1
Elasticity
2
3
Chapter 2 The Basic theory of the Plane Problem
§ 2-11 Stress function.Inverse solution method and semi-inverse method
§ 2-1 Plane stress problem and plane strain problem
§ 2-2 Differential equation of equilibrium
§ 2-3 The stress on the incline.Principal stress
§ 2-4 Geometrical equation.The displacement of the rigid body
§ 2-5 Physical equation
§ 2-6 Boundary conditions
§ 2-7 Saint-Venant’s principle
§ 2-8 Solving the plane problem according to the displacement
§ 2-9 Solving the plane problem according to the stress.Compatible
equation
§ 2-10 The simplification under the circumstances of ordinary physical
force
Exercise Lesson
4
第二章 平面问题的基本理论
§ 2-11 应力函数逆解法与半逆解法
§ 2-1 平面应力问题与平面应变问题
§ 2-2 平衡微分方程
§ 2-3 斜面上的应力主应力
§ 2-4 几何方程刚体位移
§ 2-5 物理方程
§ 2-6 边界条件
§ 2-7 圣维南原理
§ 2-8 按位移求解平面问题
§ 2-9 按应力求解平面问题。相容方程
§ 2-10 常体力情况下的简化
习题课
5
1.Plane stress problem
§ 2-1 Plane stress problem and plane strain problem
In actual problem,it is strictly saying that any elastic body whose external
force for suffering is a space system of forces is generally the space
object.However,when both the shape and force circumstance of the elastic
body for investigating have their own certain characteristics.As long as the
abstraction of the mechanics is handled together with appropriate
simplification,it can be concluded as the elasticity plane problem,
The plane problem is divided into the plane stress problem and plane strain
problem,
Equal thickness lamella bears the
surface force that parallels with plate
face and don’t change along the
thickness.At the same time,so does the
volumetric force,
σz = 0 τzx = 0 τzy = 0 Fig.2－ 1
6
一、平面应力问题
§ 2-1 平面应力问题与平面应变问题
在实际问题中，任何一个弹性体严格地说都是空间物体，
它所受的外力一般都是空间力系。但是，当所考察的弹性体
的形状和受力情况具有一定特点时，只要经过适当的简化和
力学的抽象处理，就可以归结为弹性力学平面问题。
平面问题分为平面应力问题和平面应变问题。
等厚度薄板，板边承受平
行于板面并且不沿厚度变化的
面力，同时体力也平行于板面
并且不沿厚度变化。
σ z = 0 τ zx = 0 τ zy = 0
图 2－ 1
7
x
y
Characteristics,
1) The dimension of length and breadth is far larger than that of
thickness,
2) The force along the plate face for suffering is the face force in
parallel with plate face,and along the thickness even,the volumetric force
is in parallel with plate force and doesn’t change along the thickness,and
has no external force function on the surface front and back of the flat
panel,
0?z?Attention,Plane stress problem ?z =0,but,this is contrary
to plane strain problem,
8
x
y
特点,
1) 长、宽尺寸远大于厚度
2) 沿板边受有平行板面的面力，且沿厚度均布，体力
平行于板面且不沿厚度变化，在平板的前后表面上
无外力作用。
问题相反。
0?z?注意：平面应力问题 ?z =0，但,这与平面应变
9
2.Plane strain problem
Very long column bears the face force in parallel with plate face and doesn’t
change along the length on the column face,at the same time,so does the
volumetric force,
εz = 0 τzx = 0 τzy = 0
x
Fig,2－ 2
For example:dam,circular cylinder piping by the internal air pressure and
long level laneway etc,
0?z?Attention,Plane strain problem?z = 0,but,this is contrary
to plane stress problem,
x
y
P
10
二、平面应变问题
很长的柱体，在柱面上承受平行于横截面并且不沿长度
变化的面力，同时体力也平行于横截面并且不沿长度变化。
ε z = 0 τ zx = 0 τ zy = 0
x
图 2－ 2
如：水坝、受内压的圆柱管道和长水平巷道等。
0?z?注意平面应变问题 ?z = 0，但
问题相反。
，这恰与平面应力
x
y
P
11
§ 2-2 Differential Equation of Equilibrium
Whether plane stress problem or plane strain problem,is the research problem in
plane xy,all the physics quantity has nothing to do with z,
Discuss below the correlation between any point stress and volumetric force
when the object is placed in the state of equilibrium,and lead an equilibrium
differential equation from here.From the lamella shown in Fig.2-1,we take out a
small and positive parallelepiped PABC,and take for an unit length in the
directional dimension in z,
y?
o x
y
dyyyy ??? ??
x?
dxxxx ??? ??xy?
dxxxyxy ??? ??
yx?
dyyyxyx ??? ??
P A
B C
X
Y
D
Fig.2－ 3
),( yxxx ?? ?
dx
Establishing the function of the positive
stress force in an unit on the left side
is,the coordinate on the right side
x gets the increment,the positive stress on
the face is,spreading the formula
above will be Taylor’s series,
),( ydxxx ??
n
n
x
n
x
x
xx
dx
x
yx
n
dx
x
yx
dx
x
yxyxydxx
)(),(
!
1)(),(
!2
1
),(),(),(
2
2
2
?
????
?
??
?
?
????
??
???
?
12
§ 2-2 平衡微分方程
无论平面应力问题还是平面应变问题，都是在 xy平面内研究问题,
所有物理量均与 z无关。
下面讨论物体处于平衡状态时，各点应力及体力的相互关系，并
由此导出平衡微分方程。从图 2－ 1所示的薄板取出一个微小的正平行
六面体 PABC(图 2－ 3），它在 z方向的尺寸取为一个单位长度。
y?
o x
y
dyyyy ??? ??
x?
dxxxx ??? ??xy?
dxxxyxy ??? ??
yx?
dyyyxyx ??? ??
P A
B C
X
Y
D
图 2－ 3
),( yxxx ?? ? x
dx
设作用在单元体左侧面上的正
应力是,右侧面上坐标
得到增量,该面上的正应力为
,将上式展开为泰勒级
数,
),( ydxxx ??
n
n
x
n
x
x
xx
dx
x
yx
n
dx
x
yx
dx
x
yx
yxydxx
)(
),(
!
1
)(
),(
!2
1
),(
),(),(
2
2
2
?
?
???
?
?
?
?
?
?
???
??
?
??
?
13
After omitting small quantity of the two rank and above the two rank,can get
,at the same time,,,are get the state of stress from the
drawing show,dxx yxyx xx ??? ),(),( ?? y? xy? yx?
While considering the volumetric force to the plane stress state,still prove
mutual and equal theory of shearing strength.Regard the center D and straight
line in parallel with the shaft of z as the moment shaft,list the equilibrium
equation of the moment shaft, ? ?0DM
0
2
1
2
1)(
2
1
2
1)(
??????
?
?
??
?????
?
?
?
dy
dx
dy
dxdy
y
dx
dy
dx
dydx
x
yx
yx
yx
xy
xy
xy
?
?
?
?
?
?
The both sides of the formula above divide get,dxdy
dyydxx yxyxxyxy ??????? ???? 2121
Cause 0,0 ?? dydx,Omitting small quantity isn’t accounted,can get,
yxxy ?? ?
14
略去二阶及二阶以上的微量后便得 同样,,
都一样处理，得到图示应力状态。 dxx
yxyx x
x ?
?? ),(),( ?? y? xy? yx?
对平面应力状态考虑体力时，仍可证明剪应力互等定理。以通过中
心 D并平行于 z轴的直线为矩轴，列出力矩的平衡方程, ? ?0DM
0
2
1
2
1)(
2
1
2
1)(
??????
?
?
??
?????
?
?
?
dy
dx
dy
dxdy
y
dx
dy
dx
dydx
x
yx
yx
yx
xy
xy
xy
?
?
?
?
?
?
将上式的两边除以 得到,dxdy
dyydxx yxyxxyxy ??????? ???? 2121
令 0,0 ?? dydx,即略去微量不计，得,
yxxy ?? ?
15
Deduce the equilibrium differential equation of the plane stress
problem below,list the equilibrium equation to the unit,
011
1)(11)(
:0
????????
??
?
?
???????
?
?
?
??
dydxXdx
dxdy
y
dydydx
x
F
yx
yx
yxx
x
x
x
?
?
??
?
?
011
1)(11)(
:0
????????
??
?
?
???????
?
?
?
??
dydxYdy
dydx
x
dxdxdy
y
F
xy
xy
xyy
y
y
y
?
?
??
?
?
16
下面推导平面应力问题的平衡微分方程，对单元体列平
衡方程,
011
1)(11)(
:0
????????
??
?
?
???????
?
?
?
??
dydxXdx
dxdy
y
dydydx
x
F
yx
yx
yxx
x
x
x
?
?
??
?
?
011
1)(11)(
:0
????????
??
?
?
???????
?
?
?
??
dydxYdy
dydx
x
dxdxdy
y
F
xy
xy
xyy
y
y
y
?
?
??
?
?
17
Sorting them gets,
0
0
??
?
?
?
?
?
??
?
?
?
?
?
Y
xy
X
yx
xyy
yxx
??
??
These two differential equation include three unknown
functions,Therefore,deciding the problem of the
stress weight is exceedingly and statically determinate;And still
must consider the deformation and displacement,then the problem
can be solved,
For the plane strain problem,the faces front and back still have
But they do not affect completely the establishes of the equation
above.So the equation above applies two kinds of plane problem
alike,
z?
yxxyyx ???? ?,,
18
整理得,
0
0
??
?
?
?
?
?
??
?
?
?
?
?
Y
xy
X
yx
xyy
yxx
??
??
这两个微分方程中包含着三个未知函数 。因此
决定应力分量的问题是超静定的；还必须考虑形变和位移，才能
解决问题。
对于平面应变问题,虽然前后面上还有,但它们完全不影
响上述方程的建立。所以上述方程对于两种平面问题都同样适用。
z?
yxxyyx ???? ?,,
19
§ 2-3 The stress on the Inclined Plane.Principal stress
1.The stress on the inclined plane
Having known the stress weight of any point P inside the
elastic body,we try to get the stress which pass the point P on the arbitrarily
inclined cross section.From neighborhood of point P taking a plane AB,which is
in parallel with the inclined plane above,and draws a small set square or three
column PAB on two planes which pass point P and have perpendicularity in the
shaft of x and y.When the plane AB approaches point P infinitely,the mean
stress on the plane AB will become the stress on the inclined plane above,
yxxyyx ???? ?,,
Establish the length of the face AB in the
plane xy is dS,N is the exterior normal
direction,and its direction cosine is,
myNlxN ?? ),c o s (,),c o s (
P A
B
xy?
x?
y?
N?
yx?
N?
NX
NY
S Ny
x
Fig.2－ 4
o
20
§ 2-3 斜面上的应力、主应力
一、斜面上的应力
已知弹性体内任一点 P处的应力分量,求经
过该点任意斜截面上的应力。为此在 P点附近取一个平面 AB，
它平行于上述斜面，并与经过 P点而垂直于 x轴和 y轴的两个平
面划出一个微小的三角板或三棱柱 PAB。当平面 AB与 P点无限
接近时，平面 AB上的应力就成为上述斜面上的应力。
yxxyyx ???? ?,,
设 AB面在 xy平面内的长度为 dS，
厚度为一个单位长度,N 为该面的外
法线方向，其方向余弦为,
myNlxN ?? ),c o s (,),c o s (
P A
B
xy?
x?
y?
N?
yx?
N?
NX
NY
S Ny
x
图 2－ 4
o
21
The projection of the whole stress on the inclined plane AB is XN
and YN respectively along with the shaft of x and y.From the PAB
equilibrium term can get,? ?0
xF
m d Sl d SdSX yxxN ?? ??
Divide and get,dS
yxxN mlX ?? ??
Same from and get,? ?0yF
xyyN lmY ?? ??
The positive stress on the inclined plane AB,from the
projection can get,
N?
xyyxNNN lmmlmYlX ???? 222 ?????
The shearing strength on the inclined plane AB,from the
projection can get,
N?
xyxyNNN mllmmXlY ???? )()( 22 ??????
22
斜面 AB上全应力沿 x轴及 y轴的投影分别为 XN和 YN。由 PAB
的平衡条件 可得,? ?0xF
m d Sl d SdSX yxxN ?? ??
除以 即得,dS
yxxN mlX ?? ??
同样由 得出,? ?0yF
xyyN lmY ?? ??
斜面 AB上的正应力,由投影可得,N?
xyyxNNN lmmlmYlX ???? 222 ?????
斜面 AB上的剪应力,由投影可得,
N?
xyxyNNN mllmmXlY ???? )()( 22 ??????
23
3.Principal stress
If the shearing stress of some inclined plane through point P is
equal to zero,then the positive stress of that inclined plane calls a
principal stress of point P,but that inclined plane calls the main
plane of the stress at point P,and the normal direction of that
inclined plane calls the main direction of the stress at point P,
1.The size of the principal stress
22
2
1 )
2(2 xy
yxyx ?????
?
? ?????
???
?
???
?
2.The direction of the principal stress is in the perpendicularity
with for each other,
1?
2?
24
二、主应力
如果经过 P点的某一斜面上的切应力等于零，则该斜面
上的正应力称为 P点的一个主应力，而该斜面称为 P点的一个
应力主面，该斜面的法线方向称为 P点的一个应力主向。
1.主应力的大小
22
2
1 )
2(2 xy
yxyx ?????
?
? ?????
???
?
???
?
2.主应力的方向
与 互相垂直。 1? 2?
25
§ 2-4 Geometrical Equation,
The Displacement of the Rigid Body
In plane problem,every point inside the elastic body can produce the arbitrarily
directional displacement.Take an unit PAB through any point P inside the elastic
body,such as Fig.2-5 show.After the elastic body suffers force,the point P,A,B
move to the point P′,A′,B′respectively,
P
o x
y
A
B
P?
A?
B?
u
v
dxxuu ???
dyyvv ???
dyyuu ???
dxxvv ???
?
?
Fig.2－ 5
一,The positive strain at point P
x
u
dx
udx
x
uu
x ?
????
??
?
)(
?
Here because of small deformation,
PA for causing stretch and shrink from
the y direction displacement v is the
small quantity of a high rank and this
small quantity may be omitted,
26
§ 2-4 几何方程、刚体位移
在平面问题中，弹性体中各点都可能产生任意方向的位移。
通过弹性体内的任一点 P，取一单元体 PAB，如图 2-5所示。弹性
体受力以后 P,A,B三点分别移动到 P′, A′, B′ 。
P
o x
y
A
B
P?
A?
B?
u
v
dxxuu ???
dyyvv ???
dyyuu ???
dxxvv ???
?
?
图 2－ 5
一,P点的正应变
在这里由于小变形，由 y
方向位移 v所引起的 PA的伸缩
是高一阶的微量，略去不计。
x
u
dx
udx
x
uu
x ?
????
??
?
)(
?
27
The same can get,
y
v
y ?
???
2.Shearing strain at point P
y
u
x
v
xy ?
??
?
???? ???
The corner of the line
segment PA,
x
v
dx
vdx
x
vv
?
????
??
?
)(
?
The same can get the corner
of the line segment PB,
y
u
?
???
