:f ,, I
rr. I
:
$
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F
-.f :.
ljl
--;:;',::
4
{1
,+ ll
Solution4.J
-:ri:, !:,
ll.
RDI'RI'IUI.I { F.llifl::i.r
o Ll. ertrBSi:E ,|], errrirlS;1 -.rjr, Erljltri'!:i jJ *Ll, r.jr[1[:rlt3S
E :rS !.l ftIl .l ft tl r::'fl Fl l 1l
,:li|l rlr - 1.rlr. 5 L r_1 ,, 5 *1 | n . 5
I
t]
B
0 fi
IJ E
APPENDIX:
SOLUTIOXS
TC PR.lcr:z
PRABLEUS
Solution4.4
Solution 4.5
UNDERSTANDING STRUCTURAL ENA-LYSIs Hl:lOE:
il t'E:FLEr::TI nll ,it1t"t:'
a
I
1 i. i} 3 i' 1 5 , ;r 5 :i t.:r . i:$ i fi
: 4 5
HEtlUF:E
1 i:
-|:|.
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f
:l
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.+
:
1
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11. !l
rj 11.ErEr::l E" [r15 er. il} l B. l'lr;i:
ljl
R!:I RI. :'l FnELiE {P;ftl - "i1"t
f
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!:1
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,_nr,, F1$ E
{:r,,8811,+ir4 a
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'l l :l l l "{.F " 3 -'l E t. ;1
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--*E -"+{1,,lt 4E' :l
!r" I .-; , ;1
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5
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5 tr
-.:1. I
6
Solution4.7
E
FII:ITFITI [IN ,iFrHn$]
;, t:4::
t il !
-"lt!r " 1
NHFLEI:"T'IIFI t l'll'l l
:i
I,IEI.lBEFJ
ti: " 1
E
:, 46,,1 l. l5; l
t; 7
FrEl.i[rII'lCi t'11f1E!l1 I t.::t.lt.l]
14. 4
EI
1 I
E !]. EllSeiSL: FJ.g85lE€l 0. A6+3:rj1 iJ. 1i:tE:5:i4
tjl
{} o
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Fr:rTFlTlrll.l 'i RFio:_i:!
:i{3 --li',i
1+. +
5
:l
:-iHEFIFI F*F;I:E ,::t|l.l r
1.+,+ -1+,,1
:i;E - llEl
* 1'+.'+ l.,+,,+
F{NNE
l Llr:
,.1er"45J
I 3 4
Solution 4.6
N E FLE I::T I.IH LH tl :
:
;,;i
E
E iJ
;i
".,:r, ;! ;r ,, :
l, ;;l - iJ ., ;l
il 14"4
APPENDIX:
SOLU?IOHS
lt
PRrc?1:E
PRCBLEMS
265
Solution 4.6
Solution 4.7
l*t'r 266
UNDERSTANDING STRUC?URAL'XA'YS.'.9
H'-rDE 1
l
: 3
I
:;
0 1 t. 4 * r; 1il, iitil 1 i r, 5 r3
cl I
!i ,, ;.f f r:l
tlE BrFl
11],:IFlL. j,'i FCrFil:E r.l:f.l:]
Z NEFLEI::TII']I.] ,i ntl r E
E']TETI I]I.I i Efin5:r
- 1"+.riSr::
E rl, ritr:1:f 74F B " tlrg?t4t B"rJrtlrrF 13"Ijr,-J 1i 1r
* I ;+ ,, rl:ti:j - 14, ri S E l
t:l,,E[]rit:_l1fi F, $r-.1::r;"8?
a _11.
J34
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ErEt'trrIfttr l.lr":rflr l,;h.l,l l'11
1 ;1
EI F
a
*:1!J trrj'
[l
--:ltl 3lJ
11
- -ler TIE
ti
[J
*:Ji] :iLl
i1
::0
*E;l:l
:J
_1 ,1
.+ t; t; { t:
Solution4.8
H IIEFLET:|IlOH ( f.t:
H {:i
ra1 L-r
::lil *iifi ll
tll
t:r
-.:{r
'- I Sril B rl Li
u {1
0
li1
268
trf f i
/ r *l >olutron ). )
so,u,ion56 F
W" *
E*
SOLUTIAT;S
APPENDIX: ' hdptet
5
a'
t
et- bjtI
q
7a
?=ac l :'E
''- :
:
PR ABLEI4S
Graphical solutions to Practice Problems shown in fiSs' )2 and 33'
ru.
Solution 6.1
rT' Solution 6.2
iJ4z'r'"65$"*
Solution 5.3
j
"--'1?,t tcuov I
270
UNDERS?ANDI NG STRUCTURAL A]VATYsf S
fua,t-.\*a-a,.
