Applied Mathematics and Mechanics (English Edition, Vol.7, No.l, Jan. 1986)
APPROXIMATE
Published by SUT, Shanghai, China
SOLUTION FOR BENDING OF RECTANGULAR PLATES Kantorovich-Galerkin's Method
WangLei(ZE
i~)
LiJia-bao(~)
(Hunan University, Changsha)
(Received Jan. 3, 1981 Communicated by Chien Wei-zang) Abstract This paper derives the cubic spline beam function from the generalized beam differential equation and obtains the solution of the discontinuous polynomial under concentrated loads, concentrated moment and uniform distributed by using delta function. By means of Kantorovich method of the partial differential equation of elastic plates which is transformed by the generalized function ( 6 function and ~r function), whether concentrated load, concentratedmoment, uniform distributed load or smail-square Ioad can be shown as the discontinuous polynomial deformed curve in the x-direction and the ydirection. We change the partial differential equation into the ordinary equation by using Kantorovich method and then obtain a good approximate solution by using Glerkin's method. In this paper there 'are more calculation examples involving elastic plates with various boundary-conditions, various loads and various section plates, and the classical differential problems such as cantilever plates are shown.
I. I n t r o d u c t i o n
Kantorovich. Kralove presented Kantorovich approximate variational approach to deal with the functional variation of multivariable functions. In our country Chien, W. Z. has noticed very much Kantorovich-method. In his work ~ the theory is described in detail and rich calculation examples are contained. The calculation examples of elastic plates include such boundary conditions as four clamped edges, three clamped edges and one edge free, or three edges free and another clamped. But the loads are limited to uniform distributed load and the beam function simply satisfies such conditions as two clamped ends or one end clamped and the other free. Hence, the application range is smaller. This paper derives the cubic spline function starting with the generalized beam differential equation and it is possible to obtain the solution of the discontinuous polynomial under uniform distributed load, abrupt load, concentrated load and concentrated bending moment by using 6 function and ~ function. This spline function is used as the beam function in the directions of x and y to deal with the functional variation problems of two varibles functions, which changes a partial differential equation into an ordinary differential equation, next, the approximate solution can be obtained by using Galerkin's method. In order to deal with the term of load of a partial differential equation, this paper derives the intergral formulas of product of ~ function and ~ function times any continuous functionJ(x), therefore, the
87
88
Wang Lei and Li Jia-bao
approximate solution of bending of rectangular plates with any boundary condition under uniform distributed load, small-square load, concentrated load, concentrated moment and line load can be obtained. Hu Hal-chart t21 has introduced Kantorovich method and defined the ordinary differential equation system of the finite strip to solve the plane-stress problems by means of Kantorovich method. His work is new as compared with the solution of Y. K. Cheung's finite strip method. Because of the importance of Kantorovich method in mechanics, scholars of the world have been paying attention to it's development and various practice applications. So it is significant for this paper to make further development. The relatively rich calculation examples in this paper can be taken as the supplement to Kantorovich method. II. D e r i v a t i o n
of Spline Function
Let us start with the generalized beam differential equation. N-I
E l - d4w ~ d x-
~
(2.1)
P,6(x--x,)
i-I
the difference of this statement from ordinary beams lies in the term of right-hand side. Where P~ is the concentrated load exerted on the beam (for i= 1, 2 ..... N-l), 6(x--x~) is delta function. Integrating (2.1) once gives: N -
E I d3w dx 3 =c3+
1
-~
5~ P , c s ( x - - x , )
(2.2)
i-I
statement (2.2) has a definite mechanical meaning, that is, the shear chart has sudden change. cr(:c--x~) is sigma function or step function. Integrating (2.1) twice gives: N-I
F.I d2w.l_ --c:~+cj + V' p,(x--x~) t
(2 3)
Integrating (2.1) three times gives:
EI~
= c t 4-c~.x + c 3 x ~
(9
t)
Integrating (2.1) four times gives: 9 N..1
x~
Llw.=Co+C~X+C~--+c3 "
" 2!
x~ . \,
--~ 3!
