1
Mechanics of Materials
2
3
§ 3–1 SUMMARY
§ 3–2 EXTERNAL TORQUE OF A TRANSMISSION SHAFT INTERNAL
TORQUE AND DIAGRAMS
§ 3–3 TORSION OF THIN-WALLED HOLLOW SHAFTS
§ 3–4 STRESSES OF CIRCULAR SHAFTS IN TORSION · STRENGTH
CONDITIONS
§ 3–5 DEFORMATIONS OF A CIRCULAR SHAFT IN TORSION · RIGIDITY
CONDITIONS
§ 3–6 STATICALLY INDETERMINATE PROBLEMS OF CIRCULAR
SHAFTS IN TORSION
§ 3–7 STAIN ENERGY OF CIRCULAR SHAFTS IN TORSION
§ 3–8 STRESSES AND DEFORMATIONS OF NONCIRCULAR SHAFTS IN
FREE TORSION
§ 3–9 STRESSES OF OPENED AND CLOSED RODS WITH THIN WALLED
SECTIONS IN FREE TORSION
CHAPER 3 TORSION
4
§ 3–1 概述
§ 3–2 传动轴的外力偶矩 · 扭矩及扭矩图
§ 3–3 薄壁圆筒的扭转
§ 3–4 等直圆杆在扭转时的应力 · 强度条件
§ 3–5 等直圆杆在扭转时的变形 · 刚度条件
§ 3–6 等直圆杆的扭转超静定问题
§ 3–7 等直圆杆在扭转时的应变能
§ 3–8 非圆截面等直杆在自由扭转时的应力和变形
§ 3–9 开口和闭口薄壁截面杆在自由扭转时的应力
第三章 扭 转
5
§ 3–1 SUMMARY
Shaft,In engineering the members of which deformations are
mainly torsion,Such as transmission shafts in machines,drill rods
in oil-drilling rigs etc,
Torsion,Resultant of the external forces is a force couple and
its acting plane is perpendicular to the axis of the shaft,Under
this case the deformation of the rod is torsion,
A B O
m m
? O
B
A ?
6
§ 3–1 概 述
轴,工程中以扭转为主要变形的构件。如：机器中的传动轴,
石油钻机中的钻杆等。
扭转,外力的合力为一力偶，且力偶的作用面与直杆的轴线
垂直，杆发生的变形为扭转变形。
A B O
m m
? O
B
A ?
7
The angle of twist (?),The angle of rotation of one section
with respect to another,
Shearing strain (? ),The change of a right angle
between two straight lines,
m m
? O
B
A ?
8
扭转角（ ?）,任意两截面绕轴线转动而发生的角位移。
剪应变（ ?）,直角的改变量。
m m
? O
B
A ?
9
Practical examples in engineering
Transmission shaft
10
工
程
实
例
11
§ 3–2 EXTERNAL TORQUE OF A TRANSMISSION SHAFT
INTERNAL TORQUE AND ITS DIAGRAM
1,External torque of a transmission shaft The relation
between the transmission power,revolution and external torque
of the transmission shaft,
m)( k N559 ?? nP.m
m)( k N0 2 47 ?? nP.m
m)( k N1217 ?? nP.m
where,P - power,unit,kilowatt (kW)
n – rotational speed,unit,r/min or（ rpm）
where,P - power,unit,horsepower (HP)
n – rotational speed,unit,r/min or（ rpm）
where,P - power,unit,kilowatt (kW)
n – rotational speed,unit,r/min or（ rpm）
1PS=735.5N·m/s,1HP=745.7N·m/s,1kW=1.36PS
12
§ 3–2 传动轴的外力偶矩 · 扭矩及扭矩图
一、传动轴的外力偶矩
传递轴的传递功率、转数与外力偶矩的关系,
m)( k N559 ?? nP.m
m)( k N0 2 47 ?? nP.m
m)( k N1217 ?? nP.m
其中,P — 功率，千瓦（ kW）
n — 转速，转 /分（ rpm）
其中,P — 功率，马力（ PS）
n — 转速，转 /分（ rpm）
其中,P — 功率，马力（ HP）
n — 转速，转 /分（ rpm）
1PS=735.5N·m/s,1HP=745.7N·m/s,1kW=1.36PS
13
3) Sign conventions for internal torque,
, T”will be considered to be positive when the relation
between its turning direction and the out normal line obeys the
right hand rule,otherwise it will be considered to be negative,
2,Internal torque and its diagram
1) Internal torque,The moment of internal forces acting in arbitrary section
of the member in torsion,Designated by―T‖,
2) Determine the internal torque
by the method of section,
m m
m T
mT
mTm
x
?
???? 0,0
x
14
3 扭矩的符号规定,
, T”的转向与截面外法线方向满足右手螺旋规则为正，
反之为负。
二、扭矩及扭矩图
1 扭矩,构件受扭时，横截面上的内力偶矩，记作,T‖。
2 截面法求扭矩
m m
m T
mT
mT
m
x
?
??
??
0
0
x
15
4) Internal torque diagram,Sketch that expresses the law of
change of the torque in each cross section along the axis,
Purpose
① the law of change of the torque ；
② Value of |T|max and the position of its section
strength calculation（ critical section）
x
T
?
16
4 扭矩 图,表示沿杆件轴线各横截面上扭矩变化规律的图线。
目
的
① 扭矩变化规律；
② |T|max值及其截面位置 强度计算（危险截面）。
x
T
?
