1
Elasticity
2
3
§ 10-1 The Specific Energy of Deformation and Strain
Energy of Elastic Body
§ 10-3 The Variational Method of Displacement
§ 10-4 The Variational Equations of Stress and the
Variational Method of Stress
§ 10-2 The Variational Equation of Displacement and
The Principle of Minimum Potential Energy
Chapter 10 The Energy
Principles and theVariational Calculus
4
第十章 能量原理与变分法
§ 10-1 弹性体的变形比能与形变势能
§ 10-3 位移变分法
§ 10-4 应力变分方程与应力变分方法
§ 10-2 位移变分方程与极小势能原理
5
The problems of the elastic theory need a series of partial differential
equations,It always meet with difficulties in math,So we have to seek for
approximate solution,The approximate solution of variational calculus is a
method in common use,In math,variational problems are the problems to
solve the limit of the functional,In elastic mechanics,the functional is the
energy(work) of elasticity problems,And the variational calculus is to
calculate the extremum of the energy(work),We can yield the solution of
the elastic problems when calculating the extremum,The direct method of
variation problems facilitate us to obtain the approximate solution,
We will give the expressions which calculate the potential energy of
strain at first in this chapter,We will mainly introduce the variational method
of displacement and that of stress through the relation of work and energy,
6
弹性理论问题需要解一系列偏微分方程组，并满足
边界条件，这在数学上往往遇到困难。因此需要寻求近
似的解法。变分法的近似解法是常用的一种方法。在数
学上，变分问题是求泛函的极限问题。在弹性力学里，
泛函就是弹性问题中的能量（功），变分法是求能量
（功）的极值，在求极值时得到弹性问题的解，变分问
题的直接法使我们比较方便地得到近似解。
本章首先给出计算形变势能的表达式。利用功与能
的关系，主要介绍了位移变分法和应力变分法。
7
§ 10-1 The Specific Energy of
Deformation and Strain Energy of Elastic Body
? ? ? ? ? ?? ?? ?2222221 1222 1 xyzxyzyxxzzyzyxEU ?????????????? ??????????
The specific energy expressed by the components of stress
I,The specific energy of deformation
? ?xyxyzxzxyzyzzzyyxxU ???????????? ?????? 211
Under the complex stress conditions,suppose that the elastic body is
imposed all the six components of stress,
According to the principle of conservation of energy,the value of the strain
energy has no relation to the sequence of the forces imposed on the elastic
body,but is completely decided by the final value of the stress and strain,
Thus we obtain the strain energy density or specific energy of the elastic body,
xyzxyzzyx ??????,,,,,
8
§ 10-1 弹性体的变形比能与形变势能
一 变形比能
在复杂应力状态下，设弹性体受有全部六个应力
分量 。根据能量守恒定理，形变
势能的多少与弹性体受力的次序无关，而完全确定于
应力及形变的最终大小。从而有弹性体的形变势能密
度或比能,
xyzxyzzyx ??????,,,,,
? ?xyxyzxzxyzyzzzyyxxU ???????????? ?????? 211
? ? ? ? ? ?? ?? ?2222221 1222 1 xyzxyzyxxzzyzyxEU ?????????????? ??????????
比能用应力分量表示
9
The specific energy expressed by the components of strain,
? ? ? ? ? ???
?
??
? ??????
???
2222222
1 2
1
2112 xyzxyzzyxe
EU ??????
?
?
?
Where
zyxe ??? ???
Therefore,we have the partial differentiation of specific energy by
components of stress
z
z
y
y
x
x
UUU ?
????? ??
??
?
??
?
? 111,,
xy
xy
zx
zx
yz
yz
UUU ?
????? ??
??
?
??
?
? 111,,
10
比能用应变分量表示
? ? ? ? ? ???
?
??
? ??????
???
2222222
1 2
1
2112 xyzxyzzyxe
EU ??????
?
?
?
其中
zyxe ??? ???
因此，我们有比能对应力分量的偏导
z
z
y
y
x
x
UUU ?
????? ??
??
?
??
?
? 111,,
xy
xy
zx
zx
yz
yz
UUU ?
????? ??
??
?
??
?
? 111,,
11
xy
xy
zx
zx
yz
yz
UUU ?
????? ??
??
?
??
?
? 111,,
z
z
y
y
x
x
UUU ?
????? ??
??
?
??
?
? 111,,
And the partial differentiation of specific energy by components of strain
II,Strain energy
Since the components of stress and components of strain and then the
specific energy are all the function of coordinates,the strain
energy, of the whole elastic body is,1UU
???? d x d y d zUU 121
Substituting the three kinds of expressing forms of specific energy
yields three kinds of integrating forms of strain energy
12
xy
xy
zx
zx
yz
yz
UUU ?
????? ??
??
?
??
?
? 111,,
z
z
y
y
x
x
UUU ?
????? ??
??
?
??
?
? 111,,
比能对应变分量的偏导
二 形变势能
由于应力分量和形变分量，进而比能 都是位置
坐标的函数，所以整个弹性体的形变势能 为,1
U
U
???? d x d y d zUU 121
将比能的三种表达形式代入，得形变势能的三种
积分形式
13
zyx
y
u
x
v
x
w
z
u
z
v
y
w
z
w
y
v
x
u
z
w
y
v
x
uE
U
ddd
2
1
2
1
2
1
21)1(2
222
2222
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Substituting the geometric equations can also yield the strain energy
expressed by displacement
? ???? ?????? d x d y d zU xyxyzxzxyzyzzzyyxx ????????????21
? ? ? ? ? ? d x d y d ze
EU
xyzxyzzyx??? ??
?
??
? ??????
???
2222222
2
1
2112 ???????
?
?
? ? ? ??
? ?? ?? d x d y d z
E
U
xyzxyz
yxxzzyzyx
222
222
12
2
2
1
????
??????????
????
??????? ???
14
zyx
y
u
x
v
x
w
z
u
z
v
y
w
z
w
y
v
x
u
z
w
y
v
x
uE
U
ddd
2
1
2
1
2
1
21)1(2
222
2222
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将几何方程代入，形变势能还可用位移分量来表示
? ???? ?????? d x d y d zU xyxyzxzxyzyzzzyyxx ????????????21
? ? ? ? ? ? d x d y d ze
EU
xyzxyzzyx??? ??
?
??
? ??????
???
2222222
2
1
2112 ???????
?
?
? ? ? ??
? ?? ?? d x d y d z
E
U
xyzxyz
yxxzzyzyx
222
222
12
2
2
1
????
??????????
????
??????? ???
15
I.Variation and its properties,
We have learned in higher mathematics the concept of differentiation
which is the increment of variable,Then what is the variation? Variation is
the increment of function,often denoted by d,Variation has the following
properties,
? ??
?
?
?
?
?
???
Suu d S
u
xx
u
wuwu
dδδ
δδ
δδ)(δ
§ 10-2 The Variational Equation of Displacement
and The Principle of Minimum Potential Energy
16
一 变分及其性质
高等数学我们学过微分的概念，微分是变量的
增量。那么什么是变分呢？变分是函数的增量，通
常用 δ 表示。变分具有以下的性质,
? ??
?
?
?
?
?
???
Suu d S
u
xx
u
wuwu
dδδ
δδ
δδ)(δ
§ 10-2 位移变分方程与极小势能原理
17
II,The variational equation of displacement
Suppose that the elastic body is under the status of equilibrium when
imposed some external forces,It undergoes the true displacement u,v,w,
which satisfy the equations of equilibrium and satisfy the boundary
conditions of displacement and the stress boundary conditions expressed by
the displacement,Now suppose that the components of displacement
undergo minute changes(virtual displacement) du,dv,dw which the
boundary conditions of displacement allow,Now the extermal forces do
virtual work during the virtual displacement which should be equal to the
increment of the functional of the strain energy,
? ? ? ?dSwZvYuXd x d y d zwZvYuXU ??? ?? ?????? ddddddd
This equation is the so-called variational equation of displacement,where
X,Y,Z are the components of body forces,and the are the
components of the surface faces,ZYX,,
18
二 位移变分方程
设弹性体在一定外力作用下，处于平衡状态，发
生的真实位移为 u,v,w，它们满足位移分量表示的平
衡方程，并满足位移边界条件和用位移表示的应力边
界条件。现在假设位移分量发生了位移边界条件所容
许的微小改变（虚位移） δ u, δ v,δ w，这时外力
在虚位移上作虚功，虚功应和变形能泛函的增加相等，
即
? ? ? ?dSwZvYuXd x d y d zwZvYuXU ??? ?? ?????? ddddddd
这个方程就是所谓位移变分方程。其中 X,Y,Z为体力分
量,为面力分量。 ZYX,,
19
III,The principle of the minimum potential energy
Because the virtual displacement is very small,the value and the direction
of the external forces can be regarded as invariable during the virtual
displacement,and only its acting points are changed,According to the
properties of the variation,the variational equation of displacement can be
rewrite as,
? ? ? ?? ? 0??????? ??? ?? dSwZvYuXd x d y d zZwYvXuUd
? ? ? ?dSwZvYuXd x d y d zZwYvXuV ??? ?? ???????
Suppose the potential energy of external forces is,
? ? 0?? VUdThen
The meaning of this equation is that under the action of given external
forces,among all groups of displacements that satisfy the boundary conditions
of displacement,the group of displacement that exist actually should make
the total potential energy a minimum,If considering the second order variation,
we can further analyze it and prove that this extremum is a minimum for the
stable equilibrium status,So this equation is also called the principle of
minimum potential energy,
20
三 极小势能原理
由于虚位移是微小的，因此在虚位移的过程中，
外力的大小和方向可以当做保持不便，只是作用点有
了改变。利用变分的性质，位移变分方程可改写为,
? ? ? ?? ? 0??????? ??? ?? dSwZvYuXd x d y d zZwYvXuUd
? ? ? ?dSwZvYuXd x d y d zZwYvXuV ??? ?? ???????
? ? 0?? VUd
设外力势能为
则
该式的意义是：在给定的外力作用下，在满足位
移边界条件的各组位移中，实际存在的一组位移应使
总势能为极值。如果考虑二阶变分，进一步的分析证
明，对于稳定平衡状态，这个极值是极小值。因此，
该式又称为极小势能原理。
21
Apparently,the displacement existing exactly should satisfy the
equations of equilibrium and the boundary conditions of stress expressed by
displacement besides boundary conditions of displacement,Now we can see
also that the displacement existing exactly still satisfy the variational
equations of displacement besides boundary conditions of displacement,
Furthermore,through calculating,we can deduce differential equations of
equilibrium and boundary conditions of stress expressed by displacement
from variational equations of displacement,So it is obvious that variational
equations of displacement can substitute for differential equations of
equilibrium and boundary conditions of stress,
22
显然，实际存在的位移，除了满足位移边界条件
以外，还应当满足位移表示的平衡方程和应力边界条
件；现在又看到，实际存在的位移，除了满足位移边
界条件外，还满足位移变分方程。而且，通过运算，
还可以从位移变分方程导出用位移表示的平衡微分方
程和应力边界条件。于是可见：位移变分方程可以代
替平衡微分方程和应力边界条件。
23
§ 10-3 The Variational Method of Displacement
??? ??????
m
mm
m
mm
m
mm wCwwvBvvuAuu 000,,
where u0,v0,w0 are undetermined coefficients whose boundary values
equal to the known displacement on the boundaries,um, vm,wm are the
set functions whose boundary conditions equal to zero,Am,Bm,Cm are
undetermined coefficients,by whose variation the variation of
displacement are realized,
We set expressions of components of displacement that satisfy the
boundary conditions of displacement,which contain several undetermined
coefficients,Then we decide these coefficients according to the principle
of minimum potential energy,Take the expression of components of
displacement as following,
I,The Ritz method
24
§ 10-3 位移变分法
??? ??????
m
mm
m
mm
m
mm wCwwvBvvuAuu 000,,
其中 u0,v0,w0 为设定的函数，它们的边界值等于边界
上的已知位移； um, vm,wm 为边界值等于零的设定
函数,Am,Bm,Cm为待定的系数，位移的变分由它们
的变分来实现。
先设定满足位移边界条件的位移分量的表达式，
其中包含若干个待定的系数，再根据极小势能原理，
决定这些系数。取位移分量的表达式如下,
一 瑞次法
25
??? ??? m mmm mmm mm CwwBvvAuu δ,δδ,δδδ
? ??? )δδδ(δ m
m
m
m
m
m
CCUBBUAAUU ??????
The variation of strain energy is
The variation of potential energy of external forces is
? ??
? ???
???
????
m
mmmmmm
m
mmmmmm
SCwZBvYAuX
zyxCZwBYvAXuV
d)δδδ(
ddd)δδδ(δ
The variations of components of displacement are
26
??? ??? m mmm mmm mm CwwBvvAuu δ,δδ,δδδ
? ??? )δδδ(δ m
m
m
m
m
m
CCUBBUAAUU ??????
应变能的变分为
外力势能的变分为
? ??
? ???
???
????
m
mmmmmm
m
mmmmmm
SCwZBvYAuX
zyxCZwBYvAXuV
d)δδδ(
ddd)δδδ(δ
位移分量的变分是
27
yields
Above system of linear algebraic equations includes 3m equations,
After solving these equations,we introduce the results into the
expressions of the components of displacement to obtain the approximate
solutions of the components of displacement,This method is called the
Ritz method,
Substituting into
?????
?????
?????
??
??
??
SwZzyxZw
C
U
SvYzyxYv
B
U
SuXzyxXu
A
U
mm
m
mm
m
mm
m
dddd
dddd
dddd
?
?
?
?
?
?
? ? 0?? VUd
28
中，得到
上面是个数为 3m的线性代数方程组，求解后，代回
位移分量的表达式，得到位移分量的近似解。这种
方法称为瑞次法。
代入
?????
?????
?????
??
??
??
SwZzyxZw
C
U
SvYzyxYv
B
U
SuXzyxXu
A
U
mm
m
mm
m
mm
m
dddd
dddd
dddd
?
?
?
?
?
?
? ? 0?? VUd
29
II,The Galerkin method
If we regard variation as the function of the components of strain,
then
d x d y d zUUd x d y d zUU yz
yz
x
x
?????? ???????? ????????? ?? d??d??dd 111
Because
.,,,;,,,11 ???? vzwyuxUU yzxyz
yz
x
x
ddd?dd???? ???????????????
so
??? ?????? ????????? ?????????? d x d y d zvzwyuxU yzx ?? dd?d?d
Applying the Ostrogradsky—Gauss formula to the first term,we have
30
二 伽辽金法
将变分看做形变分量的函数，则
d x d y d zUUd x d y d zUU yz
yz
x
x
?????? ???????? ????????? ?? d??d??dd 111
由于
.,,,;,,,11 ???? vzwyuxUU yzxyz
yz
x
x
ddd?dd???? ???????????????
所以
??? ?????? ????????? ?????????? d x d y d zvzwyuxU yzx ?? dd?d?d
应用奥高公式，对上式中的第一项，我们有
31
? ?
?? ???
?????? ???
?
?
??
?
?
?
?
?
?
?
?
u d x d y d z
x
u d Sl
u d x d y d z
x
d x d y d zu
x
u d x d y d z
x
x
x
x
xx
d
?
d?
d
?
d?d?
We arrange other terms likewise,then
Substituting the above expression into the variational equation and
combining the similar terms yields,
? ? ? ? ? ?? ?
d x d y d zw
yxz
v
xzy
u
zyx
dSwmlnvlnmunmlU
yzzxzxyyzyzxxyx
yzzxzxyyzyzxxyx
???
??
?
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?????????
d
???
d
???
d
???
d???d???d???d
32
? ?
?? ???
?????? ???
?
?
??
?
?
?
?
?
?
?
?
u d x d y d z
x
u d Sl
u d x d y d z
x
d x d y d zu
x
u d x d y d z
x
x
x
x
xx
d
?
d?
d
?
d?d?
对于其余各项也进行同样的处理，则
? ? ? ? ? ?? ?
d x d y d zw
yxz
v
xzy
u
zyx
dSwmlnvlnmunmlU
yzzxzxyyzyzxxyx
yzzxzxyyzyzxxyx
???
??
?
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?????????
d
???
d
???
d
???
d???d???d???d
将上式代入位移变分方程，并归项得
33
? ??
? ? ? ? ? 0?????????
????
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???
dSwZmlnvYlnm
uXnmld x d y d zwZ
yxz
vY
xzy
uX
zyx
yzzxzxyyzy
zxxyx
yzzxz
xyyzyzxxyx
d???d???
d???d
???
d
???
d
???
If the boundary conditions of stress can be satisfied,then the above
equation can be simplified as
0?
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???
d x d y d zwZ
yxz
vY
xzy
uX
zyx
yzzxz
xyyzyzxxyx
d
???
d
???
d
???
This is the equation that the variation of displacement should satisfy when
the components of displacement satisfy the boundary conditions of
displacement and that of stress,It is called Galerkin variational equation,
34
? ??
? ? ? ? ? 0?????????
????
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???
dSwZmlnvYlnm
uXnmld x d y d zwZ
yxz
vY
xzy
uX
zyx
yzzxzxyyzy
zxxyx
yzzxz
xyyzyzxxyx
d???d???
d???d
???
d
???
d
???
如果应力边界条件得到满足，则上式简化为
0?
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?
?
?
?
?
?
???
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
???
d x d y d zwZ
yxz
vY
xzy
uX
zyx
yzzxz
xyyzyzxxyx
d
???
d
???
d
???