Thus
28
同理可求得,
y
v
y ?
???
二,P点的切应变
y
u
x
v
xy ?
??
?
???? ???
线段 PA的转角,
x
v
dx
vdx
x
vv
?
????
??
?
)(
?
同理可得线段 PB的转角,
y
u
?
???
所以
29
Therefore get the geometrical equation of the plane problem
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
y
u
x
v
y
v
x
u
xy
y
x
?
?
?
From the geometrical equation above,when the displacement
weight of the object is completely certain,the deformation
weight is completely certain,unique weight can not be made sure
thoroughly,
30
因此得到平面问题的几何方程,
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
y
u
x
v
y
v
x
u
xy
y
x
?
?
?
由几何方程可见，当物体的位移分量完全确定时，形变
分量即可完全确定。反之，当形变分量完全确定时，位移分
量却不能完全确定。
31
§ 2-5 The Physical Equation
In the isotropy of the complete elasticity,the relation between the
deformation weight and the stress weight is established according
to the Hooke’s law as follows,
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
???
???
???
xyxy
zxzx
yzyz
yxzz
xzyy
zyxx
G
G
G
E
E
E
??
??
??
?????
?????
?????
1
1
1
)]([
1
)]([
1
)]([
1
32
§ 2-5 物理方程
在完全弹性的各向同性体内，形变分量与应力分量之间的
关系根据虎克定律建立如下,
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
???
???
???
xyxy
zxzx
yzyz
yxzz
xzyy
zyxx
G
G
G
E
E
E
??
??
??
?????
?????
?????
1
1
1
)]([
1
)]([
1
)]([
1
33
Inside the formula,the E is a modulus of elasticity;the G is a
stiffness modulus;the u is a poisson ratio.The relation of three ones
above,
)1(2 ???
EG
1.The physics equation of the plane stress problem
?
?
?
?
?
?
?
?
?
?
?
??
??
xyxy
xyy
yxx
E
E
E
?
?
?
????
????
)1(2
)(
1
)(
1
)( yxz E ???? ???
And have,
34
式中,E为弹性模量； G为刚度模量； 为泊松比。三者
的关系,
?
)1(2 ???
EG
一、平面应力问题的物理方程
?
?
?
?
?
?
?
?
?
?
?
??
??
xyxy
xyy
yxx
E
E
E
?
?
?
????
????
)1(2
)(
1
)(
1
)( yxz E ???? ???
且有,
35
2.The physics equation of the plane strain problem
?
?
?
?
?
??
?
?
?
?
?
?
?
?
?
?
?
?
?
?
xyxy
xyy
yxx
E
E
E
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
)1(2
)
1
(
1
)
1
(
1
2
2
3.The transformation relation of the relation type between the
stress strain and the plane strain,
The relation type of the
plane stress,
?
?
?
?
?
?
?
?
?
?
?
??
??
xyxy
xyy
yxx
E
E
E
?
?
?
????
????
2
1
)(
1
)(
1
36
二、平面应变问题的物理方程
?
?
?
?
?
??
?
?
?
?
?
?
?
?
?
?
?
?
?
?
xyxy
xyy
yxx
E
E
E
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
)1(2
)
1
(
1
)
1
(
1
2
2
三、平面应力的应力应变关系式与平面应变的关系式之间的
变换关系
将平面应力中的关系式,
?
?
?
?
?
?
?
?
?
?
?
??
??
xyxy
xyy
yxx
E
E
E
?
?
?
????
????
2
1
)(
1
)(
1
37
For change
?
?
?
?
?
?
?
?
1
1 2
E
E
Can get the relation type
in the plane strain,
?
?
?
?
?
??
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
xyxy
xyy
yxx
E
E
E
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
)1(2
1
1
1
1
2
2
Because of the similarity of this
kind,while solving plane strain
problem,the corresponding equation
of the plane problem and the elastic
constant in the answer can be
exchanged as above,can get the
solution of the homologous plane
strain problem,
38
作代换
?
?
?
?
?
?
?
?
1
1 2
E
E
就可得到平面应变中的
关系式,
?
?
?
?
?
??
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
xyxy
xyy
yxx
E
E
E
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
)1(2
1
1
1
1
2
2
由于这种相似性，在解平面
应变问题时，可把对应的平面应
力问题的方程和解答中的弹性常
数进行上述代换，就可得到相应
的平面应变问题的解。
39
§ 2-6 Boundary Conditions
When the object is placed in the state of equilibrium,its internal state of
stress at all point should satisfy the equilibrium differential equation and also
satisfy the boundary term on the boundary,
According to the difference of the boundary condition,the elasticity problem
is divided into the displacement boundary problem,stress boundary problem
and mixed boundary problem,
1.Displacement Boundary Term
When the displacement has been known on the boundary,the
displacement of the point on the object boundary and the equal term of the
fixed displacement should be established.For example,if making the
boundary of the fixed displacement is,and have(on the ),uSuS
uus ? vvs ?Among them,and means the displacement weight on the
boundary,however,and is the coordinate function we have know the
boundary,
su usv
v
40
§ 2-6 边界条件
当物体处于平衡状态时,其内部各点的应力状态应满足
平衡微分方程；在边界上应满足边界条件。
按照边界条件的不同，弹性力学问题分为位移边界问题、
应力边界问题和混合边界问题。
一、位移边界条件
当边界上已知位移时，应建立物体边界上点的位移与
给定位移相等的条件。如令给定位移的边界为,则有
（在 上）,
uS
uS
uus ? vvs ?
其中 和 表示边界上的位移分量，而 和 在边界上
是坐标的已知函数。 su
usv v
41
2.Stress boundary term
When the boundary of the object is given to surface force,then the stress of
the object on the boundary should satisfy the equilibrium term of forces with
the equilibrium of the surface force,
??
???
??
??
Ylm
Xml
sxysy
syxsx
)()(
)()(
??
??
Among them,and are the surface force weights and,,,
are the stress weights on the boundary,
X Y sx)(? sy)(?
sxy)(? syx)(?
When the boundary face is in perpendicularity in shaft x,stress boundary
term can be changed briefly into,
YX sxysx ???? )(,)( ??
When the boundary face is in perpendicularity in shaft y,stress boundary
term can be changed briefly into,
XY syxsy ???? )(,)( ??
42
二、应力边界条件
当物体的边界上给定面力时，则物体边界上的应力应满
足与面力相平衡的力的平衡条件。
??
???
??
??
Ylm
Xml
sxysy
syxsx
)()(
)()(
??
??
其中 和 为面力分量,、,, 为边界上的应
力分量。
X Y sx)(? sy)(? sxy)(? syx)(?
当边界面垂直于 轴时，应力边界条件简化为,x
YX sxysx ???? )(,)( ??
当边界面垂直于 轴时，应力边界条件简化为,y
XY syxsy ???? )(,)( ??
43
3.Mixed boundary condition
1.The displacement has been known on a part of boundaries of the object,the
result of which have the displacement boundary term,the boundaries of other parts
have the surface force we have know.And then there should be stress boundary
term and displacement boundary term respectively on two parts of the
boundaries.The left surface of the cantilever contains displacement boundary
term,such as shown in Fig.2-6,
??
?
??
??
0
0
vv
uu
s
s
Top and bottom surface contains
stress boundary term,
??
?
?
?
???
???
0)(
0)(
sy
syx
Y
X
?
?
The right surface contains stress
boundary term,
??
?
?
?
??
??
0)(
)(
sxy
sx
Y
qX
?
?
l
q
x
y
o
2h
2h
Fig.2-6
44
三、混合边界条件
1.物体的一部分边界上具有已知位移，因而具有位移边界条
件，另一部分边界上则具有已知面力。则两部分边界上分别
有应力边界条件和位移边界条件。如图 2-6，悬臂梁左端面
有位移边界条件,
??
?
??
??
0
0
vv
uu
s
s
上下面有应力边界条件,
??
?
?
?
???
???
0)(
0)(
sy
syx
Y
X
?
?
右端面有应力边界条件,
??
?
?
?
??
??
0)(
)(
sxy
sx
Y
qX
?
?
l
q
x
y
o
2h
2h
图 2-6
45
2.On the same boundary,there are not only stress boundary term but
displacement boundary term.Coupler sustains the boundary term,such
as shown in Fig.2-7,
??
?
?
?
??
??
0)(
0
sxy
s
Y
uu
?
The alveolus boundary term shown
in Fig.2-8,
?
?
?
??
??
0)(
0
sx
s
X
vv
?
o x
y
Fig.2-7
x
y
o
Fig.2-8
46
2.在同一边界上，既有应力边界条件又有位移边界条件。
如图 2-7连杆支撑边界条件,
??
?
?
?
??
??
0)(
0
sxy
s
Y
uu
?
如图 2-8齿槽边界条件,
?
?
?
??
??
0)(
0
sx
s
X
vv
?
o x
y
图 2-7
x
y
o
图 2-8
47
§ 2-7 Saint-Venant Principle
1.Saint-Venant’s Principle
If transforming a small part of the surface force on the boundary into the
surface force that has equal effect but different distribution(The main vector is
equal,so is the main quadrature to the same point as well),and then the
distribution of the stress force nearby will have prominent changes,but the
influence from the distant place can not be accounted,
2.Give Examples
Establishing the component of the column forms,the centroid of area in cross
sections of both ends suffers the tensible force which is equal in size but
contrary in direction,such as shown in Fig.2-9a.If transforming an or both ends
of tensile force into the force at the same effect as the static force,such as
shown in Fig.2-9b or Fig.2-9c,the distribution of stress force drawn only by
broken line has prominent changes,whereas,the influence of the rest parts can
not be accounted.If changing both ends of tensile force into that of uniform
distribution again,the gathering degree is equal to P/A and among them A is the
cross-section area of the component,such as shown in Fig.2-9d,there is still the
stress close to both ends under the noticeable influence,
48
§ 2-7 圣维南原理
一,圣维南原理
如果把物体的一小部分边界上的面力，变换为分布不同
但静力等效的面力（主矢量相同，对于同一点的主矩也相
同），那么，近处的应力分布将有显著的改变，但是远处所
受的影响可以不计。
二,举例
设有柱形构件，在两端截面的形心受到大小相等而方向
相反的拉力,如图 2-9a。如果把一端或两端的拉力变换为
静力等效的力，如图 2-9b或 2-9c，只有虚线划出的部分的应
力分布有显著的改变，而其余部分所受的影响是可以不计的。
如果再将两端的拉力变换为均匀分布的拉力，集度等于,
其中 为构件的横截面面积，如图 2-9d，仍然只有靠近两端
部分的应力受到显著的影响。
P
AP/ A
49
P P
2/P
2/P
2/P
2/P
P 2/P
2/P
AP/AP/
PP
Fig.2-9
(a)
(b)
(c)
(d)
(e)
Under the four kinds of
circumstances above,parts of
distribution of stress force distant
from both ends have no marked
difference,
Attention,
The application of the Saint-
Venant’s principle is by no
means separated from the term
of Equal Effect of Static Force,
50
P P
2/P
2/P
2/P
2/P
P 2/P
2/P
AP/AP/
PP
图 2-9
(a)
(b)
(c)
(d)
(e)
在上述四种情况下，离开两
端较远的部分的应力分布，并
没有显著的差别。
注意,
应用圣维南原理，绝不能
离开, 静力等效, 的条件。
51
§ 2-8 Solving the Plane Problem
according to the displacement
There are three kinds of basic methods to solve the problem in elasticity:the
solution to the problem according to displacement,stress force and admixture,
While solving problems using displacement method,we regard displacement
weight as the basic function unknown.After getting displacement weight from
only including the differential equation and boundary term of the displacement
weight,then get the deformation weight using geometrical equation,therefore,get
the stress weight with the physics equation,
1.Plane Stress Problem
?
?
?
?
?
?
?
?
?
?
?
??
??
xyxy
xyy
yxx
E
E
E
?
?
?
????
????
)1(2
)(
1
)(
1
In plane stress problem,the physics
equation is,
52
§ 2-8 按位移求解平面问题
在弹性力学里求解问题，有三种基本方法：按位移求解、
按应力求解和混合求解。
按位移求解时，以位移分量为基本未知函数，由一些只
包含位移分量的微分方程和边界条件求出位移分量以后，再
用几何方程求出形变分量，从而用物理方程求出应力分量。
一、平面应力问题
?
?
?
?
?
?
?
?
?
?
?
??
??
xyxy
xyy
yxx
E
E
E
?
?
?
????
????
)1(2
)(
1
)(
1
在平面应力问题中，物理方程为,
53
From three formulas above mentioned to solve the stress weight,can get,with the
substitution of geometrical equation,we can get the elasticity equation,
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
xyxy
xyy
yxx
E
E
E
?
?
?
???
?
?
???
?
?
)1(2
)(
1
)(
1
2
2
Again equilibrium differential equation with substitution in formula(a),simplification
hereafter,can get,
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
)(
)1(2
)(
1
)(
1
2
2
y
u
x
vE
x
u
y
vE
y
v
x
uE
xy
y
x
?
?
?
?
?
?
?
?
（ a)
This is the equilibrium differential equation to mean with the displacement,ie,when
solving the plane stress problem according to displacement method,we adopt a
basic differential equation for needs,
?
?
?
?
?
?
?
??
??
??
?
?
??
?
?
?
?
??
??
??
?
?
??
?
?
?
?
0)
2
1
2
1
(
1
0)
2
1
2
1
(
1
2
2
2
2
2
2
2
2
2
2
2
2
Y
yx
u
x
v
y
vE
X
yx
v
y
u
x
uE
??
?
??
?
（ 1）
54
由上列三式求解应力分量，得,
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
xyxy
xyy
yxx
E
E
E
?
?
?
???
?
?
???
?
?
)1(2
)(
1
)(
1
2
2 将几何方程代入，得弹性方程,
再将式（ a）代入平衡微分方程，简化以后，即得,
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
)(
)1(2
)(
1
)(
1
2
2
y
u
x
vE
x
u
y
vE
y
v
x
uE
xy
y
x
?
?
?
?
?
?
?
?
（ a)
这是用位移表示的平衡微分方程，也就是按位移求解平面应力
问题时所需用的基本微分方程。
?
?
?
?
?
?
?
??
??
??
?
?
??
?
?
?
?
??
??
??
?
?
??
?
?
?
?
0)
2
1
2
1
(
1
0)
2
1
2
1
(
1
2
2
2
2
2
2
2
2
2
2
2
2
Y
yx
u
x
v
y
vE
X
yx
v
y
u
x
uE
??
?
??
?
（ 1）
55
The stress boundary term with substitution in formula(a),simplification hereafter,
can get,
?
?
?
??
?
?
?
?
?
?
?
??
?
?
?
?
?
?
?
?
?
?
?
?
??
?
?
?
?
?
?
?
Y
y
u
x
v
l
x
u
y
v
m
E
X
x
v
y
u
m
y
v
x
u
l
E
ss
ss
])(
2
1
)([
1
])(
2
1
)([
1
2
2
?
?
?
?
?
?