17
Example 1 A transmission shaft is rotating at n =300r/min,Knowing its
input power is P1=500kW,and its output are P2=150kW,P3=150kW，
P4=200kW,Try to plot the internal torque diagram,
n
A B C D
m2 m3 m1 m4 Solution：① Calculate
the external torque
m)15.9( kN
300
500
9.55
55.9
1
1
????
?
n
P
m
m)( k N 7843001509, 5 5559 232 ??????,nP.mm
m)( k N 3763002009, 5 5559 44 ?????,nP.m
18
[例 1]已知：一传动轴,n =300r/min，主动轮输入 P1=500kW，
从动轮输出 P2=150kW,P3=150kW,P4=200kW，试绘制扭矩
图。
n
A B C D
m2 m3 m1 m4
解：①计算外力偶矩
m)15,9( kN
30 0
50 0
9,55559 1
1
??
???
n
P
.m
m)( k N 7843001509, 5 5559 232 ??????,nP.mm
m)( k N 3763002009, 5 5559 44 ?????,nP.m
19
n
A B C D
m2 m3 m1 m4 1
1
2
2 3
3
② Determine the internal torque（ suppose it is positive）
mkN78.4
0,0
21
21
?????
????
mT
mTm
x
mkN569784784(
,0
322
322
?????????
???
.)..mmT
mmT
mkN37.6
,0
42
43
???
???
mT
mT
x
20
n
A B C D
m2 m3 m1 m4 1
1
2
2 3
3
② 求扭矩（扭矩按正方向设）
mkN78.4
0,0
21
21
?????
????
mT
mTm
x
mkN569784784(
,0
322
322
?????????
???
.)..mmT
mmT
mkN37.6
,0-
42
43
???
??
mT
mT
x
21
③ Plot the internal torque diagram
mkN 569m a x ??,T
Each section in segment BC is a critical section,
x
T
n
A B C D
m2 m3 m1 m4
4.78
9.56
6.37
?
–
–
22
③ 绘制扭矩图
mkN 569m a x ??,T
BC段为危险截面。
x
T
n
A B C D
m2 m3 m1 m4
4.78
9.56
6.37
?
–
–
23
§ 3–3 TORSION OF
THE THIN-WALLED HOLLOW SHAFTS
1,Experiment,
1),Preparation,
① Plot the longitudinal lines
and circumference lines；
② To act a pair of external
torque m。
Hollow round shaft with thin wall,
010
1 rt ? （ r0,average radius） Thickness of the wall
24
§ 3–3 薄壁圆筒的扭转
薄壁圆筒,壁厚
010
1 rt ?
（ r0,为平均半径）
一、实验,
1.实验前,
① 绘纵向线，圆周线；
②施加一对外力偶 m。
25
2,After deformation,
① The circumference lines
do not change；
② The longitudinal lines
are changed into slants 。
3.Conclusions,
① Shape,size and distance of the circumference lines on the shaft surface
do not change, while rotating with respect to one another along the axis of
the shaft。
② All the longitudinal lines revolve through a same angle ? 。
③ All squares drawn on the shaft surface warp into rhombs with same sizes。
26
2.实验后,
① 圆周线不变；
②纵向线变成斜直线。
3.结论：① 圆筒表面的各圆周线的形状、大小和间距均未改
变，只是绕轴线作了相对转动。
② 各纵向线均倾斜了同一微小角度 ? 。
③ 所有矩形网格均歪斜成同样大小的平行四边形。
27
? ?
a
c d
dx
b
dy
?′
?′
? ① No normal stress
② There are only the shearing stress
? perpendicular to the radius at each
point on the cross section,Magnitude
of the shearing stresses are the same on
the same section,directions of them
coincide with that of the internal
torque,
4.) Relation between ? and ?, ? ?
L
RS
RL
??
???
??
??
o
A small cubic element is shown in the figure,
28
? ?
a
c d
dx
b
dy
?′
?′
?
① 无正应力
②横截面上各点处，只产
生垂直于半径的均匀分布的剪
应力 ?,沿周向大小不变，方
向与该截面的扭矩方向一致。
4,? 与 ? 的关系,
?
?
L
R
RL
???
???
??
??
微小矩形单元体如图所示,
29
2,Magnitude of the shearing stress ? in the hollow shaft with thin wall,
?
?
tA
T
tr
T
TtrrArS
TrA
A
A
2 2
2d o
d
0
2
0
000
0
??
???????
???
?
?
?
?
???
?
A0,Area of the circle with an average radius
30
二、薄壁圆筒剪应力 ? 大小,
?
?
tA
T
tr
T
TtrrAr
TrA
A
A
2 2
2d
d
0
2
0
000
0
???
?????????
????
?
?
???
?
A0：平均半径所作圆的面积。
31
3,Theorem of conjugate shearing
stresses,
)( h u s,
0
aT
d x d ytd x d yt
m
z
??
??
??
??????
??
Formula (a) is called theorem of conjugate shearing stresses,
It indicates shearing stresses always exist on mutually
perpendicular plane and occur in equal and opposite pairs
and point,perpendicularly,either toward or away from the
intersection line of the planes,
a
c d
dx
b
? ? dy
?′
?′
t
z
?
32
三、剪应力互等定理,
??
??
??
??????
??
