Framework methods for orthotropic plates

Framework methods for orthotropic plates

0045-7949186 $3.00 + .oO 0 1986 Pergamon Press Ltd. Compurers & Srnr~rrrr~s Vol. 22. No. 2, PP. 99-l 13. 1986 Printed in Great Britain. FRAMEWORK De...

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0045-7949186 $3.00 + .oO 0 1986 Pergamon Press Ltd.

Compurers & Srnr~rrrr~s Vol. 22. No. 2, PP. 99-l 13. 1986 Printed in Great Britain.

FRAMEWORK Department

METHODS

FOR ORTHOTROPIC

PLATES

R. HUSSEIN University of Evansville, Evansville, IN 47714, U.S.A.

of Civil Engineering,

and M. MORSI Riyadh, Saudi Arabia (Received

14 June

1984)

Abstract-Equivalent frameworks are presented to simulate orthotropic plates in extension or flexure. The structural properties of the framework models are obtained such that the actual and equivalent systems behave in the same manner. The real values of axial, flexural and torsional rigidities are considered in deriving formulas for these properties. There are no restrictions imposed on the values of Poisson’s ratios. The stress distributions in orthotropic plates are calculated from the nodal displacements obtained. The models developed are applicable to a vast number of isotropic or orthotropic plate problems. The examples used to elaborate on these models are solved on a mini computer. Micro computers and programmable calculators could also be used. The finite element method is compared with the models presented and excellent agreements are found.

that the stiffnesses of the actual and substitute structures should be the same. The models developed have no restrictions on Poisson’s ratio effect. Nevertheless, those models are applicable only for plates having constant axial and flexure rigidities and made of isotropic materials. If the problem being solved is, for example, wood wall girder, reinforced concrete girder, steel deck plate or corrugated plate, then the models available cannot be used to find a solution. This paper presents two different framework models representing rectangular elements of an orthotropic plate under extension or flexure. Based on the analogies that exist between the behavior of plate elements and the substitute structures, the properties of the elements in the latter ones are derived. In the first model, which is applicable to a plate in extension, a rectangular cell is assembled from four rigidly connected beams and two truss elements along the diagonals. The second system is assembled from six grid elements, rigidly connected. In the present work, the chord elements are not necessarily made of the same material as for the side or diagonal members; the plate also may have different axial and flexural rigidities in the principal directions; and Poisson’s ratio effect is taken into account. As a consequence, orthotropic plates can be solved now using micro computers or even pocket calculators without a need for a special purpose computer program.

NOTATION cross-sectional area of a beam or strut; flexural rigidity of a plate; torsional rigidity of a plate;

Young’s modulus; shear modulus; torsional rigidity of a grid element; shear flow or torque intensity; plate element thickness; moment of inertia of a beam; moment intensity; shear force; axial rigidity of a plate; shear rigidity of a plate; an element in stiffness matrix; displacement components along the X- and Y-axes, respectively; plate deflection; coordinate axes; displacement; element side lengths; diagonal length of equivalent system; Poisson’s ratio; rotations; I, 2, 3, . . , subscripts denote different directions as etc. shown in Figs. 1 and 2; c, d subscripts denote chord and diagonal, respectively; s subscripts denote side member; {D} nodal displacement vector; {F} nodal loads vector; [S] I2 x 12 structural stiffness matrix INTRODUCTION The analysis

of plates in extension or flexure has been presented in the literature by Yettram and Husain [4, 51. Two different framework models were derived to represent a rectangular element of a plate in extension and in flexure. Their first model was composed of two chord plane beams rigidly connected to two similar side beams. Two struts having axial rigidity only were used between the four corners. In the second model, four grid elements were connected rigidly with another two elements along the diagonals. The properties of strut, beam and grid elements were derived from the requirement

