A theory of latticed plates and shells

and (p are shown in Fig. 1.3. In this figure the deformed fibre is onwardly displaced in such a way that points M and M" to which point M has sheared coincide. Displacements change linearly according to shell's thickness. Designating the distance from the point to the medical surface as z (positively if the point lies on the external normal to the surface) we can write: uz = u + (E ( £ nu + -I- kl2v)z, vz = v + + (E (£22 + + knu)z, 22 wzz — W —w W+ + (£33 (£33 -- \)z. \)z.

1.1.2b. Medium bending. In this case ff < 1 (i = 1,2). If, in addition, the shell is shallow or during deformation divides into shallow parts, we have the relationships: £1 = eu + 7?/2,

£2 = e22 + 72/ 2 .

w

= en + e2i + 7i72

(1.6) (1.6)

for the tangential deformation components and d7l MdA dA dA \\\ 1 ((f,dl2d dl2 dB 11 f(adyi,dA &n 1 dB dB \\ l7 K2 = •72 A + ^ 7 l)1 B + 12 AB \ -AB{ -d0 d^' ) = -AB( -^<>P d0 )' 1\_dw 3iu 11 fd-fi // n a07 72 2 dA \\ \dw 1\dw dw OA = T 7l T = 7ril = 7 = 2= = Bd0 Bd/3 = -ABV^-W)' -AB^-dfi'")^ AB IIdc? d Ada' c ? l272 £^ dB -IacT' Bd0 ~~

Kl

(ȣ

"£-

( L(1.7) 7)

— for the bending deformation components. Formula (1.4) becomes simpler and can be written as: 2V> = -26 + (e 2y>. (e,122 + e 2 i)Icos cos2

(1.8) (1.8)

6

Chapter 1. Reticulated Shell Theory:

Equations

1.1.2c. Weak bending (linear theory). In this case the turn angles are small compared to unity. Leaving only linear terms relative to the displacement in (1.3) we find £i = exi,, ee22 = e 22, w w==e ie2!2 + e 22i,i, 1 9-yi d-fi 72 dA #7i 72 dA 73 /Ci — K\2e2l K\2e2i — T T K\ — — Ki£2 — 7AB ^ ^ T 7 , A-7"5j da d/3' AB dp A da 11_dj/2 #72 71 9B dB 7i K2 = k2ei - kntu -—-—1 #72 — 71■ .5 B :. dp AB da B dp AB da B a/? AB 8A da , , 1 dy 7, dj22 7i d,4 71 3>1 r = *1 1 2 £ l - ,* 1 e 2 1 A -1-#72 ^ . +— ^ . da AB dp' r = fc^-^-^ +J L -

,. „. (1.9)

Literature has several formulae of the type (1.9) for the bending deformation components due to the fact that the small terms containing the products of the tangential deformation components to the curvature of the shell's middle surface can be added to the right-hand side of these formulae: these terms do not affect evaluation of the errors based on the hypothesis of retaining the normal element. For instance, if we add the component ki2u>/2 in terms for «i and K2, and also the value 2fciC2i + £12(^2 — £ i) f° r T to the right-hand side of formulae (1.9), we obtain (assuming kl2 = — l/R12): w 1 dv. du vv dA dA w + Add A da + AB3/J ~AB dp ? Ada"^ ABdP+#~Rx Ri' 1 dv dv u dB dB w v> u £2 = + , da ++ R~2 B^() AB~fa dp AB BB dj) ~RR~2' AB 9o7 2 V. A A d /u\ B B d 0_ /v\ 2w 2w v d_ /u\ A 8 B d /v\ 2w U U + da '~ BBdJ)\A) A ~A R\2 dp Bd/3\A'Altet\B~)~R^ Ada'\B~)~R^ 1 d / 1 dw dw U V d_ V 1 d f 1 dw u v \\ A da + Y+l~RT2j A da *' '" Ad^\Ad^ = R\2 "' Afa\Ada~ R~l~Rr2) d_ 1 aw v 1 dA u 1 d_(Bv) + + { ( { Ti ABdB\ dB Bdp R R ) 2ABR [d0 3 / T ' ' 8a ''\'' dp +RT dp 2ABR AB Rl2. 22 l2 u U da U 2 dw V \_d_ v_ u U \ 1 d_ ( l_dw K2 : '~ BdJ) Bdp ~Rl2u) B \Bd0 B dp + R R dp\BdpR~2 2 ~R2Rl2 1 dB_ V 1 Dw u _J1 [dad_(Bv)(Bv) ~ Wd_(Au)>\ ' + + + ~AB ABda{ da* V A Ada R ) 2ABR da da Rr U R\2l2 j 2ABR U l V }_d_(_}_ 1 dw <^,J!_ Jf_\ 1 dA (1 dw _u_ dA 1 d_ dw }_d_(_}_ n)2) ~ if R~22) ' £l

=

These formulae are given in t h e work [14, p p . 53, 54].

1.1.

Anisotropic Shell Theory: Basic Equations

7

The book [76, p. 207] presents another variant of these relationships in lines of curvature (Rn = 0). Here the three formulae of (1.10) take the form Kl

dki u dki v w 1 3 / 1 dw\ 1 dA dw 2 ~do~A ~da~A ++ ~d0~B ~dp~~B ~~ B2B~2 ~Ada~ Ada~ \A~da~J~ \A~da~J~~AB ~AB2d0d0' 80 80' dk2 u dk2 v w 1 9 d 1 dB dBdw dw d_// 11 dw\ da A ++ dp Ji '"' 'd^A 'd^A R\ B dp^B~By ^B~W B JdP, AW da da '"' ~Bdp\Bdfi)~^B'd^d^' 2~Bd^\Bdfi)~AW'd^d^' (ki-k2)\A A d / u \ B d /v\' = 2 [Bd0 \AJ ~ Ada Ada~ \~B~). \B)_ B 1 / d2w 1 dA dw 1 }_9B^dw\ AB\dad0~ AB \dad0~ Ad/3lfo~B"da~dp)' Ad0fa~B~da~d0J'

"

2 K2

"

T

((1.11)

'

From (1.4) we have for the linear problem 2rj> — 28 w cos2y 2ip++(e(e2 2-£i)sin2y>. — £1) sin 2 ==-26 ++ u>cos

1.1.3

(1.12) (1.12)

C o n s t i t u t i v e equations for anisotropic shells

We now introduce constitutive equations corresponding to the classical anisotropic shell theory. If the shell's material at each point has only one plane of elastic symmetry parallel to the plane tangent to the middle surface the constitutive equations become [2]: Nl S

= C\\£\ + CU£2 + Cl6<*>, N 6lV, ^22 = Ci2E\ + C22e. £22 ++ CC226lV, = Cied + C26£e2 + Ceeui, H = £>J6KI + D26K2 + D^r,

Mxx M

= =

- ((£D> ,I iI K K 22 ++ DD1166TT ) , / cI , + £> D1212/c

M22

=

-(D 122K\/c, + D22K2 —(D\

+ D26DT). 26T).

(1.13)

If the material is orthotropic and the main directions of elasticity coincide with those of the coordinates, Eqs. (1.13) become simpler owing to the condition: Cie = C16 — CC2626=—-Die D\e=—DD262e=—0. 0.

(1.14)

For shells made from isotropic material, besides (1.14) the following conditions should be fulfilled: C n„ C

2

= C22 C22 = Cutv 2C«6/(1 - v) ») = Ehf(l Eh/(1 - vv%), — —C 12/v = 2C««/(1

Dnn11 = D22 = D D,u2/v /v = Dee/(l-v)— D

V

= Eh /\2(l-v Eh /\2(l-v ), 2), 3

32

(1.15)

where E and t; are respectively Young's modulus and Poisson's ratio for the material.

8

Chapter 1. Reticulated Shell Theory: Equations

1.2

C o n s t i t u t i v e E q u a t i o n s in t h e Shell T h e o r y

Reticulated

1.2.1

Constitutive equations for the rods of reticulated shells

1.2.1a. Deformation of a reticulated shell's rods. We assume that a rod's deformation is equal to that of the line coinciding with this rod's axis in the calculation model. We fix one of the families (the i-th family, 1 < i < n) of the shell's rods. The position of the axes of this family of rods on the shell's middle surface is characterized by angle (in Fig. 1.3

;, K' = KXC] C] -IT*

=

S,C,(»C2 — K\) Ki) + T TCOs2
(1-16)

Here and further St = sin y>,, Si

C; = cosy;,-.

The change of the rod's axis of curvature in the plane tangent to the shell's middle can be derived from the formula d_ V, (1.17) -V,V< dp, 9a

«--"*

^=ii+m)-

Value V'i, depending on the problem is calculated from relationships (1.4), (1.8) or (1.12) in which angles

must be replaced by y>; and fy. 1.2.1b. Forces and moments in the reticulated shell's rods. Let us assume that one of the main central axes of the rods' cross sections coincides with the direction of the normal to the shell's middle surface. The positive directions of forces and moments in the i-th family of the rod's cross section are shown in Fig. 1.4. Their dependence on the deformation components is assumed as N: N; M; G' H' QT Q: 5*

= EiFie'i, E^e;, = —EJJUK", -EiJuK-, -EiJu^, —EiJ = — £; J22iK ;K;, i, = — —GiJsiT?, G;J3,T;*, = -V.A/;, -V.A/-, = = -ViG*. -V;G*.

(1.18)

1.2. Constitutive

9

Equations

Figure 1.4: Here Fi, J n , J^, J3; are the area, main central inertia moments and inertia moment during torsion of the rod's cross section; Ej and d are Young's modulus and elasticity modulus respectively during the material's displacement. Generally the rods will be twisted naturally. Strictly speaking the influence on every force and moment of all deformation components should be taken into account. But it may be neglected, because the initial relative torsion of the reticulated shell's rods is small. It should be noted that in the work [43] the value of bending moment G* unlike in formulae (1.18) is determined not by the change of the rod's axis of curvature K° but the angle value xj>i. Hence transversal forces S" are also determined in various ways according to the defermation of the calculation model's middle surface. As this is the value of the rod's flexure in the plane tangent to the shell's middle surface, according to this work, does not increase the order of the system of differential equations, hence the additional boundary conditions (see Sec. 1.2.7) occurring when solving the boundary problems in the reticulated shell theory cannot be fulfilled. One should bear in mind that formulae (1.18) make it possible to determine forces and moments in the shell's rods if the outer load is applied to the nodes of the lattice. When a load is not applied at the nodes, the rod's stressed state due to the local load distribution along their axes should also be determined.

1.2.2

C o n s t i t u t i v e equations for a calculation m o d e l

1.2.2a. Determination ofthe calculation model's stressed state from the rod's forces and moments. Let us assume that the rod's forces N',Q', S" and moments M', G*, Hi are distributed continuously across the calculation model's cross section. Then the formulae of forces and moments (Fig. 1.2) for the reticulated shell's calculation model having n families of rods are: n n

n

E

JVi = £(A?«?-S?-«ej)/a,, ^2 = £W*< + s;Sic)/a„ i1== il1 n

D

1=1 ii ==il nn

C - + s?c?)/ S:c*)/aiaj, 5S2 = ££(/V? W s ,SlC c -,--S*s?)/a,- ST*!)/*,) Si = £ W W* w i•1== 1li

1=1 it ==li

10

Chapter 1. Reticulated Shell Theory: Equations nn

nn

E

Qi = E
E

Q* = £
it = = ii nnn

i=i »=i

nn

M2 == E< ^^(M( M - 5 ? --//* SH-s^/a, ,c i )/a i ,

E<

M, = £ W c ? + ^ . ^ ) / « . - . i=l i=i

zfj tf,

=

E 2 ^G J 7j7, /'/a n „ E

=

t. = = il (

C

c

a

nn

E(

A//;;55l lCCi i--t /ff; ;Cc??.2))//aa„„ --£^((A nn

M„ A/I,

«=1 i=l i=l

nn

tf2 = - ^ £ ( M * 3 , c ++//'HT i/ Si)/ai, 5 ,)/a.-, 1=1 1=1

nn

E

Af22,, = M = -- ] ^2 T G *G'si/ai, Sl/a„

ii = = ii

(1.19)

ii ==ii i=i

where a, is the distance between the i-th family of rods. Note that the first six formulae in (1.19) change into the last six if we replace values Ari,-/V ,S,,Si,Q N1,N2,S12,S,,Q (1.20) 2 ,Q,,Q 2 ,/v-*, 2,N^,S' 1,Q'1 £,*,
Ml,M22-H-l,-H Mi,M H2x,M ,-Hu,-M M2„M;, -H-,G'. -//,',(3*. 2„M',2„M-, 2,Mi„-M

(1.21)

1.2.2b. Constitutive equations 1) General case. By substituting values (1.18) in Eq. (1.19) and taking Eq. (1.16) into account we obtain the following constitutive equations for the calculation model in this reticulated shell theory: n

#i

E

^23,Ci/c-, = Cn£i + Ci 2 £e 2 + (7i6ui - ^3,Ci/c-, ii = = ii nn

E + Cfxu CfxU + E ^2 ^ CJK-, c?*4

c /c N e22 ++ C C2626W w++ "^2 ^T s 'w - 4i. JV22 = C21£i + C22 22£

i=l 1=1 :=1 n

5i

= C6i£i + C62£2

i=i n

E

S2 = = C6i£i + C62£2 + Ceev C66W - 22 Ss^lKKi> <' M, = M M22 = = H = Hxx =

i1== 1i i=i

-[(£>„ + A ' „ ) K1 1 + + ((DD1122--//SS::II22))««22 + ( 2 D 1 6 - / C 1 6 ) T ] , A'„)K -[(D K ) + (D22 + )K + (2D26 + )T), -[(D21 +K K22 + K K26 21 - K21 21)Kl Kl*1 + (D22 22)K22 + (2D26 26)T), (D - Ki^m + (D 62 + +^ K£>)K + (Dee + tf&V. (D6l 6l - *£>)«, + (£>62 2 ' ) «22 + (A» + tf&V.

H (Dee- - ffj?)r, K™)T, tf22 = (An (£>61 + K™)KI ffj?)lt, + (D62 - Kg>)K A£')K2 2++(A* n

E

nn

E

= -E^c.'c?, -£/?««?, M M* #««?. Ml4u = J2l?*iK°2, = £ ti = = li

1=1

i=i

(1.22)

1-2. Constitutive

11

Equations

Here n n

nn

n

1=1 = ii ii =

1ii = ==i1\

c C « == Cee KiSiCJ, C C„ = ]£*.•<$, £E Kid C» ft, == 2_^ £*<«?<$, C1616==E tf.s.c?, 5£3 Ki*i i> E i1=1 = li i= n

n n

cC26 == E ^x;/o?c / f , 3 ? ci,, C - ^ C j i , E Dn £ > c j , A2 = C /2 = ^E / , s M , Z) = E ^7, , f, Dn == E>?, E £22 £ > a j^, /->26 j = Dji, = 5E 3^*'^f " ADii^Da, £ = = E Y. Die^Y, A~n 22 , CjSj A'll = = A~ /f = = A"i K\1 = = A" A" l = = E CfJj c,, C, , A"j = A" = 5E 3 C,-SjC,A"i6 C, ,c, cos2y,, « S = K$ = E J2C cl K^ cos2^. ^ = = j^C X)C] fcos2 E C cos2, 53Ci J2 ?cos2 K$ = Ki? =E£tCis>c,^ Kg= = ^ ^ E C #.-*?, c2222 == ££>.4,

26

*->i) — *-'&■>

it = i 1=1 nn

1=1 = ii it =

nn

n n

66

i1== 1li n n

S C

it==ll 1=1

0

1

1=1 i"=i 1=1

i;==li

i=l

2222

2

22]

n

6

t1=1 =l 1=1

nn

IiS

22

16

n n

t

i=l 1=1 ■=1

g

26

i1=1 =i ■=1

"61

l

—

Ag2

—

/

.3 s „C iSti i C ;t ,

1=1 i=l 1=1 n

ti = il

»=i

1° = EiJafoi, EiJa/at, M I?