0
故
d x d ytd x d yt
m
z
上式称 为剪应力互等定理 。
该定理表明,在单元体相互垂直的两个平面上，剪应
力必然成对出现，且数值相等，两者都垂直于两平面的交
线，其方向则共同指向或共同背离该交线。
a
c d
dx
b
? ? dy
?′
?′
t
z
?
33
4,Hooke’s law of shear,
There are only shearing stresses and no normal
stresses on the four side planes of the element,which is
called pure-shear stressed state。
l
34
四、剪切虎克定律,
单元体的四个侧面上只有剪应力而无正应力作用，这
种应力状态称为 纯剪切应力状态。
l
35
?
?
T=m
?
?? ?
)( ) 2(
0 R
LtA
T
??
?
??
?
Hooke’s law in shear,The shearing stress is directly
proportional to the shearing strain if the shearing stress does
not exceed smaller than the proportional sheer limit of the
material(τ ≤τp),
36
?
?
T=m
?
?? ?
)( ) 2(
0 R
LtA
T
??
?
??
?
剪切虎克定律,当剪应力不超过材料的剪切比例极限
时（ τ ≤τp)，剪应力与剪应变成正比关系。
37
?? ?? G
In above formula G is a elastic constant of material,It is called modulus of
elasticity in shearing,as ? has no dimension,G has the same dimension with ?,
Values of G for different materials can be determined by experiments,Value of G for
steel is about 80GPa,
Modulus of elasticity in shearing,Young’s modulus and Possion’s ratio are
three elastic constants,They indicate elastic properties of materials,For the
isotropic materials there is the following relation between the three constants.（ See
later chapters about deduction of the formula）,
It is obvious that we can find out the third if any two of the three
elastic constants are known,
)1(2 ??
?
E
G
38
?? ?? G
式中,G是材料的一个弹性常数，称为剪切弹性模量，因 ? 无
量纲，故 G的量纲与 ? 相同，不同材料的 G值可通过实验确定，钢
材的 G值约为 80GPa。
剪切弹性模量、弹性模量和泊松比是表明材料弹性性质的三
个常数。对各向同性材料，这三个弹性常数之间存在下列关系
（推导详见后面章节）,
可见，在三个弹性常数中，只要知道任意两个，第三个量
就可以推算出来。
)1(2 ???
EG
39
§ 3–4 STRESSES OF CIRCULAR SHAFTS IN
TORSION · STRENGTH CONDITIONS
Stress on the cross
section of the round shaft
① Geometric deformations
② Physical relations
③ Statics
1),After deformation the cross
sections still remain planes;
2),No elongation or contraction
along the axis；
3),Longitudinal lines are still
parallel to each other,
1,Observation of the
torsional test of the
straight circular rod,
40
§ 3–4 等直圆杆在扭转时的应力 · 强度条件
等直圆杆横截面应力
① 变形几何方面
②物理关系方面
③静力学方面
1,横截面变形后
仍为平面；
2,轴向无伸缩；
3,纵向线变形后仍为平行。
一、等直圆杆扭转实验观察,
41
2,Stresses on the cross section of the round shaft in torsion,
1),Geometric deformation
relation,
xx
GG
d
d
dtg
1 ????
??
?????
xd
d ???
? ?
The shearing strain ?? at an arbitrary point is
directly proportional to the distance ? from the
center of the section to the point,
xd
d? —— of the variation rate twisting angle along the length,
42
二、等直圆杆扭转时横截面上的应力,
1,变形几何关系,
xx
GG
d
d
dtg
1 ????
??
?????
xd
d ???
? ?
距圆心为 ? 任一点处的 ??与该点到圆心的距离 ?成正比。
xd
d?
—— 扭转角沿长度方向变化率。
43
2),Physical relation,
Hooke’s law,
Substituting it into the preceding formula we get,
?? ?? G
xGxGG d
d
d
d ??????
?? ??????
xG d
d ???
? ?
44
2,物理关系,
虎克定律,
代入上式得,
?? ?? G
xGxGG d
d
d
d ??????
?? ??????
xG d
d ???
? ?
45
3),Static relation,
A
x
GA
x
G
AT
AA
A
d
d
d
d
d
d
d
22
?
??
?
?? ?
????
????
AI Ap d2???
Let
xGI T p d
d ??
pGI
T
x ? d
d ?
After substituting into the physical relation
xG d
d ???
? ?
pI
T ??
?
??we get
O
τp ?
dA
46
3,静力学关系,
A
x
G
A
x
G
AT
A
A
A
d
d
d
d
d
d
d
2
2
?
?
?
?
??
?
??
??
????
AI Ap d2???
令
xGI T p d
d ??
pGI
T
x ? d
d ?
代入物理关系式 得,
xG d
d ???
? ?
pI
T ??
?
??
O
τp ?
dA
47
pI
T ??
?
?? —Formula to calculate the shearing stress of any point in the
section with distance ? to the center,
4),Discussions on the formula,
① It can be only applied to the isotropic,linear-elastic straight-
circular rods with small deformations,
② In the above formula,T—Torque in the cross section,It is
determined from the external torques by the method of section,
? — Distance of the point to the center of the circle,
Ip—Polar moment of inertia of section,It is a purely
geometric quantity without any physical meaning,
48
pI
T ??
?