EXTENSIONOF A RECTANGULARELEMENT OF ORTHOTROPIC

AN

PLATE

Consider now a plate element subject in turn to stress flows TX, T, and H as shown in Figs. l(a), l(e) and l(g). The displacements which resulted Si (i = 1, 2, 3 and 4) are:

s,+, x

99

R. HUSSEINand M. MCIRSI

(2)

in which A. kh = the side lengths of the plate element;

(3)

(4)

S, = the axial rigidity of the plate element in the X-direction: = E,h where h is the plate thickness: S, = same as S, but in the Y-direction:

(5)

= E,h

Y

L

(a)

-2

X

258, \I

n

u

fi

------

-8

-:_

z; -l -I

IT-I

-L--B---m

(cl

Fig. 1. Orthotropic

plate element in extension and its equivalent framework model

Framework methods for orthotropic

S xy

= =

’ &&&Ad

Ex, E, = Young’s modulus of the plate material in X- and Y-directions, respectively; Gxy

=

&Ad + r’E,E,A,A,

the shear rigidity of the plate elements; Gxyh

shear modulus X-Y plane;

of the plate material in

ORTHOTROPIC

+ k’E,&A,Ad’

FOR

In a similar way, when the framework model is subjected to a force k Ty applied in the Y-direction, the displacements resulted are obtained as:

r’E,A, + k’EdAd + E,&A,Ad + k’E,EdA,Ad’

’ r’E,E,A,A, 1 & = -k’h=T 2 y

EdAd ’ r3E,E,AcA,

+ E,EdAcAd

+ k’E,EdA,Ad’

(15)

F

=

(8)

The equilibrium

in which

displacements in X- and tions, respectively.

The equilibrium are

equations

+ ;Fd)

= khH.

(17)

side and diagonal

kX, rX = the length of the chord and diagonal members, respectively; =

(16)



r2h

equation of these forces is:

2(Q.

F,, F,, Fd = forces in chord, members;

force, Fd, in the

2EdAdho

d

.,=~($+),

(14)

When the model is subjected to the forces shown in Fig. l(h), the shear force, QS, of the side members is

The axial tension or compressive diagonal members is:

87

(13)

PLATES IN EXTENSION

The substitute structure for the orthotropic plate element is shown in Fig. l(b). The chord beams have flexural and axial rigidities equal to E,Z, and E,A,, respectively. These properties for the side members are E,Z, and E,A,. The diagonals have no flexural rigidity and their axial rigidity is equal to &Ad. The behavior of the framework model is investigated next when subjected in turn at its nodes to the resultant forces on the plate element. Consider, for example, the model in Fig. l(d). The forces in the chord, side and diagonal elements are:

86,

(12)

I& = ;kh’T,

vyxr vxy = Poisson’s ratios of the plate material.

EXTENSION OF THE PLANE-FRAMEWORK

101

plates

Y-direc-

of the framework model

Equations yield:

(15) and (16) in conjunctuib

&IO= t r3k2h4H

6r’E,Z,

with eqn (17)

1 . + k2h2EdAd

The equilibrium of the moments acting on any of the four corners of the framed system results in: E,.Z,. = kE,Z,.

F,ckFd=;hTx, r F, - ;Fd=

(19)

(9) As a consequence, 0.

(10)

1

eqn (18) may be written as: 1

6,(, = - r3k3h4H . 4 6r’E,.Z,. + k3h2EdA,, Substitution yields:

(18)

(20)

of eqns (6)-(g) in eqns (9) and (10) Equation (20) may also be obtained by studying the equilibrium of the forces in the vertical direction.

86 = ;kh’T,

’ E,EdA,Ad 1 S7 = -k2X2T 2 x

r’E,A, + EdAd + r’E,E,A,A, + k3E,EdAsAd’

(11)

PROPERTIES OF THE FRAMEWORK

MEMBERS

The displacements of the plate element, Fig. l(a), and its substitute frame, Fig. l(b), must be the same for keeping the same behavior in both systems.

