1
Elasticity
2
3
§ 5-4 The single-valued condition of stress and
displacement in multiply connected region
§ 5-3 Complex-variable representation of boundary
condition
§ 5-2 Complex-variable representation of stress and
displacement
§ 5-1 Complex-variable representation of stress function
§ 5-6 Problem of infinite plane including hole
§ 5-5 Situation of infinite multiply connected body
Chapter 5 Complex-variable
Methods for Plane Elasticity
4
§ 5-4 多连通域内应力与位移的单值条件
§ 5-3 边界条件的复变函数表示
§ 5-2 应力和位移的复变函数表示
§ 5-1 应力函数的复变函数表示
§ 5-6 含孔口的无限大板问题
§ 5-5 无限大多连体的情形
第五章 平面问题的复变函数法
5
Chapter 5 Complex-variable
Methods for Plane Elasticity
When solving plane problems by Cartesian coordinates or polar
coordinates,the boundary of object is straight line or circular
arc,To other boundary,for example ellipse,hyperbola,non-
concentric circles and so on,we need use different curvilinear
coordinates,Applying complex-variable can predigest these
problems,
In this chapter,we just introduce the simple application of
complex-variable in elasticity,
6
第五章 平面问题的复变函数法
直角坐标及极坐标求解平面问题，所涉及的物
体边界是直线或圆弧形。对于其他一些边界，例如
椭圆形、双曲形、非同心圆等就要用不同的曲线坐
标。应用复变函数可使该类问题得以简化。本章只
限于介绍复变函数方法在弹性力学中的简单应用。
7
§ 5-1 Complex-variable representation of stress function
In chapter 2,we have proved,in plane problems,there
is a stress function φ that is biharmonic function of
position coordinates,if body force is constant,i.e,
04 ?? ?
i,1
i,1
??
?
?
?
?
?
?
?
?
?
?
?
y
z
x
z
y
z
x
z
Now introduce complex variable z= x＋ iy and z＝ x－ iy to
replace real variable x and y,Noticing,
8
§ 5-1 应力函数的复变函数表示
在第二章中已经证明, 在平面问题里, 如
果体力是常量, 就一定存在一个应力函数 φ,
它是位置坐标的重调和函数, 即
04 ?? ?
现在, 引入复变数 z= x＋ iy和 z＝ x－ iy以代替实
变数 x 和 y。 注意
i,1
i,1
??
?
?
?
?
?
?
?
?
?
?
?
y
z
x
z
y
z
x
z
9
We find the transformation are
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
???
?
???
)i(
)(
zzy
z
zy
z
zy
zzx
z
zx
z
zx
zyxzyx ?
??
?
??
?
?
?
??
?
??
?
? ?????? 2i,2i
furthermore,
???? 22
2
2
2
2
)(,)( zzyzzx ?????????????????
10
可以得到变换式
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
???
?
???
)i(
)(
zzy
z
zy
z
zy
zzx
z
zx
z
zx
zyxzyx ?
??
?
??
?
?
?
??
?
??
?
? ?????? 2i,2i
进而
???? 22
2
2
2
2
)(,)( zzyzzx ?????????????????
11
zzyx ??
??
?
??
?
??? ???? 2
2
2
2
2
2 4
Pzz ?????? ??
2
2 4
Let
So we can transform the function
04 ?? ? )(022
4
azz ??? ? ?
as
0)( 2224 ??????? P??
For
02 ?? P
12
zzyx ??
??
?
??
?
??? ???? 2
2
2
2
2
2 4
Pzz ?????? ??
2
2 4
令
于是可将方程式
04 ?? ? )(022
4
azz ??? ? ?
变换成为
0)( 2224 ??????? P??
由 02 ?? P
13
It is obvious known P is harmonic function which can be
obtained by real part of analytical function,Suppose f(z) as
analytical function and let
))()((
2
1 zfzfP ??
Pzz ?????? ??
2
2 4
For
)])()((21[4141
2
zfzfPzz ?????? ?
)('4)( zzf ??
))(')('(21
2
zzzz ??? ?????
Let
yields
thus
14
可知,P是调和函数可由解析函数的实部得到。设
f(z)为解析函数，可令
))()((21 zfzfP ??
Pzz ?????? ??
2
2 4由
)])()((21[4141
2
zfzfPzz ?????? ?
)('4)( zzf ??
))(')('(21
2
zzzz ??? ?????
令
得
则
15
))()(')('(21 zzzzzz ???? ?????
Then integrating with respect to z yields
))(d)()()((21 zgzzzzzz ? ???? ????
)(d)( zzz? ? ??
Let
)(')( zz ?? ?i.e,
))()()()((21 zgzzzzz ???? ????
Integrating the above equation with respect to yields z
thus
16
))()(')('(2
1 zzzzz
z ???
? ???
?
?
再对 z积分，得到
))(d)()()((21 zgzzzzzz ? ???? ????
)(d)( zzz? ? ??令
)(')( zz ?? ?即
))()()()((21 zgzzzzz ???? ????
将上式对 积分，得到 z
则
17
Notice the biharmonic function on the left side of the above
equation is a real function,It is obvious that the four terms
on the right side must be conjugate two and two,The first
two terms is conjugate,and the next two terms should be also
conjugate,
)()( zzg ??
Let
we obtain the famous gusa formula
])()()()([21 zzzzzz ????? ????
it can be also written as
)]()([Re zzz ??? ??
18
注意上式左边的重调和函数 φ 是实函数, 可见该
式右边的四项一定是两两共轭, 前两项已经是共
轭的, 后两项也应是共轭的,
)()( zzg ??
令
即得有名的古萨公式
])()()()([21 zzzzzz ????? ????
也可以写成
)]()([Re zzz ??? ??
19
So in plane problems when body force is constant,stress
function υ can be represented by two analytical functions of
complex variable z,?(z) and ?(z),named K-M function,So
solving plane problems is just selecting K-M function
properly and determining any constant in them according to
boundary condition,
20
于是可见，在常量体力的平面问题中，应力函数 φ
总可以用复变数 z的两个解析函 ?(z)和 ?(z)来表示，
称为 K-M 函数。而求解各个具体的平面问题，可归结
为适当地选择这两个解析函数，并根据边界条件决定其
中的任意常数。
21
§ 5-2 Complex-variable
representation of stress and displacement
According to the relation between stress component
and stress function,
yx
x
y
xy
y
x
??
?
??
?
?
?
?
?
?
?
?
?
?
?
?
2
2
2
2
2
One Complex-variable representation of stress component
22
§ 5-2 应力和位移的复变函数表示
根据应力分量和应力函数的关系
yx
x
y
xy
y
x
??
?
??
?
?
?
?
?
?
?
?
?
?
?
?
2
2
2
2
2
一 应力分量的复变函数表示
23
We find Complex-variable representation of stress
component,
zzxy
yx
??
??
?
??
?
??? ????? 24
2
2
2
2
))()()()((21 zzzzzz ????? ????
for
)('Re4])(')('[2 zzzyx ????? ????
yields
hence,for
24
可得到应力分量的复变函数表示
zzxy
yx
??
??
?
??
?
??? ????? 24
2
2
2
2
))()()()((21 zzzzzz ????? ????
由
)('Re4])(')('[2 zzzyx ????? ????
可得
而由
25
2
2
4
i
2
i2
2
2
2
2
i2
2
z
yx
yxyx
xyxy
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
??
?
?
?
?
?
?
?
???
?
?
???
???
)]('')(''[2i2 zzzxyxy ????? ????yields
)](')(''[2i2 zzzxyxy ????? ????or
26
2
2
4
i
2
i2
2
2
2
2
i2
2
z
yx
yxyx
xyxy
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
??
?
?
?
?
?
?
?
???
?
?
???
???
)]('')(''[2i2 zzzxyxy ????? ????可得
)](')(''[2i2 zzzxyxy ????? ????或
27
Only given ?(z) andψ (z),we can divide the right side of
above equation into imaginary part and real part,from
imaginary part we get τxy,from real part we get σy-σx,
)('Re4])(')('[2 zzzyx ????? ????
)](')(''[2i2 zzzxyxy ????? ????
and
is complex-variable representation of stress component,Of
course by building equations,σx,σy, τxy can be represented
by ?(z) and ψ (z) respectively,but that will make equations
become lengthiness and it’s not convenient to use,
28
只要已知 ?(z)及 ψ (z),就可以把上述公式右
边的虚部和实部分开, 由虚部得出 τ xy,由实部得
出 σ y-σ x。
)('Re4])(')('[2 zzzyx ????? ????
)](')(''[2i2 zzzxyxy ????? ????和
就是应力分量的复变函数表示 。 当然也可以建立
公式, 把 σ x,σ y, τ xy三者分开用 ?(z)和 ψ (z)
来表示, 但那些公式将比较冗长, 用起来很不方
便 。
29
Two Complex-variable representation of displacement component
Assuming plane stress problems,according to
geometrical equation and physical equation
yields
yyxyxx
uE ??????? )1()( ??????
?
?
xyxxyy
vE ??????? )1()( ??????
?
?
xyy
u
x
vE ?
? ??
??
?
?
? )()1(2
2
2
2
2
''
)1(])()([2
)1(])()([2
x
zz
x
x
zz
x
u
E
?
?
???
?
?
?
?
?
????
?
?
?
???
?
???
30
二 位移分量的复变函数表示
假定为平面应力问题 。 由几何方程及物理方程
yyxyxx
uE ??????? )1()( ??????
?
?
xyxxyy
vE ??????? )1()( ??????
?
?
xyy
u
x
vE ?
? ??
??
?
?
? )()1(2
2
2
2
2
''
)1(])()([2
)1(])()([2
x
zz
x
x
zz
x
u
E
?
?
???
?
?
?
?
?
????
?
?
?
???
?
???可得
31
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
???
?
???
)i(
)(
zzy
z
zy
z
zy
zzx
z
zx
z
zx
for
)(')()( zzzzz ??? ??????
and notice
])()([)()(
])()([])()([
'' zzz
z
z
z
zz
zz
zz
x
????
????
??
?
?
?
?
?
?
??
?
?
?
?
?
?
?
?
?
?
??
?
?
])()([)()(
])()([])()([i
'' zzz
z
z
z
zz
zz
zz
y
????
????
??
?
?
?
?
?
?
??
?
?
?
?
?
?
?
?
?
?
??
?
?
?similarly
yields
32
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
???
?
???
)i(
)(
zzy
z
zy
z
zy
zzx
z
zx
z
zx
由于
)(')()( zzzzz ??? ??????
并注意到
])()([)()(
])()([])()([
'' zzz
z
z
z
zz
zz
zz
x
????
????
??
?
?
?
?
?
?
??
?
?
?
?
?
?
?
?
?
?
??
?
?
])()([)()(
])()([])()([i
'' zzz
z
z
z
zz
zz
zz
y
????
????
??
?
?
?
?
?
?
??
?
?
?
?
?
?
?
?
?
?
??
?
?
?同理
可得
33
Successive integration of the above two equations with respect to
x and y,lead to
Where f1 and f2 are arbitrary functions,Substituting above
equations into the following equation
2
2
2
2
''
)1(])()([i2
)1(])()([2
y
zz
x
y
zz
y
v
E
?
?
???
?
?
??
?
?
????
?
?
?
???
?
???
?
?
?
??
?
?
?
?
?
?????
?
?
?
????
)()1(])()(i[2
)()1(])()([2
2
1
xf
y
zzEv
yf
x
zzEu
?
???
?
???
yxy
u
x
vE
xy ??
????
?
??
?
?
?
??
?
2
)()1(2
34
2
2
2
2
''
)1(])()([i2
)1(])()([2
y
zz
x
y
zz
y
v
E
?
?
???
?
?
??
?
?
????
?
?
?
???
?
???
将上两式分别对 x及 y积分，得
?
?
?
??
?
?
?
?
?
?????
?
?
?
????
)()1(])()(i[2
)()1(])()([2
2
1
xf
y
zzEv
yf
x
zzEu
?
???
?
???
其中的 f1及 f2为任意函数 。 将上式代入式
yxy
u
x
vE
xy ??
????
?
??
?
?
?
??
?
2
)()1(2
35
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
???
?
???
)i(
)(
zzy
z
zy
z
zy
zzx
z
zx
z
zx
for
])()([i)()(i
])()([i])()([
''
zzz
z
z
z
zz
zz
zz
y
????
????
???
?
?
?
?
?
?
?
?
?
?
?
??
?
?
?
?
?
?
?
?
?
?
??
?
?
])()([i)()(i
])()([i])()([i
'' zzz
z
z
z
zz
zz
zz
x
????
????
????
?
?
?
?
?
?
?
?
?
?
??
??
?
?
?
?
?
?
?
?
?
?
???
?
?
?
36
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
???
?
???
)i(
)(
zzy
z
zy
z
zy
zzx
z
zx
z
zx
由于
])()([i)()(i
])()([i])()([
''
zzz
z
z
z
zz
zz
zz
y
????
????
???
?
?
?
?
?
?
?
?
?
?
?
??
?
?
?
?
?
?
?
?
?
?
??
?
?
])()([i)()(i
])()([i])()([i
'' zzz
z
z
z
zz
zz
zz
x
????
????
????
?
?
?
?
?
?
?
?
?
?
??
??
?
?
?
?
?
?
?
?
?
?
???
?
?
?
