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.