1
Elasticity
2
3
Chapter 9 Torsion
§9-1 The Torsion of Equal Section Pole
§9-2 The Torsion of Elliptic Section Pole
§9-3 Membrane assimilation
§9-4 The Torsion of Rectangular Section Pole
§9-5 The Torsion of Ringent Thin Cliff Pole
4
第九章 扭 转
§ 9-1 等截面直杆的扭转
§ 9-2 椭圆截面杆的扭转
§ 9-3 薄膜比拟
§ 9-4 矩形截面杆的扭转
§ 9-5 开口薄壁杆件的扭转
5
Material mechanics has solved the torsion problems of round
section pole,but it can’t be used to analyze the torsion problems
of non-round section pole,For the torsion of any section pole,it
is a relatively simple spatial problem,According to the
characteristic of the problem,this chapter first gives the
differential functions and boundary conditions,which the stress
function should satisfy of solving the torsion problems,Then,in
order to solve the torsion problems of relatively complex section
pole,we introduction the method of membrane assimilation,
6
材料力学解决了圆截面直杆的扭转问题，但对非
圆截面杆的扭转问题却无法分析。对于任意截面杆的
扭转，这本是一个较简单的空间问题，根据问题的特
点，本章首先给出了求解扭转问题的应力函数所应满
足的微分方程和边界条件。其次，为了求解相对复杂
截面杆的扭转问题，我们介绍了薄膜比拟方法。
7
§ 9-1 The Torsion of Equal Section Pole
1,Stress Function
A equal section straight pole,ignoring
the body force,is under the action of
torsion M at its two end planes,Take one
end as the xy plane,as shown in fig,The
other stress components are zero except
for the shear stress τzx,τzy
0???? xyzyx ????
Substitute the stress components
and body forces X=Y=Z=0 into the
equations of equilibrium,we get
x
M
M
o
y
z
8
§ 9-1 等截面直杆的扭转
一 应力函数
设有等截面直杆，体力不计，
在两端平面内受扭矩 M作用。取杆
的一端平面为 xy面，图示。横截
面上除了切应力 τ zx,τ zy以外，
其余的应力分量为零
0???? xyzyx ????
将应力分量及体力 X=Y=Z=0代入平
衡方程，得
x
M
M
o
y
z
9
)()( zyzx yx ?????? ??
From the first two equations,we
know,τzx,τzy are functions of only x
and y,they have nothing to do with z,
From the third formula,
0,0,0 ???? yxzz zyzxyzxz ????????????
?
?
?
?
?
?
?
?
?
??
?
?
?
?
?
?
?
?
??
?
?
?
?
?
?
?
?
??
?
?
?
?
?
?
?
?
0
0
0
Z
yxz
Y
xzy
X
zyx
yzxzz
xyzyy
zxyxx
???
???
??? Annotation, the differential equations of equilibrium for spatial problems are,
According to the theory of differential
equations,there must exist a function
??x,y?,from it
,yxz ???? ? xzy ???? ??
The function ??x,y? is called stress
function of torsion problems,
?a?
10
)()( zyzx yx ?????? ??
根据前两方程可见,τ zx,τ zy只是
x和 y的函数，与 z无关，由第三式
0,0,0 ???? yxzz zyzxyzxz ????????????
?
?
?
?
?
?
?
?
?
??
?
?
?
?
?
?
?
?
??
?
?
?
?
?
?
?
?
??
?
?
?
?
?
?
?
?
0
0
0
Z
yxz
Y
xzy
X
zyx
yzxzz
xyzyy
zxyxx
???
???
???
注：空间问题平衡微分方程
根据微分方程理论，一定存在一
个函数 ??x,y?，使得
,yxz ???? ? xzy ???? ??
函数 ??x,y?称为 扭转问题的应力函
数 。
?a?
11
0)1(
0)1(
0)1(
2
2
2
2
2
2
?
??
??
???
?
??
??
???
?
??
??
???
yx
xz
zy
xy
zx
yz
??
??
?? 0(
0)1(
0)1(
2
2
2
2
2
2
2
2
2
?
??
?
?
?
??
???
?
?
??
???
z
y
x
z
y
x
?
??
??
The note,when the body
force is zero,the
compatibility equations in
terms of stress components
for spatial problems are
Substitute stress components into
the compatibility equations which
ignoring the body force,we can see,
the first three formulas and the last
are satisfied,the other two formulas
demand
0
0
2
2
??
??
?
?
?
?
?
?
y
x
Namely
C?? ?2 ?b?
12
0)1(
0)1(
0)1(
2
2
2
2
2
2
?
??
??
???
?
??
??
???
?
??
??
???
yx
xz
zy
xy
zx
yz
??
??
?? 0)1(
0)1(
0)1(
2
2
2
2
2
2
2
2
2
?
?
??
???
?
?
??
???
?
?
??
???
z
y
x
z
y
x
??
??
??
注：体力为零时，空间问题
应力分量表示的相容方程 将应力分量代入不计体力的
相容方程，可见：前三式及最后
一式得到满足，其余二式要求
0
0
2
2
??
??
?
?
?
?
?
?
y
x
即
C?? ?2 ?b?
13
2,Boundary conditions
On the profiles of the pole,substitute
n=0 and surface force components which
are zero into the boundary conditions,we
get that the first two formulas can always
be satisfied,the third formula demands,
The note:the stress boundary
conditions for spatial problems
are,
? ? ? ? ? ?
? ? ? ? ? ?
? ? ? ? ? ? Zmln
Ylnm
Xnml
syzsxzsz
sxyszysy
szxsyxsx
???
???
???
???
???
???
0)()( ?? szyszx ml ??
0??
?
??
?
??
???
?
???
?
ss x
myl ??????
Namely
Being at the boundary,
s
xm
s
yl
d
d,
d
d ???
14
二 边界条件
在杆的侧面上，将 n=0，及面
力分量为零代入边界条件，可见前
两式总能满足，而第三式要求
注：空间问题应力边界条件
? ? ? ? ? ?
? ? ? ? ? ?
? ? ? ? ? ? Zmln
Ylnm
Xnml
syzsxzsz
sxyszysy
szxsyxsx
???
???
???
???
???
???
0)()( ?? szyszx ml ??
0??
?
??
?
??
???
?
???
?
ss x
myl ??????
即
由于在边界上
s
xm
s
yl
d
d,
d
d ???
15
0
d
d
d
d
d
d ???
?
??
?
??
???
?
???
?
ss
x
xs
y
y ss
?
?
??
?
??
Then we have
This illuminates that at the boundary of the cross-section,
the stress functionφ is a constant,Because the stress
components don’t change when the stress function subtracts
a constant,we can suppose when it is a single successional
section(solid pole),
0?s?
?c?
At the arbitrary end of the pole,the shear stress composes
torsion
16
0
d
d
d
d
d
d ???
?
??
?
??
???
?
???
?
ss
x
xs
y
y ss
?
?
??
?
??
于是有
说明在横截面的边界上，应力函数 φ 为常量，由于应力
函数减一个常数，应力分量不受影响，因此在单连通截
面（实心杆）时可设
0?s?
?c?
在杆的任一端，剪应力合成为扭矩
17
x
x
xyy
y
yx
yx
x
x
y
yyxxyM zyzx
dddd
dd)(dd)(
? ? ? ?
????
?
?
?
?
?
??
?
?
?
?
?
????
??
??
??
???
???
?????
?
?
?????
?
?
xxxxx
x
x
yyyyy
y
y
BBAA
BBAA
ddd
ddd
????
?
????
?
Integral step by step,and notice that
φ equals to zero at the boundary
At last we get
?? ? Myx dd2 ?
?d?
18
x
x
xyy
y
yx
yx
x
x
y
yyxxyM zyzx
dddd
dd)(dd)(
? ? ? ?
????
?
?
?
?
?
??
?
?
?
?
?
?????
??
??
??
???
???
?????
?
?
?????
?
?
xxxxx
x
x
yyyyy
y
y
BBAA
BBAA
ddd
ddd
????
?
????
?
分步积分，并注意 φ 在边界上为零
最后得到
?? ? Myx dd2 ?
?d?
19
3,Displacement Formula
According to the relations of stresses,strains and displacements,
we get
0
0
0
?