)

)

n

iC iC

u ¥ i)

i==ll i1=1 n

S

W

1=1 r=i i=i

K,t = = a ^ VVa^V,{E,J ^ J w / c ?2,^), ), K

(1.23)

where A"; = £,,\F,/at, /, = EiJu/o-i, Ci = GjJ 3 j/ai. Parameters A",, /,, 7°, C, are relationships of the corresponding rod's rigidity characteristics to the distance between their axes. The dependence Ki6 = /Q, — K^ should be noted. The calculation model's cross section with J 2l ^ 0, unlike the usual continuous shell have bending moments Mt, and A/2j, acting on the plane tangent to the middle surface (Fig. 1.2). As shown in Sees. 3.1.2 and 3.3.3 this increases the order of the system of differential equations and fully corresponds to the increased in the number of boundary conditions (see Sec. 1.2.7). 2) Particular cases. In practical calculations one can assume that J 2 ; = 0, if this does not result in a geometrically changed calculation model. This is similar to girder calculations: the rods' rigid connection at the joints is considered as hinged. As for stresses occurring due to the rods' bending tangential to the plane's middle surface their values can be obtained from the known deformation of the middle surface. Further we assume J 2 , = 0 everywhere, except in Sees. 1.2.7, 3.1.2, 3.3.3 where the numerical evaluation of the error due to this assumption is given for specific problems. Thus parameters If, K; become zero. In this case some constitutive Eqs. (1.22) can

12

Chapter 1. Reticulated Shell Theory:

Equations

be substantially simplified. Precisely for tangential forces M

=

Cnei + C12£2 + Ci Ci66u;, uj,

N22

= =

02161 + 02262 C/2262 + O26W, C/2i£i C2ew,

S 5

==

0C66ll ei £ l ++ 062£2 C62e2 ++ O66W, CeeW,

(1.24)

where 5 = S\ = S2. Moreover, from the last two equations (1.22) it follows that Mi, = M2. = 0. We now present the calculating model's constitutive equations for particular cases of the lattice (in each, in accordance with the previous statements, J 2l = 0). 1. In torsion the rods' rigidity is assumed zero (J3l- = 0). The constitutive equations have the form as in (1.13). For particular cases of the lattice they will coincide with those of anisotropic shells whose material has an elastic symmetrical plane parallel to the plane tangential to the middle surface. In addition, considering Eq. (1.14) they reduce to constitutive equations of shells made from orthotropic material provided that coordinates a, f$ and the normal to the middle surface are the main directions of elasticity: N\ Af, M,

C\52£2, £ 2 , N2 = Cn2iei Ceeu, = CU£i + o\ ei + C22£2, S = = Cee^, = -(ZJnK! -(£>!,«! + Z), Di2n2 *2),2 ), M2 = =-{D -(D KI ,/c, +D + D n K ), H = DD T.T. (1.25) 2X 2 22 22 22 2 ) , 6666

Now we give some possible variants of the shells' lattices which satisfy conditions (1.14). It is sufficient to consider the first two such conditions as, in accordance with Eq. (1.23), the latter two are obtained from the former by replacing K, by /,. If n = 2, only the following two cases are possible:
=

K2

_ =

sin 2y> 2ip33 s'm((fi sin(y 33 -—yip sin(y> i^) sin 2y?i sin(y>2 — <^i) sin(y>2 + s?<^i) sin 2y>3 sin(y>i - y? tp33)) sin(y>i s\n(ipi + +

ip\) 2 +2 +

In the latter case the domain of possible values 0. Evidently, the lattice will have orthotropic properties if it is presented as a sum of simpler lattices each of which is orthotropic. 2. The shell consists of three families of rods (n = 3). All the rods have identical cross sections and material and their axes form equilateral triangles at the shell's middle surface: Fi = F; Ju = Ju Ju E{ = = E, G; d = = G, F JJz, Ei 3, = J 3, f>\ = f>, 3 = I^ + 2 T / 3 . 1

a, = = a, a, a,

(* == M ), (1 1,3), (1.26) (1.26)

1.2. Constitutive

Equations

13

In this case 3 3

E E ? s

1i■=i == 1i 3

33

33

=E E ? = §' E E s ? c ' = 8'

5
E E srf< 1t = = l 1 = 11

9

c

i11=== 11i 3

3

1i = 1i 1=1

SiCi S i
E E

°- -

11i ==11l

The latter relationships allow us to present the constitutive equation in the fol­ lowing form:

9EF ( 9EF „ 9EF 9EF ( 1 \\ 1 \ £ N 8a £ £l, + 3£J 8a £2£2++ 3£l£i 9EF 3fiJi(l+7) 9EF H = Hx = H2 = T, S 4a = 8aIT"' 3£J , 30.7,(3 , 11 - 777 Kj \ 1 (3 + 7) ( M, = «! + Ml

Nl =

=

M M 2

=

-"ST ^ll

3 'J' J ' ^>=="ST s rl ( 33 Jj '•

fa8a

("'

3EM3 3£J,(3 + + 7J)) / 8a

3 ++ 7 " * ) ' , 11 - 77 ^ Ra «2 3+ 7

(1.27)

where GJ3/EJ1. The result agrees with those given in the work [10] which studied the shells' stability precisely with such a lattice. Formulae (1.27) may be presented in the form (5.70). Comparing these formulae with those for continuous shells we find that in this particular case we can use constitutive equations for isotropic shells made from ma­ terial with modulus of elasticity E\ in the formulae for tangential forces N\, N2, S values h = F/a, fi = 1/3; must be taken for the shell's thickness and Poisson's ratio; in the formulae for bending and torsion the values of cylindrical rigidity and Poisson's ration are considered equal D = 3EJ 3 £ J1,(3(3 + + 7-r)/8a, )/8a,

u"22== ((1l - 77))// ( 3 ++ 7). 7)-

When EJ\ » GJ3 formulae (1.27) may be brought to correspond with formulae in the isotropic shell theory; constitutive Eq. (1.27) will coincide with Eq. (1.13) under conditions (1.14), (1.15), if the values of the shell's thickness, Young's modulus and Poisson's ratio of its material are as in [62] h = 2s/zJjF, 2^3Ji/F,

EF E'=^-y/3F/J E' = ^VMVU u 6a oa

v= ^ = 1/3.

(1.28)

When studying shells with the described lattice, Eq. (1.28) make it possible to use the isotropic shell theory and solve many problems on its basis.

14

Chapter 1. Reticulated Shell Theory:

Equations

Figure 1.5: 3. The shell has a rhombic lattice (n = 2). Such a lattice is obtained from the one shown in Fig. 1.5 if we assume no third and fourth families of rods and that the rods of both families are identical and aj = a2 = a. Remember, that the suffix i, denoting the t-th rod's family of lattice with i = 1 for all values except forces and moments, may be omitted. Thus in the case of this lattice ip = ipi = —p2, c = C\ = cos<^i, s — s% = sini^i etc. The constitutive equations for this rhombic lattice take the form: iVj = Ni = N2cot cot\2

, M, = -2c2[{Ic2c2 + C S 2 ) K , + (/ --C C)S2K2], Mi 2 -2S2[(I-C)C2KI KC)c Ki + (IS2 + 1

M2

=

//, HI

C CCOS2V)T, = 2s {2Ic (2Ic + Ccos2f)T. Ccos2
H2

= =

2

cCC2)K2],

2

2C (2IS s* + 2

22 2

(1.29)

It is important that linear forces Ni and N2 are interconnected. A similar connection will exist between linear bending moments Mi and M2 if the torsional rigidity of the rods is neglected (C = 0). This reveals the considerable difference between constitutive equations (1.29) and the corresponding equations in the continuous shell theory. 4. The shell's lattice consists of four families of rods (n = 4), with the rods of the first and second families being identical and ai = a2 = a,
S = Csew,

Mi Mi

(1.30)

Here: Cn = = 2Kc4I + K4, Cu

C12 = C66 C66 = =2Ks 2Ks22cc22, ,1

44

C C2222= 2Ks C22 ==2Ks

+ +K K33,,

1.2. Constitutive ft, ft2 0» ft i fox fti ft.t

1.2.3

= = = == =

Equations

15

- ( 2 //c c 4 + hu + 2Cs2c2), i ft2 = --22s«2 V c 2( (/ / - C), C), 4 2 -(21sIs* + h + 22Cs c% // sin2 2tp + 2Cc22 COS cos 2<^ 2^ ++ C C44,, 2 2 2 / ssin i n 22

j + +C 3C. 3 .

(1.31)

Assessment of the deformation c o m p o n e n t s and forces in the rods using the forces and m o m e n t s of t h e calculation model

The above constitutive equation are unambiguous dependences of forces and moments occurring in the cross section of the reticulated shell's calculation model on the deformation components of its middle surface. Now we shall study the reciprocal dependences (the calculation model's deformation components on its forces and moments). 1.2.3a. Tangential deformation components. Here we consider tangential deformation components e\, e-i, u itt Eqs. (1.24) as unknowns. The determinant of this algebraic system of equations may be presented as n

D =

E KiKi,K (ctfc}sl c| | + 22s c s c slc

Y,

k

S

2 2

i

3 i

ls

i,j,k=l ij,k=i

4 3 s3 2 2 2 2 22 22 3 3 4\ -cUfcslc k°k- - s]c s)c)s kc k - sxc]Sjc)S\).

(1.32)

We give this in simpler form. For this we fix three arbitrary integral numbers each of which is greater than zero and less than n + 1. In accordance with the possible permutations of these numbers we consecutively assign the following values to indices i,j,k in Eq. (1.32): i = u, j = m, k = v\ i = /x, j = v, k = = m; i = m, m, j = /i, [i, k = v\ i = m, j = v, k — = fi; p.; i = v, jj = = H, n, kfc== m\ = m, k = /i. m; i = v, j = The sum of the obtained six members after trigonometric transformations may be written as:

1

— K„.KmA"„[sin 2(yit -
j_11 Y,"" D = — D = — 16 Y, n

£

u,m,i/=l

''KliKmKv[siii2{V>u - pM) KllKnKl,[sin2(ipl, - Vlt)

+ 2(¥>„ll --tpm) yv)]22..

(1.33) (1.33)

16

Chapter 1. Reticulated Shell Theory:

Equations

The prime near the sum symbol means that only one member corresponds to each combination of indices fi,rn,i/ (there are six such combinations). It should be noted that value of the said determinant does not depend on the choice of the axes of coordinates on the shell's middle surface as Eq. (1.33) includes only the differences of angles >pi. 1) The number of rods' families greater than two. With all indices satisfying the condition H^m, m//ii//,, /i ^ m, n±v, ft ^ t>, m (1.34) the expression in square brackets in Eq. (1.33) will not be zero. Hence, D > 0 when n > 2, as in this case there are n(n — l)(n — 2)/6 (the number of all combinations from n various elements is three) combinations of indices /J, m, v satisfying condition (1.34). Hence, the system of algebraic equations (1.24) relative to £i, e 2 , w always has a unique solution (D > 0). This result is natural as with n > 2 the reticulated shell consists of a system of rods, rigid in the plane tangential to the middle surface: the homogeneous system of Eqs. (1.24) can only have a trivial solution. We can show that the solution of this system of equations is £l £1

=

£2 £2

=

111-^1+112^2 111-^1+012^2 + 1135, 1 2 1 ^ 1 + 0 2 2 ^ 2 + 1235,

ui (d =

0 3 1! ^ ,1 + 01 3 2 ^ 2 + 1335, 0335,

(1.35)

where a

O " "i"l

= = =

a 2 Ol2 '

==

E

i,;=i i,j=l

i113 n = = 22 °a-i2

1l n-- K K s s2 2 sin2 2 2D E E KiKjS ips), 2D ' i l j ts)sm (ifi (Vi -- ¥>;),

==

l1 n" <>D E

E

KiK SiCiS cCiSjCj s i n 2

t>,jj == ii 1l n-

i

ii

( ^ .
+ ip^siiSivi ipj), ^*>i j (¥>.+¥'j)sin (¥>,- ~-
-jTp E E -jp ~2D 1

A

5 sin

2

■,i=i

n

-

% *i} E i }<%

2D E

K K KiKi<

tt,j=i j=i t,j=i 1 n"

123 123 = =

1 -- = = jj pp E E

"33 =

1 n" 2D ^2 E

E

i,i=i i,i=i ■J=I

E

2D i,i=i

cc

KiK c c KiKic*ci

KiK

i*i

i

sin2

sin2

( v^ .' 1_- Vi)> ¥>j).

sin sin(^'t

2 (^' + + Vi) Vj) sin sin2(<^,((^,- -- y>j), y>j),

( v . + Vi)sin2(v3, -
ay = a_,,.

i,i=i

2) The number of rod's families is two. It follows from Eq. (1.33) that if at least one of the three conditions ft = rn, fi = u, m = %>, is satisfied, the corresponding

1.2. Constitutive

17

Equations

member of the sum included into it is equal to zero. Hence, D = 0 when n = 2 as in this case the combination of indices Eq. (1.34) is excluded. Thus, the system of Eqs. (1.24) with Nx = N2 = S = 0 allows non-trivial solutions. This means that the calculation model, inadequately fixed over the contour, will be geometrically changeable. The calculation model's geometrical changeability is discussed in Sec. 1.2.6. In this case the following dependence exists between the tangential forces N\, N2, S: NxXs3xXss22 + N2cxc2 -- 5sin(y>i Ssin(y>i + V2) y 2 ) = 0, 0, (1.36) which is a simultaneous condition for a non-homogeneous system of Eqs. (1.24) relative to ei, £21 w If (pi +
sSjS SlS Nir i -C C22 - CuNj xs222 22A X2NN 2 2 u> + 22

i tp2) sin 2 (^i ip22),)' sin(v?i -I- ip KxK (
£i

=

e

ei

?.

(1.37)

where u; is arbitrary. With ipi + ip2 = 0 2 ,, CeeN\ CM^I — —CieS C\eS Cl6S CMNI — £1 = -£2 tan ru> (p ++ — „—.. ., „ . . —£2tan 4 K sin sm' 2ip JCi Bin28 2

xKJfaC* 2c

C NXx CuS CX6 XXS-—C XeN a> = — „—. - . 4 KA\/Gc sin2 22-p' 2^ xK2c* sin

,, QO, (1.38) (1.38)

^

'

Here value £2 <s arbitrary, tp = tpx. Hence, the calculation model for a shell not fixed over the contour, if the lattice is formed from two families of rods, is geometrically changeable as its middle surface can deform without forces being applied. 1.2.3b. Assessment of bending deformation components. Here we determine bending deformation components KX, K2, T using the systems of equations Mt

= -[{D11 + Ku)K1 + (D,2-Ku)K212)^2 + (2016(2Dle-Kl6)r},

M2

=

Hx

(D61l-K -K^)K, = (De 6\ )K, V i1 + (D62 +

-[{D2X +K -[(D K2X )KX n n)Kt )

{D22 + (D 22 + K22)K2

Ki\2))K2 Ki

{2D26 + (2D 626)T], 26 + ^ 2K ) (D66 + KKg Kg))T, )r, + (Dee w

•"'66

H + (Dee Kif)r, H22 = = (D (D6l6l + + Kff) tf<2)Kl)/c. + + (D ( A62 * - Kg)^ K$)K2 «a + (Dee -- K^)T,

(1.39)

included in (1.22). If E, Ju 3> C,: J3,, the rod's torsion rigidity can be neglected if this does not lead to the calculation model's geometrical changeability. The introduced error was numerically determined when considering specific problems (for instance, in Sec. 3.4.4f). Then the values of torsion moments in the reticulated shell's rods can be found from H" = GiJ&T* where the calculated values r,* become known (determined according to'Eq. (1.16)).

18

Chapter 1. Reticulated Shell Theory:

Equations

Assuming J 3 ; = 0, we obtain from (1.39) M,

=

-(£>,i*e, + D12K2

+ 2 D2D i 616rT), ),

M22 = 2DWT), = -(£>2i«i -(D21K1 + + DnK2 D22K+2-h2D 26r), H = D6lKi + K + Deer, K H = Asi>ci + D D62 K + Deer, 62 22 2

(1.40)

where / / = / / , = # 2 . From a comparison of the system of Eqs. (1.24) (relative to ei, e2, u) and Eq. (1.40) (relative to ej, e2, u>) and Eq. (1.40) (relative to e\, e2, UJ) and Eq. (1.40) (relative to K\, K2, T) it follows that the first changes to the second if we replace values ei,£ ei,C2,u,N (1.41) 2,u>,N uN 2,S,FiS,Ft i uN a,S,Fi correspondingly by KUuK ,2T,-MU-M1,—M2,H,J ,li-H, K2i,2T,—M 2,H,JU.

Jn.

(1.42)

Hence all the above concerning the system of Eq. (1.24) is valid for the system of Eq. (1.40). In particular, Eq. (1.33) (for arbitrary value n), (1.35) (for n > 2), (1.36)-(1.38) (for n = 2) if replacing Eq. (1.41) by Eq. (1.42) is valid. In this case when the torsion rod's rigidity cannot be disregarded (J3, ^ 0), parameters K\, K2, T are determined from the system (1.39) containing four equations. We can prove that the determinant of the system of the three first equations (1.39) is not zero. Hence, this system with Mi = M2 = H1 = H2 = 0 allows only a trivial solution. The bending and torsion moments are connected with the dependence at which the system (1.39) is simultaneous and has unique solution. 1.2.3c. Forces and moments in the reticulated shell's rods 1) Statically determinable cases. If dependencies reciprocal to Eq. (1.19) are unambiguous they can be used to determine the forces and moments in the reticulated shell's rods by calculation model's known stressed state (a statically determinable problem). We now examine such cases. If the lattice is formed by three families of rods (n = 3) the solution of the first three equations from Eq. (1.19) may be written as (remember, S\ = S2 = S as

st = 0)

DN{ sin(tp3 -

i) + N2cxc2 sin(y>2 - y>i) + S{c\ - c\)]a3, where D = cxc2 sin(y32 - 1 - ip3) + c2c3 sin(<^3 - tp2). We can show that value D does not become zero if the axes of the rods of one of the families have no general tangent with those of other families (there is no degeneration when n < 3).

1.2. Constitutive

Equations

19

W h e n

2, a = aj = 02, these formulae have a simpler form DQN£ * 1 D00N N222 N; N;

= = =

[-Niss3 sin(v? sin(v3 + y>3) + N N2CC3 sin(y? ++ y> ^ 33)) ++ S(cl S(cl -- cc2c22)]a, )]a, 2cc3 sin(y? [Niss [Niss33 sin(

- - 2 2 2 2 a3(NlS2-N a32(N c2)(4-c lS -N)-\ 2c )(4-c )-\

(1.43)

where D0 = (c§ — c2) sin 2i^. From (1.19) it follows that the formulae for the rods bending moments if their torsion of rigidity is neglected can be derived from Eq. (1.43) for the particular type of lattice by replacing N', Nj, S by M', Mj, —H correspondingly (i = 1,2,3; j = 1,2). After this we find D00M[ M[ = [-Miss [-M!ss 33sin(¥> sin(i^ + +
(1.43')

Now we consider t h e case when n = 2 and all t h e rods' rigidity characteristics re including J 2 t, aare other than zero. Then the number of unknowns when assessing forces and moments occurring in the shell's rods for each system included in Eq. (1.19) is equal to the number of equations. The solution of these systems of equations is Ni Ni N22 N 5S,, S22 „.