IO2

R. HUSSEINand M. M~KSI

Thus

= the flexural rigidities of the orthotropic

D,

plate per unit length in the X-direction; 6, = 86,

(21)

62 = 87,

(22)

83 = 6s,

(23)

84 = 69.

(24)

65 = 610.

(25)

Substitution of eqns (1 l), (12), (13), (14) and (18) in eqns (21)-(25), and solving for the unknowns E,A,, E,Z,, E,A,, E,I, and EdAd results in: E,.A,.

= ; &Xl1 : A

E,&

= S, z

vuxf . x> >1

k2-v,,y 1 _

v,, u,.,

= E,Z where I is the moment of inertia of the plate per unit length; = same as D, but in the Y-direction;

D,

= E,.I. The angle of twist, 05, of the plate element when subjected to a uniform torque of intensity His given by (j 5

kAH 2 D,,’

(36)

(26)

in which (27)

1

D,,y = the torsional

rigidity of the orthotropic

plate; EdAd

E,I,

r’h v,y = S, 2k 1 - v,\- v,; S,,

= bk2h’

S, vyx 1 - vxy vyx > ’

-

( E,I,

S,,

= hkA’

-

i

(28)

Sx vxy

1 - VXYvyx > .

In eqns (26)-(30) the following relations sidered Exv,,

= Eyvun

or

Sxv,,, = S?v,,.

= G,J.

(29) FLEXUREOF THE GRID-FRAMEWORK MODEL (30)

are con-

(31)

It is worth mentioning here that, in the particular cases when the plate and the framework systems are made of the same isotropic material Ic = 1 or Poisson’s ratio equals f, the present development reduces to the results available in the literature 11, 3, 4, 51.

The plate element shown in Fig. 2(a) is simulated by six grid members rigidly connected at their ends as shown in Fig. 2(b). The chord beams are of length kh ; have flexural and torsion rigidities equal to E, I, and GJ, , respectively. These properties for the side beams are A, E,I, and G,J,, The diagonals are of length rh and have flexural rigidity ErlId with no torsion rigidity. The structural stiffness matrix of the grid-framework model is assembled from the stiffness matrices of its members as (37)

1Fl = ISI {Dl. in which

FLEXUREOFARECTANGULARELEMENTOFAN ORTHOTROPICPLATE When the orthotropic plate element shown in Fig. 2(a) is subjected in turn to a uniform moment M, and M,, Figs. 2(b) and 2(c), the resulting rotations are :

IF.1 = the nodal loads vector; 1Ml.r MI,. P, Mzx Mz,. Qz M3, M3,, Q3 M4.y = Mdv Qd}‘, where Mix and Mi_y are the applied nodal moments about X- and Y- axes, respectively: Qj is the nodal vertical force in the Z-direction (i = a node number = 1, 2. 3 and 4);

0, = kAM,,lD,,

(32)

e2 = vyx A M,lD,,

(33)

{D) =

03 = AM,tD,,

(34)

=

04 = v,,khMxlD,,

(35)

in which M,, M, = unidirectional moment about X- and Yaxes, respectively; 8; = the rotation resulted in the plate element (i = 1, 2, 3 and 4);

the nodal displacement

vector;

{0,, 81, WI 82x 02y M’2 83x 03J M’3 bx @4? %I’, where Ri, and 8iy are the rotations resulted about the X- and Y- axes, respectively; u’i is the lateral deflection in the Z-direction of node i (i = 1, 2, 3 and 4):

[Sl = the 12 x 12 structural stiffness matrix, the elements of which are given in Appendix

I.