这就是位移分量满足位移边界条件及应力边界条件时，
位移变分所应满足的方程，称为伽辽金变分方程。
35
??? ??????
m
mm
m
mm
m
mm wCwwvBvvuAuu 000,,
If we take the expression of the components of displacement as following
??? ???
m
mm
m
mm
m
mm CwwBvvAuu δ,δδ,δδδ
so that the boundary conditions of displacement and that of stress are both
satisfied, Then we substitute the variation of displacement
into the Galerkin equation,and we obtain
d x d y d zwZ
yxz
C
d x d y d zvY
xzy
B
d x d y d zuX
zyx
A
m
yzzxz
m
m
m
xyyzy
m
m
m
zxxyx
m
m
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
? ???
? ???
? ???
???
d
???
d
???
d
36
??? ??????
m
mm
m
mm
m
mm wCwwvBvvuAuu 000,,
若取位移分量的表达式如下,
??? ???
m
mm
m
mm
m
mm CwwBvvAuu δ,δδ,δδδ
使得位移边界条件和应力边界条件都得到满足，则将
位移变分
代入伽辽金方程，就得到
d x d y d zwZ
yxz
C
d x d y d zvY
xzy
B
d x d y d zuX
zyx
A
m
yzzxz
m
m
m
xyyzy
m
m
m
zxxyx
m
m
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
? ???
? ???
? ???
???
d
???
d
???
d
37
Because are arbitrary,their coefficients should be zero
separately,So we obtain
mmm CBA ddd,,
0
0
0
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
???
???
???
d x d y d zwZ
yxz
d x d y d zvY
xzy
d x d y d zuX
zyx
m
yzzxz
m
xyyzy
m
zxxyx
???
???
???
Expressing the components of stress of the above three equations by the
components of strain through the physical equations,then expressing
them by the components of displacement through the geometric
equations,and simplifying yields
38
由于
mmm CBA ddd,,
的任意性，它们的系数应当分别为零
于是得
0
0
0
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
???
???
???
d x d y d zwZ
yxz
d x d y d zvY
xzy
d x d y d zuX
zyx
m
yzzxz
m
xyyzy
m
zxxyx
???
???
???
将上列三方程中的应力分量通过物理方程用形变分量表
示，再通过几何方程用位移分量表示，简化后即得,
39
? ?
? ?
? ?
0
21
1
12
0
21
1
12
0
21
1
12
2
2
2
??
?
?
?
?
?
?
?
?
?
?
?
?
?
?
??
?
?
??
??
?
?
?
?
?
?
?
?
?
?
?
?
?
?
??
?
?
??
??
?
?
?
?
?
?
?
?
?
?
?
?
?
?
??
?
?
??
???
???
???
d x d y d zwZw
z
eE
d x d y d zvYv
y
eE
d x d y d zuXu
x
eE
m
m
m
??
??
??
Thus we obtain the linear system of equations about undetermined
coefficient of the function of displacement,After solving the equations,we
substitute the results into the expression of the components of displacement,
then we obtain the approximate solution of the components of displacement,
This method is called the Galerkin method,
What we should pay attention to is that the components of displacement
solved through the variational method of displacement can achieve high
precision if we take only a few terms,However,the stresses solved through
this method are not precise,So we must take more terms to obtain
sufficiently precise stress,
40
? ?
? ?
? ?
0
21
1
12
0
21
1
12
0
21
1
12
2
2
2
??
?
?
?
?
?
?
?
?
?
?
?
?
?
?
??
?
?
??
??
?
?
?
?
?
?
?
?
?
?
?
?
?
?
??
?
?
??
??
?
?
?
?
?
?
?
?
?
?
?
?
?
?
??
?
?
??
???
???
???
d x d y d zwZw
z
eE
d x d y d zvYv
y
eE
d x d y d zuXu
x
eE
m
m
m
??
??
??
这样就得到位移函数待定常数的线性方程组，求解
后，代回位移分量的表达式，得到位移分量的近似解。
这种方法称为伽辽金法。
要注意的是：用位移变分法求位移分量，只须取几项
就可达到较高的精度，然而由此求出的应力却很不精确。
为了求得的应力充分精确，必须取更多项。
41
III,Application and example
If we apply the variational method of displacement to the plane problems,
the Ritz method and the Galerkin method will both be simplified,For these
two kinds of plane problems,because we need not consider the displacement
w in z direction,and u and v don’t change along with the change of z
coordinate,so the the expression of the components of displacement can be
set as
?? ????
m
mm
m
mm vBvvuAuu 00,
???
???
??
??
SvYyxYv
B
U
SuXyxXu
A
U
mm
m
mm
m
ddd
ddd
?
?
?
?
When using the Ritz method,in order to decide the coefficient Am and
Bm,we take a unit length in z direction,then we simply apply the two
equations as follows to solve the linear system of equations,
42
三 应用与举例
将位移变分法应用于平面问题，瑞次法和伽辽金法
都将得到简化。由于两种平面问题都不必考虑 z 方向的
位移 w,且 u 和 v 都不随坐标 z 而变，所以位移分量的
表达式可设为
?? ????
m
mm
m
mm vBvvuAuu 00,
在采用瑞次法时，为了决定系数 Am及 Bm,在 z方向
取一个单位长度，只须应用如下二式来求解线性方程组
???
???
??
??
SvYyxYv
B
U
SuXyxXu
A
U
mm
m
mm
m
ddd
ddd
?
?
?
?
43
Where the strain energy expressed by the components of displacement can
be simplified as follows,For plane strain problems
? ? d x d yy
u
x
v
y
v
x
u
y
v
x
uEU ??
?
?
?
?
?
?
?
?
???
?
???
?
?
??
?
??
???
?
???
?
?
???
?
??
?
?
?
??
???
?
???
?
?
??
?
?
??
?
2222
2
1
112 ?
?
?
For plane stress problems
? ? d x d yy
u
x
v
y
v
x
u
y
v
x
uEU ??
?
?
?
?
?
?
?
?
???
?
???
?
?
??
?
???
?
?
?
??
???
?
???
?
?
???
?
??
?
?
?
?
?
?
222
2 2
12
12
??
?
When using the Galerkin method,for plane strain problems,we should
use the two equations as follows to solve the linear system of equations
? ?
? ?
0
21
1
12
0
21
1
12
2
2
??
?
?
?
?
?
???
?
?
??
?
?
??
?
?
??
??
?
?
?
?
?
???
?
?
??
?
?
??
?
?
??
??
??
d x d yvYv
y
eE
d x d yuXu
x
eE
m
m
??
??
44
其中形变势能用位移分量表示形式简化如下，平面应变问题
? ? d x d yy
u
x
v
y
v
x
u
y
v
x
uEU ??
?
?
?
?
?
?
?
?
???
?
???
?
?
??
?
??
???
?
???
?
?
???
?
??
?
?
?
??
???
?
???
?
?
??
?
?
???
2222
2
1
112 ?
?
?
平面应力问题
? ? d x d yy
u
x
v
y
v
x
u
y
v
x
uEU ??
?
?
?
?
?
?
?
?
???
?
???
?
?
??
?
???
?
?
?
??
???
?
???
?
?
???
?
??
?
?
?
?
??
222
2 2
12
12
??
?
在采用伽辽金法时，对于平面应变问题，要应用如下二式
来求解线性方程组
? ?
? ?
0
21
1
12
0
21
1
12
2
2
??
?
?
?
?
?
???
?
?
??
?
?
??
?
?
??
??
?
?
?
?
?
???
?
?
??
?
?
??
?
?
??
??
??
d x d yvYv
y
eE
d x d yuXu
x
eE
m
m
??
??
45
For plane stress problems,the equations set to be solved is
0
2
1
2
1
1
0
2
1
2
1
1
2
2
2
2
2
2
2
2
2
2
2
2
??
?
?
?
?
?
???
?
?
??
?
?
??
??
?
?
??
?
?
?
?
??
?
?
?
?
?
???
?
?
??
?
?
??
??
?
?
??
?
?
?
?
??
??
d x d yvY
yx
u
x
v
y
vE
d x d yuX
yx
v
y
u
x
uE
m
m
??
?
??
?
The calculation workload of the Galerkin method is small,But its
requirement to the function of displacement is high,Besides satisfying the
boundary conditions of displacement,it needs that the stresses solved by the
function of displacement satisfy the boundary conditions of stress,In particular
case,for example,the problem has only boundary conditions of displacement
and no boundary conditions of stress,It means also that the boundary conditions
of stress are satisfied,It is very convenient to use the Galerkin method here,
The key of the Ritz method is to find the function of displacement that
satisfy all the boundary conditions,but this kind of function is always hard to
find,especially in the case that the boundaries are not regular,Therefore,the
application of the Ritz method is extremely restricted on this aspect,
46
对于平面应力问题，要求解的方程组如下
0
2
1
2
1
1
0
2
1
2
1
1
2
2
2
2
2
2
2
2
2
2
2
2
??
?
?
?
?
?
???
?
?
?
?
?
?
??
??
?
?
??
?
?
?
?
??
?
?
?
?
?
???
?
?
?
?
?
?
??
??
?
?
??
?
?
?
?
??
??
d x d yvY
yx
u
x
v
y
vE
d x d yuX
yx
v
y
u
x
uE
m
m
??
?
??
?
伽辽金方法的计算工作量较小，但对位移函数的要
求较高，除了要求满足位移边界条件外，还要求根据位
移函数求得的应力应满足应力边界条件。在特殊情况，
如仅有位移边界，而无应力边界，这也表示着应力边界
条件得到满足，这时用伽辽金方法十分方便。
瑞次法的要点是要找到满足全部边界条件的位移函
数，而这种函数一般仍然难以找到，尤其在边界不规整
的情况下。所以瑞次方法的应用在这一点上受到极大的
限制。
47
Example 1,For the thin plate as shown in the figure,ignoring the body
forces,please evaluate the displacement,
Solution,We assume the displacements
yBvBvxAuAu 111111,????
which satisfy the boundary conditions of the
displacement boundaries(left and underside),
??
??
??
?
?
??
?
?
?
?
?
?
??
?
?
?
?
?
?
?
?
?
?
?
?
?
???
?
?
??
?
?
?
?
??
?
?
?
?
?
?
?
?
?
yx
y
u
x
v
yx
y
v
x
u
y
v
x
uE
U
dd
2
1
dd2
)1(2
2
22
2
?
?
?
Under the conditions of plane stress
? ? ???? a b d x d yBABAEU 0 0 1121212 )2()1(2 ??
We have
x
y
q1
q2
b
a
o
48
例 1：如图所示的薄板，不计体力，求薄板的位移
解,设位移
yBvBvxAuAu 111111,????
它们是满足位移边界（左边和下边）
的边界条件的。
??
??
??
?
?
??
?
?
?
?
?
?
??
?
?
?
?
?
?
?
?
?
?
?
?
?
???
?
?
??
?
?
?
?
??
?
?
?
?
?
?
?
?
?
yx
y
u
x
v
yx
y
v
x
u
y
v
x
uE
U
dd
2
1
dd2
)1(2
2
22
2
?
?
?
在平面应力状态下
? ? ???? a b d x d yBABAEU 0 0 1121212 )2()1(2 ??
可得
x
y
q1
q2
b
a
o
49
)2()1(2 1121212 BABAE a bU ?? ????
i.e,
abqBUabqAU 2
1
1
1
,???? ????
We have
?? ?? svYBUsuXAU d,d 1
1
1
1 ?
?
?
?From
abqAB
E a b
abqBA
E a b
2112
1112
)22(
)1(2
)22(
)1(2
???
?
???
?
?
?
?
?
i.e,
E
qqB
E
qqA 12
1211,
?? ??????
yE qqvxE qqu 1221,?? ??????
Solving them yields
50
)2()1(2 1121212 BABAE a bU ?? ????即
abqBUabqAU 2
1
1
1
,???? ????
可得
?? ?? svYBUsuXAU d,d 1
1
1
1 ?
?
?
?由
abqAB
E a b
abqBA
E a b
2112
1112
)22(
)1(2
)22(
)1(2
???
?
???
?
?
?
?
?
即
E
qqB
E
qqA 12
1211,
?? ??????
yE qqvxE qqu 1221,?? ??????
解得
51
Example 2,As shown in the figure,a rectangular
thin plate with the width of 2a and the height of b
is fixed on the left,right and at bottom,The
displacement of its top is given as
???
?
???
? ????
2
2
1,0 axvu ?
Ignoring the body force,please evaluate the
displacement and stress of the thin plate,
Solution,Set the coordinate axis as shown in the
figure,Suppose the components of displacement are
?????? ???
?
?
???
? ??
b
y
b
y
a
x
a
xAu 11
2
2
1
?????? ???
?
?
???
? ??
???
?
???
? ???
b
y
b
y
a
xB
b
y
a
xv 111
2
2
12
2
?
a a
b b
o x
y
η
52
例 2：如图所示，宽为 2a而高度为 b
的矩形薄板，左右两边及下边均被
固定，而上边的位移给定为
???
?
???
? ????
2
2
1,0 axvu ?
不计体力，试求薄板的位移和应力。
解,取坐标轴如图所示。设位移
分量为
?????? ???
?
?
???
? ??
b
y
b
y
a
x
a
xAu 11
2
2
1
?????? ???
?
?
???
? ??
???
?
???
? ???
b
y
b
y
a
xB
b
y
a
xv 111
2
2
12
2
?
a a
b b
o x
y
η
53
which can satisfy the boundary conditions of displacement,i.e,
? ? ? ? ? ? 0,0,0 0 ??? ???? byyax uuu
? ? ? ? ? ? ??
?
?
???
? ?????
???? 2
2
0 1,0,0 a
xvvv
byyax ?
There is not boundary conditions of stress in this problem,So we can
consider that since the given displacements satisfy the boundary conditions
of displacement,then they satisfy all the boundary conditions,So we can
employ the Galerkin method to solve this problem to simplify the
calculation,
Notice that the body forces X=Y=0 and m=1,so the Galerkin
equations become
0
2
1
2
1
1
0
2
1
2
1
1
1
0
2
2
2
2
2
2
1
0
2
2
2
2
2
2
???
?
?
?
?
?
?
??
??
?
?
??
?
?
?
?
???
?
?
??
?
?
??
??
?
?
??
?
?
?
?
? ?
? ?
?
?
d x d yv
yx
u
x
v
y
vE
d x d yu
yx
v
y
u
x
uE
a
a
b
a
a
b
??
?
??
?
54
可以满足位移边界条件，即
? ? ? ? ? ? 0,0,0 0 ??? ???? byyax uuu
? ? ? ? ? ? ??
?
?
???
? ?????
???? 2
2
0 1,0,0 a
xvvv
byyax ?
在该问题中，并没有应力边界条件，因此可以认
为所设位移既然满足了位移边界条件，也就满足了全
部边界条件，这就可以应用伽辽金法求解，使数学运
算比较简单一些。
注意体力 X=Y=0而 m = 1，伽辽金方程成为
0
2
1
2
1
1
0
2
1
2
1
1
1
0
2
2
2
2
2
2
1
0
2
2
2
2
2
2
???
?
?
?
?
?
?
??
??
?
?
??
?
?
?
?
???
?
?
??
?
?
??
??
?
?
??
?
?
?
?
? ?
? ?
?
?
d x d yv
yx
u
x
v
y
vE
d x d yu
yx
v
y
u
x
uE
a
a
b
a
a
b
??
?
??
?
55
Substituting the second derivative of the components of every displacement
.21
22
,2131
,1
2
,
22
,
2
,
6
1
2
2
2
1
2
2
2
2
1
2
2
2
2
2
1
22
2
3
3
2
1
2
2
2
2
2
1
2
2
?
?
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b
y
a
x
ab
B
a
x
abyx
v
b
y
a
x
ab
A
yx
u
a
x
b
B
y
v
b
y
b
y
a
B
b
y
ax
v
a
x
a
x
b
A
y
u
b
y
b
y
a
x
a
A
x
u
?
?
and
???
?
???
? ?
???
?
???
? ??
???
?
???
? ?
???
?
???
? ??
2
2
2
2
12
2
3
3
1 1,b
y
b
y
a
xv
b
y
b
y
a
x
a
xu
into the Galerkin equations,integrating and solving them yields
? ?
? ?
? ?
? ??
??
?
??
??
??
??
??
1216
15,
12042
135
2
211
b
a
B
b
a
b
aA
56
将位移分量的各二阶导数
.21
22
,2131
,1
2
,
22
,
2
,
6
1
2
2
2
1
2
2
2
2
1
2
2
2
2
2
1
22
2
3
3
2
1
2
2
2
2
2
1
2
2
?
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???
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?
???
?
?
?
?
?
?
?
?
?
?
???
?
?
b
y
a
x
ab
B
a
x
abyx
v
b
y
a
x
ab
A
yx
u
a
x
b
B
y
v
b
y
b
y
a
B
b
y
ax
v
a
x
a
x
b
A
y
u
b
y
b
y
a
x
a
A
x
u
?
?
以及
???
?
???
? ?
???
?
???
? ??
???
?
???
? ?
???
?
???
? ??
2
2
2
2
12
2
3
3
1 1,b
y
b
y
a
xv
b
y
b
y
a
x
a
xu
代入伽辽金方程，进行积分并求解得
? ?
? ?
? ?
? ??
??
?
??
??
??
??
??
1216
15,
12042
135
2
211
b
a
B
b
a
b
aA
57
In order to be simple,let b=a and ?=0.2,Substituting A1 and B1 into the
given function of displacement yields
?
?
?
?
?
?
?
?
???
?
?
?
?
?
?
???
??
?
?
??
?
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???
?
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??
?
?
??
2
2
2
2
2
2
3
3
2 2 7.07 7 3.01
7 2 4.0
a
y
a
y
a
x
v
a
y
a
y
a
x
a
x
u
?
?
Employing the geometric equations and the physical equations,we can solve
the components of stress through the above equations
?
?