G\1 „.

G'22 Q. Q.

Q\11 „.. Q 22

Ql

l l = ai(c!S ai(c!S2Ni sic2N N22 + SiSiSt S1S2S1 - cic cic2S225)sm~ (22), 22Ni+c iC2N 2)s\n~{y>i 2N 2Sx 2-c^s xc2S2) 2SsirC* __11 = ai(siS ai(s1s22Ni + cxc2N22 - s i cc 22 55 ii - cis 22SS22)sin )sin (¥J (¥>22 ~- sin{

If in these formulae we assume t h a t

0.5a[Ar, +NW 0.5a[Ni
Q\ Qha

0.5a(Q,OMQtc-^QiS- ),

= =

1

11 G;, OMM1.C-1 1- 1 T M2.SMi.*G ; I222 =- OMMuc),), G\,

(1.44)

20

Chapter

1. Reticulated

Shell Theory:

Equations

From Eqs. (1.19) we conclude t h a t formulae for t h e rod's bending and m o m e n t s for t h e given lattice knowing the calculation model's stressed s t a t e can be derived from Eq. (1.44), if we replace values N', S', Ni, Si, correspondingly by M", — H", Mi, —Hi (i = 1,2). T h e n we obtain M'ltj = 0.5a[A/! + M Mlj Mi2 T [Hx t a n tp + H2 cot tp)],
(1.44')

Hl2 = 0.5a[tf, + H Hi2 ± (Afi (Mi t a n tp - M2 cot tp)].

If t h e lattice consists of two families of rods and t h e hypothesis J2l- = 0 ( 5 * = 0) is accepted, the calculation model's linear forces N\, N2, S will b e connected with Eq. (1.36) by expressing a simultaneous condition for t h e first three equations from (1.19) containing two unknown longitudinal forces in rods N{ and N2". T h e solution of these equations at tpx + tp2 ^ 0 is 2

w. JVJ 1

[N (yv.a -- N-2 NNMhi c\)gx lS\ 22cpg (Ni4 =_ 22 ' ~ c22 - c\

. N N; /V 2

,_

N.

■

w.

•

~

= -

c j —- c

C

^

[N Mxs\

l

°2

2c\)g h N2Nc])a .2

-

'

(145) (1.45)

rcc 222 _- ccr. 2l2 2

l

When tpx +
«-(M)l«-(*+?)£• *-(*-?)& *-(*-?)§• JV,- =

5'

^1 c

«1

iv2- =

2c

Ss Si

Ni

(1.46)

2c'

where tp = tp\. If we neglect the rod's torsional rigidity (J& = 0), H" = 0, we then derive formula for calculating bending moments in t h e shell's rods from Eqs. (1.45), (1.46) after replacing values ZVf, Nt, S by M?, Mit -H (i = 1,2). This yields 22 _ {M (Afis A^c?.)^ M24)a, (A/.s xs\. - M 2cl)ax

Mr '"l 1 = — ~~

22 rc_2 C,l

-

2 2 2 cc c22

''>

_ (M ( Axx/s\ , s 2 -— s\

Af = 2 2* ~ —

M

o 2 °2

—

M2c\)a c\)g2 c

1'

22

l

if i^i + tp2 / 0, and

«-(*-?) I- *-(*+?)!■ Ml1 =

Mt_

V cc

H £l sa ,/ 2c 2c'

A/22* =

A^

V c

//

02

W 22cc '

with f = fi = —tp2. 2) Statically indeterminate cases. If the shell's lattice is such t h a t t h e number of equations in the system (1.19) is less than t h e number of t h e unknown, t h e problem of finding forces or moments in the shells' rods using t h e known calculation model's stressed s t a t e becomes statically indeterminable. Nevertheless, this problem can be solved rather simply: first we must find t h e deformation components of the calculation model's middle surface (see Eqs. (1.32), (1.33)) according to its stressed state and then use formulae (1.16) a n d (1.18).

1.2. Constitutive

Equations

21

Here is an example of finding the forces and moments in the rods of the lattice shown in Fig. 1.5 from the knowledge of calculation model's stressed state. We assume that rods (i = 1,2) of the first two families are identical and omit index i(i = 1) in all designated values referring to the first family of rods, except forces and moments. The constitutive equations for shells with such a lattice have the form as Eq. (1.30). Using relations (1.35) we obtain the following expressions for the tangential deformation components: 4 2 2 Dei = £>£, = [2Ni(2Ks [2AT,(2AV + + K3) K3K)-KN 2v\Ks2c2, 3)-KN 2sin 2sm2
where in accordance with (1.33) D = [{K3c4 + K4s4)I< + K3K4/2]KK sin2 2
i,2 Nl NNl

= [(K3Nic N1c2j + K4N2s K44N22s22)K°±S(Ksm2
(1.47)

Here K° = [2K{K3c44 + K4s44) ) + /C3/C4]"1; in the first of these formulae the sign in front of the second term is assumed negative. If we express bending deformation components K\, K2, T in constitutive Eq. (1.22) through linear moments Mi, M2, Hi of the calculation model (the constitutive equation for torsion moment H2 will be satisfied) and then use Eqs. (1.16) and (1.18) we obtain the following formulae for the values of bending and torsion moments occurring in the reticulated shell's rods: Ml, MZt2 Ml Ml4 H'xa

2 {[(/3c + (I4s2.2 + 2C {[(73C + 2Cs2c2)Mi + 2Cs2c2)M2]I0 =F#i(/ sin 2ip =F77,(/ 2}G.73,3, = {±[(2/s I33)Mi (21c + Ih)M 2HifaS?cos 2
=

7/3 77* = where

— 2G3J33Hi0l1 -2GaJaHi&,

, H' HI4=2G = 2G Hifc, 4J 34Hi/331 , 4J34

70 = [27(73c4 + 7 + I733I744 + + 2Cs22cc22(2I (27 + I733 + + 7,)]" I44ss44)) + A)]" 1 ;

and /?3i is the same as in constitutive Eq. (1.30) referring to the given lattice.

(1.48)

22

Chapter 1. Reticulated Shell Theory:

Equations

Figure 1.7:

1.2.4

Constitutive equations for an oblique-angled system of coordinates

Let a', /?' be an arbitrary oblique-angled system of coordinates at the middle surface relative to the system of coordinates in lines of curvature a, /? as shown in Fig. 1.6. Positive directions of linear forces and moments in the oblique-angled system of coordinates are shown in Fig. 1.7. The following formulae are derived from the equilibrium condition of an element of the calculation model's middle surface (S' = S[ = S'2): N[sinx N[ sin x == N1sm20 9 + N2 cosN22cos e 20-Ssin20, 22 2 = Nis\n N = JVisinXA + + JV2cos A + Ssin2A, 5sin2A, Ni&inx 2smx S'sinx = 5'sinx — AfjSinx = M2 sin x = tfjsinx X = tfjsinx = H'2s'mx =

Ni sin Asin# — —N N22cos\cos0 cos A cos 5++ Ssm(0 Ssin(0— —X), A), Nism\sm0 {M\ cos A + Hi sin A)sin0 + (M2 sin A + H2cos A)\)cos6, [Mi cos 6, H\ sin 0) 6) sin A + (M2 sin 06 — H2 cos 0) 6) cos A, (Mi cos 0 — Hi {M2 - Mi) sin 0cos 0 + Hi sin2 0[M 6- H2 cos2 0, 2 2 (M2 — Mi)s'm A cos A — Hi sinsin A —cos H22cos2 A.

These formulae which take Eqs. (1.22) and (1.37) into account as well as the relations given in the book [14, p. 61]: £isinx wsinx

= =

e'i cos Asinfl + u/sin A sin 0 + e'2sin Acos0, — eicos(A - 0) - u / s i n ( A — 0) + e2cos(A — 0),

1.2. Constitutive

Equations

23

e£ 2 sinx /K!sinx ^sinx rsinx

= e'i sin Acos0 — u/cos A cos 5 + e'2cos A sin 9, = K' «'iI sin 2 0 + T'COS(A — 9) + /c'2sin2 A, = — "i sin 0 cos 0 — r'sin(A — 9) + /c'2sin A cos A,

K2 sin x

=

K'I cos COS22 9 — — T' COS(A — 6) 9) + + K'22 COS cos2 A

(deformation components for the oblique-angled system of coordinates are primed) may be presented in the following constitutive equations:

JVf'Wx = CPZ c ^ + +c ^ + cpV JV<'>sin2x = CJP^+CJpu/ (j

22

A/^ >sin M sin x

=

(i M), (i = =M ),

-(D^K^ -(D[ K\ K[ + + D^K^ D< V2 + + D^T Z ) 3 V)) J)

;

J 1

(j =T~i). =174).

(1.49)

Here (1) Ar (1) Ar 11

M ' '' M'

A'' 2 » = AT Af' 3 ) = = 5S', ', = JV,', N[, NM N^,2) NW 22 = M[, A/;, M< M< »» = M M 22)) A/( A/<3»3 ) = i/;, i/;,

3) C[ Cf =

n n

^A A Ti a. aK, ,^^ ) E 1

M M^« = = ff H'2,2,

(; = U ; f c = T73), (j=T73;fc I73),

n n

D& = E( Y,lIih"tP(i) + ' W ) (i = 174; A: = 173), an

= ej e| ' = c 2 cos Asin0 Asinfl + s 2 sin Acos0 — SjC,cos(A — 9), 0),


ee

. ! — cc-f2 sin A cos cos 99 + + ss 22 cos cos AA sin sin 00 ++ S,-CJ s.Cj cos(A cos(A — — 69), ),

= d]
d\ ' = (si cos 9 — c, sin 0) 2 , 6., = 6u 63,

C], c3i

=

-2
Vi

d] ' = d; = (c, sin A + -(- «< cos A) 2 , 62, = 2ff<4) = sin 2 (0 + ^ , ) ,

), 2e| 2ej

= 2e,' = sin 2{ipi - 9), = 2s\n(9(0 - A - X-2ipi). 2ipi).

44))

= - sin 2(A V,), 2(A + ¥>■),

c2i = sin2(A + y>,),

One should bear in mind that when determining angles A, 9,
=

6l, =

1] ej'> = -2«,< 3 > = c 2 , 4i> =

4 =

4 2 = 42' = ! 2 » == 22g^ = 6b22i, = =
3) 4) 2) 2a3i3 l = 63; = -cu - c „ == cen = 2<, 2, = C3,- = C& = 22cos2<^,. c o s 2
Here constitutive Eq. (1-49) transform into Eqs. (1.22) and (1.37).

24

1.2.5

Chapter

1. Reticulated

Shell Theory:

Equations

More complex version of the constitutive equations

In t h e theory of reticulated shells just as in the theory of constitutuous shells various versions of constitutive equations may be used. Various variants of constitutive equations in t h e theory of thin elastic isotropic shells is discussed in literature [14, pp. 80-82]. For this t h e version (1.39), (1.24) as the simplest may be recommended. However these constitutive equations have certain drawbacks if applied to problems in t h e reticulated shell theory. One of t h e m is t h a t t h e sixth equation in (1.1) is, in t h e general case, fulfilled approximately. Now we obtain another version of t h e calculation model's constitutive equation. Let us assume t h a t coordinates a and /? coincide with t h e main lines of curvature of the calculation model's middle surface. T h e n t h e said equation for t h e linear problem is SSr-Sii+ Hth Hxh - H H2k22k2== 0.0. (1.50) 1-S2 Now we introduce small terms to constitutive Eq. (1.24) making it possible to strictly fulfil Eq. (1.50): Ni

=

Cuet

S\

=

CeiBi Ceeu + CeiEi + Ce C&2e2e22 -f Ce^ui

+ C12£2 e2 + Cl6u,

+[(Dn S2

= =

N2 = C 2 i£i + C22 2 2e£22 + C 22 6^, 6^,

+ / $ > ) « , + (D62 - K$)K2')*» + (Dee -

*£')■K^)r}k2,

Cei£i + t/6 2 £ 2 + Ceeu Ceew + +[{Dn + 4 ! J ) « i + (D62 + K$)K2')*2 + (Dee + K^)r]h. K^)r]h.

(1.51)

Equations (1.39) and (1.51) have another important specific feature: we can show t h a t they can be presented as a linear transformation y = Tx where y

=

xi

=

(JVi,JVa 1 $(N UN2,SUS2,M UM2,HUH 2)', 1 ,&,Mi,Af a ,fr„J5r a )*, 2 1 , 2 1 2 , 2 1 ( £e(«1 », »,«( ),«( ),T( »,r( )£)'e ),u;( W , « (',u;( ),wW ,*W ,rW , T «»)', }'1 1«W

(1.52)

with a symmetrical matrix T (asterisk means a transposition). This transformation is deduced from Eqs. (1.39), (1.51) if we consider energy deformation components of t h e middle surface according to formulae [14, pp. 79, 80]. £<*> = £<*)

£,,

££<(22»> = £ 2 ,

11

/c'' ' K

=

-KU :1>

1

=

T - hiu/2, fclW/2,

T* '

2 2 2W<11> = -2^< -2a,< 2u>< » >== uw;,

2

«< ) = - K 2 , r<2> = -r

+ kk2u>/2. 2uij2.

Just as in t h e anisotropic shell theory [15, pp. 67-69] we can prove t h a t Kirchhoff's uniqueness theorem and Betti's reciprocity principal hold in constitutive Eqs. (1.39), (1.51) in t h e reticulated shell theory in t h e case of their geometrical stability. Naturally, this does not mean t h a t with other constitutive equations t h e solution will not be unique (here the proof method used may not have validity). Note t h a t in t h e case of t h e calculation model's isotropy formulae (1.26) are valid with J3i = 0 constitutive Eqs. (1.39), (1.51) change t o equations suggested in t h e works [3] and [35] with parameters of the continuous shell (1.28).

1.2. Constitutive

1.2.6

Equations

25

Study of the geometrical stability of the reticulated shell's calculation model. Deformation energy

The calculation model is geometrically stable if displacement W of the shell's middle surface as a rigid body satisfies the condition that no deformation energy is present, i.e. W = 0: £l U> = = K\ K\ = =
1

W = IV = ii [f fj'ABdad/3 fj"ABdad0 2/ /

(1.54)

G G

where J' is a doubled deformation energy value J' = JV,e, + N Af22ee22 -Su- 5w - Mi«j - M2K2 + + (Hi (//, + + H2)T,

(1.55)

G is an area of the shell's middle surface. Using Eq. (1.22) with J2i = 0 we obtain from Eq. (1.55) J*

=

Cue] + C22 t\ + Cxiu? + 2(Ci2£iE £i£ 2 + CjeEjo; deeju; + C C2626e£22uju 22£^ + (0n + + K KUn)K] (D22 ++ KK2222)K\ +(0n )i4 + (0« )K\ ++ (2A* (20„ ++ AJT<1' JT<1' 6 6 -+2{(Dl2 + + /C, XM)KIKJ (20,6 2 )K,/C 2 + (2£> 16 - K16)KXT

+ (2D 2D26 26 -

1 Kg))r K £)T27 Kff

^ 1K616)K2T]

or, if we consider (1.23) n

J'

E

= ^[i
+ /,(/cic,2 + K (c22S522 + r sin 2v?t')2 + Cj(/C22s,Cj S,Cj — — KiS.C, 2tfii)22}.}. +C,(/c Kis.c,++7"rCOS cos 2(pi)

(1.56)

The last formula can be made simpler on the basis of (1.16): n

E

2 7 jr* = = £(tf,£* J2(K<£? ++ 7W -<2 ++cc,n ^f)-). 1= = 11 1

( L57 )

Since the rigidity characteristics of rods Kj, /,, Ct- are not negative from (1.57) it follows that J' > 0. Hence from (1.54) we conclude that the necessary and sufficient condition for the calculation model's geometrical stability is the fulfilment of (1.53) at J" = 0. Let J* = 0. Then from (1.56) 2 e\j = 0 at A Kic U / 0, (K 2 — -/ci)sin2(^ Ki) sin 2yjj++2"rcos2<^ 2T COS 2
(1.58)

26

Chapter 1. Reticulated Shell Theory: Equations

Hence, if we consider that the rigidity characteristics of the shell's rods are not zero we obtain a homogeneous system from 3n algebraic equations concerning e%, e2, u, Ki, K2, T. System (1.58) will have In equations if we assume d = 0 (i = l,n). For a geometrically stable calculation model we may obtain three solutions: 1. System (1.58) has only a trivial solution (1.53). 2. Systems (1.58) provides a solution different from (1.53) but the deformation continuity equations are not satisfied. The continuous equations can be considered as the integration conditions when plotting displacements according to preset deformations. 3. System (1.58) allows a decision different from (1.53) satisfying the deformation's equation. At the same time displacements corresponding to this solution are excluded by the boundary conditions (only displacements of the rigid entirety are possible). In the latter case in contrast to the first two the unfixed calculation model will be geometrically variable. In the latter two instances the displacement problem can be solved without the deformation continuity equations substituting relations (1.3) in system (1.58). It is well known that here deformation continuity equations are satisfied. This approach to solving the problem is simpler if we want to know whether the calculation model's fixing conditions assure its geometrical stability. Let us consider cases when the shell consists of two (n = 2) or more than two (n > 2) families of rods. 1.2.6a. The number of rod families greater than two. Let us fix three different families of rods »' = fit i = TO, i = v. Assume for these families that KijLQ, Ki?0,

IijtO Ii^O

{i = fi,m,v). n,m,is).