~i - I
i~
(x-x~) 3T
(2.5)
For the integral of the fight hand side of eq. (2.1), we introduce the concept of "discontinuous polynomial." For any positive integral number k defines. j (x--x,)'
( for xL~O)
( x - x , ) ~_ 0 For k = O, we define:
( for x < 0 )
Approximate Solution for Bending of Rectangular Plates 0
(X--Xl)O=cr(x--xl) =
f
l
(
for x<.v,)
(
for x = x , )
89
t ( for x'>.'.~ ) For simplisity we only give the figure for xi = xl shown as Fig. 1
(x-xO~_
]tO
(x-x,);
Y.
i [~/2 Xl
t----x,_--__4
" I
Fig. I
Fig. 3
Fig. 2
obviously the discontinuous polynomial is a step function. For k = I define .f
0
( for
x~xt)
(x-x~)l+
(x--xt) 1 ( for x~>xl) shown as Fig. 2 For k = 2 define
0
(.v-x~)-' = {
(x--xl) ~
( for x<~xl) ( for x~>xl)
shown as Fig. 3 For k = 3 define 0
(
(x-xl)~ =
k
(x--x,)S
(
for
x~xl)
(
for
x>xl)
Where, " + " is called discontinuous sign and above four statements are known as halfdiscontinuous polynomial or interrupt polynomial. The figures of (x-- x~ ) h. (for k = 0, 1, 2, 3) are shown as Fig. 1, 2, 3, 4. This kind of discontinuous polynomial is the important component of the spline function, which can be represented by the unit step function, namely tO
(x-x,) ?
-~r
~.___:/e . . . . . ~_.
t/e__ -4
9 .{
Fig. 4
(x-xO;
Fig. 5
= x~(x-xl)~
For concentrated moment, the generalized differential equation is to be written:
90
Wang Lei and Li Jia-bao N-1
d'w -- \ ~ M~d'(x--x~) E I m dx 4
(2.6)
z..,.a i-I
the first integration gives: N
EI
daw
d-3-U-- = c ~ +
- I
X_~ M,c3(x-x,)
(2.7)
i-I
the second integration gives: N- 1
d:"w
t2I~=c:+cj+
\,..~ M , a ( x - - x , )
(2.8)
i-I
where a(x--x~) is just (x--x~) ~ it has definite mechanical meaning, that is, the moment chart has a sudden change. the third integration of eq. (2.6) gives:
El
=c~+c~x+c3---~+
~
M,(x--a',)'
(2.9)
INI
and the fourth integration of eq. (2.6) gives:
lilW=Co + q X + C 2 ~x c23
,r ' " --~C+ \~ ' M, 9
(x-x,):
,-t
21
(2.10)
Compare (2.6) with (2.7), we observe that the right term in statement (2.7) decreases one power. Let us take an example to illustrate the generalized beam differential equation. Try to derive the deform curve of beam. Suppose a simply supported beam (Fig. 5) is supported by a concentrated bending moment M at the central point. From statement (2.5), using the boundary conditions x=0
w=O
dZw dx z = 0 ;
x=l
dZw
dx z = 0
w=0
W e can find four integral constants as follows
co=O
cz=O
c1=
-•4 Mol
e3--
Mo
I
thus obtain the deflection curve:
E I w - - - - - ~ ( lZx--4x 3)
( O~ x ~ l )
It is the same as the result given in ~ Strench of Materials ~ (P. 640) by Timoshenko. Here is another case. A beam with the left end simply supported and the right end clamped beam a concentrated bending moment M and a concentrated load P. The generalized beam differential equation is written as
Approximate Solution for Bending of Rectangular Plates
%
1
~ _ _ ..t/3_____f . . . . t/3 _ .. _f._. Fig. 6
91
-ff=l
t/ ~- ----d ~ L ~ U T ' T T ~ ~ _ _ ~ Fig. 7
El- d'Wdx, = M (
x - 3/--) +P3. ( x - 2
J
_ A - relative displacement
~g. 8
I)
(2.11)
After integration yields: ~?