Similar to the present analysis of plates in extension. the grid-framework

is subjected

in turn to re-

103

Framework methods for orthotropic plates

KX

tl

(a)

0

\I

(b)

*I

MY

62 J

(d)

(cl

Fig. 2. Orthotropic plate element in flexure and its equivalent grid-framework modei.

suitant moments XM, and khM, as shown in Figs. 2(d) and 2(f). In the first loading case shown in Fig. 2(d), the load and displacement vectors are given by

M,US11 + s14 - sn % = (S22 - &s + s2s - 5211)(511 - SllO) - (S12 - SlId(S21

{E)= $h {OM, 00 -My OOMy 00 -My O}=,

" =

-Myh(S12 (s22 -

-

s25

SlIOf

+

s28

-

(SI2

-

-

Sd + -

(40)

St4

-

S17’

-

s17'

S210)

- S111)

sZ,d(sl, SIld(S21

+ -

s14

S210)

1781

\--,

(0)

=

tie7

060@7

-ego

-87

060

-0,

(41)

-e60}T,

(39) Substituting eqns (38) and (39) in eqn (37) and sofving the first two equations in 86 and 07 yieIds:

in which Sij = an element in the structuraf stiffness matrix [S](i > J).

104

R.

HUSSEIN and M. MORSI

When the grid-framework is loaded as shown in Fig. 2(f), the load and displacement vectors are

knowns E,I,, obtain:

{F} = -~kA{M,OOM,OO

E

I

s

-M,OO {D} = --~{~8e!?oes

-M,OO}‘,

E,.I,. =

09 0 -0s

-09 o}T.

(43)

The rotations 0s and O9are obtained in this case as: kAM.z(S22 es

=

(sll

+

-

-

s14

-

s,7

S211)

-

(Sl2

S25 + S28 -

s,lO)(s22 -

=

s

(42)

--ego -0x

E,I,.,

-

EdId =

EdId, G,J,

D, A (k’ 2k (1 -

and G,J, one could

- u,,) v,, u,.,)’

(55)

D, A (1 - k’v,r,)

(56)

2 (1 - II.,. u,,) D, v,,, r3A 2k (1 -

(57)

v,, v~,)’

S21d -

S,,,)(S,,

s25 -

+

(58)

s28'

S210)

(44) -kAM, ey

=

(Sl,

+

-

s14

-

S211)

-

(Sz, -

SllO)(S22 (SIZ

-

(59)

S210) -

s25

+

SIII)(SZI

S28

-

In eqns (55)-(59) the following relation ered

.

S210)

is consid-

(45) D, v,,

The behavior of the grid-framework is also studied when subjected to nodal torques as shown in Fig. 2(h). The nodal force and displacement vectors in this case are {F} = -fh{H

-kHO

It should be mentioned here that the solutions available in the literature, for the particular cases of isotropic material with yXv = vyX = 4 or zero and when k = 1, can be obtained from eqns (55)-(59).

-H

-kH

0 H kH 0 -H

kH O}T,

(46) NUMERICAL

{D} = -

-e,,ff-w4}T.

-e,,ooo?o

i 000

(47) Equations (46) and (47) in conjunction with the first and third equations of eqn (37) yield: elo =

AH/2 - S14 - S,,o + S,l,lk + (~,,2/~312)(~310 -S3,0

+

’ -

S3lJk)

(49)

10.

SSlZ

PROPERTIES

(48)

S3,Jke

w‘j =

= D,. u,,.

OF THE GRID-FRAMEWORK

MEMBERS

The behavior of the element of an orthotropic plate and the grid-framework model will be the same if the corresponding displacements are identical. Thus 8, =

e6r

(50)

e2

=

e7.

(51)

e3

=

es,

(52)

e4

=

e9,

(53)

es

=

elo.

(54)

OF ORTHOTROPHIC

PLATES

IN EXTENSION

An orthotropic plate in extension or flexure may be idealized into equivalent framework or grid systems which behave in the same manner under the applied loads. The model developed may be solved by standard computer programs or for a certain extent by pocket calculators. The distributions of the stresses in the plate can be calculated from the displacements obtained from the equivalent systems. In the following, two examples are solved with the present development. The first example is for a square wall girder under a uniform inplane stress applied at its upper edge. The second example deals with the same wall but with an isotropic material. Example I

Consider a square wall girder of side length equal to 200 in. (508 cm) and subjected to an inplane uniform stress of 1 psi (0.007 MN/m’) at its upper edge. The width of the support at both sides of the wall is equal to & of the total span. The wall thickness is taken to be equal to one unit and its material properties are E,

By substituting eqns (32)-(36); (40), (41), (44), (45) and (48) in eqns (50)-(54) and solving for the un-

EVALUATION

= 4 x lo6 psi;

G,, = 1.8 x lo6 psi;

E, = 5

x

v,, = 0.25;

lo6 psi; u,, = 0.20.