(1.59)

The values of the other rigidity characteristics of the rods forming the shell's lattice are arbitrary. Then from (1.58) we obtain c*£! + s\e2 + s,C(W = 0, cjei c K cj/ci rsin 2(fi = 0 (i (t = /i,TO, f). l i -|+ sS\K ?*a2 + Tsin2y>j m, v).

(1.60)

Hence, each of the triplet values E\, e2, ui and KU K2, 2T is the solution of the same homogeneous system of three linear algebraic equations. The determinant of this system of equations D = 0.25[sin 2(y>„ - M) - y>m + sin 2(y>m - tpv)] is therefore not zero. That is why the solution of system (1.60) can only be trivial, i.e. when J" = 0 there is always (1.53). Thus, if n > 2 it suffices to satisfy condition (1.59) to ensure the calculation model's geometrical stability even if it is not fixed.

1.2.

Constitutive

Equations

27

W h e n using t h e momentless theory for calculating reticulated shells we assume K{ / 0, /,■ = Ci = 0 (i = T~n). Due to this with zero deformation energy only the tangential deformation components (ei = £2 = u; = 0) become zero, hence t h e middle surface of t h e shell's calculation model can have small flextures. T h u s , to assure the calculation momentless model's geometrical stability it is necessary and sufficient that it has fastenings excluding small flextures of t h e middle surface (it must be rigid). 1.2.6b. The case when the number of rod families is two. We now examine the condition when t h e rods' rigidity characteristics are not zero: A-,/0, KijQ,

7,^0,

djtO Q?0

(i = = 1,2). l,2).

(1.61)

From (1.58) we obtain c 2 £! -r c]c\ -|- sfe2 + + SjCiU s;Cib} = 2

cc-«i Ki + +S*K S*K + rsin¥>, rsin,; + 2T cos2ifii

= =

0, 0

== 0.0.

(i (i = 1,2), 1,2),

(1.62)

T h e determinant of t h e first three homogeneous equations (1.62) as regards K\, K2, T is sin2(y>i — y>2) hence, non-vanishing. It is equal to zero only when t h e two families of rods degenerate into one. Hence, K\ = K2 = = T == 0. 0. Ki=K2

(1.63)

T h e tangential deformation components are assessed by solving a system of two equation from (1.62) and three continuity deformation equations which take (1.63) into account. T h e reticulated shell's calculation model will be geometrically stable if the said system of five equations as regards £1, £21 w >s either compatible or has no solutions satisfying t h e displacement conditions of the shell's fixture (only displacements of a rigid entirety are possible). Evidently if t h e calculation model is geometrically changeable it will still be such when (1.61) is not valid: system (1.62) will contain less equations. Note t h a t geometrically stable calculation models are used for researches although geometrically changeable calculation models also have applications. In t h e latter instance t h e action of external forces on t h e displacements not inducing forces or m o m e n t s in t h e calculation model must be zero. 1.2.6c. Examples. Let us consider t h e geometrical stability of a calculation model of circular cylindrical shells with lattices formed by two families of rods. 1) Axes of one family of rods coincide with the generatrices and the other with parallels of t h e circular cylindrical shell's middle surface (<^i = 0, tp2 = T / 2 ) . T h e r o d s ' rigidity characteristics satisfy conditions (1.61).

28

Chapter 1. Reticulated Shell Theory: Equations

The first two equations from (1.62) take the form ex = e2 = 0. Besides, there is also (1.63). Consequently, all deformation components except u are zero. Using formulae (1.10) we write

+ +W K - »■ » ===°' »-°> da = ITa °' ldP+W du

d2w 0a22 da* da

=

„

dv dv

„ '

2 iu d,2w dv _ „ dh 2 W~dp~' a/? dp dP W~ dP dp"'

(1.64) "•«> d2w W dv _ „ dadP ~ da dadP oadp dadp da da~ ~

,„ . . .

'

(1.65)

.A y . . are distances along the generatrix Here a = x/R, P = y/R, where x and generatri and parallel, R! is is the the radius radius of of the the shell's shell's middle middle surface surface cross cross section. section. lution of this system of five differential equations with three functions functioi has The solution the form

uH = = 4>(P), 4,(0), v = (Ci (d sin sinP/? P+ + + Co C C22cosP)a cos P)a P)c + + CC33sinp sin p + +C C44cosP cos /? + + C C00,, [C2 s\nP-C PP P,P, w = (C sin P fi -— C, C\ cos cosP)a-C P)a — cos P + +C44Csin /3, xcosP)i 3 cos 3 CC$ 4 sin

(1.66)

where Ci (i = 0,4) are arbitrary constants. With this the deformation is C1sinP + d^/dP, Rw = Ci sin P + + C Ci2cosP cos P + #/<*/?,

(1.67)

and the turn angles Rfi = Ci cos P — C2 sin /?, P,

Rj22 = = C0. 7?7

Thus, the unfixed calculation model is geometrically changeable: in this with J" = 0 non-zero displacements UJ are possible. Let us find the conditions of fixing the contour (a = at\, a 2 ; P = Pi, /32) so that the calculation model is geometrically stable. From (1.67) it follows that for this it is necessary and sufficient to satisfy the conditions CisinP C -|- dcosP Ci cos P + dip/sp dip/sP = 0. (1.68) (1.68; x sin P + Since from the functions of displacements only u depends on if a n ^ u = i)(P) then from (1.68) it follows that one of the shell's longitudinal cross sections need to have connections preventing displacements of the middle surface points along its generatrices. Here we assume u = 0 (a = = const).

(1.69)

The condition u = 0 can be replaced by u = const, and v = 0 by v = const. This would result in the calculation model's possible displacement as a rigid entirety. True, from (1.68) we find that solution (1.66) corresponds to such displacement if

29

1.2. Constitutive Equations

tp = C\ cos/? — C2sinP + Cs. Here with account of (1.66) we find that ip = 0. Then from (1.68) C\ = C2 = 0. This is possible in one variant of the boundary conditions in addition to (1.69): 1) 7i = 2) v = 3) w = 4) v = 55 )) vv = = = 66 )) »» = 7) to to = = 7) gS)v )v = = 99 )) vv = =

0 0 0

(Q = const); (a = (a = a !a,uaa22));; (a == Qi,a (Q Qi,a22););

0 00 00 00 00 00

= ct\); ai); ((a = a = a up a,,/? (a = = aaO; (a 0; (a = aia,p„ 0 (a = (a i,0 (a = = aa,,/? (0 = /?,); (/3 = A ) ;

w ID = = 0 (a — a 2 ); = = A p, uAP2));; w= = 00 (/? (0 = £ „, & w = A A )) ;; = j0,,0,); = pup2); = u> = = = A A )) ;; u; = 00 (/3 (0 = w = 0 (a = a , , 0 «» = 0 (a = a , , 0 = =

(Q (a

ft); ft); 0ft). 2 ).

Boundary conditions 5), 6), 7) can be accepted when points 0 = 0\, 0 = 02 do not lie on one diameter of the middle surface cross section; boundary conditions 8) and 9)—if the normals to the middle surface passing through points P = A and P = A are not orthogonal. 2) A circular cylindrical shell formed by a rhombic lattice (tp\ = —ip2 = V = const). Assume inequality (1.61). From the first two equations (1.62) we obtain c2£i + s\£2 ± SiCiui = 0 hence Cj£i + s\e2 = 0, ij} = 0 or considering (1.10), ddu u 2,?du

, (dv /'0u du

\\

„

C?F ++ 5 ,dp + =0 aa- 'UH '

dv da

du_ du du

+ dP~

^ ^=°-

.

„.

(L70)

To these two equations we must add three equations (1.65) expressing conditions (1.63). Solution of system (1.65) and (1.70) will be uu = - [Ci sin P + C2 cos 0P + (C5a2 + C6a) tan 2

(1.71)

Here C, (i = 1,8) are arbitrary constants. The non-zero deformation components will only be £i and e 2 : + C6, £i = - e j2 ttan a n2 V (1.72) Re2 = 2Cs5aa + v?. Turn angles corresponding to solution (1.71) are: fl7l = - ( C , sin /9 0 + C2 cos£ cos P + 2C S ),

#72 «72 = C C88..

30

Chapter 1. Reticulated Shell Theory:

Equations

From (1.72) it follows that for the calculation model's geometrical stability it is necessary and sufficient that it has fastenings at which C 5 = Ce = 0 (a calculation model without fastening is geometrically changeable). Here are some variants of the boundary conditions making the calculation model geometrically stable: 1) w 1) w 2) w 3) 71 4)7i

= = = = =

00 0 0 0

(a (a (a (a (a

= = = = =

Q,,o 22 ); Q,,Q ); const); 71 = 0 (0 = const); const); u = 0 (0 = const); const); u = 0 (a = a i , a 2 ) .

3) The problem's conditions are the same but the inequalities (1.61) are different: tf,/0, AT./O,

Ji/0,

C, C,- = 0

(1.73)

(i = 1,2). 1,2).

This often occurs in calculations (the rods' torsional rigidity is small compared to the bending rigidity). With the rods' rigidity characteristics (1.73), from (1.58) we obtain two identical equations concerning e\, £2, w and Ki, K 2 , IT. c\e\ + sje 2 ± siC\U3 = 0, C\K\ +S C\KI + 2S12KK2 ± 2S\C\T == 0.0. 2±2SIC 1T

(1.74)

If dependences between the deformation components and displacements are assumed as £l

Kl

=

"' ~"

du 1 fdv fdv /dt; \ 1_ du £i= R^da R\jp + W)' 2 2 11 da2u> w __ J_ _dLw ^^ 12 222 K 2 _ 2 K 2 da''' Wdp*' Rfl "hR?da d0 fl da d£2'2'

11_ fdu du dv\ + 80"* R{d0 da-)> da j 22 ww 11 ddu> T T 22 ~ ~~R?dad0' RR dad0' )ad0'

U=

( (1.75) 1

'

from (1.74) we obtain Eqs. (1.70) and 2 22 u 2d w 2td w 2 C 2 + S 2

dada~

d0W~~

_ = 0,

w dd22w = 0. dad/3 = ~daj0

We can prove that the solution of the system of these four equations with three functions u,v,w has the form: 22 u = = [Mv [Mv + + 0) 0) + + Mv-0) Mv-0) +C C33r,r,33-C -C2J] + Csv]tanip, Csv]tanip, 2J] + «?-/?) + 3i 2 3 3 -Mli(v + c-Cifi 0 - C,{? ++Cc4A0, 0) ++Mv-0)-C Mv-fi)-c3030-C 0, + 0) 31p vv = -Mv

w = 3C3(02-T, -ri22))V2) + 2Cl100 + 2C2V -(C
+ CCbb),),

(1.76)

where r\ = atany>; rpi(x), ^ 2 (y) are random functions, d(i constants.

= 175) are random

1.2. Constitutive Equations

31

With this solution the deformation components in accordance with (1.75) are identical (the stroke means a derivative) &2 2

R K2

22 - 2C -(V>i + 0022 + 3C = -{i>[ 3C377 rt2r)~- CC &)' 37; - 2C 5),

= =

-—66C3, C3,

22

«! «i = = -K —/cj tan tp, ip, 2 tan

tan2 2ip, y?, t == - -£ e2 2tan

£e,

uix! = =T T= = 0. 0.

(1.77)

Thus, if the calculation model is unfixed it is geometrically changeable. Let a portion of boundary conditions expressing the fastening of the shell's transversal edges, have the from v = (T? = = 0,»/.), 0,»7.), =w w= = 00 (r? (1.78) where n. = (/ tan tp)/R, I is the shell's length. We can prove that in this instance in solution (1.77) we must assume 01 = Xl Xi + X2,

02 = "Xi -Xi + X2,

C, = 0

(2 (i = 175), 1,5),

where xi, X2 are arbitrary periodical odd and even functions with period 2r]'. Here solution (1.76) takes the form « = \Xi(l + P)-xAv-P) •ifo-/») + X7(ri + 0) + X7(ri-P)]tan
» = -\xi(v + P) + xi(n-P)+X2(n + 0)-X2(ri-P)], P)-X2(ri-P)],

w w Then from Then from

= 0. = 0. (1.77) it follows that two deformation components remain non-vanishing: (1.77) it follows that two deformation components remain non-vanishing:

^2 ei ei

= -Ix[{v -{x'i(r) + P)-X'i(ri-P) P)-x[(v-0) = = — —£2tan £2tan22y>. y>.

+ X2(r) P) ++ X2(v-P)}, X2(v-P)}, X2(v ++P)

So we can conclude that the fastenings (1.78) are insufficient to assume the calculation model's geometrical stability. To achieve this we can supplement conditions (1-78) with conditions such that u = v = 0 with f) = const (and obtain Xi = X2 = 0). The same refers to the calculation of a cylindrical shell.

1.2.7

Boundary conditions

If we neglect the rods' torsion of rigidity on the plane tangential to the shell's middle surface (J21 = 0, t = 1, n), the order of the systems of differential equations describing the stressed-deformed states of the reticulated and continuous shells coincide, as well as the formulations of the problems corresponding boundary conditions. When the said rods' rigidity is not zero the order of the system of differential equations in the reticulated shell theory increases from eight to twelve. In this more general case, six and not four boundary conditions must be set in the problem.

32

Chapter 1. Reticulated Shell Theory: Equations

Figure 1.8: Similarly, in the plane problem for reticulated plates the order of the system of differential equations increases from four to eight and for its solution four and not two boundary conditions must be used. We now consider the type of two additional boundary conditions needed for reticulated shells' calculations when J2i j= 0. Of practical interest is the instance when the shell's lattice consists of two families of rods as only in this case will it be necessary to consider the rods' torsional rigidity in the plane tangential to the shell's middle surface. Hence, we assume n = 2. For certainty we formulate additional boundary conditions at that section of the contour of the calculation model's middle surface coinciding with coordinate 0 = 0oA few examples of most typical boundary conditions follow. 1. The rods are linked together along the shell's contour with a hinge which allows the rods of each of the two families to freely rotate around the normal to the middle surface (Fig. 1.8a). In this case G\ = G2 = 0, on the contour which according to (1.19) provides the following two additional boundary conditions: Mu = M22a. = 0 at Mi,

0 = 00Oa.

(1.79)

2. The rods are rigidly hinged along the contour: the turn angles of the two families of rods around the normal to the shell's middle surface on its contour are equal. Besides, the linear bending moment at the plane tangential to the shell's middle surface has a preset value on the contour. Here one of the two additional boundary conditions becomes

M22,, = M°,(a) M°,{a) at 0 = 0fa, O,

(1.80)

where M°,(a) is a preset function. The second additional boundary condition is obtained by equalising the turn angles around the normal to the shell's middle surface of the two families of rods [cos 2ipi — COS 2

(cos2y?i w(cos2 cos2^2) (£2 - £i)(sin2<^i —sin sin2y> sin2y> 2) 2+

(1.81) (1-81)

This condition follows from (1.12). The shell's support scheme when M°, = 0 (condition (1.80) is homogeneous) is shown in Fig. 1.8b.

1.3. More Precise Constitutive Equations

33

3. The two families' rods along the shell's contour are rigidly fixed as regards their turning around the normal to the middle surface (Fig. 1.8c). The two additional boundary conditions in accordance with (1.12) have the form (* = 1,2) — ucos2tpi ++ (£2-ei)sm2v,= (£2—£i)sin2vi = 00 at at 00 == 00OO. . -2626++ucos2i = —ipi, the boundary conditions (1.81) and (1.82) take the form When

i = —ipi, the boundary conditions (1.81) and (1.82) take the form £2 — £i = 0 when 0 = 0O, (1.81') e 2 - e , = 0 when 0 = 0O, (1.81') (1.82') £2 — £i = u;cos2y> — 28 = 0 when 0 = 0O. e 2 - £ i = u> cos 2y? - 2<5 = 0 when 0 = 0o(1.82') 4. The two families' rods are rigidly interconnected and linearly-elastically linked with the contour as regards the turn angle around the normal to the shell's middle surface. In this instance one of the additional boundary conditions should be as in (181) and the other has the form M2, = fc(a)V>i ^(0)^1 = fc(a)V>2 k(a)il>2 at

0 = 0Q ,

where k(ct) is the rigidity function of the elastic fastening, $i, 4>2 a r e turn angles around the normal to the middle surface of two families of rods. These angles, depending on the particular boundary problem are assessed according to formulae (1.4), (1.8) or (112) in which V> and f> should be replaced by ipi and ,. Note, that other additional conditions can also be formulated easily. In a particular case when the axis of one of the two families' rods coincides with the calculation model's contour only one and not two additional boundary condition can thus be formulated which fully corresponds to the type of the system of equations (see Sec. 3.1.2a).

1.3

More Precise Constitutive Equations in the Reticulated Shell Theory

The constitutive equations in the reticulated shell theory (1.22) have been derived from the classical rod deformation theory. In many cases this leads to substantial errors. Hence, we form more precise constitutive equations.

