Approximate solution for bending of rectangular plates Kantorovich-Galerkin's method

Applied Mathematics and Mechanics (English Edition, Vol.7, No.l, Jan. 1986)

Published by SUT, Shanghai, China

A P P R O X I M A T E S O L U T I O N FOR B E N D I N G OF R E C T A N G U L A R P L A T E S

Kantorovich-Galerkin 's M e t h o d

WangLei(ZE i~) L i J i a - b a o ( ~ )

(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 y-

direction. 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. Introduct ion

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 o f S p l i n e F u n c t i o n

Let us start with the generalized beam differential equation.

N - I

E l - d4w d x ~ - ~ P , 6 ( x - - x , ) (2.1)

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 - 1

E I d3w -~ d x 3 = c 3 + 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:

E I ~ = c t 4-c~.x +c3x ~ (9 t)

Integrating (2.1) four times gives:

�9 N . . 1 x ~ x ~ . \ , ( x - x ~ )

L l w . = C o + C ~ X + C ~ - - + c 3 - - ~ i~ " " 2! 3! ~ 3T i - I

(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 x < 0 )

For k = O, we define:

Approximate Solution for Bending of Rectangular Plates 89

0 ( for x<.v , )

l (X--Xl)O=cr(x--xl) = f ( for x = x , )

t ( for x'>.'.~ )

For simplisity we only give the figure for xi = xl shown as Fig. 1

(x-xO~_ ]tO

i [~/2

(x-x,);

Y . X l " I

t----x,_--__4

Fig. I 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)

Fig. 3

shown as Fig. 2

For k = 2 define

0 ( for x<~xl) (.v-x~)-' = {

(x--xl) ~ ( for x~>xl)

shown as Fig. 3

For k = 3 define

0 ( for x~xl ) (

( x - x l ) ~ = k (x--x,)S ( for x>xl)

Where, " + " is called discontinuous sign and above four statements are known as half-

discontinuous 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,) ?

- ~.___:/e . . . . . ~_. t/e__ -4 -~r �9 .{

Fig. 4 Fig. 5

( x - x O ; = x~(x-xl)~

For concentrated moment, the generalized differential equation is to be written:


E I m N - 1

d'w -- \~ M~d'(x--x~) d x 4 z..,.a i - I

(2.6)

the first integration gives:

N - I

E I daw X_~ d-3-U-- = c ~ + M,c3(x-x,)

i - I

(2.7)

the second integration gives: N - 1

d:"w t 2 I ~ = c : + c j + \,..~ M , a ( x - - x , ) 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:

( 2 . 8 )

E l =c~+c~x+c3- - -~+ ~ M,(x- -a ' , ) ' INI

(2.9)

and the fourth integration of eq. (2.6) gives:

x 2 ,r ' " ( x - x , ) : lilW=Co + q X + C 2 ~ c 3 \ ' - -~C+ ~ M, (2.10) �9 ,-t 21

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

dZw dZw x = 0 w=O dx z =0 ; x = l w = 0 dx z = 0

We can find four integral constants as follows

-•4 Mo co=O cz=O c1= Mol e3-- 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


% 1 ~ _ _ ..t/3_____f . . . . t/3 _ .. _f._.

Fig. 6

- f f = l J

t / ~- ----d ~ L ~ U T ' T T ~ ~ _ _ _ - ~ A - relative displacement

Fig. 7 ~g. 8

El- d'Wdx, = M ( x - 3/-- ) +P3. ( x - 2 I ) ( 2 . 1 1 )

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 - ~ + B ~=F- 3V+-~ F -+ A

with

a l to 1 I X s X 4 ~3 --d~ = T t 6 ~ - - 1 5 F + 1 o~- + - / / )

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:

x s 10 x ' 10 -~:~ x w = l~ 3 I' + 3 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.


thus, integrating 5( .~ ..... -, ) following equations

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 , ) d x d x = ( x - - x , ) }

k + 1 times and ignoring all integral constants, we obtain the

If 0~.r~,(a , we can prove:

[~ o

,~(-v--.':l)dx=~r(:,--xl) = c r ( a ) - - a ( 0 ) =1 �9 0 0

The formula of integral by parts is

(~ b (b f ( x ) g ' ( x ) d x = f ( x ) g(x) -- g ( x ) f ' ( . v ) d x (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 , ) -- cr(x--xz)f ' ( .v)dx a a

According to the properties of cr function, above equation becomes:

f ( x ) a ' ( x - - x , ) d x = / ( b ) a ( b - - x , ) - - /'(x)dx=/(b)--/(x) r X l X 1

=f (b ) --~- f (b) - - f (Xl) ] = f ( x , )

Hence, we can define the generalized derivative $ ( x - - x I ) of a ( x - - x t) as such a function

which satisfies the following integral.