37
and yields
x
xf
y
yf
d
)(d
d
)(d 11 ??
Thus we can find the displacement of rigid body
f1(y)＝ u0－ ωy,f2(x)＝ v 0＋ ωx
We get
)(
d
d
)(
d
d
)1(2
)(
d
d
)1(])()([i2
)(
d
d
)1(])()([2
21
2
2
2
1
2
xf
x
yf
yyx
xf
xyx
zz
x
u
yf
yyx
zz
y
u
x
v
y
u
E
??
??
?
???
?
??
?
???
?
?
?
?
??
?
???
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
???
?
???
38
)(
d
d
)(
d
d
)1(2
)(
d
d
)1(])()([i2
)(
d
d
)1(])()([2
21
2
2
2
1
2
xf
x
yf
yyx
xf
xyx
zz
x
u
yf
yyx
zz
y
u
x
v
y
u
E
??
??
?
???
?
??
?
???
?
?
?
?
??
?
???
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
???
?
???
从而得到
x
xf
y
yf
d
)(d
d
)(d 11 ??
于是得到刚体位移
f1(y)＝ u0－ ωy,f2(x)＝ v 0＋ ωx
故有
39
If neglecting displacement of rigid body,we have
)i)(1()(4)i( yxzvuE ????????? ????
zyx ?
??
?
??
?
? ??? 2i
for
])()()()([21 zzzzzz ????? ????
yields
)()()(
)(')()(2i
'
'
zzzz
zzzz
zyx
???
???
???
???
???
?
?
?
?
?
?
?
?
40
若不计刚体位移，则有
zyx ?
??
?
??
?
? ??? 2i由式
])()()()([21 zzzzzz ????? ????
)()()(
)(')()(2i
'
'
zzzz
zzzz
zyx
???
???
???
???
???
?
?
?
?
?
?
?
?得到
)i)(1()(4)i( yxzvuE ????????? ????
41
let the result back substitution,and on the two side divide 1+ν
yields
)()()(13)i(1 ' zzzzvuE ?????? ???????
This is complex-variable representation of displacement
component,If ?(z) and ψ (z) are given,we can divide the real
part and imaginary part of the right side of the above formula,
and u and v can be solved,
The above formula is educed in plane stress problem,To
plane strain problem,we need replace E with E/(1－ 2) and
with /(1－ )。
? ?
??
42
将结果回代,并两边除以 1+ 得
这就是位移分量的复变函数表示。若已知 ?(z)及
ψ (z)，就可以将该式右边的实部和虚部分开，
从而得出 u和 v。
)()()(13)i(1 ' zzzzvuE ?????? ???????
?
上述公式是针对平面应力情况导出的。对于
平面应变情况，须将式中的 E改换为 E/(1－ 2),
改换为 /(1－ )。 ?
?
??
43
§ 5-3 Complex-variable Representation
Of Boundary Condition
To evaluate φ of every crunode in boundary,we need
apply boundary condition of stress,i.e.,
Yml
Xml
yxy
xyx
??
??
??
??
yxxy xyyx ??
???
?
??
?
?? ?????? 2
2
2
2
2
,,and
Substitution into the
above equation,gives
Y
x
m
yx
l
X
yx
m
y
l
???
?
?
??
?
?
?
?
???
?
?
??
?
?
??
?
?
???
?
?
??
?
?
??
?
???
?
?
??
?
?
?
?
2
22
2
2
2
??
??
44
§ 5-3 边界条件的复变函数表示
为了求得边界上各结点处的 φ 值，须要应用应
力边界条件，即,
Yml
Xml
yxy
xyx
??
??
??
??
yxxy xyyx ??
???
?
??
?
?? ?????? 2
2
2
2
2
,,而
代入上式，即得,
Y
x
m
yx
l
X
yx
m
y
l
???
?
?
??
?
?
?
?
???
?
?
??
?
?
??
?
?
???
?
?
??
?
?
??
?
???
?
?
??
?
?
?
?
2
22
2
2
2
??
??
45
As the figure shown
l=cos(N,x)=dy/ds,
m=cos(N,y)=-dx/ds,
So the above equation can be
rewritten as,
Y
xs
x
yxs
y
X
yxs
x
ys
y
???
?
?
?
?
?
?
?
?
???
?
?
?
?
?
?
??
?
?
???
?
?
?
?
?
?
??
?
???
?
?
?
?
?
?
?
?
2
22
2
2
2
d
d
d
d
d
d
d
d
??
??
YxsXys ??????? ??????
?
?
???
?
?
? ??
d
d,
d
d
Thus,yields
46
由图可见,
l=cos(N,x)=dy/ds,
m=cos(N,y)=-dx/ds,
于是, 前式可改写为,
Y
xs
x
yxs
y
X
yxs
x
ys
y
???
?
?
?
?
?
?
?
?
???
?
?
?
?
?
?
??
?
?
???
?
?
?
?
?
?
??
?
???
?
?
?
?
?
?
?
?
2
22
2
2
2
d
d
d
d
d
d
d
d
??
??
YxsXys ??????? ??????
?
?
???
?
?
? ??
d
d,
d
d由此得,
47
Suppose A is a fix point in the boundary,and B is a arbitrary
point,so the composition of forces from A to B can be obtained
by integrating of the above equation with respect to s from A to
B,
B
A
B
A
B
A
B
A
yx
yxxy
s
xys
sYXPP
??
?
?
??
?
?
?
?
?
?
?
????
?
?
??
?
?
?
?
?
?
?
?
??
?
?
??
?
?
?
?
?
?
?
???? ??
????
??
iii
di
d
d
dii
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
???
?
???
)i(
)(
zzy
z
zy
z
zy
zzx
z
zx
z
zx
Substituting this formula
48
设 A是边界上的固定点,B为任意一点，则从 A到 B
边界上的合力，可用上式从 A点到 B点对 s积分得
到,
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
???
?
???
)i(
)(
zzy
z
zy
z
zy
zzx
z
zx
z
zx
将式
B
A
B
A
B
A
B
A
yx
yxxy
s
xys
sYXPP
??
?
?
??
?
?
?
?
?
?
?
????
?
?
??
?
?
?
?
?
?
?
?
??
?
?
??
?
?
?
?
?
?
?
???? ??
????
??
iii
di
d
d
dii
49
into the above equation,and rearrangement yields
BAyx zzzzPP ])()(')([ii ??? ?????
Adding a complex constant into stress function,which
doesn’t influence the stress,So we can let A of stress
function as zero,and then σ in the boundary gives
])()(')([ii ??????? ????? yx PP
)()(')()i(i ??????? ???? yx PP
or
This is boundary condition of stress,
50
代入，整理得,
BAyx zzzzPP ])()(')([ii ??? ?????
把应力函数加上一个复常数, 并不影响应力 。
因此, 可把应力函数 A处的值设为零, 于是对于
边界上的 σ 有
])()(')([ii ??????? ????? yx PP
)()(')()i(i ??????? ???? yx PP或
这就是应力边界条件。
51
To boundary condition of displacement
)()()(13)i(1 ' zzzzvuE ?????? ???????
ss vvuu ??,
Substituting them into the following equation
We can obtain complex-variable representation of boundary
condition of displacement in plane stress problem,
)i(1)()()(13 ' ss vuEzzzz ??????? ??????
To plane strain problem,we need replace E with E/(1－ 2) and with /(1－ )。
? ?
?
?
52
对于位移边界条件
ss vvuu ??,
将其代入下式
即得平面应力情况下位移边界条件的复变函数表示
)i(1)()()(13 ' ss vuEzzzz ??????? ??????
)()()(13)i(1 ' zzzzvuE ?????? ???????
对于平面应变，须将式中的 E 改换为 E/(1－ 2)，
改换为 /(1－ )。 ?? ?
?
53
§ 5-4 The single-valued condition of stress
and displacement in multiply connected region
When stress is determined,the stress function can still a
arbitrary linear function,so the K-M function is not determined
completely,so to the simply connected region,the K-M
function can be determined by selecting the suitable coordinate,
But to the multiply connected region,it is still a problem,In this
section,the condition of the K-M function satisfied single-
valued in multiply connected region is discussed,
Suppose there is a multiply connected region that has a
interior boundary C,and in the interior boundary C the external
force vector is given,Generally multiform function is
logarithmic function,we suppose
54
§ 5-4 多连通域内应力与位移的单值条件
应力确定后，应力函数仍可差一个任意的线性函数，
这时 K-M函数并未完全确定,对于单连通区域，可以通过选
取适当坐标系等办法，使得 K-M函数完全确定 ;但对于多连
通区域仍不能完全确定,本节讨论 K-M函数在多连通区域内
满足单值的条件。
设有多连通区域，有一内边界 C，设在边界 C上的外力
矢量已给定。通常的多值函数是对数函数，我们设
55
)()l n ()(
)()l n ()(
zzzBz
zzzAz
fkk
fkk
??
??
???
???
D
C
Where zk is a arbitrary point in the interior boundary,?f andψf are
single-valued analytical function(holomorphic function),and Ak
and Bk are constants gives,
56
)()l n ()(
)()l n ()(
zzzBz
zzzAz
fkk
fkk
??
??
???
???
D
C
这里 zk为内部边界内的任意一点,?f和 ψ f为单值的
解析函数（全纯函数），而 Ak, Bk为常数,
57
kkkkkk BA 'i',i ???? ????
The derivative of above function is single-valued,but it is multivalued
by itself,while z move around the circum once,the value of ln(zk) have
a increment of 2πi,so the increments of ?(z) and ψ (z) is 2πi Ak and
2πiBk respectively,and according to the following formula,the master
vector of stresses
)]()(')([ii zzzzYXBA ??? ??????
the master vector of stress(around the whole boundary) is on the
left side,and the increment is on the right side,
ykxkkk PPBA i)(2 ???? ?
58
kkkkkk BA 'i',i ???? ????
前面的函数的导数是单值的，但他们本身是多值
的，当 z绕周边一周时，函数值 ln(zk)产生一个增
量 2π i,于是 ?(z)和 ψ (z)的增量分别是 2π i Ak
和 2π iBk,这时应力主矢量按照公式
)]()(')([ii zzzzYXBA ??? ??????
左边将得到应力主矢量 (沿整个边界），右边得到
一增量,
ykxkkk PPBA i)(2 ???? ?
59
While according to the following formula,displacement
will obtain increment,and according to single-valued the
increment should be zero,
ykxkkk PPBA i)(2 ???? ?
and
yields
)1(2
i
)1(2
i
??
?
?
?
?
?
?
?
??
ykxk
k
ykxk
k
PP
B
k
PP
A
)13(0 ???? ????? kk BA
)()()(13)i(1 ' zzzzvuE ?????? ???????
60
这时位移按照公式
)()()(13)i(1 ' zzzzvuE ?????? ???????
也将得到增量，根据单值性这个增量应为零,
)13(0 ???? ????? kk BA
ykxkkk PPBA i)(2 ???? ?
结合
可得到
)1(2
i
)1(2
i
??
?
?
?
?
?
?
?
??
ykxk
k
ykxk
k
PP
B
k
PP
A
61
)()ln (
)1(2
i
)(
)()ln (
)1(2
i
)(
f
f
zzz
PP
z
zzz
PP
z
k
ykxk
k
ykxk
?
??
??
?
??
?
??
?
?
?
??
?
?
??
thus
While there is m interior boundary,let
)()l n (
)1(2
i
)(
)()l n (
)1(2
i
)(
f
1
f
1
zzz
PP
z
zzz
PP
z
m
k
k
ykxk
m
k
k
ykxk
?
??
??
?
??
?
??
?
?
?
??
?
?
??
?
?
?
?
62
)()ln (
)1(2
i
)(
)()ln (
)1(2
i
)(
f
f
zzz
PP
z
zzz
PP
z
k
ykxk
k
ykxk
?
??
??
?
??
?
??
?
?
?
??
?
?
??
于是
当有 m个内边界时，取
)()l n (
)1(2
i
)(
)()l n (
)1(2
i
)(
f
1
f
1
zzz
PP
z
zzz
PP
z
m
k
k
ykxk
m
k
k
ykxk
?
??
??
?
??
?
??
?
?
?
??
?
?
??
?
?
?
?
63
§ 5-5 Situation of infinite multiply connected body
When exterior boundary of multiply connected body
approach infinite farness,this multiply connected body become
infinite multiply connected body,besides the above conditions,we
need consider the ultimate situation of infinite farness,
Regards origin of coordinate as the centre of circle,draw
a big enough circle sR,which include all interior boundary,To
an arbitrary point in the elastomer,but beyond sR,gives
??
??
?
?
?
?
?
????
?
?
?
?
?
????
z
z
z
z
z
z
z
z
zzz kkkk
ln
2
1
ln1lnln)ln (
2
?
the analytical function beyond sR
64
§ 5-5 无限大多连体的情形
当多连体的外边界趋于无限远时，该多连体成为无
限大的多连体，除上述条件外，还需考虑无限远的极限
情况。
以坐标原点为圆心，作充分大的圆周 sR，将所有的
内边界包围在其内，对于 sR之外，弹性体之内的任意一
点，可得到
??
??
?
?
?
?
?
????
?
?
?
?
?
????
z
z
z
z
z
z
z
z
zzz kkkk
ln
2
1
ln1lnln)ln (
2
?