x~ + _ _ ~ ( (
I =
P
.3
For beam with two free ends the statement of deflection is to be derived by means of fifth power spline function as well as by ordinary means: Coasidering the following polynomial: x6/l 0 - 3xS/l~+ 5x'/214 which has satisfied the conditions that both the bending moment and the shear are equal to zero at two ends. In the calculation of elastic plates, the beam function must satisfy the condition that the vertical relative displacement is one i.e. A = 1 and the rotating angle is zero at the central point shown as Fig. 7. Hence, the beam function is written in this case: X e
X 5
5 ~4
~=F- - 3V+-~F -+ A ~ + B with alto
1 I
Xs
--d~ = T t 6 ~ - -
X4 ~3 1 5 F + 1o~- + - / / )
For x = 1]2, w = 1 (relative displacement at central point) For x = i/2, dw/dx = 0, A = - i/2, B = 75/64. For such condition as one end simply supported and the other free, shown as Fig. 8 the beam function is written as:
w=
xs l~
10 3
x' I'
10 + 3
-~:~ x 13 + l
which satisfies the conditions that both the deflection and the bending moment at the left end are zero and the bending moment and shear at the right end are zero. In addition the condition that the displacement at the right end is two is satisfied. Above two sorts of static-beam are unequilibrum. Hence, it is impossible to find the absolute displacement. Thus, we have to call the displacement to equal any constant which is called the relative displacement. However, the concept of relative displaoement is different from that in the structure mechanics. In the calculation of bending plates, we need only to learn the beam function (which is relative displac~aent) in the x-direction and the y-direction, and then we can obtain the absolute displacement by means of the variational approach. Integrals of generalized function $ ( x - - x , ) and integrating by Parts.
92
Wang Lei and Li Jia-bao Integrating
(3(x--.r~)
once we obtain:
[a(x--x,)dx= ( x - x , )
0
Integrating once again and not considering the integral constant we obtain:
i~(x--x,)dxdx=(x--x,)} thus, integrating 5( .~..... -, ) following equations
If 0~.r~,(a
k + 1 times and ignoring all integral constants, we obtain the
, we can prove:
[~,~(-v--.':l)dx=~r(:,--xl)
9
o
= c r ( a ) - - a ( 0 ) =1
0
0
The formula of integral by parts is (~
f(x)g'(x)dx=f(x)
g(x)
b
--
(b g(x)f'(.v)dx
(2 13)
Suppose fix) to be a smooth enough continuous function and g(x) a generalized function, above equation becomes:
bf(x)~V(x--x,)dx=fc~(x--x,) a
--
cr(x--xz)f'(.v)dx
a
According to the properties of
cr
function, above equation becomes:
/'(x)dx=/(b)--/(x)
f(x)a'(x--x,)dx=/(b)a(b--x,)-r
Xl
X 1
=f(b) --~-f(b) - - f ( X l ) ] = f ( x , ) Hence, we can define the generalized derivative which satisfies the following integral.
$ ( x - - x I)
of
a ( x - - x t)
(o<.,
as such a function
(,..i,)
where, fix) is continuous at place x = xlThis statement has definite mechanical meaning. Ordinary use of Concentrated Load. The fight hand side of equation (2.14) can be understood as the work done by unit concentrated load. The generalized function 8(x--x1) can be taken derivative again and function ~'(x--xt ) is also a generatized function. From statement (2.14) we obtain:
~bS' (x--x,) f ( xld.,:= -- f' ( x) tl
(2.15)
Approximate Solution for Bending of Rectangular Plates
,.Cx~<~b ,
similarly proved if then
(3 ~ ( x - - x ~ ) f ( x ) d x = ( - - 1 ) " .
93
?~(x--.v~)f " ( x ) d x = ( - - 1 ) ~ f
a
.