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???
a
y
a
x
a
x
a
y
a
y
a
x
a
E
a
y
a
y
a
x
a
y
a
x
a
E
a
y
a
y
a
x
a
y
a
x
a
E
xy
y
x
21302.0189.0644.0
31302.0473.0305.01
31754.0095.0161.01
3
3
2
2
2
2
2
2
2
2
2
2
2
2
2
2
?
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?
58
为简单起见，取 b=a而 μ=0.2,将 A1和 B1代入所设位移函数得
?
?
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??
2
2
2
2
2
2
3
3
2 2 7.07 7 3.01
7 2 4.0
a
y
a
y
a
x
v
a
y
a
y
a
x
a
x
u
?
?
应用几何方程及物理方程，可有上式求得应力分量
?
?
?
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a
y
a
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a
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a
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a
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a
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a
E
a
y
a
y
a
x
a
y
a
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a
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a
y
a
y
a
x
a
y
a
x
a
E
xy
y
x
21302.0189.0644.0
31302.0473.0305.01
31754.0095.0161.01
3
3
2
2
2
2
2
2
2
2
2
2
2
2
2
2
?
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?
?
59
§ 10-4 The Variational Equations of
Stress and the Variational Method of Stress
Suppose that there is an arbitrary elastic body which is in equilibrium
under the action of equilibrium,Letσij be the components of stress that exist
exactly,which satisfy the differential equations of equilibrium and the
boundary conditions of stress,also satisfy the compatible equations,Now
we assume that the body forces don’t change,while the components of
stress undergo small changeδσij,which is so-called the virtual stress or the
variation of stress,so that the components of stress become σij +δσij,，
Suppose they only satisfy the differential equations of equilibrium and the
boundary conditions of stress,
I,The variational equations of stress
Since the two groups of equations both satisfy the differential equations
of equilibrium under the action of the same body forces,the variation of the
components of stress must satisfy the equations of equilibrium without body
forces,i.e,
60
§ 10-4 应力变分方程应力变分方法
设有任一弹性体，在外力的作用下处于平衡。命
σ ij为实际存在的应力分量，它们满足平衡微分方程
和应力边界条件，也满足相容方程。现在，假想体力
不变,而应力分量发生了微小的变化 δσij,即所谓虚应
力或应力的变分，使应力分量成为 σ ij +δ σ ij,设它
们只满足平衡微分方程和应力边界条件。
一 应力变分方程
既然两组应力分量都满足同样体力作用下的平衡
微分方程，应力分量的变分必然满足无体力时的平衡
方程，即
61
0δδδ
0δδδ
0δδδ
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
yzxzz
xyzyy
zxyxx
yxz
yzy
zyx
???
???
???
(a)
Znml
Ynml
nml
zzyzx
yzyxy
zxxyx
δδδδ
δδδδ
Xδδδδ
???
???
???
???
???
???
(b)
At the same time,at the boundaries that the displacements are given(the
surface forces are not possible to give),the variation of the components of
stress must go with the variation of the components of surface
forces, According to the requirement of the boundary conditions of
stress,at the boundaries,the variation of the components of stress must satisfy ZYX ddd,,
62
0δδδ
0δδδ
0δδδ
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
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?
?
?
?
yzxzz
xyzyy
zxyxx
yxz
yzy
zyx
???
???
???
(a)
Znml
Ynml
nml
zzyzx
yzyxy
zxxyx
δδδδ
δδδδ
Xδδδδ
???
???
???
???
???
???
(b)
同时，在位移给定的边界上（面力不可能给定），应力分
量的变分必然伴随着面力分量的变分 。 根据应力
边界条件的要求，应力分量的变分在边界上必须满足
ZYX ddd,,
63
Owing to the variation of the components of stress,the strain energy
must have the corresponding variation,We regard strain energy as the
function of the components of stress,then the variation of the strain energy
should be
zyx
UU
zyxUU
yz
yz
x
x
ddd.,,δ.,,δ
dddδδ
11
1
???
???
?
?
?
?
?
?
?
?
?
?
?
??
?
?
?
?
?
?
?
?
Introducing the following equation
.,,
,,,
111
111
xy
xy
zx
zx
yz
yz
z
z
y
y
x
x
UUU
UUU
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
64
由于应力分量的变分，形变势能必有相应的变分。
把形变势能看做应力分量的函数，则形变势能的变分应
为
zyx
UU
zyxUU
yz
yz
x
x
ddd.,,δ.,,δ
dddδδ
11
1
???
???
?
?
?
?
?
?
?
?
?
?
?
??
?
?
?
?
?
?
?
?
将下式代入
.,,
,,,
111
111
xy
xy
zx
zx
yz
yz
z
z
y
y
x
x
UUU
UUU
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
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?
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?
?
65
? ?
zyx
z
v
y
w
x
u
zyxU
yzx
yzyzxx
ddd.,,δ.,,δ
ddd.,,δ.,,δδ
???
???
?
?
?
?
?
?
?
?
???
?
?
??
?
?
?
?
?
?
?
??
?
?
?
????
??
????
zyxxuSluzyxxu xxx ddd)δ(dδdddδ ???????? ?????? ???
and introducing the geometric equations yields
According to the integration by parts and the Ostrogradsky—Gauss
formula,we arrange every term on the right of the above equation,for
example,
We finally arrive at
66
? ?
zyx
z
v
y
w
x
u
zyxU
yzx
yzyzxx
ddd.,,δ.,,δ
ddd.,,δ.,,δδ
???
???
?
?
?
?
?
?
?
?
???
?
?
??
?
?
?
?
?
?
?
??
?
?
?
????
??
????
zyxxuSluzyxxu xxx ddd)δ(dδdddδ ???????? ?????? ???
再将几何方程代入，得
根据分步积分和奥－高公式，对上式右边的各项进行处理，例如
最后可得
67
zyx
zyx
u
SnmluU
zxyxx
zxxyx
ddd
δδδ
d])δδδ([δ
???
??
?
?
?
?
?
?
?
?
???
?
?
??
?
?
?
?
?
?
?
?
?
?
?
????
?
?
???
???
Then introducing the equations (a) and (b) gives
?? ??? SZwYvXuU d)δδδ(δ
This is the so-called the variational equation of stress(the Castigliano
variational equation),The right part of the equation represents the work done
by the variation of the surface forces during the true displacement,We can
see from here that due to the variation that happen to the stress,the variation
of strain energy equals to the work done by the variation of surface forces
during the true displacement.。
68
zyx
zyx
u
SnmluU
zxyxx
zxxyx
ddd
δδδ
d])δδδ([δ
???
??
?
?
?
?
?
?
?
?
???
?
?
??
?
?
?
?
?
?
?
?
?
?
?
????
?
?
???
???
再将式（ a）和（ b）代入，即得
?? ??? SZwYvXuU d)δδδ(δ
这就是所谓应力变分方程 (卡斯提安诺变分方程 )。方程的右
边代表面力的变分在实际位移上所做的功。由此可见，由于
应力发生的变分，形变势能的变分等于面力的变分在实际位
移上所做的功。
69
If the surface forces are given at a certain part of boundary,then the
surface forces at this part of boundary can’t have variation,
so, Accordingly the corresponding integrating terms on the
right of variational equation of stress become zero,If the displacements are
given to be zero at a certain part of boundary,then the corresponding
integrating terms on the right of variational equation of stress also become
zero,Therefore,the integration on the right of the variational equation of
stress need only be conducted at such boundaries where the surface forces
aren’t given and the given displacements aren’t equal to zero,
0??? ZYX ddd
II,The principle of minimum complementary energy
Rewrite the variational equation of stress as
0d)δδδ(δ ???? ?? SZwYvXuU
Because at the boundaries where integration is needed,the displacements
are given and keep invariable during the variation,the above equation can
be rewrote as
70
如果在某一部分边界上，面力是给定的，则该部分边
界上的面力不能有变分，于是,而应力变分
方程右边的相应积分项成为零；如果在某一部分边界上，
给定的位移等于零，则应力变分方程右边的相应积分项也
成为零。因此，应力变分方程右边的积分，只须在这样的
边界上进行：面力没有给定，而给定的位移又不等于零。
0??? ZYX ddd
二 极小余能原理
将应力变分方程改写为
0d)δδδ(δ ???? ?? SZwYvXuU
由于在需要积分的边界上，位移是给定的，在变分过
程中保持不变，所以上式可以改写为
71
The expression in the square brackets is called the complementary
energy of the elastic body,Hence,among all groups of stresses that satisfy
the equations of equilibrium and the boundary conditions of stress,the
complementary energy of the elastic body with the group of stresses existing
exactly becomes minimum,If we consider the second-order variation,we
can prove that this extremum is a minimum,So the above conclusion is
called the principle of minimum complementary energy,
? ? 0d)(-Uδ ????? SZwYvXu
We notice previously that the stresses existing exactly should satisfy the
compatible equations besides the differential equations of equilibrium,Now we
see also that the stresses existing exactly satisfy the variational equations of
stress besides the differential equations of equilibrium and the boundary
conditions of stress,Furthermore,we can deduce the compatible conditions
from the variational equations of stress through calculating,So we can see that
the variational equations of stress can substitute for the compatible conditions,
72
中括号内的表达式称为弹性体的余能。因此，在满足
平衡方程和应力边界条件的各组应力中间，实际存在的一
组应力应使弹性体的余能成为极值。如果考虑二阶变分，
可以证明这个极值是极小值，所以上述结论称为极小余能
原理。
? ? 0d)(-Uδ ????? SZwYvXu
以前看到，实际存在的应力，除了满足平衡微分方
程以外，还应当满足相容方程。现在又看到，实际存在
的应力，除了满足平衡微分方程和应力边界条件以外，
还满足应力变分方程。而且，通过运算，还可以从应力
变分方程导出相容条件。于是可见，应力变分方程可以
代替相容条件。
73
III,The variational method of stress
Set the expressions of the components of stress and let them satisfy the
equations of equilibrium and the boundary conditions of stress,which
contain several undetermined coefficients,Then we can decide these
coefficients according to the variational equations of stress,The
components of stress can be generally set as
Where Am are m independent coefficients,(σij)0 is the set function which
satisfies the differential equations of equilibrium and the boundary
conditions of stress,(σij)m is the set function which satisfies,the
differential equations of equilibrium and the boundary conditions of stress
without the action of body forces and surface forces”,Thus,σij satisfies the
differential equations of equilibrium and the boundary conditions of stress
at all time no matter how to determine the value of the coefficient Am,
Notice that the variation of stress is only realized by the variation of the
coefficient Am,
mm ijmijij A )()( 0 ??? ???
(c)
74
三 应力变分法
设定应力分量的表达式，使其满足平衡方程和应力边
界条件，但其中包含若干待定系数，然后根据应力变分方
程决定这些系数。应力分量一般可设为
其中 Am为互不依赖的 m个系数。 (σij)0是满足平衡微分
方程和应力边界条件的设定函数,(σij)m 是满足, 无体力
和面力作用时的平衡微分方程和应力边界条件, 的设定函
数。这样，不论系数 Am如何取值,σij总能满足平衡微分方
程和应力边界条件。
注意：应力的变分只是由系数 Am的变分来实现。
mm ijmijij A )()( 0 ??? ???
(c)
75
If either the surface forces are given or the displacements equal to zero
at any part of boundary of the elastic body,then the variational equation of
stress is simplified as
Apparently,the strain energy U is the quadratic function of Am,So the
equation (d) will be the simple equation of Am,There are altogether m such
equations which can even be used to solve the coefficient Am,Then
substitute it into the equation (c) to solve the components of stress,
0?Ud
0???
mA
Ui.e,(d)
If the displacements are given but not equal to zero at a certain part of
boundary,we must apply variational equations directly at this part of
boundary,i.e,
?? ??? SZwYvXuU d)δδδ(δ
76
如果在弹性体的每一部分边界上，不是面力被给定，
便是位移等于零，则应力变分方程简化为
显然，形变势能 U是 Am的二次函数，因而 (d)式将是 Am的一
次方程。这样的方程共有 m 个，恰好可以用来求解系数 Am，
回代 (c)式，求得应力分量。
0?Ud
0???
mA
U
即 (d)
如果在某一部分边界上，位移是给定的，但并不等
于零，则在这一部分边界上须直接应用变分方程，即
?? ??? SZwYvXuU d)δδδ(δ
77
Here,u,v,w are known,The integration only includes this part of
boundary,Substituting the relations between the variation of surface forces
and the variation of stress
zzyzx
yzyxy
zxxyx
nmlZ
nmlY
nml
???
???
???
δδδδ
δδδδ
δδδXδ
???
???
???
into the integration on the right of the equation will result in
m
m
m ABSZwYvXu d??? ??? d)δδδ(
where Bm is a constant,On the other hand,the left of the equation
m
m
AAUU dd ? ???
78
在这里,u,v,w是已知的，积分只在该部分边界。将面
力的变分与应力的变分两者之间的关系，即
zzyzx
yzyxy
zxxyx
nmlZ
nmlY
nml
???
???
???
δδδδ
δδδδ
δδδXδ
???
???
???
代入方程的右边积分后，将得出如下的结果,
m
m
m ABSZwYvXu d??? ??? d)δδδ(
其中 Bm是常数。另一方面，方程的左边
m
m
AAUU dd ? ???
79
So we can get
m
m
BAU ???
Last the type is still a power of the Am distance,have the m altogether,and the
each Am is with each other not related,as a result can get all Am,returning to the
generation( c) type,getthe stress dint weight,
At applied should the dint become to divide the method,to make enactment
of should the dint weight since satisfy should the dint boundary condition,and then
satisfy the square distance of equilibrium differential calculus,this is usually very
difficult.But,exist in the problem of some types should the dint function,and use
should the dint function mean of should the dint weight satisfy the square distance
of equilibrium differential calculus again.At this time,we only the beard enactment
should the expression type of the dint function,make it to of should the dint weight
can satisfy should the dint boundary condition,difficulty reduced consumedly,
80
因而得
m
m
BAU ???
上式仍然是 Am的一次方程，总共有 m个，且各个 Am是互不
相关的，因而可以求出所有的 Am,回代 (c)式，求得应力
分量。
在应用应力变分法时，要使设定的应力分量既满足应
力边界条件，又满足平衡微分方程，这往往是很困难的。
但是，在某些类型的问题中存在着应力函数，而且用应力
函数表示的应力分量又能满足平衡微分方程。这时，我们
就只须设定应力函数的表达式，使它给出的应力分量能满
足应力边界条件，困难就大大减少了。
81
???
m
mmA ??? 0
Where Am are m independent coefficients,The stresses given byφ0
satisfy the exact actual boundary conditions of stress,And the stresses given
by φm satisfy the boundary conditions of stress without surface forces,
Because the number of the components of stress is large,they are hard
to determine,We often use the stress function method,
In the plane stress problems,if the components of body forces are
constants,then stress function exist,We set the stress function as
IV,The stress function method
Under the status of plane stress,the strain energy expressed by the
components of stress is
? ??? ????? yxEU xyyxyx dd)1(222 1 222 ???????
82
???
m
mmA ??? 0
其中 Am为互不依赖的 m个系数。 φ0给出的应力满
足实际的应力边界条件,φm 给出的应力满足无面力时
的应力边界条件。
由于应力分量的数量较多，确定起来有困难，通
常用应力函数方法。
在平面应力问题中，如果体力分量为常数，则存
在应力函数。将应力函数设定为
四 应力函数方法
在平面应力状态，用应力分量表示的形变势能为
? ??? ????? yxEU xyyxyx dd)1(222 1 222 ???????
83
If we consider simply connected body,and it is the problem of
boundary conditions of stress,the components of stress have no relation to ?
and it can be set as zero,Thus two types of plane problems can both be
simplified as
? ??? ??? yxEU xyyx dd22 1 222 ???
Express it through the function of stress
?? ??
?
?
?
?
?
?
???
?
???
?
??
??
???
?
???
?
?
?
??
???
?
???
?
?
?
?? yx
yx
Yy
x
Xx
yE
U dd2
2
1
222
2
22
2
2 ???
For the plane strain problems
? ?? ?? ?d x d yEU xyyxyx?? ?????? 222 22121 ????????
84
如果考虑单连体，且是应力边界问题，应力分量
应与 ?无关，可设其为零。则两类平面问题皆简化为
? ??? ??? yxEU xyyx dd22 1 222 ???
用应力函数表示为
?? ??
?
?
?
?
?
?
???
?
???
?
??
??
???
?
???
?
?
?
??
???
?
???
?
?
?
?? yx
yx
Yy
x
Xx
yE
U dd2
2
1
222
2
22
2
2 ???
对于平面应变问题
? ?? ?? ?d x d yEU xyyxyx?? ?????? 222 22121 ????????
85
Because the surface forces can’t have variation in the problems of stress
boundaries,the variational equation is simplified as dU=0,So the coefficients
satisfy
0???
mA
U
The above equation is a linear system of equations,After solving Am,we
obtain the approximate solution of the function of stress and finally obtain all
the components of stress,
Introducing the expression of the function of stress yields
0dd2
dd
22
2
2
2
2
2
2
2
2
???
?
?
?
?
?
?
??
?
?
?
??
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
???
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
??
??
yx
yxAyx
yx
xA
Yy
xyA
Xx
y
m
mm
??
????
86
在应力边界问题中，因为面力不能有变分，所以变分
方程简化为 δ U=0,因此系数应满足
0???
mA
U
上式为线性方程组，求解 Am后，得到应力函数的近似
解，最后得到各应力分量。
将应力函数的表达式代入，得
0dd2
dd
22
2
2
2
2
2
2
2
2
???
?
?
?
?
?
?
??
?
?
?
??
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
???
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
??
??
yx
yxAyx
yx
xA
Yy
xyA
Xx
y
m
mm
??
????