?
?
?
?
?
?
?
?
z
w
y
v
x
u
0
1
1
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
??
?
?
?
?
?
y
u
x
v
yGx
w
z
u
xGz
v
y
w
?
?
After integral,we get
Kxzzxvv
Kyzyzuu
xz
zy
????
????
??
??
0
0
20
三 位移公式
根据应力、应变、位移的关系可以得到
0
0
0
?
?
?
?
?
?
?
?
?
z
w
y
v
x
u
0
1
1
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
??
?
?
?
?
?
y
u
x
v
yGx
w
z
u
xGz
v
y
w
?
?
积分后得到
Kxzzxvv
Kyzyzuu
xz
zy
????
????
??
??
0
0
21
Where,K denotes the torsion angle per unit length,Ignoring
the displacement of the rigid body,we get,
Substitute them into the above first two formulas at the right
The above two formulas can be used to solve displacement
components w。,
K x zv
K y zu
?
?? ?e?
Kx
xGy
w
Ky
yGx
w
?
?
?
??
?
?
?
?
?
?
?
?
?
?
1
1
?f?
22
其中 K表示杆的单位长度内的扭转角,不计刚体位移
代入前面右边前两式
上两式可用来求出位移分量 w。
K x zv
K y zu
?
?? ?e?
Kx
xGy
w
Ky
yGx
w
?
?
?
??
?
?
?
?
?
?
?
?
?
?
1
1
?f?
23
Differentiating the above two formulas with
respect to y and x,then subtracting these two,
we get,
Gk22 ??? ?
Obviously the above
formula
C?? ?2
Where C=-2GK,
Obviously,in order to seek the solution of the torsion
problems,we only need to find the stress function ?,We
make it satisfy the equations ?b?,?c? and ?d?,then we
solve the stress components from formula ?a? and give the
value of the displacement components from formulas ?e?
and ?f?,
24
上两式分别对 y和 x求导，再相减，得
Gk22 ??? ?
可见前面公式 ?b?中
C?? ?2
的 C=-2GK,
显然，为了求得扭转问题的解，只须寻出应力函数
?，使它满足方程 ?b?,?c?和 ?d?，然后由 ?a?式求出应力
分量，由式 ?e? 和 ?f?给出位移分量的值。
25
§ 9-2 The Torsion of Elliptic Section Pole
The semi-major axes and semi-minor
axes of the elliptic are a and b respectively,
its boundary function is,
012
2
2
2
??? byax
The stress function equals to zero at the
boundary,so we fetch
?
?
??
?
? ??? 1
2
2
2
2
b
y
a
xm?
Substitute it into C?? ?
2
?1?
x
y
a
b
o
26
§ 9-2 椭圆截面杆的扭转
椭圆的半轴分别为 a和 b，其边界
方程为
012
2
2
2
??? byax
应力函数在边界上应等于零，故取
?
?
??
?
? ??? 1
2
2
2
2
b
y
a
xm?
代入 C?? ?
2
?1?
x
y
a
b
o
27
We get
Then we have C
ba
ba
m
C
b
m
a
m
)(2
22
22
2
22
?
?
??
Substitute it formula (1),we get
?
?
??
?
? ??
?? 1)(2 2
2
2
2
22
22
b
y
a
xC
ba
ba?
?? ? Myx dd2 ?
Form
28
得
求得 C
ba
ba
m
C
b
m
a
m
)(2
22
22
2
22
?
?
??
回代入 ?1?式 得
?
?
??
?
? ??
?? 1)(2 2
2
2
2
22
22
b
y
a
xC
ba
ba?
?? ? Myx dd2 ?
由
29
MyxyxybyxxaCba ba ???? ?????? )dddd1dd1( 222222
22
We can get
33
22 )(2
ba
MbaC
?
???
Then we get
?
?
??
?
? ???? 1
2
2
2
2
b
y
a
x
ab
M
?
?
At last we have
30
MyxyxybyxxaCba ba ???? ?????? )dddd1dd1( 222222
22
可得
33
22 )(2
ba
MbaC
?
???
于是得
?
?
??
?
? ???? 1
2
2
2
2
b
y
a
x
ab
M
?
?
最后得
31
We get the final solutions,
xba MyabM yzxz 33 2,2 ???? ????
xy zyyzzxxz ?
????
?
???? ?????,
And from
The total shear stress at any point of the cross-section is
? ? 2
1
4
2
4
2
2
1
22 2
??
?
?
??
?
?
????
b
y
a
x
ab
M
zyzx ????
32
最后得到解答
xba MyabM yzxz 33 2,2 ???? ????
xy zyyzzxxz ?
????
?
???? ?????,
于是由
横截面上任意一点的合剪应力是
? ? 2
1
4
2
4
2
2
1
22 2
??
?
?
??
?
?
????
b
y
a
x
ab
M
zyzx ????
33
§ 9-3 Membrane Assimilation
From the example of the last section,we know,for the simple equal section
pole of elliptic,we just give the calculating expression of shear stress at the
cross-section,we haven’t pointed out the position and direction of the maximum
shear stress at the section; but for the poles with not too complex section such as
rectangular and thin cliff,it is considerably difficult to solve its precise solution,
let alone the relatively complex section pole,For this reason,we introduce the
method of membrane assimilation,This method is built at the basis of similative
of the mathematic relation between the torsion problem of pole and elasticity
membrane which is under the action of equal side pressure and exaggerates tight
around,
Supposing there are even membrane,spreading it at the boundary which is
equal to or proportionate to the section of the torsion pole,When under the
action of small even pressure on the profile,the inner of the membrane will
produce even tensility,each point on membrane will occur small uprightness
angle change along z direction as shown in fig,
34
§ 9-3 薄膜比拟
由上节的例子可以看出，对于椭圆形这种简单等截面
直杆，我们给出了横截面上剪应力的计算表达式，但却没
有指出截面最大剪应力的位置及其方向；而对于矩形、薄
壁杆件这些截面并不复杂的柱体，要求出其精确解都是相
当困难的，更不用说较复杂截面的杆件了。为了解决较复
杂截面杆件的扭转问题，特提出薄膜比拟法。该方法是建
立在柱体扭转问题与受均匀侧压力而四周张紧的弹性薄膜
之间数学关系相似的基础上。
设有一块均匀薄膜，张在与扭转杆件截面相同或成比
例的边界上。当在侧面上受着微小的均匀压力时，在薄膜
内部将产生均匀的张力，薄膜的各点将发生图示 z 方向微
小的垂度。
35
Fetch a small segment abcd of the
membrane,as shown in fig,Its projection on
the xy plane is a rectangle,which side lengths
are dx and dy respectively,Suppose the pull
of the membrane per unit width is T,then
from the condition of equilibrium along z
direction,we get,? ?0Z
0??
?
?
???
?
?
??
?
?
?
?
?
?
?
?
?
?
??
?
?
?
?
?
?
?
?
?
?
q d x d y
y
z
T d xdy
y
z
z
y
T d x
x
z
T d ydx
x
z
z
x
T d y
After predigestion,we get
02
2
2
2
????
?
?
???
?
?
??
?
? q
y
z
x
zT
y
Tdx
Tdy
dx
dy
a
b
d
c
x
y
T
T
x
z
q
o
36
取薄膜的一个微小部分 abcd
图示，它在 xy面上的投影是一个
矩形，矩形的边长分别是 dx和 dy。
设薄膜单位宽度上的拉力为 T，
则由 z方向的平衡条件 得 ? ?0Z
0??
?
?
???
?
?
??
?
?
?
?
?
?
?
?
?
?
??
?
?
?
?
?
?
?
?
?
?
q d x d y
y
z
T d xdy
y
z
z
y
T d x
x
z
T d ydx
x
z
z
x
T d y
简化后得
02
2
2
2
????
?
?
???
?
?
??
?
? q
y
z
x
zT
Tdx
Tdy
dx
dy
a
b
d
c
x
y
T
T
x
z
q
o
y
37
Namely
T
qz ??? 2
Moreover,obviously the uprightness angle of the membrane at
the boundary equals to zero,namely
0?sz
For q/T is a constant,the above two formulas can be rewritten
as
0,012 ???
?
?
???