Solution6.4
fi''iTiinF
e_
ta
'N{ Solution5.5
L^.^
Solution 5.6a
APPENDIX:
SOLUTIONS TA PRACTICE PROBLEMS
[T
.CA'''
27r Solution 6.6b
o4n"ne.ri _[-[
r . ,-. Conputer sofutians
These solutions are to be used as the basis of practice lor Chapters 7 and 9, The computer solution to the structures shown in fiSs 55 and J6 is given in the following pages. The input data are shown below.
5
I tla
2
-b;@;
I9l,,l. robabton seLLlat^a.41 8 mat
27 2
UNDERSTANDING STRUCTURAL ANALYSI S
li!trE
li trIF L,.Ef:TI fl.| ,:tltrt:r
1
tjl
t:1 tal r;l t:l
; -a
|::]
ll tll
r'rfl,18Eti
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r,1 6
I
LI ri:1
ra1 rl
,.+ :;
il r:1
d:
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13. 5
70
Solution I
rl i:,, '1::J ltl
IiHE RF] ,il F-[]E:|:::H ':t:.l 'l :1 - :[:r. ll 1.1
-t.itr":i J:r
48. q
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3 -:l :16 *:llil
:lli
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F
APPENDIX:
utrtrE 1 I 3 ti
{; ;: llEtlFt[:E
n
Sc.iilfIo.s
271
PROBLEMS
= lsFLErlTr
}.: LIEFLEiTIAfl i Htl ..
L-rt-l 't H.l
R{rTf{TIilll t FrFlrrgt g
tr E
F - 1-:r8
t1 t1 E
g. !;:E'Jl B
fJ " 0Er&&48 -0, Er0nldo --r, ErErr!01& Lir" qErE:lEr']1
E1
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i H E H F: FIIF][:F: ,i ti H .r
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ll r:iF.:r:E { t::t.l.)
1
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tj
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-rjt ' I riJ. i 1:1' I
-.11. s *:i ;.
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4
Solution 2
274
UNDE RSTANDI N G STRUCTURAL EUA'YSJ.S
HLrnE
1 l
+ 5
tlEI'tUEt;:
I
H L:IEFLEI::TI I]J,] 'i f, .t ;r
:
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tjj
rlj. it;lti
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Solution 3
flr:rnE 1 I
li {lFJ:LI:aT I trl.l r.t.lt,l.r t:l l:r, ll!,,+ J." t;Eir 1 " 5,:i i"
ri i'lE H f:rEl;l
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iI
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So.lution4
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a
4.
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Solution 3
Solution 4
276
UNDE RSTANDI N C STRUCTU RAL .4TA'YsIS l l r - lF E
1 J
:;: nEFL..ELiTI Uf..l i:t lt1.:
::: frH Fl- l:
I L-rl] '-:T ,:.1.1t 1r
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Solution 5
FIONE
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r :j
4 5
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: J
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4
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1 !
Solution 5
':l
5
ej, ;l
I
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l':l
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fi
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2 78
UNDE RS T AN DI N G S T RI]C?I] R AL,qIT4IYJ]: na
d
.t.1ass
.e thod
gtj o-
Ihe qualitative practice problems shown in fig. 4l har.e beer g:\ei. loac and dim ens ion v alues and analy s ed b y c o m p u t e r . T h e a r p h a - n u ne r r c o r r e s u r r s are given on the Jollowing page. The section of the members are conslant tor each problem and are ds follows:
= 3 000 x l ou mmu "* = 5 0 00x
106 mmq
= 1 4 kN /mm, = l 0 kN /mm,
z,
t
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@
APPENDIX:
t!0nE
]
SOA'?I(XS
M
NEFLE.:TITX'I i ltli:
L
PRrcTICE
PROBLEAS
:': Rr]TFr'T I lrf.l ,:t;:t1L-rF l,
f
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6
rl
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279
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A, Btrr:lf::Jft
.-R, E$t 1r:_i
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E:E:!f_'I l.t6 flnfiEt.lT {:lri:HlIl
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I - -1 b , , I
:-
,- .;:;i . :t
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L]
56. 6
li-l" iJ
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al
o
tr
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0 Solution 8.1
tl-lllE
t
t'EFJLE*T1r:rf.l { t{t'l1
tr
L
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Solution 8.2
UNDEP,STANDI N G STRUCTURAL EIVAIYSiS Chapter Graphical
l1:
The 'lieid
so lu tio n s
ljne
analVsis
to p r a ctice
af reinforced
p r o b lems
fa d . * | ,r t I
concrei.e
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faa-or
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70.
