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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:
U1[xxyyzzxyxyyzyzzxzx]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,zandxy,yz,zx are the corresponding strain components.
According to the basic assumption of thin plates,
z0,yzzx0 (2-2)
then Eq.(2-1) can be simplified as,
U1[xxyyxyxy]dxdydz (2-3)
2the geometric equation and physical equation of plates are
Geometric equation:
- 1 - 2wxz2x2wyz2yxywhere
w is the deflection of the plate.
Physical equation:
(2-4)
2w2zxyEz2w2wx2221xyEz2w2wy (2-5)
12y2x2Ez2wxy1xySubstituting Eq.(2-4) and Eq.(2-5) in Eq.(2-3), then we have,
2222Ewww322Uz(w)2(1)dxdydz
222(12)xyxyV2h/2222Ewww222(w)2(1)dxdyzdz
2222(1)sxyxyh/2that is
2222D22wwwU(w)2(1)2dxdy (2-6)
22sxyxywhere
EEh32D2/2zdz12(12)
2(12)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/22w2w2w2I2dxdy
2xyxy22232ww3wwwww2dxdy
222xyxxyxyxxy
- 2 - w2ww2wdxdy
2xxyyxxywith Green equation in advanced mathematics
P(x,y)Q(x,y)dxdyy
xQ(x,y)dxP(x,y)dywe can simplify the equation above as
w2ww2wIdxdy
2xxyxywhere 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
w0. Thus
I0, Eq.(2-6) can continuously simplified as
xUD22(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
2wdx0,20
yalso on the edge where
y is a constant
w0
xthus, we also obtain
I0, 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
Wqwdxdy (2-8)
Now, suppose the deflection
w as series modality
nwcmwm (2-9)
m1where 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(UW)0, we obtain
- 3 - Vwhere
V is the total potential energy.
Uqwmdxdycm0
m1cmnBecause the variation
cm is arbitrary, so from the equation above, we obtain
Uqw1dxdy0c1Uqw2dxdy0c2 (2-10)
Uqwndxdy0cnThis 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,(n1,2,,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)
r2x2y2,tg1then, we have
y (2-11)
xryrxsin
cos
yrxrxcosysin2
2yrrxrrwith these equations, we obtain
wwrwwsinw
cosxrxxrr
- 4 - wwrwwcosw
sinyryyrrand
2wsinwsinwcoscos
x2rrrr2w2sincos2w
cosr2rr22sinw
rr2sincoswsin22w
222rr2wcoswcoswsinsin
y2rrrr2w2sincos2w
sinr2rr2c2osw
rr2sincoswcos22w
222rr2wsinwcoswcossin
xyrrrr2wcos2sin22wsincosw
sincos
2rrrrrcos2sin2wsincos2w
rr2According to these results, immediately we obtain
2w2w2w1w12ww222
xyrrrr222Rotating 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
rddr, we obtain the expression of strain energy in polar coordinates:
2222D22w1w1w1w1wU(w)2(1)22rdrd
222rrrrrrr(2-12)
- 5 - Now, the potential energy of the external force can be expressed as
Wqwrdrd (2-13)
Then, we obtain the system of
n linear equations regarding the coefficients of the
deflection of a thin plate in polar coordinates
Uqw1rdrd0c1Uqw2rdrd0c2 (2-14)
Uqwnrdrd0cnThinking the problem of bending of a circular plate with symmetric anises, Eq.(2-12) can be
simplified as
d2w21dw2dwd2wUDr2dr (2-15)
22drrdrdrdraccordingly Eq.(2-14) reduce to
U2qw1rdrc1U2qw2rdrc2 (2-16)
U2qwnrdrcnIf all the edges of the circular plate are fixed, at the edges we have
second term of the integration in Eq.(2-15) becomes
2dw21dw2dwd2w1dw0
drdr2dr2ddr2drdrrarbdw0. In that case, the
drat last Eq.(2-15) can be simplified as
d2w21dw2UDr2dr (2-17)
drrdr
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:
w0x (3-1)
wwhen yb,w0,0ywhen xa,w0,Assuming the deflection function as
wc(x2a2)2(y2b2)2 (3-2)
where
c is a undetermined coefficient. Obviously, the equation above fits all the boundary
conditions.
From Eq.(3-2), we obtain,
2w222224c(3xa)(yb)
2x2w4c(x2a2)2(3y2b2)
2y22222222222w4c(3xa)(yb)(xa)(3yb)
Substituting these equations in Eq.(2-7), taking notice of the symmetric nature of the problem.
We obtain
U32Dc20a2222222222(3xa)(yb)(xa)(3yb)dxdy
0b163844Da5b5c2a4a2b2b4
15757Secondly, by using Eq.(2-10), we have
32768554422Dabcaabb4157574q0
a0
b0(x2a2)2(y2b2)2dxdy0- 7 - that is
32768554422256Dabcaabb4q0a5b5
15757225so the coefficient
c can be determined
7q044224caabb (3-4)
128D7thus, substituting Eq.(3-4) in Eq.(3-3) we obtain the approximate solution
117q04w(x2a2)2(y2b2)2a4a2b2b4 (3-5)
128D7Finally, let’s check the precision of this solution. Take a square plate for example, we have
ab, 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%.
- 8 -
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