?? —横截面上距圆心为 ?处任一点剪应力计算公式。
4,公式讨论,
① 仅适用于各向同性、线弹性材料，在小变形时的等圆截面
直杆。
② 式中,T—横截面上的扭矩，由截面法通过外力偶矩求得。
? —该点到圆心的距离。
Ip—截面极惯性矩，纯几何量，无物理意义。
49
unit,mm4,m4。
AI Ap d2???
③ The formula is deduced from the solid circular shaft,but it
can also be applied to the hollow circular shaft,Values of their Ip
are different,
4
4
2
0
2
2
10
32
d2
d
D.
D
AI
D
Ap
??
? ????
??
?
????
?
a,For a solid circular section,
D ?
d?
O
50
单位,mm4,m4。
AI Ap d2???
③ 尽管由实心圆截面杆推出，但同样适用于空心圆截面杆,
只是 Ip值不同。
4
4
2
0
2
2
10
32
d2
d
D.
D
AI
D
Ap
??
? ????
??
?
????
?
a,对于实心圆截面,
D ?
d?
O
51
b,For a hollow circular section,
)1(10)1(
32
)(
32
d2
d
444
4
44
2
2
2
2
??
?
?
????
?
????
??
?
????
??
D.
D
dD
AI
D
d
Ap
)( Dd??
d D O
?
d?
52
b,对于空心圆截面,
)1(10)1(
32
)(
32
d2
d
444
4
44
2
2
2
2
??
?
?
????
?
????
??
?
????
??
D.
D
dD
AI
D
d
Ap
)( Dd??
d D O
?
d?
53
④ Distribution of the shearing stresses
（ solid section） （ hollow section）
In engineering the members with hollow section are widely
used to increase the strength,save materials and decrease the
weight of structures,
54
④ 应力分布
（实心截面） （空心截面）
工程上采用空心截面构件：提高强度，节约材料，重量轻,
结构轻便，应用广泛。
55
⑤ Determine the maximum shearing stress,
pI
T ??
?
??From
when
m a x,2 ??? ? ???
dR
)
2
e t (
2
2
h e n
m a x
d
IWl
W
T
d
I
T
I
d
T
T
p
t
p
p
???
?
??
tW
T?
m a x?
Wt — section factor in torsion（ section
factor modulus in torsion ),geometric
quantity,unit,mm3 or m3
a,For the solid circular section,
33 2016 D.DRIW
pt ??? ?
b,For the hollow circular section,
)-(12016)1( 4343 ??? D.DRIW pt ????
56
⑤ 确定最大剪应力,
pI
T ??
?
??
由 知：当
m a x,2 ??? ? ???
dR
)
2
(
2
2
m a x
d
IW
W
T
d
I
T
I
d
T
p
t
p
p
???
?
?? 令?
tW
T?
m a x?
Wt — 抗扭截面系数（抗扭截面模量）,
几何量，单位,mm3或 m3。
对于实心圆截面,
33 2016 D.DRIW
pt ??? ?
对于空心圆截面,
)-(12016)1( 4343 ??? D.DRIW pt ????
57
3,Stress on the inclined plane of the round shaft in torsion
Lower-carbon steel
specimen,
Rupture along the cross
section,
Cast-iron specimen,
Rupture along the helix
direction with an
approximate inclined
angle 45?,
So the stresses on the inclined plane must be analyzed,
58
三、等直圆杆扭转时斜截面上的应力
低碳钢试件,
沿横截面断开。
铸铁试件,
沿与轴线约成 45?的
螺旋线断开。
因此还需要研究斜截面上的应力。
59
1,Stressed element at point M
is shown in Fig.(b),
(a)
M
(b)
′ ′
(c)
2,Stresses on the inclined
section； Take a free body as
shown in Fig.(d),
(d)
?
′
?
x
60
1,点 M的应力单元体如图 (b),
(a)
M
(b)
′ ′
(c)
2,斜截面上的应力；
取分离体如图 (d),
(d)
?
′
?
x
61
(d)
?
′
?
x
n
t
Conventions for the angle of rotation,
Positive direction of the axis
turns to the out normal line of section
countclockwise：,+‖
clockwise：,–‖
From the equilibrium equations,
0) c o ss i nd() s i nc o sd(d ; 0 ?????? ??????? ? AAAF n
0) s i ns i nd() c o sc o sd(d ; 0 ?????? ??????? ? AAAF t
?? ??
We have,
?????? ?? 2c o s ; 2s i n ???
62
(d)
?
′
?
x
n
t
转角 α规定,
x轴正向转至截面外法线 逆时针：为,+‖
顺时针：为,–‖
由平衡方程,
0) c o ss i nd() s i nc o sd(d ; 0 ?????? ??????? ? AAAF n
0) s i ns i nd() c o sc o sd(d ; 0 ?????? ??????? ? AAAF t
?? ??
解得,
?????? ?? 2c o s ; 2s i n ???
63
?????? ?? 2c o s ; 2s i n ???
Analysis,As ? = 0°,
???? ??? ?? m a x00,0
As ? = 45°,
0,45m i n45 ???? ?? ????
As ? = – 45°,
0,45m a x45 ??? ???? ????
As ? = 90°,
???? ????? ?? m a x9090,0
′
45°
It is found that for a circular shaft in
torsion the maximum shearing stress occurs on
transverse and longitudinal sections,and that the
maximum tensile and compressive normal
stresses occur on the inclined sections oriented at
? = ? 45?,From these conclusions we can
explain the forementioned rupture phenomenon,
64
?????? ?? 2c o s ; 2s i n ???