87
Because it is a approximate solution,the components of tress can’t
satisfy the compatible conditions precisely,and the components of strain
solved from the components of stress can’t satisfy the deformation
compatibility conditions precisely too,and we can’t solve the components
of displacement according to the geometric equation,
We can solve the coefficient Am, Consequently we decide the stress
function ?,through which we solve all the components of stress,
The key of the stress function method is to find the stress function that
satisfy all the boundary conditions,which is still hard to find,especially in
the case that the boundaries are not regular,So the application of stress
method is extremely restricted on this aspect,
88
由于是近似解，应力分量不能精确满足相容条
件，由应力分量求得的应变分量也不能精确满足变
形协调条件，不能根据几何方程求得位移分量。
由上式即可解得系数 Am。从而确定应力函数 ?，
再由应力函数 ?求得各应力分量。
应力函数法的要点是要找到满足全部边界条件
的应力函数，而这种函数一般仍然难以找到，尤其
在边界不规整的情况下。所以应力方法的应用在这
一点上受到极大的限制。
89
Example 3,Suppose there is a rectangular thin plate whose body forces are
ignored,Its two opposite sides are subjected to tensile forces with the
distribution of parabola,whose maximum intensity is q,as shown in the
figure,Please solve the stress of the plate,
Solution,In the coordinates as shown in
the figure,the boundary conditions are,
? ? ? ?
? ? ? ? 0,0
0,1
2
2
??
???
?
?
??
?
?
??
????
????
byxybyy
axxyaxx b
y
q
??
??
Take
2
2
22
2
2
2
12
2
2 11
612 ?
?
?
?
???
? ?
???
?
???
? ??
???
?
???
? ??
b
y
a
xqbA
b
yyq?
x
y
a a
b
b
q
90
例题 3：设有矩形薄板，体力不计，在两对边上受有按抛物
线分布的拉力，其最大集度为 q, 图示。试求板的应力。
解：在图示坐标下，边界条
件是
? ? ? ?
? ? ? ? 0,0
0,1
2
2
??
???
?
?
??
?
?
??
????
????
byxybyy
axxyaxx b
y
q
??
??
取
2
2
22
2
2
2
12
2
2 11
612 ?
?
?
?
???
? ?
???
?
???
? ??
???
?
???
? ??
b
y
a
xqbA
b
yyq?
x
y
a a
b
b
q
91
???
?
???
? ??
2
2
2
0 612 b
yyq?Then from we can obtain the components of stress
? ? ? ? ? ? 0,0,1 0
2
02
0
2
02
2
2
0
2
0 ???
????
?
??
???
?
???
? ??
?
??
yxxb
yq
y xyyx
??????
which satisfy the boundary conditions,And the stresses corresponding to
2
2
22
2
2
2
1 11 ???
?
???
? ?
???
?
???
? ??
b
y
a
xqb?
can satisfy the boundary conditions without surface forces,
Substitute ? and the body forces X=Y=0 into the following equation
0dd2
dd
22
2
2
2
2
2
2
2
2
???
?
?
?
?
?
?
??
?
?
?
??
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
???
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
??
??
yx
yxAyx
yx
xA
Yy
xyA
Xx
y
m
mm
??
????
92
???
?
???
? ??
2
2
2
0 612 b
yyq?
则由 给出的应力分量
? ? ? ? ? ? 0,0,1 0
2
02
0
2
02
2
2
0
2
0 ???
????
?
??
???
?
???
? ??
?
??
yxxb
yq
y xyyx
??????
满足边界条件。而由 2
2
22
2
2
2
1 11 ???
?
???
? ?
???
?
???
? ??
b
y
a
xqb?
所对应的应力能满足无面力时的边界条件。
将 ?以及体力 X=Y=0代入下式中
0dd2
dd
22
2
2
2
2
2
2
2
2
???
?
?
?
?
?
?
??
?
?
?
??
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
???
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
??
??
yx
yxAyx
yx
xA
Yy
xyA
Xx
y
m
mm
??
????
93
Integrating and simplifying yields
076449256764 14
4
2
2
???
?
?
???
? ?? A
a
b
a
b
In order to be simple,let a=b,then 0425.0
1 ?A
Substitute it into ?,and let a=b,then solve the components of stress
22
2
2
22
2
2
2
2
2
2
2
2
2
2
2
2
2
2
2
2
1168.0
1
3
117.0
3
1117.01
a
xy
a
y
a
x
q
yx
a
y
a
x
q
x
a
y
a
x
q
a
y
q
y
xy
y
x
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
???
??
?
??
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
???
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
??
?
?
?
?
?
?
?
?
??
?
?
?
?
?
?
?
?
?
94
进行积分，并简化以后得
076449256764 14
4
2
2
???
?
?
???
? ?? A
a
b
a
b
为简单起见，令 a=b，则 0425.0
1 ?A
代入 ?中，并令 a=b，再求应力分量得
22
2
2
22
2
2
2
2
2
2
2
2
2
2
2
2
2
2
2
2
1168.0
1
3
117.0
3
1117.01
a
xy
a
y
a
x
q
yx
a
y
a
x
q
x
a
y
a
x
q
a
y
q
y
xy
y
x
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
???
??
?
??
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
???
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
??
?
?
?
?
?
?
?
?
??
?
?
?
?
?
?
?
?
?
95
Exercise 10.1 A square thin plate in vertical plane,with the length of side of
2a,is fixed at its four sides and is only subjected to the action of gravity,Let
?=0,Take the expressions of the components of displacement as
?
?
?
?
?
?
?
?
?????
?
?
?
?
?
?
???
?
?
?
?
?
?
??
?
?
?
?
?
?
?
?
?????
?
?
?
?
?
?
???
?
?
?
?
?
?
??
?
?
2
2
32
2
212
2
2
2
2
2
32
2
212
2
2
2
11
11
a
y
B
a
x
BB
a
y
a
x
v
a
y
A
a
x
AA
a
y
a
x
a
y
a
x
u
Solve the components of displacement(take the term A1 and B1) the
components of stress with the Ritz method,
a a
a
a x
y
Solution,When we take the term A1 and B1 only
???
?
???
? ?
???
?
???
? ??
???
?
???
? ?
???
?
???
? ??
2
2
2
2
122
2
2
2
1 11,11 a
y
a
xBv
a
xy
a
y
a
xAu
96
练习 10.1 铅直平面内的正方形薄板，边长为 2a，四边固
定，只受重力作用，设,试取位移分量的表达式为
?
?
?
?
?
?
?
?
?????
?
?
?
?
?
?
???
?
?
?
?
?
?
??
?
?
?
?
?
?
?
?
?????
?
?
?
?
?
?
???
?
?
?
?
?
?
??
?
?
2
2
32
2
212
2
2
2
2
2
32
2
212
2
2
2
11
11
a
y
B
a
x
BB
a
y
a
x
v
a
y
A
a
x
AA
a
y
a
x
a
y
a
x
u
用瑞次法求解位移分量 (取 A1项及 B1项 )及应力分量。
a a
a
a x
y
0??
解,当只取 A1项及 B1项时,
???
?
???
? ?
???
?
???
? ??
???
?
???
? ?
???
?
???
? ??
2
2
2
2
122
2
2
2
1 11,11 a
y
a
xBv
a
xy
a
y
a
xAu
97
gYX ???,0
The strain energy
d x d y
y
u
x
v
y
v
x
uEU a a? ?
?
?
?
?
?
?
?
?
???
?
???
?
?
??
?
??
???
?
???
?
?
???
?
??
?
?
?
??
0 0
222
2
1
2
Now we evaluate and,
1A
U
?
?
1B
U
?
?
?
?
?
?
?
?
??
?
?
?
?
?
?
??
?
?
??
?
?
?
?
?
?
?
?
?
??
?
?
??
?
?
?
?
?
?
?
???
?
?
??
?
?
?
?
?
?
?
?
??
?
?
?
?
?
?
?
?
?
?
?
?
?
?
? ?
37
6
75
32
22
2
11
0 0
1111
BAE
d x d y
y
u
x
v
Ay
u
x
v
y
v
Ay
v
x
u
Ax
uE
A
U a a
?
?
?
?
?
?
??
?
?
?
?
?
?
??
?
?
??
?
?
?
?
?
?
?
?
?
??
?
?
??
?
?
?
?
?
?
?
???
?
?
??
?
?
?
?
?
?
?
?
??
?
?
?
?
?
?
?
?
?
?
?
?
?
?
? ?
5
6
45
32
22
2
1
1
0 0
1111
A
B
E
d x d y
y
u
x
v
By
u
x
v
y
v
By
v
x
u
Bx
uE
B
U a a
98
gYX ???,0
形变势能
d x d y
y
u
x
v
y
v
x
uEU a a? ?
?
?
?
?
?
?
?
?
???
?
???
?
?
??
?
??
???
?
???
?
?
???
?
??
?
?
?
??
0 0
222
2
1
2
现计算 和
1A
U
?
?
1B
U
?
?
?
?
?
?
?
?
??
?
?
?
?
?
?
??
?
?
??
?
?
?
?
?
?
?
?
?
??
?
?
??
?
?
?
?
?
?
?
???
?
?
??
?
?
?
?
?
?
?
?
??
?
?
?
?
?
?
?
?
?
?
?
?
?
?
? ?
37
6
75
32
22
2
11
0 0
1111
BAE
d x d y
y
u
x
v
Ay
u
x
v
y
v
Ay
v
x
u
Ax
uE
A
U a a
?
?
?
?
?
?
??
?
?
?
?
?
?
??
?
?
??
?
?
?
?
?
?
?
?
?
??
?
?
??
?
?
?
?
?
?
?
???
?
?
??
?
?
?
?
?
?
?
?
??
?
?
?
?
?
?
?
?
?
?
?
?
?
?
? ?
5
6
45
32
22
2
1
1
0 0
1111
A
B
E
d x d y
y
u
x
v
By
u
x
v
y
v
By
v
x
u
Bx
uE
B
U a a
99
When applying the Ritz method,we expect
???
???
??
?
?
??
?
?
dSvYd x d yYv
B
U
dSuXd x d yXu
A
U
mm
m
mm
m
From
0
1
???AU
2
2
2
0 0 2
2
1 9
16114 agd x d y
a
y
a
xg
B
U a a ?? ?
???
?
???
? ?
???
?
???
? ??
?
? ? ?
Solving it yields
E
gaB
E
gaA 2
1
2
1 533
225,
1066
175 ?? ??
100
在用瑞次法时，要求
???
???
??
?
?
??
?
?
dSvYd x d yYv
B
U
dSuXd x d yXu
A
U
mm
m
mm
m
由
0
1
???AU
2
2
2
0 0 2
2
1 9
16114 agd x d y
a
y
a
xg
B
U a a ?? ?
???
?
???
? ?
???
?
???
? ??
?
? ? ?
解之得
E
gaB
E
gaA 2
1
2
1 533
225,
1066
175 ?? ??
101
Therefore
2
2
2
2
2
2
2
2
2
11
5 3 3
2 2 5
11
1 0 6 6
1 7 5
a
a
y
a
x
E
g
v
xy
a
y
a
x
E
g
u
?
?
?
?
?
?
?
?
???
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?
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?
?
?
??
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???
?
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?
?
??
?
?
?
2
2
2
2
2
2
1
533
450
1
3
1
1 0 6 6
175
a
x
gy
y
v
E
y
a
y
a
x
g
x
u
E
y
x
??
??
?
?
?
?
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?
????
?
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???
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??
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??
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?
?
2
2
2
2
2
2
1
533
2253
11
2 1 3 2
175
2
a
y
gx
a
y
a
x
gx
x
v
y
uE
xy
??
?
102
所以
2
2
2
2
2
2
2
2
2
11
5 3 3
2 2 5
11
1 0 6 6
1 7 5
a
a
y
a
x
E
g
v
xy
a
y
a
x
E
g
u
?
?
?
?
?
?
?
?
???
?
?
?
?
?
?
??
?
?
?
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???
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??
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???
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?
??
?
?
?
2
2
2
2
2
2
1
533
450
1
3
1
1 0 6 6
175
a
x
gy
y
v
E
y
a
y
a
x
g
x
u
E
y
x
??
??
?
?
?
?
?
?
?
?
????
?
?
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???
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??
??
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?
2
2
2
2
2
2
1
533
2253
11
2 1 3 2
175
2
a
y
gx
a
y
a
x
gx
x
v
y
uE
xy
??
?
103
Exercise 10.2 Show that the freely supported beam is subjected to uniform
load q, The length of the beam is l,The equation of equilibrium of the beam
is
Assume that the equation of line of
deflection of the beam is
04
4
?? qdx wdEI
l
xnaw
n
n
?s i n
5,3,1
?
?
?
?
Solve the line of deflection of the beam with the Galerkin method,
q
x
y
l
Solution,The boundary conditions of the beam is
? ? ? ? 0,00 ?? ?? lxx ww
So the equation of line of deflection satisfies the boundary conditions,
104
练习 10.2 简支梁受均布荷载 q，梁长为 (图示 )。梁的平衡
方程为
设梁的挠曲线方程为
04
4
?? qdx wdEI
l
l
xnaw
n
n
?s i n
5,3,1
?
?
?
?
试用伽辽金法求梁的挠曲线。
解,梁的边界条件为
? ? ? ? 0,00 ?? ?? lxx ww
故挠曲线方程满足边界条件。
q
x
y
l
105
The equation of equilibrium of the beam is
04
4
?? qdx wdEI
Substituting it into the Galerkin equation yields
0s ins in
0
4
?
?
?
?
?
?
?
?
?
??
?
??
?
?? dx
l
xnq
l
xn
l
nE I al
n
???
After integrating we arrive at the undetermined coefficient
EIn
qla
n 55
44
??
So the equation of line of deflection is
l
xn
nEI
qlw
n
?
?
s i n14
5,3,1
55
4
?
?
?
?
?
106
梁的平衡方程为 0
4
4
?? qdx wdEI
代入伽辽金方程，得
0s ins in
0
4
?
?
?
?
?
?
?
?
?
??
?
??
?
?? dx
l
xnq
l
xn
l
nE I al
n
???
积分后得待定系数
EIn
qla
n 55
44
??
故挠曲线方程为
l
xn
nEI
qlw
n
?
?
s i n14
5,3,1
55
4
?
?
?
?
?
107
Apparently,the assumed equation of line of deflection of the beam
3322 xaxaw ??
Evaluate the coefficients a2 and a3 with the principle of minimum energy,
The strain energy is
Solution,Assume that the line of deflection of the beam is
3322 xaxaw ??
? ? 0,0
0
0 ???
??
?
??
?
?
x
x dx
dwwIts boundary condition is
Exercise 10.3 The cantilevered beam is
subjected to the action of the concentrated
force P at free end as shown in the figure,
Evaluate the maximum deflection with the
principle of minimum energy,
P
l
x
y
? ? dxxaaEIdx
dx
wdEIU ll 2
0 32
2
0 2
2
62
22 ??
????
?
?
???
??
satisfy the boundary conditions,
108
显然所设梁的挠曲线方程 满足边界条件。 3
322 xaxaw ??
由极小势能原理求系数 a2和 a3 。形变势能为
解,设梁的挠曲线为 3
322 xaxaw ??
? ? 0,0
0
0 ???
??
?
??
?
?
x
x dx
dww其边界条件为
练习 10.3 悬臂梁在自由端受集中
力 P作用，如图所示。试用极小势
能原理求最大挠度。
P
l
x
y
? ? dxxaaEIdx
dx
wdEIU ll 2
0 32
2
0 2
2
62
22 ??
????
?
?
???
??
109
The potential energy of the external forces is
? ? ? ?3322 lalaPwPV lx ????? ?
Then the total potential energy is
? ? ? ?33222
0 32
622 lalaPdxxaaEIVU L ??????? ?
Apply the principle of minimum potential energy
0
0
3
3
2
2
?
?
??
?
?
??
??
a
a
a
a
dd
d
Then ? ?
06242 2
0 322
????? ?? ? PldxxaaEIa L
110
外力势能为
? ? ? ?3322 lalaPwPV lx ????? ?
则总势能为
? ? ? ?33222
0 32
622 lalaPdxxaaEIVU L ??????? ?
应用极小势能原理
0
0
3
3
2
2
?
?
??
?
?
??
??
a
a
a
a
dd
d
则 ? ?
06242 2
0 322
????? ?? ? PldxxaaEIa L
111
? ? 062122 3
0 323
????? ?? ? PldxxaaxEIa L
Integrating yields
EI
Pl
lala
EI
Pl
lala
6
2
2
32
2
2
32
2
2
32
??
??
From the above two equations we can obtain
EI
Pa
EI
Pla
6,2 32 ???
So the line of deflection is
The value of the maximum deflection is
?????? ?? 236 2 xxEIPlw
EI
Plw
3
3
m a x ?
112
? ? 062122 3
0 323
????? ?? ? PldxxaaxEIa L
积分得
EI
Pl
lala
EI
Pl
lala
6
2
2
32
2
2
32
2
2
32
??
??
由上述两方程解得
EI
Pa
EI
Pla
6,2 32 ???
故挠曲线为
最大挠度值为
?????? ?? 236 2 xxEIPlw
EI
Plw
3
3
m a x ?