This is the stress force boundary to mean with the displacement,ie,we adopt
the boundary term of the stress force when solving the plane stress problem
according to displacement method,
（ 2）
Sum up,when solving the plane stress problem according to displacement
method,we should make the displacement weight satisfy differential
equation(1) and combine to satisfy displacement boundary term or stress
boundary term or stress boundary term(2) on the boundary,After getting
displacement weight,we can get the deformation weight with geometrical
equation and then get the stress force weight with the physics equation,
2.Plane strain problem
?
??
? ???? 1,1 2
EE
Make the substitution between and in each equation of the plane
strain problem,
E ?
56
将（ a）式代入应力边界条件，简化以后，得,
?
?
?
??
?
?
?
?
?
?
?
??
?
?
?
?
?
?
?
?
?
?
?
?
??
?
?
?
?
?
?
?
Y
y
u
x
v
l
x
u
y
v
m
E
X
x
v
y
u
m
y
v
x
u
l
E
ss
ss
])(
2
1
)([
1
])(
2
1
)([
1
2
2
?
?
?
?
?
?
这是用位移表示的应力边界条件，也就是按位移求解平面应
力问题时所用的应力边界条件。
（ 2）
总结起来，按位移求解平面应力问题时，要使得位移分
量满足微分方程（ 1），并在边界上满足位移边界条件或应
力边界条件（ 2）。求出位移分量以后，用几何方程求出形
变分量，再用物理方程求出应力分量。
二、平面应变问题
?
??
? ???? 1,1 2
EE
只须将平面应力问题的各个方程中 和 作代换,E ?
57
§ 2-9 Solving the Plane Problem According
to the Stress Force.Compatible Equantion
While solving the plane problem according to the displacement,we must
combine two partial differential equation of the second ranks to solve the
problem,this is very difficult on the mathematics,But while solving the plane
problem according to the stress force,we can avoid this difficulty and so what
we adopt more is to get the solution according to the stress force,
While getting the solution according to the stress force,we regard stress
weight as the basic function unknown.After getting displacement weight from
only including the differential equation and boundary term of displacement
weight,then get the deformation weight using physics equation,therefore,get
the displacement weight with geometrical equation,
Compatible Equation
From geometrical equation of the
plane problem,
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
y
u
x
v
y
v
x
u
xy
y
x
?
?
?
58
§ 2-9 按应力求解平面问题。相容方程
按位移求解平面问题时，必须求解联立的两个二阶偏微分
方程，这在数学上是相当困难的。而按应力求解弹性力学平面
问题，则避免了这个困难，故更多采用的是按应力求解。
按应力求解时，以应力分量为基本未知函数，由一些只包
含应力分量的微分方程和边界条件求出应力分量以后，再用物
理方程求出形变分量，从而用几何方程求出位移分量。
相容方程
由平面问题的几何方程,
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
y
u
x
v
y
v
x
u
xy
y
x
?
?
?
59
Can get,
yxx
v
y
u
yxxy
v
yx
u
xy
xyyx
??
??
?
??
?
?
??
??
??
??
??
??
?
??
?
? ??? 22
2
3
2
3
2
2
2
2
)(
ie,
yxxy
xyyx
??
??
?
??
?
? ??? 2
2
2
2
2
This relation type calls the deformation moderates equation or
compatible equation,
1.Compatible equation in plane stress force
))(1())(( 2
2
2
2
y
Y
x
X
yx yx ?
??
?
?????
?
??
?
? ???
2.Compatible equation in plane strain force
)(1 1))(( 2
2
2
2
y
Y
x
X
yx yx ?
??
?
?
?????
??
?
?
???
60
可得,
yxx
v
y
u
yxxy
v
yx
u
xy
xyyx
??
??
?
??
?
?
??
??
??
??
??
??
?
??
?
? ??? 22
2
3
2
3
2
2
2
2
)(
即,
yxxy
xyyx
??
??
?
??
?
? ??? 2
2
2
2
2
这个关系式称为形变协调方程或相容方程。
（一）平面应力问题的相容方程
))(1())(( 2
2
2
2
y
Y
x
X
yx yx ?
??
?
?????
?
??
?
? ???
（二）平面应变问题的相容方程
)(1 1))(( 2
2
2
2
y
Y
x
X
yx yx ?
??
?
?
?????
??
?
?
???
61
While solving the plane problem according to the stress force,
the stress weight should not only satisfy both the equilibrium
differential equation and compatible equation,but satisfy the
stress boundary term on the boundary whether is a plane stress
problem or plane strain problem,
62
按应力求解平面问题时，无论是平面应力问题还是平面
应变问题，应力分量除了满足平衡微分方程和相容方程外，
在边界上还应当满足应力边界条件。
63
§ 2-10 The Simplification Under the
Circumstances of Ordinary Physical Force
Under the circumstances of ordinary physical force,the compatible equation
of two kinds of plane problems is simplified as,
0))(( 2222 ??????? yxyx ??
yx ?? ?
yx ?? ?
2?
Therefore,under the circumstances of ordinary physical force,should
satisfy Laplace differential equation(in harmony with equation),should
be harmonic functions.Represent with the mark,the formula above
can be simplified as,
2
2
2
2
yx ?????
0)(2 ??? yx ??Conclusion
In the stress boundary problem of single connection if two elastic bodies have
the same boundary shape and suffer the external force of the same distribution,
and then stress force distribution,,should be the same whether the
materials of two elastic bodies are same or not and whether they are under the
plane stress circumstances or under the plane strain circumstances(Two kinds of
the stress force weight in the plane problem,the deformation and the
displacement are uncertainly the same),
x? y? xy?
z?
64
§ 2-10 常体力情况下的简化
常体力下，两种平面问题的相容方程都简化为,
0))(( 2
2
2
2
??????? yxyx ??
可见，在常体力的情况下,应当满足拉普拉斯微分方程
（调和方程）,应当是调和函数。
用记号 代表,上式简写为,
2
2
2
2
yx ?
??
?
?
yx ?? ?
yx ?? ?
2? 0)(2 ???
yx ??
结论
在单连体的应力边界问题中，如果两个弹性体具有相同的
边界形状，并受到同样分布的外力，那么，不管这两个弹性体
的材料是否相同，也不管它们是在平面应力情况下或是在平面
应变情况下，应力分量,, 的分布是相同的（两种平
面问题中的应力分量,以及形变和位移，却不一定相同）。
x? y? xy?
z?
65
Inference 2
When measuring the above stress weight of the structure or component with
the method of experiment,we can make the model using the material of the
convenient measurement in order to replace original structure or component
materials of the inconvenient measurement; we also can adopt structure or
component of long column shape under the plane strain circumstances,
Inference 3
Under the circumstance of constant volumetric force,for the stress boundary
problem of single connection,we can charge the function of the volumetric
force as that of the surface force in order to solve the problem and experiment
measurement,
Inference 1
The stress weight,,that is solved according to any object is also
applicable to the object which has the same boundary and other materials
suffering the same external force; The stress weight that is solved according to
plane stress problem is also applicable to the object which has the same boundary
and the same external force under the plane strain circumstances,
x? y? xy?
66
推论 2
在用实验方法测量结构或构件的上述应力分量时，可以用
便于量测的材料来制造模型，以代替原来不便于量测的结构或
构件材料；还可以用平面应力情况下的薄板模型，来代替平面
应变情况下的长柱形的结构或构件。 推论 3
常体力的情况下，对于单连体的应力边界问题，还可以
把体力的作用改换为面力的作用，以便于解答问题和实验量
测。
推论 1
针对任一物体而求出的应力分量,,,也适用于
具有同样边界并受有同样外力的其它材料的物体；针对平面应
力问题而求出的这些应力分量，也适用于边界相同、外力相同
的平面应变情况下的物体。
x? y? xy?
67
§ 2-11 Stress Function.Inverse
Solution Method and Semi-Inverse Method
1.Stress function
While solving the stress boundary problem according to the stress force and
when the volumetric force is the constant quantity,the stress weight,,
should satisfy the equilibrium differential equation,x? y? xy?
?
?
?
?
?
?
?
??
?
?
?
?
?
??
?
?
?
?
?
0
0
Y
xy
X
yx
xyy
xyx
??
??
（ a）
And compatible equation
0))(( 2
2
2
2
??????? yxyx ??
（ b)
The solution to the equation (a) includes two parts,arbitrarily a particular
solution and the general solution to the following homogeneous differential
equation,
68
§ 2-11应力函数、逆解法与半逆解法
一、应力函数
按应力求解应力边界问题时，在体力为常量的情况下，应
力分量,, 应当满足平衡微分方程,
x? y? xy?
?
?
?
?
?
?
?
??
?
?
?
?
?
??
?
?
?
?
?
0
0
Y
xy
X
yx
xyy
xyx
??
??
（ a）
以及相容方程
0))(( 2
2
2
2
??????? yxyx ??
（ b)
方程（ a）的解包含两部分：任意一个特解和下列齐次
微分方程的通解。
69
The particular solution is,
Rewrite the former equation inside the homogeneous
differential equation (c)as,
)( xyx yx ?? ??????
According to the differential equation theory,it is certain to
exist some function,make,),( yxA
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
0
0
xy
yx
xyy
xyx
??
??
（ c）
0,,????? xyyx YyXx ??? （ d）
x
A
y
A
xy
x
?
?
??
?
?
?
?
? （ e）
（ f）
70
特解取为,
将齐次微分方程（ c）中前一个方程改写为,
)( xyx yx ?? ??????
根据微分方程理论，一定存在某一个函数,使得,),( yxA
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
0
0
xy
yx
xyy
xyx
??
??
（ c）
0,,????? xyyx YyXx ??? （ d）
x
A
y
A
xy
x
?
?
??
?
?
?
?
? （ e）
（ f）
71
Similarly rewrite the second equation
inside (c) as,)( xyy xy ?? ??????
It is certain to exist some function as well,make,),( yxB
y
B
x
B
xy
y
?
?
??
?
?
?
?
? （ g）
（ h）
From the formula (f)
and (h),can get,
y
B
x
A
?
??
?
?
Thus,it is certain to exist some function,make,),( yx?
x
B
y
A
?
?
?
?
?
?
?
? （ i）
（ j）
72
同样将（ c）中的第二个方程改写为,
)( xyy xy ?? ??????
也一定存在某一个函数,使得,),( yxB
y
B
x
B
xy
y
?
?
??
?
?
?
?
? （ g）
（ h）
由式（ f）及（ h）得,
y
B
x
A
?
??
?
?
因而一定存在某一个函数,使得,),( yx?
x
B
y
A
?
?
?
?
?
?
?
? （ i）
（ j）
73
Make the formula (i) substitute to (e),(j)to(g),and (i)to(f),then get the general
solution,
yxxy xyyx ??
???
?
??
?
?? ?????? 2
2
2
2
2,,（ k）
Make the general solution (k)plus the particular solution(d),then get the
whole solution of the differential equation(a),
The function calls the stress function of the plane problem and also calls the
Array stress function,
In order that the stress weight(1) can also satisfy the compatible equation(b),
make formula (1)substitute formula (b),then get,
?
yxYyxXxy xyyx ??
????
?
???
?
?? ?????? 2
2
2
2
2,,（ 1）
0))(( 22222222 ????????????? YyxXxyyx ??
The formula above can be simplified,
0))(( 22222222 ??????????? yxyx ??
74
将式（ i）代入（ e），式（ j）代入（ g），并将式（ i）代入
（ f），即得通解,
yxxy xyyx ??
???
?
??
?
?? ?????? 2
2
2
2
2,,（ k）
将通解（ k）与特解（ d）叠加，即得微分方程（ a）的全
解,
函数 称为平面问题的应力函数，也称为艾瑞应力函数。 ?
yxYyxXxy xyyx ??
????
?
???
?
?? ?????? 2
2
2
2
2,,（ 1）
为了应力分量（ 1）同时也能满足相容方程（ b），将（ 1）
代入式（ b），即得,
0))(( 22222222 ????????????? YyxXxyyx ??
上式可简化为,
0))(( 22222222 ??????????? yxyx ??
75
Or spreading the formula is,
02 4422444 ?????? ???? yyxx ???
Further simplification is,04 ?? ?
（ 2）
2.Inverse solution method and semi-inverse method
Inverse solution,the first step is to set up multiform stress function which
satisfy the compatible equation(2),and get the stress weight with the formula(1),
then investigate according to the stress boundary term.On the elastic body in
every kind of shape,these stress weights correspondence in what kinds of surface
force,from which we know that the stress function for setting up can solve what
kinds of problem,
?
While solving the stress boundary problem according to stress force,if the
volumetric force is constant quantity,we may only consult the differential
equation(2) to solve the stress function,and then get the stress weight with
the formula(1),but these stress weights should satisfy the stress boundary term
on the boundary,
?
The basic step of inverse solution method,
76
或者展开为,
02 4422444 ?????? ???? yyxx ???
进一步简写为,04 ?? ?
（ 2）
二、逆解法与半逆解法
逆解法,先设定各种形式的、满足相容方程（ 2）的应力函
数,用公式（ 1）求出应力分量，然后根据应力边界条件来
考察，在各种边界形状的弹性体上，这些应力分量对应于什么
样的面力，从而得知所设定的应力函数可以解决什么问题。
?
按应力求解应力边界问题时，如果体力是常量，就只须由
微分方程（ 2）求解应力函数,然后用公式（ 1）求出应力
分量，但这些应力分量在边界上应当满足应力边界条件。
?
逆解法基本步骤,
77
Semi-inverse method,Aiming at the problem for requesting solution,according
to boundary shape of elastic body and force circumstance for suffering,supposing
the partial and the whole stress weight as a certain form function,from which
conclude stress function,then come to investigate whether this stress function
satisfy compatible equation or not.And the original stress weight for supposing
and the rest stress weight from this stress function are whether to satisfy stress
boundary term and single worth term in displacement or not.If both compatible
equation and all aspects of terms can be satisfied;it is natural to get the accurate
answer;if some aspect can not be satisfied,we should establish assumption on the
other hand and investigate it again,
Setting up
?
Solution to stress
weight
Solution to surface
force(Resultant
force)
Solution to
surface
force(Resultant
force)
Substitution
Substitution
Type(1) Stress
boundary
term
Maki
ng
sure
The basic steps of semi-inverse method,
Setting up Getting the
accurate solution
Educing the stress type of
expression
Satisfying the
boundary term
Satisfying
04 ??? Yes Yes
No No
Type(1) stress
Boundary term
?
78
半逆解法,针对所要求解的问题，根据弹性体的边界形状和受
力情况，假设部分或全部应力分量为某种形式的函数，从而推
出应力函数,然后来考察，这个应力函数是否满足相容方程，
以及，原来所假设的应力分量和由这个应力函数求出的其余应
力分量，是否满足应力边界条件和位移单值条件。如果相容方
程和各方面的条件都能满足，自然就得出正确的解答；如果某
一方面不能满足，就要另作假设，重新考察。
?
设定 ? 求出 应力分量 求出 面力（合力） 解决 什么问题
代入 代入
式（ l） 应力边
界条件
确定
半逆解法基本步骤,
设定 ? 导出 应力表达式 得到 正确解答 满足 边界条件 满足
04 ??? 是 是
否 否
式（ l） 应力边
界条件
79
Exercise Lesson for the Basic
Theory of the Plane Problem
Solution,From the mechanics of materials,the
bending moment through point P on the cross-
section is,
yxlh qhyxlqI yM
z
z
x
3
33
3 2)
1
12(
6 ???????