I
ffii
Solution I l. I
affi*e" .vq4'tat., ef , PqrvLA
etf trJ,/gfh
ef--
E1+. 1tu4h.\ \ 5.t -
P4"d,
S olut i o nI1 .2
vaais ol rtla'twt,
E{+ I*JA".
r4d. + ,*a./-lt\
S olut i o nI1 .3
.
Solution I I.4
Index
a, distribution lactor, 157 Arbitrary sway deflection, 158 Area-Moments, 150 Artilicial restraint' 124' 129 Ax es , 6 Axial shortening, 40 Bars, number lor statical determinacy, lJ Beam, built-in, 24 Bound, lower , 188 unique, 188, 201 upper , 186' 187 Cantilever beam' 48 pr opped, 25 Carry-over factor' I l9 Clerk-Maxwell, Theorems of Reciprocal dellections' 109, l3l, 133, 226 Collapse, plastic, 179 frames, 190 mechanism, 182' 188, 190' 243 Compatibility, equations ot, 100' 102, 104, 106 Component forces, l7 Computer output, 135 Continuous beamr three sPan, 46 two span, 45 Contraflexure, 44 Convention, diagrammatic, 9' 41, 59 plotting, I l,l3 Deflected shape ol beams, 13, l6 of indeterminate beafis, 44' 46,47 ol indeterminate lrames' 62, 65' 66' 68 of statically determinate lrames, 20 Dellection, using Virtual Work, 84, 89 oJ indeterminate structurest 9J Determinacy, condition oJ' 22 Determinate structures, 97
Distributionfactor, a, I57 Effective depth of reinlorcement,204 ElJective length of yield line,207 Elastic analysis,196 Elasticity, modulusof, 85 E qui l i bri um,3, 5,6,23 equationsol, 3,23' l3l frames, l7 internal, 9, ll, 25 node,57 Factor ol safety, 135, Fixed-endmoinents,158, 159' 172, 174 Flexibility method,comparisonwith stilfness, I32 matrix, 109 pinnedframes, ll I ri gi d frames,97' I16' 151,225 Free-bodydiagram' 5 Grids, 137 G, modulusol rigidity, 88' 143, 144 Hinge,internal l5' 53' 55, 6l' 70 plastic, 180 Hyperstatic,28 Indetetminecy,kinematical,I 28, 141,144 statical, 3, 23' 28 Instantaneous centret 197 J, torsionalconstant,88, 143, 144 K, relative stiflness, Linear elasticity' 179 Load, moment' 48, 49 uniformly distributed'50' 5l Lower bound,188
UNDERSTANDING STRUCTURAL AA'AIYSIS
Matrix, Jlexibility, .l0T stiJ J nes s ,I 32, 145 Mechanism, 33, 183, l9O, 193 p s uedo, 33 Modulus of elasticity, 85 of ri8idity, 88, 143, 144 , section, IEJ Mohr's Moment Area Method, 150 Moment distribution, 149, l6l Moment, load, 48, 49 sw ay , 157 with sloping members, 170 Neutral axis, 85 Pinned structures, l3 Plastic analysis, 178 collapse, 179 hin ge, 180 modulus, I92 Portal lrames, 67 -'f ropped cantilever, 43 Qualitative analysis, 5, 38, 57 Reactions, 8, 15, 17, 24, 30, 36, 40, 45, 58, t64 / Redundancy, 3, 28 Reinlorced concrete, ultimate strength in bending, 205 Reinforcement, 204, 20J Release, 23,29, 129 Resultant, I7 Rigidity, internal, 28 Secondary elfects, 40 Section modulus, 185 Serviceability state, 7-5 Settlement, l20r 155 Shape factor, l8l Shear forces, 9, I53
Shearing force diagrams, 9, l4 Simultaneous equations, 97 Statically determinate structures, 3, 23 Statically indeterminate structures, 22 Statics, J Jtrllness, l/ absolute, I15, 155 matrix, l3l, I3J, 145 relative, lJ7, 160 torsion, I42 StilJness method of analysis, I l4 Stiffness method of analysis - comparison with flexibility, .132 Strain energy, 7J Superposition, 41, 54, 58, 100 Supports,24 S\yay,63,7A, 167 Sway moments, 168 three-dimensional structures, 3J, 138 Torsion,88, 139 Tree structures, J0, 36 Ultimate load analysis, 179 theory,206 Upper bound, 186 V i r t u a l Wo r k , 7 5 , 8 5 , 1 8 6 displacements, 77, 186, l9I,2Oq forces,76,84,87 moment,85,93 Virtual work applied to redundant structures, 9J yield line analysis, 202, 206 Yield line analysis, 202 Yield line, hoqRins, 2.1j Yield strain, i79 moment, 179, lg0 stress, 180