???
???
?
???
??
s
z
q
Tz
q
T ?a?
And the differential equation and the boundary condition which
the stress function satisfies are,
? ? 0,22 ???? sGk ??
38
即
T
qz ??? 2
此外，薄膜在边界上的垂度显然等于零，即
0?sz
由于 q/T为常量，所以以上两式可改写为
0,012 ???
?
?
???
???
???
?
???
??
s
z
q
Tz
q
T
?a?
而应力函数所满足的微分方程和边界条件为
? ? 0,22 ???? sGk ??
39
Where Gk is also a constant,so they can be rewritten as,
02,0122 ??
?
??
?
????
?
??
?
??
sGkGk
?? ?b?
Comparing formula?b? with formula?a?,we see that
and are all determined by the same differential equation
and boundary condition,so they inevitably have the same
solution,Then we have,
Gk2
?
zqT
q
Tz
Gk ?2
?
Namely
Tq
Gk
z /
2?? ?c?
40
其中 Gk也是常量，故也可改写为
02,0122 ??
?
??
?
????
?
??
?
??
sGkGk
?? ?b?
将式 ?b?与式 ?a?对比，可见 与 决定于同样的微
分方程和边界条件，因而必然具有相同的解答。于是有
Gk2
? z
q
T
q
Tz
Gk ?2
?
即
Tq
Gk
z /
2?? ?c?
41
Suppose the volume between membrane and the boundary plane
is V,and we notice that
?? ? Myx dd2 ?
Then we have
?? ?? ??? G T kqMd x d yG T kqz d x d yV 42 ?
Thereby we have
Tq
Gk
V
M
/
2
2 ?
?d?
From
xy zyyzzxxz ?
????
?
???? ?????,
Moreover,we get
Tq
Gk
y
z
zx /
2/ ?
?
??
?e?
42
设薄膜及其边界平面之间的体积为 V，并注意到
?? ? Myx dd2 ?
则有 ?? ?? ???
G T k
qMd x d y
G T k
qz d x d yV
42 ?
从而有
Tq
Gk
V
M
/
2
2 ?
?d?
由
xy zyyzzxxz ?
????
?
???? ?????,
又可得
Tq
Gk
y
z
zx /
2/ ?
?
??
?e?
43
Adjust the pressure q of which the membrane is under,and
make the rights of formulas ?c?,?d?,?e? equal to one,then we
can gain some conclusions as follows,
?1? The stress function ? of wringed pole equals to the uprightness
angle of the membrane
?2? The torsion M which wringed pole received equals to two times
of the volume between the membrane and the boundary plane,
?3? The shear stress at some point and along arbitrary direction of
the wringed pole just equals to the slope at the counterpoint and
along perpendicular direction of the membrane,
Thus it can be seen,the maximum
shear stress at cross-section of the elliptic
section wringed pole exists at two end
points of the semi-minor axes,its direction
is parallel to the semi-major axes,
x
y
a
b
o
44
调整薄膜所受的压力 q，使得 ?c?,?d?,?e?三式等号
的右边为 1，则可得出如下结论,
?1? 扭杆的应力函数 ?等于薄膜的垂度 z。
?2? 扭杆所受的扭矩 M等于该薄膜及其边界平面之间
的体积的两倍。
?3? 扭杆横截面上某一点处的、沿任意方向的剪应力，
就等于该薄膜在对应点处的、沿垂直方向的斜率。
由此可见，椭圆截面扭杆横截
面上的最大剪应力发生在短轴的两
端点处，方向平行于长轴。
x
y
a
b
o
45
§ 9-4 The Torsion of Rectangular Section Pole
一 The Torsion of Narrow and Long Rectangular Section Pole
Suppose the side lengths of the rectangular section are a and
b,If a is large than b(as shown in fig),we call it narrow and
long rectangle,From the membrane assimilation,we deduce
that the stress function ? almost doesn’t change along with x
at most cross-section,then we have,
dy
d
yx
??? ?
?
??
?
?,0
Then C?? ?2
can be written as
Cdyd ?2
2?
y
a
x b
o
46
§ 9-4 矩形截面杆的扭转
一 狭长矩形截面杆的扭转
设矩形截面的边长为 a和 b (图示 ) 。
若 a?b, 则称为狭长矩形。由薄膜比拟
可以推断，应力函数 ?在绝大部分横截
面上几乎不随 x变化，于是有
dy
d
yx
??? ?
?
??
?
?,0
则 C?? ?2
成为
Cdyd ?2
2?
y
a
x b
o
47
The stress components are,
0
6
3
?
?
?
??
??
?
?
?
x
y
ab
M
y
zy
zx
?
?
?
?
From the membrane assimilation,
we know,the maximum shear
stress exists at the long side of the
rectangular section,Its direction is
parallel to x axis,and its value is
? ? 2
2m a x
3
ab
Mb
yzx ?? ????
?? ? Myx dd2 ?
After integral,and notice that
on the boundary
? ? 02 ??? by?
We get
???
?
???
? ??
42
2
2 byC?
Substitute ? into
After integral,we get
3
6
ab
MC ??
So
Gab
M
G
C
K
y
b
ab
M
3
2
2
3
3
2
4
3
???
??
?
?
??
?
?
???
48
?? ? Myx dd2 ?
积分，并注意在边界上
? ? 02 ??? by?
即得
???
?
???
? ??
42
2
2 byC?
将 ?代入
积分后得
3
6
ab
MC ??
故
Gab
M
G
C
K
y
b
ab
M
3
2
2
3
3
2
4
3
???
?
?
?
?
?
?
?
?
???
应力分量为
0
6
3
?
?
?
??
??
?
?
?
x
y
ab
M
y
zy
zx
?
?
?
?
由薄膜比拟可知，最
大剪应力发生在矩形截面
的长边上，方向平行于 x
轴，其大小为
? ? 2
2m a x
3
ab
Mb
yzx ?? ????
49
2,Pole with Rectangular Section
At the basis of the stress function for narrow and long
rectangular section pole,we choose the stress function for
any rectangular section pole as follow
b
ym
b
xmchAybGk
m
m
??? c o s
4,5,3,1
2
2
?
?
?
???
?
?
???
? ??
?
Substitute into the differential function,
Gk22 ??? ?
? ? ? ? 0,0 22 ?? ???? byax ??
And make ? satisfy the boundary conditions,
50
二 矩形截面杆
在狭矩形截面扭杆应力函数的基础上，取任意矩形
截面杆应力函数为
b
ym
b
xmchAybGk
m
m
??? c o s
4,5,3,1
2
2
?
?
?
???
?
?
???
? ??
?
代入微分方程
Gk22 ??? ?
? ? ? ? 0,0 22 ?? ???? byax ??
并使 ?满足边界条件
51
We get
???
?
???
? ????
? 4
c o s
2
2
2
,5,3,1
byGk
b
ym
b
amchA
m
m
??
?
Spread the right of the above formula into the progression
of at the range of y∈ ?-b/2,b/2?,then compare the
coefficient of both sides,we get,b
ym?cos
? ?
b
am
chm
Gkb
A
m
m
2
81
33
2
2
1
?
?
?
?
?
Substitute Am into ?,we get the certain stress function,
? ?
?
?
?
?
?
?
?
?
?
?
?
??? ?
?
?
?
?,5,3,1 3
2
1
3
2
2
2
2
c o s1
8
4 m
m
b
am
chm
b
ym
b
xm
ch
b
y
b
Gk
?
??
?
?
52
得到
???
?
???
? ????
? 4
c o s
2
2
2
,5,3,1
byGk
b
ym
b
amchA
m
m
??
?
将上式右边在 y∈ ?-b/2,b/2?区间展为 的级
数，然后比较两边的系数，得 b
ym?cos
? ?
b
am
chm
Gkb
A
m
m
2
81
33
2
2
1
?
?
?
?
?
将 Am代入 ?，得确定的应力函数
? ?
?
?
?
?
?
?
?
?
?
?
?
??? ?
?
?
?
?,5,3,1 3
2
1
3
2
2
2
2
c o s1
8
4 m
m
b
am
chm
b
ym
b
xm
ch
b
y
b
Gk
?
??
?
?