1.3.1

Allowance for transverse shear, cross-section warping and transverse deformation of rods

If the reticulated shell's rods are made from a composite material or have a large cross-section height (the shell is not thin) constitutive equations (1.22) should be

34

Chapter 1. Reticulated Shell Theory: Equations

specified and we make them more precise. For this the method suggested in the work [68] for a rectangular rod must generalized for a curvilinear rod. Assume that the shallow curvilinear rod has an elongated rectangular cross section in the direction of the normal to the shell's middle surface and its material is isotropic. Assuming also that during the shell's deformation the rod has a flat stressed state we write the known equation of Hook's generalized law as: _ <7i <7, — - VOi vo-i (T\ V02

£l =

_ <Ji cr2

£2=

Ei —ET>

vGi 12. vG\

_ T

U3=

£Ei E2T XE\'

G-G

(1.83) (L83)

Then, just as in the classical shell theory, the first right-hand term in the formula for £2 is neglected. We express the rod's deformation components through longitudinal U(x, z) and transverse W(x, 2) displacements £l

W W Ro RQ + + z'z'

__dU_ dU_ dV = dx

_ dW dW dz ' a*'

_dU_ _dV_ dU_ dz

U! =

£2 =

dW dW dx ' dx'

(1.84)

where RQ is the radius of curvature of the rod's axis. Coordinates x, z are along the rod's axis and the external normal. In the first approximation we assume that the rod's stresses are calculated according to the strength of materials' formulae N M N M = Zz F ' °* -F--J ' ^F~T'' y <

H

Q ffh* Q * 9. h

T = T=

2 Zl)-\ j 2J{T2J

Z

A a\ )-

/,(1.85) 00 (L85)

Here N, Q are longitudinal and tranverse forces; M a bending moment; h, F, J are height, area and the main central moment of inertia of the rod's cross section respectively. Using the equilibrium equations in the plane elasticity theory dr dot ^1 + ^ = 0 dx ' dz dz boundary condition oi{—h/2) = 0, and the second formulae in (1.85) we obtain: 02

(z3z*_ h2 ~ V 33 " 4 Z

V

h3\ \_dQ 1 dQ 12 2J 12J U dx' dx'

, v (1.86)

Hence, in particular, it follows that dQ/dx = — 6CT 2 (/I/2). From the second formulae (1.83), (1.84) and the first formula in (1.85) after integration within the range from 0 up to z we find ,,,. „,/

, %

W(x,z) = w +

v" (MM , E 2J'

N \\

Tz)

, (1.87)

1.3. More Precise Constitutive Equations

35

Here and further W(x,0) = w, E = E\. Similarly, from the third formulae (1.83), (1.84) and the second formula in (1.85) taking (1.87) into account we have

_9_

... . dw Q fh2 U(x,,) = u - zz - + £j ( j Tx 2GJ 4

z3 \ v (z2 dN Q 3\ j j + - ( - - + g,» j , 6J* 3; k2J

(1.88)

where u = U(x,0). From formulae (1.87), (1.88) it follows that the displacement vector components change non-linearly according to the cross section height of the rods. We now make the formula for cr\ more precise using (1.85). We write the first formula in (1.83) as <7j = Ee\ + vcr2. Consequently, taking (1.84), (1.85) and (1.87), (1.88) into account we derive 22 „du ^du „.GJ _ (T*~j) 2 2

,,

2

xx

2

i i ^ — - — - —- )^ 1 £^ ^ J \ 33~ 8* 8 24JJ 2 4 ) \
+

Z

4 ++

E vV (M F {M ,u> + rr* % ++ z' flo * i?o + z Jo + 2 V2J

+•= Here <Ti = a. Functions

fh/2 yA/2 yA/2

vz d N

22 2F dx 'i 2F dx N N \ FF2*jy) • 2

(1.89) (1.89)

fh/2 yft/2

yft/2

N adz, N = = bb' /II adz, M M = = -b -b J-h/2 crzdz, crzdz, J-h/2 J-k/2 J-h/2 J-h/2 J-k/2 where b is the rod's cross section width, with consideration of (1.89) are assessed according to formulae vh dQ 2
N = EFe-

V v

2Ro

v/, I 2

'V, l2Rc 12i?o - - - * " -T T4 £
rft' M = = - E-EJKJ t - ^ M

+

4 A v,v 4 99 ++ 5l5* 2r'

:

22GG - " 2G

K = —

(
whCTe

rr ==

and £

=


r

w to to Ro

E£ _F_

9J ==

UV

2

Jx

(1.90)

"•'»'

(L91

»

w \\ w

Rg

are the rod's axis deformation components. We note that for the isotropic material 3 = 1. When deriving formulae (1.90) it was assumed that 2

/ h\2

UHo,J <
vh22

2

cPN/dx PNIdx 2 Af

"24 ^ v — « L -2T

36

Chapter 1. Reticulated

Shell Theory:

Equations

Formula (1.89), taking (1.90) into account may be written as:

i)

2 3 N M z zz^\ \ dQ M _N__M_ Ig (h*z I (h z Z+ a°' = z+ lx' ~ J-T F Jj{-20-j)diJ V 20 3 ) dx' F~ T

When deriving this formula it was assumed that uh/gR EJ uh If( uh rk\dQ rh\dQ uEJ rh\ K. 9R~ dx

_

_^f

(1.92) (L92)

J^ '^1 ,

N = £ f £ __( 4 + _j -£-_«. N = £ F £ _ _ ( 4 + _j-£-_«.

„ „, (1.93)

(1.93)

From (1.87) it follows that assuming v = 0 t h e deflection function depends only on the longitudinal coordinate, consequently the theory neglects t h e transverse defor­ mation. If in addition we consider that t h e material's shear modulus tends to infinity we obtain t h e classical theory's results. T h e results obtained for a particular case when t h e rod's curvature is zero coincide with those in the work [68]. This work proves that in many cases t h e more precise theory's solutions either coincide with those for t h e plane elasticity theory or are close t o them. It also testifies that the classical theory's solutions and those obtained according t o Timoshenko's theory [74] can cause significant errors. We also mention that a review of works on making Timoshenko's theory more precise is given in [67]. We now consider t h e mean curve of a section of t h e rod 90 = h/2) - U(x, -h/2)\/h. = 6(x) = [U{x, [U(x, ft/2) Then, using formula (1.88) we obtain Q = kGF{6 kGF(6 + + dw/dx), dw/dx), where

(1.94) (1.94)

2 1=^. ~k=2-^GjE" 2 + uG/E

(1.95) (L95)

2 + vG/E' Formula (1.93) taking (1.94) into account, become vh2 vh22

M

N N

= =

=

rh22kk„„/d0 d0_ EJ V vhk„„( d rh r> \dx + dxdx2 3J, 2Ro 8 \ RoJ

EjK+

uh ™ ™ \2RoEFe

~ 2*r - ir

( x) U *?) ■

,

,

(1.96) (L96)

When using these formulae for practical calculations they can be simplified and written as: //lit dO ■Pw_ 1T( Tx" dx2 rh2k 'M_ — ( cf dx2 , +S ( (1.97) 97)

»- --?-K)(£+£0. »-»-^"K)(S+S)-

u . -„._^ (-

).

,

37 37

1.3. More Precise Constitutive Equations

Formulae (1.94), (1.97) are more precise constitutive equations for a plane form of deformation of a flat curvilinear rod having a rectangular cross section. From these equations for the reticulated shell's rod we obtain NT = EiFtf E^e: M At;'

=

- ^ G , F , ( 4 + rr.h^V^ A W V.-w), v, )V,W + V.-u,), 8 _'

-EiJuK' -EiJuK-

- 2*^6 '^!&-GiFiVi(0T '^B-GiFiViW iV, 4

(1.98)

ViWiW),), +• V Viii

is a n where k' = fc,c? + fc2s?; 0* = 0,c, + 0 2 s t ; V. V, is an operator in (1.17); 8U 62— are the mean curve angles of the normal in planes /? = const and a = const. In these formulae functions £*, /c* are determined through the shell's middle surface deformation components according to formulae (1.16). We obtain the reticulated shell's constitutive equations by substituting formulae (1.98) into

n1

N> = t^r,

2

n

N2 = ±S1N:,

n

S = ±aS-^N:,

' E .=i "• .=i■I*°« .=i ' SS M, = Mr = E ±CC1M JM;, = = ±± 1M 1M:,:, H =HH=-±fM?, = -±S-^M; u222 = ±i * A :, M -E is"*" ■K,

5 =

t=i

1=1

a

—v S,Cj

aa

■ i=l i ==i i ''

aa

i=l ,=i < n S s

i=l i=i

1=1

aa

<

«> ==E±fQh Qx ^ ; . Q2Qi =E E^QI = a± i.Ql i=; i=i

n ■

a

'

i=i a'

(1.99) (i.99)

<—i

These formulae disregard the rods' torsional rigidity and bending in the plane tangent to the shell's middle surface.

1.3.2

Allowance for the rods' non-linear-elastic deformation

We now obtain constitutive equations in the reticulated shell theory when the rods' material is non-linearly elastic. Assume during the rods' deformation that Bernoulli's hypothesis is satisfied. Then from formulae (1.16) for the fibre of a rod positioned distance z from the middle surface we obtain ,c? + te22s\ (nicf ++KK2Ssin2ipi)z. 2
(1.100) (1.100)

The rod's axial force and bending moment are derived from formulae

N' oldF, M' M:==- -J a'zdF,, N'= = IJ f WF, a:dF„ jI a'izdf <^F t, JF,

m

where a* 7* = =
(1.101)

38

Chapter 1. Reticulated Shell Theory: Equations

The constitutive equations are obtained taking (1.100), (1.101) in formulae (1.99) into account. Note that the constitutive Eqs. (1.22) with J 2 ; = 0 and the lattice's stable parameters with the main terms of effective characteristics for simple frameworks resulting from averaging processes in periodic media [6], [18], [24].

Chapter 2 DECOMPOSITION M E T H O D 2.1 2.1.1

Solution of Equations and Boundary Value Problems by the Decomposition Method Decomposition method

This method was suggested in the work [49] and the basis of it is contained in the following. Let us assume we must find solution y = {yi{x),..., j/ m (x)} of the boundary value problem

Li{y) = fi(x), WV) /<(*). i!' = = l,...,m, l,--.,m, x a: e 6 ft; IAv) = = « ( * )x)> , j' =i = l,...,r, 'i(y)
ft;

(2.1) (2.1) (2.2) (2.2)

where Z; and lj are respectively the operators of the equations and the boundary conditions, the fi(x) and
Li = Y, 2 J Lik* Lik,

(2.3) (2-3)

k=\ *=1

where some of the terms in the operators La, may not occur in £,-, and let us introduce the notation (2.4) Lik{y) = y?(«). /?(«). (2.4) Lik{y) yfw. From Eqs. (2.1), (2.3) and (2.4) it follows that k=h

=E

^(*) = H / ?f(«), ( * ) . i = l,...,m. *=1 *=1

39

(2.5)

40

Chapter 2. Decomposition

Next, let us add following h auxiliary problems for yk = {yk{x),. • •, Vm(x)}

Lik(yukk)) = = /*(*), f,k(x), £»(»*) f?(* l

*x € e f tl , i = l,...,m» l , . . . ) m , i f c = k-l,...,h; l,...,A;

i(y") Vj(*)> *x e rr] ikt e rr,, j =»«,... = ilk,...,i 'i(v*) »»i(*)i ,i rkrk. Y) = Vj(x),

Method :

(2.6) (2.7) (2.7)

Depending on the value of k in conditions (2.7), either iik = I or i/* = 0, / = 1 , . . . , r, and the boundary condition corresponding to the case j = 0 is omitted. The conditions (2.7) are chosen so that at each fixed point of the contour V the conditions (2.2) are satisfied by at least one solution yk. The following theorem holds. THEOREM. If there exist solutions of problem (2.1), (2.2), they coincide with the solutions which are common to all the problems (2.6), (2.7): y*=yk,

k = 2,...,h. 2,...,h.

(2.8)

In fact, in this case boundary conditions (2.2) will be satisfied, as a consequence of the selection of the conditions (2.7) indicated above. Furthermore, Eqs. (2.1) will be satisfied as well, since Eqs. (2.3), (2.5) and (2.6) hold. On the other hand, it is easy to see that a solution y is a solution of each of the h problems (2.6), (2.7) as well. The theorem is proved. Consequently, the task of solving the boundary value problem (2.1), (2.2) may be replaced by that finding solutions of the auxiliary problems (2.6), (2.7), containing mxh unknown functions fk(x) with the addition of the m x h conditions (2.5), (2.8) on the solutions.

2.1.2

Merits of the method

In the method proposed here, the formation of the auxiliary problems may be carried out in infinitely many ways. The nonuniqueness of the decomposition of the original problem is a consequence of two causes. The first consists in the arbitrariness of the choice of the operators La,, which may be one-dimensional as well as multidimensional, and, moreover, may contain terms not included in the operators L, of the original problem. The second cause consists in the arbitrariness of the choice of the boundary conditions of the auxiliary problems: an auxiliary problem may contain either no boundary conditions at all, or all boundary conditions (2.2), or any part of these conditions (if a solution of the original problem exists, then in solving each of the auxiliary problems all the boundary conditions (2.2) may be satisfied, due to the arbitrariness of the right-hand sides of Eqs. (2.6)). It is only necessary that at each point of the contour the conditions (2.2) be satisfied, when taking the union of the boundary conditions of all the auxiliary problems. The considerable degrees of arbitrariness in the decomposition of the problem provides wide possibilities of choosing the auxiliary problems so as to facilitate the construction of the desired solution.

2.2. Application of the Decomposition Method

41

The decomposition method makes it possible to rather easily obtain highly accurate formulae for the maximum function and eigenvalues of the boundary value problems. When using basis functions there is no need to satisfy the boundary conditions since these functions are used not for solution representation in series form but functions /*(x). This assures a high rate of convergence. When using numerical methods for solving the problems, functions /*(x) are taken as the basic unknown ones. The grid approximation of these functions guarantees a high accuracy of results on rather sparse grids. In conclusion, let us observe that the L, of the original problem (2.1), (2.2) may not only be differential operators, but also more general operators.

2.2 2.2.1

A p p l i c a t i o n of t h e D e c o m p o s i t i o n M e t h o d for Particular Problems Analytical solutions

2.2.1a. Precise solution of a non-linear differential equation with partial derivatives. We now solve the quasilinear quasilinear ainerenuai differential equation Dive tne equation )(du/dx + 2(au + bxy)(du/dx bxy)(dujdx + du/dy)du/dy) + (x + y)bu = = 9q^x) -f qq2(y), l (x) + 2(y),

(2.9) (2.9)

where a, b are constants. We introduce two equations (2au1 + bxy)du dui1/dx+ Idx 4 byui (2au22 + + bxy)du y)du2/dy + - bxu2

= f{x,y) + + qi(x), = -f{x, y) + q2(y).

Having equated these equations' solutions [19] we determine function f(x,y). result for u = u\ = u2 we obtain

As a

au2 + bxyu = tp(x - y) + + / qi(x)dx + / q2(y)dy, where ifi(z) is an arbitrary function. The result can be generalized for a multi-dimensional case and when the righthand side of Eq. (2.9) has a general form. 2.2.1b. Approximate solutions of boundary value problems a) We need to determine deflection w of a uniformly stretched square membrane at the central point subjected to a uniform transverse load q. The boundary value problem has the form: d2w/dx2 + 8d)22w/dy2 == -q/o-6; -q/ffS; q/aS; (2.10) w(x,±a) w(x, ±
w(±a, w(±a,y)y) = 0,

(2.11)

42

Chapter 2. Decomposition

Method

where 6 is the membrane's thickness; a the tensile stress; 2a the length of the contour's side. We introduce notations a = x/a, /3 = y/a, w = qa2v/Scr. Then problem (2.10), (2.11) can be written in a dimensionless form: d2v/da2 2■+ Fv/dp ay&**+aty80 aV3« ^ / ¥2 ■■== --ii;; a

v(a,±l) = v(a,±\) = 0; 0;

v(±l,/3) v{±\,P) = = 0. 0.

(2.12)

(2.13)

According to the method of solution we examine two (h = 2) additional boundary value problems d V "/ /ddaa 22 d2vW/d/3 vW/d022

= =

/f(a,P) ( a , / ? ) - 0 , 5 ; „i')n v ( l ) (±l,/3) = 0; 2 (a, < ±1) = 0, - / ( a *, /, 03 )) --U 0 ,, 55 ;; u< v, <(2)>2)((a,

(2.14) (2.15)

where f(a,j3) is an unknown function. If we obtain the solution of problems (2.14), (2.15) and equate them (function / ( a , /?) will be assessed) then we have the sought solution 2 „„ == „(>) „(') ,<•> = = u< u< ,< 22>.).).

(2.16)

When solving additional boundary value problems it is useful, if possible, to preliminary assess the properties of unknown functions included on the right-hand side of the equations. Thus, in the given problem from the symmetry and condition (2.16) it follows that the right-hand side of (2.15) must be derived from the right-hand side of (2.14) by substituting o <—> /?. Hence /(a,/?) /(«,/?) = - / ( / ? , a ) .

(2.17)

In addition, from the parity of function v along a and /? we have the parity of f(a, 0) according to these arguments. Due to the boundary conditions at a = ± 1 and j3 = ±1 the left-hand side of Eqs. (2.14) and (2.15) correspondingly become zero as functions differentiated twice along a and /?, become zero. Hence, / ( a , l1) ) = - / ( !l , / ? ) = 0.5.

(2.18)

After the double integration of Eq. (2.14) we find „(D - Jdajf(a,P)da /2(/?). <^(/J). u' 1 ' = Jda[f( a
(2.19) (2.19)

Using boundary condition (2.14) and assuming
[/*,

/(a,/?)
2.2.