'~ (x,)
(2.16)
a
where f " ( x ) is continuous at x=x~, otherwise, the right hand side of statement (2.16) has no meaning. Thus, we can define the function ~, "'(x--x~ ) by means of ~quation (2.16). In this way 8<"(x--xL) is also the generalized function. What follows is a theorem. Assume J ( x - - x , ) to be the step function defined in the interval [a, b] with discontinuity points x~ (i= 1, 2 .... n) with ( z = x o ~ x , . C x . : ~ . . . . - ~ x , = b , In addition, assumeflx) to be any continuous function in [a, b], the derivative of which f ' ( x ) can be integrated, then, the following formula \'
Iba(x--x~)f'(x)da=crf
~
Cl
[cr(x~)]f(x)
(2 17) o
t-I
is true. is jumpmg value offlx,) at place x = x , Where Ecr(x~)]=cy(x,+ 0 ) - , r ( x , - 0 ) Proof: Since in [a, b] or(x-x~ ) can be represented as the linear combination of the unit step function i.e. n
,:~(x-x,)= ~
Taking derivative to
~7(x--x~)
Er
and noting that n
\:~ Ef(.v,) -!,~(_v-.v, )
,T' ( x - x, ) =,~( , : - x, ) =
(2.18)
Thus, using integrating by parts we obtain:
Ec c ( x - - x , ) f ' ( x ) d x = c r f )"-- f ~ T ' ( x - - x , ) f ( x ,) a
Since
.v) ,t.v= ~
"
tx
(2.19)
a
[,~(.x', ) ] I bJ(_,.)<'~Ix-- v d .v
9 r
n
= E
[c~(:<,)~/(x,)
(2.2o)
i-I
Substitution of above statement into equation (2.19) yields formula (2.17). The theorem is proved. Formula (2. l 5) has definite mechanical meaning, that is, the right hand side of formula (2. I 5) is the integral of concentrated bending moment times the generalized function <'5'( x-- x, } timesflx) and the right-hand side of formula (2.17) can be understood as the work done by unit bending moment. The integral of the step function times the continuous functionflx) in formula (2.17) is often used in where the section of plates varies suddenly.
94
Wang Lei and Li Jia-bao
III. Kantorovich-Galerkin's Approximate Solution Kantorovich-Kralove presented Kantorovich approximate variational approach to deal with the variational problems of multiple variables function. Themethodistochoosethefunctionsequence ,/'~ ( x~ x .....x, , ) ( f o r k = l , 2 .... m) which satisfies the boundary conditions and write the approximate solution of variation problems as follows: tr~
II(x~.x:,.
.x,)=
'> ' i I . , ( x , ) , / . ( x ,
x., ... x ,
,)
-I
where Ak(x,) is undetermined functions of the x,. Substituting I I ( x , . x : - . . . x , ) into the functional, the origional functional II(tt) of function 11 (-~,.-L:,....x,) becomes the functional of functions d~ (x,,) st:. ( x , ) .... , ~t,, ( x , ) which is written ~s Now, the question becomes to choose such ..t,..4 . . . . . , A m ( x , ) that let I[*(..I, ..4........-1,,~ reach the extreme value. The procedure of finding the extreme value of 17 ~ (A ~..... A,,) is to find Euler equations of .-t~ ( x,, ). .-t: ( x,, ) . . . . . . 4 ~,C.v, ) and the concerned boundary conditions by means of variation. These Euler equations are generally ordinary differential equations. In this way, the origional partial differential equation containing multiple variables becomes the ordinary equation containing single variable, which is the essentiality of Kantorovich method. Setting m~r and taking lira?t, we can obtain the exact solution under certain conditions. Let m be a limited number, then we obtain the approximate solution by such method. The functional of rectangul',r elastic plates with clamped or simply supported edges is:
[1=~~[" {1]~(x.!,)(\:~,o)':--qw
~ ,/,'dT./
(?.1)
JO.O
The functional of rectangular plates with free edges is: -,, .-t, 1 , t 0 ~~ O~0] :: ox- +~.71 1 / = 9Io . o T l ~ ( ~ " " ,/~" ,.,.,,-.---~ - 2(IOZ~L)
a
I O"~' O:zc
,
~_-di,.~-
,~,# - -
b
where D(x. y) is variable bending stiffness which is the function of x and y and q is a uniform distributed load. Concentrated load, concentrated bending moment or small square lead is represented by the generalized function. I) Concentrated load (Fig. 9).