53
From the membrane assimilation,we know,the maximum
shear stress exists at midpoint of the long side of the rectangular
section? if a≥b?
? ?
?
?
?
?
?
?
?
?
?
?
?
?
??
?
?
?
?
?
?
?
?
?
?
??
?
?
?
???
???
?5,3,1 2
2
2
,0
2
,0m a x
2
18
1
m
b
yx
b
yxzx
b
am
chm
G k b
y
??
?
??
Where the wring angle k is obtained from
54
由薄膜比拟可以推断，最大剪应力发生在矩形截面长
边的中点 ?若 a≥b?
? ?
?
?
?
?
?
?
?
?
?
?
?
?
??
?
?
?
?
?
?
?
?
?
?
??
?
?
?
???
???
?5,3,1 2
2
2
,0
2
,0m a x
2
18
1
m
b
yx
b
yxzx
b
am
chm
G k b
y
??
?
??
其中扭角 k 由
55
?? ? Myx dd2 ?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
? ?,5,3,1
55
3 264
3
1
m m
b
am
th
a
b
Gab
M
k
?
?
56
?? ? Myx dd2 ?
求得
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
? ?,5,3,1
55
3 264
3
1
m m
b
am
th
a
b
Gab
M
k
?
?
57
§ 9-5 The Torsion of Ringent Thin Cliff Pole
Actually we always face ringent thin cliff poles from engineer
problems,such as angle iron,trough,I –shaped iron and so on,
The cross-sections of these thin cliff poles are always composed
of narrow rectangle which has the equal width,Whatever straight
or bent,from membrane assimilation,we know,if only the narrow
rectangle has the same length and width,then the torsion and the
shear stress at the cross-section of two wringed pole are almost
the same values,
a1
b1
a2
a1
a1
b2
a3
a2
a1
a3
b2
58
§ 9-5 开口薄壁杆件的扭转
实际工程上经常遇到开口薄壁杆件，例如角钢、槽钢、
工字钢等，这些薄壁件其横截面大都是由等宽的狭矩形组
成。无论是直的还是曲的，根据薄膜比拟，只要狭矩形具
有相同的长度和宽度，则两个扭杆的扭矩及其横截面剪应
力没有多大差别。
a1
b1
a2
a1
a1
b2
a3
a2
a1
a3
b2
59
Supposing ai and bi denote the length and the width of the i
narrow rectangle of the cross-section for the wringed pole,Mi
denotes the torsion which the rectangular section is undergone,
M denotes the torsion of all the cross-section,?I denotes the
shear stress near the midpoint of the long side of the rectangle,
k denotes the wrest angle of the wringed pole,From the result
of the narrow rectangle,we get,
3
2
3
3
ii
i
ii
i
i
bGa
M
k
ba
M
?
??
From the later formula,we get
3
3
ii
i
bG k aM ?
60
设 ai 及 bi 分别表示扭杆横截面的第 i 个狭矩形的长度
和宽度,Mi表示该矩形截面上承受的扭矩, M表示整个横
截面上的扭矩, ?i代表该矩形长边中点附近的剪应力,k代
表扭杆的扭角。则由狭矩形的结果，得
3
2
3
3
ii
i
ii
i
i
bGa
M
k
ba
M
?
??
由后一式得
3
3
ii
i
bG k aM ?
61
Also
3
3 iii ba
GkMM ?? ??
So we have,
MbabaM
ii
ii
i ?? 3
3
Consequently we have,
?
?
?
?
3
3
3
3
ii
ii
i
i
baG
M
k
ba
Mb
?
It is noticeable that,the shear stress of the midpoint of the
long side of the narrow rectangle is considerably precise,
However,because of the existence of stress concentration,the
local shear stress maybe is more larger than the mentioned at
the joint of two narrow rectangle,
62
而
3
3 iii ba
GkMM ?? ??
故有
MbabaM
ii
ii
i ?? 3
3
从而有
?
?
?
?
3
3
3
3
ii
ii
i
i
baG
M
k
ba
Mb
?
值得注意的是：由上述公式给出的狭矩形长边中点
的剪应力已相当精确，然而，由于应力集中的存在，两
个狭矩形的连接处，可能存在远大于此的局部剪应力。
63
Exercise 9.1 One wringed pole with the cross-section of
equilateral triangle has its high of a,the coordinate is shown as
fig,The three sides AB,OA,OB of the triangle satisfy equations,
? ?? ?? ?yxyxaxm 33 ?????
.03;03 ???? yxyx ;0?? ax Please prove the stress function satisfies any
condition,and solve the maximum shear stress and twisty angle
Solution,substitute into the equations of compatibility ?
C?? ?2
We get Cam ?4
Namely
a
Cm
4?
o
x
y
B
A
a
64
习题 9.1 有一根高为 a 的等边三角形截面扭杆，坐标如图
所示。三角形三条边 AB,OA,OB的方程分别为,
.03;03 ???? yxyx
? ?? ?? ?yxyxaxm 33 ?????;0?? ax
试证应力函数
能满足一切条件，并求出最大剪应力及扭角。
解,将 代入相容方程 ?
C?? ?2
得 Cam ?4
即
a
Cm
4?
o
x
y
B
A
a
65
Thereupon ? ?? ?? ?yxyxax
a
C 33
4 ?????
The wringed pole hasn’t hole,obviously satisfies the boundary
conditions on profile,From the boundary condition at the
end of the pole,
?
0?s?
?? ? Md x d y?2
We get ? ?? ?? ?? ?
?
????a
x
x Mdyyxyxaxdxa
C
0
3
3
332
MaC 4 330??
Consequently ? ?? ?? ?
yxyxaxa M 332 315 5 ??????
66
故 ? ?? ?? ?yxyxax
a
C 33
4 ?????
扭杆无孔洞,显然满足侧面边界条件,由杆端
部边界条件
? 0?s?
?? ? Md x d y?2
得 ? ?? ?? ?
? ?? ????a x x MdyyxyxaxdxaC 0 3
3
332
MaC 4 330??
从而得 ? ?? ?? ?
yxyxaxa M 332 315 5 ??????
67
The shear stresses
? ?32
5
5
323
2
315
)(
345
yaxx
a
M
x
yax
a
M
y
yz
xz
???
?
?
??
??
?
?
?
?
?
?
?
For equilateral triangle,from the membrane assimilation,we
know:the maximum shear stress exists at 0,?? yax
Namely ? ?
3
22
5m a x 2
31523
2
315
a
Maa
a
M ????
The twisty angle per unit length is,
4
315
2 Ga
M
G
CK ???
68
剪应力
? ?32
5
5
323
2
315
)(
345
yaxx
a
M
x
yax
a
M
y
yz
xz
???
?
?
??
??
?
?
?
?
?
?
?
对于等边三角形，由薄膜比拟法知，最大剪应力发生在
0,?? yax 处，即
? ? 3225m a x 2 315232 315 a Maaa M ????
单位长度上的扭角
4
315
2 Ga
M
G
CK ???
69
Exercise 9.2 consider a pole with equal section,Its twisty angle
per unit length is K,the shear elasticity module is, A
function,where are real constants,
is a function under affirmance,Solve the question,what is the
conditions must satisfy,then can be viewed as
stress function of the torsion problems?
? ? 2co s rfAr ???? ? ??
G
??,,A
? ??f
? ???? fA,,,?
Substitute into the equation,we get
Solution,if the given function can be used for the stress
function of the torsion problems,if must satisfy the equations of
compatibility,,
GK22 ??? ? ?
? ? ? ? ? ? GKffrA 24c o s ''222 ????? ? ?????? ?
This is the condition need satisfying,? ???? fA,,,
70
习题 9.2 等截面杆单位长度上的扭转角为 K,剪切弹性
模量为,若函数,其中 为
实常量,为待定函数。试问,满足什么
条件时,可作为扭转问题的应力函数？
? ? 2co s rfAr ???? ? ??G ??,,A
? ??f ? ???? fA,,,
?
答,若题意所给函数能作为扭转问题的应力函数，则该
函数必须满足相容方程
GK22 ??? ?
将 代入后得 ?
? ? ? ? ? ? GKffrA 24c o s ''222 ????? ? ?????? ?