)6(
3
lqxM z ??
（ 1）
)(
36
0
22
3
2
3 xfyxlh
q
y d yx
lh
q
dy
x
yx
x
xy
xyx
???
?
?
??
?
?
?
?
?
?
? ?
?
?
??Equilibrium differential equation in
substitution,get,（ 2）
[Exercise 1] The upper part of the cantilever suffers
the load of the line distribution show in figure
according to the type of expression in the mechanics
of materials,then lead the type of expression of
and with the equilibrium differential equation,x? y?xy? o
y
x
l
q
P
2/h
2/h
）（ 1??
80
,平面问题的基本理论, 习题课
解,由材料力学知，过 点横截面
上的弯矩为,
P
yxlh qhyxlqI yM
z
z
x
3
33
3 2)
1
12(
6 ???????
)6(
3
lqxM z ??
（ 1）
)(
36
0
22
3
2
3 xfyxlh
q
y d yx
lh
q
dy
x
yx
x
xy
xyx
???
?
?
??
?
?
?
?
?
?
? ?
?
?
??代入平衡微分方程,（ 2）
[练习 1] 悬臂梁上部受线形分布载荷，
如图所示。试根据材料力学中 的
表达式，再用平衡微分方程导出
和 的表达式。
x?
y?
xy?
o
y
x
l
q
P
2/h
2/h
）（ 1??
81
Make use of the boundary term above and underneath and make sure f(x),
Make the type (3) substitute the second type of the equilibrium differential equation
and get,
xhyhy
lh
q
x
l
q
xg
xgxyhy
lh
q
dy
x
y
h
y
y
xy
y
)34(
2
2
)(,0)(
)()34(
2
323
3
2
23
3
????
???
????
?
?
??
?
?
?
?
?
?
（ 4）
)4(
4
3
4
3
)(,0)(
22
3
2
2
2
hy
lh
qx
x
lh
q
xf
xy
h
y
xy
??
???
??
?
?
（ 3）
Attention,Type (1),(3),(4) expressed only are perhaps stress weight form the
state if having right solution,and they still need to satisfy the compatible
equation expressed with stress force,
82
利用上、下面边界条件确定 )(xf
将式（ 3）代入平衡微分方程中的第二式，得,
xhyhy
lh
q
x
l
q
xg
xgxyhy
lh
q
dy
x
y
h
y
y
xy
y
)34(
2
2
)(,0)(
)()34(
2
323
3
2
23
3
????
???
????
?
?
??
?
?
?
?
?
?
（ 4）
)4(
4
3
4
3
)(,0)(
22
3
2
2
2
hy
lh
qx
x
lh
q
xf
xy
h
y
xy
??
???
??
?
?
（ 3）
注意,式（ 1）、（ 3）、（ 4）表达的仅是静力可能的应力分
量，若为正确解答，则还需满足以应力表示的相容方程。
83
A
B
xy
o
0?x?
0?xy?
0?y?
0?yx?
[Exercise 2] In the plane object shown in
figure,both angle A and angle B are right one
and the surface on the boundary nearby doesn’t
suffer the external force.Try to explain the
stress state on the point A and B,
Solution,Because the boundary near point A doesn’t
suffer the external force,the stress weight
on this point should satisfy the following
boundary term,
0)()()( ??? AxyAyAx ???
Ie,point A is placed in zero stress state,But point B is placed in the vertex of the
cave angle,each surface of differential unit for taking from this point is not the
boundary surface,therefore,the stress weight on it is unknown and not
definitely zero.From the theory analysis,we know that the stress force on point
B of the cave angle is infinite big,
84
A
B
xy
o
0?x?
0?xy?
0?y?
0?yx?
[练习 2] 如图 所示为平面物体，角
和角 均为直角，其附近边界表面
均不受外力，试说明, 两点的
应力状态。
A
B
A B
解,由于 点附近边界不受外力，该
点的应力分量应满足如下边界条
件,
0)()()( ??? AxyAyAx ???
即 点处于零应力状态。而 点处
于凹角的顶点，该点所取的微分单
元体的各个面均不是边界面，因此，
其上的应力分量是未知的，未必为
零，由理论分析知，凹角处 点的
应力趋于无限大。
A
A B
B
85
[Exercise 3] Try to write the displacement boundary term (Using the rectangular
coordinates) of each plane object in the form,Among them,point 0 in the second
figure doesn’t move and the level line segment through point 0 has no rotation,
Solution,The boundary term in each displacement in the form,
h x
y
o
h x
y
o
0,0,
2
,0
0,
2
,0
????
????
vu
h
yx
u
h
yx
0,0,0
0,0
?????
??
x
vvu
yx
86
[练习 3] 试写出表中所示各平面物体的位移边界条件（用直角
坐标），其中第二图中 点不动，过 点的水平线段无转动。 o o
解,各位移边界条件见表所列。
h x
y
o
h x
y
o
0,0,
2
,0
0,
2
,0
????
????
vu
h
yx
u
h
yx
0,0,0
0,0
?????
??
x
vvu
yx
87
h x
y
o
l
h x
y
o
l
?
0,
2
,
0,0,0,
0,0,0,0
???
????
????
v
h
ylx
vuylx
vuyx
0s inc o s,0,
0,0,0,0
????
????
?? vuylx
vuyx
88
h x
y
o
l
h x
y
o
l
?
0,
2
,
0,0,0,
0,0,0,0
???
????
????
v
h
ylx
vuylx
vuyx
0s inc o s,0,
0,0,0,0
????
????
?? vuylx
vuyx
89
[Solution 4] Whether are some kinds of suffering force shown in figure
the plane problem or not? If it is,then plane stress problem or still plane
strain problem?
R
O
q
x
y
q
h
z
y
o
R>>h
a)
h
Q
Q
O z
y
x
y
R O
R>>h
b)
90
[练习 4] 图所示的几种受力体是否为平面问题？若是，则是
平面应力问题，还是平面应变问题？
R
O
q
x
y
q
h
z
y
o
R>>h
a)
h
Q
Q
O z
y
x
y
R O
R>>h
b)
91
R<<l
p
R
O
p
y
x O
p
z
L
p
y
c)
Figure a) is the plane stress problem.The load carries the
perpendicular function in board surface shown in Fig.b),so
it is the bending problem of lamella,The load carries the
perpendicular function in board edge shown in Fig.c).The
load and the cross-section have no variety along the shaft
of Z and R<<1,so it is the plane strain problem,
Solution,
92
R<<l c)
图 a)所示为平面应力问题。图 b）所示荷载垂直作用
于板面，故为薄板弯曲问题。图 c）所示荷载作用于板
边，荷载及横截面沿 z轴无变化，且 R<<L，故为平面
应变问题。
解,
p
R
O
p
y
x O
p
z
L
p
y
93
[Exercise 5] The lamella bar suffers the even tensile force
function in the direction of y shown in figure.Try to prove
there is no existence of the stress force on the top point A
of outstanding part in the center board,
q
2N
O
C
B
A
y
q
x 2?
1?
1N
Solution,It can be seen as plane stress problem.Both AB
and AC are free boundary(and ),and
have no surface force function,ie,
Substituting boundary term has,
21 ?? ?
0?? YX
1111 s i n,c o s ?? ?? ml
The boundary of AB,
( 1 )
0σs i n ατc o s α
0τs i n ασc o s α
y1xy1
xy1x1
??
?
?
?
??
??
94
[练习 5] 如图所示薄板条在 y方向受均匀拉力
作用，试证明在板中间突出部分的尖端 A处无
应力存在。
解, 本题可视为平面应力问题,AB和 AC都
是自由边界（且 ），无面
力作用，即,。代入边
界条件有,
21 ?? ?
0?? YX
AB边界,
1111 s i n,c o s ?? ?? ml
( 1 )
0σs i n ατc o s α
0τs i n ασc o s α
y1xy1
xy1x1
??
?
?
?
??
??
q
2N
O
C
B
A
y
q
x 2?
1?
1N
95
The boundary of AC,
12
122
s i n
c o sc o s
?
??
??
??
m
l
)2(
0s i nc o s
0s i nc o s
11
11
??
?
?
?
??
??
yxy
xyx
????
????
Because A is placed in the boundary of AB and AC,it needs to satisfy type(1) and
type (2) at the same time,From it,can get,the problem has
been proved,0??? xyyx ???
[Exercise 6] Figure a) is a rectangle cross-section dam,whose right side suffers the
pressure force of the static water,The top section suffers the function of the
concentrated force P,Try to write a stress boundary term of dam and fix edge needn’t
be considered,
96
AC边界,
12
122
s i n
c o sc o s
?
??
??
??
m
l
)2(
0s i nc o s
0s i nc o s
11
11
??
?
?
?
??
??
yxy
xyx
????
????
由于 A同处于 AB,AC边界，因此，需同时满足式（ 1）和式（ 2）
，由此解得：,问题得证。
0??? xyyx ???
[练习 6] 图 a）所示为一矩形截面水坝，其右侧面受静水压力。
顶部受集中力 P作用。试写出水坝的应力边界条件，固定边不必考
虑。
97
P x
? O
y
y?
2h
h h
a)
P x
? O
y
y?
xy?
x?
y?
yx?
b)
P
?x O
y?
yx?
y
c)
Solution,1.List the stress boundary term,
(1) The left boundary,)1( 0)(,0)( ??
?? hxxyhxx ??
(2) The right boundary,
)2( 0)(,)( ??? ???? hxxyyhxx ???
98
解,1、列出应力边界条件
（ 1）左边界,
)1(0)(,0)( ?? ?? hxxyhxx ??
（ 2）右边界,
)2(0)(,)( ??? ???? hxxyyhxx ???
P x
? O
y
y?
2h
h h
a)
P x
? O
y
y?
xy?
x?
y?
yx?
b)
P
?x O
y?
yx?
y
c)
99
Top End Part,
)5(c o s)(
)4(s in
2
)(
)3(s in)(
0
0
0
??
??
??
Pdx
h
Px d x
Pdx
y
h
h
xy
h
h
yy
h
h
yy
?
??
??
?
?
?
?
?
?
?
?
?
[Exercise 7] The structure in the figure is consist of two kinds of different
materials.Try to get solution to displacement and stress answer with the function
of load q in the vertical distribution of equality (Establishing h,a,l,
has been known),
qEE,,,,2211 ??
1.Adopt the solution to displacement.Because this structure is placed in
the pressuring state with double direction of equality(Both stress and
strain are the constant quantity),therefore,we can suppose its
displacement as a line function and now divide two fields on top and
underneath to express,
Solution,
100
（ 3）上端部,
)5(c o s)(
)4(s in
2
)(
)3(s in)(
0
0
0
??
??
??
Pdx
h
Px d x
Pdx
y
h
h
xy
h
h
yy
h
h
yy
?
??
??
?
?
?
?
?
?
?
?
?
[练习 7] 图所示结构由两种不同材料构成。试求其在竖向均布
荷载 q作用下的位移和应力解答（设 h,a,L,均已
知）。
qEE,,,,2211 ??
1、采用位移解法。由于此结构处于双向均匀受压状态
（应力、应变为常量），因此，可假设其位移是线性函数，
现分上、下两区域表达为,
解,
101
The Part ABCD,
)1(,11111111 fycxdvcybxau ??????
The Part CDEF,
)2(,22222222 fycxdvcybxau ??????
Obviously,the type(1) and type(2) can satisfy Lamei equation of the plane stress
circumstance,
2.Consider the displacement constraints and
continuous term of transformation,
ayay
hyhylxlx
vv
vuuu
??
??????
?
????
)()(
0)(,0)(,0)(,0)(
21
2221
From it,we can get,
)(,0,,0
0
2111222
222111
haeaefdhefd
cbacba
????????
??????
102
ABCD部分,
)1(,11111111 fycxdvcybxau ??????
CDEF部分,
)2(,22222222 fycxdvcybxau ??????
显然式（ 1）、式（ 2）能满足平面应力情况下的拉梅方程式。
2、考虑位移约束和变形连续条件,
ayay
hyhylxlx
vv
vuuu
??
??????
?
????
)()(
0)(,0)(,0)(,0)(
21
2221
由此解得,
)(,0,,0
0
2111222
222111
haeaefdhefd
cbacba
????????
??????
103
q
x
B
D
a
h
F
L L
y
O A
C
E
11 ?E
22 ?E
)4()(,0
)3()()(,0
222
2111
hyevu
haeayevu
???
?????
104
)4()(,0
)3()()(,0
222
2111
hyevu
haeayevu
???
?????
q
x
B
D
a
h
F
L L
y
O A
C
E
11 ?E
22 ?E
105
)6(
0,0
1
,
1
)(
1
1
,
1
)(
1
)5(
0,,0
0,,0
2111
2
2
2
2
21
2
1
1
111
2
1
1
1
2
2
2
22
21
2
1
11
111
2
1
1
1
2222
1111
?
?
?
?
?
?
?
?
?
?
?
???
?
?
?
??
?
?
?
?
?
??
?
?
?
?
?
?
?
???
???
xyxyxy
yxyy
xyxx
xyyx
xyyx
G
e
E
e
EE
e
E
e
EE
e
e
???
?
?
?
???
?
?
?
?
?
?
?
???
?
?
???
???
106
)6(
0,0
1
,
1
)(
1
1
,
1
)(
1
)5(
0,,0
0,,0
2111
2
2
2
2
21
2
1
1
111
2
1
1
1
2
2
2
22
21
2
1
11
111
2
1
1
1
2222
1111
?
?
?
?
?
?
?
?
?
?
?
???
?
?
?
??
?
?
?
?
?
??
?
?
?
?
?
?
?
???
???
xyxyxy
yxyy
xyxx
xyyx
xyyx
G
e
E
e
EE
e
E
e
EE
e
e
???
?
?
?
???
?
?
?
?
?
?
?
???
?
?
???
???
107
3.Consider the stress boundary term and stress continuous term (The surface of CD
is a smooth contact),
0)()(,)()(
0)(,)(
2121
0101
???
???
????
??
ayxyayxyayyayy
yxyyy q
????
??
From it,we can get,
qEeqEe
2
22
2
1
21
1
1,1 ?? ??????
4.The displacement and the stress weight are,
)(
1
,0
)(
1
)(
1
,0
2
2
2
22
2
2
2
1
2
1
11
hyq
E
vu
haq
E
ayq
E
vu
?
?
???
?
?
??
?
???
?
??
108
3、考虑应力边界条件和应力连续条件（ CD面为光滑接触）,
0)()(,)()(
0)(,)(
2121
0101
???
???
????
??
ayxyayxyayyayy
yxyyy q
????
??
由此解得,
qEeqEe
2
22
2
1
21
1
1,1 ?? ??????
4、位移及应力分量为,
)(
1
,0
)(
1
)(
1
,0
2
2
2
22
2
2
2
1
2
1
11
hyq
E
vu
haq
E
ayq
E
vu
?
?
???
?
?
??
?
???
?
??
109
0,,
0,,
2222
1111
?????
?????
xyyx
xyyx
qq
qq
????
????
110
0,,
0,,
2222
1111
?????
?????
xyyx
xyyx
qq
qq
????