在 sR之外的解析函数
65
)()l n (
)1(2
i
)(
)()l n (
)1(2
i
)(
f
1
f
1
zzz
PP
z
zzz
PP
z
m
k
k
ykxk
m
k
k
ykxk
?
??
??
?
??
?
??
?
?
?
??
?
?
??
?
?
?
?
so
)(ln
)1(2
i
)(
)(ln
)1(2
i
)(
f
*
f
*
zz
PP
z
zz
PP
z
yx
yx
?
??
??
?
??
?
?
?
?
?
?
?
?
??
It can be written as
Where Px,Py are the sum of surface forces in m boundary,
66
)()l n (
)1(2
i
)(
)()l n (
)1(2
i
)(
f
1
f
1
zzz
PP
z
zzz
PP
z
m
k
k
ykxk
m
k
k
ykxk
?
??
??
?
??
?
??
?
?
?
??
?
?
??
?
?
?
?
于是
)(ln
)1(2
i
)(
)(ln
)1(2
i
)(
f
*
f
*
zz
PP
z
zz
PP
z
yx
yx
?
??
??
?
??
?
?
?
?
?
?
?
?
??
可写为
其中 Px,Py为 m个边界上沿 x,y方向的面力之和。
67
Expand holomorphic function φ*f andψ*f in Multiply connected
region by luolang series,
?
?
??
??
??
??
?
?
n
n
n
n
zbz
zaz
)(
)(
f
*
f
*
?
?
?
??
??
?? ??
?
?
?
?
?
?
?
?
?
?
?
???
)(2
1
)(12
i1
)(12
i
2
11 n
n
n
n
yxyx
yx
zazan
z
PP
z
PP
????
??so
For components of stresses in infinite farness is finite,the
coefficients of n≥2 is zero,
)2(0,0 ??? naa nn
68
将多连通区域内的全纯函数 φ *f和 ψ *f展开为罗郎级数,
?
?
??
??
??
??
?
?
n
n
n
n
zbz
zaz
)(
)(
f
*
f
*
?
?
?
??
??
?? ??
?
?
?
?
?
?
?
?
?
?
?
???
)(2
1
)(12
i1
)(12
i
2
11 n
n
n
n
yxyx
yx
zazan
z
PP
z
PP
????
??于是
由于在无穷远处的应力分量应该是有限的，级数中
n≥2 的系数应为零。
)2(0,0 ??? naa nn
69
)](')(''[2i2 zzzxyxy ????? ????
Similarity,from
for components of stress in infinite farness is finite,so
)2(,0 ?? nb n
)()()()(
)()()()(
0f11
1
11f
*
0f
1
f
*
zzizbziz
zzizaziz
n
n
n
n
n
n
??????
??????
??????
??????
?
?
??
?
?
??
?
?
Where neglecting the constant terms that have no relation
to stresses,
70
)](')(''[2i2 zzzxyxy ????? ????
同样从
中，由于在无穷远处的应力分量应该是有限的，故有
)2(,0 ?? nb n
)()()()(
)()()()(
0f11
1
11f
*
0f
1
f
*
zzizbziz
zzizaziz
n
n
n
n
n
n
??????
??????
??????
??????
?
?
??
?
?
??
?
?
其中略去了和应力无关的常数项。
71
)(ln
)1(2
i
)(
)(ln
)1(2
i
)(
f
*
f
*
zz
PP
z
zz
PP
z
yx
yx
?
??
??
?
??
?
?
?
?
?
?
?
?
??so
)()i()i()(
)()i()i()(
0f11
1
11f
*
0f
1
f
*
zzzbzz
zzzazz
n
n
n
n
n
n
??????
??????
??????
??????
?
?
??
?
?
??
?
?
Where β is no relation to stress calculate,it can be regarded as
zero,and
?
?
??
?
?
??
?
?
?
?
1
0f
1
0f
)(
)(
n
n
n
n
n
n
zbz
zaz
?
?
72
)(ln
)1(2
i
)(
)(ln
)1(2
i
)(
f
*
f
*
zz
PP
z
zz
PP
z
yx
yx
?
??
??
?
??
?
?
?
?
?
?
?
?
??于是
)()i()i()(
)()i()i()(
0f11
1
11f
*
0f
1
f
*
zzzbzz
zzzazz
n
n
n
n
n
n
??????
??????
??????
??????
?
?
??
?
?
??
?
?
其中 β与应力计算无关，可取为零，而
?
?
??
?
?
??
?
?
?
?
1
0f
1
0f
)(
)(
n
n
n
n
n
n
zbz
zaz
?
?
73
?
??
?
?? ???
?
?
?
?
?
?
?
?
?
?
?
???
1
)(24
1
)(12
i1
)(12
i
2
n
n
n
n
n
yxyx
yx
zazan
z
PP
z
PP
?
????
??At this time
When z→∞, fields
??? 4?? yx
Similarity,when z→∞,for
)](')(''[2i2 zzzxyxy ????? ????
fields
11 i22i2 ????? ???? xyxy
From above equation,we can obtain the corresponding coefficient
and we can also find in infinite farness the distributing of stress is
symmetrical,
74
?
??
?
?? ???
?
?
?
?
?
?
?
?
?
?
?
???
1
)(24
1
)(12
i1
)(12
i
2
n
n
n
n
n
yxyx
yx
zazan
z
PP
z
PP
?
????
??这时
当 z→∞ 时，可得 ??? 4??
yx
同样当 z→∞ 时，由
)](')(''[2i2 zzzxyxy ????? ????
可得
11 i22i2 ????? ???? xyxy
从中可求得相应的系数，并可以看到在无限远处，应力的分
布是均匀的。
75
)(
2
)()(
4
)()(
1
1
??
???
?
???
?
xy
xy
yx
??
??
?
??
?
coefficients
)(
)(i
2
)()(
ln
)1(2
i
)(
)(
4
)()(
ln
)1(2
i
)(
0f
xy
0f
z
zz
PP
z
zzz
PP
z
xyyx
yxyx
?
?
??
??
??
?
??
??
?
?
?
?
?
?
?
?
??
???
?
?
?
?
?
???
?
?
?
??
thus
76
)(
2
)()(
4
)()(
1
1
??
???
?
???
?
xy
xy
yx
??
??
?
??
?
系数
)(
)(i
2
)()(
ln
)1(2
i
)(
)(
4
)()(
ln
)1(2
i
)(
0f
xy
0f
z
zz
PP
z
zzz
PP
z
xyyx
yxyx
?
?
??
??
??
?
??
??
?
?
?
?
?
?
?
?
??
???
?
?
?
?
?
???
?
?
?
??
则
77
§ 5-6 Problem of infinite plane including hole
Regards origin of coordinate as the centre of circle,draw a
big enough circle sR,which include all interior boundary,so to
an arbitrary point in the elastomer but beyond sR,we have
)(ln
)1(2
i
)(
)(ln
)1(2
i
)(
f
*
f
*
zz
PP
z
zz
PP
z
yx
yx
?
??
??
?
??
?
?
?
?
?
?
?
?
??
78
§ 5-6 含孔口的无限大板问题
以坐标原点为圆心，作充分大的圆周 sR，将所有的内
边界包围在其内，对于 sR之外，弹性体之内的任意一点，
可得到
)(ln
)1(2
i
)(
)(ln
)1(2
i
)(
f
*
f
*
zz
PP
z
zz
PP
z
yx
yx
?
??
??
?
??
?
?
?
?
?
?
?
?
??
79
?
?
??
?
?
??
?
?
???
??
1
11f
*
1
f
*
)i()(
)(
n
n
n
n
n
n
zbzz
zazz
???
??
?
?
?
?
?
??
?
?
??
?
?
??
??
?
?
?
???
?
?
?
??
?
?
??
1
)2(
2
1
)1(
11
1
)1(
)1(
1
)1(2
i
)(''
i
1
)1(2
i
)('
1
)1(2
i
)('
n
n
n
yx
n
n
n
yx
n
n
n
yx
znna
z
PP
z
nzb
z
PP
z
nza
z
PP
z
??
?
??
??
??
?
??
?
80
?
?
??
?
?
??
?
?
???
??
1
11f
*
1
f
*
)i()(
)(
n
n
n
n
n
n
zbzz
zazz
???
??
?
?
?
?
?
??
?
?
??
?
?
??
??
?
?
?
???
?
?
?
??
?
?
??
1
)2(
2
1
)1(
11
1
)1(
)1(
1
)1(2
i
)(''
i
1
)1(2
i
)('
1
)1(2
i
)('
n
n
n
yx
n
n
n
yx
n
n
n
yx
znna
z
PP
z
nzb
z
PP
z
nza
z
PP
z
??
?
??
??
??
?
??
?
81
?
?
?
?
?
??
?
?
?
?
?
?
??
?
?
1
)1(
0
0
)("
)('
)('
k
k
k
k
k
k
k
k
k
kzAz
zBz
zAz
?
?
?
rewritten as
1100 ii ???? ???? BA
where
)2()1()1(
)1(2
i
)1(2
i
11
11
???????
?
?
?
?
?
??
?? kbkBakA
PP
B
PP
A
kkkk
yxyx
??
?
??
82
?
?
?
?
?
??
?
?
?
?
?
?
??
?
?
1
)1(
0
0
)("
)('
)('
k
k
k
k
k
k
k
k
k
kzAz
zBz
zAz
?
?
?
改写为
1100 ii ???? ???? BA
其中
)2()1()1(
)1(2
i
)1(2
i
11
11
???????
?
?
?
?
?
??
?? kbkBakA
PP
B
PP
A
kkkk
yxyx
??
?
??
83
To the points in the
boundary of hole ?ieaz ?
?
?
?
?
?
??
?
?
?
?
?
??
??
?
?
0
i2i
0
i
0
i
e)("e
e)('
e)('
1
1
1
k
kk
kC
k
kk
kC
k
kk
kC
kaAzz
aAz
aAz
??
?
?
?
?
?
?
?
?
?
???
?
?
?
?
???
???
?
0
i)2(
2
i1
1
2i
0
0
)2i(2i
eee
e)('e
1
k
kk
k
k
kk
kC
aBaBB
aBz
???
??
?
)20( ?? ??
84
对于孔边上的点 ?ieaz ?
?
?
?
?
?
??
?
?
?
?
?
??
??
?
?
0
i2i
0
i
0
i
e)("e
e)('
e)('
1
1
1
k
kk
kC
k
kk
kC
k
kk
kC
kaAzz
aAz
aAz
??
?
?
?
?
?
?
?
?
?
???
?
?
?
?
???
???
?
0
i)2(
2
i1
1
2i
0
0
)2i(2i
eee
e)('e
1
k
kk
k
k
kk
kC
aBaBB
aBz
???
??
?
)20( ?? ??
85
Substituting the above equations into the following equation
???? ??????????? 2ie)](')(''[)(')('i ?????
We can obtain the stress boundary condition of series forms of
circle boundary in polar coordinate,
Assuming the external force is known and spread it by
Fourier series,
?
?
?
?
???
??
??
?
?
??
?
??
?
?
2
0
i
i
de)i(
2
1
ei
k
k
k
k
k
ppC
Cpp
86
将上列各式代入
???? ??????????? 2ie)](')(''[)(')('i ?????
就得到极坐标下圆周边界上的级数形式的应力边界条件。
设周边上的外力为已知，并将其展开为傅氏级数
?
?
?
?
???
??
??
?
?
??
?
??
?
?
2
0
i
i
de)i(
2
1
ei
k
k
k
k
k
ppC
Cpp
87
?? ????????? 2i e)](')(''[)(')('e i
k
k
kC ?????
?
???
?
????
?
?
??
?
?
?
?
??
?
?
?
?
?
??
??
???
???
???
0
i-)2(
2
i1
1
2i
0
0
i
0
i
0
ii
eee
eeee
k
kk
k
k
kk
k
k
kk
k
k
kk
k
k
k
k
k
aBaBB
kaAaAaAC
???
????
By comparing coefficient of eik? and e-ik?,fields
12
11
02
2
00
C
a
B
a
A
C
a
B
AA
??
???
88
?? ????????? 2i e)](')(''[)(')('e i
k
k
kC ?????
?
???
?
????
?
?
??
?
?
?
?
??
?
?
?
?
?
??
??
???
???
???
0
i-)2(
2
i1
1
2i
0
0
i
0
i
0
ii
eee
eeee
k
kk
k
k
kk
k
k
kk
k
k
kk
k
k
k
k
k
aBaBB
kaAaAaAC
???
????
比较两边 eik?和 e-ik?的系数，可得
12
11
02
2
00
C
a
B
a
A
C
a
B
AA
??
???
89
1
)1(
)3(
2
2
202
2
???
?
??
??
??
?
kC
a
B
a
Ak
kC
a
A
CB
a
A
kk
k
k
k
kk
k
1100 ii ???? ???? BA
For the stresses condition in infinite farness,yields
)(,2 )()(,4 )()( 11 ?????????? xyxyyx ????????
90
1
)1(
)3(
2
2
202
2
???
?
??
??
??
?
kC
a
B
a
Ak
kC
a
A
CB
a
A
kk
k
k
k
kk
k
1100 ii ???? ???? BA
由无限远处的应力条件，可得
)(,2 )()(,4 )()( 11 ?????????? xyxyyx ????????
91
For the single-valued condition of displacement yields
011 ?? BA?
12
11 C
a
B
a
A ??
and
We find
?