此即为 需满足的条件。 ? ???? fA,,,
71
72
Elasticity
2
3
Chapter 9 Torsion
§9-1 The Torsion of Equal Section Pole
§9-2 The Torsion of Elliptic Section Pole
§9-3 Membrane assimilation
§9-4 The Torsion of Rectangular Section Pole
§9-5 The Torsion of Ringent Thin Cliff Pole
4
第九章 扭 转
§ 9-1 等截面直杆的扭转
§ 9-2 椭圆截面杆的扭转
§ 9-3 薄膜比拟
§ 9-4 矩形截面杆的扭转
§ 9-5 开口薄壁杆件的扭转
5
Material mechanics has solved the torsion problems of round
section pole,but it can’t be used to analyze the torsion problems
of non-round section pole,For the torsion of any section pole,it
is a relatively simple spatial problem,According to the
characteristic of the problem,this chapter first gives the
differential functions and boundary conditions,which the stress
function should satisfy of solving the torsion problems,Then,in
order to solve the torsion problems of relatively complex section
pole,we introduction the method of membrane assimilation,
6
材料力学解决了圆截面直杆的扭转问题，但对非
圆截面杆的扭转问题却无法分析。对于任意截面杆的
扭转，这本是一个较简单的空间问题，根据问题的特
点，本章首先给出了求解扭转问题的应力函数所应满
足的微分方程和边界条件。其次，为了求解相对复杂
截面杆的扭转问题，我们介绍了薄膜比拟方法。
7
§ 9-1 The Torsion of Equal Section Pole
1,Stress Function
A equal section straight pole,ignoring
the body force,is under the action of
torsion M at its two end planes,Take one
end as the xy plane,as shown in fig,The
other stress components are zero except
for the shear stress τzx,τzy
0???? xyzyx ????
Substitute the stress components
and body forces X=Y=Z=0 into the
equations of equilibrium,we get
x
M
M
o
y
z
8
§ 9-1 等截面直杆的扭转
一 应力函数
设有等截面直杆，体力不计，
在两端平面内受扭矩 M作用。取杆
的一端平面为 xy面，图示。横截
面上除了切应力 τ zx,τ zy以外，
其余的应力分量为零
0???? xyzyx ????
将应力分量及体力 X=Y=Z=0代入平
衡方程，得
x
M
M
o
y
z
9
)()( zyzx yx ?????? ??
From the first two equations,we
know,τzx,τzy are functions of only x
and y,they have nothing to do with z,
From the third formula,
0,0,0 ???? yxzz zyzxyzxz ????????????
?
?
?
?
?
?
?
?
?
??
?
?
?
?
?
?
?
?
??
?
?
?
?
?
?
?
?
??
?
?
?
?
?
?
?
?
0
0
0
Z
yxz
Y
xzy
X
zyx
yzxzz
xyzyy
zxyxx
???
???
??? Annotation, the differential equations of equilibrium for spatial problems are,
According to the theory of differential
equations,there must exist a function
??x,y?,from it
,yxz ???? ? xzy ???? ??
The function ??x,y? is called stress
function of torsion problems,
?a?
10
)()( zyzx yx ?????? ??
根据前两方程可见,τ zx,τ zy只是
x和 y的函数，与 z无关，由第三式
0,0,0 ???? yxzz zyzxyzxz ????????????
?
?
?
?
?
?
?
?
?
??
?
?
?
?
?
?
?
?
??
?
?
?
?
?
?
?
?
??
?
?
?
?
?
?
?
?
0
0
0
Z
yxz
Y
xzy
X
zyx
yzxzz
xyzyy
zxyxx
???
???
???
注：空间问题平衡微分方程
根据微分方程理论，一定存在一
个函数 ??x,y?，使得
,yxz ???? ? xzy ???? ??
函数 ??x,y?称为 扭转问题的应力函
数 。
?a?
11
0)1(
0)1(
0)1(
2
2
2
2
2
2
?
??
??
???
?
??
??
???
?
??
??
???
yx
xz
zy
xy
zx
yz
??
??
?? 0(
0)1(
0)1(
2
2
2
2
2
2
2
2
2
?
??
?
?
?
??
???
?
?
??
???
z
y
x
z
y
x
?
??
??
The note,when the body
force is zero,the
compatibility equations in
terms of stress components
for spatial problems are
Substitute stress components into
the compatibility equations which
ignoring the body force,we can see,
the first three formulas and the last
are satisfied,the other two formulas
demand
0
0
2
2
??
??
?
?
?
?
?
?
y
x
Namely
C?? ?2 ?b?
12
0)1(
0)1(
0)1(
2
2
2
2
2
2
?
??
??
???
?
??
??
???
?
??
??
???
yx
xz
zy
xy
zx
yz
??
??
?? 0)1(
0)1(
0)1(
2
2
2
2
2
2
2
2
2
?
?
??
???
?
?
??
???
?
?
??
???
z
y
x
z
y
x
??
??
??
注：体力为零时，空间问题
应力分量表示的相容方程 将应力分量代入不计体力的
相容方程，可见：前三式及最后
一式得到满足，其余二式要求
0
0
2
2
??
??
?
?
?
?
?
?
y
x
即
C?? ?2 ?b?
13
2,Boundary conditions
On the profiles of the pole,substitute
n=0 and surface force components which
are zero into the boundary conditions,we
get that the first two formulas can always
be satisfied,the third formula demands,
The note:the stress boundary
conditions for spatial problems
are,
? ? ? ? ? ?
? ? ? ? ? ?
? ? ? ? ? ? Zmln
Ylnm
Xnml
syzsxzsz
sxyszysy
szxsyxsx
???
???
???
???
???
???
0)()( ?? szyszx ml ??
0??
?
??
?
??
???
?
???
?
ss x
myl ??????
Namely
Being at the boundary,
s
xm
s
yl
d
d,
d
d ???
14
二 边界条件
在杆的侧面上，将 n=0，及面
力分量为零代入边界条件，可见前
两式总能满足，而第三式要求
注：空间问题应力边界条件
? ? ? ? ? ?
? ? ? ? ? ?
? ? ? ? ? ? Zmln
Ylnm
Xnml
syzsxzsz
sxyszysy
szxsyxsx
???
???
???
???
???
???
0)()( ?? szyszx ml ??
0??
?
??
?
??
???
?
???
?
ss x
myl ??????
即
由于在边界上
s
xm
s
yl
d
d,
d
d ???
15
0
d
d
d
d
d
d ???
?
??
?
??
???
?
???
?
ss
x
xs
y
y ss
?
?
??
?
??
Then we have
This illuminates that at the boundary of the cross-section,
the stress functionφ is a constant,Because the stress
components don’t change when the stress function subtracts
a constant,we can suppose when it is a single successional
section(solid pole),
0?s?
?c?
At the arbitrary end of the pole,the shear stress composes
torsion
16
0
d
d
d
d
d
d ???
?
??
?
??
???
?
???
?
ss
x
xs
y
y ss
?
?
??
?
??
于是有
说明在横截面的边界上，应力函数 φ 为常量，由于应力
函数减一个常数，应力分量不受影响，因此在单连通截
面（实心杆）时可设
0?s?
?c?
在杆的任一端，剪应力合成为扭矩
17
x
x
xyy
y
yx
yx
x
x
y
yyxxyM zyzx
dddd
dd)(dd)(
? ? ? ?
????
?
?
?
?
?
??
?
?
?
?
?
????
??
??
??
???
???
?????
?
?
?????
?
?
xxxxx
x
x
yyyyy
y
y
BBAA
BBAA
ddd
ddd
????
?
????
?
Integral step by step,and notice that
φ equals to zero at the boundary
At last we get
?? ? Myx dd2 ?
?d?
18
x
x
xyy
y
yx
yx
x
x
y
yyxxyM zyzx
dddd
dd)(dd)(
? ? ? ?
????
?
?
?
?
?
??
?
?
?
?
?
?????
??
??
??
???
???
?????
?
?
?????
?
?
xxxxx
x
x
yyyyy
y
y
BBAA
BBAA
ddd
ddd
????
?
????
?
分步积分，并注意 φ 在边界上为零
最后得到
?? ? Myx dd2 ?
?d?
19
3,Displacement Formula
According to the relations of stresses,strains and displacements,
we get
0
0
0
?