Application

of the Decomposition

Method

43

T h u s , solution (2.19) can be presented as

1 l -1c r „('> J°dtJf(t,0)dt+ -^-. vW == J°dtJf(t 1P)dt+ -^-. f(t,W 4 2

[^

(2.20) (2.20)

Similarly (considering condition (2.17))

22 t a dt ] = 1-^-. t„< J0^/(*,«)* dtjf(t,a)dt+ ",<<2>'>:==/(, dtff( > ) + - /--A

2 21 ((2.21) - )

Substituting Eqs. (2.20), (2.21) into condition (2.16) we derive an integral equation for t h e function / ( a , / ? ) . Now we obtain t h e problem's approximate solution. Considering t h e parity of functions / ( a , / ? ) along a and /? and also (2.17) into account we assume s 2 2i f(a,p) -p )- /3&\ J[a,p) f(a,/3) 12c(a U rS ) ,/?) == l2c(a

(2.22)

where c is t h e constant to) bbe e found. From Eqs. (2.20)-(2.22) we obtain tu/ >' = = „<» (11

w

vvmJ „( )

== =

( a 44 --6 a6 2a /2 302 + Cfl2 6 / ? 2 - l ) c + ( l -- a 22))/ / 4 , 2

( , 34 4-- 66Q a 2 2/ /3 ? 2 + fi,v 6 a_2 - l ) + ( l - / ? 22 ) / 4 . 0 3 4 - 6 a 2 / ? 2 + 6 a 2 - l ) cC+ ( l -l -/ ?^ 2 )) //4 .

(2.23) (2.23)

Using these formulae we find

Using these formulae we find w0 = v(0,0) w0 = v(0,0)

(1) {2) = vv™{0 (0,0) = vW(0. (0,0) = v ( 1 ) (0,0) = Vv{2)(0,0)

= 0,25 - c. = 0,25 - c.

(2.24) (2.24)

To d e t e r m i n e constant c we a d a p t t h e obtained results (2.23) to condition (2.16). Table 2.1 lists t h e calculation results according to formula (2.24) and their deviations in per cent from t h e accurate solution given in column 2 with t h e following six methods of assessing value c: 1) i ; ( , ) ( l / 2 , 0 ) = v< 2 >(l/2,0) )(l/2,0) (column 4) 2) t;<'>(l,0) u<'>(l,0) = U< «<22>(1,0) >(1,0) (column 5) 3) » W ( 1 , 1 / 2 ) = U < 2 > ( 1 , 1 / 2 ) (column 6) 4) E q u a t i n g coefficients in solutions (2.23) with ao?2 and /? 2 (column 7). 5) Satisfied on average are boundary conditions for u' 1 ' with /? = 1 and for u' 2 'with a = 1 (column 8), i.e. t h e condition:

f „<»;

\)dct = 0. f/ vv((11))((aa>, ,, \)da \)da = = 0. Jo Jo 6) Minimization of t h e functional according t o c

J(c)=

1 ) dadc jfi:[ 2I~~ / ( Q , ,1)1 i

44

Chapter

1 2 u>0 0.295 To

0.675

Ho 0.141

Table 2.1: 5 0.300 1.7% 0.700 3.7% 0.142 0.7%

3 0.312 5.8% 0.625 7.4% 0.139 1.4%

4 0.293 0.7% 0.674 0.1% 0.140 0.7%

2. Decomposition

6 0.289 2.0% 0.656 2.8% 0.138 2.1%

7 0.292 1.0% 0.667 1.2% 0.139 1.4%

Method

8 0.302 2.4% 0.708 4.9% 0.142 0.7%

T h e m e t h o d of determining c is based on property (2.18) and can b e used for u' 1 ', u' 2 ' solutions. T h e result obtained agreed with t h e previous one (column 8). From Table 2.1 it follows t h a t all t h e solutions are close t o t h e accurate one. T h e third column in Table 2.1 gives an approximate solution of the problem under review, obtained using Ritz' method [30]. Comparison with Ritz' m e t h o d favours the decomposition m e t h o d . b) We now investigate t h e torsion of a rod with a square cross section. T h e b o u n d a r y value problem for the stress function ^> can be written as d2iPldx j>/dx22 + il>/dx + d2V tP/dy / d y2 2 == = -2G9\ -2G6; rp/dy 2G6;

il>(±a, ta,y)y) == 00, iip{±a,y)

ij>(x,±a) ip(x, = 0; i , ±±a) a) =

(2.25)

where 9 is t h e rod's twist angle per unit length; G t h e material's shear modulus; 2a t h e length of a side of t h e cross section. T h e greatest tangential stress m a x r and torsional moment / / are found using formulae

-'--(2)... m a xxTr == - f ^

\dxj \dxj

|

xx = a v

=0

;

—xc " H-==' 4iff J/ T/ J-^-xdxdy. ox O * Jo Jo Jo Jo

„=0

OX

(2.26)

Designating a = x/a, /? = y/a, ip = 2a2G0v, the boundary value problem (2.25) may be written as (2.12), (2.13) and formulae (2.26) have t h e form max 2aG0r0O;o;\ max T T = 2aG6T 2QG'0T where TO = -

fdv\ \da)

■ =i ' 0=0

4

H = 8a8a4G0H G6Ho0,

tfo =

rj

o Jo

dv

-adaa ad/3.

dat

(2.27)

T h e approximate value of function v is given in two forms (2.23) in t h e previous problem. T h e y coincide on t h e diagonal of t h e rod's cross section a = /3. From t h e auxiliary boundary value problems (2.14) and (2.15) it follows t h a t functions t*'1' and tiW precisely satisfy t h e boundary conditions with a = ± 1 and /3 = ± 1 respectively. T h u s we take v « u' 1 ' with a > /? and v « u' 2 ' with a < /?. Consequently, formulae

2.2.

Application

of the Decomposition

Method

45

(2.27) become

* To

— {da {-»r)) =

*=>

; Ho

p= o p= o

=L U I'M 3

8yW xda Qda - + da (

it,' 1 '

l da -a*-Aadadp.)df}-

-da~

The calculation results using these formulae and different methods of assessing the constant c (see the above example) are tabulated in Table 2.1 (columns 4-8) with their deviations as a percentage of the precise solution (column 2). The third column in Table 2.1 gives the approximate solution using Ritz' method [30]. It is seen from Table 2.1 that the decomposition method gives a highly accurate solution of this problem. Assuming v = u' 1 ' within the entire integration domain in formula (2.27) the result for H0 will hardly change. c) Let us consider the torsion of a beam with an elliptical cross section and a central elliptical cavity. The problem reduces to solving a differential equation in the double connected domain of the beam's cross section dhp_

d24>

2

dy*

dx

(2.28)

26G,

satisfying the following boundary conditions on its external (Ti) and internal (T2) contours (2.29) T/>(r\) V»(r 0, ^;(T V(r 0 ( r 222) = const, *<£ -£-ds = ---2n. 2 0J 1. /-(r.)a) == o, ^d8 = JT, on Here 6 is the relative twist angle of the bar, G the material's shear modulus, d/dn a derivative of the normal to contour r 2 , fi the area limited to this contour. We introduce notations

f

Q =x <*=-, a

p= = e ==0Ga eGa2v v, P h jl T== e, '4> v> ^ = S' 13=1, ^

ac

_ y_ be' be

a

=

b

2

9Ga2v,

( 2 - 3 °)

(2.30)

where a, 6 and ac, be are semiaxes of contours Ti and r 2 respectively. Then equation (2.28) becomes d2v _ d2v 2c2. (2.31) 2 2 + da dP dp2 ~

4

Contours T[ and T 2 derivedd from equations rT[:a ; :2a2 + +4 p?22 == c~\ c-2,

2 T r 2 ::aaa222 + 4 /3 = 1l

r=

(2.32)

correspond to contours T\ and T 2 . W i t h (2.30) b o u n d a r y conditions (2.29) can be written as

V ( r ' , )= V(T\) = 00,, (ri)

V(r' V(T' (r'22)) == const,

/^ It —-

* — 7 == -2ir. -2ir.
(2.33)

46

Chapter 2. Decomposition

Method

Thus, the original boundary value problem (2.28), (2.29) was transformed to boundary value problem (2.31), (2.33). In accordance with the decomposition method we examine two auxiliary boundary value problems. The first auxiliary boundary value problem: d2vt g l2 = / ( ao,P)-c , / ? ) 2-,c 2 , da ~~

w,(ri) (rj) = 0.

(2.34)

The second auxiliary problem: 2 9?v V2 _ , , a, 22 2 ee2 2 ^22 = e = -f(a,0)-c *,0)~c ,, d/3 ~ /(Q,/?) C

W ~

i

" '

(2.35)

(2.36)

2(r (T'222)) = =const, 2:(T'

v = vi V Vi = vv22..

(2.37)

Problem (2.34)-(2.37) can be solved using various known methods. We shall use a direct method, when function /(a,/?) is expanded into a series according to a certain system of linearly independent basis functions in the domain limited by contours T\ and T2. Here the construction of basis functions is much simpler than when using other known methods as these functions need not satisfy any boundary conditions. Another advantage is that with the same number of terms the decomposition method ensures a higher accuracy as in this case not the sought function but its derivatives are approximated. It should be noted also that the constructed system of basis functions can be used for solving various problems irrespective of the types of operators, provided their domains of assessment coincide. Assume that /(a,/?) = k (k =const). Then the solution of boundary value problem (2.34) has the form «i ( a22 + /3 »i = 0.5(Jfc 0.5(fc: --c c2 2))(a 0/ ?1 22-- cc22))..

(2.38)

The solution of Eq. (2.35) k + c22 v {a) + ,P v23 = Vl yj,(a) + l3 P

(2.39)

in which from the parity condition v2 according to /?, f2{a) = 0 must obey boundary conditions (2.36). It is easy to prove that these conditions using function ip\(a) can

2.2.

Application of the Decomposition Method

47

be fulfilled to any preset accuracy (for which, for example, yi(a) may be presented as two terms one of which is constant). As for condition (2.37) it can be satisfied only approximately because of value k. Here we only evaluate the maximum modulus tangential stress value max|r| = T.. And in this case it is not necessary to determine function ¥>i(a). It is known (also from the physical concepts) that r. occurs at the beam's cross section points having coordinates x = 0, y = ±6 (for definiteness assume a > b). Hence, _ djP __9Ctf 9Ctf OGa2 dv dv r. = (2.40) T (240) '~ be dp ' ~ dy dy **===coo be be OP dp °j o===0oo • »= = l/c \h

_ k

y=b

In accordance with this formula, on the basis of Eq. (2.37) we can write the condition for assessing; the fc as M R ; constant L U i i M d i u iA, ao dv-i dvi

dvi2 dv

-dJ W W W w ~~~dp ==

W th

'

1

a

0 ° == °^ ^ = -cc

Now, using (2.38) and (2.39) we obtain k = fjvjc2; thus, assuming in (2.40)

8v dv dv dv\ dv22 dvx dv? dP == ~dp~ dp~ dp ~dp~ == ~dp~ ~dj)

w

we find

a2b r. = 2 0 G — 2 - r =2. + b'b ' a' + The obtained formula agrees with /ith the known precise solutio, solution. In the works [12], [27], [50] the decomposition method was used to obtain analytical solutions of specific linear and non-linear engineering problems having sufficient accuracy for practical purposes.

2.2.2

Numerical solutions

We show one of the possibilities of using the decomposition method for an approximate numerical solution of Dirichlet's problem for Poisson's two-dimensional equation in a rectangular domain. Let us examine the following problem: 2

dd2Uu dx2

2

dd22Uu +

V

U =

..

,

(2.41)

/(*,3>)>

0 with

i = 0, a;

y = Q,b. 0,b.

(2.42) (2.42)

For the approximate solution of problem (2.41), (2.42) we study two auxiliary boundary value problems 7 dd*uw UW

- dx &2 r

1i

+ i/C2^x^ = f(^y) +

(2.43)

48

Chapter 2. Decomposition

Method

(l) [/O U [/<■> :== 00 with with xx == 0,a, {/< 1 d9 £/< (/<22) > a..2 ZJjL= - -V(x,!/) v ( x , y ) + i2/ ( xx,y), ,y),

(2.44) (2.44)

22

{2) 2

U > --= t/<

(2.45)

(2.45)

(2.46) (2.46)

= 0,6, 0,b, 0 with with yy == 0,6,

where i^(x, y) is the unknown auxiliary function. According to the decomposition method, provided we have the solution of problem (2.41), (2.42) it agrees with those of problem (2.43), (2.44) and (2.45), (2.46): 2 (1) : = r/(a) 7(i) [/ >.. U == [/<» £/ = [/( C/(2)

(2.47) (2.47)

Solution UW of problem (2.43), (2.44) may be written as: 1) ("(/, y)*, £/«(*, £/(x,y) = [/(■'(O, t/<«)(0,y) t/<«)(0, y) + + x 77((1) (0,y) (0,y) i> i>Wm{t,y)dt, (t,y)dt, (0, + f *P>

/o Jo

where

=f

( l1 ) V-("| *«(.,») tf>«(x, y) = = jT fQ L Lx,y) x , y)l + + «fa, 777<"(0, (0, y) y) == ((~pj ^ ) "(0, + ii/(x, |/ j( x , Vy)) «fa,

v 9x

) -)U

^. .

Q

We rewrite t/' 1 ' after substituting the expression for >r rp^Hx,y): tp^\x,y):

I 1'

(,) ) (,) (0, y) +■ xx <( 111 »(0, UM(x, t/C>(0 ( y) + U<% V™{x,y)y) =: £/ (0,y) *777 >(0,y) + J' [ ' dt * [J

where

X
>

(2.48) (248)

f'


-J? J:

2 (2) (2) i/«, x,s)ds t/< >(x, y) = C/ / ( 2 ) ((x, x , 0) + yi y7(% (*, 0) - J'dt J ll-Fy(x, y), 2 (2)

t/< >(x, y) = U™(x, 0) + y 7 where

(x, 0) -fdtf


Jo

2

/d£/< >>> (2) 2) (*>0) = (j~j (j~j 7<( 2>(x,0) (x,l = 77

V 9y J v=o

,,

F„(x,y) Fy(x,y)==j j **dtdt ^v(x,y) ./n

Jo

Assume that between the uniform grid's adjacent nodes x„ = = nhx, ym = mAy,

= 0,1,...,/V, n = 0,l,...,N,N N + 1,\, m = 0 , 1 , . . . , M , M + 1,

J'f(x,SS)ds. j'f(x, )ds.

(2.49) (2.49)

2.2.

49

Application of the Decomposition Method

the unknown function ip(x,y) changes linearly. Then from (2.48) and (2.49) we obtain t/<» «£!

n - O

nh xlal + Aj hi ( s| + ^ ) ^VOrr. = t ^ + nfc.TSJ

»-»

/,2

E »-o

j

A n ,m I FFx,nm +hl im + im nm +■ -F + ■hi x V E ^( n " -- i)ip OVim + -f

(2.50)

1=1

rrCO + «2 a£2 = U^

(2)

?AV

aO

mh^-hlf^^iy^ ^ 3 2 71-1

¥>n0

H

m m _- 1

2

m _ i /i A2 li -i)Vn. --A A 22, £ \](m-j) n j - -f O Vnm + -F„,„ m .

-sJEo

in,

(2.51) (2.51)

Using condition (2.47) in the grid's internal nodes

7
= f/(2)

B = 1,JV,

m1 = 1 ,

M,

we derive N x M equations:

uiOm rr(l)

nh -v (1)

«

n

" x70m

'G •-?) /1

1 \


\

■■'

/

n -n— 1

?

hi &> t

_J

Pim - - OW"

hi/i2 , + i1 *z,nm

.=1

m-A 2 rnh gg( (mm- -j> = 1"n jn j )^ni yTnO n$ 0 +m ^AE< *{ 2 J mh

u

2 fcS A

J=I

1i

-—^Vnm y¥>mn + + ^ X-fy.nm, m ,

2 *•""

= l,N, l,Af, = nTl = l,JV,

m = =TTl=l,M. J U .. m l ,, M

(2.52) (2.52)

Functions ^v(e, V(°.») g^ ' ^ ■ 2^°' £ ' ^y).

(2.53) (2-53)

Similarly using boundary conditions (2.42) and Eq. (2.43) we obtain

4>

rt*,0) = - i / ( » , 0 ) , (p(x,0)

4'

?(x,6) = - i //((zx,,& „(x,6) 6) .

(2.54)

50

Chapter 2. Decomposition

Method

Now we rewrite Eq. (2.52) taking boundary conditions (2.44), (2.46), (2.53), (2.54) into account: n-l

ro—1 m—1

1=1 1=1

Jj = =l

{ im h m "-M£! ^ 7 ^ - mhrJ mVrio ++ hl Xl(" l XI( ~~ ifo**++ 51=1 > -~ ^ + hlYiyn-i)H j=i = =

n =

^

1, A, l,N,

z—g—p.. V{fnm

6

n " -- 1l \ /om 1 ^ 1 + I rn-l\ /n0 1 m+ 1 f/ fo m +++ h J+ „ 1 /nO jFl „ ' i , ™+ + --Tj, °"> 2*' V~) '" 22^^"""""" 2 > (3 ( 3 + TJ1 T~) / n ° ~ ~ 22Fx-™ m = 1, M.

=i\

1

h -2 ~2** *[3(3 (§ ++ — ^l") ~2** ^l") 7

L2 »2 _I_ 1I 1,2 L2 L2

If we add M Eq. (2.50) on the border x = a (n = N + 1) and N equations (2.51) on the border y = b (m = M + 1) to the obtained system we get a system of 7VM -f A7 + M linear algebraic equation with unknown values fnm{n = 1,N,

m = TJT), 7£>(n = MO, 7£>(m = Mf).