????
111
112
Elasticity
2
3
Chapter 2 The Basic theory of the Plane Problem
§ 2-11 Stress function.Inverse solution method and semi-inverse method
§ 2-1 Plane stress problem and plane strain problem
§ 2-2 Differential equation of equilibrium
§ 2-3 The stress on the incline.Principal stress
§ 2-4 Geometrical equation.The displacement of the rigid body
§ 2-5 Physical equation
§ 2-6 Boundary conditions
§ 2-7 Saint-Venant’s principle
§ 2-8 Solving the plane problem according to the displacement
§ 2-9 Solving the plane problem according to the stress.Compatible
equation
§ 2-10 The simplification under the circumstances of ordinary physical
force
Exercise Lesson
4
第二章 平面问题的基本理论
§ 2-11 应力函数逆解法与半逆解法
§ 2-1 平面应力问题与平面应变问题
§ 2-2 平衡微分方程
§ 2-3 斜面上的应力主应力
§ 2-4 几何方程刚体位移
§ 2-5 物理方程
§ 2-6 边界条件
§ 2-7 圣维南原理
§ 2-8 按位移求解平面问题
§ 2-9 按应力求解平面问题。相容方程
§ 2-10 常体力情况下的简化
习题课
5
1.Plane stress problem
§ 2-1 Plane stress problem and plane strain problem
In actual problem,it is strictly saying that any elastic body whose external
force for suffering is a space system of forces is generally the space
object.However,when both the shape and force circumstance of the elastic
body for investigating have their own certain characteristics.As long as the
abstraction of the mechanics is handled together with appropriate
simplification,it can be concluded as the elasticity plane problem,
The plane problem is divided into the plane stress problem and plane strain
problem,
Equal thickness lamella bears the
surface force that parallels with plate
face and don’t change along the
thickness.At the same time,so does the
volumetric force,
σz = 0 τzx = 0 τzy = 0 Fig.2－ 1
6
一、平面应力问题
§ 2-1 平面应力问题与平面应变问题
在实际问题中，任何一个弹性体严格地说都是空间物体，
它所受的外力一般都是空间力系。但是，当所考察的弹性体
的形状和受力情况具有一定特点时，只要经过适当的简化和
力学的抽象处理，就可以归结为弹性力学平面问题。
平面问题分为平面应力问题和平面应变问题。
等厚度薄板，板边承受平
行于板面并且不沿厚度变化的
面力，同时体力也平行于板面
并且不沿厚度变化。
σ z = 0 τ zx = 0 τ zy = 0
图 2－ 1
7
x
y
Characteristics,
1) The dimension of length and breadth is far larger than that of
thickness,
2) The force along the plate face for suffering is the face force in
parallel with plate face,and along the thickness even,the volumetric force
is in parallel with plate force and doesn’t change along the thickness,and
has no external force function on the surface front and back of the flat
panel,
0?z?Attention,Plane stress problem ?z =0,but,this is contrary
to plane strain problem,
8
x
y
特点,
1) 长、宽尺寸远大于厚度
2) 沿板边受有平行板面的面力，且沿厚度均布，体力
平行于板面且不沿厚度变化，在平板的前后表面上
无外力作用。
问题相反。
0?z?注意：平面应力问题 ?z =0，但,这与平面应变
9
2.Plane strain problem
Very long column bears the face force in parallel with plate face and doesn’t
change along the length on the column face,at the same time,so does the
volumetric force,
εz = 0 τzx = 0 τzy = 0
x
Fig,2－ 2
For example:dam,circular cylinder piping by the internal air pressure and
long level laneway etc,
0?z?Attention,Plane strain problem?z = 0,but,this is contrary
to plane stress problem,
x
y
P
10
二、平面应变问题
很长的柱体，在柱面上承受平行于横截面并且不沿长度
变化的面力，同时体力也平行于横截面并且不沿长度变化。
ε z = 0 τ zx = 0 τ zy = 0
x
图 2－ 2
如：水坝、受内压的圆柱管道和长水平巷道等。
0?z?注意平面应变问题 ?z = 0，但
问题相反。
，这恰与平面应力
x
y
P
11
§ 2-2 Differential Equation of Equilibrium
Whether plane stress problem or plane strain problem,is the research problem in
plane xy,all the physics quantity has nothing to do with z,
Discuss below the correlation between any point stress and volumetric force
when the object is placed in the state of equilibrium,and lead an equilibrium
differential equation from here.From the lamella shown in Fig.2-1,we take out a
small and positive parallelepiped PABC,and take for an unit length in the
directional dimension in z,
y?
o x
y
dyyyy ??? ??
x?
dxxxx ??? ??xy?
dxxxyxy ??? ??
yx?
dyyyxyx ??? ??
P A
B C
X
Y
D
Fig.2－ 3
),( yxxx ?? ?
dx
Establishing the function of the positive
stress force in an unit on the left side
is,the coordinate on the right side
x gets the increment,the positive stress on
the face is,spreading the formula
above will be Taylor’s series,
),( ydxxx ??
n
n
x
n
x
x
xx
dx
x
yx
n
dx
x
yx
dx
x
yxyxydxx
)(),(
!
1)(),(
!2
1
),(),(),(
2
2
2
?
????
?
??
?
?
????
??
???
?
12
§ 2-2 平衡微分方程
无论平面应力问题还是平面应变问题，都是在 xy平面内研究问题,
所有物理量均与 z无关。
下面讨论物体处于平衡状态时，各点应力及体力的相互关系，并
由此导出平衡微分方程。从图 2－ 1所示的薄板取出一个微小的正平行
六面体 PABC(图 2－ 3），它在 z方向的尺寸取为一个单位长度。
y?
o x
y
dyyyy ??? ??
x?
dxxxx ??? ??xy?
dxxxyxy ??? ??
yx?
dyyyxyx ??? ??
P A
B C
X
Y
D
图 2－ 3
),( yxxx ?? ? x
dx
设作用在单元体左侧面上的正
应力是,右侧面上坐标
得到增量,该面上的正应力为
,将上式展开为泰勒级
数,
),( ydxxx ??
n
n
x
n
x
x
xx
dx
x
yx
n
dx
x
yx
dx
x
yx
yxydxx
)(
),(
!
1
)(
),(
!2
1
),(
),(),(
2
2
2
?
?
???
?
?
?
?
?
?
???
??
?
??
?
13
After omitting small quantity of the two rank and above the two rank,can get
,at the same time,,,are get the state of stress from the
drawing show,dxx yxyx xx ??? ),(),( ?? y? xy? yx?
While considering the volumetric force to the plane stress state,still prove
mutual and equal theory of shearing strength.Regard the center D and straight
line in parallel with the shaft of z as the moment shaft,list the equilibrium
equation of the moment shaft, ? ?0DM
0
2
1
2
1)(
2
1
2
1)(
??????
?
?
??
?????
?
?
?
dy
dx
dy
dxdy
y
dx
dy
dx
dydx
x
yx
yx
yx
xy
xy
xy
?
?
?
?
?
?
The both sides of the formula above divide get,dxdy
dyydxx yxyxxyxy ??????? ???? 2121
Cause 0,0 ?? dydx,Omitting small quantity isn’t accounted,can get,
yxxy ?? ?
14
略去二阶及二阶以上的微量后便得 同样,,
都一样处理，得到图示应力状态。 dxx
yxyx x
x ?
?? ),(),( ?? y? xy? yx?
对平面应力状态考虑体力时，仍可证明剪应力互等定理。以通过中
心 D并平行于 z轴的直线为矩轴，列出力矩的平衡方程, ? ?0DM
0
2
1
2
1)(
2
1
2
1)(
??????
?
?
??
?????
?
?
?
dy
dx
dy
dxdy
y
dx
dy
dx
dydx
x
yx
yx
yx
xy
xy
xy
?
?
?
?
?
?
将上式的两边除以 得到,dxdy
dyydxx yxyxxyxy ??????? ???? 2121
令 0,0 ?? dydx,即略去微量不计，得,
yxxy ?? ?
15
Deduce the equilibrium differential equation of the plane stress
problem below,list the equilibrium equation to the unit,
011
1)(11)(
:0
????????
??
?
?
???????
?
?
?
??
dydxXdx
dxdy
y
dydydx
x
F
yx
yx
yxx
x
x
x
?
?
??
?
?
011
1)(11)(
:0
????????
??
?
?
???????
?
?
?
??
dydxYdy
dydx
x
dxdxdy
y
F
xy
xy
xyy
y
y
y
?
?
??
?
?
16
下面推导平面应力问题的平衡微分方程，对单元体列平
衡方程,
011
1)(11)(
:0
????????
??
?
?
???????
?
?
?
??
dydxXdx
dxdy
y
dydydx
x
F
yx
yx
yxx
x
x
x
?
?
??
?
?
011
1)(11)(
:0
????????
??
?
?
???????
?
?
?
??
dydxYdy
dydx
x
dxdxdy
y
F
xy
xy
xyy
y
y
y
?
?
??
?
?
17
Sorting them gets,
0
0
??
?
?
?
?
?
??
?
?
?
?
?
Y
xy
X
yx
xyy
yxx
??
??
These two differential equation include three unknown
functions,Therefore,deciding the problem of the
stress weight is exceedingly and statically determinate;And still
must consider the deformation and displacement,then the problem
can be solved,
For the plane strain problem,the faces front and back still have
But they do not affect completely the establishes of the equation
above.So the equation above applies two kinds of plane problem
alike,
z?
yxxyyx ???? ?,,
18
整理得,
0
0
??
?
?
?
?
?
??
?
?
?
?
?
Y
xy
X
yx
xyy
yxx
??
??
这两个微分方程中包含着三个未知函数 。因此
决定应力分量的问题是超静定的；还必须考虑形变和位移，才能
解决问题。
对于平面应变问题,虽然前后面上还有,但它们完全不影
响上述方程的建立。所以上述方程对于两种平面问题都同样适用。
z?
yxxyyx ???? ?,,
19
§ 2-3 The stress on the Inclined Plane.Principal stress
1.The stress on the inclined plane
Having known the stress weight of any point P inside the
elastic body,we try to get the stress which pass the point P on the arbitrarily
inclined cross section.From neighborhood of point P taking a plane AB,which is
in parallel with the inclined plane above,and draws a small set square or three
column PAB on two planes which pass point P and have perpendicularity in the
shaft of x and y.When the plane AB approaches point P infinitely,the mean
stress on the plane AB will become the stress on the inclined plane above,
yxxyyx ???? ?,,
Establish the length of the face AB in the
plane xy is dS,N is the exterior normal
direction,and its direction cosine is,
myNlxN ?? ),c o s (,),c o s (
P A
B
xy?
x?
y?
N?
yx?
N?
NX
NY
S Ny
x
Fig.2－ 4
o
20
§ 2-3 斜面上的应力、主应力
一、斜面上的应力
已知弹性体内任一点 P处的应力分量,求经
过该点任意斜截面上的应力。为此在 P点附近取一个平面 AB，
它平行于上述斜面，并与经过 P点而垂直于 x轴和 y轴的两个平
面划出一个微小的三角板或三棱柱 PAB。当平面 AB与 P点无限
接近时，平面 AB上的应力就成为上述斜面上的应力。
yxxyyx ???? ?,,
设 AB面在 xy平面内的长度为 dS，
厚度为一个单位长度,N 为该面的外
法线方向，其方向余弦为,
myNlxN ?? ),c o s (,),c o s (
P A
B
xy?
x?
y?
N?
yx?
N?
NX
NY
S Ny
x
图 2－ 4
o
21
The projection of the whole stress on the inclined plane AB is XN
and YN respectively along with the shaft of x and y.From the PAB
equilibrium term can get,? ?0
xF
m d Sl d SdSX yxxN ?? ??
Divide and get,dS
yxxN mlX ?? ??
Same from and get,? ?0yF
xyyN lmY ?? ??
The positive stress on the inclined plane AB,from the
projection can get,
N?
xyyxNNN lmmlmYlX ???? 222 ?????
The shearing strength on the inclined plane AB,from the
projection can get,
N?
xyxyNNN mllmmXlY ???? )()( 22 ??????
22
斜面 AB上全应力沿 x轴及 y轴的投影分别为 XN和 YN。由 PAB
的平衡条件 可得,? ?0xF
m d Sl d SdSX yxxN ?? ??
除以 即得,dS
yxxN mlX ?? ??
同样由 得出,? ?0yF
xyyN lmY ?? ??
斜面 AB上的正应力,由投影可得,N?
xyyxNNN lmmlmYlX ???? 222 ?????
斜面 AB上的剪应力,由投影可得,
N?
xyxyNNN mllmmXlY ???? )()( 22 ??????
23
3.Principal stress
If the shearing stress of some inclined plane through point P is
equal to zero,then the positive stress of that inclined plane calls a
principal stress of point P,but that inclined plane calls the main
plane of the stress at point P,and the normal direction of that
inclined plane calls the main direction of the stress at point P,
1.The size of the principal stress
22
2
1 )
2(2 xy
yxyx ?????
?
? ?????
???
?
???
?
2.The direction of the principal stress is in the perpendicularity
with for each other,
1?
2?
24
二、主应力
如果经过 P点的某一斜面上的切应力等于零，则该斜面
上的正应力称为 P点的一个主应力，而该斜面称为 P点的一个
应力主面，该斜面的法线方向称为 P点的一个应力主向。
1.主应力的大小
22
2
1 )
2(2 xy
yxyx ?????
?
? ?????
???
?
???
?
2.主应力的方向
与 互相垂直。 1? 2?
25
§ 2-4 Geometrical Equation,
The Displacement of the Rigid Body
In plane problem,every point inside the elastic body can produce the arbitrarily
directional displacement.Take an unit PAB through any point P inside the elastic
body,such as Fig.2-5 show.After the elastic body suffers force,the point P,A,B
move to the point P′,A′,B′respectively,
P
o x
y
A
B
P?
A?
B?
u
v
dxxuu ???
dyyvv ???
dyyuu ???
dxxvv ???
?
?
Fig.2－ 5
一,The positive strain at point P
x
u
dx
udx
x
uu
x ?
????
??
?
)(
?
Here because of small deformation,
PA for causing stretch and shrink from
the y direction displacement v is the
small quantity of a high rank and this
small quantity may be omitted,
26
§ 2-4 几何方程、刚体位移
在平面问题中，弹性体中各点都可能产生任意方向的位移。
通过弹性体内的任一点 P，取一单元体 PAB，如图 2-5所示。弹性
体受力以后 P,A,B三点分别移动到 P′, A′, B′ 。
P
o x
y
A
B
P?
A?
B?
u
v
dxxuu ???
dyyvv ???
dyyuu ???
dxxvv ???
?
?
图 2－ 5
一,P点的正应变
在这里由于小变形，由 y
方向位移 v所引起的 PA的伸缩
是高一阶的微量，略去不计。
x
u
dx
udx
x
uu
x ?
????
??
?
)(
?
27
The same can get,
y
v
y ?
???
2.Shearing strain at point P
y
u
x
v
xy ?
??
?
???? ???
The corner of the line
segment PA,
x
v
dx
vdx
x
vv
?
????
??
?
)(
?
The same can get the corner
of the line segment PB,
y
u
?
???