?
?
?
?
?
?
?
?
z
w
y
v
x
u
0
1
1
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
??
?
?
?
?
?
y
u
x
v
yGx
w
z
u
xGz
v
y
w
?
?
After integral,we get
Kxzzxvv
Kyzyzuu
xz
zy
????
????
??
??
0
0
20
三 位移公式
根据应力、应变、位移的关系可以得到
0
0
0
?
?
?
?
?
?
?
?
?
z
w
y
v
x
u
0
1
1
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
??
?
?
?
?
?
y
u
x
v
yGx
w
z
u
xGz
v
y
w
?
?
积分后得到
Kxzzxvv
Kyzyzuu
xz
zy
????
????
??
??
0
0
21
Where,K denotes the torsion angle per unit length,Ignoring
the displacement of the rigid body,we get,
Substitute them into the above first two formulas at the right
The above two formulas can be used to solve displacement
components w。,
K x zv
K y zu
?
?? ?e?
Kx
xGy
w
Ky
yGx
w
?
?
?
??
?
?
?
?
?
?
?
?
?
?
1
1
?f?
22
其中 K表示杆的单位长度内的扭转角,不计刚体位移
代入前面右边前两式
上两式可用来求出位移分量 w。
K x zv
K y zu
?
?? ?e?
Kx
xGy
w
Ky
yGx
w
?
?
?
??
?
?
?
?
?
?
?
?
?
?
1
1
?f?
23
Differentiating the above two formulas with
respect to y and x,then subtracting these two,
we get,
Gk22 ??? ?
Obviously the above
formula
C?? ?2
Where C=-2GK,
Obviously,in order to seek the solution of the torsion
problems,we only need to find the stress function ?,We
make it satisfy the equations ?b?,?c? and ?d?,then we
solve the stress components from formula ?a? and give the
value of the displacement components from formulas ?e?
and ?f?,
24
上两式分别对 y和 x求导，再相减，得
Gk22 ??? ?
可见前面公式 ?b?中
C?? ?2
的 C=-2GK,
显然，为了求得扭转问题的解，只须寻出应力函数
?，使它满足方程 ?b?,?c?和 ?d?，然后由 ?a?式求出应力
分量，由式 ?e? 和 ?f?给出位移分量的值。
25
§ 9-2 The Torsion of Elliptic Section Pole
The semi-major axes and semi-minor
axes of the elliptic are a and b respectively,
its boundary function is,
012
2
2
2
??? byax
The stress function equals to zero at the
boundary,so we fetch
?
?
??
?
? ??? 1
2
2
2
2
b
y
a
xm?
Substitute it into C?? ?
2
?1?
x
y
a
b
o
26
§ 9-2 椭圆截面杆的扭转
椭圆的半轴分别为 a和 b，其边界
方程为
012
2
2
2
??? byax
应力函数在边界上应等于零，故取
?
?
??
?
? ??? 1
2
2
2
2
b
y
a
xm?
代入 C?? ?
2
?1?
x
y
a
b
o
27
We get
Then we have C
ba
ba
m
C
b
m
a
m
)(2
22
22
2
22
?
?
??
Substitute it formula (1),we get
?
?
??
?
? ??
?? 1)(2 2
2
2
2
22
22
b
y
a
xC
ba
ba?
?? ? Myx dd2 ?
Form
28
得
求得 C
ba
ba
m
C
b
m
a
m
)(2
22
22
2
22
?
?
??
回代入 ?1?式 得
?
?
??
?
? ??
?? 1)(2 2
2
2
2
22
22
b
y
a
xC
ba
ba?
?? ? Myx dd2 ?
由
29
MyxyxybyxxaCba ba ???? ?????? )dddd1dd1( 222222
22
We can get
33
22 )(2
ba
MbaC
?
???
Then we get
?
?
??
?
? ???? 1
2
2
2
2
b
y
a
x
ab
M
?
?
At last we have
30
MyxyxybyxxaCba ba ???? ?????? )dddd1dd1( 222222
22
可得
33
22 )(2
ba
MbaC
?
???
于是得
?
?
??
?
? ???? 1
2
2
2
2
b
y
a
x
ab
M
?
?
最后得
31
We get the final solutions,
xba MyabM yzxz 33 2,2 ???? ????
xy zyyzzxxz ?
????
?
???? ?????,
And from
The total shear stress at any point of the cross-section is
? ? 2
1
4
2
4
2
2
1
22 2
??
?
?
??
?
?
????
b
y
a
x
ab
M
zyzx ????
32
最后得到解答
xba MyabM yzxz 33 2,2 ???? ????
xy zyyzzxxz ?
????
?
???? ?????,
于是由
横截面上任意一点的合剪应力是
? ? 2
1
4
2
4
2
2
1
22 2
??
?
?
??
?
?
????
b
y
a
x
ab
M
zyzx ????
33
§ 9-3 Membrane Assimilation
From the example of the last section,we know,for the simple equal section
pole of elliptic,we just give the calculating expression of shear stress at the
cross-section,we haven’t pointed out the position and direction of the maximum
shear stress at the section; but for the poles with not too complex section such as
rectangular and thin cliff,it is considerably difficult to solve its precise solution,
let alone the relatively complex section pole,For this reason,we introduce the
method of membrane assimilation,This method is built at the basis of similative
of the mathematic relation between the torsion problem of pole and elasticity
membrane which is under the action of equal side pressure and exaggerates tight
around,
Supposing there are even membrane,spreading it at the boundary which is
equal to or proportionate to the section of the torsion pole,When under the
action of small even pressure on the profile,the inner of the membrane will
produce even tensility,each point on membrane will occur small uprightness
angle change along z direction as shown in fig,
34
§ 9-3 薄膜比拟
由上节的例子可以看出，对于椭圆形这种简单等截面
直杆，我们给出了横截面上剪应力的计算表达式，但却没
有指出截面最大剪应力的位置及其方向；而对于矩形、薄
壁杆件这些截面并不复杂的柱体，要求出其精确解都是相
当困难的，更不用说较复杂截面的杆件了。为了解决较复
杂截面杆件的扭转问题，特提出薄膜比拟法。该方法是建
立在柱体扭转问题与受均匀侧压力而四周张紧的弹性薄膜
之间数学关系相似的基础上。
设有一块均匀薄膜，张在与扭转杆件截面相同或成比
例的边界上。当在侧面上受着微小的均匀压力时，在薄膜
内部将产生均匀的张力，薄膜的各点将发生图示 z 方向微
小的垂度。
35
Fetch a small segment abcd of the
membrane,as shown in fig,Its projection on
the xy plane is a rectangle,which side lengths
are dx and dy respectively,Suppose the pull
of the membrane per unit width is T,then
from the condition of equilibrium along z
direction,we get,? ?0Z
0??
?
?
???
?
?
??
?
?
?
?
?
?
?
?
?
?
??
?
?
?
?
?
?
?
?
?
?
q d x d y
y
z
T d xdy
y
z
z
y
T d x
x
z
T d ydx
x
z
z
x
T d y
After predigestion,we get
02
2
2
2
????
?
?
???
?
?
??
?
? q
y
z
x
zT
y
Tdx
Tdy
dx
dy
a
b
d
c
x
y
T
T
x
z
q
o
36
取薄膜的一个微小部分 abcd
图示，它在 xy面上的投影是一个
矩形，矩形的边长分别是 dx和 dy。
设薄膜单位宽度上的拉力为 T，
则由 z方向的平衡条件 得 ? ?0Z
0??
?
?
???
?
?
??
?
?
?
?
?
?
?
?
?
?
??
?
?
?
?
?
?
?
?
?
?
q d x d y
y
z
T d xdy
y
z
z
y
T d x
x
z
T d ydx
x
z
z
x
T d y
简化后得
02
2
2
2
????
?
?
???
?
?
??
?
? q
y
z
x
zT
Tdx
Tdy
dx
dy
a
b
d
c
x
y
T
T
x
z
q
o
y
37
Namely
T
qz ??? 2
Moreover,obviously the uprightness angle of the membrane at
the boundary equals to zero,namely
0?sz
For q/T is a constant,the above two formulas can be rewritten
as
0,012 ???
?
?
???
???