?
? ????? 1,1
1
1
1
1
CaBCaA
for
kk
k
k
k
kk
k
C
a
B
a
Ak
kC
a
A
CB
a
A
??
?
??
?
??
??
2
2
202
2
)1(
)3(
92
由位移的单值条件有
011 ?? BA?
12
11 C
a
B
a
A ??
及
可求得
?
?
? ????? 1,1
1
1
1
1
CaBCaA
再由
kk
k
k
k
kk
k
C
a
B
a
Ak
kC
a
A
CB
a
A
??
?
??
?
??
??
2
2
202
2
)1(
)3(
93
)3()1(
)3(
)2(),(
22
2
00
2
220
2
2
????
??
????
???
kCaAakB
kCaA
CAaBCBaA
k
k
kk
k
k
k
yields
By now,all coefficients have been solved,
For example,assuming the uniform pressure around hole is p and
the stress of infinite farness is zero,
94
)3()1(
)3(
)2(),(
22
2
00
2
220
2
2
????
??
????
???
kCaAakB
kCaA
CAaBCBaA
k
k
kk
k
k
k
可求得
至此，全部系数均已求出。
例 设孔周边为均匀压力 p，无限远处的应力为零。
95
)0(0,
i
0
0,
0
00
????
???
??
???
kCpC
ppp
BA
pp
k
??
??
?
Thus
)3(,0
0
2
2
211
???
?
???
kBA
paB
ABA
kk
So we get
96
)0(0,
i
0
0,
0
00
????
???
??
???
kCpC
ppp
BA
pp
k
??
??
?
则有
)3(,0
0
2
2
211
???
?
???
kBA
paB
ABA
kk
于是可求得
97
z
pa
z
z
2
)('
0)('
?
?
?
?
0,
2
0,,
2
2
2
2
2
??
????
??
????
?
?
?
?
?
?
u
G
pa
u
papa
Finally,we get
According to the above means,the general problems of infinite
plane including hole can be solved,
98
z
pa
z
z
2
)('
0)('
?
?
?
?
0,
2
0,,
2
2
2
2
2
??
????
??
????
?
?
?
?
?
?
u
G
pa
u
papa
最后得到
根据上述方法，圆孔口无限大板的一般问题都可以得到解决。
99
Exercise 5.1 Try to check-up the following Complex-variable
(1)
(2)
? ? ? ? zqzzqz 2,4 ??? ??
? ? ? ? i q zzz ?? ??,0
Solution,The fundamental formula gives
? ? ? ?az xy ??? ??'Re4
(1) Substituting ? ? ? ? zqzzqz
2,4 ??? ??
into (a),(b)
? ? ? ?? ? ? ?bizzz xyxy ????? 2'2 ????
100
练习 5.1 试考察下列复变函数所解决的问题
(1)
(2)
? ? ? ? zqzzqz 2,4 ??? ??
? ? ? ? i q zzz ?? ??,0
解, 基本公式为
? ? ? ?az xy ??? ??'Re4
? ? ? ?? ? ? ?bizzz xyxy ????? 2'2 ????
(1) 将 ? ? ? ? zqzzqz
2,4 ??? ??
分别代入 (a),(b)式
101
yields
xyq ?? ??
xyxy iq ??? 2????
associate with the above two equations,
yields
0,0,??? xyyx q ???
The given function can solve the problems that the rectangular
sheet is under uniform pulling force q in axis x,As the figure
5.1(a) shown,
(2) Substituting ? ? ? ? i q zzz ?? ??,0 into (a) and (b),yields
xyxy iiq ??? 22 ???
x y q q
Figure 5.1(a)
yx ?? ??0
102
得
xyq ?? ??
xyxy iq ??? 2????
联立求解以上两式,得
0,0,??? xyyx q ???
所给的函数可以解决矩形薄板在 x方向受均布拉力 q的
问题,如图 5.1(a)所示
(2) 将 ? ? ? ? i q zzz ?? ??,0 代入 (a),(b)两式,得
xyxy iiq ??? 22 ???
x y q q
图 5.1(a)
yx ?? ??0
103
associate with the above two equations,
yields
qxyyx ??? ???,0,0
The given function can solve the
pure shear problem of rectangular
sheet,As the figure 5.1(b) shown,
q q x y
图 5.1(b)
Exercise 5.2 As the figure shown,Try to prove we can use
complex-variable to solve the
pure bending problem of beam of rectangular cross-section,
? ? ? ? 22 88 zIiMzzIiMz ??? ??
Where I is the inertia moment of cross section of beam,M is
the moment of flexion,
M
y
x z y Solution,
The fundamental formula gives
104
联立求解以上两式,得
qxyyx ??? ???,0,0
所给的函数可以解决矩形薄
板受纯剪切问题,如图 5.1(b)示,
q q x y
图 5.1(b)
练习 5.2 如图所示,试证矩形截面梁的纯弯曲问题可用
如下的复变函数求解,
? ? ? ? 22 88 zIiMzzIiMz ??? ??
其中 I为梁截面的惯矩,M为作用的弯矩,
M
y
x z y 解,
基本公式为
105
? ? ? ?1Re4 ' xyz ??? ??
? ? ? ?? ? ? ?222 ' xyxy izzz ????? ????
? ? ? ? ? ? ? ? ? ?3' dSYiXizzzz ? ???? ???
Substituting ? ? ? ?zz ??,into (1),(2)
For (1),yields
? ? xyiyxIiM ?? ???
?
?
??
? ??
4Re4
i.e,
? ?4xyIMy ?? ??
106
? ? ? ?1Re4 ' xyz ??? ??
? ? ? ?? ? ? ?222 ' xyxy izzz ????? ????
? ? ? ? ? ? ? ? ? ?3' dSYiXizzzz ? ???? ???
将 ? ? ? ?zz ??,代入 (1),(2)式
由 (1)式得
? ? xyiyxIiM ?? ???
?
?
??
? ??
4Re4
即
? ?4xyIMy ?? ??
107
or
? ?5
0 ??
?
?
?
?
???
xy
xyI
My
?
??
xyxy iI
My ??? 2????
For equation (2) yields,
? ? xyxy iz
I
iM
I
iMiyx ??? 2
44
2 ????
?
?
??
? ??
?
??
?
? ??
i.e,
equations(4) and (5),yields
108
或
? ?5
0 ??
?
?
?
?
???
xy
xyI
My
?
??
xyxy iI
My ??? 2????
由 (2)式得
? ? xyxy iz
I
iM
I
iMiyx ??? 2
44
2 ????
?
?
??
? ??
?
??
?
? ??
即
将 (4),(5)式联立求得
109
00 ?? ? xyyx IMy ???
Verified the boundary conditions (3)
On the flank,0,0 ?? YX so ? ? 0??? dSYiXi
for ? ?
2
8 zI
iMz ???
yields
? ?
? ?
? ?
2'
'
'
4
4
4
z
I
iM
zz
z
I
iM
z
z
I
iM
z
?
?
??
?
?
?
110
00 ?? ? xyyx IMy ???
验证边界条件 (3)
在侧面, 0,0 ?? YX 所以 ? ? 0??? dSYiXi
由 ? ?
2
8 zI
iMz ???
得
? ?
? ?
? ?
2'
'
'
4
4
4
z
I
iM
zz
z
I
iM
z
z
I
iM
z
?
?
??
?
?
?
111
for ? ?
2
8 zI
iMz ?? yields ? ? 2
8 zI
iMz ???
so ? ? ? ? ? ?
0848 222' ??????? zIiMzIiMzIiMzzzz ???
we find equation (3) is identical equation,
The answer are stress state of
beam cross-section when it’s pure bending,
00 ?? ? xyyx IMy ???
112
由 ? ?
2
8 zI
iMz ?? 得 ? ? 2
8 zI
iMz ???
故 ? ? ? ? ? ?
0848 222' ??????? zIiMzIiMzIiMzzzz ???
即 (3)式恒成立,
由解答 所表示的是一个
纯弯时,梁横截面上的应力状态,
00 ?? ? xyyx IMy ???
113
Exercise 5.3 Try to educe formula of the stresses components in
polar coordinate that is described by Complex-variable and ??z? ? ?z?
? ?
? ? ? ?? ? ???
?
?????
???
i
rr
r
ezzzi
z
2'''
'
22
Re4
????
??
Solution,because in plane problems
tc o n sryx t a n???? ???? ?
so ? ?zr ’??? ? Re4??
Moreover in plane problems,we have
???????? 2s in2c o s22 xyyxyxr ?????
114
练习 5.3 试导出用复变函数 及 表示极坐标中
应力分量的公式
??z? ? ?z?
? ?
? ? ? ?? ? ???
?
?????
???
i
rr
r
ezzzi
z
2'''
'
22
Re4
????
??
解, 因为在平面问题中
常数???? ryx ???? ?
所以 ? ?zr ’??? ? Re4??
又因为在平面问题中,有
???????? 2s in2c o s22 xyyxyxr ?????
115
???
??
?
???
????
?
?
?
2c o s2s i n
2
2s i n2c o s
22
xy
yx
r
xy
yxyx
?
?
??
?
?
?
?
?
? ?? ? ? ?
? ?
?
??
???
???????
???
??
???
????
???
????
???
i
xyxy
xyyx
xy
yx
xy
yxyx
xy
yxyx
rr
ei
iii
i
i
2
2
2s in2c o s22s in2c o s
2c o s2s in
2
2
2s in2c o s
22
2s in2c o s
22
2
???
??????
?
?
?
?
?
?
?
?
?
?
??
?
?
?
?
?
?
?
?
?
???
thus
116
???
??
?
???
????
?
?
?
2c o s2s i n
2
2s i n2c o s
22
xy
yx
r
xy
yxyx
?
?
??
?
?
?
?
?
? ?? ? ? ?
? ?
?
??
???
???????
???
??
???
????
???
????
???
i
xyxy
xyyx
xy
yx
xy
yxyx
xy
yxyx
rr
ei
iii
i
i
2
2
2s in2c o s22s in2c o s
2c o s2s in
2
2
2s in2c o s
22
2s in2c o s
22
2
???
??????
?
?
?
?
?
?
?
?
?
?
??
?
?
?
?
?
?
?
?
?
???
则
117
for
so
? ? ? ?? ?zzzi xyxy '''22 ????? ????
? ? ? ?? ? ??? ????? irr ezzzi 2'''22 ????
Exercise 5.4 Try to apply the formulae
? ?
? ? ? ?? ? ???
?
?????
???
i
rr
r
ezzzi
z
2'''
'
22
Re4
????
??
for educe the foumula of
stresses components that the half plane body is under concentrated
force in the boundary,
? ? ? ? zPzzPz ln2,ln2 ???? ???
118
因为
所以
? ? ? ?? ?zzzi xyxy '''22 ????? ????
? ? ? ?? ? ??? ????? irr ezzzi 2'''22 ????
练习 5.4 试用公式
? ?
? ? ? ?? ? ???
?
?????
???
i
rr
r
ezzzi
z
2'''
'
22
Re4
????
??
由 导出半平面体在边界上
受集中力作用时的应力分量公式,
? ? ? ? zPzzPz ln2,ln2 ???? ???
119
r ?
y
r
o
P Solution,for
yields
for ? ?
? ?1
c o s22
1
Re
2
Re4
22
'
r
P
yx
xP
z
P
z
r
?
??
?
???
?
??
?
???
?
?
?
?
?
?
????
? ? ? ?
? ?
z
P
z
z
P
z
z
P
z
1
2
1
2
,
1
2
'
2
'''
??
?????
?
?
?
?
?
?
? ? ? ? zPzzPz ln2,ln2 ???? ???
120
r ?
y
r
o
P 解,由
得
? ? ? ? zPzzPz ln2,ln2 ???? ???
? ? ? ?
? ?
z
P
z
z
P
z
z
P
z
1
2
1
2
,
1
2
'
2
'''
??
?????
?
?
?
?
?
?
因为 ? ?
? ?1
c o s22
1
Re
2
Re4
22
'
r
P
yx
xP
z
P
z
r
?
??
?
???
?
??
?
???
?
?
?
?
?
?
????
121
? ? ? ?? ?
?
?
?
??
?
??
?????
i
i
i
rr
e
z
zzP
e
z
P
z
P
z
ezzzi
2
2
2
2
2'''
1
2
1
2
2
22
?
?
?
?
?
?
?
?
????
????
and
2c o s,
??
? ?
ii
i eerez
??
??
so
r
Pi
rr
?
???? ??
c o s22 ???
122
? ? ? ?? ?
?
?
?
??
?
??
?????
i
i
i
rr
e
z
zzP
e
z
P
z
P
z
ezzzi
2
2
2
2
2'''
1
2
1
2
2
22
?
?
?
?
?
?
?
?
????
????
而
2c o s,
??
? ?
ii
i eerez
??
??
所以
r
Pi
rr
?
???? ??
c o s22 ???
123
i.e,
? ?2c o s2 rPr ???? ? ??
? ?30??? r
For（ 1）、（ 2）、（ 3） fields
0
0
c o s2
?
?
??
?
?
?
?
?
?
?
r
r
r
P
124
即
? ?2c o s2 rPr ???? ? ??
? ?30??? r
由（ 1）、（ 2）、（ 3）式得
0
0
c o s2
?
?
??
?
?
?
?
?
?