I
"j"l'-~,,~(x--,,),~(,~--71)d.~d,/ o
(:~ .;")
d
2) Small square load (Fig. 10): -
-
ix:i~zqwd.\'d., n
(3.4)
~71
3) Uniform distributed load on strip region (Fig. II):
(3.5)
Approximate Solution for Bending of Rectangular Plates
95
Lx" .
i
t-
i .... r
I/,
T i-
__
L,.
'
i
Ii
,1
-
12
.......
I ~9
o'~ x
I
_ _
]- ....
---i
Fig. 9
Fig. 10
Fig. I1
4) Line load (Fig. 12):
--
Ia I b l~Zd~3(:/'-rl)d'"d/: ,,
(3.6)
e
Example 1 Rectangular plate with three simply supported edges
and one clamped edge under uniform distributed load. Try to find the deflection of the middle point. (See Fig. 13)
,/x
1~
/"
L--- . . . . . .
Fig. 12
Fig. 13
Fig.
I~
. . . . . .
14
Solution: Suggest using approximate beam function '
. Y
Substituting (3.7) into (3.8) yields
/7=
"1.--2- 0.1/3015bu":(.'c)--2:-0.:',12:~,;71 ,~l")~"(:) 7 "91 "b:' u (.r):
1- - 0 . 15bqu(.r)~ d.r
After variation calculus substituting simply supported boundary condition w = 0 and ~' = 0 as x = 0 and x = a into Euler equation
yields:
d]'"
d
:h~
d.v
.:91" 0~'
d" da'"
dI'" du"
--
f}
96
Wang Lei and Li Jia-bao
O.03015buW--2• Simplified gives: uv--22-74314~=u"+2;tS.,~0597~,u--l.97512
/~ = 0
(3.9)
Now we solve the ordinary differential equation by Galerkin's method. Suppose X4
-V:~
X
where ct, c, ..... is undetermined coefficient. Taking the first term cl and substituting u(x) into Galerkin's equation gives: I [ (u'--22"74344"~zu"+238"80597b'u--4 99 7 5 1 2 ~ - - ) ( - ~4 - z - - C"xs - + ~)dx=
0
After integration arranging yields: cl
4.8 (~1'~ +11.0t59.1
_ +11.75059
=0.995024" D
If a = b, then cl = 0.0360561qadD Thus the deflection equation of square plates is:
w=cl
~
a:~
a l\
a --5 a3 + 3
At p l a ~ x = a/2, y = a]2. the deflection of middle point is w=0.0360361
aq~
•
5 i 1-~-•
qa 4 D
The error of the result is very small compared with the classical solution. As we choose the beem function with one clamped end and the other simply supported or two simply supported ends, we will encounter the follo~ving types of integrals.
which need not calculating and can be found out in the attached Table 1 at the end of this paper. E x * m p l e l The conditions are the same as above example but uniform distributed load is changed into line uniform distributed load. (See Fig. 14) Solution: The functional equation is the same as (3.8) but the term of load should be changed as
flI2
PS(.x--2).VYdxd~,=P
IIYdyX
~
(
10)
Still using equation (3.8) but must changing the beam function so that it satisfies the load and boundary conditions i.e. 1,> .(
,~3 \
,,(.~) = t o T - - 4 7 ) ( c , consider the following ~ni,::~ra1~:
+c,.x+ c;x~ 4- ... )
(3.11)
Approximate Solution for Bending of Rectangular Plates .