???
?
???
??
s
z
q
Tz
q
T ?a?
And the differential equation and the boundary condition which
the stress function satisfies are,
? ? 0,22 ???? sGk ??
38
即
T
qz ??? 2
此外，薄膜在边界上的垂度显然等于零，即
0?sz
由于 q/T为常量，所以以上两式可改写为
0,012 ???
?
?
???
???
???
?
???
??
s
z
q
Tz
q
T
?a?
而应力函数所满足的微分方程和边界条件为
? ? 0,22 ???? sGk ??
39
Where Gk is also a constant,so they can be rewritten as,
02,0122 ??
?
??
?
????
?
??
?
??
sGkGk
?? ?b?
Comparing formula?b? with formula?a?,we see that
and are all determined by the same differential equation
and boundary condition,so they inevitably have the same
solution,Then we have,
Gk2
?
zqT
q
Tz
Gk ?2
?
Namely
Tq
Gk
z /
2?? ?c?
40
其中 Gk也是常量，故也可改写为
02,0122 ??
?
??
?
????
?
??
?
??
sGkGk
?? ?b?
将式 ?b?与式 ?a?对比，可见 与 决定于同样的微
分方程和边界条件，因而必然具有相同的解答。于是有
Gk2
? z
q
T
q
Tz
Gk ?2
?
即
Tq
Gk
z /
2?? ?c?
41
Suppose the volume between membrane and the boundary plane
is V,and we notice that
?? ? Myx dd2 ?
Then we have
?? ?? ??? G T kqMd x d yG T kqz d x d yV 42 ?
Thereby we have
Tq
Gk
V
M
/
2
2 ?
?d?
From
xy zyyzzxxz ?
????
?
???? ?????,
Moreover,we get
Tq
Gk
y
z
zx /
2/ ?
?
??
?e?
42
设薄膜及其边界平面之间的体积为 V，并注意到
?? ? Myx dd2 ?
则有 ?? ?? ???
G T k
qMd x d y
G T k
qz d x d yV
42 ?
从而有
Tq
Gk
V
M
/
2
2 ?
?d?
由
xy zyyzzxxz ?
????
?
???? ?????,
又可得
Tq
Gk
y
z
zx /
2/ ?
?
??
?e?
43
Adjust the pressure q of which the membrane is under,and
make the rights of formulas ?c?,?d?,?e? equal to one,then we
can gain some conclusions as follows,
?1? The stress function ? of wringed pole equals to the uprightness
angle of the membrane
?2? The torsion M which wringed pole received equals to two times
of the volume between the membrane and the boundary plane,
?3? The shear stress at some point and along arbitrary direction of
the wringed pole just equals to the slope at the counterpoint and
along perpendicular direction of the membrane,
Thus it can be seen,the maximum
shear stress at cross-section of the elliptic
section wringed pole exists at two end
points of the semi-minor axes,its direction
is parallel to the semi-major axes,
x
y
a
b
o
44
调整薄膜所受的压力 q，使得 ?c?,?d?,?e?三式等号
的右边为 1，则可得出如下结论,
?1? 扭杆的应力函数 ?等于薄膜的垂度 z。
?2? 扭杆所受的扭矩 M等于该薄膜及其边界平面之间
的体积的两倍。
?3? 扭杆横截面上某一点处的、沿任意方向的剪应力，
就等于该薄膜在对应点处的、沿垂直方向的斜率。
由此可见，椭圆截面扭杆横截
面上的最大剪应力发生在短轴的两
端点处，方向平行于长轴。
x
y
a
b
o
45
§ 9-4 The Torsion of Rectangular Section Pole
一 The Torsion of Narrow and Long Rectangular Section Pole
Suppose the side lengths of the rectangular section are a and
b,If a is large than b(as shown in fig),we call it narrow and
long rectangle,From the membrane assimilation,we deduce
that the stress function ? almost doesn’t change along with x
at most cross-section,then we have,
dy
d
yx
??? ?
?
??
?
?,0
Then C?? ?2
can be written as
Cdyd ?2
2?
y
a
x b
o
46
§ 9-4 矩形截面杆的扭转
一 狭长矩形截面杆的扭转
设矩形截面的边长为 a和 b (图示 ) 。
若 a?b, 则称为狭长矩形。由薄膜比拟
可以推断，应力函数 ?在绝大部分横截
面上几乎不随 x变化，于是有
dy
d
yx
??? ?
?
??
?
?,0
则 C?? ?2
成为
Cdyd ?2
2?
y
a
x b
o
47
The stress components are,
0
6
3
?
?
?
??
??
?
?
?
x
y
ab
M
y
zy
zx
?
?
?
?
From the membrane assimilation,
we know,the maximum shear
stress exists at the long side of the
rectangular section,Its direction is
parallel to x axis,and its value is
? ? 2
2m a x
3
ab
Mb
yzx ?? ????
?? ? Myx dd2 ?
After integral,and notice that
on the boundary
? ? 02 ??? by?
We get
???
?
???
? ??
42
2
2 byC?
Substitute ? into
After integral,we get
3
6
ab
MC ??
So
Gab
M
G
C
K
y
b
ab
M
3
2
2
3
3
2
4
3
???
??
?
?
??
?
?
???
48
?? ? Myx dd2 ?
积分，并注意在边界上
? ? 02 ??? by?
即得
???
?
???
? ??
42
2
2 byC?
将 ?代入
积分后得
3
6
ab
MC ??
故
Gab
M
G
C
K
y
b
ab
M
3
2
2
3
3
2
4
3
???
?
?
?
?
?
?
?
?
???
应力分量为
0
6
3
?
?
?
??
??
?
?
?
x
y
ab
M
y
zy
zx
?
?
?
?
由薄膜比拟可知，最
大剪应力发生在矩形截面
的长边上，方向平行于 x
轴，其大小为
? ? 2
2m a x
3
ab
Mb
yzx ?? ????
49
2,Pole with Rectangular Section
At the basis of the stress function for narrow and long
rectangular section pole,we choose the stress function for
any rectangular section pole as follow
b
ym
b
xmchAybGk
m
m
??? c o s
4,5,3,1
2
2
?
?
?
???
?
?
???
? ??
?
Substitute into the differential function,
Gk22 ??? ?
? ? ? ? 0,0 22 ?? ???? byax ??
And make ? satisfy the boundary conditions,
50
二 矩形截面杆
在狭矩形截面扭杆应力函数的基础上，取任意矩形
截面杆应力函数为
b
ym
b
xmchAybGk
m
m
??? c o s
4,5,3,1
2
2
?
?
?
???
?
?
???
? ??
?
代入微分方程
Gk22 ??? ?
? ? ? ? 0,0 22 ?? ???? byax ??
并使 ?满足边界条件
51
We get
???
?
???
? ????
? 4
c o s
2
2
2
,5,3,1
byGk
b
ym
b
amchA
m
m
??
?
Spread the right of the above formula into the progression
of at the range of y∈ ?-b/2,b/2?,then compare the
coefficient of both sides,we get,b
ym?cos
? ?
b
am
chm
Gkb
A
m
m
2
81
33
2
2
1
?
?
?
?
?
Substitute Am into ?,we get the certain stress function,
? ?
?
?
?
?
?
?
?
?
?
?
?
??? ?
?
?
?
?,5,3,1 3
2
1
3
2
2
2
2
c o s1
8
4 m
m
b
am
chm
b
ym
b
xm
ch
b
y
b
Gk
?
??
?
?
52
得到
???
?
???
? ????
? 4
c o s
2
2
2
,5,3,1
byGk
b
ym
b
amchA
m
m
??
?
将上式右边在 y∈ ?-b/2,b/2?区间展为 的级
数，然后比较两边的系数，得 b
ym?cos
? ?
b
am
chm
Gkb
A
m
m
2
81
33
2
2
1
?
?
?
?
?
将 Am代入 ?，得确定的应力函数
? ?
?
?
?
?
?
?
?
?
?
?
?
??? ?
?
?
?
?,5,3,1 3
2
1
3
2
2
2
2
c o s1
8
4 m
m
b
am
chm
b
ym
b
xm
ch
b
y
b
Gk
?
??
?
?