The wall is analyzed first by the equivalent planeframework method and secondly by the finite element method. The actual values of the material properties are used.

Framework methods for orthotropic

The plane-framework idealization of the wall girder is shown in Fig. 3. The applied inplane stress at the upper edge is replaced by equivalent static concentrated nodal loads. The properties of the framework members are calculated from eqns (26)(30) as: E,A,

(Ul E* ux = 2h( 1 - VWVrx)[

= 31.579 x 1061b

UxYY(zIl -

u2

-

v3

= 42.105 x lo6 lb

+v

x (0.189 x lo6 kN),

(u1+

u2

-

YX

x (0.134 x lo6 kN), x

TV

=

lo6 lb in.2

x (14.864 x lo6 kN cm2), E,Z, = 510.367

x

Gxy (UI 2h [

-

-

u2

-

u3

according

-

u4)

k +

u3

-

u3

+

v4)

7J2 -

u4)

,

u3

+

z14)

(61)

~4)

1

+(~l+~2-v3-~4)

(60)

1

k ,

lo6 lb in.2

(62)

in which

x (14.864 x lo6 kN cm2). By assuming that E, = E, = Ed = lo6 psi (6914.286 MN/m’), it follows: A, = 31.579 in2 (204.376 cm2), A, = 42.105 in.2 (272.500 cm2), Ad = 29.773 in.2 (192.688 cm2), I, = I, = 510.367 in.4 (21242.946 cm4). By following the direct stiffness method, the wall girder is solved on the mini computer PerkinElmer. The stress components u,, uy and T,~ are

Ui, Vi = the displacement in the X- and Y-directions, respectively, at node i (i = 1, 2, 3 and 4). The numbers of nodes are shown in Fig. l(a). The stress distributions at different cross-sections along the wall girder are shown in Figs. 4-6. The finite element method is also used to analyze the wall girder. The constant strain triangle element is used and the wall is treated as a plain stress problem. The finite element mesh for half the structure is shown in Fig. 7. The displacement and stress components at different locations in the wall are

‘2P

Fig. 3. Plane-framework

@2

k

EdAd = 29.773 x lo6 lb

E,Z, = 510.367

+

EY (VI uy = 2h( 1 - Vxyuy*) [

(0.142 x lo6 kN),

E,A,

calculated from the nodal displacements to the following equations:

+

x

105

plates

idealization of a wall girder.

106

R. HUSSEINand M. MORSI

I :/ i

1.0

.-

I I I SECTION AT X =

.la

SECTION

=

.3a

IO-’

Psi

stress

Fig. 4. The distribution of the normal stress (T, in the orthotropic

AT

unit

X

Y

100.0

-60.0 - 20.0 20.0 60.0 100.0 100.0 100.0 100.0 100.0 100.0 60.0 20.0 - 20.0 - 60.0 - 100.0

100.0 100.0 100.0 100.0 80.0 60.0 40.0 20.0 0.0 0.0 0.0 0.0 0.0 0.0

Horizontal displacement (IO3 in.) Framework 0.773 0.727 2.261 3.431 3.857 2.854 1.906 I.111 0.497 0.0 0.0 0.0 0.0 0.0 0.0

Finite element 0.961 0.394 2.170 3.480 4.080 3.070 2.110 1.270 0.585 0.0 0.0 0.0 0.0 0.0 0.0

is

wall girder.