113
114
Elasticity
2
3
§ 10-1 The Specific Energy of Deformation and Strain
Energy of Elastic Body
§ 10-3 The Variational Method of Displacement
§ 10-4 The Variational Equations of Stress and the
Variational Method of Stress
§ 10-2 The Variational Equation of Displacement and
The Principle of Minimum Potential Energy
Chapter 10 The Energy
Principles and theVariational Calculus
4
第十章 能量原理与变分法
§ 10-1 弹性体的变形比能与形变势能
§ 10-3 位移变分法
§ 10-4 应力变分方程与应力变分方法
§ 10-2 位移变分方程与极小势能原理
5
The problems of the elastic theory need a series of partial differential
equations,It always meet with difficulties in math,So we have to seek for
approximate solution,The approximate solution of variational calculus is a
method in common use,In math,variational problems are the problems to
solve the limit of the functional,In elastic mechanics,the functional is the
energy(work) of elasticity problems,And the variational calculus is to
calculate the extremum of the energy(work),We can yield the solution of
the elastic problems when calculating the extremum,The direct method of
variation problems facilitate us to obtain the approximate solution,
We will give the expressions which calculate the potential energy of
strain at first in this chapter,We will mainly introduce the variational method
of displacement and that of stress through the relation of work and energy,
6
弹性理论问题需要解一系列偏微分方程组，并满足
边界条件，这在数学上往往遇到困难。因此需要寻求近
似的解法。变分法的近似解法是常用的一种方法。在数
学上，变分问题是求泛函的极限问题。在弹性力学里，
泛函就是弹性问题中的能量（功），变分法是求能量
（功）的极值，在求极值时得到弹性问题的解，变分问
题的直接法使我们比较方便地得到近似解。
本章首先给出计算形变势能的表达式。利用功与能
的关系，主要介绍了位移变分法和应力变分法。
7
§ 10-1 The Specific Energy of
Deformation and Strain Energy of Elastic Body
? ? ? ? ? ?? ?? ?2222221 1222 1 xyzxyzyxxzzyzyxEU ?????????????? ??????????
The specific energy expressed by the components of stress
I,The specific energy of deformation
? ?xyxyzxzxyzyzzzyyxxU ???????????? ?????? 211
Under the complex stress conditions,suppose that the elastic body is
imposed all the six components of stress,
According to the principle of conservation of energy,the value of the strain
energy has no relation to the sequence of the forces imposed on the elastic
body,but is completely decided by the final value of the stress and strain,
Thus we obtain the strain energy density or specific energy of the elastic body,
xyzxyzzyx ??????,,,,,
8
§ 10-1 弹性体的变形比能与形变势能
一 变形比能
在复杂应力状态下，设弹性体受有全部六个应力
分量 。根据能量守恒定理，形变
势能的多少与弹性体受力的次序无关，而完全确定于
应力及形变的最终大小。从而有弹性体的形变势能密
度或比能,
xyzxyzzyx ??????,,,,,
? ?xyxyzxzxyzyzzzyyxxU ???????????? ?????? 211
? ? ? ? ? ?? ?? ?2222221 1222 1 xyzxyzyxxzzyzyxEU ?????????????? ??????????
比能用应力分量表示
9
The specific energy expressed by the components of strain,
? ? ? ? ? ???
?
??
? ??????
???
2222222
1 2
1
2112 xyzxyzzyxe
EU ??????
?
?
?
Where
zyxe ??? ???
Therefore,we have the partial differentiation of specific energy by
components of stress
z
z
y
y
x
x
UUU ?
????? ??
??
?
??
?
? 111,,
xy
xy
zx
zx
yz
yz
UUU ?
????? ??
??
?
??
?
? 111,,
10
比能用应变分量表示
? ? ? ? ? ???
?
??
? ??????
???
2222222
1 2
1
2112 xyzxyzzyxe
EU ??????
?
?
?
其中
zyxe ??? ???
因此，我们有比能对应力分量的偏导
z
z
y
y
x
x
UUU ?
????? ??
??
?
??
?
? 111,,
xy
xy
zx
zx
yz
yz
UUU ?
????? ??
??
?
??
?
? 111,,
11
xy
xy
zx
zx
yz
yz
UUU ?
????? ??
??
?
??
?
? 111,,
z
z
y
y
x
x
UUU ?
????? ??
??
?
??
?
? 111,,
And the partial differentiation of specific energy by components of strain
II,Strain energy
Since the components of stress and components of strain and then the
specific energy are all the function of coordinates,the strain
energy, of the whole elastic body is,1UU
???? d x d y d zUU 121
Substituting the three kinds of expressing forms of specific energy
yields three kinds of integrating forms of strain energy
12
xy
xy
zx
zx
yz
yz
UUU ?
????? ??
??
?
??
?
? 111,,
z
z
y
y
x
x
UUU ?
????? ??
??
?
??
?
? 111,,
比能对应变分量的偏导
二 形变势能
由于应力分量和形变分量，进而比能 都是位置
坐标的函数，所以整个弹性体的形变势能 为,1
U
U
???? d x d y d zUU 121
将比能的三种表达形式代入，得形变势能的三种
积分形式
13
zyx
y
u
x
v
x
w
z
u
z
v
y
w
z
w
y
v
x
u
z
w
y
v
x
uE
U
ddd
2
1
2
1
2
1
21)1(2
222
2222
?
?
?
?
??
?
?
??
?
?
???
?
?
?
?
?
????
?
?
??
?
?
??
?
?
?
?
?
?
???
?
?
??
?
?
??
?
?
?
?
?
???
?
?
??
?
?
???
?
?
??
? ???
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
Substituting the geometric equations can also yield the strain energy
expressed by displacement
? ???? ?????? d x d y d zU xyxyzxzxyzyzzzyyxx ????????????21
? ? ? ? ? ? d x d y d ze
EU
xyzxyzzyx??? ??
?
??
? ??????
???
2222222
2
1
2112 ???????
?
?
? ? ? ??
? ?? ?? d x d y d z
E
U
xyzxyz
yxxzzyzyx
222
222
12
2
2
1
????
??????????
????
??????? ???
14
zyx
y
u
x
v
x
w
z
u
z
v
y
w
z
w
y
v
x
u
z
w
y
v
x
uE
U
ddd
2
1
2
1
2
1
21)1(2
222
2222
?
?
?
?
??
?
?
??
?
?
???
?
?
?
?
?
????
?
?
??
?
?
??
?
?
?
?
?
?
???
?
?
??
?
?
??
?
?
?
?
?
???
?
?
??
?
?
???
?
?
??
? ???
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
将几何方程代入，形变势能还可用位移分量来表示
? ???? ?????? d x d y d zU xyxyzxzxyzyzzzyyxx ????????????21
? ? ? ? ? ? d x d y d ze
EU
xyzxyzzyx??? ??
?
??
? ??????
???
2222222
2
1
2112 ???????
?
?
? ? ? ??
? ?? ?? d x d y d z
E
U
xyzxyz
yxxzzyzyx
222
222
12
2
2
1
????
??????????
????
??????? ???
15
I.Variation and its properties,
We have learned in higher mathematics the concept of differentiation
which is the increment of variable,Then what is the variation? Variation is
the increment of function,often denoted by d,Variation has the following
properties,
? ??
?
?
?
?
?
???
Suu d S
u
xx
u
wuwu
dδδ
δδ
δδ)(δ
§ 10-2 The Variational Equation of Displacement
and The Principle of Minimum Potential Energy
16
一 变分及其性质
高等数学我们学过微分的概念，微分是变量的
增量。那么什么是变分呢？变分是函数的增量，通
常用 δ 表示。变分具有以下的性质,
? ??
?
?
?
?
?
???
Suu d S
u
xx
u
wuwu
dδδ
δδ
δδ)(δ
§ 10-2 位移变分方程与极小势能原理
17
II,The variational equation of displacement
Suppose that the elastic body is under the status of equilibrium when
imposed some external forces,It undergoes the true displacement u,v,w,
which satisfy the equations of equilibrium and satisfy the boundary
conditions of displacement and the stress boundary conditions expressed by
the displacement,Now suppose that the components of displacement
undergo minute changes(virtual displacement) du,dv,dw which the
boundary conditions of displacement allow,Now the extermal forces do
virtual work during the virtual displacement which should be equal to the
increment of the functional of the strain energy,
? ? ? ?dSwZvYuXd x d y d zwZvYuXU ??? ?? ?????? ddddddd
This equation is the so-called variational equation of displacement,where
X,Y,Z are the components of body forces,and the are the
components of the surface faces,ZYX,,
18
二 位移变分方程
设弹性体在一定外力作用下，处于平衡状态，发
生的真实位移为 u,v,w，它们满足位移分量表示的平
衡方程，并满足位移边界条件和用位移表示的应力边
界条件。现在假设位移分量发生了位移边界条件所容
许的微小改变（虚位移） δ u, δ v,δ w，这时外力
在虚位移上作虚功，虚功应和变形能泛函的增加相等，
即
? ? ? ?dSwZvYuXd x d y d zwZvYuXU ??? ?? ?????? ddddddd
这个方程就是所谓位移变分方程。其中 X,Y,Z为体力分
量,为面力分量。 ZYX,,
19
III,The principle of the minimum potential energy
Because the virtual displacement is very small,the value and the direction
of the external forces can be regarded as invariable during the virtual
displacement,and only its acting points are changed,According to the
properties of the variation,the variational equation of displacement can be
rewrite as,
? ? ? ?? ? 0??????? ??? ?? dSwZvYuXd x d y d zZwYvXuUd
? ? ? ?dSwZvYuXd x d y d zZwYvXuV ??? ?? ???????
Suppose the potential energy of external forces is,
? ? 0?? VUdThen
The meaning of this equation is that under the action of given external
forces,among all groups of displacements that satisfy the boundary conditions
of displacement,the group of displacement that exist actually should make
the total potential energy a minimum,If considering the second order variation,
we can further analyze it and prove that this extremum is a minimum for the
stable equilibrium status,So this equation is also called the principle of
minimum potential energy,
20
三 极小势能原理
由于虚位移是微小的，因此在虚位移的过程中，
外力的大小和方向可以当做保持不便，只是作用点有
了改变。利用变分的性质，位移变分方程可改写为,
? ? ? ?? ? 0??????? ??? ?? dSwZvYuXd x d y d zZwYvXuUd
? ? ? ?dSwZvYuXd x d y d zZwYvXuV ??? ?? ???????
? ? 0?? VUd
设外力势能为
则
该式的意义是：在给定的外力作用下，在满足位
移边界条件的各组位移中，实际存在的一组位移应使
总势能为极值。如果考虑二阶变分，进一步的分析证
明，对于稳定平衡状态，这个极值是极小值。因此，
该式又称为极小势能原理。
21
Apparently,the displacement existing exactly should satisfy the
equations of equilibrium and the boundary conditions of stress expressed by
displacement besides boundary conditions of displacement,Now we can see
also that the displacement existing exactly still satisfy the variational
equations of displacement besides boundary conditions of displacement,
Furthermore,through calculating,we can deduce differential equations of
equilibrium and boundary conditions of stress expressed by displacement
from variational equations of displacement,So it is obvious that variational
equations of displacement can substitute for differential equations of
equilibrium and boundary conditions of stress,
22
显然，实际存在的位移，除了满足位移边界条件
以外，还应当满足位移表示的平衡方程和应力边界条
件；现在又看到，实际存在的位移，除了满足位移边
界条件外，还满足位移变分方程。而且，通过运算，
还可以从位移变分方程导出用位移表示的平衡微分方
程和应力边界条件。于是可见：位移变分方程可以代
替平衡微分方程和应力边界条件。
23
§ 10-3 The Variational Method of Displacement
??? ??????
m
mm
m
mm
m
mm wCwwvBvvuAuu 000,,
where u0,v0,w0 are undetermined coefficients whose boundary values
equal to the known displacement on the boundaries,um, vm,wm are the
set functions whose boundary conditions equal to zero,Am,Bm,Cm are
undetermined coefficients,by whose variation the variation of
displacement are realized,
We set expressions of components of displacement that satisfy the
boundary conditions of displacement,which contain several undetermined
coefficients,Then we decide these coefficients according to the principle
of minimum potential energy,Take the expression of components of
displacement as following,
I,The Ritz method
24
§ 10-3 位移变分法
??? ??????
m
mm
m
mm
m
mm wCwwvBvvuAuu 000,,
其中 u0,v0,w0 为设定的函数，它们的边界值等于边界
上的已知位移； um, vm,wm 为边界值等于零的设定
函数,Am,Bm,Cm为待定的系数，位移的变分由它们
的变分来实现。
先设定满足位移边界条件的位移分量的表达式，
其中包含若干个待定的系数，再根据极小势能原理，
决定这些系数。取位移分量的表达式如下,
一 瑞次法
25
??? ??? m mmm mmm mm CwwBvvAuu δ,δδ,δδδ
? ??? )δδδ(δ m
m
m
m
m
m
CCUBBUAAUU ??????
The variation of strain energy is
The variation of potential energy of external forces is
? ??
? ???
???
????
m
mmmmmm
m
mmmmmm
SCwZBvYAuX
zyxCZwBYvAXuV
d)δδδ(
ddd)δδδ(δ
The variations of components of displacement are
26
??? ??? m mmm mmm mm CwwBvvAuu δ,δδ,δδδ
? ??? )δδδ(δ m
m
m
m
m
m
CCUBBUAAUU ??????
应变能的变分为
外力势能的变分为
? ??
? ???
???
????
m
mmmmmm
m
mmmmmm
SCwZBvYAuX
zyxCZwBYvAXuV
d)δδδ(
ddd)δδδ(δ
位移分量的变分是
27
yields
Above system of linear algebraic equations includes 3m equations,
After solving these equations,we introduce the results into the
expressions of the components of displacement to obtain the approximate
solutions of the components of displacement,This method is called the
Ritz method,
Substituting into
?????
?????
?????
??
??
??
SwZzyxZw
C
U
SvYzyxYv
B
U
SuXzyxXu
A
U
mm
m
mm
m
mm
m
dddd
dddd
dddd
?
?
?
?
?
?
? ? 0?? VUd
28
中，得到
上面是个数为 3m的线性代数方程组，求解后，代回
位移分量的表达式，得到位移分量的近似解。这种
方法称为瑞次法。
代入
?????
?????
?????
??
??
??
SwZzyxZw
C
U
SvYzyxYv
B
U
SuXzyxXu
A
U
mm
m
mm
m
mm
m
dddd
dddd
dddd
?
?
?
?
?
?
? ? 0?? VUd
29
II,The Galerkin method
If we regard variation as the function of the components of strain,
then
d x d y d zUUd x d y d zUU yz
yz
x
x
?????? ???????? ????????? ?? d??d??dd 111
Because
.,,,;,,,11 ???? vzwyuxUU yzxyz
yz
x
x
ddd?dd???? ???????????????
so
??? ?????? ????????? ?????????? d x d y d zvzwyuxU yzx ?? dd?d?d
Applying the Ostrogradsky—Gauss formula to the first term,we have
30
二 伽辽金法
将变分看做形变分量的函数，则
d x d y d zUUd x d y d zUU yz
yz
x
x
?????? ???????? ????????? ?? d??d??dd 111
由于
.,,,;,,,11 ???? vzwyuxUU yzxyz
yz
x
x
ddd?dd???? ???????????????
所以
??? ?????? ????????? ?????????? d x d y d zvzwyuxU yzx ?? dd?d?d
应用奥高公式，对上式中的第一项，我们有
31
? ?
?? ???
?????? ???
?
?
??
?
?
?
?
?
?
?
?
u d x d y d z
x
u d Sl
u d x d y d z
x
d x d y d zu
x
u d x d y d z
x
x
x
x
xx
d
?
d?
d
?
d?d?
We arrange other terms likewise,then
Substituting the above expression into the variational equation and
combining the similar terms yields,
? ? ? ? ? ?? ?
d x d y d zw
yxz
v
xzy
u
zyx
dSwmlnvlnmunmlU
yzzxzxyyzyzxxyx
yzzxzxyyzyzxxyx
???
??
?
?
?
?
?
?
?
?
??
?
?
??
?
?
?
?
?
?
?
?
?
?
???
?
?
??
?
?
?
?
?
?
?
?
?
?
???
?
?
??
?
?
?
?
?
?
?
?
?
?
?
?????????
d
???
d
???
d
???
d???d???d???d
32
? ?
?? ???
?????? ???
?
?
??
?
?
?
?
?
?
?
?
u d x d y d z
x
u d Sl
u d x d y d z
x
d x d y d zu
x
u d x d y d z
x
x
x
x
xx
d
?
d?
d
?
d?d?
对于其余各项也进行同样的处理，则
? ? ? ? ? ?? ?
d x d y d zw
yxz
v
xzy
u
zyx
dSwmlnvlnmunmlU
yzzxzxyyzyzxxyx
yzzxzxyyzyzxxyx
???
??
?
?
?
?
?
?
?
?
??
?
?
??
?
?
?
?
?
?
?
?
?
?
???
?
?
??
?
?
?
?
?
?
?
?
?
?
???
?
?
??
?
?
?
?
?
?
?
?
?
?
?
?????????
d
???
d
???
d
???
d???d???d???d
将上式代入位移变分方程，并归项得
33
? ??
? ? ? ? ? 0?????????
????
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
??
???
dSwZmlnvYlnm
uXnmld x d y d zwZ
yxz
vY
xzy
uX
zyx
yzzxzxyyzy
zxxyx
yzzxz
xyyzyzxxyx
d???d???
d???d
???
d
???
d
???
If the boundary conditions of stress can be satisfied,then the above
equation can be simplified as
0?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
???
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
???
d x d y d zwZ
yxz
vY
xzy
uX
zyx
yzzxz
xyyzyzxxyx
d
???
d
???
d
???
This is the equation that the variation of displacement should satisfy when
the components of displacement satisfy the boundary conditions of
displacement and that of stress,It is called Galerkin variational equation,
34
? ??
? ? ? ? ? 0?????????
????
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
??
???
dSwZmlnvYlnm
uXnmld x d y d zwZ
yxz
vY
xzy
uX
zyx
yzzxzxyyzy
zxxyx
yzzxz
xyyzyzxxyx
d???d???
d???d
???
d
???
d
???