53
From the membrane assimilation,we know,the maximum
shear stress exists at midpoint of the long side of the rectangular
section? if a≥b?
? ?
?
?
?
?
?
?
?
?
?
?
?
?
??
?
?
?
?
?
?
?
?
?
?
??
?
?
?
???
???
?5,3,1 2
2
2
,0
2
,0m a x
2
18
1
m
b
yx
b
yxzx
b
am
chm
G k b
y
??
?
??
Where the wring angle k is obtained from
54
由薄膜比拟可以推断，最大剪应力发生在矩形截面长
边的中点 ?若 a≥b?
? ?
?
?
?
?
?
?
?
?
?
?
?
?
??
?
?
?
?
?
?
?
?
?
?
??
?
?
?
???
???
?5,3,1 2
2
2
,0
2
,0m a x
2
18
1
m
b
yx
b
yxzx
b
am
chm
G k b
y
??
?
??
其中扭角 k 由
55
?? ? Myx dd2 ?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
? ?,5,3,1
55
3 264
3
1
m m
b
am
th
a
b
Gab
M
k
?
?
56
?? ? Myx dd2 ?
求得
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
? ?,5,3,1
55
3 264
3
1
m m
b
am
th
a
b
Gab
M
k
?
?
57
§ 9-5 The Torsion of Ringent Thin Cliff Pole
Actually we always face ringent thin cliff poles from engineer
problems,such as angle iron,trough,I –shaped iron and so on,
The cross-sections of these thin cliff poles are always composed
of narrow rectangle which has the equal width,Whatever straight
or bent,from membrane assimilation,we know,if only the narrow
rectangle has the same length and width,then the torsion and the
shear stress at the cross-section of two wringed pole are almost
the same values,
a1
b1
a2
a1
a1
b2
a3
a2
a1
a3
b2
58
§ 9-5 开口薄壁杆件的扭转
实际工程上经常遇到开口薄壁杆件，例如角钢、槽钢、
工字钢等，这些薄壁件其横截面大都是由等宽的狭矩形组
成。无论是直的还是曲的，根据薄膜比拟，只要狭矩形具
有相同的长度和宽度，则两个扭杆的扭矩及其横截面剪应
力没有多大差别。
a1
b1
a2
a1
a1
b2
a3
a2
a1
a3
b2
59
Supposing ai and bi denote the length and the width of the i
narrow rectangle of the cross-section for the wringed pole,Mi
denotes the torsion which the rectangular section is undergone,
M denotes the torsion of all the cross-section,?I denotes the
shear stress near the midpoint of the long side of the rectangle,
k denotes the wrest angle of the wringed pole,From the result
of the narrow rectangle,we get,
3
2
3
3
ii
i
ii
i
i
bGa
M
k
ba
M
?
??
From the later formula,we get
3
3
ii
i
bG k aM ?
60
设 ai 及 bi 分别表示扭杆横截面的第 i 个狭矩形的长度
和宽度,Mi表示该矩形截面上承受的扭矩, M表示整个横
截面上的扭矩, ?i代表该矩形长边中点附近的剪应力,k代
表扭杆的扭角。则由狭矩形的结果，得
3
2
3
3
ii
i
ii
i
i
bGa
M
k
ba
M
?
??
由后一式得
3
3
ii
i
bG k aM ?
61
Also
3
3 iii ba
GkMM ?? ??
So we have,
MbabaM
ii
ii
i ?? 3
3
Consequently we have,
?
?
?
?
3
3
3
3
ii
ii
i
i
baG
M
k
ba
Mb
?
It is noticeable that,the shear stress of the midpoint of the
long side of the narrow rectangle is considerably precise,
However,because of the existence of stress concentration,the
local shear stress maybe is more larger than the mentioned at
the joint of two narrow rectangle,
62
而
3
3 iii ba
GkMM ?? ??
故有
MbabaM
ii
ii
i ?? 3
3
从而有
?
?
?
?
3
3
3
3
ii
ii
i
i
baG
M
k
ba
Mb
?
值得注意的是：由上述公式给出的狭矩形长边中点
的剪应力已相当精确，然而，由于应力集中的存在，两
个狭矩形的连接处，可能存在远大于此的局部剪应力。
63
Exercise 9.1 One wringed pole with the cross-section of
equilateral triangle has its high of a,the coordinate is shown as
fig,The three sides AB,OA,OB of the triangle satisfy equations,
? ?? ?? ?yxyxaxm 33 ?????
.03;03 ???? yxyx ;0?? ax Please prove the stress function satisfies any
condition,and solve the maximum shear stress and twisty angle
Solution,substitute into the equations of compatibility ?
C?? ?2
We get Cam ?4
Namely
a
Cm
4?
o
x
y
B
A
a
64
习题 9.1 有一根高为 a 的等边三角形截面扭杆，坐标如图
所示。三角形三条边 AB,OA,OB的方程分别为,
.03;03 ???? yxyx
? ?? ?? ?yxyxaxm 33 ?????;0?? ax
试证应力函数
能满足一切条件，并求出最大剪应力及扭角。
解,将 代入相容方程 ?
C?? ?2
得 Cam ?4
即
a
Cm
4?
o
x
y
B
A
a
65
Thereupon ? ?? ?? ?yxyxax
a
C 33
4 ?????
The wringed pole hasn’t hole,obviously satisfies the boundary
conditions on profile,From the boundary condition at the
end of the pole,
?
0?s?
?? ? Md x d y?2
We get ? ?? ?? ?? ?
?
????a
x
x Mdyyxyxaxdxa
C
0
3
3
332
MaC 4 330??
Consequently ? ?? ?? ?
yxyxaxa M 332 315 5 ??????
66
故 ? ?? ?? ?yxyxax
a
C 33
4 ?????
扭杆无孔洞,显然满足侧面边界条件,由杆端
部边界条件
? 0?s?
?? ? Md x d y?2
得 ? ?? ?? ?
? ?? ????a x x MdyyxyxaxdxaC 0 3
3
332
MaC 4 330??
从而得 ? ?? ?? ?
yxyxaxa M 332 315 5 ??????
67
The shear stresses
? ?32
5
5
323
2
315
)(
345
yaxx
a
M
x
yax
a
M
y
yz
xz
???
?
?
??
??
?
?
?
?
?
?
?
For equilateral triangle,from the membrane assimilation,we
know:the maximum shear stress exists at 0,?? yax
Namely ? ?
3
22
5m a x 2
31523
2
315
a
Maa
a
M ????
The twisty angle per unit length is,
4
315
2 Ga
M
G
CK ???
68
剪应力
? ?32
5
5
323
2
315
)(
345
yaxx
a
M
x
yax
a
M
y
yz
xz
???
?
?
??
??
?
?
?
?
?
?
?
对于等边三角形，由薄膜比拟法知，最大剪应力发生在
0,?? yax 处，即
? ? 3225m a x 2 315232 315 a Maaa M ????
单位长度上的扭角
4
315
2 Ga
M
G
CK ???
69
Exercise 9.2 consider a pole with equal section,Its twisty angle
per unit length is K,the shear elasticity module is, A
function,where are real constants,
is a function under affirmance,Solve the question,what is the
conditions must satisfy,then can be viewed as
stress function of the torsion problems?
? ? 2co s rfAr ???? ? ??
G
??,,A
? ??f
? ???? fA,,,?
Substitute into the equation,we get
Solution,if the given function can be used for the stress
function of the torsion problems,if must satisfy the equations of
compatibility,,
GK22 ??? ? ?
? ? ? ? ? ? GKffrA 24c o s ''222 ????? ? ?????? ?
This is the condition need satisfying,? ???? fA,,,
70
习题 9.2 等截面杆单位长度上的扭转角为 K,剪切弹性
模量为,若函数,其中 为
实常量,为待定函数。试问,满足什么
条件时,可作为扭转问题的应力函数？
? ? 2co s rfAr ???? ? ??G ??,,A
? ??f ? ???? fA,,,
?
答,若题意所给函数能作为扭转问题的应力函数，则该
函数必须满足相容方程
GK22 ??? ?
将 代入后得 ?
? ? ? ? ? ? GKffrA 24c o s ''222 ????? ? ?????? ?
此即为 需满足的条件。 ? ???? fA,,,
71
72