Table 1. Comparison of the framework and finite element solutions for the displacement orthotropic wall girder Distance from origin (in.)

x

components

of the

Vertical displacement (IO’ in.) Framework - 1.833 - 3.296 - 4.369 - 5.239 - 6.043 - 6.032 - 6.036 - 6.053 -6.071 - 6.079 -- 5.314 - 3.627 -4.112 -3.x75 3.808

Finite element

- 1.840 ~ 3.320 4.390 ~ 5.260 - 6.060 - 6.040 - 6.030 - 6.030 - 6.040 - 6.050 -5.280 - 4.600 ~ 4.090 - 3.850 3.760

Framework methods for orthotropic

-

section section section

.-.

0.

I.0

E.0

3.0

at at at

4.0

107

plates

x = .+l a x = .3a x= .6a

5.0

NORMAL

6.0

STRESS ( 10-Z

7.0

9.0

8.0

IO.

C,I,

PSI)

Fig. 5. The distribution of the normal stress cry in the orthotropic

A,

compared with the results of the framework model in Tables 1 and 2. It is seen that both results are in very good agreement.

wall girder.

= A,

Ad = 7.54

=

8.0 in.2, in.2,

I, = I, = 88.89 in.4. Example 2

Should the wall girder in the previous example and the members of the framework model be made of an isotropic material with uXY= uYX= 0.25, E, = 4 x lo6 psi the properties of the system in Fig. 3 become

The results in this case based on the present model are shown in Figs. 8-12. Tables 3 and 4 compare the solutions of the finite element and framework methods. It is seen that excellent agreement exists between the two models.

Table 2. Comparison of the framework and finite element solutions for the stress components wall girder Distance from origin (in.)

Framework method (IO3 psi)

in the orthotropic

Finite element method (IO’ psi)

X

Y

0;

I+

7,

c-7

u>

10.0 30.0 10.0 30.0 10.0 30.0 10.0 30.0

0.0 0.0 40.0 40.0 80.0 80.0 -40.0 -40.0

- 256.50 -217.51 - 152.96 - 128.97 - 101.45 - 82.84 -310.00 - 297.00

-734.13 - 825.63 -908.24 - 943.49 - 984.74 -991.34 -411.08 - 577.73

63.58 172.66 31.49 82.02 11.63 28.39 96.72 286.86

- 272.50 -231.50 - 159.60 - 132.70 - 96.73 - 78.25 - 307.00 -310.25

- 730.00 -831.00 - 909.00 - 944.50 - 982.75 - 990.50 -410.25 - 565 .OO

T.Y> 67.28 182.50 31.325 81.53 10.33 26.75 95.08 290.00

R.

‘*

HUSSEIN and M. MORSE

section section section

at x = .+l a at x = .5 a at x =; .S a

A----!& ‘\

t

II

-..-

SHEAR

8TREBS (Id2

Fig. 6.

7~~

PSf)

The distribution of the shear stress 7,, in the orthotropic wall girder.

EQUIVALENT

NUMERICAL EVALUATION

LOAD

FOR ORTHOTROPIC

PLATES

IN FLEXURE

Consider the square orthotropic clamped plate shown in Fig. 13 subjected to a uniform distributed load of 1.O psi. The principal directions of natural orthotropy are taken parallel to the plate edges. The flexural rigidities of this plate are taken as: D, = 10” lb in.; L), = 0.5

x

IO’ lb in.,

D,,, = 0.512 x 10’ lb in.; v,, = u~.~= 0.30. The plate is divided into g~de-framework as shown in Fig. 13. The properties of the members in the equivalent system are calculated from eqns (55)(59) as E,f,

= 4.581

x

IO6 lbs in.‘,

,!?‘,I, = 11.126 x IO6 lb in.‘. E,Jtl= 5.553 GJ, Fig.

7.Finite

element

idealization

for

half the wall girder.

x

lo6 lb in.‘.

= G,J,.= 8.861x IO’ Ibin.“.