如果应力边界条件得到满足，则上式简化为
0?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
???
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
???
d x d y d zwZ
yxz
vY
xzy
uX
zyx
yzzxz
xyyzyzxxyx
d
???
d
???
d
???
这就是位移分量满足位移边界条件及应力边界条件时，
位移变分所应满足的方程，称为伽辽金变分方程。
35
??? ??????
m
mm
m
mm
m
mm wCwwvBvvuAuu 000,,
If we take the expression of the components of displacement as following
??? ???
m
mm
m
mm
m
mm CwwBvvAuu δ,δδ,δδδ
so that the boundary conditions of displacement and that of stress are both
satisfied, Then we substitute the variation of displacement
into the Galerkin equation,and we obtain
d x d y d zwZ
yxz
C
d x d y d zvY
xzy
B
d x d y d zuX
zyx
A
m
yzzxz
m
m
m
xyyzy
m
m
m
zxxyx
m
m
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
? ???
? ???
? ???
???
d
???
d
???
d
36
??? ??????
m
mm
m
mm
m
mm wCwwvBvvuAuu 000,,
若取位移分量的表达式如下,
??? ???
m
mm
m
mm
m
mm CwwBvvAuu δ,δδ,δδδ
使得位移边界条件和应力边界条件都得到满足，则将
位移变分
代入伽辽金方程，就得到
d x d y d zwZ
yxz
C
d x d y d zvY
xzy
B
d x d y d zuX
zyx
A
m
yzzxz
m
m
m
xyyzy
m
m
m
zxxyx
m
m
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
? ???
? ???
? ???
???
d
???
d
???
d
37
Because are arbitrary,their coefficients should be zero
separately,So we obtain
mmm CBA ddd,,
0
0
0
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
???
???
???
d x d y d zwZ
yxz
d x d y d zvY
xzy
d x d y d zuX
zyx
m
yzzxz
m
xyyzy
m
zxxyx
???
???
???
Expressing the components of stress of the above three equations by the
components of strain through the physical equations,then expressing
them by the components of displacement through the geometric
equations,and simplifying yields
38
由于
mmm CBA ddd,,
的任意性，它们的系数应当分别为零
于是得
0
0
0
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
???
???
???
d x d y d zwZ
yxz
d x d y d zvY
xzy
d x d y d zuX
zyx
m
yzzxz
m
xyyzy
m
zxxyx
???
???
???
将上列三方程中的应力分量通过物理方程用形变分量表
示，再通过几何方程用位移分量表示，简化后即得,
39
? ?
? ?
? ?
0
21
1
12
0
21
1
12
0
21
1
12
2
2
2
??
?
?
?
?
?
?
?
?
?
?
?
?
?
?
??
?
?
??
??
?
?
?
?
?
?
?
?
?
?
?
?
?
?
??
?
?
??
??
?
?
?
?
?
?
?
?
?
?
?
?
?
?
??
?
?
??
???
???
???
d x d y d zwZw
z
eE
d x d y d zvYv
y
eE
d x d y d zuXu
x
eE
m
m
m
??
??
??
Thus we obtain the linear system of equations about undetermined
coefficient of the function of displacement,After solving the equations,we
substitute the results into the expression of the components of displacement,
then we obtain the approximate solution of the components of displacement,
This method is called the Galerkin method,
What we should pay attention to is that the components of displacement
solved through the variational method of displacement can achieve high
precision if we take only a few terms,However,the stresses solved through
this method are not precise,So we must take more terms to obtain
sufficiently precise stress,
40
? ?
? ?
? ?
0
21
1
12
0
21
1
12
0
21
1
12
2
2
2
??
?
?
?
?
?
?
?
?
?
?
?
?
?
?
??
?
?
??
??
?
?
?
?
?
?
?
?
?
?
?
?
?
?
??
?
?
??
??
?
?
?
?
?
?
?
?
?
?
?
?
?
?
??
?
?
??
???
???
???
d x d y d zwZw
z
eE
d x d y d zvYv
y
eE
d x d y d zuXu
x
eE
m
m
m
??
??
??
这样就得到位移函数待定常数的线性方程组，求解
后，代回位移分量的表达式，得到位移分量的近似解。
这种方法称为伽辽金法。
要注意的是：用位移变分法求位移分量，只须取几项
就可达到较高的精度，然而由此求出的应力却很不精确。
为了求得的应力充分精确，必须取更多项。
41
III,Application and example
If we apply the variational method of displacement to the plane problems,
the Ritz method and the Galerkin method will both be simplified,For these
two kinds of plane problems,because we need not consider the displacement
w in z direction,and u and v don’t change along with the change of z
coordinate,so the the expression of the components of displacement can be
set as
?? ????
m
mm
m
mm vBvvuAuu 00,
???
???
??
??
SvYyxYv
B
U
SuXyxXu
A
U
mm
m
mm
m
ddd
ddd
?
?
?
?
When using the Ritz method,in order to decide the coefficient Am and
Bm,we take a unit length in z direction,then we simply apply the two
equations as follows to solve the linear system of equations,
42
三 应用与举例
将位移变分法应用于平面问题，瑞次法和伽辽金法
都将得到简化。由于两种平面问题都不必考虑 z 方向的
位移 w,且 u 和 v 都不随坐标 z 而变，所以位移分量的
表达式可设为
?? ????
m
mm
m
mm vBvvuAuu 00,
在采用瑞次法时，为了决定系数 Am及 Bm,在 z方向
取一个单位长度，只须应用如下二式来求解线性方程组
???
???
??
??
SvYyxYv
B
U
SuXyxXu
A
U
mm
m
mm
m
ddd
ddd
?
?
?
?
43
Where the strain energy expressed by the components of displacement can
be simplified as follows,For plane strain problems
? ? d x d yy
u
x
v
y
v
x
u
y
v
x
uEU ??
?
?
?
?
?
?
?
?
???
?
???
?
?
??
?
??
???
?
???
?
?
???
?
??
?
?
?
??
???
?
???
?
?
??
?
?
??
?
2222
2
1
112 ?
?
?
For plane stress problems
? ? d x d yy
u
x
v
y
v
x
u
y
v
x
uEU ??
?
?
?
?
?
?
?
?
???
?
???
?
?
??
?
???
?
?
?
??
???
?
???
?
?
???
?
??
?
?
?
?
?
?
222
2 2
12
12
??
?
When using the Galerkin method,for plane strain problems,we should
use the two equations as follows to solve the linear system of equations
? ?
? ?
0
21
1
12
0
21
1
12
2
2
??
?
?
?
?
?
???
?
?
??
?
?
??
?
?
??
??
?
?
?
?
?
???
?
?
??
?
?
??
?
?
??
??
??
d x d yvYv
y
eE
d x d yuXu
x
eE
m
m
??
??
44
其中形变势能用位移分量表示形式简化如下，平面应变问题
? ? d x d yy
u
x
v
y
v
x
u
y
v
x
uEU ??
?
?
?
?
?
?
?
?
???
?
???
?
?
??
?
??
???
?
???
?
?
???
?
??
?
?
?
??
???
?
???
?
?
??
?
?
???
2222
2
1
112 ?
?
?
平面应力问题
? ? d x d yy
u
x
v
y
v
x
u
y
v
x
uEU ??
?
?
?
?
?
?
?
?
???
?
???
?
?
??
?
???
?
?
?
??
???
?
???
?
?
???
?
??
?
?
?
?
??
222
2 2
12
12
??
?
在采用伽辽金法时，对于平面应变问题，要应用如下二式
来求解线性方程组
? ?
? ?
0
21
1
12
0
21
1
12
2
2
??
?
?
?
?
?
???
?
?
??
?
?
??
?
?
??
??
?
?
?
?
?
???
?
?
??
?
?
??
?
?
??
??
??
d x d yvYv
y
eE
d x d yuXu
x
eE
m
m
??
??
45
For plane stress problems,the equations set to be solved is
0
2
1
2
1
1
0
2
1
2
1
1
2
2
2
2
2
2
2
2
2
2
2
2
??
?
?
?
?
?
???
?
?
??
?
?
??
??
?
?
??
?
?
?
?
??
?
?
?
?
?
???
?
?
??
?
?
??
??
?
?
??
?
?
?
?
??
??
d x d yvY
yx
u
x
v
y
vE
d x d yuX
yx
v
y
u
x
uE
m
m
??
?
??
?
The calculation workload of the Galerkin method is small,But its
requirement to the function of displacement is high,Besides satisfying the
boundary conditions of displacement,it needs that the stresses solved by the
function of displacement satisfy the boundary conditions of stress,In particular
case,for example,the problem has only boundary conditions of displacement
and no boundary conditions of stress,It means also that the boundary conditions
of stress are satisfied,It is very convenient to use the Galerkin method here,
The key of the Ritz method is to find the function of displacement that
satisfy all the boundary conditions,but this kind of function is always hard to
find,especially in the case that the boundaries are not regular,Therefore,the
application of the Ritz method is extremely restricted on this aspect,
46
对于平面应力问题，要求解的方程组如下
0
2
1
2
1
1
0
2
1
2
1
1
2
2
2
2
2
2
2
2
2
2
2
2
??
?
?
?
?
?
???
?
?
?
?
?
?
??
??
?
?
??
?
?
?
?
??
?
?
?
?
?
???
?
?
?
?
?
?
??
??
?
?
??
?
?
?
?
??
??
d x d yvY
yx
u
x
v
y
vE
d x d yuX
yx
v
y
u
x
uE
m
m
??
?
??
?
伽辽金方法的计算工作量较小，但对位移函数的要
求较高，除了要求满足位移边界条件外，还要求根据位
移函数求得的应力应满足应力边界条件。在特殊情况，
如仅有位移边界，而无应力边界，这也表示着应力边界
条件得到满足，这时用伽辽金方法十分方便。
瑞次法的要点是要找到满足全部边界条件的位移函
数，而这种函数一般仍然难以找到，尤其在边界不规整
的情况下。所以瑞次方法的应用在这一点上受到极大的
限制。
47
Example 1,For the thin plate as shown in the figure,ignoring the body
forces,please evaluate the displacement,
Solution,We assume the displacements
yBvBvxAuAu 111111,????
which satisfy the boundary conditions of the
displacement boundaries(left and underside),
??
??
??
?
?
??
?
?
?
?
?
?
??
?
?
?
?
?
?
?
?
?
?
?
?
?
???
?
?
??
?
?
?
?
??
?
?
?
?
?
?
?
?
?
yx
y
u
x
v
yx
y
v
x
u
y
v
x
uE
U
dd
2
1
dd2
)1(2
2
22
2
?
?
?
Under the conditions of plane stress
? ? ???? a b d x d yBABAEU 0 0 1121212 )2()1(2 ??
We have
x
y
q1
q2
b
a
o
48
例 1：如图所示的薄板，不计体力，求薄板的位移
解,设位移
yBvBvxAuAu 111111,????
它们是满足位移边界（左边和下边）
的边界条件的。
??
??
??
?
?
??
?
?
?
?
?
?
??
?
?
?
?
?
?
?
?
?
?
?
?
?
???
?
?
??
?
?
?
?
??
?
?
?
?
?
?
?
?
?
yx
y
u
x
v
yx
y
v
x
u
y
v
x
uE
U
dd
2
1
dd2
)1(2
2
22
2
?
?
?
在平面应力状态下
? ? ???? a b d x d yBABAEU 0 0 1121212 )2()1(2 ??
可得
x
y
q1
q2
b
a
o
49
)2()1(2 1121212 BABAE a bU ?? ????
i.e,
abqBUabqAU 2
1
1
1
,???? ????
We have
?? ?? svYBUsuXAU d,d 1
1
1
1 ?
?
?
?From
abqAB
E a b
abqBA
E a b
2112
1112
)22(
)1(2
)22(
)1(2
???
?
???
?
?
?
?
?
i.e,
E
qqB
E
qqA 12
1211,
?? ??????
yE qqvxE qqu 1221,?? ??????
Solving them yields
50
)2()1(2 1121212 BABAE a bU ?? ????即
abqBUabqAU 2
1
1
1
,???? ????
可得
?? ?? svYBUsuXAU d,d 1
1
1
1 ?
?
?
?由
abqAB
E a b
abqBA
E a b
2112
1112
)22(
)1(2
)22(
)1(2
???
?
???
?
?
?
?
?
即
E
qqB
E
qqA 12
1211,
?? ??????
yE qqvxE qqu 1221,?? ??????
解得
51
Example 2,As shown in the figure,a rectangular
thin plate with the width of 2a and the height of b
is fixed on the left,right and at bottom,The
displacement of its top is given as
???
?
???
? ????
2
2
1,0 axvu ?
Ignoring the body force,please evaluate the
displacement and stress of the thin plate,
Solution,Set the coordinate axis as shown in the
figure,Suppose the components of displacement are
?????? ???
?
?
???
? ??
b
y
b
y
a
x
a
xAu 11
2
2
1
?????? ???
?
?
???
? ??
???
?
???
? ???
b
y
b
y
a
xB
b
y
a
xv 111
2
2
12
2
?
a a
b b
o x
y
η
52
例 2：如图所示，宽为 2a而高度为 b
的矩形薄板，左右两边及下边均被
固定，而上边的位移给定为
???
?
???
? ????
2
2
1,0 axvu ?
不计体力，试求薄板的位移和应力。
解,取坐标轴如图所示。设位移
分量为
?????? ???
?
?
???
? ??
b
y
b
y
a
x
a
xAu 11
2
2
1
?????? ???
?
?
???
? ??
???
?
???
? ???
b
y
b
y
a
xB
b
y
a
xv 111
2
2
12
2
?
a a
b b
o x
y
η
53
which can satisfy the boundary conditions of displacement,i.e,
? ? ? ? ? ? 0,0,0 0 ??? ???? byyax uuu
? ? ? ? ? ? ??
?
?
???
? ?????
???? 2
2
0 1,0,0 a
xvvv
byyax ?
There is not boundary conditions of stress in this problem,So we can
consider that since the given displacements satisfy the boundary conditions
of displacement,then they satisfy all the boundary conditions,So we can
employ the Galerkin method to solve this problem to simplify the
calculation,
Notice that the body forces X=Y=0 and m=1,so the Galerkin
equations become
0
2
1
2
1
1
0
2
1
2
1
1
1
0
2
2
2
2
2
2
1
0
2
2
2
2
2
2
???
?
?
?
?
?
?
??
??
?
?
??
?
?
?
?
???
?
?
??
?
?
??
??
?
?
??
?
?
?
?
? ?
? ?
?
?
d x d yv
yx
u
x
v
y
vE
d x d yu
yx
v
y
u
x
uE
a
a
b
a
a
b
??
?
??
?
54
可以满足位移边界条件，即
? ? ? ? ? ? 0,0,0 0 ??? ???? byyax uuu
? ? ? ? ? ? ??
?
?
???
? ?????
???? 2
2
0 1,0,0 a
xvvv
byyax ?
在该问题中，并没有应力边界条件，因此可以认
为所设位移既然满足了位移边界条件，也就满足了全
部边界条件，这就可以应用伽辽金法求解，使数学运
算比较简单一些。
注意体力 X=Y=0而 m = 1，伽辽金方程成为
0
2
1
2
1
1
0
2
1
2
1
1
1
0
2
2
2
2
2
2
1
0
2
2
2
2
2
2
???
?
?
?
?
?
?
??
??
?
?
??
?
?
?
?
???
?
?
??
?
?
??
??
?
?
??
?
?
?
?
? ?
? ?
?
?
d x d yv
yx
u
x
v
y
vE
d x d yu
yx
v
y
u
x
uE
a
a
b
a
a
b
??
?
??
?
55
Substituting the second derivative of the components of every displacement
.21
22
,2131
,1
2
,
22
,
2
,
6
1
2
2
2
1
2
2
2
2
1
2
2
2
2
2
1
22
2
3
3
2
1
2
2
2
2
2
1
2
2
?
?
?
?
?
?
???
??
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
??
??
?
?
?
?
?
?
?
?
?
???
?
?
?
?
?
?
?
?
?
?
???
?
?
?
?
?
?
?
?
?
?
???
?
?
?
?
?
?
?
?
?
?
???
?
?
b
y
a
x
ab
B
a
x
abyx
v
b
y
a
x
ab
A
yx
u
a
x
b
B
y
v
b
y
b
y
a
B
b
y
ax
v
a
x
a
x
b
A
y
u
b
y
b
y
a
x
a
A
x
u
?
?
and
???
?
???
? ?
???
?
???
? ??
???
?
???
? ?
???
?
???
? ??
2
2
2
2
12
2
3
3
1 1,b
y
b
y
a
xv
b
y
b
y
a
x
a
xu
into the Galerkin equations,integrating and solving them yields
? ?
? ?
? ?
? ??
??
?
??
??
??
??
??
1216
15,
12042
135
2
211
b
a
B
b
a
b
aA
56
将位移分量的各二阶导数
.21
22
,2131
,1
2
,
22
,
2
,
6
1
2
2
2
1
2
2
2
2
1
2
2
2
2
2
1
22
2
3
3
2
1
2
2
2
2
2
1
2
2
?
?
?
?
?
?
???
??
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
??
??
?
?
?
?
?
?
?
?
?
???
?
?
?
?
?
?
?
?
?
?
???
?
?
?
?
?
?
?
?
?
?
???
?
?
?
?
?
?
?
?
?
?
???
?
?
b
y
a
x
ab
B
a
x
abyx
v
b
y
a
x
ab
A
yx
u
a
x
b
B
y
v
b
y
b
y
a
B
b
y
ax
v
a
x
a
x
b
A
y
u
b
y
b
y
a
x
a
A
x
u
?
?
以及
???
?
???
? ?
???
?
???
? ??
???
?
???