分析,当 ? = 0° 时,
???? ??? ?? m a x00,0
当 ? = 45° 时,
0,45m i n45 ???? ?? ????
当 ? = – 45° 时,
0,45m a x45 ??? ???? ????
当 ? = 90° 时,
???? ????? ?? m a x9090,0
′
45°
由此可见：圆轴扭转时，在横截
面和纵截面上的剪应力为最大值；在
方向角 ? = ? 45?的斜截面上作用有最
大压应力和最大拉应力。根据这一结
论，就可解释前述的破坏现象。
65
4,Strength calculation of the round shaft in torsion
Strength condition,
For the round shaft
with equal sections,
][m a x ?? ?
][m a x ??
tW
T
([?] is called permissible
shearing stress.)
Three aspects of strength calculations,
① Check the strength,
② Design the dimension
of the section,
③ Calculate the permissible load,
][m a xm a x ?? ??
tW
T
][
m a x
?
TW
t ?
][m a x ?tWT ?
?
?
?
?
?
?
?
?
?
?
? ）（
4
3
3
1
16
16
?
?
?
D
h o l l o w
Ds o l i d
W
t
66
四、圆轴扭转时的强度计算
强度条件,
对于等截面圆轴,
][m a x ?? ?
][m a x ??
tW
T
([?] 称为许用剪应力。 )
强度计算三方面,
① 校核强度,
② 设计截面尺寸,
③ 计算许可载荷,
][m a xm a x ?? ??
tW
T
][
m a x
?
TW
t ?
][m a x ?tWT ?
??
?
?
?
??
?
?
?
? ）（空：
实：
4
3
3
1
16
16
?
?
?
D
D
W
t
67
Example 2 Power is 150kW,speed of rotation is 15.4r/s for the rotator of
a motor as shown in the figure.the Permissible shearing stress is [?]=30MPa,Try
to check its strength,
n
NmT
BC ?2
10 3???
m)( k N551
m)(N
4151432
10150
3
??
?
??
?
?
.
..
T
m
Solution,① Determine its
torques and plot the torque diagram
② Calculate and check the
shearing stress strength
③ Strength of this shaft satisfies request
D3 =135 D2=75 D1=70
A B C
m m
x
][M P a23
16070
10551
3
3
m a x ??? ???
???
.
.
W
T
t
68
[例 2] 功率为 150kW，转速为 15.4转 /秒的电动机转子轴如图,
许用剪应力 [?]=30M Pa,试校核其强度。
n
NmT
BC ?2
10 3???
m)( k N551
m)(N
4151432
10150
3
??
?
??
?
?
.
..
T
m
解：①求扭矩及扭矩图
② 计算并校核剪应力强度
③ 此轴满足强度要求。
D3 =135 D2=75 D1=70
A B C
m m
x
][M P a23
16070
10551
3
3
m a x ??? ???
???
.
.
W
T
t
69
§ 3–5 DEFORMATION OF A CIRCULAR SHAFT IN TORSION
RIGIDITY CONDITIONS
1,Deformation in torsion
From the formula
pGI
T
x ? d
d ?
We can determine the relative angle ? of twist
between two end sections of the rod with the length l,
）c o n s t a n t is ( i f d d
0
T
GI
Tl
x
GI
T
p
l
p
???? ???
70
§ 3–5 等直圆杆在扭转时的变形 · 刚度条件
一、扭转时的变形
由公式
pGI
T
x ? d
d ?
知,长为 l一段杆两截面间相对扭转角 ? 为
值不变）若 (
d d
0
T
GI
Tl
x
GI
T
p
l
p
?
???? ??
71
2,The angle of twist per unit length?,
( r a d / m )
d
d
pGI
T
x
??
?
? / m )(
1 8 0
d
d ????
?
??
pGI
T
x
or
3,Rigidity condition
? ? ( r a d / m ) m a xm a x ?? ??
pGI
T
? ? / m )(
1 8 0
m a xm a x ???? ?
?
?
pGI
T
or
G Ip indicates the capacity that the dimension and material
properties of the section resist torsional deformations,It is called
the torsional rigidity of the circular shaft,
[? ] is called the permissible angle of twist per unit length of the shaft,
72
二、单位长度扭转角 ?,
( r a d / m ) dd
pGI
T
x ??
??
/ m )( 1 8 0 dd ???? ???
pGI
T
x
或
三、刚度条件
或
GIp反映了截面尺寸和材料性能抵抗扭转变形的能力，称为
圆轴 的抗扭刚度 。
[? ]称为许用单位长度扭转角。
? ? ( r a d / m ) m a xm a x ?? ??
pGI
T
? ? / m )( 1 8 0 m a xm a x ???? ?
?
?
pGI
T
73
Three aspects of rigidity calculations,
① Check the rigidity,
② Design the dimension
of the section,
③ Calculate the
permissible load,
? ? m a x ?? ?
] [
m a x
?G
T
I p ?
] [ m a x ?pGIT ?
Sometimes,these conditions may be considered to select materials.,
74
刚度计算的三方面,
① 校核刚度,
② 设计截面尺寸,
③ 计算许可载荷,
? ? m a x ?? ?
] [
m a x
?G
T
I p ?
] [ m a x ?pGIT ?