By following

the direct

stiffness

method,

the

Framework methods for orthotropic

109

plates

SECTION AT x =.Za SECTION AT x 8.40

---

U X I& IN.)

DISPLACEMENT,

Fig. 8. The distribution of the displacement

in X-direction of the isotropic wall girder.

.1

P

.c

;

- ---

SECTION AT x =.6a

-

SECTION AT x = a

-4

-.4

-.f

-14

0.0

2.

4.

DISPLACEMENT,

Fig. 9. The distribution of the displacement

6.

8.

V X I& I NJ

in Y-direction of the isotropic wall girder.

R. HUSSEIN and M. M~RSI

-4.

-2.

.O

NORMAL

Fig. 10. The distribution

nodal deflections

STRESS

of the normal

are found as:

2. aj, x 16

stress

4.

( PSI )

cry in the isotropic

The analytical (2) is

wall girder.

solution for the maximum deflection

WI = Wj = W7 = Wp = 0.057 in., Ws = 0.1563 in

W2 = W8 = 0.095 in.. W, = W6 = 0.089 in.,

The discrepancy between the present model and the analytical solution is 4%. The accuracy of the so-

W5 = 0.1501 in.

-.e -1.0 .O

2. NORMAL

Fig.

Il. The distribution

6.

4. STRESS

of the normal

5,,

8.

10.

f 10-2 Psi)

stress

v’, in the isotropic

wall girder.

Framework methods for orthotropic

111

plates

-.6

-.6

00

2.

SHEAR

4.

STRESS

6.

T,~

6.

10.

12.

(4 x 10-2 PSI 1

Fig. 12. The distribution of the shear stress rxy in the isotropic wall girder.

Table 3. Comparison of the framework and finite element solutions for the displacement isotropic wall girder Horizontal displacement (10’ in.)

Distance from origin (in.) X

Y

Framework

Finite element

100.0 100.0 100.0 100.0 100.0 80.0 60.0 40.0 20.0 0.0 0.0 0.0 0.0 0.0 0.0

- 60.0 - 20.0 20.0 60.0 100.0 100.0 100.0 100.0 100.0 100.0 60.0 20.0 - 20.0 -60.0 - 100.0

0.120 0.156 0.349 0.486 0.545 0.420 0.297 0.186 0.088 0.0 0.0 0.0 0.0 0.0 0.0

0.158 0.125 0.342 0.492 0.568 0.443 0.318 0.203 0.098 0.0 0.0 0.0 0.0 0.0 0.0

components

Vertical displacement (lo2 in.) Framework

Finite element

-2.29 - 4.08 -5.37 -6.44 - 7.44 -7.42 - 7.42 -7.42 -7.43 -7.43 - 6.47 -5.59 -4.92 -4.61 -4.52

-2.27 -4.06 -5.35 -6.41 -7.41 -7.38 - 7.360 -7.35 -7.35 - 7.35 - 6.39 -5.51 -4.85 -4.52 -4.40

Table 4. Comparison of the framework and finite element solutions for the stress components girder Distance from origin (in.)

of the

Framework method

in the isotropic wall

Finite element method

X

Y

0, (psi)

cry (psi)

T,, (psi)

ox (psi)

0,. (usi)

T*,.(usi)

10.0 30.0 10.0 30.0 10.0 30.0 10.0 30.0

0.0 0.0 40.0 40.0 80.0 80.0 -40.0 -40.0

-271.94 - 230.07 - 147.98 - 124.23 -82.23 - 66.87 - 350.69 - 330.28

- 762.99 - 846.02 - 924.50 - 954.06 - 988.56 -994.22 -442.17 - 604.07

61.29 164.49 27.21 71.72 8.80 22.88 100.79 294.95

- 289.00 - 243.50 154.48 - 127.58 - 76.30 -61.45 -351.25 - 349.50

- 762.75 - 854.00 -926.50 - 955.75 - 987.25 - 993.25 - 444.00 - 596.25

64.53 173.15 26.90 69.68 7.86 20.27 98.78 298.75

K. HUSSEINand M. MORN

100 IN.

\1

0

NODE

Fig. 13. An equivalent

lution can be considerably finer grid.