? ?
???
?
???
? ??
2
2
2
2
12
2
3
3
1 1,b
y
b
y
a
xv
b
y
b
y
a
x
a
xu
代入伽辽金方程，进行积分并求解得
? ?
? ?
? ?
? ??
??
?
??
??
??
??
??
1216
15,
12042
135
2
211
b
a
B
b
a
b
aA
57
In order to be simple,let b=a and ?=0.2,Substituting A1 and B1 into the
given function of displacement yields
?
?
?
?
?
?
?
?
???
?
?
?
?
?
?
???
??
?
?
??
?
?
???
?
?
??
?
?
??
2
2
2
2
2
2
3
3
2 2 7.07 7 3.01
7 2 4.0
a
y
a
y
a
x
v
a
y
a
y
a
x
a
x
u
?
?
Employing the geometric equations and the physical equations,we can solve
the components of stress through the above equations
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
??
?
?
?
?
?
?
?
?
??
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
???
?
?
?
?
?
?
?
?
?
?
?
?
?
?
???
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
???
?
?
?
?
?
?
?
?
?
?
?
?
?
?
???
a
y
a
x
a
x
a
y
a
y
a
x
a
E
a
y
a
y
a
x
a
y
a
x
a
E
a
y
a
y
a
x
a
y
a
x
a
E
xy
y
x
21302.0189.0644.0
31302.0473.0305.01
31754.0095.0161.01
3
3
2
2
2
2
2
2
2
2
2
2
2
2
2
2
?
?
?
?
?
?
58
为简单起见，取 b=a而 μ=0.2,将 A1和 B1代入所设位移函数得
?
?
?
?
?
?
?
?
???
?
?
?
?
?
?
???
??
?
?
??
?
?
???
?
?
??
?
?
??
2
2
2
2
2
2
3
3
2 2 7.07 7 3.01
7 2 4.0
a
y
a
y
a
x
v
a
y
a
y
a
x
a
x
u
?
?
应用几何方程及物理方程，可有上式求得应力分量
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
??
?
?
?
?
?
?
?
?
??
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
???
?
?
?
?
?
?
?
?
?
?
?
?
?
?
???
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
???
?
?
?
?
?
?
?
?
?
?
?
?
?
?
???
a
y
a
x
a
x
a
y
a
y
a
x
a
E
a
y
a
y
a
x
a
y
a
x
a
E
a
y
a
y
a
x
a
y
a
x
a
E
xy
y
x
21302.0189.0644.0
31302.0473.0305.01
31754.0095.0161.01
3
3
2
2
2
2
2
2
2
2
2
2
2
2
2
2
?
?
?
?
?
?
59
§ 10-4 The Variational Equations of
Stress and the Variational Method of Stress
Suppose that there is an arbitrary elastic body which is in equilibrium
under the action of equilibrium,Letσij be the components of stress that exist
exactly,which satisfy the differential equations of equilibrium and the
boundary conditions of stress,also satisfy the compatible equations,Now
we assume that the body forces don’t change,while the components of
stress undergo small changeδσij,which is so-called the virtual stress or the
variation of stress,so that the components of stress become σij +δσij,，
Suppose they only satisfy the differential equations of equilibrium and the
boundary conditions of stress,
I,The variational equations of stress
Since the two groups of equations both satisfy the differential equations
of equilibrium under the action of the same body forces,the variation of the
components of stress must satisfy the equations of equilibrium without body
forces,i.e,
60
§ 10-4 应力变分方程应力变分方法
设有任一弹性体，在外力的作用下处于平衡。命
σ ij为实际存在的应力分量，它们满足平衡微分方程
和应力边界条件，也满足相容方程。现在，假想体力
不变,而应力分量发生了微小的变化 δσij,即所谓虚应
力或应力的变分，使应力分量成为 σ ij +δ σ ij,设它
们只满足平衡微分方程和应力边界条件。
一 应力变分方程
既然两组应力分量都满足同样体力作用下的平衡
微分方程，应力分量的变分必然满足无体力时的平衡
方程，即
61
0δδδ
0δδδ
0δδδ
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
yzxzz
xyzyy
zxyxx
yxz
yzy
zyx
???
???
???
(a)
Znml
Ynml
nml
zzyzx
yzyxy
zxxyx
δδδδ
δδδδ
Xδδδδ
???
???
???
???
???
???
(b)
At the same time,at the boundaries that the displacements are given(the
surface forces are not possible to give),the variation of the components of
stress must go with the variation of the components of surface
forces, According to the requirement of the boundary conditions of
stress,at the boundaries,the variation of the components of stress must satisfy ZYX ddd,,
62
0δδδ
0δδδ
0δδδ
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
yzxzz
xyzyy
zxyxx
yxz
yzy
zyx
???
???
???
(a)
Znml
Ynml
nml
zzyzx
yzyxy
zxxyx
δδδδ
δδδδ
Xδδδδ
???
???
???
???
???
???
(b)
同时，在位移给定的边界上（面力不可能给定），应力分
量的变分必然伴随着面力分量的变分 。 根据应力
边界条件的要求，应力分量的变分在边界上必须满足
ZYX ddd,,
63
Owing to the variation of the components of stress,the strain energy
must have the corresponding variation,We regard strain energy as the
function of the components of stress,then the variation of the strain energy
should be
zyx
UU
zyxUU
yz
yz
x
x
ddd.,,δ.,,δ
dddδδ
11
1
???
???
?
?
?
?
?
?
?
?
?
?
?
??
?
?
?
?
?
?
?
?
Introducing the following equation
.,,
,,,
111
111
xy
xy
zx
zx
yz
yz
z
z
y
y
x
x
UUU
UUU
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
64
由于应力分量的变分，形变势能必有相应的变分。
把形变势能看做应力分量的函数，则形变势能的变分应
为
zyx
UU
zyxUU
yz
yz
x
x
ddd.,,δ.,,δ
dddδδ
11
1
???
???
?
?
?
?
?
?
?
?
?
?
?
??
?
?
?
?
?
?
?
?
将下式代入
.,,
,,,
111
111
xy
xy
zx
zx
yz
yz
z
z
y
y
x
x
UUU
UUU
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
65
? ?
zyx
z
v
y
w
x
u
zyxU
yzx
yzyzxx
ddd.,,δ.,,δ
ddd.,,δ.,,δδ
???
???
?
?
?
?
?
?
?
?
???
?
?
??
?
?
?
?
?
?
?
??
?
?
?
????
??
????
zyxxuSluzyxxu xxx ddd)δ(dδdddδ ???????? ?????? ???
and introducing the geometric equations yields
According to the integration by parts and the Ostrogradsky—Gauss
formula,we arrange every term on the right of the above equation,for
example,
We finally arrive at
66
? ?
zyx
z
v
y
w
x
u
zyxU
yzx
yzyzxx
ddd.,,δ.,,δ
ddd.,,δ.,,δδ
???
???
?
?
?
?
?
?
?
?
???
?
?
??
?
?
?
?
?
?
?
??
?
?
?
????
??
????
zyxxuSluzyxxu xxx ddd)δ(dδdddδ ???????? ?????? ???
再将几何方程代入，得
根据分步积分和奥－高公式，对上式右边的各项进行处理，例如
最后可得
67
zyx
zyx
u
SnmluU
zxyxx
zxxyx
ddd
δδδ
d])δδδ([δ
???
??
?
?
?
?
?
?
?
?
???
?
?
??
?
?
?
?
?
?
?
?
?
?
?
????
?
?
???
???
Then introducing the equations (a) and (b) gives
?? ??? SZwYvXuU d)δδδ(δ
This is the so-called the variational equation of stress(the Castigliano
variational equation),The right part of the equation represents the work done
by the variation of the surface forces during the true displacement,We can
see from here that due to the variation that happen to the stress,the variation
of strain energy equals to the work done by the variation of surface forces
during the true displacement.。
68
zyx
zyx
u
SnmluU
zxyxx
zxxyx
ddd
δδδ
d])δδδ([δ
???
??
?
?
?
?
?
?
?
?
???
?
?
??
?
?
?
?
?
?
?
?
?
?
?
????
?
?
???
???
再将式（ a）和（ b）代入，即得
?? ??? SZwYvXuU d)δδδ(δ
这就是所谓应力变分方程 (卡斯提安诺变分方程 )。方程的右
边代表面力的变分在实际位移上所做的功。由此可见，由于
应力发生的变分，形变势能的变分等于面力的变分在实际位
移上所做的功。
69
If the surface forces are given at a certain part of boundary,then the
surface forces at this part of boundary can’t have variation,
so, Accordingly the corresponding integrating terms on the
right of variational equation of stress become zero,If the displacements are
given to be zero at a certain part of boundary,then the corresponding
integrating terms on the right of variational equation of stress also become
zero,Therefore,the integration on the right of the variational equation of
stress need only be conducted at such boundaries where the surface forces
aren’t given and the given displacements aren’t equal to zero,
0??? ZYX ddd
II,The principle of minimum complementary energy
Rewrite the variational equation of stress as
0d)δδδ(δ ???? ?? SZwYvXuU
Because at the boundaries where integration is needed,the displacements
are given and keep invariable during the variation,the above equation can
be rewrote as
70
如果在某一部分边界上，面力是给定的，则该部分边
界上的面力不能有变分，于是,而应力变分
方程右边的相应积分项成为零；如果在某一部分边界上，
给定的位移等于零，则应力变分方程右边的相应积分项也
成为零。因此，应力变分方程右边的积分，只须在这样的
边界上进行：面力没有给定，而给定的位移又不等于零。
0??? ZYX ddd
二 极小余能原理
将应力变分方程改写为
0d)δδδ(δ ???? ?? SZwYvXuU
由于在需要积分的边界上，位移是给定的，在变分过
程中保持不变，所以上式可以改写为
71
The expression in the square brackets is called the complementary
energy of the elastic body,Hence,among all groups of stresses that satisfy
the equations of equilibrium and the boundary conditions of stress,the
complementary energy of the elastic body with the group of stresses existing
exactly becomes minimum,If we consider the second-order variation,we
can prove that this extremum is a minimum,So the above conclusion is
called the principle of minimum complementary energy,
? ? 0d)(-Uδ ????? SZwYvXu
We notice previously that the stresses existing exactly should satisfy the
compatible equations besides the differential equations of equilibrium,Now we
see also that the stresses existing exactly satisfy the variational equations of
stress besides the differential equations of equilibrium and the boundary
conditions of stress,Furthermore,we can deduce the compatible conditions
from the variational equations of stress through calculating,So we can see that
the variational equations of stress can substitute for the compatible conditions,
72
中括号内的表达式称为弹性体的余能。因此，在满足
平衡方程和应力边界条件的各组应力中间，实际存在的一
组应力应使弹性体的余能成为极值。如果考虑二阶变分，
可以证明这个极值是极小值，所以上述结论称为极小余能
原理。
? ? 0d)(-Uδ ????? SZwYvXu
以前看到，实际存在的应力，除了满足平衡微分方
程以外，还应当满足相容方程。现在又看到，实际存在
的应力，除了满足平衡微分方程和应力边界条件以外，
还满足应力变分方程。而且，通过运算，还可以从应力
变分方程导出相容条件。于是可见，应力变分方程可以
代替相容条件。
73
III,The variational method of stress
Set the expressions of the components of stress and let them satisfy the
equations of equilibrium and the boundary conditions of stress,which
contain several undetermined coefficients,Then we can decide these
coefficients according to the variational equations of stress,The
components of stress can be generally set as
Where Am are m independent coefficients,(σij)0 is the set function which
satisfies the differential equations of equilibrium and the boundary
conditions of stress,(σij)m is the set function which satisfies,the
differential equations of equilibrium and the boundary conditions of stress
without the action of body forces and surface forces”,Thus,σij satisfies the
differential equations of equilibrium and the boundary conditions of stress
at all time no matter how to determine the value of the coefficient Am,
Notice that the variation of stress is only realized by the variation of the
coefficient Am,
mm ijmijij A )()( 0 ??? ???
(c)
74
三 应力变分法
设定应力分量的表达式，使其满足平衡方程和应力边
界条件，但其中包含若干待定系数，然后根据应力变分方
程决定这些系数。应力分量一般可设为
其中 Am为互不依赖的 m个系数。 (σij)0是满足平衡微分
方程和应力边界条件的设定函数,(σij)m 是满足, 无体力
和面力作用时的平衡微分方程和应力边界条件, 的设定函
数。这样，不论系数 Am如何取值,σij总能满足平衡微分方
程和应力边界条件。
注意：应力的变分只是由系数 Am的变分来实现。
mm ijmijij A )()( 0 ??? ???
(c)
75
If either the surface forces are given or the displacements equal to zero
at any part of boundary of the elastic body,then the variational equation of
stress is simplified as
Apparently,the strain energy U is the quadratic function of Am,So the
equation (d) will be the simple equation of Am,There are altogether m such
equations which can even be used to solve the coefficient Am,Then
substitute it into the equation (c) to solve the components of stress,
0?Ud
0???
mA
Ui.e,(d)
If the displacements are given but not equal to zero at a certain part of
boundary,we must apply variational equations directly at this part of
boundary,i.e,
?? ??? SZwYvXuU d)δδδ(δ
76
如果在弹性体的每一部分边界上，不是面力被给定，
便是位移等于零，则应力变分方程简化为
显然，形变势能 U是 Am的二次函数，因而 (d)式将是 Am的一
次方程。这样的方程共有 m 个，恰好可以用来求解系数 Am，
回代 (c)式，求得应力分量。
0?Ud
0???
mA
U
即 (d)
如果在某一部分边界上，位移是给定的，但并不等
于零，则在这一部分边界上须直接应用变分方程，即
?? ??? SZwYvXuU d)δδδ(δ
77
Here,u,v,w are known,The integration only includes this part of
boundary,Substituting the relations between the variation of surface forces
and the variation of stress
zzyzx
yzyxy
zxxyx
nmlZ
nmlY
nml
???
???
???
δδδδ
δδδδ
δδδXδ
???
???
???
into the integration on the right of the equation will result in
m
m
m ABSZwYvXu d??? ??? d)δδδ(
where Bm is a constant,On the other hand,the left of the equation
m
m
AAUU dd ? ???
78
在这里,u,v,w是已知的，积分只在该部分边界。将面
力的变分与应力的变分两者之间的关系，即
zzyzx
yzyxy
zxxyx
nmlZ
nmlY
nml
???
???
???
δδδδ
δδδδ
δδδXδ
???
???
???
代入方程的右边积分后，将得出如下的结果,
m
m
m ABSZwYvXu d??? ??? d)δδδ(
其中 Bm是常数。另一方面，方程的左边
m
m
AAUU dd ? ???
79
So we can get
m
m
BAU ???
Last the type is still a power of the Am distance,have the m altogether,and the
each Am is with each other not related,as a result can get all Am,returning to the
generation( c) type,getthe stress dint weight,
At applied should the dint become to divide the method,to make enactment
of should the dint weight since satisfy should the dint boundary condition,and then
satisfy the square distance of equilibrium differential calculus,this is usually very
difficult.But,exist in the problem of some types should the dint function,and use
should the dint function mean of should the dint weight satisfy the square distance
of equilibrium differential calculus again.At this time,we only the beard enactment
should the expression type of the dint function,make it to of should the dint weight
can satisfy should the dint boundary condition,difficulty reduced consumedly,
80
因而得
m
m
BAU ???
上式仍然是 Am的一次方程，总共有 m个，且各个 Am是互不
相关的，因而可以求出所有的 Am,回代 (c)式，求得应力
分量。
在应用应力变分法时，要使设定的应力分量既满足应
力边界条件，又满足平衡微分方程，这往往是很困难的。
但是，在某些类型的问题中存在着应力函数，而且用应力
函数表示的应力分量又能满足平衡微分方程。这时，我们
就只须设定应力函数的表达式，使它给出的应力分量能满
足应力边界条件，困难就大大减少了。
81
???
m
mmA ??? 0
Where Am are m independent coefficients,The stresses given byφ0
satisfy the exact actual boundary conditions of stress,And the stresses given
by φm satisfy the boundary conditions of stress without surface forces,
Because the number of the components of stress is large,they are hard
to determine,We often use the stress function method,
In the plane stress problems,if the components of body forces are
constants,then stress function exist,We set the stress function as
IV,The stress function method
Under the status of plane stress,the strain energy expressed by the
components of stress is
? ??? ????? yxEU xyyxyx dd)1(222 1 222 ???????
82
???
m
mmA ??? 0
其中 Am为互不依赖的 m个系数。 φ0给出的应力满
足实际的应力边界条件,φm 给出的应力满足无面力时
的应力边界条件。
由于应力分量的数量较多，确定起来有困难，通
常用应力函数方法。
在平面应力问题中，如果体力分量为常数，则存
在应力函数。将应力函数设定为
四 应力函数方法
在平面应力状态，用应力分量表示的形变势能为
? ??? ????? yxEU xyyxyx dd)1(222 1 222 ???????
83
If we consider simply connected body,and it is the problem of
boundary conditions of stress,the components of stress have no relation to ?
and it can be set as zero,Thus two types of plane problems can both be
simplified as
? ??? ??? yxEU xyyx dd22 1 222 ???
Express it through the function of stress
?? ??
?
?
?
?
?
?
???
?
???
?
??
??
???
?
???
?
?
?
??
???
?
???
?
?
?
?? yx
yx
Yy
x
Xx
yE
U dd2
2
1
222
2
22
2
2 ???
For the plane strain problems
? ?? ?? ?d x d yEU xyyxyx?? ?????? 222 22121 ????????
84
如果考虑单连体，且是应力边界问题，应力分量
应与 ?无关，可设其为零。则两类平面问题皆简化为
? ??? ??? yxEU xyyx dd22 1 222 ???