有时，还可依据此条件进行选材。
75
Example 3 Uniformly distributed couple m=20Nm/m acted a circular shaft
with length L=2m is shown in the figure,If the ratio of the inside diameter and the
outside diameter of the shaft is ? =0.8,G=80GPa,the permissible shearing
stress is [?]=30MPa,Try to design the outside diameter of the shaft； As
[?]=2o/m,try to check the rigidity of the shaft and determine the angle of rotation
of the right-end section,
Solution,① Design the outside diameter of the rod
][
m a x
?
TW
t ?
116D 4
3
）（ ?? ??tW
3
1
4
m a x
][ 1
16
?
?
?
?
?
?
?
?
??? ）（
T
D
76
[例 3]长为 L=2m 的圆杆受均布力偶 m=20Nm/m 的作用，如图，
若杆的内外径之比为 ? =0.8, G=80GPa,许用剪应力
[?]=30MPa，试设计杆的外径；若 [?]=2o/m,试校核此杆的刚
度，并求右端面转角。
解,①设计杆的外径
][
m a x
?
TW
t ?
116D 4
3
）（ ?? ??tW
3
1
4
m a x
][ 1
16
?
?
?
?
?
?
?
?
??? ）（
T
D
77
3
1
4
m a x
][ 1
16
?
?
?
?
?
?
?
?
??? ）（
T
D
40Nm
x
T
Substituting the values we
get,
D ? 0.0226m。
② Check the rigidity according
to the rigidity condition in
torsion
??
1 8 0m a x
m a x ??
PGI
T
78
3
1
4
m a x
][ 1
16
?
?
?
?
?
?
?
?
??? ）（
T
D
40Nm
x
T
代入数值得,
D ? 0.0226m。
② 由扭转刚度条件校核刚度
??
1 8 0m a x
m a x ??
PGI
T
79
40Nm
x
T
??
1 8 0m a x
m a x ??
PGI
T ? ?
?
??
??
???
??? m
D
/89.1
)1(1080
1804032
4429
?
③ Angle of twist of the right-end section,
）ad( 033.0
)4(
102040 2
0
2
2
00
r
xx
GI
dx
GI
x
dx
GI
T
PP
L
P
?
??
?
?? ???
80
40Nm
x
T
??
1 8 0m a x
m a x ??
PGI
T ? ??
??
??
???
??? ?891
11080
1804032
4429,)(D
③ 右端面转角 为,
弧度）( 0330
4
102040 2
0
22
00
.
)xx(
GI
dx
GI
x
dx
GI
T
PP
L
P
?
???
?
????
81
Example 4 It is Required that the speed of rotation of the shaft is n = 500 r /
min,input power is N1 = 500 HP,output power is respectively N2 = 200HP
and N3 = 300HP in the design of a transmission shaft,Knowing,G=80GPa,
[? ]=70MPa,[? ]=1o/m,Try to determine,① Diameter d1 of segment AB and
diameter d2 of segment BC？ ② If select the same diameter for the whole shaft,
how much is the diameter？ ③ How to
arrange the driver and the driven pulleys
is reasonable？
Solution,① Torque diagram is
shown in the lower figure of the
shaft for the structure shown in the
upper figure,From the strength
condition we get,
500 400
N1 N3 N2
A C B
T x
–7.024 – 4.21
(kNm)
m)( k N0247 ?? nN.m
82
[例 4] 某传动轴设计要求转速 n = 500 r / min，输入功率 N1 = 500
马力,输出功率分别 N2 = 200马力及 N3 = 300马力，已知：
G=80GPa, [? ]=70M Pa,[? ]=1o/m,试确定,
① AB 段直径 d1和 BC 段直径 d2？
② 若全轴选同一直径，应为多少？
③ 主动轮与从动轮如何安排合理？
解,① 图示状态下,扭矩如
图,由强度条件得,
500 400
N1 N3 N2
A C B
T x
–7.024 – 4.21
(kNm)
m)( k N0247 ?? nN.m
83
][16
31
?
? TdW
t ??
? ? mm4671070143
42101616
3
632,.
Td ?
??
?????
??
] [ 32
4
?
?
G
TdI
p ??
? ? mm801070143
7 0 2 41616
3
6
31 ?
??
?????
.
Td
??
From the rigidity condition,
500 400
N1 N3 N2
A C B
T
x
–7.024 –4.21
(kNm)
84
][16
31
?
? TdW
t ??
? ? mm4671070143
42101616
3
632,.
Td ?
??
?????
??
] [ 32
4
?
?
G
TdI
p ??
? ? mm801070143
7 0 2 41616
3
6
31 ?
??
?????
.
Td
??
由刚度条件得,
500 400
N1 N3 N2
A C B
T
x
–7.024 –4.21
(kNm)
85
mm474
11080143
180421032
] [
32
4
9242,.G
Td ?
???
??????
??
mm84
11080143
180702432
] [
32
4
9241 ????
???????
.G
Td
??
? ? ? ? mm75 mm85 21 ?? d,dSum up,
② If select the same diameter for
the whole shaft ? ? ? ? mm851 ?? dd
86
mm474
11080143
180421032
] [
32
4
9242,.G
Td ?
???
??????
??
mm84
11080143
180702432
] [
32
4
9241 ????
???????
.G
Td
??
