基于能量法的近似方法-里茨法

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2023年12月12日发(作者：)

板壳力学平时作业

Approximate Method Based on the Energy

Methods——Ritz Method

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[注]：[参考书目]：刘人怀.《板壳力学》（超星）.机械工业出版社. 1990.10.

第四章第一、二节

Approximate Method Based on the Energy Methods

——Ritz Method

1. Summary

Solution which strictly fits the differential equation and the corresponding boundary

conditions for a deflection curved surface of a plate is the exact solution. Obviously, for a plate

with simple geometry shape, uniform load and simply supported edges, we can get this kind of

exact solution. But for a more complex problem maybe we can not get the exact solution. Here,

the approximate methods or value methods will be the effective way.

Among the commonly used approximate methods, the classical Ritz method is more

commonly used. Ritz method is also called the energy method. The basic train of thought is that

after exactly adapt to the displacement boundary conditions, loosen some qualification for the

differential equation of the deflection curved surface. Thereby we can obtain the approximate

solution. This method discusses plates’ balanced problem with the energy point of view. The core

is the minimum potential energy principle. Under the given external force, among all groups of

displacements which fit the boundary conditions, there’s only one group of displacement can also

adapt to the balance equation, and it can make the addition of strain potential energy and external

force potential energy be minimum.

2. The Strain Energy of Plates

During the mechanics of machinery course we have deduced the expression of potential

energy of the elastic body in the rectangular coordinates. It can be expressed as follows:

U1[xxyyzzxyxyyzyzzxzx]dxdydz (2-1)

2where U is the potential energy of the whole elastic body,

x,y,zand

xy,yz,zx are stress

components,

x,y,zandxy,yz,zx are the corresponding strain components.

According to the basic assumption of thin plates,

z0,yzzx0 (2-2)

then Eq.(2-1) can be simplified as,

U1[xxyyxyxy]dxdydz (2-3)

2the geometric equation and physical equation of plates are

Geometric equation:

- 1 - 2wxz2x2wyz2yxywhere

w is the deflection of the plate.

Physical equation:

 (2-4)

2w2zxyEz2w2wx2221xyEz2w2wy (2-5)

12y2x2Ez2wxy1xySubstituting Eq.(2-4) and Eq.(2-5) in Eq.(2-3), then we have,

2222Ewww322Uz(w)2(1)dxdydz

222(12)xyxyV2h/2222Ewww222(w)2(1)dxdyzdz

2222(1)sxyxyh/2that is

2222D22wwwU(w)2(1)2dxdy (2-6)

22sxyxywhere

EEh32D2/2zdz12(12)

2(12)his called flexural rigidity. And the integral area in Eq.(2-6) is carried on the whole surface of the

plate.

Eq.(2-6) is the strain energy expression of small deflection of thin plates.

Inspect the second term of Eq.(2-6), it can be written as

h/22w2w2w2I2dxdy

2xyxy22232ww3wwwww2dxdy

222xyxxyxyxxy

- 2 - w2ww2wdxdy

2xxyyxxywith Green equation in advanced mathematics

P(x,y)Q(x,y)dxdyy

xQ(x,y)dxP(x,y)dywe can simplify the equation above as

w2ww2wIdxdy

2xxyxywhere the integral is carried on all the edges of the plate.

When all the edges of the plate are fixed, no matter what shape are the edges, we obtain

w0. Thus

I0, Eq.(2-6) can continuously simplified as

xUD22(w)dxdy (2-7)

2sIf there is a rectangular thin plate whose edges are all simply supported, then on the edge

where coordinate

x is a constant, we obtain

2wdx0,20

yalso on the edge where

y is a constant

w0

xthus, we also obtain

I0, the strain energy of the plate can be calculated with Eq.(2-7).

Obviously, when the boundary conditions of a rectangular plate is simply supported or fixed,

Eq.(2-7) expresses its strain energy.

Secondly, for a thin plate is only subjected to a transverse uniform load

q, the potential

energy of the external force can be expressed as

Wqwdxdy (2-8)

Now, suppose the deflection

w as series modality

nwcmwm (2-9)

m1where every function

wm is chosen a expression which fit the deflect surface while also fit the

boundary conditions, and

cn are undetermined coefficients.

Substituting Eq.(2-6) and Eq.(2-7) in the expression of minimum potential energy principle

V(UW)0, we obtain

- 3 - Vwhere

V is the total potential energy.