Function tp values in the grid's internal nodes and 7' 1 ', 7 " ' values of partial derivatives of function U on the border of the domain obtained when solving the resulted system make it possible to use (2.50) or (2.51) to calculate the original problem's (2.41), (2.42) solution at any internal point of the grid. For problem (2.41), (2.42) calculations were performed with the right term f(x, y) = — s i n ^ s i n ^ . The solution obtained for the grid was compared with the accurate one '%2 it2YX . JW . iry uU(x,y) { x y )== + sin1 —s,nsin — . a b The linear system of algebraic equations was solved using the Gaussian method with columnwise pivoting. Calculations were performed on three grids with division of the rectangular domain's sides in 2, 4 and 8 parts (N = M = 1, 3, 7 respectively) for the domains with the b/a sides' relationship 1 and 3. In all cases relative errors

>

{^

¥)

T T-

\uiP-u,\

max J —rj—. e = ma \Uii\ i,j ' \Uij\

(2.55)

did not exceed 10~16. In all the variants, the maximum in (2.55) was reached in the grid's internal nodes adjacent to the border of the domain. In the centre of the domain the error was minimum. An algorithm for solving this problem with a non-uniform grid has also been developed. In the work [22] the decomposition method is used for a numerical solution of composite problems for parabolic equations and elliptic boundary value problems such as: biharmonic, Dirichlet's problems for Laplace's three-dimensional equation and the boundary value problem with boundary conditions of the first type of the three-dimensional elasticity theory. All this reduces to the successive solution of smaller dimensional problems using grid methods. The order of the error is assessed and this proves the method's reliability.

Chapter 3 STATICS 3.1

Plane Problem

This chapter examines a plane problem of a rectangular plate with different lattice structures. Here the bending rigidity of a plate's lattice consisting of two families of rods is considered.

3.1.1

A plate with more than two families of rods

We refer the plate's middle plane to the Cartesian system of coordinates x, y. The differential equilibrium equations have the form dNt SiVj dS 95 ax "+ ay +

dN22 dS 3^2 as x, „ ay ax + Y = dy + dx

„

v

^ * = °' -di ^

^

°-

(3.1) (31)

With the deformation components corresponding to the plane problem With the deformation components corresponding to the plane problem £i

au du ax dx' ox

£2 =

au dv ay ay

du au ay dy ay

w = du

dv ov dv dx' ox ox

we may present the constitutive equation (1.24) as n n

= J2 Y,Ki4Vi{ciU Siv), AT, ^ # ; jsr«?Vj(cj« c ? VfVi(. j ( c , u + 3,-v), a,t;), JV, Nx =

E t=i 1=1 n n

E

fvf(.

Siv), AT22 = J^Kts^iiau £ /r,-5?Vj(c,1l /r,-5?Vj(c,H + 3<W), N i=l n n

E^

5S = ^ i OKiSjCiVijCjU . - C i:,V,( V i f Cl cu + Siu). S = ^" == ^" KiSjCiVijCjU + SJV). SJV). ii = = llli i=

51

. . (3.2)

52

Chapter 3.

Statics

Operator or Vj Vj iiintroduced for performing the differentiation towards the tangent to the axis of the i-th rod family has the form d d_ d .<;.- d V; V, = Ci-£- + Si—. ox 'dy' oy Note that value CiU + s,v is the displacement of a point on the middle surface towards the tangent to the axis of the i-th rod family. Using the obtained constitutive equation and (3.1) we find the following system of equations for the displacements: Li ln(u) (v) + + = 0, 0, Li122(t>) 4- X -' = l!"! +L

LM«) L-22 0. 12{u) + 22{v) ( t + Y = 0. + Ln(v)

(3.3) (3.3)

Here L,j are linear differential operators:

i, n =. Ln

~ ~ Li 1 2 —~_ Lu Ln

2

a + r d2 +c°d r d" +2Cw2 r 6 Ad2 +r°d Cu Cl2 lj,2 2 ++ l T A + dx' 2Cw~dltfy~ CudT*+ + Cl2df2 + l a5x ^ ++ '333 5y^ a?

oT*+ ~dltfy~ dfa2 a^ a? Q2 d d2 d^0 d , JL Ci2 +or * -i-r +r° +r° Cl6 16 +2Cu + C26 2 3 + C r dT> dx* d^ dy~ dX~ *dy'' 5

oT>+2Cudx%, dxdy + C26dy~2 + 3 a ^ + 2 9 ^ ' d Q2 &2 d d_ ^ - +■+■C° +C°VC +2C 167T~Z +C 1 ■WZ—T 'ay' L22 - Ci2-^+2C26-^-y oxoy + C22dyi+C2-^ dx + C< dy, where additional notations are adopted Lu

~

Cl6

z, r L22 _• C

dCu/dx dCu/dx C° = dC + dC dCie/dy, /dy,— if C° =■dC /dx + dC dC /dy, /dy, u/dx u/dx l2 2626 C? = ■ *~i6 /dx + Ku/dy /dy, C° = dC dCis/dx - dC dCdC /dy. C° dCi66/dx + dC /dy. /dy. 12/dy, 22 C3° =-- dCi l2 26/dx + 22 22 *o/

(3.4) (3.4)

If the lattices parameters are constant (each family of the lattices rods can be characterized by their own parameters independently on the coordinate parameters), the differential operators of the system (3.3) will take a simpler form: n

! 2

2 2 r.. — L L„ = = Y,K.c £ t f , c Vl Vv ,

=£

2

u

i=l i=l

v

I'

n n

= £*

! 22

2 L = X>,s2V Vv ,,

L 222 2 = £ K , S 1i = =1l

2

n

£

2

L ^ ^ t f . ^tf.s.CiV*. V . v 2 ..

L 12 = £

(3.5) (3.5)

1=1 i1== 1l

Now we consider a plate with three families of rods. Assume that the lattice's parameters are constant f = f\ = —¥>2> V3 = 0, a = ai = a 2 = 2ca 3 , F = F\ = F2, E = Ei = E2 = E3. ~p r an b o b t a i n e d from f r n m the t.hp one n n p shown s h n w n in i n Fig. P i i r 1.5 1 Fk provided nrr\\i\At>A t.h. Such a lattice can bee obtained that it has no third family of rods (the fourth family of rods is taken for the third). Note that the parameters without subscript i will refer to the first family of rods i.e. » = 1. For instance s = Si = siny>].

3.1.

Plane Problem

53

In this case t h e reticulated plates' constitutive equations are formulae for from (1.30) where: 4

2 2

Cuu — C = = 2Kc 2Kc +K K33,, Cu ''
C\ C C112 = = Cee Cee = = 2Ks 2Ks Ks2c2c, 22 — ?66 —

N,N2,S

4

,

4 4 C C22222 == 2Ks Ks 2Ks . .

For t h e given t y p e of lattice t h e differential operators (3.5) in t h e system of equations (3.3) become:

£i nnn

= =

4

r+

\c44KK(2c (2c K+ +

^£-K )£- +2s c K^, 3

2 2

2 2 K3)£2 2+2s c K^,

Y iL2 = MK^Ux2 + tantrtrjfe), *dy ) 2

i 2222 2 —

2

2

iLll2 2

= =

d K' ssin i n22 2 2
Assuming Y = 0, and also u = -(a3a/EF)L„(9), 3/£F)i22(*),

v = (a3/EF)L„(9) (a3/EF)L12(Hf),

t h e system (3.3) can be reduced t o an equation

KK

d4

d +K +K+K+K =x \$ x= X, dx dv 4

( €* "& ^ °^*w)* °S)^ >' 2

2

(36) (3.6)

where A", K A auuaai2, ai2,i 2 , x, = a

— a22aa22na, 12, KK22 =

K = an
and 33 =63 = 03 4+ "a nn = 03 +
2 l2 12 = aa12 — ss c,c,

3 ip, a22 , 22 = s tan <

6 d33 = = t3F3/F. f.

Let us consider a particular instance when part of t h e boundary conditions has t h e form 0,1, u = 5 S = 0 when wh x = = 0, /, where / is t h e size of t h e plate in t h e direction of axis x. T h e n it is convenient t o search for t h e solution of t h e differential equation (3.6) as expressed in t h e series CO 00

] P v Jipmm((y) j / ) ssin iin i n AX Ammmxx ** (( *x ,, yy )) = "^2 m-1 m—1

rmr/l), (\m, m = rmr/l),

(3.7) (3.7)

where each t e r m satisfies these conditions. Assessment of functions ipm(y), corresponding t o t h e solution of homogeneous Eq. (3.6) at X = 0, reduces t o solving t h e characteristic equation 4 2 4 2 s«J.-2A^(o -2\22nm(a (a22/a /ai)s \4Ja2 = 0, S nm-2X l)sl+X m/ « ,Ja )£ + +Al/«? = 0, / m • mi 1 2 2 2 a, = a, = {Kl/Kyl (Kl/Kl , K^K'K*)-*' . KW , , 2 a 2 = K° . 3(K°K° 2)-

54

Chapter 3.

Statics

Figure 3.1: The type of roots of this equation is determined by the relationship between values if and S3. They can be real, complex or purely imaginary. The domains of various types of roots (imaginary, complex and real marked 1, 2, 3) are shown in Fig. 3.1. The multiple roots corresponding to the curve dividing domains 2 and 3 having equation y> = arccos(0.5<53 ), are: Sm Sm

1/22 2 33 ZAmmS5 - [(5h3 + c )< = )c]'/ . = ±A

The equation of the curve separating domain 1 and 2 has the form

^m Sm

33 11/2 = ±a ±i\„, ±am cotvp(l+.53C)) ./ 2. cot (fi( \-s3c- y' m cotvp(l+.53C-

Forces in each of the three families of rods are determined from the calculation model's known stressed state by the formulae fNt v l = "u *-\a

±

a

s

-) -c)2s' i?

«3 = "*~{c

'NL_

a s 2 J2' 2'

Njc"

(3.8)

derived from (1.43) assuming y>3 = 0, a = 2c
3.1.2

A plate with two families of rods

First we consider the rods' rigidity in bending in the plane of the plate. Using formulae (1.5), (1.12) and (3.2), we obtain the turn angle of the i-th family rod's axis around the normal to the shell's middle surface:

( -4

fl d

4>i =

Ci

-Sis

f) ri \

dyj

CiV — SiU).

{c v s u)

Yx- %) ' - ' -

To change the curvature of this family of rod's axis in the plane of the plate in accordance with (1.17) we find

u (

d2

dd22

2

d \\ 2O

\V — SiU) K*? = + s in2y ' = ~ \c'd7dx2 2~* ' m 2 i f i i a)xdy ^ + A S<W2){CiV"SiU)-

(3.9)

3.1. Plane Problem

55

Now using formulae (1.22) assuming n = 2, the plate's constitutive equations may be written easily. The calculation model's stressed state allows assessment of forces and moments in discrete rods of the plate according to formulae in Sec. 1.2.3c. 3.1.2a. A plate with a rectangular lattice. We fix the coordinate axes so that Vi = 0,(^2 = T / 2 . Then from relationships (1.22), (3.2) and (3.9) we obtain formulae (whose validity is evident)

= *£, N

du , dv Nt = N = *K * * 2 2== =K*2 K£, * Ni Nl dy' Ox ox oy q _= ,o93V v ^ u S2 _— = '20d?U Sx dx 3 ' dy3' M Muu --

£ 7?2 1

= ^dx~fi.,2' '

M2 Mi, =- I2f

'

d2U 22

dfdy ''

where If = EJ2,/a, (i = 1,2). With these linear force values the constitutive equations for the reticulated shell's calculation model dN, dNi ^L ox dx dx become

+

dS2 ^uo2 oy dy ay

2

„ v + XX == 0,0,

f

dN dN22 ^ oy dy dy

|

22

,d2dv u

dS 95,t v ++^ 0 7T ox ox ox

n

(3.10) (3.10)

4

d*u __ _ r dd vv 0d v -I°0l-d*u A, K: 22 /? 'dx* dy = W ~hoy~ ' {lW ~ ' ~dT* ~ ~YW

Kl

(3.11)

Thus for the plate with such a lattice the system of equations during displacements breaks down into two independent equations. A calculation example. The requirement is to ascertain the influence the rods' bending rigidity has on the reticulated plate's stressed state in the plane of the plate having initial data (6 is the size of the plate towards axis y) I = 26, X = Y = 0 and the following boundary conditions (p = const): u = v = Mu = 0 when x = 0, v = M\, = 0 when x = I, u = v = M2, = 0 when y = 0, b.

Ni = — Pi K

These conditions correspond to the plate's hinged-fixed support along three sides; a tensioning uniformly distributed load of intensity p is applied to the fourth side (x = I) normal to the edge of the plate. Note, that the rods' axes coincide with the calculation model's contour only when one additional boundary condition is met ensuing from the rods' rigidity in bending in the plane of the plate (see Sec. 1.2.7).

56

Chapter 3.

Statics

As the second differential equation (3.11) as regards functions v is homogeneous as well as the boundary conditions for v and Mu we obtain v = NN2 2:= Si Mi. = 0. St = Mu Hence, the first family of rods is subject to longitudinal forces and the second— only to bending in the plane of the plate. The solution for the function u is 00 OO

( ) ^y^y, E> „(x)sinA

uu(x,y) ( x i y) = E

Um z sin

m

nw/b), (Am = "»*•/&), (*m

m=l

of which each term satisfies the boundary conditions when y = 0, 6. Substituting this series in the first of equations (3.11) we obtain homogeneous second order differential equations for functions um(x) which must be solved with account of the boundary conditions x = 0 and x = I. The first of these conditions is homogeneous (u m (0) = 0), and the second is homogeneous only at precise m values (the edge load p =const after its expansion into a trigonometric series according to functions sin Xmy yields non-vanishing terms with odd values of m). Hence, only terms with odd m will remain in the expansion of the solution. For the case when the two families cross sections are identical and a.\ = a2 = a, we write the final solution for the forces and bending moments in the reticulated shell's rods: AT; = apN°, oN°, S; S'' 22 = -alpM° -alpM°,. = -apS%, ~a* «, G\ '2 =tpMl-s. ft In these formulae oo 4 4 °°

1 _1

——-/) cosh(s„ cosh(ss x)sin x) sin E-. >sh(s ■K *—* mcosh(s ()

N° N° = - }]

m

7T

m = l oo

E

v

m m

Xmmy,

m

i

4r2A °° 1 m = 1,3,...), g -sinh(s m x)cos Amy S£ = 4 r 2 AmY—J1—xh(s —— (m = l,3,...), ml)- s i n h ( s m x ) c o s A m y 4r 2 1 M°, -Y] ——-sinh(s Ysinh(s i)sinAm j/, y, MS. == j mmi)sinA IT ir *—' mcosh(5 m ()

S°2

E

2 2 21 / 2 2 : where ^smm = r2l\2m,r2 = {hlFpyi (h/Fl )*' = lIUO ^ , A =--l/b. l/b. ( J 2 / FFpyn / ) 0""2 r°, The dependences of the non-dimensional values lues ./V°( N?t ./V°(0; V,°(0;6/2) 6/2) ,and A/°a(/; 6/2) 6/2) con those of the non-dimensional radius of inertia of the rod's cross section with wit relationship of the plate's sides / = 26 are given in Fig. 3.2 (curves 1 and 2 respectively). From this figure it follows that the parameter value r° substantially effects the plate's stressed state.

3.1.2b. A plate with a rhombic lattice. Assume that ip = a2 ■
3.1.

Plane Problem

57

Figure 3.2: Using (1.22), (3.2) and (3.9) we obtain the following constitutive equations for the plate with this lattice 2 2 2 TV, N N^ N, = Nj v(DW _ j yJV<M2> V « ttan a ip i V>p+ ^ 22 = = N^tan ,\ ++i N^(2)N^\ 2 S, == 5<" ++ 5( S<2>>, , _ s^i -ft-sW-S&W?, S^'i t a n ' p , oi = s2 = 2 /0 A ^td v d0 22 » • 22„ „ 52u . . \ i 2<^ - 2 2Sm2 2 22ip) Mu = Mu ic * +1- ^ S 1 = Y{ dz-i dy- tdy*- d^-ysm2ip)'

?(

M M2.

10 L *92u P

d2u . 2n n22:p dx2*

„o d2v . .2 \ 2ip) dxdy

(312) (3.12)

- = T ^'%dv+ S ^ ^ - ^ ^ S =

2

u

In these equations yd)

:

TV<2>

=

c(2)2 S< >

=

/iTsin22v du 2 a~cotcot ip -yTy) 2 Ox z 2
\

2 2^ ' U ^3 + S

dy

o .22

:

°^ ^< 3U^"" 2

dx dy

o 2 2^ ^

>c ^ 2 dxdy

K sin2 2y 2

S (i)

23V

■'£)

Hi

dx

du\ dy)

°sin 2 2^,

°^ ^ ^ \ rr° O • 2 11 , c"
C<

V

s dx J

Forces and moments occurring in rods are assessed at the known stressed state of the plate's calculation model according to formulae (1.44). Equilibrium equations (3.10) taking formulae (3.12) into account result in the following system of differential equations for the displacements: IX 2X 2X + £LLli1(v)+ , ((v) p ) + r r r r2, r - = 0, L£n(u) 0, ll(u) + 12 K sin 2y> lip 2y> 1Y 2Y T..Ji,\ + 4- L /.„. Lu(u) = = 0 0, 22(v)++ 2 2OJ K sin zip

(3.13)

58

Chapter

3.