Thus
28
同理可求得,
y
v
y ?
???
二,P点的切应变
y
u
x
v
xy ?
??
?
???? ???
线段 PA的转角,
x
v
dx
vdx
x
vv
?
????
??
?
)(
?
同理可得线段 PB的转角,
y
u
?
???
所以
29
Therefore get the geometrical equation of the plane problem
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
y
u
x
v
y
v
x
u
xy
y
x
?
?
?
From the geometrical equation above,when the displacement
weight of the object is completely certain,the deformation
weight is completely certain,unique weight can not be made sure
thoroughly,
30
因此得到平面问题的几何方程,
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
y
u
x
v
y
v
x
u
xy
y
x
?
?
?
由几何方程可见，当物体的位移分量完全确定时，形变
分量即可完全确定。反之，当形变分量完全确定时，位移分
量却不能完全确定。
31
§ 2-5 The Physical Equation
In the isotropy of the complete elasticity,the relation between the
deformation weight and the stress weight is established according
to the Hooke’s law as follows,
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
???
???
???
xyxy
zxzx
yzyz
yxzz
xzyy
zyxx
G
G
G
E
E
E
??
??
??
?????
?????
?????
1
1
1
)]([
1
)]([
1
)]([
1
32
§ 2-5 物理方程
在完全弹性的各向同性体内，形变分量与应力分量之间的
关系根据虎克定律建立如下,
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
???
???
???
xyxy
zxzx
yzyz
yxzz
xzyy
zyxx
G
G
G
E
E
E
??
??
??
?????
?????
?????
1
1
1
)]([
1
)]([
1
)]([
1
33
Inside the formula,the E is a modulus of elasticity;the G is a
stiffness modulus;the u is a poisson ratio.The relation of three ones
above,
)1(2 ???
EG
1.The physics equation of the plane stress problem
?
?
?
?
?
?
?
?
?
?
?
??
??
xyxy
xyy
yxx
E
E
E
?
?
?
????
????
)1(2
)(
1
)(
1
)( yxz E ???? ???
And have,
34
式中,E为弹性模量； G为刚度模量； 为泊松比。三者
的关系,
?
)1(2 ???
EG
一、平面应力问题的物理方程
?
?
?
?
?
?
?
?
?
?
?
??
??
xyxy
xyy
yxx
E
E
E
?
?
?
????
????
)1(2
)(
1
)(
1
)( yxz E ???? ???
且有,
35
2.The physics equation of the plane strain problem
?
?
?
?
?
??
?
?
?
?
?
?
?
?
?
?
?
?
?
?
xyxy
xyy
yxx
E
E
E
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
)1(2
)
1
(
1
)
1
(
1
2
2
3.The transformation relation of the relation type between the
stress strain and the plane strain,
The relation type of the
plane stress,
?
?
?
?
?
?
?
?
?
?
?
??
??
xyxy
xyy
yxx
E
E
E
?
?
?
????
????
2
1
)(
1
)(
1
36
二、平面应变问题的物理方程
?
?
?
?
?
??
?
?
?
?
?
?
?
?
?
?
?
?
?
?
xyxy
xyy
yxx
E
E
E
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
)1(2
)
1
(
1
)
1
(
1
2
2
三、平面应力的应力应变关系式与平面应变的关系式之间的
变换关系
将平面应力中的关系式,
?
?
?
?
?
?
?
?
?
?
?
??
??
xyxy
xyy
yxx
E
E
E
?
?
?
????
????
2
1
)(
1
)(
1
37
For change
?
?
?
?
?
?
?
?
1
1 2
E
E
Can get the relation type
in the plane strain,
?
?
?
?
?
??
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
xyxy
xyy
yxx
E
E
E
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
)1(2
1
1
1
1
2
2
Because of the similarity of this
kind,while solving plane strain
problem,the corresponding equation
of the plane problem and the elastic
constant in the answer can be
exchanged as above,can get the
solution of the homologous plane
strain problem,
38
作代换
?
?
?
?
?
?
?
?
1
1 2
E
E
就可得到平面应变中的
关系式,
?
?
?
?
?
??
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
xyxy
xyy
yxx
E
E
E
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
)1(2
1
1
1
1
2
2
由于这种相似性，在解平面
应变问题时，可把对应的平面应
力问题的方程和解答中的弹性常
数进行上述代换，就可得到相应
的平面应变问题的解。
39
§ 2-6 Boundary Conditions
When the object is placed in the state of equilibrium,its internal state of
stress at all point should satisfy the equilibrium differential equation and also
satisfy the boundary term on the boundary,
According to the difference of the boundary condition,the elasticity problem
is divided into the displacement boundary problem,stress boundary problem
and mixed boundary problem,
1.Displacement Boundary Term
When the displacement has been known on the boundary,the
displacement of the point on the object boundary and the equal term of the
fixed displacement should be established.For example,if making the
boundary of the fixed displacement is,and have(on the ),uSuS
uus ? vvs ?Among them,and means the displacement weight on the
boundary,however,and is the coordinate function we have know the
boundary,
su usv
v
40
§ 2-6 边界条件
当物体处于平衡状态时,其内部各点的应力状态应满足
平衡微分方程；在边界上应满足边界条件。
按照边界条件的不同，弹性力学问题分为位移边界问题、
应力边界问题和混合边界问题。
一、位移边界条件
当边界上已知位移时，应建立物体边界上点的位移与
给定位移相等的条件。如令给定位移的边界为,则有
（在 上）,
uS
uS
uus ? vvs ?
其中 和 表示边界上的位移分量，而 和 在边界上
是坐标的已知函数。 su
usv v
41
2.Stress boundary term
When the boundary of the object is given to surface force,then the stress of
the object on the boundary should satisfy the equilibrium term of forces with
the equilibrium of the surface force,
??
???
??
??
Ylm
Xml
sxysy
syxsx
)()(
)()(
??
??
Among them,and are the surface force weights and,,,
are the stress weights on the boundary,
X Y sx)(? sy)(?
sxy)(? syx)(?
When the boundary face is in perpendicularity in shaft x,stress boundary
term can be changed briefly into,
YX sxysx ???? )(,)( ??
When the boundary face is in perpendicularity in shaft y,stress boundary
term can be changed briefly into,
XY syxsy ???? )(,)( ??
42
二、应力边界条件
当物体的边界上给定面力时，则物体边界上的应力应满
足与面力相平衡的力的平衡条件。
??
???
??
??
Ylm
Xml
sxysy
syxsx
)()(
)()(
??
??
其中 和 为面力分量,、,, 为边界上的应
力分量。
X Y sx)(? sy)(? sxy)(? syx)(?
当边界面垂直于 轴时，应力边界条件简化为,x
YX sxysx ???? )(,)( ??
当边界面垂直于 轴时，应力边界条件简化为,y
XY syxsy ???? )(,)( ??
43
3.Mixed boundary condition
1.The displacement has been known on a part of boundaries of the object,the
result of which have the displacement boundary term,the boundaries of other parts
have the surface force we have know.And then there should be stress boundary
term and displacement boundary term respectively on two parts of the
boundaries.The left surface of the cantilever contains displacement boundary
term,such as shown in Fig.2-6,
??
?
??
??
0
0
vv
uu
s
s
Top and bottom surface contains
stress boundary term,
??
?
?
?
???
???
0)(
0)(
sy
syx
Y
X
?
?
The right surface contains stress
boundary term,
??
?
?
?
??
??
0)(
)(
sxy
sx
Y
qX
?
?
l
q
x
y
o
2h
2h
Fig.2-6
44
三、混合边界条件
1.物体的一部分边界上具有已知位移，因而具有位移边界条
件，另一部分边界上则具有已知面力。则两部分边界上分别
有应力边界条件和位移边界条件。如图 2-6，悬臂梁左端面
有位移边界条件,
??
?
??
??
0
0
vv
uu
s
s
上下面有应力边界条件,
??
?
?
?
???
???
0)(
0)(
sy
syx
Y
X
?
?
右端面有应力边界条件,
??
?
?
?
??
??
0)(
)(
sxy
sx
Y
qX
?
?
l
q
x
y
o
2h
2h
图 2-6
45
2.On the same boundary,there are not only stress boundary term but
displacement boundary term.Coupler sustains the boundary term,such
as shown in Fig.2-7,
??
?
?
?
??
??
0)(
0
sxy
s
Y
uu
?
The alveolus boundary term shown
in Fig.2-8,
?
?
?
??
??
0)(
0
sx
s
X
vv
?
o x
y
Fig.2-7
x
y
o
Fig.2-8
46
2.在同一边界上，既有应力边界条件又有位移边界条件。
如图 2-7连杆支撑边界条件,
??
?
?
?
??
??
0)(
0
sxy
s
Y
uu
?
如图 2-8齿槽边界条件,
?
?
?
??
??
0)(
0
sx
s
X
vv
?
o x
y
图 2-7
x
y
o
图 2-8
47
§ 2-7 Saint-Venant Principle
1.Saint-Venant’s Principle
If transforming a small part of the surface force on the boundary into the
surface force that has equal effect but different distribution(The main vector is
equal,so is the main quadrature to the same point as well),and then the
distribution of the stress force nearby will have prominent changes,but the
influence from the distant place can not be accounted,
2.Give Examples
Establishing the component of the column forms,the centroid of area in cross
sections of both ends suffers the tensible force which is equal in size but
contrary in direction,such as shown in Fig.2-9a.If transforming an or both ends
of tensile force into the force at the same effect as the static force,such as
shown in Fig.2-9b or Fig.2-9c,the distribution of stress force drawn only by
broken line has prominent changes,whereas,the influence of the rest parts can
not be accounted.If changing both ends of tensile force into that of uniform
distribution again,the gathering degree is equal to P/A and among them A is the
cross-section area of the component,such as shown in Fig.2-9d,there is still the
stress close to both ends under the noticeable influence,
48
§ 2-7 圣维南原理
一,圣维南原理
如果把物体的一小部分边界上的面力，变换为分布不同
但静力等效的面力（主矢量相同，对于同一点的主矩也相
同），那么，近处的应力分布将有显著的改变，但是远处所
受的影响可以不计。
二,举例
设有柱形构件，在两端截面的形心受到大小相等而方向
相反的拉力,如图 2-9a。如果把一端或两端的拉力变换为
静力等效的力，如图 2-9b或 2-9c，只有虚线划出的部分的应
力分布有显著的改变，而其余部分所受的影响是可以不计的。
如果再将两端的拉力变换为均匀分布的拉力，集度等于,
其中 为构件的横截面面积，如图 2-9d，仍然只有靠近两端
部分的应力受到显著的影响。
P
AP/ A
49
P P
2/P
2/P
2/P
2/P
P 2/P
2/P
AP/AP/
PP
Fig.2-9
(a)
(b)
(c)
(d)
(e)
Under the four kinds of
circumstances above,parts of
distribution of stress force distant
from both ends have no marked
difference,
Attention,
The application of the Saint-
Venant’s principle is by no
means separated from the term
of Equal Effect of Static Force,
50
P P
2/P
2/P
2/P
2/P
P 2/P
2/P
AP/AP/
PP
图 2-9
(a)
(b)
(c)
(d)
(e)
在上述四种情况下，离开两
端较远的部分的应力分布，并
没有显著的差别。
注意,
应用圣维南原理，绝不能
离开, 静力等效, 的条件。
51
§ 2-8 Solving the Plane Problem
according to the displacement
There are three kinds of basic methods to solve the problem in elasticity:the
solution to the problem according to displacement,stress force and admixture,
While solving problems using displacement method,we regard displacement
weight as the basic function unknown.After getting displacement weight from
only including the differential equation and boundary term of the displacement
weight,then get the deformation weight using geometrical equation,therefore,get
the stress weight with the physics equation,
1.Plane Stress Problem
?
?
?
?
?
?
?
?
?
?
?
??
??
xyxy
xyy
yxx
E
E
E
?
?
?
????
????
)1(2
)(
1
)(
1
In plane stress problem,the physics
equation is,
52
§ 2-8 按位移求解平面问题
在弹性力学里求解问题，有三种基本方法：按位移求解、
按应力求解和混合求解。
按位移求解时，以位移分量为基本未知函数，由一些只
包含位移分量的微分方程和边界条件求出位移分量以后，再
用几何方程求出形变分量，从而用物理方程求出应力分量。
一、平面应力问题
?
?
?
?
?
?
?
?
?
?
?
??
??
xyxy
xyy
yxx
E
E
E
?
?
?
????
????
)1(2
)(
1
)(
1
在平面应力问题中，物理方程为,
53
From three formulas above mentioned to solve the stress weight,can get,with the
substitution of geometrical equation,we can get the elasticity equation,
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
xyxy
xyy
yxx
E
E
E
?
?
?
???
?
?
???
?
?
)1(2
)(
1
)(
1
2
2
Again equilibrium differential equation with substitution in formula(a),simplification
hereafter,can get,
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
)(
)1(2
)(
1
)(
1
2
2
y
u
x
vE
x
u
y
vE
y
v
x
uE
xy
y
x
?
?
?
?
?
?
?
?
（ a)
This is the equilibrium differential equation to mean with the displacement,ie,when
solving the plane stress problem according to displacement method,we adopt a
basic differential equation for needs,
?
?
?
?
?
?
?
??
??
??
?
?
??
?
?
?
?
??
??
??
?
?
??
?
?
?
?
0)
2
1
2
1
(
1
0)
2
1
2
1
(
1
2
2
2
2
2
2
2
2
2
2
2
2
Y
yx
u
x
v
y
vE
X
yx
v
y
u
x
uE
??
?
??
?
（ 1）
54
由上列三式求解应力分量，得,
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
xyxy
xyy
yxx
E
E
E
?
?
?
???
?
?
???
?
?
)1(2
)(
1
)(
1
2
2 将几何方程代入，得弹性方程,
再将式（ a）代入平衡微分方程，简化以后，即得,
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
)(
)1(2
)(
1
)(
1
2
2
y
u
x
vE
x
u
y
vE
y
v
x
uE
xy
y
x
?
?
?
?
?
?
?
?
（ a)
这是用位移表示的平衡微分方程，也就是按位移求解平面应力
问题时所需用的基本微分方程。
?
?
?
?
?
?
?
??
??
??
?
?
??
?
?
?
?
??
??
??
?
?
??
?
?
?
?
0)
2
1
2
1
(
1
0)
2
1
2
1
(
1
2
2
2
2
2
2
2
2
2
2
2
2
Y
yx
u
x
v
y
vE
X
yx
v
y
u
x
uE
??
?
??
?
（ 1）
55
The stress boundary term with substitution in formula(a),simplification hereafter,
can get,
?
?
?
??
?
?
?
?
?
?
?
??
?
?
?
?
?
?
?
?
?
?
?
?
??
?
?
?
?
?
?
?
Y
y
u
x
v
l
x
u
y
v
m
E
X
x
v
y
u
m
y
v
x
u
l
E
ss
ss
])(
2
1
)([
1
])(
2
1
)([
1
2
2
?
?
?
?
?
?