(o<.,<b) (,..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) (2.15) t l


similarly proved if ,.Cx~<~b ,

then

(3 ~ ( x - - x ~ ) f ( x ) d x = ( - - 1 ) " ?~(x--.v~)f " ( x ) d x = ( - - 1 ) ~ f '~ (x , ) (2.16) . a . 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

I b a ( x - - x ~ ) f ' ( x ) d a = c r f (2 17) \ ' [cr(x~)]f(x) ~ o

C l t - I

is true.

Where

Proof: Since in [a, b]

function i.e.

Ecr(x~)]=cy(x,+ 0 ) - , r ( x , - 0 ) is jumpmg value offlx,) at place x=x , or(x-x~ ) can be represented as the linear combination of the unit step

n

, : ~ ( x - x , ) = ~ Er

Taking derivative to ~7(x--x~) and noting that

,T' ( x - x, ) =,~( , : - x, ) =

Thus, using integrating by parts we obtain:

n

\:~ Ef(.v,) -!,~(_v-.v, )

Since

(2.18)

E )" f c c ( x - - x , ) f ' ( x ) d x = c r f -- ~ T ' ( x - - x , ) f ( x tx (2.19) ) a , a

" I b .v) ,t.v= ~ [,~(.x', ) ] J(_,.)<'~I x-- 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.


III. Kantorovich-Galerkin's Approximate Solution

Kantorovich-Kralove presented Kantorovich approximate variational approach to deal with

the variational problems of multiple variables function.

T h e m e t h o d i s t o c h o o s e t h e f u n c t i o n s e q u e n c e ,/'~ ( 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~

I I ( 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 I I ( t t ) 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 . . . . . ,Am(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 ) J O . O

The functional of rectangular plates with free edges is: -,, .-t, 1 , t 0 ~~ O~0] :: I O"~' O:zc

1 / = Io o T l ~ ( ~ " , /~" , , - . ---~ - , - - �9 . " ,.,. o x - + ~ . 7 1 2 ( I - ~_-di,.~- ,~,#

OZ~L) a 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):

3)

ix: i~z - - qwd.\'d., n ~71

Uniform distributed load on strip region (Fig. II):

( 3 . 4 )

(3.5)


t-

12 - I i

.

, 1

I x o'~

_ _ ]- . . . .

Lx"

i . . . . T i- i r I / , _ _ L , . ' I i

. . . . . . . ~9 - - - i

Fig. 9 Fig. 10 Fig. I1

4) Line load (Fig. 12):

I a I b - - , , e l~Zd~3(:/'-rl)d'"d/: (3.6)

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 - - - . . . . . . I~ . . . . . .

Fig. 12 Fig. 13 Fig. 1 4

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")~"(:)

"91 1 15bqu(.r)~ d.r 7 "b:' u ( . r ) : - - 0 .

After variation calculus substituting simply supported boundary condition w = 0 and ~' = 0 as x = 0

and x = a into Euler equation

d]'" d .:91" d" dI'" - - f }

:h~ d.v 0~' da'" du"

yields:


O . 0 3 0 1 5 b u W - - 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 9 7 5 1 2 ~ - - ) ( - ~ ~4 "xs ~ ) d x = �9 - z - - C - + 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 = c l ~ a: ~ a l \ a --5 a3 + 3

At p l a~ x = a/2, y = a]2. the deflection of middle point is

aq~ 5 i q a 4 w=0.0360361 • 1-~-• 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

f l I 2 P S ( . x - - 2 ) . V Y d x d ~ , = P I I Y d y X ~ ( 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 , +c, .x+ c;x~ 4 - ... ) (3 .11)

consider the following ~ni,::~ra 1~:


. Q I l

"0

and use the value of the term of load at place x = 1/2

( |

u:'18 , 2 [T.-V ".\ 'dx-- I ,8 - 0 a

( x _ . , d P x . 3 u 1 aS- ) x = 2

2

Thus integrating after variation calculus yields:

= P x 1

b @ ~ u P I).030156• a~ ; 2: (~.:',12857I• 1.8 + 7 . 2 • . . . . 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

( x ' x:' "~" ) (c ,+c . , .<+c~ .~+ ...) (3 .13) u ( x ) = 2 u' - - 5 - - ~ ; - + 3 - ~ -

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.~) ~-

~a I o q +238.80597 • 0.030i~,~- - - 4 .975 } .~ f i • 0.15. = 0

For square plate, a = b. then cs =O.033619qa' /D The deflection of the middle point i.-,

q ~ ,~,~ , q u 4 1 | () t ) l l . ) l ~ , L 2 . _ Z }

c0=0.0,~,,01~-- D - • :.: - i -= . -

]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

9 8 W a n g L e i a n d L i J i a - b a o

f h e n

. . . . i' ~, !

!

t - - b . . . . 1

x

a , ---" z-

Fig t5 Fig. 16

: . ' - ~ ( " ) '~ - - - i - ~ ; 6 : - -

, ) ( , l I ) l~ ,~ "~ 1,1 "

I[---_--_-_/) I ' o " - . > b e e q - - = - - - ' - l , q O ( l - - , ~ )

I x

l,

l i i i ~ i

17t ] I I ] D~! t + (ru)/s

b

Fig. 17

i .

@

f

(z

( ,3 .11)

1 8 2 qb - - - - . " : i , ' - - 2 1 / i

, )

,'~i/.-=_ _ l i P ' I :,, I "[

u u" 7, 61 #~()uc lx . ~ 0 1 " 7 " , ' - - 6 0 ( 6 - - 7 n ) - ~ i - - - u r b - - 2 1 qb\ ,0 ' ; 1

+ ~ [ 1 ) ] I" t) ,, 3 ?) - 13( 1 2 - - 1 Z / t ) : , ' - -

Eu le r e q u a t i o n in this case is

u u" 36.! :51) 1 - ~ 6 0 ( 6 - - 7 H ) - - - ~ 4 - 9 36.1 u r b _ 2 1 qb

9 - 7 y - = 6

T h e end c o n d i t i o n o f u is

182 Oul " t l ' " - - - - -0

9 , ,,)

( 3 . 1 5 )

1 ,''> . -~ l:';[',~t ~ / ~ .... /, . - 7

at p lace .c = 0 and x -:.: ,.z

A s s u m e two o p p o s i t e edges s imply s u p p o r t e d , then

(ZY- - ' - ~ ( . v ) = 2 :~ + ( c : + c . : x + . . . )

S u b s t i t u t i n g tL---- 0. :; , we o b t a i n

O c Sl~l',~() (l lqoo56~--~-- 60 ".. Z 9 ' . - . ( - - ( I 183676) 1

�9 " ' - " " ( l b

X6 t b I qab - - - x t , , , I - - = } = 2 1 -. fl 9 _ _ _

,) - . ( , . �9 ~ 1 )

F o r s q u a r e pla te , a ~ b, then 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 edge is

- o :,, ~L , o12628;, i ) "" 16 x ' 3 ~ 0 011g?,!) - - - J)

Approx ima te Solution for Bending of Rectangular Plates 99

E x a m p l e 5 Calcula t ion 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

d isplacement 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 . 1 6 )

In subst i tut ion the following integrals are needed. A s s u m e "

then

}~" - - [ ,ttl 4 ,I/:~ .U -.-t---Or-- 2 ~ + T )

f7 (,+,r),.,,,,= I j- ~ ".\d~/=--': 1So62 l

u

J'i l

( ' f ) 1 + 7 i , l , , t= ().!)/1 il

Substi tut ing each value of integrals and performing variat ion yields.