?
r
r
r
P
125
126
Elasticity
2
3
§ 5-4 The single-valued condition of stress and
displacement in multiply connected region
§ 5-3 Complex-variable representation of boundary
condition
§ 5-2 Complex-variable representation of stress and
displacement
§ 5-1 Complex-variable representation of stress function
§ 5-6 Problem of infinite plane including hole
§ 5-5 Situation of infinite multiply connected body
Chapter 5 Complex-variable
Methods for Plane Elasticity
4
§ 5-4 多连通域内应力与位移的单值条件
§ 5-3 边界条件的复变函数表示
§ 5-2 应力和位移的复变函数表示
§ 5-1 应力函数的复变函数表示
§ 5-6 含孔口的无限大板问题
§ 5-5 无限大多连体的情形
第五章 平面问题的复变函数法
5
Chapter 5 Complex-variable
Methods for Plane Elasticity
When solving plane problems by Cartesian coordinates or polar
coordinates,the boundary of object is straight line or circular
arc,To other boundary,for example ellipse,hyperbola,non-
concentric circles and so on,we need use different curvilinear
coordinates,Applying complex-variable can predigest these
problems,
In this chapter,we just introduce the simple application of
complex-variable in elasticity,
6
第五章 平面问题的复变函数法
直角坐标及极坐标求解平面问题，所涉及的物
体边界是直线或圆弧形。对于其他一些边界，例如
椭圆形、双曲形、非同心圆等就要用不同的曲线坐
标。应用复变函数可使该类问题得以简化。本章只
限于介绍复变函数方法在弹性力学中的简单应用。
7
§ 5-1 Complex-variable representation of stress function
In chapter 2,we have proved,in plane problems,there
is a stress function φ that is biharmonic function of
position coordinates,if body force is constant,i.e,
04 ?? ?
i,1
i,1
??
?
?
?
?
?
?
?
?
?
?
?
y
z
x
z
y
z
x
z
Now introduce complex variable z= x＋ iy and z＝ x－ iy to
replace real variable x and y,Noticing,
8
§ 5-1 应力函数的复变函数表示
在第二章中已经证明, 在平面问题里, 如
果体力是常量, 就一定存在一个应力函数 φ,
它是位置坐标的重调和函数, 即
04 ?? ?
现在, 引入复变数 z= x＋ iy和 z＝ x－ iy以代替实
变数 x 和 y。 注意
i,1
i,1
??
?
?
?
?
?
?
?
?
?
?
?
y
z
x
z
y
z
x
z
9
We find the transformation are
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
???
?
???
)i(
)(
zzy
z
zy
z
zy
zzx
z
zx
z
zx
zyxzyx ?
??
?
??
?
?
?
??
?
??
?
? ?????? 2i,2i
furthermore,
???? 22
2
2
2
2
)(,)( zzyzzx ?????????????????
10
可以得到变换式
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
???
?
???
)i(
)(
zzy
z
zy
z
zy
zzx
z
zx
z
zx
zyxzyx ?
??
?
??
?
?
?
??
?
??
?
? ?????? 2i,2i
进而
???? 22
2
2
2
2
)(,)( zzyzzx ?????????????????
11
zzyx ??
??
?
??
?
??? ???? 2
2
2
2
2
2 4
Pzz ?????? ??
2
2 4
Let
So we can transform the function
04 ?? ? )(022
4
azz ??? ? ?
as
0)( 2224 ??????? P??
For
02 ?? P
12
zzyx ??
??
?
??
?
??? ???? 2
2
2
2
2
2 4
Pzz ?????? ??
2
2 4
令
于是可将方程式
04 ?? ? )(022
4
azz ??? ? ?
变换成为
0)( 2224 ??????? P??
由 02 ?? P
13
It is obvious known P is harmonic function which can be
obtained by real part of analytical function,Suppose f(z) as
analytical function and let
))()((
2
1 zfzfP ??
Pzz ?????? ??
2
2 4
For
)])()((21[4141
2
zfzfPzz ?????? ?
)('4)( zzf ??
))(')('(21
2
zzzz ??? ?????
Let
yields
thus
14
可知,P是调和函数可由解析函数的实部得到。设
f(z)为解析函数，可令
))()((21 zfzfP ??
Pzz ?????? ??
2
2 4由
)])()((21[4141
2
zfzfPzz ?????? ?
)('4)( zzf ??
))(')('(21
2
zzzz ??? ?????
令
得
则
15
))()(')('(21 zzzzzz ???? ?????
Then integrating with respect to z yields
))(d)()()((21 zgzzzzzz ? ???? ????
)(d)( zzz? ? ??
Let
)(')( zz ?? ?i.e,
))()()()((21 zgzzzzz ???? ????
Integrating the above equation with respect to yields z
thus
16
))()(')('(2
1 zzzzz
z ???
? ???
?
?
再对 z积分，得到
))(d)()()((21 zgzzzzzz ? ???? ????
)(d)( zzz? ? ??令
)(')( zz ?? ?即
))()()()((21 zgzzzzz ???? ????
将上式对 积分，得到 z
则
17
Notice the biharmonic function on the left side of the above
equation is a real function,It is obvious that the four terms
on the right side must be conjugate two and two,The first
two terms is conjugate,and the next two terms should be also
conjugate,
)()( zzg ??
Let
we obtain the famous gusa formula
])()()()([21 zzzzzz ????? ????
it can be also written as
)]()([Re zzz ??? ??
18
注意上式左边的重调和函数 φ 是实函数, 可见该
式右边的四项一定是两两共轭, 前两项已经是共
轭的, 后两项也应是共轭的,
)()( zzg ??
令
即得有名的古萨公式
])()()()([21 zzzzzz ????? ????
也可以写成
)]()([Re zzz ??? ??
19
So in plane problems when body force is constant,stress
function υ can be represented by two analytical functions of
complex variable z,?(z) and ?(z),named K-M function,So
solving plane problems is just selecting K-M function
properly and determining any constant in them according to
boundary condition,
20
于是可见，在常量体力的平面问题中，应力函数 φ
总可以用复变数 z的两个解析函 ?(z)和 ?(z)来表示，
称为 K-M 函数。而求解各个具体的平面问题，可归结
为适当地选择这两个解析函数，并根据边界条件决定其
中的任意常数。
21
§ 5-2 Complex-variable
representation of stress and displacement
According to the relation between stress component
and stress function,
yx
x
y
xy
y
x
??
?
??
?
?
?
?
?
?
?
?
?
?
?
?
2
2
2
2
2
One Complex-variable representation of stress component
22
§ 5-2 应力和位移的复变函数表示
根据应力分量和应力函数的关系
yx
x
y
xy
y
x
??
?
??
?
?
?
?
?
?
?
?
?
?
?
?
2
2
2
2
2
一 应力分量的复变函数表示
23
We find Complex-variable representation of stress
component,
zzxy
yx
??
??
?
??
?
??? ????? 24
2
2
2
2
))()()()((21 zzzzzz ????? ????
for
)('Re4])(')('[2 zzzyx ????? ????
yields
hence,for
24
可得到应力分量的复变函数表示
zzxy
yx
??
??
?
??
?
??? ????? 24
2
2
2
2
))()()()((21 zzzzzz ????? ????
由
)('Re4])(')('[2 zzzyx ????? ????
可得
而由
25
2
2
4
i
2
i2
2
2
2
2
i2
2
z
yx
yxyx
xyxy
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
??
?
?
?
?
?
?
?
???
?
?
???
???
)]('')(''[2i2 zzzxyxy ????? ????yields
)](')(''[2i2 zzzxyxy ????? ????or
26
2
2
4
i
2
i2
2
2
2
2
i2
2
z
yx
yxyx
xyxy
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
??
?
?
?
?
?
?
?
???
?
?
???
???
)]('')(''[2i2 zzzxyxy ????? ????可得
)](')(''[2i2 zzzxyxy ????? ????或
27
Only given ?(z) andψ (z),we can divide the right side of
above equation into imaginary part and real part,from
imaginary part we get τxy,from real part we get σy-σx,
)('Re4])(')('[2 zzzyx ????? ????
)](')(''[2i2 zzzxyxy ????? ????
and
is complex-variable representation of stress component,Of
course by building equations,σx,σy, τxy can be represented
by ?(z) and ψ (z) respectively,but that will make equations
become lengthiness and it’s not convenient to use,
28
只要已知 ?(z)及 ψ (z),就可以把上述公式右
边的虚部和实部分开, 由虚部得出 τ xy,由实部得
出 σ y-σ x。
)('Re4])(')('[2 zzzyx ????? ????
)](')(''[2i2 zzzxyxy ????? ????和
就是应力分量的复变函数表示 。 当然也可以建立
公式, 把 σ x,σ y, τ xy三者分开用 ?(z)和 ψ (z)
来表示, 但那些公式将比较冗长, 用起来很不方
便 。
29
Two Complex-variable representation of displacement component
Assuming plane stress problems,according to
geometrical equation and physical equation
yields
yyxyxx
uE ??????? )1()( ??????
?
?
xyxxyy
vE ??????? )1()( ??????
?
?
xyy
u
x
vE ?
? ??
??
?
?
? )()1(2
2
2
2
2
''
)1(])()([2
)1(])()([2
x
zz
x
x
zz
x
u
E
?
?
???
?
?
?
?
?
????
?
?
?
???
?
???
30
二 位移分量的复变函数表示
假定为平面应力问题 。 由几何方程及物理方程
yyxyxx
uE ??????? )1()( ??????
?
?
xyxxyy
vE ??????? )1()( ??????
?
?
xyy
u
x
vE ?
? ??
??
?
?
? )()1(2
2
2
2
2
''
)1(])()([2
)1(])()([2
x
zz
x
x
zz
x
u
E
?
?
???
?
?
?
?
?
????
?
?
?
???
?
???可得
31
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
???
?
???
)i(
)(
zzy
z
zy
z
zy
zzx
z
zx
z
zx
for
)(')()( zzzzz ??? ??????
and notice
])()([)()(
])()([])()([
'' zzz
z
z
z
zz
zz
zz
x
????
????
??
?
?
?
?
?
?
??
?
?
?
?
?
?
?
?
?
?
??
?
?
])()([)()(
])()([])()([i
'' zzz
z
z
z
zz
zz
zz
y
????
????
??
?
?
?
?
?
?
??
?
?
?
?
?
?
?
?
?
?
??
?
?
?similarly
yields
32
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
???
?
???
)i(
)(
zzy
z
zy
z
zy
zzx
z
zx
z
zx
由于
)(')()( zzzzz ??? ??????
并注意到
])()([)()(
])()([])()([
'' zzz
z
z
z
zz
zz
zz
x
????
????
??
?
?
?
?
?
?
??
?
?
?
?
?
?
?
?
?
?
??
?
?
])()([)()(
])()([])()([i
'' zzz
z
z
z
zz
zz
zz
y
????
????
??
?
?
?
?
?
?
??
?
?
?
?
?
?
?
?
?
?
??
?
?
?同理
可得
33
Successive integration of the above two equations with respect to
x and y,lead to
Where f1 and f2 are arbitrary functions,Substituting above
equations into the following equation
2
2
2
2
''
)1(])()([i2
)1(])()([2
y
zz
x
y
zz
y
v
E
?
?
???
?
?
??
?
?
????
?
?
?
???
?
???
?
?
?
??
?
?
?
?
?
?????
?
?
?
????
)()1(])()(i[2
)()1(])()([2
2
1
xf
y
zzEv
yf
x
zzEu
?
???
?
???
yxy
u
x
vE
xy ??
????
?
??
?
?
?
??
?
2
)()1(2
34
2
2
2
2
''
)1(])()([i2
)1(])()([2
y
zz
x
y
zz
y
v
E
?
?
???
?
?
??
?
?
????
?
?
?
???
?
???
将上两式分别对 x及 y积分，得
?
?
?
??
?
?
?
?
?
?????
?
?
?
????
)()1(])()(i[2
)()1(])()([2
2
1
xf
y
zzEv
yf
x
zzEu
?
???
?
???
其中的 f1及 f2为任意函数 。 将上式代入式
yxy
u
x
vE
xy ??
????
?
??
?
?
?
??
?
2
)()1(2
35
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
???
?
???
)i(
)(
zzy
z
zy
z
zy
zzx
z
zx
z
zx
for
])()([i)()(i
])()([i])()([
''
zzz
z
z
z
zz
zz
zz
y
????
????
???
?
?
?
?
?
?
?
?
?
?
?
??
?
?
?
?
?
?
?
?
?
?
??
?
?
])()([i)()(i
])()([i])()([i
'' zzz
z
z
z
zz
zz
zz
x
????
????
????
?
?
?
?
?
?
?
?
?
?
??
??
?
?
?
?
?
?
?
?
?
?
???
?
?
?
36
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
???
?
???
)i(
)(
zzy
z
zy
z
zy
zzx
z
zx
z
zx
由于
])()([i)()(i
])()([i])()([
''
zzz
z
z
z
zz
zz
zz
y
????
????
???
?
?
?
?
?
?
?
?
?
?
?
??
?
?
?
?
?
?
?
?
?
?
??
?
?
])()([i)()(i
])()([i])()([i
'' zzz
z
z
z
zz
zz
zz
x
????
????
????
?
?
?
?
?
?
?
?
?
?
??
??
?
?
?
?
?
?
?
?
?
?
???
?
?
?