This is the stress force boundary to mean with the displacement,ie,we adopt
the boundary term of the stress force when solving the plane stress problem
according to displacement method,
（ 2）
Sum up,when solving the plane stress problem according to displacement
method,we should make the displacement weight satisfy differential
equation(1) and combine to satisfy displacement boundary term or stress
boundary term or stress boundary term(2) on the boundary,After getting
displacement weight,we can get the deformation weight with geometrical
equation and then get the stress force weight with the physics equation,
2.Plane strain problem
?
??
? ???? 1,1 2
EE
Make the substitution between and in each equation of the plane
strain problem,
E ?
56
将（ a）式代入应力边界条件，简化以后，得,
?
?
?
??
?
?
?
?
?
?
?
??
?
?
?
?
?
?
?
?
?
?
?
?
??
?
?
?
?
?
?
?
Y
y
u
x
v
l
x
u
y
v
m
E
X
x
v
y
u
m
y
v
x
u
l
E
ss
ss
])(
2
1
)([
1
])(
2
1
)([
1
2
2
?
?
?
?
?
?
这是用位移表示的应力边界条件，也就是按位移求解平面应
力问题时所用的应力边界条件。
（ 2）
总结起来，按位移求解平面应力问题时，要使得位移分
量满足微分方程（ 1），并在边界上满足位移边界条件或应
力边界条件（ 2）。求出位移分量以后，用几何方程求出形
变分量，再用物理方程求出应力分量。
二、平面应变问题
?
??
? ???? 1,1 2
EE
只须将平面应力问题的各个方程中 和 作代换,E ?
57
§ 2-9 Solving the Plane Problem According
to the Stress Force.Compatible Equantion
While solving the plane problem according to the displacement,we must
combine two partial differential equation of the second ranks to solve the
problem,this is very difficult on the mathematics,But while solving the plane
problem according to the stress force,we can avoid this difficulty and so what
we adopt more is to get the solution according to the stress force,
While getting the solution according to the stress force,we regard stress
weight as the basic function unknown.After getting displacement weight from
only including the differential equation and boundary term of displacement
weight,then get the deformation weight using physics equation,therefore,get
the displacement weight with geometrical equation,
Compatible Equation
From geometrical equation of the
plane problem,
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
y
u
x
v
y
v
x
u
xy
y
x
?
?
?
58
§ 2-9 按应力求解平面问题。相容方程
按位移求解平面问题时，必须求解联立的两个二阶偏微分
方程，这在数学上是相当困难的。而按应力求解弹性力学平面
问题，则避免了这个困难，故更多采用的是按应力求解。
按应力求解时，以应力分量为基本未知函数，由一些只包
含应力分量的微分方程和边界条件求出应力分量以后，再用物
理方程求出形变分量，从而用几何方程求出位移分量。
相容方程
由平面问题的几何方程,
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
y
u
x
v
y
v
x
u
xy
y
x
?
?
?
59
Can get,
yxx
v
y
u
yxxy
v
yx
u
xy
xyyx
??
??
?
??
?
?
??
??
??
??
??
??
?
??
?
? ??? 22
2
3
2
3
2
2
2
2
)(
ie,
yxxy
xyyx
??
??
?
??
?
? ??? 2
2
2
2
2
This relation type calls the deformation moderates equation or
compatible equation,
1.Compatible equation in plane stress force
))(1())(( 2
2
2
2
y
Y
x
X
yx yx ?
??
?
?????
?
??
?
? ???
2.Compatible equation in plane strain force
)(1 1))(( 2
2
2
2
y
Y
x
X
yx yx ?
??
?
?
?????
??
?
?
???
60
可得,
yxx
v
y
u
yxxy
v
yx
u
xy
xyyx
??
??
?
??
?
?
??
??
??
??
??
??
?
??
?
? ??? 22
2
3
2
3
2
2
2
2
)(
即,
yxxy
xyyx
??
??
?
??
?
? ??? 2
2
2
2
2
这个关系式称为形变协调方程或相容方程。
（一）平面应力问题的相容方程
))(1())(( 2
2
2
2
y
Y
x
X
yx yx ?
??
?
?????
?
??
?
? ???
（二）平面应变问题的相容方程
)(1 1))(( 2
2
2
2
y
Y
x
X
yx yx ?
??
?
?
?????
??
?
?
???
61
While solving the plane problem according to the stress force,
the stress weight should not only satisfy both the equilibrium
differential equation and compatible equation,but satisfy the
stress boundary term on the boundary whether is a plane stress
problem or plane strain problem,
62
按应力求解平面问题时，无论是平面应力问题还是平面
应变问题，应力分量除了满足平衡微分方程和相容方程外，
在边界上还应当满足应力边界条件。
63
§ 2-10 The Simplification Under the
Circumstances of Ordinary Physical Force
Under the circumstances of ordinary physical force,the compatible equation
of two kinds of plane problems is simplified as,
0))(( 2222 ??????? yxyx ??
yx ?? ?
yx ?? ?
2?
Therefore,under the circumstances of ordinary physical force,should
satisfy Laplace differential equation(in harmony with equation),should
be harmonic functions.Represent with the mark,the formula above
can be simplified as,
2
2
2
2
yx ?????
0)(2 ??? yx ??Conclusion
In the stress boundary problem of single connection if two elastic bodies have
the same boundary shape and suffer the external force of the same distribution,
and then stress force distribution,,should be the same whether the
materials of two elastic bodies are same or not and whether they are under the
plane stress circumstances or under the plane strain circumstances(Two kinds of
the stress force weight in the plane problem,the deformation and the
displacement are uncertainly the same),
x? y? xy?
z?
64
§ 2-10 常体力情况下的简化
常体力下，两种平面问题的相容方程都简化为,
0))(( 2
2
2
2
??????? yxyx ??
可见，在常体力的情况下,应当满足拉普拉斯微分方程
（调和方程）,应当是调和函数。
用记号 代表,上式简写为,
2
2
2
2
yx ?
??
?
?
yx ?? ?
yx ?? ?
2? 0)(2 ???
yx ??
结论
在单连体的应力边界问题中，如果两个弹性体具有相同的
边界形状，并受到同样分布的外力，那么，不管这两个弹性体
的材料是否相同，也不管它们是在平面应力情况下或是在平面
应变情况下，应力分量,, 的分布是相同的（两种平
面问题中的应力分量,以及形变和位移，却不一定相同）。
x? y? xy?
z?
65
Inference 2
When measuring the above stress weight of the structure or component with
the method of experiment,we can make the model using the material of the
convenient measurement in order to replace original structure or component
materials of the inconvenient measurement; we also can adopt structure or
component of long column shape under the plane strain circumstances,
Inference 3
Under the circumstance of constant volumetric force,for the stress boundary
problem of single connection,we can charge the function of the volumetric
force as that of the surface force in order to solve the problem and experiment
measurement,
Inference 1
The stress weight,,that is solved according to any object is also
applicable to the object which has the same boundary and other materials
suffering the same external force; The stress weight that is solved according to
plane stress problem is also applicable to the object which has the same boundary
and the same external force under the plane strain circumstances,
x? y? xy?
66
推论 2
在用实验方法测量结构或构件的上述应力分量时，可以用
便于量测的材料来制造模型，以代替原来不便于量测的结构或
构件材料；还可以用平面应力情况下的薄板模型，来代替平面
应变情况下的长柱形的结构或构件。 推论 3
常体力的情况下，对于单连体的应力边界问题，还可以
把体力的作用改换为面力的作用，以便于解答问题和实验量
测。
推论 1
针对任一物体而求出的应力分量,,,也适用于
具有同样边界并受有同样外力的其它材料的物体；针对平面应
力问题而求出的这些应力分量，也适用于边界相同、外力相同
的平面应变情况下的物体。
x? y? xy?
67
§ 2-11 Stress Function.Inverse
Solution Method and Semi-Inverse Method
1.Stress function
While solving the stress boundary problem according to the stress force and
when the volumetric force is the constant quantity,the stress weight,,
should satisfy the equilibrium differential equation,x? y? xy?
?
?
?
?
?
?
?
??
?
?
?
?
?
??
?
?
?
?
?
0
0
Y
xy
X
yx
xyy
xyx
??
??
（ a）
And compatible equation
0))(( 2
2
2
2
??????? yxyx ??
（ b)
The solution to the equation (a) includes two parts,arbitrarily a particular
solution and the general solution to the following homogeneous differential
equation,
68
§ 2-11应力函数、逆解法与半逆解法
一、应力函数
按应力求解应力边界问题时，在体力为常量的情况下，应
力分量,, 应当满足平衡微分方程,
x? y? xy?
?
?
?
?
?
?
?
??
?
?
?
?
?
??
?
?
?
?
?
0
0
Y
xy
X
yx
xyy
xyx
??
??
（ a）
以及相容方程
0))(( 2
2
2
2
??????? yxyx ??
（ b)
方程（ a）的解包含两部分：任意一个特解和下列齐次
微分方程的通解。
69
The particular solution is,
Rewrite the former equation inside the homogeneous
differential equation (c)as,
)( xyx yx ?? ??????
According to the differential equation theory,it is certain to
exist some function,make,),( yxA
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
0
0
xy
yx
xyy
xyx
??
??
（ c）
0,,????? xyyx YyXx ??? （ d）
x
A
y
A
xy
x
?
?
??
?
?
?
?
? （ e）
（ f）
70
特解取为,
将齐次微分方程（ c）中前一个方程改写为,
)( xyx yx ?? ??????
根据微分方程理论，一定存在某一个函数,使得,),( yxA
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
0
0
xy
yx
xyy
xyx
??
??
（ c）
0,,????? xyyx YyXx ??? （ d）
x
A
y
A
xy
x
?
?
??
?
?
?
?
? （ e）
（ f）
71
Similarly rewrite the second equation
inside (c) as,)( xyy xy ?? ??????
It is certain to exist some function as well,make,),( yxB
y
B
x
B
xy
y
?
?
??
?
?
?
?
? （ g）
（ h）
From the formula (f)
and (h),can get,
y
B
x
A
?
??
?
?
Thus,it is certain to exist some function,make,),( yx?
x
B
y
A
?
?
?
?
?
?
?
? （ i）
（ j）
72
同样将（ c）中的第二个方程改写为,
)( xyy xy ?? ??????
也一定存在某一个函数,使得,),( yxB
y
B
x
B
xy
y
?
?
??
?
?
?
?
? （ g）
（ h）
由式（ f）及（ h）得,
y
B
x
A
?
??
?
?
因而一定存在某一个函数,使得,),( yx?
x
B
y
A
?
?
?
?
?
?
?
? （ i）
（ j）
73
Make the formula (i) substitute to (e),(j)to(g),and (i)to(f),then get the general
solution,
yxxy xyyx ??
???
?
??
?
?? ?????? 2
2
2
2
2,,（ k）
Make the general solution (k)plus the particular solution(d),then get the
whole solution of the differential equation(a),
The function calls the stress function of the plane problem and also calls the
Array stress function,
In order that the stress weight(1) can also satisfy the compatible equation(b),
make formula (1)substitute formula (b),then get,
?
yxYyxXxy xyyx ??
????
?
???
?
?? ?????? 2
2
2
2
2,,（ 1）
0))(( 22222222 ????????????? YyxXxyyx ??
The formula above can be simplified,
0))(( 22222222 ??????????? yxyx ??
74
将式（ i）代入（ e），式（ j）代入（ g），并将式（ i）代入
（ f），即得通解,
yxxy xyyx ??
???
?
??
?
?? ?????? 2
2
2
2
2,,（ k）
将通解（ k）与特解（ d）叠加，即得微分方程（ a）的全
解,
函数 称为平面问题的应力函数，也称为艾瑞应力函数。 ?
yxYyxXxy xyyx ??
????
?
???
?
?? ?????? 2
2
2
2
2,,（ 1）
为了应力分量（ 1）同时也能满足相容方程（ b），将（ 1）
代入式（ b），即得,
0))(( 22222222 ????????????? YyxXxyyx ??
上式可简化为,
0))(( 22222222 ??????????? yxyx ??
75
Or spreading the formula is,
02 4422444 ?????? ???? yyxx ???
Further simplification is,04 ?? ?
（ 2）
2.Inverse solution method and semi-inverse method
Inverse solution,the first step is to set up multiform stress function which
satisfy the compatible equation(2),and get the stress weight with the formula(1),
then investigate according to the stress boundary term.On the elastic body in
every kind of shape,these stress weights correspondence in what kinds of surface
force,from which we know that the stress function for setting up can solve what
kinds of problem,
?
While solving the stress boundary problem according to stress force,if the
volumetric force is constant quantity,we may only consult the differential
equation(2) to solve the stress function,and then get the stress weight with
the formula(1),but these stress weights should satisfy the stress boundary term
on the boundary,
?
The basic step of inverse solution method,
76
或者展开为,
02 4422444 ?????? ???? yyxx ???
进一步简写为,04 ?? ?
（ 2）
二、逆解法与半逆解法
逆解法,先设定各种形式的、满足相容方程（ 2）的应力函
数,用公式（ 1）求出应力分量，然后根据应力边界条件来
考察，在各种边界形状的弹性体上，这些应力分量对应于什么
样的面力，从而得知所设定的应力函数可以解决什么问题。
?
按应力求解应力边界问题时，如果体力是常量，就只须由
微分方程（ 2）求解应力函数,然后用公式（ 1）求出应力
分量，但这些应力分量在边界上应当满足应力边界条件。
?
逆解法基本步骤,
77
Semi-inverse method,Aiming at the problem for requesting solution,according
to boundary shape of elastic body and force circumstance for suffering,supposing
the partial and the whole stress weight as a certain form function,from which
conclude stress function,then come to investigate whether this stress function
satisfy compatible equation or not.And the original stress weight for supposing
and the rest stress weight from this stress function are whether to satisfy stress
boundary term and single worth term in displacement or not.If both compatible
equation and all aspects of terms can be satisfied;it is natural to get the accurate
answer;if some aspect can not be satisfied,we should establish assumption on the
other hand and investigate it again,
Setting up
?
Solution to stress
weight
Solution to surface
force(Resultant
force)
Solution to
surface
force(Resultant
force)
Substitution
Substitution
Type(1) Stress
boundary
term
Maki
ng
sure
The basic steps of semi-inverse method,
Setting up Getting the
accurate solution
Educing the stress type of
expression
Satisfying the
boundary term
Satisfying
04 ??? Yes Yes
No No
Type(1) stress
Boundary term
?
78
半逆解法,针对所要求解的问题，根据弹性体的边界形状和受
力情况，假设部分或全部应力分量为某种形式的函数，从而推
出应力函数,然后来考察，这个应力函数是否满足相容方程，
以及，原来所假设的应力分量和由这个应力函数求出的其余应
力分量，是否满足应力边界条件和位移单值条件。如果相容方
程和各方面的条件都能满足，自然就得出正确的解答；如果某
一方面不能满足，就要另作假设，重新考察。
?
设定 ? 求出 应力分量 求出 面力（合力） 解决 什么问题
代入 代入
式（ l） 应力边
界条件
确定
半逆解法基本步骤,
设定 ? 导出 应力表达式 得到 正确解答 满足 边界条件 满足
04 ??? 是 是
否 否
式（ l） 应力边
界条件
79
Exercise Lesson for the Basic
Theory of the Plane Problem
Solution,From the mechanics of materials,the
bending moment through point P on the cross-
section is,
yxlh qhyxlqI yM
z
z
x
3
33
3 2)
1
12(
6 ???????