用应力函数表示为
?? ??
?
?
?
?
?
?
???
?
???
?
??
??
???
?
???
?
?
?
??
???
?
???
?
?
?
?? yx
yx
Yy
x
Xx
yE
U dd2
2
1
222
2
22
2
2 ???
对于平面应变问题
? ?? ?? ?d x d yEU xyyxyx?? ?????? 222 22121 ????????
85
Because the surface forces can’t have variation in the problems of stress
boundaries,the variational equation is simplified as dU=0,So the coefficients
satisfy
0???
mA
U
The above equation is a linear system of equations,After solving Am,we
obtain the approximate solution of the function of stress and finally obtain all
the components of stress,
Introducing the expression of the function of stress yields
0dd2
dd
22
2
2
2
2
2
2
2
2
???
?
?
?
?
?
?
??
?
?
?
??
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
???
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
??
??
yx
yxAyx
yx
xA
Yy
xyA
Xx
y
m
mm
??
????
86
在应力边界问题中，因为面力不能有变分，所以变分
方程简化为 δ U=0,因此系数应满足
0???
mA
U
上式为线性方程组，求解 Am后，得到应力函数的近似
解，最后得到各应力分量。
将应力函数的表达式代入，得
0dd2
dd
22
2
2
2
2
2
2
2
2
???
?
?
?
?
?
?
??
?
?
?
??
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
???
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
??
??
yx
yxAyx
yx
xA
Yy
xyA
Xx
y
m
mm
??
????
87
Because it is a approximate solution,the components of tress can’t
satisfy the compatible conditions precisely,and the components of strain
solved from the components of stress can’t satisfy the deformation
compatibility conditions precisely too,and we can’t solve the components
of displacement according to the geometric equation,
We can solve the coefficient Am, Consequently we decide the stress
function ?,through which we solve all the components of stress,
The key of the stress function method is to find the stress function that
satisfy all the boundary conditions,which is still hard to find,especially in
the case that the boundaries are not regular,So the application of stress
method is extremely restricted on this aspect,
88
由于是近似解，应力分量不能精确满足相容条
件，由应力分量求得的应变分量也不能精确满足变
形协调条件，不能根据几何方程求得位移分量。
由上式即可解得系数 Am。从而确定应力函数 ?，
再由应力函数 ?求得各应力分量。
应力函数法的要点是要找到满足全部边界条件
的应力函数，而这种函数一般仍然难以找到，尤其
在边界不规整的情况下。所以应力方法的应用在这
一点上受到极大的限制。
89
Example 3,Suppose there is a rectangular thin plate whose body forces are
ignored,Its two opposite sides are subjected to tensile forces with the
distribution of parabola,whose maximum intensity is q,as shown in the
figure,Please solve the stress of the plate,
Solution,In the coordinates as shown in
the figure,the boundary conditions are,
? ? ? ?
? ? ? ? 0,0
0,1
2
2
??
???
?
?
??
?
?
??
????
????
byxybyy
axxyaxx b
y
q
??
??
Take
2
2
22
2
2
2
12
2
2 11
612 ?
?
?
?
???
? ?
???
?
???
? ??
???
?
???
? ??
b
y
a
xqbA
b
yyq?
x
y
a a
b
b
q
90
例题 3：设有矩形薄板，体力不计，在两对边上受有按抛物
线分布的拉力，其最大集度为 q, 图示。试求板的应力。
解：在图示坐标下，边界条
件是
? ? ? ?
? ? ? ? 0,0
0,1
2
2
??
???
?
?
??
?
?
??
????
????
byxybyy
axxyaxx b
y
q
??
??
取
2
2
22
2
2
2
12
2
2 11
612 ?
?
?
?
???
? ?
???
?
???
? ??
???
?
???
? ??
b
y
a
xqbA
b
yyq?
x
y
a a
b
b
q
91
???
?
???
? ??
2
2
2
0 612 b
yyq?Then from we can obtain the components of stress
? ? ? ? ? ? 0,0,1 0
2
02
0
2
02
2
2
0
2
0 ???
????
?
??
???
?
???
? ??
?
??
yxxb
yq
y xyyx
??????
which satisfy the boundary conditions,And the stresses corresponding to
2
2
22
2
2
2
1 11 ???
?
???
? ?
???
?
???
? ??
b
y
a
xqb?
can satisfy the boundary conditions without surface forces,
Substitute ? and the body forces X=Y=0 into the following equation
0dd2
dd
22
2
2
2
2
2
2
2
2
???
?
?
?
?
?
?
??
?
?
?
??
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
???
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
??
??
yx
yxAyx
yx
xA
Yy
xyA
Xx
y
m
mm
??
????
92
???
?
???
? ??
2
2
2
0 612 b
yyq?
则由 给出的应力分量
? ? ? ? ? ? 0,0,1 0
2
02
0
2
02
2
2
0
2
0 ???
????
?
??
???
?
???
? ??
?
??
yxxb
yq
y xyyx
??????
满足边界条件。而由 2
2
22
2
2
2
1 11 ???
?
???
? ?
???
?
???
? ??
b
y
a
xqb?
所对应的应力能满足无面力时的边界条件。
将 ?以及体力 X=Y=0代入下式中
0dd2
dd
22
2
2
2
2
2
2
2
2
???
?
?
?
?
?
?
??
?
?
?
??
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
???
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
??
??
yx
yxAyx
yx
xA
Yy
xyA
Xx
y
m
mm
??
????
93
Integrating and simplifying yields
076449256764 14
4
2
2
???
?
?
???
? ?? A
a
b
a
b
In order to be simple,let a=b,then 0425.0
1 ?A
Substitute it into ?,and let a=b,then solve the components of stress
22
2
2
22
2
2
2
2
2
2
2
2
2
2
2
2
2
2
2
2
1168.0
1
3
117.0
3
1117.01
a
xy
a
y
a
x
q
yx
a
y
a
x
q
x
a
y
a
x
q
a
y
q
y
xy
y
x
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
???
??
?
??
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
???
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
??
?
?
?
?
?
?
?
?
??
?
?
?
?
?
?
?
?
?
94
进行积分，并简化以后得
076449256764 14
4
2
2
???
?
?
???
? ?? A
a
b
a
b
为简单起见，令 a=b，则 0425.0
1 ?A
代入 ?中，并令 a=b，再求应力分量得
22
2
2
22
2
2
2
2
2
2
2
2
2
2
2
2
2
2
2
2
1168.0
1
3
117.0
3
1117.01
a
xy
a
y
a
x
q
yx
a
y
a
x
q
x
a
y
a
x
q
a
y
q
y
xy
y
x
?
?
?
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95
Exercise 10.1 A square thin plate in vertical plane,with the length of side of
2a,is fixed at its four sides and is only subjected to the action of gravity,Let
?=0,Take the expressions of the components of displacement as
?
?
?
?
?
?
?
?
?????
?
?
?
?
?
?
???
?
?
?
?
?
?
??
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?????
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???
?
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?
?
??
?
?
2
2
32
2
212
2
2
2
2
2
32
2
212
2
2
2
11
11
a
y
B
a
x
BB
a
y
a
x
v
a
y
A
a
x
AA
a
y
a
x
a
y
a
x
u
Solve the components of displacement(take the term A1 and B1) the
components of stress with the Ritz method,
a a
a
a x
y
Solution,When we take the term A1 and B1 only
???
?
???
? ?
???
?
???
? ??
???
?
???
? ?
???
?
???
? ??
2
2
2
2
122
2
2
2
1 11,11 a
y
a
xBv
a
xy
a
y
a
xAu
96
练习 10.1 铅直平面内的正方形薄板，边长为 2a，四边固
定，只受重力作用，设,试取位移分量的表达式为
?
?
?
?
?
?
?
?
?????
?
?
?
?
?
?
???
?
?
?
?
?
?
??
?
?
?
?
?
?
?
?
?????
?
?
?
?
?
?
???
?
?
?
?
?
?
??
?
?
2
2
32
2
212
2
2
2
2
2
32
2
212
2
2
2
11
11
a
y
B
a
x
BB
a
y
a
x
v
a
y
A
a
x
AA
a
y
a
x
a
y
a
x
u
用瑞次法求解位移分量 (取 A1项及 B1项 )及应力分量。
a a
a
a x
y
0??
解,当只取 A1项及 B1项时,
???
?
???
? ?
???
?
???
? ??
???
?
???
? ?
???
?
???
? ??
2
2
2
2
122
2
2
2
1 11,11 a
y
a
xBv
a
xy
a
y
a
xAu
97
gYX ???,0
The strain energy
d x d y
y
u
x
v
y
v
x
uEU a a? ?
?
?
?
?
?
?
?
?
???
?
???
?
?
??
?
??
???
?
???
?
?
???
?
??
?
?
?
??
0 0
222
2
1
2
Now we evaluate and,
1A
U
?
?
1B
U
?
?
?
?
?
?
?
?
??
?
?
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?
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?
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?
?
?
?
?
?
? ?
37
6
75
32
22
2
11
0 0
1111
BAE
d x d y
y
u
x
v
Ay
u
x
v
y
v
Ay
v
x
u
Ax
uE
A
U a a
?
?
?
?
?
?
??
?
?
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??
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??
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??
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?
?
?
?
?
?
?
?
?
?
? ?
5
6
45
32
22
2
1
1
0 0
1111
A
B
E
d x d y
y
u
x
v
By
u
x
v
y
v
By
v
x
u
Bx
uE
B
U a a
98
gYX ???,0
形变势能
d x d y
y
u
x
v
y
v
x
uEU a a? ?
?
?
?
?
?
?
?
?
???
?
???
?
?
??
?
??
???
?
???
?
?
???
?
??
?
?
?
??
0 0
222
2
1
2
现计算 和
1A
U
?
?
1B
U
?
?
?
?
?
?
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??
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?
?
?
?
? ?
37
6
75
32
22
2
11
0 0
1111
BAE
d x d y
y
u
x
v
Ay
u
x
v
y
v
Ay
v
x
u
Ax
uE
A
U a a
?
?
?
?
?
?
??
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?
?
? ?
5
6
45
32
22
2
1
1
0 0
1111
A
B
E
d x d y
y
u
x
v
By
u
x
v
y
v
By
v
x
u
Bx
uE
B
U a a
99
When applying the Ritz method,we expect
???
???
??
?
?
??
?
?
dSvYd x d yYv
B
U
dSuXd x d yXu
A
U
mm
m
mm
m
From
0
1
???AU
2
2
2
0 0 2
2
1 9
16114 agd x d y
a
y
a
xg
B
U a a ?? ?
???
?
???
? ?
???
?
???
? ??
?
? ? ?
Solving it yields
E
gaB
E
gaA 2
1
2
1 533
225,
1066
175 ?? ??
100
在用瑞次法时，要求
???
???
??
?
?
??
?
?
dSvYd x d yYv
B
U
dSuXd x d yXu
A
U
mm
m
mm
m
由
0
1
???AU
2
2
2
0 0 2
2
1 9
16114 agd x d y
a
y
a
xg
B
U a a ?? ?
???
?
???
? ?
???
?
???
? ??
?
? ? ?
解之得
E
gaB
E
gaA 2
1
2
1 533
225,
1066
175 ?? ??
101
Therefore
2
2
2
2
2
2
2
2
2
11
5 3 3
2 2 5
11
1 0 6 6
1 7 5
a
a
y
a
x
E
g
v
xy
a
y
a
x
E
g
u
?
?
?
?
?
?
?
?
???
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?
?
??
?
?
?
2
2
2
2
2
2
1
533
450
1
3
1
1 0 6 6
175
a
x
gy
y
v
E
y
a
y
a
x
g
x
u
E
y
x
??
??
?
?
?
?
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?
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????
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?
?
?
2
2
2
2
2
2
1
533
2253
11
2 1 3 2
175
2
a
y
gx
a
y
a
x
gx
x
v
y
uE
xy
??
?
102
所以
2
2
2
2
2
2
2
2
2
11
5 3 3
2 2 5
11
1 0 6 6
1 7 5
a
a
y
a
x
E
g
v
xy
a
y
a
x
E
g
u
?
?
?
?
?
?
?
?
???
?
?
?
?
?
?
??
?
?
?
?
?
?
?
?
???
?
?
?
?
?
?
??
?
?
?
?
?
?
?
?
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?
???
?
?
?
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?
?
?
?
?
?
?
???
?
?
?
?
?
?
??
?
?
?
2
2
2
2
2
2
1
533
450
1
3
1
1 0 6 6
175
a
x
gy
y
v
E
y
a
y
a
x
g
x
u
E
y
x
??
??
?
?
?
?
?
?
?
?
????
?
?
?
?
?
?
???
?
?
?
?
?
?
??
??
?
?
??
?
?
?
?
?
?
?
?
2
2
2
2
2
2
1
533
2253
11
2 1 3 2
175
2
a
y
gx
a
y
a
x
gx
x
v
y
uE
xy
??
?
103
Exercise 10.2 Show that the freely supported beam is subjected to uniform
load q, The length of the beam is l,The equation of equilibrium of the beam
is
Assume that the equation of line of
deflection of the beam is
04
4
?? qdx wdEI
l
xnaw
n
n
?s i n
5,3,1
?
?
?
?
Solve the line of deflection of the beam with the Galerkin method,
q
x
y
l
Solution,The boundary conditions of the beam is
? ? ? ? 0,00 ?? ?? lxx ww
So the equation of line of deflection satisfies the boundary conditions,
104
练习 10.2 简支梁受均布荷载 q，梁长为 (图示 )。梁的平衡
方程为
设梁的挠曲线方程为
04
4
?? qdx wdEI
l
l
xnaw
n
n
?s i n
5,3,1
?
?
?
?
试用伽辽金法求梁的挠曲线。
解,梁的边界条件为
? ? ? ? 0,00 ?? ?? lxx ww
故挠曲线方程满足边界条件。
q
x
y
l
105
The equation of equilibrium of the beam is
04
4
?? qdx wdEI
Substituting it into the Galerkin equation yields
0s ins in
0
4
?
?
?
?
?
?
?
?
?
??
?
??
?
?? dx
l
xnq
l
xn
l
nE I al
n
???
After integrating we arrive at the undetermined coefficient
EIn
qla
n 55
44
??
So the equation of line of deflection is
l
xn
nEI
qlw
n
?
?
s i n14
5,3,1
55
4
?
?
?
?
?
106
梁的平衡方程为 0
4
4
?? qdx wdEI
代入伽辽金方程，得
0s ins in
0
4
?
?
?
?
?
?
?
?
?
??
?
??
?
?? dx
l
xnq
l
xn
l
nE I al
n
???
积分后得待定系数
EIn
qla
n 55
44
??
故挠曲线方程为
l
xn
nEI
qlw
n
?
?
s i n14
5,3,1
55
4
?
?
?
?
?
107
Apparently,the assumed equation of line of deflection of the beam
3322 xaxaw ??
Evaluate the coefficients a2 and a3 with the principle of minimum energy,
The strain energy is
Solution,Assume that the line of deflection of the beam is
3322 xaxaw ??
? ? 0,0
0
0 ???
??
?
??
?
?
x
x dx
dwwIts boundary condition is
Exercise 10.3 The cantilevered beam is
subjected to the action of the concentrated
force P at free end as shown in the figure,
Evaluate the maximum deflection with the
principle of minimum energy,
P
l
x
y
? ? dxxaaEIdx
dx
wdEIU ll 2
0 32
2
0 2
2
62
22 ??
????
?
?
???
??
satisfy the boundary conditions,
108
显然所设梁的挠曲线方程 满足边界条件。 3
322 xaxaw ??
由极小势能原理求系数 a2和 a3 。形变势能为
解,设梁的挠曲线为 3
322 xaxaw ??
? ? 0,0
0
0 ???
??
?
??
?
?
x
x dx
dww其边界条件为
练习 10.3 悬臂梁在自由端受集中
力 P作用，如图所示。试用极小势
能原理求最大挠度。
P
l
x
y
? ? dxxaaEIdx
dx
wdEIU ll 2
0 32
2
0 2
2
62
22 ??
????
?
?
???
??
109
The potential energy of the external forces is
? ? ? ?3322 lalaPwPV lx ????? ?
Then the total potential energy is
? ? ? ?33222
0 32
622 lalaPdxxaaEIVU L ??????? ?
Apply the principle of minimum potential energy
0
0
3
3
2
2
?
?
??
?
?
??
??
a
a
a
a
dd
d
Then ? ?
06242 2
0 322
????? ?? ? PldxxaaEIa L
110
外力势能为
? ? ? ?3322 lalaPwPV lx ????? ?
则总势能为
? ? ? ?33222
0 32
622 lalaPdxxaaEIVU L ??????? ?
应用极小势能原理
0
0
3
3
2
2
?
?
??
?
?
??
??
a
a
a
a
dd
d
则 ? ?
06242 2
0 322
????? ?? ? PldxxaaEIa L
111
? ? 062122 3
0 323
????? ?? ? PldxxaaxEIa L
Integrating yields
EI
Pl
lala
EI
Pl
lala
6
2
2
32
2
2
32
2
2
32
??
??
From the above two equations we can obtain
EI
Pa
EI
Pla
6,2 32 ???
So the line of deflection is
The value of the maximum deflection is
?????? ?? 236 2 xxEIPlw
EI
Plw
3
3
m a x ?
112
? ? 062122 3
0 323
????? ?? ? PldxxaaxEIa L
积分得
EI
Pl
lala
EI
Pl
lala
6
2
2
32
2
2
32
2
2
32
??
??
由上述两方程解得
EI
Pa
EI
Pla
6,2 32 ???
故挠曲线为
最大挠度值为
?????? ?? 236 2 xxEIPlw
EI
Plw
3
3
m a x ?
113
114