Q
97
(|
Il
2 [T.-V
u:'18 , "0
".\'dx--
I ,8
0
-
a
and use the value of the term of load at place x = 1/2 (x_ Px.3u
.,d 1 aS- ) x = 2 = P x
1
2
Thus integrating after variation calculus yields: I).030156•
b @ a~ ; 2: (~.:',12857I• 1.8
~
+7.2•
u
P .... t ) . 1 ? ; 6 - ~ -
For square plate, a = b. et-0.01706p,t '/D. The deflection of the middle point is I ) a :~
l
1 ~ a :~
~1.f)1766--73- • 1 x - ~ - = 0 . 0 0 , 1 4 1 5 ~
]Example II A rectangular plate with two adjacent edges simply supported and another two adjacent edges clamped bears of uniform distributed load. Try to find the deflection of the middle point. (See Fig. 15) Solution" Suppose _ !/a ,/z .~
Similarly to Example I we obtain Euler equation, then solve the ordinary differ~,ntial equation by means of Galerkin equation. Suppose u ( x ) = ( 2 x' u' - - 5 - -x:' ~ ; - + 3 - ~"~" - ) ( c , + c . , . < + c ~ . ~ + ...)
(3.13)
Taking the first term cj and substituting the above statement into Galerkin's equation gives:
i
a
(1
'~/
N4
Looking up the attached Table 1 we obtain the values of integrals, the above statement becomes
[
o
c~ 4 . 8 x 0 . 1 5 - ~ - - 2 2 . 7 1 3 4 . I •
(--d.3428~,7.~)
+238.80597 • 0.030i~,~~a I - - 4 . 9 7 5 } . ~o fqi • For square plate, a = b. then
cs
=O.033619qa'/D ,~,~ ,
qu 4
c0=0.0,~,,01~--D - •
+!~-
0.15. = 0
The deflection of the middle point i.-, 1
|
:.: -i-=
() t ) l l . ) l ~ , L 2 . _ Zq}~
.
-
]Example 4 An elastic plate whh two opposite edges simply supported ~.nd one ,:,~gc ~,an.~ and another free. (See Fig. 16)Given Poisso~ r~tio !~=f~. " Solution
Suppose
98
Wang
Lei and
Li Jia-bao
Ix
.
.
.
i'
.
liii~i
x
i
a
---"
,
.
z-
@
f
!
~,
(z
!
t--
b. . . .
1
]II
17t
]
D~! t + (ru)/s b F i g t5
Fig. 17
Fig. 16
:.'-~(")
'~ - - -
i-~;
(,3.11)
:--
6
fhen (,l
,)
I
I[---_--_-_/) I 182
----.":i,'-,)
)
' o
l~
,~
qb
21
"~
1,1 "
"-.>beeq--=---'-l,qO(l--,~)
l,
/ i
I . ~ 0 1 u" 7 " , ' - - 6 0 ( 6 - - 7 n ) - ~ u" i -7,- -61u r b - - 2 1 ,0 "[
,'~i/.-=_ _ liP' :,, I
'
+~[1)
] I"
t)
#~()uclx qb\
;1
,,
Oul"
182
13( 1 2 - - 1 Z / t ) : , ' - -
3 ?) -
9
tl'"
, ,,)
-----0
E u l e r e q u a t i o n in this c a s e is
u
u"
:51) 1 - ~ 6 0 ( 6 - - 7 H )
---~4-
36.! 36.1 9 urb_21 9
qb
(3.15)
-7 y-=6
T h e e n d c o n d i t i o n o f u is
1 ,''>
....
/,
. -~ l:';[',~t
~/~
.-7
at p l a c e .c = 0 a n d x -:.:,.z
Assume two opposite edges simply supported, then
(ZY-
~(.v)= Substituting
tL---- 0. :;
2
-'
-
(c:+c.:x+...)
:~ +
, we o b t a i n O
Sl~l',~() (l lqoo56~--~-- 60 ".. Z 9'.-.(--(I
c
9
"
X6 t ---x ,)
'
-
"
b I
(l
qab
-. fl 9___
t ,,,I--=}=21 -.