Statics

where operators have t h e form iL\\n

=

L L12

T T = T

>?Tcot L22 Z/22 = L t a n 2

-— p\T

LI-AT, — p\T,

2 -^ ^ 4 p2^ ((c2 J?2 -^ ^) 2 * •dxdy * 333J ) ' dxdy dx333dy dy dxdy 2 \ dxdy dx dxdy \ dx dy dxdy J ' 4 4 , ,2 dg s*4 d* d22 2 2 d4 2 22 4 fi d „ . d s d* L = cot cc2 ^ +(i3 2 + L = c o 2t
d ^1 ^1 d

= = dT*

dxW

2 2

s^9^t '/' 2 '

^

sdy'

here :re p\ •-= J22/F /F is tth h e square of t h e rod's cross section inertia radius during bending nl in t h e plane of t h e plate. Assuming Y = 0 and uIf == -—L/ ,22o2o2((QW U *>)),,.

t; L V- = = i li22(( * ) , = £«(•),

we reduce t h e system (3.13) to one differential equation as regards to function $

i ( i + u n4 Vx) 9 {{ *a? *a? -- dxW ^w *v ^ d 6 "- ^ rr ^5?£ a + s5s {U cr ip + 1 6 4- (7 t a nn y

cot2 cot cot

o4 9* ^

d9*4 dx dv2

o 22

2+ +4 tan tana n

2

22

9* ^9 4 dy

s

2 [ 2„

(1 ++ tan ^ (1 P2 tan V)B*

6

2

22

22

4

d6

++ 77 cc;otoo tt
t dy2

2

+ s2(1 + t a n V )

22

'

2

+pt

c4cotV

a n Vif) + p -P\ t cc 4 cot ,s +ft tan dyJ ^^^^iwv + s2(1 ^^J (( c o t(Vdx^^ 8 , 2 d* A" , 2 2 o3 ^ \ 1 . • t ann 6s a cc 4s4, 3 2, a9"+ +5 4Utan 4c 2 ++66sC A3 ai/ JJ)/ ** ~ dxW dy diW< ~ dxW ' *d?) --4c4c ^Jx9w &w " a^iv * V 5 5y V 2

4

iT2«„4 "

6

4

88

2

2

2 2

6

2

2

2

6

8

8

2X ~ A"sin Ksm222ip' 2ip'

(3.14)

From this equation it follows t h a t the assessment of t h e rods' bending rigidity in t h e plane of t h e plate (p2 ^ 0) increases its order from four to eight. As this is so the formulation of two more boundary conditions has also been considered in Sec. 1.2.7. Considering formulae (1.5) and (3.2) t h e additional b o u n d a r y conditions (1.82) expressing the absence of turn angles of both families of rods around the normal to the calculation model's middle surface for this particular plate can be written as dv du dv dv 2dv 2du ss' c cc 5■=—• 5 ^^ = -5 ^ — ■ a~ 3 ~ == 00 Ox dx dy aoxi oy 03/ aoxz ay oy

5"

when when

y yy == = yoyVo0-

(3.15) (3.15)

T h e second of these conditions coincides with (1.81'). A calculation example. To assess t h e influence of t h e rods' bending rigidity on the plate's stressed-deformed state, calculations were performed after fixing t h e following parameters: / = 26,

v3 === JT/4,

p^ pp22 == 3.19 3.19 ••110100-4-4,4,,

7Y == - 0, 0,

X X == const. const. const

3.2.

Bending

of Plates

59

T h e following b o u n d a r y conditions were assumed 3 u = = S Si1=S' = 11+5T-+ S; = 2 ■= G\-G"2G'2 =

dv U = V =dv_ U = V=

Ju du

0

=du

dv dv fir dx--dydy

= 0, /, x= 0,l,

when

^-%= ^-%==° ° When 3/=y =°'0,6. °'- b-

dx dx--dy-

Wh6n

y =

b

T h e last two of these conditions coincide with (3.15) at
dN, dNi 2 - t a n tp oy dy

dS as dx ox

n

T h e solution of this system is N\

tany), ip2{y -+f xxta.ntp) Dt atan n y itp) ) + ++ipi(y — x xt&ntp),

=

S == where ^1 and ^2

are

c tan tp)] t tantp, a n ytp, [—xp22(y(y + x1 tan ip) + 4- V'iCv i>\{y ~~ — xtan<^)] , t a n tp) f) xtanip)]

arbitrary functions of their arguments.

T h e n using formulae for forces in the rods when tp = tpx = —tp2 2N;a = (Nic~22 2JV' 2 = (NlC-

± 2 51s isin n " 1 2tp)a 2i ± 2 5 s i n " 1 2tp)a

we derived 2

Ni = ac--22Vx(; ip\{y — x: tan tan tp) tp), N{ = ac~ ipi(y — x tan tp),

tan
and consequently t h e forces Nf and N2 do not change if conditions y — x tan tp
3.2

Bending of Plates

This section examines transversal bending of rectangular reticulated shells. study, except Sec. 3.2.5, is based on Kirchhoff-Love hypotheses.

This

60

3.2.1

Chapter 3.

Statics

Differential equation for bending

Equilibrium equations for an element of the plate in the Cartesian system of coordinates has the form dQ fl fi Zr == o,0, -~oV ox + Oy oy + 2

fir

W

_W dM6 2__(

0, "5 noy V22 -= U, ox <7i ay dH2 8Ml (3.16) -« ~ (Q?ii == o0 .. (3.16) ay ax oy ox From the two last equation considering formulae From the two last equation considering formulae 2 2 2 K] »ci = -d -d22w/dx w/dx2:, K = -d w/dy d2w/dxdy w/dxdy AC 2 = " 3 W, , rr == - —d »ci = -d2w/dx2, K2 = -d2w/dy2, —d2w/dxdy and constitutive equations (1.22) we derive values of the linear transverse forces as and constitutive equations (1.22) we derive values of the linear transverse forces as d3w /<"is4 (3£>6i Oi = ( A i + Jfu) 3x3 H 'I?) )x2dy 3 a ^ +(D«« + D, A„ D62- K&) "1y dxdy2 2

+(fl»+^ - A„ - KShgfc+(D. - *#)§£ i + Ku) -h°n + Kg') ; ox + [3^'+*»> + >*+*#>]£ 9i + JW ^ + 1^1 +++) A+AKI6,|)1j(+.i)D+| ^( 6D( f66l-6M-^-]?)))xdy g]] ^^+ [-(2* a^ 2

2

5"

6

») 7 d ^W^-x^W^-^h +

3-(^i2-

))
£

+

A

«l '

d2w) «„2

3

(D D A A --+11 IKI ++n / ^ )d ^W66 ++ D ^66 D1211 22---An

dx*d\ 3 8Pw d w ±
h

[^

9w\ A A „n) ) ]0}. 0 b> ^)]g}. 3v I "

n i A6 1 +a >22 262 ++ O + 6(D 62 + a 2 ++ + [^ &<> n

(M

2

(3,7)

(3.17)

With these linear force values from the first equilibrium equation (3.16) we find the bending equation I(w) L(w)-Z (3.18) £ ( » ) ■ - Z = 0,

61

3.2. Bending of Plates where the linear differential operator is derived from formula

U«) = L(u,) = K«)

(Ifc++ * „«. )d'w 2 ( 2 * - * „ ) *3 L (fti £ ++ 2(21*-*„)*£ dx dy d*w

crui + 2(2* ) 3jgjL 16k - * +? K)~ > 2adv^2 + 2(2Z) « + K ^+ *£dxdy + ^22+ ( * dx

+(3D + +(™<*

k

^d*w

D„ + Kn) + + 2 +Ku, U >dy* "dx

D i — K\e) c [; ^_ + [ 3 A (2D61 _^ 6) + | . 3( 3 ^ + ^ ) ] dPw

tet

I' '

6

&w

ax3

&W

3Z)66 + < ) + 3*|-(2D62 + ^ ) fix ] civ'* ^ + [A di ( ]^_ + [ 3 A (2D61 _^ 6) + | . ( 3 ^ + ^ ) ff*w V \l(2L IDn + Ku) ■ 2 (D?22 + K\l, dy3 > T, ' d_

8*w

2

2

\ d .„ +h [£(*>„ £ < A , - *„)] ddxgwj V P C 61 -ff16)+ S-SD ^ ( D „ + Ku) + r^_(2D 2

-f[£(2D 6 1 -^ 6 ) + ^ ( 2 ^ + /^') + ^(2A 2 + ^,)]|j + [£(A 2 -^i.) -f[£(2D d 1 dw 6 1 -^ 6 ) + ^d ( 2 ^ + /^') 2

+

2

2fl +jr

^ -

2

(3.19)

+

(319)

»)+^ H^-

Here the following additional designations are adopted:

Here the following additional designations are adopted: n n 2 2 2 ° » = = nX ) C i osc o2s 22<^,. ff{°> = n£ ( 1- 6- 5 6c52c>2)C„ ^ K& ^0)

E =^(1 i1== 1l

0)

E< i=l

(3.20)

2

K[ - 6s}c?)Ci, Ki = J2 Ci cos 2fi. (3.20) •=i i=i Equations (3.18) for the particular case where all the coefficients are constant is Equations (3.18) for the particular case where all the coefficients are constant is given given in in the the work work [69]. [69]. Now we investigate the bending of plates with particular types of lattices whose parameters do not depend on coordinates. In this instance formulae for linear moments can be presented as n

M Mxx = Mj

E<

V:V ss , A ., > , £^ c , V ,,((//, ,cc1iVV,, H+ C ,s;A,)' 1=1

n n

;, ((/^ 55,Vc V, V- C ,A; Ai ,) ">' , 5\ V Vv, - C, ,cc A = ^£E* = = -- ]]E T c ,, V Vv.( V C ,, C c ,,A A = T (( // , Ss ;v,V ,, -- C A ,, > > ,,

Mi M22 = M

1

j i

1

1=1

#1 Hi Hi

n n

c .>=i 1==11

1 1

1

i

i

62

Chapter 3.

Statics

n

-E

s,A, H2 = -^a C,-« - £ Si V , ( /i c,-V< , c i*v,-V t + C, iA,>) i V1 -(/ 5 ,A.>, ,v.i

(3-21)

t=i i=i

where in addition to (1.17) a differentiation operator is introduced in the direction orthogonal to the tangent to the i-th family rod's axis: A, = Sid/dx sid/dx Cid/dy. A,= id/dx -— c,d/dy. id/dy.

(3.22)

Formulae (3.17) for the linear transversal forces of the plate's calculation model can be written as n

-I

,v,, ++ Ci5ls.A.) Q, = - £ vn 2 v?( (/iC)V A,)' 1i = = 1i

n

Qs ==

E

cA, 3,V, - j c i A.-)iff. -5^Vj(/ n 7?(i aiV«-C i=i

Q2 = - ^ V ^ . V . - C . c A . K

Operator (3.19) of differential equation (3.18) becomes: i=i Operator (3.19) of differential equation (3.18) becomes: n n

I

2 L(w) V?i '•Vv t + C
3.2.2

(3.23)

(3.23) (3.24)

(3-24)

A plate with a rhombic lattice

For the particular type of lattice (diagonals of the lattice's rhombs coincide with the directions of coordinates of the middle plane)

= = = = =

2

2

2 222 222 2 2 2 2c2[(Ic [(7c22 + )a )d2w/dx widx u;/ai + ( / - C c)s ) a)s addudwldy% ; w/dy / a y 22]},, + Cs22)d (I-C)s 2 2 2 2 2 2 2 2 2 2s {(I-C)c d ?d w/dxw/dx + (Is +- Cc )d wldy w/dy2 }, 2 -2c2{2IsI2s 2 - + f C Ccos2)c p)d2w/dxdy w/dxdy, s £.y>)u w/uxuy, 2 2 2 2 2 2 12scos 2y> 2s {Ccos2
(3.25)

Formulae (3.23) for Qi and Q2 become much simpler:

* - -"'('E^fis) Q. =

2jPw_\

dx

dxdyj

-- 22 ,> 7^ + + (I -3^], * dxw ,> CC [,g [,g + (I -3^], c

Q2

=

■id3™

-2s 2 / rf*?j (

c2)

3

aw ?x2dv

2cPw\

- 2 e . C [ ^ + (1 -3,.,£|]. x dy 2

(3.26)

3.2.

Bending

of Plates

63

T h e differential equation of the plate's bending (3.18) taking into account the structure of t h e rods' lattice in operator (3.24) changes to: d*w d*w d*w crw D 1^1 + DD3 1 a-2e>.. == D 2+ 2+ D 2

'\d^ ^

dxW *e^

Z Z_

(3.27) (327)

W 7-fr >W

T h e coefficients of this equation are: Dx Dx

=

2(c* + 77«o). -ya 2(c a 0 )0),, (c44 +

D2

=

9/c + J Tan). 2{s* -,aQ),

D3

=

12a00 + 27(1 7(1 - 6baa 00),) ,

4

22 2 where 7 = GJ3/EJi,aJi,a 0 0 = ss c . nd torsion mc T h e bending and moments as well as transversal forces in t h e plate's rods are derived from formulae (1.44) and (1.44'). Let us consider t h e bending of a plate with a rhombic lattice hinged-supported along t h e entire contour with transversal loading Z = q = c o n s t . T h e solution of equation (3.27) with hinged support boundary conditions

w

=

M\ = 0

when

x = 0,1,

w

=

M2 = = 0

when

y = 0,6,

may be easily obtained as an expansion into a binary trigonometric series

= E£*. 00 OO

00 OO

.

r-< r—» r—* kmSm mirxSlnsin . kwy kiry kxy si. mnx m S~^~ i n — - S ~I~^ in_ = 2^ 2^ ( ,y) = 2^ 2^ * T /T' ■ 1 k=x m=l When the external load does k=X notm=l depend on coordinates the coefficient values of h e n tat h eodd external loadequal does not depend on coordinates t h e coefficient values of thisWseries m, k are this series at odd m, k are equal 4 16o/ A\ =' l \ A 6£^ 11 (» A =_ 1 16gl* A* = : kmm m 66 22 44 * IT \ m*k* P + + DU-y^k") \ b) ' 22\
W i t h even m and k values these coefficients are zero. Using (3.25) for t h e values of linear bending and torsion moments of the reticulated plate's calculation model we derive formulae (k, m = 1 , 3 , 5 , . . . ) 9

OO

OO

=

r

r

A/2 M2

= =

Hx Hx

= =

1

>r / >c—» *—» . , 2•> „ „, , ,9,9, mirxr . A:7rj/ kwy . .. m-KX D t+ - y2 J2Y,Ak^m A k[m A 2i (+l 2- 7 ■) «7 )oa*oa 2^A2k]]ssX. inn —s — s -is ni n ——-, - ^ . m D m 0 " / t=l m=l n= l *=i 9 OO 00 OO > k=l m=l kny mirx 2 2 jr-j 9 r 00 00 > /J^i2,l]sin - r—/ r ^- Xc , n' 4 *. " ' r«,. ^ 1 - ^ 70 )*02a "0 m1 1 ++ *, 5AA, ?A sin — s s i .n -~T' £* • 2 7—imirx kny - — 2 k=X 2 2 m=X 2 ^km[2{l--y)ttom +k \ D2]sin-Y-sin-^-, d=i k=l m=X 9 T OO OO OO . °° OO V~* OO m7ra .kiry 9 x 9 i1T v-^ . mirx . Akmmk mirx --22(( 2 a 0 ++ 7 c 22 ccos o s 22^ )•—' V} ^t V} ^ ' y4tm"igcos — — cos - kiry 7-, - 2 ( 2 a 0 + 7 c c o s 2 'y>> !) — } t } Akmmk cos — — cos —67 - , *=1 t=1 *=1 m=l m=l

Mx = - —E Y,Y, Mx

^

^-^ °

=

—rb-

EE

"I

9 7 T

H2

^

OO OO

IT I

~~r

OC' OO

E-

,

2 \ 7T'7 " * V ' V * ^A max. iri «:7ri/ kiry km os cos cos2y>) fccoo s mirx ( 2Qa 0— - 7 75 s s2
T

64

Chapter 3. Table 3.1: l =b 5/= M° M?.2 u>° 1.237 4.534 1.479 1.117 4.112 1.399 10.600 9.320 5.430

Statics

46

w° 6.673 6.260 6.170

In the work [70] the constructional mechanics force method for the rod's system was used to calculate the reticulated plate's bending with y> = 7r/4 for the uniformly distributed transversal load Z = q. The rods' torsional rigidity was not considered. The calculation's results of non-dimensional values of rods' bending moments M°2 = —a _3,_1 A/,* 2 and deflection w° = a'^q^EJiW SJltu for fo: the plate's central points (the first line) with two different relationships of its sides are given in Table 3.1. In both instances the rods' lattice was very sparse (/ = 4\/2a). Nevertheless, the solution of these problems according to the stated formulae were valid (second line in the Table) which is seen from the percentage discrepancies (third line). It should be noted that as the number of support points of the rods' lattice on one of the plate's sides is increased from five (/ = 6) to six (5/ = 46) the calculation results of the two comparable methods become closer (the support points also include the plates' corners).

3.2.3

A plate with more than two families of rods

This section investigates the transversal bending of plates consisting of three or four families of rods. 3.2.3a. The number of rods' families equal to three. Let us examine a rectangular plate having the lattice shown in Fig. 1.5 where the third rod's family is absent (the rods' fourth family will be called the third). Assume that the rods' cross sections of the first two families are identical. Note that for the particular rods' lattice a = 2ca3. The calculation model's constitutive equation for this plate is derived from (1.30) if we fix coefficients /3tJ in the formulae for linear bending and torsion moments equal to fa Sii fa #22 fa 812 fa 331 fa 8*1

4 2 2 = + hI3 + 2C*. 2Cs ), -(2/c == --(2Ic ( 22/c" /c4 + 2Csicc'), 222 2 == -2s(Is -2s{Is + + Cc ), \s(Is is +Uc ) , 2 22 2 2 c-2s (Ic l-C), C), = c --(I-C), (I - C), C), = -2s>.s c\l 22 2 == /r sin sin 2y> 2i^ '+ +++2Cc 2Cc cos 2