? ? ? ? mm75 mm85 21 ?? d,d综上,
② 全轴选同一直径时 ? ? ? ?
mm851 ?? dd
87
③ the smaller the torque with the maximum absolute value in the shaft is,the more
reasonable it is,So wheel 1 must be replaced with wheel 2,After transposition,
torque diagram of the shaft is shown in the figure,Under this case,the maximum
diameter of the shaft is 75mm。
T
x
– 4.21
(kNm)
2.814
88
③ 轴上的 绝对值 最大 的 扭矩 越小越 合理，所以,1轮和 2轮应
该 换 位。 换位后,轴的扭矩如图所示,此时,轴的最大直径才
为 75mm。
T
x
– 4.21
(kNm)
2.814
89
§ 3–6 STATICALLY INDETERMINATE PROBLEMS OF
ROUND SHAFT IN TORSION
Method and Steps to solve the statically indeterminate
problems in torsion,
Write out the equilibrium equations；
Write out the geometric equation—compatibility equation of
deformation；
Find out the complementary equations when combined
geometric equations and physical equations；
Write out the physical equations；
Solve the combined equations including equilibrium equations
and complementary equations,
90
§ 3–6 等直圆杆的扭转超静定问题
解决扭转超静定问题的方法步骤,
平衡方程；
几何方程 ——变形协调方程；
补充方程：由几何方程和物理方程得；
物理方程；
解由平衡方程和补充方程组成的方程组。
91
Example 5 A circular rod is subjected to uniformly distributed a force couple as
shown in the figure,Length of the rod is L=2m,If the ratio between the inside
and outside diameter is ? =0.8,and outside diameter of the rod is D=0.0226m，
G =80GPa,Try to determine the constraint couple in the fixed ends,
Solution,① Free body
diagram of the rod is shown
in the figure,It is a one-time
statically indeterminate
problem,Its equilibrium
equation is,
02 ??? BA mmm
A B
92
[例 5]长为 L=2m 的圆杆受均布力偶 m=20Nm/m 的作用，如图，
若杆的内外径之比为 ? =0.8,外径 D=0.0226m, G=80GPa，
试求固定端反力偶。
解, ①杆的受力图如图示,
这是一次超静定问题。
平衡方程为,
02 ??? BA mmm
A B
93
② Geometric equation—compatibility equation of deformation
0?BA?
③ Summing up the physical equation and the geometric equation we can get
the complementary equation,
0
402202
00
?
?
?
?
?? ??
P
A
P
A
L
P
BA GI
m
dx
GI
xm
dx
GI
T?
mN 20 ??? Am
④ From the equilibrium equation and the complementary equation we get,
Additionally,this problem can be directly solved by using symmetry,
mN 20 ??Bm
94
② 几何方程 ——变形协调方程
0?BA?
③ 综合物理方程与几何方程，得补充方程,
040220200 ???? ????
P
A
P
AL
P
BA GI
mdx
GI
xmdx
GI
T?
mN 20 ??? Am
④ 由平衡方程和补充方程得,
另,此题可由对称性直接求得结果。
mN 20 ??Bm
95
§ 3–7 STAIN ENERGY OF ROUND SHAFTS IN TORSION
?? G?
dV)dx(d z d y )(dW ???? 2121 ??
2
2
1
2
1
d
d
d
d ??? G
V
W
V
Uu ????
1,Strain energy and its density
a
c d
dx
b
? ? dy
?′
?′
dz
z
?
x
y Infinitesimal work of the element,
Strain energy per unit volume,
96
§ 3–7 等直圆杆在扭转时的应变能
?? G?
dV)dx(d z d y )(dW ???? 2121 ??
2
2
1
2
1
d
d
d
d ??? G
V
W
V
Uu ????
一,应变能与能密度
a
c d
dx
b
? ? dy
?′
?′
dz
z
?
x
y 单元体微功,
应变比能,
97
2,Calculation of the stress in cylindric close-coiled helical
springs
1),Calculation of the stress
= +
? Q ?
T
Q
T
t
TQ W
T
A
Q ???? ???
m a x
Approximate value,
323
8
1
2
416
2
d
DP
D
d
d
P
d
PD
???
?
?
?
?
?
?
????
P
Q
T
98
二、圆柱形密圈螺旋弹簧的计算
1,应力的计算
= +
? Q ?
T
Q
T
t
TQ W
T
A
Q ???? ???
m a x
近似值,
323
8
1
2
416
2
d
DP
D
d
d
P
d
PD
???
?
?
?
?
?
?
????
P
Q
T
99
2) Strength condition of the spring wire,
? ??
?
? ?? 8 3
d
DPK
m a x
Exact value：（ Revised formula is to considered the effect of the curvature
and shearing stress of the spring）
33m a x
886150
44
14
d
DPK
d
DP
C
.
C
C
??
? ??
?
??
?
? ?
?
??
where,
d
DC ?
C
.
C
CK 6 1 50
44
14 ?
?
??
Is called the spring index,
is called the coefficient of
curvature,
100
2,弹簧丝的强度条件,
? ??
?
? ?? 8 3
d
DPK
m a x
精确值：（修正公式，考虑弹簧曲率及剪力的影响）
33m a x
886150
44
14
d
DPK
d
DP
C
.
C
C
??
? ??
?
??
?
? ?
?
??
其中,
d
DC ?
C
.
C
CK 6 1 50
44
14 ?
?
??
称为弹簧指数。
称为曲度系数。
101
3).Calculation of the displacement(energy method）
。c o n s t a n t s p r i n g t h es
64;
64
;
3
4
4
3
i
nR
Gd
K
K
P
Gd
nPR
UW
?????