37
and yields
x
xf
y
yf
d
)(d
d
)(d 11 ??
Thus we can find the displacement of rigid body
f1(y)＝ u0－ ωy,f2(x)＝ v 0＋ ωx
We get
)(
d
d
)(
d
d
)1(2
)(
d
d
)1(])()([i2
)(
d
d
)1(])()([2
21
2
2
2
1
2
xf
x
yf
yyx
xf
xyx
zz
x
u
yf
yyx
zz
y
u
x
v
y
u
E
??
??
?
???
?
??
?
???
?
?
?
?
??
?
???
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
???
?
???
38
)(
d
d
)(
d
d
)1(2
)(
d
d
)1(])()([i2
)(
d
d
)1(])()([2
21
2
2
2
1
2
xf
x
yf
yyx
xf
xyx
zz
x
u
yf
yyx
zz
y
u
x
v
y
u
E
??
??
?
???
?
??
?
???
?
?
?
?
??
?
???
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
???
?
???
从而得到
x
xf
y
yf
d
)(d
d
)(d 11 ??
于是得到刚体位移
f1(y)＝ u0－ ωy,f2(x)＝ v 0＋ ωx
故有
39
If neglecting displacement of rigid body,we have
)i)(1()(4)i( yxzvuE ????????? ????
zyx ?
??
?
??
?
? ??? 2i
for
])()()()([21 zzzzzz ????? ????
yields
)()()(
)(')()(2i
'
'
zzzz
zzzz
zyx
???
???
???
???
???
?
?
?
?
?
?
?
?
40
若不计刚体位移，则有
zyx ?
??
?
??
?
? ??? 2i由式
])()()()([21 zzzzzz ????? ????
)()()(
)(')()(2i
'
'
zzzz
zzzz
zyx
???
???
???
???
???
?
?
?
?
?
?
?
?得到
)i)(1()(4)i( yxzvuE ????????? ????
41
let the result back substitution,and on the two side divide 1+ν
yields
)()()(13)i(1 ' zzzzvuE ?????? ???????
This is complex-variable representation of displacement
component,If ?(z) and ψ (z) are given,we can divide the real
part and imaginary part of the right side of the above formula,
and u and v can be solved,
The above formula is educed in plane stress problem,To
plane strain problem,we need replace E with E/(1－ 2) and
with /(1－ )。
? ?
??
42
将结果回代,并两边除以 1+ 得
这就是位移分量的复变函数表示。若已知 ?(z)及
ψ (z)，就可以将该式右边的实部和虚部分开，
从而得出 u和 v。
)()()(13)i(1 ' zzzzvuE ?????? ???????
?
上述公式是针对平面应力情况导出的。对于
平面应变情况，须将式中的 E改换为 E/(1－ 2),
改换为 /(1－ )。 ?
?
??
43
§ 5-3 Complex-variable Representation
Of Boundary Condition
To evaluate φ of every crunode in boundary,we need
apply boundary condition of stress,i.e.,
Yml
Xml
yxy
xyx
??
??
??
??
yxxy xyyx ??
???
?
??
?
?? ?????? 2
2
2
2
2
,,and
Substitution into the
above equation,gives
Y
x
m
yx
l
X
yx
m
y
l
???
?
?
??
?
?
?
?
???
?
?
??
?
?
??
?
?
???
?
?
??
?
?
??
?
???
?
?
??
?
?
?
?
2
22
2
2
2
??
??
44
§ 5-3 边界条件的复变函数表示
为了求得边界上各结点处的 φ 值，须要应用应
力边界条件，即,
Yml
Xml
yxy
xyx
??
??
??
??
yxxy xyyx ??
???
?
??
?
?? ?????? 2
2
2
2
2
,,而
代入上式，即得,
Y
x
m
yx
l
X
yx
m
y
l
???
?
?
??
?
?
?
?
???
?
?
??
?
?
??
?
?
???
?
?
??
?
?
??
?
???
?
?
??
?
?
?
?
2
22
2
2
2
??
??
45
As the figure shown
l=cos(N,x)=dy/ds,
m=cos(N,y)=-dx/ds,
So the above equation can be
rewritten as,
Y
xs
x
yxs
y
X
yxs
x
ys
y
???
?
?
?
?
?
?
?
?
???
?
?
?
?
?
?
??
?
?
???
?
?
?
?
?
?
??
?
???
?
?
?
?
?
?
?
?
2
22
2
2
2
d
d
d
d
d
d
d
d
??
??
YxsXys ??????? ??????
?
?
???
?
?
? ??
d
d,
d
d
Thus,yields
46
由图可见,
l=cos(N,x)=dy/ds,
m=cos(N,y)=-dx/ds,
于是, 前式可改写为,
Y
xs
x
yxs
y
X
yxs
x
ys
y
???
?
?
?
?
?
?
?
?
???
?
?
?
?
?
?
??
?
?
???
?
?
?
?
?
?
??
?
???
?
?
?
?
?
?
?
?
2
22
2
2
2
d
d
d
d
d
d
d
d
??
??
YxsXys ??????? ??????
?
?
???
?
?
? ??
d
d,
d
d由此得,
47
Suppose A is a fix point in the boundary,and B is a arbitrary
point,so the composition of forces from A to B can be obtained
by integrating of the above equation with respect to s from A to
B,
B
A
B
A
B
A
B
A
yx
yxxy
s
xys
sYXPP
??
?
?
??
?
?
?
?
?
?
?
????
?
?
??
?
?
?
?
?
?
?
?
??
?
?
??
?
?
?
?
?
?
?
???? ??
????
??
iii
di
d
d
dii
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
???
?
???
)i(
)(
zzy
z
zy
z
zy
zzx
z
zx
z
zx
Substituting this formula
48
设 A是边界上的固定点,B为任意一点，则从 A到 B
边界上的合力，可用上式从 A点到 B点对 s积分得
到,
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
???
?
???
)i(
)(
zzy
z
zy
z
zy
zzx
z
zx
z
zx
将式
B
A
B
A
B
A
B
A
yx
yxxy
s
xys
sYXPP
??
?
?
??
?
?
?
?
?
?
?
????
?
?
??
?
?
?
?
?
?
?
?
??
?
?
??
?
?
?
?
?
?
?
???? ??
????
??
iii
di
d
d
dii
49
into the above equation,and rearrangement yields
BAyx zzzzPP ])()(')([ii ??? ?????
Adding a complex constant into stress function,which
doesn’t influence the stress,So we can let A of stress
function as zero,and then σ in the boundary gives
])()(')([ii ??????? ????? yx PP
)()(')()i(i ??????? ???? yx PP
or
This is boundary condition of stress,
50
代入，整理得,
BAyx zzzzPP ])()(')([ii ??? ?????
把应力函数加上一个复常数, 并不影响应力 。
因此, 可把应力函数 A处的值设为零, 于是对于
边界上的 σ 有
])()(')([ii ??????? ????? yx PP
)()(')()i(i ??????? ???? yx PP或
这就是应力边界条件。
51
To boundary condition of displacement
)()()(13)i(1 ' zzzzvuE ?????? ???????
ss vvuu ??,
Substituting them into the following equation
We can obtain complex-variable representation of boundary
condition of displacement in plane stress problem,
)i(1)()()(13 ' ss vuEzzzz ??????? ??????
To plane strain problem,we need replace E with E/(1－ 2) and with /(1－ )。
? ?
?
?
52
对于位移边界条件
ss vvuu ??,
将其代入下式
即得平面应力情况下位移边界条件的复变函数表示
)i(1)()()(13 ' ss vuEzzzz ??????? ??????
)()()(13)i(1 ' zzzzvuE ?????? ???????
对于平面应变，须将式中的 E 改换为 E/(1－ 2)，
改换为 /(1－ )。 ?? ?
?
53
§ 5-4 The single-valued condition of stress
and displacement in multiply connected region
When stress is determined,the stress function can still a
arbitrary linear function,so the K-M function is not determined
completely,so to the simply connected region,the K-M
function can be determined by selecting the suitable coordinate,
But to the multiply connected region,it is still a problem,In this
section,the condition of the K-M function satisfied single-
valued in multiply connected region is discussed,
Suppose there is a multiply connected region that has a
interior boundary C,and in the interior boundary C the external
force vector is given,Generally multiform function is
logarithmic function,we suppose
54
§ 5-4 多连通域内应力与位移的单值条件
应力确定后，应力函数仍可差一个任意的线性函数，
这时 K-M函数并未完全确定,对于单连通区域，可以通过选
取适当坐标系等办法，使得 K-M函数完全确定 ;但对于多连
通区域仍不能完全确定,本节讨论 K-M函数在多连通区域内
满足单值的条件。
设有多连通区域，有一内边界 C，设在边界 C上的外力
矢量已给定。通常的多值函数是对数函数，我们设
55
)()l n ()(
)()l n ()(
zzzBz
zzzAz
fkk
fkk
??
??
???
???
D
C
Where zk is a arbitrary point in the interior boundary,?f andψf are
single-valued analytical function(holomorphic function),and Ak
and Bk are constants gives,
56
)()l n ()(
)()l n ()(
zzzBz
zzzAz
fkk
fkk
??
??
???
???
D
C
这里 zk为内部边界内的任意一点,?f和 ψ f为单值的
解析函数（全纯函数），而 Ak, Bk为常数,
57
kkkkkk BA 'i',i ???? ????
The derivative of above function is single-valued,but it is multivalued
by itself,while z move around the circum once,the value of ln(zk) have
a increment of 2πi,so the increments of ?(z) and ψ (z) is 2πi Ak and
2πiBk respectively,and according to the following formula,the master
vector of stresses
)]()(')([ii zzzzYXBA ??? ??????
the master vector of stress(around the whole boundary) is on the
left side,and the increment is on the right side,
ykxkkk PPBA i)(2 ???? ?
58
kkkkkk BA 'i',i ???? ????
前面的函数的导数是单值的，但他们本身是多值
的，当 z绕周边一周时，函数值 ln(zk)产生一个增
量 2π i,于是 ?(z)和 ψ (z)的增量分别是 2π i Ak
和 2π iBk,这时应力主矢量按照公式
)]()(')([ii zzzzYXBA ??? ??????
左边将得到应力主矢量 (沿整个边界），右边得到
一增量,
ykxkkk PPBA i)(2 ???? ?
59
While according to the following formula,displacement
will obtain increment,and according to single-valued the
increment should be zero,
ykxkkk PPBA i)(2 ???? ?
and
yields
)1(2
i
)1(2
i
??
?
?
?
?
?
?
?
??
ykxk
k
ykxk
k
PP
B
k
PP
A
)13(0 ???? ????? kk BA
)()()(13)i(1 ' zzzzvuE ?????? ???????
60
这时位移按照公式
)()()(13)i(1 ' zzzzvuE ?????? ???????
也将得到增量，根据单值性这个增量应为零,
)13(0 ???? ????? kk BA
ykxkkk PPBA i)(2 ???? ?
结合
可得到
)1(2
i
)1(2
i
??
?
?
?
?
?
?
?
??
ykxk
k
ykxk
k
PP
B
k
PP
A
61
)()ln (
)1(2
i
)(
)()ln (
)1(2
i
)(
f
f
zzz
PP
z
zzz
PP
z
k
ykxk
k
ykxk
?
??
??
?
??
?
??
?
?
?
??
?
?
??
thus
While there is m interior boundary,let
)()l n (
)1(2
i
)(
)()l n (
)1(2
i
)(
f
1
f
1
zzz
PP
z
zzz
PP
z
m
k
k
ykxk
m
k
k
ykxk
?
??
??
?
??
?
??
?
?
?
??
?
?
??
?
?
?
?
62
)()ln (
)1(2
i
)(
)()ln (
)1(2
i
)(
f
f
zzz
PP
z
zzz
PP
z
k
ykxk
k
ykxk
?
??
??
?
??
?
??
?
?
?
??
?
?
??
于是
当有 m个内边界时，取
)()l n (
)1(2
i
)(
)()l n (
)1(2
i
)(
f
1
f
1
zzz
PP
z
zzz
PP
z
m
k
k
ykxk
m
k
k
ykxk
?
??
??
?
??
?
??
?
?
?
??
?
?
??
?
?
?
?
63
§ 5-5 Situation of infinite multiply connected body
When exterior boundary of multiply connected body
approach infinite farness,this multiply connected body become
infinite multiply connected body,besides the above conditions,we
need consider the ultimate situation of infinite farness,
Regards origin of coordinate as the centre of circle,draw
a big enough circle sR,which include all interior boundary,To
an arbitrary point in the elastomer,but beyond sR,gives
??
??
?
?
?
?
?
????
?
?
?
?
?
????
z
z
z
z
z
z
z
z
zzz kkkk
ln
2
1
ln1lnln)ln (
2
?
the analytical function beyond sR
64
§ 5-5 无限大多连体的情形
当多连体的外边界趋于无限远时，该多连体成为无
限大的多连体，除上述条件外，还需考虑无限远的极限
情况。
以坐标原点为圆心，作充分大的圆周 sR，将所有的
内边界包围在其内，对于 sR之外，弹性体之内的任意一
点，可得到
??
??
?
?
?
?
?
????
?
?
?
?
?
????
z
z
z
z
z
z
z
z
zzz kkkk
ln
2
1
ln1lnln)ln (
2
?