NUMBER

grid-framework

improved

by using a

CONCLUSIONS

Orthotropic plates in extension or flexure are represented by equivalent frameworks that simulate the actual behavior when subjected to statistically equivalent loads. The real values of axial, flexural and torsional rigidities are considered in calculating the properties of the members in the models developed. There are no restrictions imposed on the values of Poisson’s ratios. The stress distributions in the plates are calculated from the nodal displacements. The present models are applicable to a vast number of plate problems which may be solved now using micro computers or programmable tors. The examples used to elaborate

\I

model for an orthotropic

4. A. L. Yettram and M. H. Husain, Plane-framework methods for plates in extension. J. Engng Mech. Div. A.S.C.E. 92 (EMI), Proc. Paper 4695 (February 1966). 5. A. L. Yettram and M. H. Husain, Grid framework method for plates in flexure. J. Engng Me&. Div. A.S.C.E. 91 (EM3), Proc. Paper 4361 (June 1965).

APPENDIXI

The elements of the structural stiffness matrix [S] in eqn (37) The elements of the 12 x I2 symmetric structural stiffness matrix [S] in eqn (37) are given by

s II

=

s,, = s

.s+$ =

-_) /.T

SZl = 2

= -

-2s

=

-

,,)?

,’

J

=

-

2h

s,j = -

kX S,:,

=

-

f

=

-

Z S,,2/k

54

=

-

2 s*4

= 4 kh S,,,?

-

2 S&k

zx -

{ASoc

= $ A Slzz

=

-

f h S,,?

A. Hrennikoff, Solutions of problems of elasticity by the framework method. .I. Appl. Mech. ASME 63,

= + h S8h = :

(1941). S. G.

= - 3 A S,,3 = $ A Svh = f /iA’s,,?

Lekhnitskii, Anistropic Plates. Gordon & Breach, New York (1968). D. McHenry, A lattice analysis for the soluiion of stress problems. J. Inst. Civ. Engrs 21, (1943).

=

s*, = s, ,,,

$khSgq

= i kh S,h =

q

s,, , = 2 J 2k s,o, = -

=

REFERENCES

+ 4 E.,I,

4E‘lId

+

calcula-

on these models are solved with the present framework systems; the finite element and analytical methods. There is very good agreement between the results. Not only was there no need for a special purpose computer program for solving these problems, but also it was difficult to handle them by standard numerical or mathematical procedures.

clamped plate.

4kE,,I,/ ----T--

rh



kh S,,,?

Framework methods for orthotropic

113

plates

= -Q k2h2 Se, = -_

h4

E,I,

=-

7) ,

+

=-

Qk2X2 S,29 2E,I, kh

s‘j, = S,07 = - y, s,,

=

s*O4

zz-

4 A s9, = - 4 A s,3 = f h S’24

=

d A S73

=-

4 A2

s93

=

-

;

A2

S,26

=

_

f

),

s82

=

s,lS

s33

=

S66

’ --

=

=

GsJd A

s99

=

EJc +

s,06

s1212

E

k3

=- 2 Ed, A s22

=

SSS

=



s51 SE8

=

sill,

s32

=

-

&5

=

s98

=

SllS

=

-

r3 -

= f kh S119

4

s

Se,

=

s8l

=

s42

=

s43

=

s83

=

s,14

=

=

EdId r3

s72

=

SIOS

=

Sl25

=

S116

=

s117

s127

=

s,OS

=

slO9

=

0,

s92

s,2,,

k&Id

6 (Wc =27+s52

=

I

I

5

=

=

‘tk’E&i + 7 r )

+ G,J,



in which ’

kA X52 = 4 kA S53

=

’ kA s128

Sij = an element located on row i and column j of the matrix [S] where i > j.