)6(
3
lqxM z ??
（ 1）
)(
36
0
22
3
2
3 xfyxlh
q
y d yx
lh
q
dy
x
yx
x
xy
xyx
???
?
?
??
?
?
?
?
?
?
? ?
?
?
??Equilibrium differential equation in
substitution,get,（ 2）
[Exercise 1] The upper part of the cantilever suffers
the load of the line distribution show in figure
according to the type of expression in the mechanics
of materials,then lead the type of expression of
and with the equilibrium differential equation,x? y?xy? o
y
x
l
q
P
2/h
2/h
）（ 1??
80
,平面问题的基本理论, 习题课
解,由材料力学知，过 点横截面
上的弯矩为,
P
yxlh qhyxlqI yM
z
z
x
3
33
3 2)
1
12(
6 ???????
)6(
3
lqxM z ??
（ 1）
)(
36
0
22
3
2
3 xfyxlh
q
y d yx
lh
q
dy
x
yx
x
xy
xyx
???
?
?
??
?
?
?
?
?
?
? ?
?
?
??代入平衡微分方程,（ 2）
[练习 1] 悬臂梁上部受线形分布载荷，
如图所示。试根据材料力学中 的
表达式，再用平衡微分方程导出
和 的表达式。
x?
y?
xy?
o
y
x
l
q
P
2/h
2/h
）（ 1??
81
Make use of the boundary term above and underneath and make sure f(x),
Make the type (3) substitute the second type of the equilibrium differential equation
and get,
xhyhy
lh
q
x
l
q
xg
xgxyhy
lh
q
dy
x
y
h
y
y
xy
y
)34(
2
2
)(,0)(
)()34(
2
323
3
2
23
3
????
???
????
?
?
??
?
?
?
?
?
?
（ 4）
)4(
4
3
4
3
)(,0)(
22
3
2
2
2
hy
lh
qx
x
lh
q
xf
xy
h
y
xy
??
???
??
?
?
（ 3）
Attention,Type (1),(3),(4) expressed only are perhaps stress weight form the
state if having right solution,and they still need to satisfy the compatible
equation expressed with stress force,
82
利用上、下面边界条件确定 )(xf
将式（ 3）代入平衡微分方程中的第二式，得,
xhyhy
lh
q
x
l
q
xg
xgxyhy
lh
q
dy
x
y
h
y
y
xy
y
)34(
2
2
)(,0)(
)()34(
2
323
3
2
23
3
????
???
????
?
?
??
?
?
?
?
?
?
（ 4）
)4(
4
3
4
3
)(,0)(
22
3
2
2
2
hy
lh
qx
x
lh
q
xf
xy
h
y
xy
??
???
??
?
?
（ 3）
注意,式（ 1）、（ 3）、（ 4）表达的仅是静力可能的应力分
量，若为正确解答，则还需满足以应力表示的相容方程。
83
A
B
xy
o
0?x?
0?xy?
0?y?
0?yx?
[Exercise 2] In the plane object shown in
figure,both angle A and angle B are right one
and the surface on the boundary nearby doesn’t
suffer the external force.Try to explain the
stress state on the point A and B,
Solution,Because the boundary near point A doesn’t
suffer the external force,the stress weight
on this point should satisfy the following
boundary term,
0)()()( ??? AxyAyAx ???
Ie,point A is placed in zero stress state,But point B is placed in the vertex of the
cave angle,each surface of differential unit for taking from this point is not the
boundary surface,therefore,the stress weight on it is unknown and not
definitely zero.From the theory analysis,we know that the stress force on point
B of the cave angle is infinite big,
84
A
B
xy
o
0?x?
0?xy?
0?y?
0?yx?
[练习 2] 如图 所示为平面物体，角
和角 均为直角，其附近边界表面
均不受外力，试说明, 两点的
应力状态。
A
B
A B
解,由于 点附近边界不受外力，该
点的应力分量应满足如下边界条
件,
0)()()( ??? AxyAyAx ???
即 点处于零应力状态。而 点处
于凹角的顶点，该点所取的微分单
元体的各个面均不是边界面，因此，
其上的应力分量是未知的，未必为
零，由理论分析知，凹角处 点的
应力趋于无限大。
A
A B
B
85
[Exercise 3] Try to write the displacement boundary term (Using the rectangular
coordinates) of each plane object in the form,Among them,point 0 in the second
figure doesn’t move and the level line segment through point 0 has no rotation,
Solution,The boundary term in each displacement in the form,
h x
y
o
h x
y
o
0,0,
2
,0
0,
2
,0
????
????
vu
h
yx
u
h
yx
0,0,0
0,0
?????
??
x
vvu
yx
86
[练习 3] 试写出表中所示各平面物体的位移边界条件（用直角
坐标），其中第二图中 点不动，过 点的水平线段无转动。 o o
解,各位移边界条件见表所列。
h x
y
o
h x
y
o
0,0,
2
,0
0,
2
,0
????
????
vu
h
yx
u
h
yx
0,0,0
0,0
?????
??
x
vvu
yx
87
h x
y
o
l
h x
y
o
l
?
0,
2
,
0,0,0,
0,0,0,0
???
????
????
v
h
ylx
vuylx
vuyx
0s inc o s,0,
0,0,0,0
????
????
?? vuylx
vuyx
88
h x
y
o
l
h x
y
o
l
?
0,
2
,
0,0,0,
0,0,0,0
???
????
????
v
h
ylx
vuylx
vuyx
0s inc o s,0,
0,0,0,0
????
????
?? vuylx
vuyx
89
[Solution 4] Whether are some kinds of suffering force shown in figure
the plane problem or not? If it is,then plane stress problem or still plane
strain problem?
R
O
q
x
y
q
h
z
y
o
R>>h
a)
h
Q
Q
O z
y
x
y
R O
R>>h
b)
90
[练习 4] 图所示的几种受力体是否为平面问题？若是，则是
平面应力问题，还是平面应变问题？
R
O
q
x
y
q
h
z
y
o
R>>h
a)
h
Q
Q
O z
y
x
y
R O
R>>h
b)
91
R<<l
p
R
O
p
y
x O
p
z
L
p
y
c)
Figure a) is the plane stress problem.The load carries the
perpendicular function in board surface shown in Fig.b),so
it is the bending problem of lamella,The load carries the
perpendicular function in board edge shown in Fig.c).The
load and the cross-section have no variety along the shaft
of Z and R<<1,so it is the plane strain problem,
Solution,
92
R<<l c)
图 a)所示为平面应力问题。图 b）所示荷载垂直作用
于板面，故为薄板弯曲问题。图 c）所示荷载作用于板
边，荷载及横截面沿 z轴无变化，且 R<<L，故为平面
应变问题。
解,
p
R
O
p
y
x O
p
z
L
p
y
93
[Exercise 5] The lamella bar suffers the even tensile force
function in the direction of y shown in figure.Try to prove
there is no existence of the stress force on the top point A
of outstanding part in the center board,
q
2N
O
C
B
A
y
q
x 2?
1?
1N
Solution,It can be seen as plane stress problem.Both AB
and AC are free boundary(and ),and
have no surface force function,ie,
Substituting boundary term has,
21 ?? ?
0?? YX
1111 s i n,c o s ?? ?? ml
The boundary of AB,
( 1 )
0σs i n ατc o s α
0τs i n ασc o s α
y1xy1
xy1x1
??
?
?
?
??
??
94
[练习 5] 如图所示薄板条在 y方向受均匀拉力
作用，试证明在板中间突出部分的尖端 A处无
应力存在。
解, 本题可视为平面应力问题,AB和 AC都
是自由边界（且 ），无面
力作用，即,。代入边
界条件有,
21 ?? ?
0?? YX
AB边界,
1111 s i n,c o s ?? ?? ml
( 1 )
0σs i n ατc o s α
0τs i n ασc o s α
y1xy1
xy1x1
??
?
?
?
??
??
q
2N
O
C
B
A
y
q
x 2?
1?
1N
95
The boundary of AC,
12
122
s i n
c o sc o s
?
??
??
??
m
l
)2(
0s i nc o s
0s i nc o s
11
11
??
?
?
?
??
??
yxy
xyx
????
????
Because A is placed in the boundary of AB and AC,it needs to satisfy type(1) and
type (2) at the same time,From it,can get,the problem has
been proved,0??? xyyx ???
[Exercise 6] Figure a) is a rectangle cross-section dam,whose right side suffers the
pressure force of the static water,The top section suffers the function of the
concentrated force P,Try to write a stress boundary term of dam and fix edge needn’t
be considered,
96
AC边界,
12
122
s i n
c o sc o s
?
??
??
??
m
l
)2(
0s i nc o s
0s i nc o s
11
11
??
?
?
?
??
??
yxy
xyx
????
????
由于 A同处于 AB,AC边界，因此，需同时满足式（ 1）和式（ 2）
，由此解得：,问题得证。
0??? xyyx ???
[练习 6] 图 a）所示为一矩形截面水坝，其右侧面受静水压力。
顶部受集中力 P作用。试写出水坝的应力边界条件，固定边不必考
虑。
97
P x
? O
y
y?
2h
h h
a)
P x
? O
y
y?
xy?
x?
y?
yx?
b)
P
?x O
y?
yx?
y
c)
Solution,1.List the stress boundary term,
(1) The left boundary,)1( 0)(,0)( ??
?? hxxyhxx ??
(2) The right boundary,
)2( 0)(,)( ??? ???? hxxyyhxx ???
98
解,1、列出应力边界条件
（ 1）左边界,
)1(0)(,0)( ?? ?? hxxyhxx ??
（ 2）右边界,
)2(0)(,)( ??? ???? hxxyyhxx ???
P x
? O
y
y?
2h
h h
a)
P x
? O
y
y?
xy?
x?
y?
yx?
b)
P
?x O
y?
yx?
y
c)
99
Top End Part,
)5(c o s)(
)4(s in
2
)(
)3(s in)(
0
0
0
??
??
??
Pdx
h
Px d x
Pdx
y
h
h
xy
h
h
yy
h
h
yy
?
??
??
?
?
?
?
?
?
?
?
?
[Exercise 7] The structure in the figure is consist of two kinds of different
materials.Try to get solution to displacement and stress answer with the function
of load q in the vertical distribution of equality (Establishing h,a,l,
has been known),
qEE,,,,2211 ??
1.Adopt the solution to displacement.Because this structure is placed in
the pressuring state with double direction of equality(Both stress and
strain are the constant quantity),therefore,we can suppose its
displacement as a line function and now divide two fields on top and
underneath to express,
Solution,
100
（ 3）上端部,
)5(c o s)(
)4(s in
2
)(
)3(s in)(
0
0
0
??
??
??
Pdx
h
Px d x
Pdx
y
h
h
xy
h
h
yy
h
h
yy
?
??
??
?
?
?
?
?
?
?
?
?
[练习 7] 图所示结构由两种不同材料构成。试求其在竖向均布
荷载 q作用下的位移和应力解答（设 h,a,L,均已
知）。
qEE,,,,2211 ??
1、采用位移解法。由于此结构处于双向均匀受压状态
（应力、应变为常量），因此，可假设其位移是线性函数，
现分上、下两区域表达为,
解,
101
The Part ABCD,
)1(,11111111 fycxdvcybxau ??????
The Part CDEF,
)2(,22222222 fycxdvcybxau ??????
Obviously,the type(1) and type(2) can satisfy Lamei equation of the plane stress
circumstance,
2.Consider the displacement constraints and
continuous term of transformation,
ayay
hyhylxlx
vv
vuuu
??
??????
?
????
)()(
0)(,0)(,0)(,0)(
21
2221
From it,we can get,
)(,0,,0
0
2111222
222111
haeaefdhefd
cbacba
????????
??????
102
ABCD部分,
)1(,11111111 fycxdvcybxau ??????
CDEF部分,
)2(,22222222 fycxdvcybxau ??????
显然式（ 1）、式（ 2）能满足平面应力情况下的拉梅方程式。
2、考虑位移约束和变形连续条件,
ayay
hyhylxlx
vv
vuuu
??
??????
?
????
)()(
0)(,0)(,0)(,0)(
21
2221
由此解得,
)(,0,,0
0
2111222
222111
haeaefdhefd
cbacba
????????
??????
103
q
x
B
D
a
h
F
L L
y
O A
C
E
11 ?E
22 ?E
)4()(,0
)3()()(,0
222
2111
hyevu
haeayevu
???
?????
104
)4()(,0
)3()()(,0
222
2111
hyevu
haeayevu
???
?????
q
x
B
D
a
h
F
L L
y
O A
C
E
11 ?E
22 ?E
105
)6(
0,0
1
,
1
)(
1
1
,
1
)(
1
)5(
0,,0
0,,0
2111
2
2
2
2
21
2
1
1
111
2
1
1
1
2
2
2
22
21
2
1
11
111
2
1
1
1
2222
1111
?
?
?
?
?
?
?
?
?
?
?
???
?
?
?
??
?
?
?
?
?
??
?
?
?
?
?
?
?
???
???
xyxyxy
yxyy
xyxx
xyyx
xyyx
G
e
E
e
EE
e
E
e
EE
e
e
???
?
?
?
???
?
?
?
?
?
?
?
???
?
?
???
???
106
)6(
0,0
1
,
1
)(
1
1
,
1
)(
1
)5(
0,,0
0,,0
2111
2
2
2
2
21
2
1
1
111
2
1
1
1
2
2
2
22
21
2
1
11
111
2
1
1
1
2222
1111
?
?
?
?
?
?
?
?
?
?
?
???
?
?
?
??
?
?
?
?
?
??
?
?
?
?
?
?
?
???
???
xyxyxy
yxyy
xyxx
xyyx
xyyx
G
e
E
e
EE
e
E
e
EE
e
e
???
?
?
?
???
?
?
?
?
?
?
?
???
?
?
???
???
107
3.Consider the stress boundary term and stress continuous term (The surface of CD
is a smooth contact),
0)()(,)()(
0)(,)(
2121
0101
???
???
????
??
ayxyayxyayyayy
yxyyy q
????
??
From it,we can get,
qEeqEe
2
22
2
1
21
1
1,1 ?? ??????
4.The displacement and the stress weight are,
)(
1
,0
)(
1
)(
1
,0
2
2
2
22
2
2
2
1
2
1
11
hyq
E
vu
haq
E
ayq
E
vu
?
?
???
?
?
??
?
???
?
??
108
3、考虑应力边界条件和应力连续条件（ CD面为光滑接触）,
0)()(,)()(
0)(,)(
2121
0101
???
???
????
??
ayxyayxyayyayy
yxyyy q
????
??
由此解得,
qEeqEe
2
22
2
1
21
1
1,1 ?? ??????
4、位移及应力分量为,
)(
1
,0
)(
1
)(
1
,0
2
2
2
22
2
2
2
1
2
1
11
hyq
E
vu
haq
E
ayq
E
vu
?
?
???
?
?
??
?
???
?
??
109
0,,
0,,
2222
1111
?????
?????
xyyx
xyyx
????
????
110
0,,
0,,
2222
1111
?????
?????
xyyx
xyyx
????
????
111
112