在 sR之外的解析函数
65
)()l n (
)1(2
i
)(
)()l n (
)1(2
i
)(
f
1
f
1
zzz
PP
z
zzz
PP
z
m
k
k
ykxk
m
k
k
ykxk
?
??
??
?
??
?
??
?
?
?
??
?
?
??
?
?
?
?
so
)(ln
)1(2
i
)(
)(ln
)1(2
i
)(
f
*
f
*
zz
PP
z
zz
PP
z
yx
yx
?
??
??
?
??
?
?
?
?
?
?
?
?
??
It can be written as
Where Px,Py are the sum of surface forces in m boundary,
66
)()l n (
)1(2
i
)(
)()l n (
)1(2
i
)(
f
1
f
1
zzz
PP
z
zzz
PP
z
m
k
k
ykxk
m
k
k
ykxk
?
??
??
?
??
?
??
?
?
?
??
?
?
??
?
?
?
?
于是
)(ln
)1(2
i
)(
)(ln
)1(2
i
)(
f
*
f
*
zz
PP
z
zz
PP
z
yx
yx
?
??
??
?
??
?
?
?
?
?
?
?
?
??
可写为
其中 Px,Py为 m个边界上沿 x,y方向的面力之和。
67
Expand holomorphic function φ*f andψ*f in Multiply connected
region by luolang series,
?
?
??
??
??
??
?
?
n
n
n
n
zbz
zaz
)(
)(
f
*
f
*
?
?
?
??
??
?? ??
?
?
?
?
?
?
?
?
?
?
?
???
)(2
1
)(12
i1
)(12
i
2
11 n
n
n
n
yxyx
yx
zazan
z
PP
z
PP
????
??so
For components of stresses in infinite farness is finite,the
coefficients of n≥2 is zero,
)2(0,0 ??? naa nn
68
将多连通区域内的全纯函数 φ *f和 ψ *f展开为罗郎级数,
?
?
??
??
??
??
?
?
n
n
n
n
zbz
zaz
)(
)(
f
*
f
*
?
?
?
??
??
?? ??
?
?
?
?
?
?
?
?
?
?
?
???
)(2
1
)(12
i1
)(12
i
2
11 n
n
n
n
yxyx
yx
zazan
z
PP
z
PP
????
??于是
由于在无穷远处的应力分量应该是有限的，级数中
n≥2 的系数应为零。
)2(0,0 ??? naa nn
69
)](')(''[2i2 zzzxyxy ????? ????
Similarity,from
for components of stress in infinite farness is finite,so
)2(,0 ?? nb n
)()()()(
)()()()(
0f11
1
11f
*
0f
1
f
*
zzizbziz
zzizaziz
n
n
n
n
n
n
??????
??????
??????
??????
?
?
??
?
?
??
?
?
Where neglecting the constant terms that have no relation
to stresses,
70
)](')(''[2i2 zzzxyxy ????? ????
同样从
中，由于在无穷远处的应力分量应该是有限的，故有
)2(,0 ?? nb n
)()()()(
)()()()(
0f11
1
11f
*
0f
1
f
*
zzizbziz
zzizaziz
n
n
n
n
n
n
??????
??????
??????
??????
?
?
??
?
?
??
?
?
其中略去了和应力无关的常数项。
71
)(ln
)1(2
i
)(
)(ln
)1(2
i
)(
f
*
f
*
zz
PP
z
zz
PP
z
yx
yx
?
??
??
?
??
?
?
?
?
?
?
?
?
??so
)()i()i()(
)()i()i()(
0f11
1
11f
*
0f
1
f
*
zzzbzz
zzzazz
n
n
n
n
n
n
??????
??????
??????
??????
?
?
??
?
?
??
?
?
Where β is no relation to stress calculate,it can be regarded as
zero,and
?
?
??
?
?
??
?
?
?
?
1
0f
1
0f
)(
)(
n
n
n
n
n
n
zbz
zaz
?
?
72
)(ln
)1(2
i
)(
)(ln
)1(2
i
)(
f
*
f
*
zz
PP
z
zz
PP
z
yx
yx
?
??
??
?
??
?
?
?
?
?
?
?
?
??于是
)()i()i()(
)()i()i()(
0f11
1
11f
*
0f
1
f
*
zzzbzz
zzzazz
n
n
n
n
n
n
??????
??????
??????
??????
?
?
??
?
?
??
?
?
其中 β与应力计算无关，可取为零，而
?
?
??
?
?
??
?
?
?
?
1
0f
1
0f
)(
)(
n
n
n
n
n
n
zbz
zaz
?
?
73
?
??
?
?? ???
?
?
?
?
?
?
?
?
?
?
?
???
1
)(24
1
)(12
i1
)(12
i
2
n
n
n
n
n
yxyx
yx
zazan
z
PP
z
PP
?
????
??At this time
When z→∞, fields
??? 4?? yx
Similarity,when z→∞,for
)](')(''[2i2 zzzxyxy ????? ????
fields
11 i22i2 ????? ???? xyxy
From above equation,we can obtain the corresponding coefficient
and we can also find in infinite farness the distributing of stress is
symmetrical,
74
?
??
?
?? ???
?
?
?
?
?
?
?
?
?
?
?
???
1
)(24
1
)(12
i1
)(12
i
2
n
n
n
n
n
yxyx
yx
zazan
z
PP
z
PP
?
????
??这时
当 z→∞ 时，可得 ??? 4??
yx
同样当 z→∞ 时，由
)](')(''[2i2 zzzxyxy ????? ????
可得
11 i22i2 ????? ???? xyxy
从中可求得相应的系数，并可以看到在无限远处，应力的分
布是均匀的。
75
)(
2
)()(
4
)()(
1
1
??
???
?
???
?
xy
xy
yx
??
??
?
??
?
coefficients
)(
)(i
2
)()(
ln
)1(2
i
)(
)(
4
)()(
ln
)1(2
i
)(
0f
xy
0f
z
zz
PP
z
zzz
PP
z
xyyx
yxyx
?
?
??
??
??
?
??
??
?
?
?
?
?
?
?
?
??
???
?
?
?
?
?
???
?
?
?
??
thus
76
)(
2
)()(
4
)()(
1
1
??
???
?
???
?
xy
xy
yx
??
??
?
??
?
系数
)(
)(i
2
)()(
ln
)1(2
i
)(
)(
4
)()(
ln
)1(2
i
)(
0f
xy
0f
z
zz
PP
z
zzz
PP
z
xyyx
yxyx
?
?
??
??
??
?
??
??
?
?
?
?
?
?
?
?
??
???
?
?
?
?
?
???
?
?
?
??
则
77
§ 5-6 Problem of infinite plane including hole
Regards origin of coordinate as the centre of circle,draw a
big enough circle sR,which include all interior boundary,so to
an arbitrary point in the elastomer but beyond sR,we have
)(ln
)1(2
i
)(
)(ln
)1(2
i
)(
f
*
f
*
zz
PP
z
zz
PP
z
yx
yx
?
??
??
?
??
?
?
?
?
?
?
?
?
??
78
§ 5-6 含孔口的无限大板问题
以坐标原点为圆心，作充分大的圆周 sR，将所有的内
边界包围在其内，对于 sR之外，弹性体之内的任意一点，
可得到
)(ln
)1(2
i
)(
)(ln
)1(2
i
)(
f
*
f
*
zz
PP
z
zz
PP
z
yx
yx
?
??
??
?
??
?
?
?
?
?
?
?
?
??
79
?
?
??
?
?
??
?
?
???
??
1
11f
*
1
f
*
)i()(
)(
n
n
n
n
n
n
zbzz
zazz
???
??
?
?
?
?
?
??
?
?
??
?
?
??
??
?
?
?
???
?
?
?
??
?
?
??
1
)2(
2
1
)1(
11
1
)1(
)1(
1
)1(2
i
)(''
i
1
)1(2
i
)('
1
)1(2
i
)('
n
n
n
yx
n
n
n
yx
n
n
n
yx
znna
z
PP
z
nzb
z
PP
z
nza
z
PP
z
??
?
??
??
??
?
??
?
80
?
?
??
?
?
??
?
?
???
??
1
11f
*
1
f
*
)i()(
)(
n
n
n
n
n
n
zbzz
zazz
???
??
?
?
?
?
?
??
?
?
??
?
?
??
??
?
?
?
???
?
?
?
??
?
?
??
1
)2(
2
1
)1(
11
1
)1(
)1(
1
)1(2
i
)(''
i
1
)1(2
i
)('
1
)1(2
i
)('
n
n
n
yx
n
n
n
yx
n
n
n
yx
znna
z
PP
z
nzb
z
PP
z
nza
z
PP
z
??
?
??
??
??
?
??
?
81
?
?
?
?
?
??
?
?
?
?
?
?
??
?
?
1
)1(
0
0
)("
)('
)('
k
k
k
k
k
k
k
k
k
kzAz
zBz
zAz
?
?
?
rewritten as
1100 ii ???? ???? BA
where
)2()1()1(
)1(2
i
)1(2
i
11
11
???????
?
?
?
?
?
??
?? kbkBakA
PP
B
PP
A
kkkk
yxyx
??
?
??
82
?
?
?
?
?
??
?
?
?
?
?
?
??
?
?
1
)1(
0
0
)("
)('
)('
k
k
k
k
k
k
k
k
k
kzAz
zBz
zAz
?
?
?
改写为
1100 ii ???? ???? BA
其中
)2()1()1(
)1(2
i
)1(2
i
11
11
???????
?
?
?
?
?
??
?? kbkBakA
PP
B
PP
A
kkkk
yxyx
??
?
??
83
To the points in the
boundary of hole ?ieaz ?
?
?
?
?
?
??
?
?
?
?
?
??
??
?
?
0
i2i
0
i
0
i
e)("e
e)('
e)('
1
1
1
k
kk
kC
k
kk
kC
k
kk
kC
kaAzz
aAz
aAz
??
?
?
?
?
?
?
?
?
?
???
?
?
?
?
???
???
?
0
i)2(
2
i1
1
2i
0
0
)2i(2i
eee
e)('e
1
k
kk
k
k
kk
kC
aBaBB
aBz
???
??
?
)20( ?? ??
84
对于孔边上的点 ?ieaz ?
?
?
?
?
?
??
?
?
?
?
?
??
??
?
?
0
i2i
0
i
0
i
e)("e
e)('
e)('
1
1
1
k
kk
kC
k
kk
kC
k
kk
kC
kaAzz
aAz
aAz
??
?
?
?
?
?
?
?
?
?
???
?
?
?
?
???
???
?
0
i)2(
2
i1
1
2i
0
0
)2i(2i
eee
e)('e
1
k
kk
k
k
kk
kC
aBaBB
aBz
???
??
?
)20( ?? ??
85
Substituting the above equations into the following equation
???? ??????????? 2ie)](')(''[)(')('i ?????
We can obtain the stress boundary condition of series forms of
circle boundary in polar coordinate,
Assuming the external force is known and spread it by
Fourier series,
?
?
?
?
???
??
??
?
?
??
?
??
?
?
2
0
i
i
de)i(
2
1
ei
k
k
k
k
k
ppC
Cpp
86
将上列各式代入
???? ??????????? 2ie)](')(''[)(')('i ?????
就得到极坐标下圆周边界上的级数形式的应力边界条件。
设周边上的外力为已知，并将其展开为傅氏级数
?
?
?
?
???
??
??
?
?
??
?
??
?
?
2
0
i
i
de)i(
2
1
ei
k
k
k
k
k
ppC
Cpp
87
?? ????????? 2i e)](')(''[)(')('e i
k
k
kC ?????
?
???
?
????
?
?
??
?
?
?
?
??
?
?
?
?
?
??
??
???
???
???
0
i-)2(
2
i1
1
2i
0
0
i
0
i
0
ii
eee
eeee
k
kk
k
k
kk
k
k
kk
k
k
kk
k
k
k
k
k
aBaBB
kaAaAaAC
???
????
By comparing coefficient of eik? and e-ik?,fields
12
11
02
2
00
C
a
B
a
A
C
a
B
AA
??
???
88
?? ????????? 2i e)](')(''[)(')('e i
k
k
kC ?????
?
???
?
????
?
?
??
?
?
?
?
??
?
?
?
?
?
??
??
???
???
???
0
i-)2(
2
i1
1
2i
0
0
i
0
i
0
ii
eee
eeee
k
kk
k
k
kk
k
k
kk
k
k
kk
k
k
k
k
k
aBaBB
kaAaAaAC
???
????
比较两边 eik?和 e-ik?的系数，可得
12
11
02
2
00
C
a
B
a
A
C
a
B
AA
??
???
89
1
)1(
)3(
2
2
202
2
???
?
??
??
??
?
kC
a
B
a
Ak
kC
a
A
CB
a
A
kk
k
k
k
kk
k
1100 ii ???? ???? BA
For the stresses condition in infinite farness,yields
)(,2 )()(,4 )()( 11 ?????????? xyxyyx ????????
90
1
)1(
)3(
2
2
202
2
???
?
??
??
??
?
kC
a
B
a
Ak
kC
a
A
CB
a
A
kk
k
k
k
kk
k
1100 ii ???? ???? BA
由无限远处的应力条件，可得
)(,2 )()(,4 )()( 11 ?????????? xyxyyx ????????
91
For the single-valued condition of displacement yields
011 ?? BA?
12
11 C
a
B
a
A ??
and
We find
?