183676) 1 "
(,.
9~
1)
F o r s q u a r e plate, a ~ b, t h e n ci = O.O126285qa'/D T h e d i s p l a c e m e n t o f the m i d d l e p o i n t o f the free e d g e is :,,
~L,
o12628;,
-
i)
""
16
x'3~0
o
011g?,!) - - -
J)
b
A p p r o x i m a t e Solution for Bending of Rectangular Plates
99
E x a m p l e 5 Calculation of plates of variable section 9 Since D(x, y) is variable in the case of variable section, D(x. y) must be integrated along with the displacement function. An elastic plate with four edges simply supported. Suppose that its thickness varies in straight line with y and the load close so. (See Fig. 17) The displacement function is written as ~o=u(x)
(
'P
-
2
+%-) "'
'/'
(3.16)
In substitution the following integrals are needed. A s s u m e " }~" -- [
,ttl4
,I/:~
.U
-.-t---Or-- 2 ~
+T)
then
f7 (,+,r),.,,,,= I j-
~
u
".\d~/=--': 1So62 l
J'i il
l
(1'+f7 )
i,l,,t= ().!)/1
Substituting each value of integrals and performing variation yields. I I ?b 0"2'992067>bu~--o~'18562-;-u" ' > 1 0 ; ~ .b--~---u' =!1.'.1----1), Calling X4
X '
_~_)
u t .v ) = (--Tz-- -- ~'~ ~:< k
( c, + c .'," = c, x . . . . . . I
and taking the undetermined constant c/, we obtain b
c~ 0 2 9 9 2 0 0 3 x 1 8 - - ' ~
+21.6•
18562x I - - l l
18.,~-;T 6 !
t~ 1 qab 0 . 0 , 1 9 2 0 5 6 - - ~ - - t - - f ) . 1X Do
F o r square plate, a = b, then c / = O.0505542qaqD. T h e displacement of the middle point is t~c-~0
0D055.12x
R 16
q:~: I,,
/J
t', / i
1t30
W a n g Lei a n d Li Jia-bao If the load is an u n i f o r m d i s t r i b u t e d load, then
T h e term o f load is O.04qab/Do. F o r square plate, a = h, then c~ = 0.0112343qa'/D,~ The d i s p l a c e m e n t of the middle p o i n t is q,a "~ 5 l~-,, " " - - 1 6
:"-, = t J . O l 1 2 : ; l : ; .
3 x . . .16 ...
0.0010971
(la 4 --Do-
6 A cantilever plate with one clamped edge and three free edges u n d e r u n i f o r m
Example
distributed load. G i v e n Poisson radio / z = O. ;* . T r y to find the deflcction o f the middle point. This is a difficult problem. It is c o n v e n i e n t for us to use the m i n i m u m potential energy principle to solve it. Since the displacement boundary' c o n d i t i o n simply needs satisfying a n d the stress b o u n d a r y c o n d i t i o n is not necessary to be satisfied 9 The f u n c t i o n a l can be written as: w = C X Y
.-2r
'
:.
,',--! r
[
,lc.\'~,,e.xJ.,,=-,
The beam f u n c t i o n of beams with two frcc cp,!s can be chosen a s .,(f' [ r
';
9
--
,:
*.
3.17)
_
d!
with: .\ r .
1:',65a,
0
i
J. 1 2 8 5 - - ~
.
"
t2'
•
[ \'-Jx=o , 177., 9
.
~l
t n .\J/-
.
"
L/
,L t 9 c,~] I'- , ,.'" i,~
S u b s t i t u t i n g these values of integrals into the tc~wtional e q u a t i o n a n d p e r f o r m i n g v a r i a t i o n with resp,:ct to c vie!e~: ' ~ ~
' <
,.
i!--
.
9
For s c m . , : e , < ~
.,t=/.tl
2
.
,
. : = . ' ::
t,
'..
-
i
,
I,Jt):)i
'
.,
]
1 -I
l')~!i
.
.
[
k.