Uqwmdxdycm0

m1cmnBecause the variation

cm is arbitrary, so from the equation above, we obtain

Uqw1dxdy0c1Uqw2dxdy0c2 (2-10)

Uqwndxdy0cnThis is a system of

n linear equations regarding

c1,c2,,cn. Contact these equations and

solve the question we can determine the value of these coefficients

ci,(n1,2,,n). Then we

obtain the approximate expression of deflection

w. This method of calculating deflection

wbrings us to closer and closer approximations as the number

n of the terms increases, and by

taking

n infinitely large we obtain an exact

o

solution of the problem.

x



rAt more situation, we need the approximate

d

expression in polar coordinates. Naturally, we can

start with the strain energy of the three-dimensional

elastic body in column coordinates. But we’ve

y

dr

deduced the expression in rectangular coordinates, so

use the method of coordinate’s transformation

Fig.2-1 Coordinates transformation.

directly will be more convenient.

The relation of polar coordinates

r, and rectangular coordinates

x,y express as

(Fig.2-1)

r2x2y2,tg1then, we have

y (2-11)

xryrxsin

cos

yrxrxcosysin2

2yrrxrrwith these equations, we obtain

wwrwwsinw

cosxrxxrr

- 4 - wwrwwcosw

sinyryyrrand

2wsinwsinwcoscos

x2rrrr2w2sincos2w

cosr2rr22sinw

rr2sincoswsin22w

222rr2wcoswcoswsinsin

y2rrrr2w2sincos2w

sinr2rr2c2osw

rr2sincoswcos22w

222rr2wsinwcoswcossin

xyrrrr2wcos2sin22wsincosw

sincos

2rrrrrcos2sin2wsincos2w

rr2According to these results, immediately we obtain

2w2w2w1w12ww222

xyrrrr222Rotating the

x axis and

y axis to the direction of tiny element

r and

, to make



equal zero. Then substituting the equation above to Eq.(2-6), replacing the area of the tiny element

dxdy with

rddr, we obtain the expression of strain energy in polar coordinates:

2222D22w1w1w1w1wU(w)2(1)22rdrd

222rrrrrrr(2-12)

- 5 - Now, the potential energy of the external force can be expressed as

Wqwrdrd (2-13)

Then, we obtain the system of

n linear equations regarding the coefficients of the

deflection of a thin plate in polar coordinates

Uqw1rdrd0c1Uqw2rdrd0c2 (2-14)

Uqwnrdrd0cnThinking the problem of bending of a circular plate with symmetric anises, Eq.(2-12) can be

simplified as

d2w21dw2dwd2wUDr2dr (2-15)

22drrdrdrdraccordingly Eq.(2-14) reduce to

U2qw1rdrc1U2qw2rdrc2 (2-16)

U2qwnrdrcnIf all the edges of the circular plate are fixed, at the edges we have

second term of the integration in Eq.(2-15) becomes

2dw21dw2dwd2w1dw0

drdr2dr2ddr2drdrrarbdw0. In that case, the

drat last Eq.(2-15) can be simplified as

d2w21dw2UDr2dr (2-17)

drrdr

3. Application of Ritz Method

Now use Ritz method to calculate the maximum deflection of a rectangular plate that is

subjected to a transverse uniform load

q0 and with its four edges fixed. The length of two sides

of this rectangular plate are

2a and

2b. Establishing the rectangular coordinates with its origin

at the center of the plate (Fig.2-2).

- 6 - a

b

a

o

x

b

y

Fig.2-2 Rectangular plate

with its edges fixed

We can easily know the boundary conditions:

w0x (3-1)

wwhen yb,w0,0ywhen xa,w0,Assuming the deflection function as

wc(x2a2)2(y2b2)2 (3-2)

where

c is a undetermined coefficient. Obviously, the equation above fits all the boundary

conditions.

From Eq.(3-2), we obtain,

2w222224c(3xa)(yb)

2x2w4c(x2a2)2(3y2b2)

2y22222222222w4c(3xa)(yb)(xa)(3yb)

Substituting these equations in Eq.(2-7), taking notice of the symmetric nature of the problem.

We obtain

U32Dc20a2222222222(3xa)(yb)(xa)(3yb)dxdy

0b163844Da5b5c2a4a2b2b4

15757Secondly, by using Eq.(2-10), we have

32768554422Dabcaabb4157574q0

a0

b0(x2a2)2(y2b2)2dxdy0- 7 - that is

32768554422256Dabcaabb4q0a5b5

15757225so the coefficient

c can be determined

7q044224caabb (3-4)

128D7thus, substituting Eq.(3-4) in Eq.(3-3) we obtain the approximate solution

117q04w(x2a2)2(y2b2)2a4a2b2b4 (3-5)

128D7Finally, let’s check the precision of this solution. Take a square plate for example, we have

ab, and the maximum deflection must at the center, use Eq.(3-5), we have

wmaxq0a449q0a4

0.02132304DDq0a4compare this approximate solution with the exact solution

0.0202, the extent of relative

Derror is no more then

5.4%.

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