?
? ????? 1,1
1
1
1
1
CaBCaA
for
kk
k
k
k
kk
k
C
a
B
a
Ak
kC
a
A
CB
a
A
??
?
??
?
??
??
2
2
202
2
)1(
)3(
92
由位移的单值条件有
011 ?? BA?
12
11 C
a
B
a
A ??
及
可求得
?
?
? ????? 1,1
1
1
1
1
CaBCaA
再由
kk
k
k
k
kk
k
C
a
B
a
Ak
kC
a
A
CB
a
A
??
?
??
?
??
??
2
2
202
2
)1(
)3(
93
)3()1(
)3(
)2(),(
22
2
00
2
220
2
2
????
??
????
???
kCaAakB
kCaA
CAaBCBaA
k
k
kk
k
k
k
yields
By now,all coefficients have been solved,
For example,assuming the uniform pressure around hole is p and
the stress of infinite farness is zero,
94
)3()1(
)3(
)2(),(
22
2
00
2
220
2
2
????
??
????
???
kCaAakB
kCaA
CAaBCBaA
k
k
kk
k
k
k
可求得
至此，全部系数均已求出。
例 设孔周边为均匀压力 p，无限远处的应力为零。
95
)0(0,
i
0
0,
0
00
????
???
??
???
kCpC
ppp
BA
pp
k
??
??
?
Thus
)3(,0
0
2
2
211
???
?
???
kBA
paB
ABA
kk
So we get
96
)0(0,
i
0
0,
0
00
????
???
??
???
kCpC
ppp
BA
pp
k
??
??
?
则有
)3(,0
0
2
2
211
???
?
???
kBA
paB
ABA
kk
于是可求得
97
z
pa
z
z
2
)('
0)('
?
?
?
?
0,
2
0,,
2
2
2
2
2
??
????
??
????
?
?
?
?
?
?
u
G
pa
u
papa
Finally,we get
According to the above means,the general problems of infinite
plane including hole can be solved,
98
z
pa
z
z
2
)('
0)('
?
?
?
?
0,
2
0,,
2
2
2
2
2
??
????
??
????
?
?
?
?
?
?
u
G
pa
u
papa
最后得到
根据上述方法，圆孔口无限大板的一般问题都可以得到解决。
99
Exercise 5.1 Try to check-up the following Complex-variable
(1)
(2)
? ? ? ? zqzzqz 2,4 ??? ??
? ? ? ? i q zzz ?? ??,0
Solution,The fundamental formula gives
? ? ? ?az xy ??? ??'Re4
(1) Substituting ? ? ? ? zqzzqz
2,4 ??? ??
into (a),(b)
? ? ? ?? ? ? ?bizzz xyxy ????? 2'2 ????
100
练习 5.1 试考察下列复变函数所解决的问题
(1)
(2)
? ? ? ? zqzzqz 2,4 ??? ??
? ? ? ? i q zzz ?? ??,0
解, 基本公式为
? ? ? ?az xy ??? ??'Re4
? ? ? ?? ? ? ?bizzz xyxy ????? 2'2 ????
(1) 将 ? ? ? ? zqzzqz
2,4 ??? ??
分别代入 (a),(b)式
101
yields
xyq ?? ??
xyxy iq ??? 2????
associate with the above two equations,
yields
0,0,??? xyyx q ???
The given function can solve the problems that the rectangular
sheet is under uniform pulling force q in axis x,As the figure
5.1(a) shown,
(2) Substituting ? ? ? ? i q zzz ?? ??,0 into (a) and (b),yields
xyxy iiq ??? 22 ???
x y q q
Figure 5.1(a)
yx ?? ??0
102
得
xyq ?? ??
xyxy iq ??? 2????
联立求解以上两式,得
0,0,??? xyyx q ???
所给的函数可以解决矩形薄板在 x方向受均布拉力 q的
问题,如图 5.1(a)所示
(2) 将 ? ? ? ? i q zzz ?? ??,0 代入 (a),(b)两式,得
xyxy iiq ??? 22 ???
x y q q
图 5.1(a)
yx ?? ??0
103
associate with the above two equations,
yields
qxyyx ??? ???,0,0
The given function can solve the
pure shear problem of rectangular
sheet,As the figure 5.1(b) shown,
q q x y
图 5.1(b)
Exercise 5.2 As the figure shown,Try to prove we can use
complex-variable to solve the
pure bending problem of beam of rectangular cross-section,
? ? ? ? 22 88 zIiMzzIiMz ??? ??
Where I is the inertia moment of cross section of beam,M is
the moment of flexion,
M
y
x z y Solution,
The fundamental formula gives
104
联立求解以上两式,得
qxyyx ??? ???,0,0
所给的函数可以解决矩形薄
板受纯剪切问题,如图 5.1(b)示,
q q x y
图 5.1(b)
练习 5.2 如图所示,试证矩形截面梁的纯弯曲问题可用
如下的复变函数求解,
? ? ? ? 22 88 zIiMzzIiMz ??? ??
其中 I为梁截面的惯矩,M为作用的弯矩,
M
y
x z y 解,
基本公式为
105
? ? ? ?1Re4 ' xyz ??? ??
? ? ? ?? ? ? ?222 ' xyxy izzz ????? ????
? ? ? ? ? ? ? ? ? ?3' dSYiXizzzz ? ???? ???
Substituting ? ? ? ?zz ??,into (1),(2)
For (1),yields
? ? xyiyxIiM ?? ???
?
?
??
? ??
4Re4
i.e,
? ?4xyIMy ?? ??
106
? ? ? ?1Re4 ' xyz ??? ??
? ? ? ?? ? ? ?222 ' xyxy izzz ????? ????
? ? ? ? ? ? ? ? ? ?3' dSYiXizzzz ? ???? ???
将 ? ? ? ?zz ??,代入 (1),(2)式
由 (1)式得
? ? xyiyxIiM ?? ???
?
?
??
? ??
4Re4
即
? ?4xyIMy ?? ??
107
or
? ?5
0 ??
?
?
?
?
???
xy
xyI
My
?
??
xyxy iI
My ??? 2????
For equation (2) yields,
? ? xyxy iz
I
iM
I
iMiyx ??? 2
44
2 ????
?
?
??
? ??
?
??
?
? ??
i.e,
equations(4) and (5),yields
108
或
? ?5
0 ??
?
?
?
?
???
xy
xyI
My
?
??
xyxy iI
My ??? 2????
由 (2)式得
? ? xyxy iz
I
iM
I
iMiyx ??? 2
44
2 ????
?
?
??
? ??
?
??
?
? ??
即
将 (4),(5)式联立求得
109
00 ?? ? xyyx IMy ???
Verified the boundary conditions (3)
On the flank,0,0 ?? YX so ? ? 0??? dSYiXi
for ? ?
2
8 zI
iMz ???
yields
? ?
? ?
? ?
2'
'
'
4
4
4
z
I
iM
zz
z
I
iM
z
z
I
iM
z
?
?
??
?
?
?
110
00 ?? ? xyyx IMy ???
验证边界条件 (3)
在侧面, 0,0 ?? YX 所以 ? ? 0??? dSYiXi
由 ? ?
2
8 zI
iMz ???
得
? ?
? ?
? ?
2'
'
'
4
4
4
z
I
iM
zz
z
I
iM
z
z
I
iM
z
?
?
??
?
?
?
111
for ? ?
2
8 zI
iMz ?? yields ? ? 2
8 zI
iMz ???
so ? ? ? ? ? ?
0848 222' ??????? zIiMzIiMzIiMzzzz ???
we find equation (3) is identical equation,
The answer are stress state of
beam cross-section when it’s pure bending,
00 ?? ? xyyx IMy ???
112
由 ? ?
2
8 zI
iMz ?? 得 ? ? 2
8 zI
iMz ???
故 ? ? ? ? ? ?
0848 222' ??????? zIiMzIiMzIiMzzzz ???
即 (3)式恒成立,
由解答 所表示的是一个
纯弯时,梁横截面上的应力状态,
00 ?? ? xyyx IMy ???
113
Exercise 5.3 Try to educe formula of the stresses components in
polar coordinate that is described by Complex-variable and ??z? ? ?z?
? ?
? ? ? ?? ? ???
?
?????
???
i
rr
r
ezzzi
z
2'''
'
22
Re4
????
??
Solution,because in plane problems
tc o n sryx t a n???? ???? ?
so ? ?zr ’??? ? Re4??
Moreover in plane problems,we have
???????? 2s in2c o s22 xyyxyxr ?????
114
练习 5.3 试导出用复变函数 及 表示极坐标中
应力分量的公式
??z? ? ?z?
? ?
? ? ? ?? ? ???
?
?????
???
i
rr
r
ezzzi
z
2'''
'
22
Re4
????
??
解, 因为在平面问题中
常数???? ryx ???? ?
所以 ? ?zr ’??? ? Re4??
又因为在平面问题中,有
???????? 2s in2c o s22 xyyxyxr ?????
115
???
??
?
???
????
?
?
?
2c o s2s i n
2
2s i n2c o s
22
xy
yx
r
xy
yxyx
?
?
??
?
?
?
?
?
? ?? ? ? ?
? ?
?
??
???
???????
???
??
???
????
???
????
???
i
xyxy
xyyx
xy
yx
xy
yxyx
xy
yxyx
rr
ei
iii
i
i
2
2
2s in2c o s22s in2c o s
2c o s2s in
2
2
2s in2c o s
22
2s in2c o s
22
2
???
??????
?
?
?
?
?
?
?
?
?
?
??
?
?
?
?
?
?
?
?
?
???
thus
116
???
??
?
???
????
?
?
?
2c o s2s i n
2
2s i n2c o s
22
xy
yx
r
xy
yxyx
?
?
??
?
?
?
?
?
? ?? ? ? ?
? ?
?
??
???
???????
???
??
???
????
???
????
???
i
xyxy
xyyx
xy
yx
xy
yxyx
xy
yxyx
rr
ei
iii
i
i
2
2
2s in2c o s22s in2c o s
2c o s2s in
2
2
2s in2c o s
22
2s in2c o s
22
2
???
??????
?
?
?
?
?
?
?
?
?
?
??
?
?
?
?
?
?
?
?
?
???
则
117
for
so
? ? ? ?? ?zzzi xyxy '''22 ????? ????
? ? ? ?? ? ??? ????? irr ezzzi 2'''22 ????
Exercise 5.4 Try to apply the formulae
? ?
? ? ? ?? ? ???
?
?????
???
i
rr
r
ezzzi
z
2'''
'
22
Re4
????
??
for educe the foumula of
stresses components that the half plane body is under concentrated
force in the boundary,
? ? ? ? zPzzPz ln2,ln2 ???? ???
118
因为
所以
? ? ? ?? ?zzzi xyxy '''22 ????? ????
? ? ? ?? ? ??? ????? irr ezzzi 2'''22 ????
练习 5.4 试用公式
? ?
? ? ? ?? ? ???
?
?????
???
i
rr
r
ezzzi
z
2'''
'
22
Re4
????
??
由 导出半平面体在边界上
受集中力作用时的应力分量公式,
? ? ? ? zPzzPz ln2,ln2 ???? ???
119
r ?
y
r
o
P Solution,for
yields
for ? ?
? ?1
c o s22
1
Re
2
Re4
22
'
r
P
yx
xP
z
P
z
r
?
??
?
???
?
??
?
???
?
?
?
?
?
?
????
? ? ? ?
? ?
z
P
z
z
P
z
z
P
z
1
2
1
2
,
1
2
'
2
'''
??
?????
?
?
?
?
?
?
? ? ? ? zPzzPz ln2,ln2 ???? ???
120
r ?
y
r
o
P 解,由
得
? ? ? ? zPzzPz ln2,ln2 ???? ???
? ? ? ?
? ?
z
P
z
z
P
z
z
P
z
1
2
1
2
,
1
2
'
2
'''
??
?????
?
?
?
?
?
?
因为 ? ?
? ?1
c o s22
1
Re
2
Re4
22
'
r
P
yx
xP
z
P
z
r
?
??
?
???
?
??
?
???
?
?
?
?
?
?
????
121
? ? ? ?? ?
?
?
?
??
?
??
?????
i
i
i
rr
e
z
zzP
e
z
P
z
P
z
ezzzi
2
2
2
2
2'''
1
2
1
2
2
22
?
?
?
?
?
?
?
?
????
????
and
2c o s,
??
? ?
ii
i eerez
??
??
so
r
Pi
rr
?
???? ??
c o s22 ???
122
? ? ? ?? ?
?
?
?
??
?
??
?????
i
i
i
rr
e
z
zzP
e
z
P
z
P
z
ezzzi
2
2
2
2
2'''
1
2
1
2
2
22
?
?
?
?
?
?
?
?
????
????
而
2c o s,
??
? ?
ii
i eerez
??
??
所以
r
Pi
rr
?
???? ??
c o s22 ???
123
i.e,
? ?2c o s2 rPr ???? ? ??
? ?30??? r
For（ 1）、（ 2）、（ 3） fields
0
0
c o s2
?
?
??
?
?
?
?
?
?
?
r
r
r
P
124
即
? ?2c o s2 rPr ???? ? ??
? ?30??? r
由（ 1）、（ 2）、（ 3）式得
0
0
c o s2
?
?
??
?
?
?
?
?
?
?
r
r
r
P
125
126