.
.
~ ,
o ~
-
ti:~53
1
'j:'.
: . ,',.~,1I ab -- [ 06-!7;21 ..
.
' f)
j~
thenc~=0.032649qa*/D
-~ !,. di,pb~.cc~ acr~t of the middle p o i n t of the free edge of the square pla:c b; q,:J ~
:,,..=~.i;:;'~,~[.q'< ! ,. ::; ] } : > . . . . . : , a c t solution 0.11~C.,,
q d
= 0.()~t703
]--TT-
L,. the error it 17 8% . By m,zar: " ,,~ K a n t o r o v i c h -
Approximate Solution for Bending of Rectangular Plates
~
,~
~
t~-
101
~
co
o
0
2., oo
[
--
~
0
m O4
t.--
~ '~
o
~
!
~
~ c~
~
c~
I
~
~:~
~
~
~
~,
~ ,---
I
"
t..J-"~,
c,.o oo. 9
~
,.....-.
~
o
oo
.J. ,=, LO " ~
40
~
~-
4-
c.o :"t
4-
I
~l~. ~1 ~
4-
I
V/
I
4-
I
o0 o.I
E
-i 0
-I
E#
~. s
U
~
~,
o
U
?
102
Wang Lei and Li Jia-bao'
Galerkin's method, the error is dropped under 5.0%. E x a m p l e 7 An elastic plate with three simply supproted edges and one free edge.Under uniform distributed load q. Poisson ratio u = (1.3 . Try to find the deflection of the middle point. The beam function of beams with one simple supported end and the other free can be taken as: ,\--
.\" tl"
I
0 ,) ')
_v ~
+
d 4
I (l
_v :
+ - -.v
,) ,,
tl '
(I
The concerned five types of integrals can be all looked up in the attached Table 1. Substituting each value of integrals and performing variation we obtain: tl
c O.Olf12056"<:;.80.q6 - -b' -
i '~'..,ql~ :;'.'(--I/
185676)>,'0.9811321--a l__t,v
+ I . 8,< 1 0503-I" . aO-~7-> . =-1.I>~ . 0 185711:< 1:3 '2~6 l~a b - =o.8333:o3xO.2i~-~b For square plates, a = b, c=O.0210976qa*/D. The deflection of the middle point of the free edge of the square plates is: qd ~ rt,,. = O. 0 2 1 0 9 7 6 - - 7 7 •
7)
qd ~ ;< '2 = O. 013186 [-~-
C o m p a r e d w i t h the exact s o l u t i o n O.O1286qa'/D. the e r r o r is o n l y 2.47%.
References
[ 1 ] Chien, W. Z., Variational Methods and Finite Elements, Science Press (1980). (in Chinese) [ 2 ] Hu, H. C., Variational Principles in Elasticity and Their Applications, Science Press ( 1981 ). (in Chinese) [ 3 ] Beijing Mechanics Institute, The Science Academy of China: The Buckling Stability of Sandwich and its Vibration, Science Press (1977). (in Chinese) [ 4 ] Timoshenko, S. P., Theory of Plates and Shells, McGraw-Hill Book Company, New York (1960). [ 5 ] Timoshen k o. S. P., Strength of Materials, Science Press (1978. 3). (in Chinese) [ 6 ] Cheung Y. K., Finite Strip Method of Structure Analysis (1976). [ 7 ] Xu, Z. L., Elastic Theory, People's Education Press (1979.8) (in Chinese) [ 8 ] Hu Hai-Chang, Suggestions on the application of finite element method with examples of plane stress problems in elasticity, Acta Mechanica Solida Sinica (1981. 1). [ 9 ] Wang. Lei, Trial function and weighted residuals method, Journal ofHunan UniversiO' (1981. 1). [10] Wang. Lei, On middle-thick plates and try-functions, Engineering Mechanics, (1984. 1). [11] Wang. Lei. Analysis of boundary integrate method for middle tliick plates, Computational Structural ~Iechanics and Applications (1985. 2).