C a l l i n g

I I ?b 0"2'992067>bu~--o~'18562-;-u" ; ' > 1 0 ~ .b--~---u' =!1.'.1----1),

X 4 X ' _ ~ _ ) u t .v ) = (--Tz-- - - ~'~ ~:< k ( c, + c .'," = c, x . . . . . . I

and taking the undetermined constant c/, we obtain

b c~ 0 2992003x 1 8 - - ' ~ 18562x I - - l l 18.,~-;T 6 !

t~ 1 qab + 2 1 . 6 • 0 .0 ,192056- -~ - - t - - f ) . 1X

Do

For square plate, a = b, then c / = O.0505542qaqD. The displacement of the middle point is

R q:~: t ~ c - ~ 0 0D055.12x 16 I,, /J

t ' , / i

1t30 Wang Lei and Li J ia-bao

If the load is an un i form dis t r ibuted load, then

The term of load is O.04qab/Do. For square plate, a = h, then c~ = 0.0112343qa'/D,~

The displacement of the middle point is

q,a "~ 5 3 (la 4 :"-, = t J . O l 1 2 : ; l : ; . l~-,, " " - - 16 x . . . . . . 16 0 . 0 0 1 0 9 7 1 - -Do-

E x a m p l e 6 A canti lever plate with one clamped edge and three free edges under uni form

distr ibuted load. Given Poisson radio /z= O. ;* . Try to find the deflcction of the middle point .

This is a difficult problem. It is convenient for us to use the m i n i m u m potent ial energy principle

to solve it. Since the displacement boundary ' condi t ion simply needs satisfying and the stress

bounda ry condi t ion is not necessary to be satisfied�9 The funct ional can be writ ten as: w= C X Y

[ . - 2 r ' : . , ' , - - ! ,lc.\ '~,,e.xJ., ,=-, r

The beam funct ion of beams with two frcc cp,!s can be chosen a s

. , ( f '

r ' ; �9 * .

[ - - _

,: d! 3 . 1 7 )

with:

.\ r 1 : ' , 6 5 a , i . J . 1 2 8 5 - - ~ . 0 " t 2 '

[ \ ' -Jx=o , 177., • �9 ~ l . " L /

t . \ J / - t c,~] I'- .'" i,~ �9 , , , L

. n

Subst i tu t ing these values of integrals into the tc~wtional equa t ion and performing var ia t ion

resp,:ct to c vie!e~:

' ~ ~ ' < i ! - - 2 ' . . I , J t ) : ) i . , 1 - I l ' ) ~ ! i . . . . [ k. 1 ti:~53

with

] ' j : ' . ' f )

, . . - , ' o , . . . , ~ - �9 . t, i : . ,',.~,1I ab -- [ 06-!7;21 ~ j~

For s c m . , : e , < ~ . , t = / . t l . := . ' :: thenc~=0.032649qa*/D

-~ !,. di,pb~.cc~ acr~t of the middle point of the free edge of the square pla:c b;

q,:J ~ q d

:,,..=~.i;:;'~,~[.q'< ! ,. ::; ] } = 0.()~t703 ]--TT-

: > . . . . . : , a c t solut ion 0.11~C.,, L,. the error it 17 8% . By m,zar: " ,,~ Kantorov ich-


~ , ~ ~ t ~- ~

o

[ - -

~ m

t .--

~ ~ ~ ~ ' ~ o ~ c~ ~ ,--- I

! c ~ I ~ "

t . . J - " ~ ,

�9 ~ ~ ~:~ ~ ~

,.....-. o ~ ~, ,=,

0 ~ ~ - c.o

o . I

4-

I

o0

oo

~l~. ~1 ~

I V /

:"t

I

0

O4

c,.o oo.

.J.

LO "~

4-

I

4-

?

co

0

2., oo

4-

4-

-I

E

0

E # ~ . s U ~ ~ ,

-i

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:

.\" I 0 _v ~ I ( l _v : .v , \ - - + + - - t l " , ) d 4 ,, ') , ) t l ' (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:

c O.Olf12056"<:; .80.q6 - - - t l

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 ~ 7) qd ~ rt,,. = O. 0210976 - -77 • ;< '2 = O. 013186 [-~-

Compared wi th the exact so lu t ion O.O1286qa'/D. the error is on ly 2.47%.

R e f e r e n c e s

[ 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).

Documents

Approximate solution for bending of rectangular plates Kantorovich-Galerkin's method