?
?PW 21 ?
Work of the external force,
Strain energy,
AL
I
T
G
VUU
pVVV
d
2
1
d
2
1
d
2
?
?
?
?
?
?
??????
?
??
? ?
pp GI
PRRn
L
I
T
G 2
2
2
1 22 ???
102
3.位移的计算 (能量法）
为弹簧常数。
64; 64 ; 3
4
4
3
nR
GdK
K
P
Gd
nPRUW ????? ?
?PW 21 ?
外力功,
变形能,
AL
I
T
G
VUU
pVVV
d
2
1
d
2
1
d
2
?
?
?
?
?
?
??????
?
??
? ?
pp GI
PRRn
L
I
T
G 2
2
2
1 22 ???
103
Example 6 Mean diameter of a close-coiled helical spring is
D=125mm,wire diameter is d =18mm,It is subjected to a couple
of tensile forces with magnitude P=500N,Try to determine the
exact value and approximate value of the maximum shearing stress；
If G =82GPa,in order to make the deformation of the spring be
6mm,how many turns should the spring have at least？
Solution,① Approximate value of the maximum shearing stress,
M P a329
0180
50012508
1
1252
18
8
1
2
3
3m a x
.
.
.
)(
d
DP
)
D
d
(
?
?
??
?
?
?
??
?
?
?
104
[例 6] 圆柱形密圈螺旋弹簧的平均直径为,D=125mm，簧丝直
径为,d =18mm，受拉力 P=500N 的作用,试求最大剪应力
的近似值和精确值；若 G =82GPa，欲使弹簧变形等于 6mm,
问：弹簧至少应有几圈？
解,① 最大剪应力的近似值,
M P a329
0180
50012508
1
1252
18
8
1
2
3
3m a x
.
.
.
)(
d
DP
)
D
d
(
?
?
??
?
?
?
??
?
?
?
105
② Exact value of the maximum shearing stress,
091615044 14 ; 631518125,C.CCK.dDC ??????? ?
M P a2330180 500125080918 33m a x,...dDPK ?? ????? ???
③ The number of turns of the spring,
66
1 2 505 0 064
1018826
64 3
64
3
4
.
.PR
Gdn ?
??
????? ??
（ turns）
106
② 最大剪应力的精确值,
091615044 14 ; 631518125,C.CCK.dDC ??????? ?
M P a2330180 500125080918 33m a x,...dDPK ?? ????? ???
③ 弹簧圈数,
66
1 2 505 0 064
1018826
64 3
64
3
4
.
.PR
Gdn ?
??
????? ??
（圈）
107
§ 3–8 STRESSES AND DEFORMATIONS OF NONCIRCULAR
SHAFTS IN FREE TORSION
Noncircular straight shaft,Here the assumptions of plane can not be
applied,That is,cross-sections of the rod do not remain plane due to warping,
Therefore the preceding formulas about stress and deformation for the straight
circular shaft can not be applied either,This kind of problem must be solved by
the method in elastic mechanics,
108
§ 3–8 非圆截面等直杆在自由扭转时的应力和变形
非圆截面等直杆,平面假设不成立。即各截面发生翘曲不保
持平面。因此，由等直圆杆扭转时推出的应力、变形公式不
适用，须由弹性力学方法求解。
109
1,Free torsion,Warping of each cross section of the rod in torsion
is not restricted and the magnitude is completely same.,
2,Restricted torsion,Warping of each cross section of the rod in
torsion is restricted and the magnitude is different.,
3,shearing stress on the section of the
rectangular rod,
h ?
b
h
? 1 T
? max
Attention！ b
1),Distribution of the shearing
stress is shown in the figure,
（ Corner point,center of the section,
middle points of the long and short sides）
110
一,自由扭转,杆件扭转时，横截面的翘曲不受限制，任意两相
邻截面的翘曲程度完全相同。
二,约束扭转,杆件扭转时，横截面的翘曲受到限制，相邻截面
的翘曲程度不同。
三,矩形杆横截面上的剪应力,
h ?
b
h
? 1 T
? max
注意！ b
1,剪应力分布如图,
（角点、形心、长短边中点）
111
2),Maximum shearing stress and angle of twist per unit length
m a x1 ??? ?
h ?
b
h
? 1 T
? max
attention！ b
m a xm a x
tW
T
??
3 b
tW ??
where,
4 bI
t ??
,
tGI
T??
where,It—equivalent polar
moment of inertia,
112
2,最大剪应力及单位扭转角
m a x1 ??? ?
h ?
b
h
? 1 T
? max
注意！ b
m a xm a x
tW
T
?? 3 btW ??
其中,
4 bI
t ??
,
tGI
T??
其中,It—相当极惯性矩。
113
hbtWw
W
T
t
2m a x
m a x,h e r e ?? ??
Attention！ There are the following definitions for W t and It in most
teaching books and engineering handbooks,
hbIw
GI
T
t
t
3,h e r e,?? ??
m a x1 ??? ?
3
1; ) 10 ( ??? ??
b
h
In reference to the table about ? and ? we must pay attention to
which expression the ? and ? are corresponding to,
h ?
b
h
? 1 T
? max
Attention！ b For the narrow rectangular sections
114
hb
t
W
W
T
t
2m a x
m a x, ?? ?? 其中
注意！ 对于 W t 和 It, 多数教材与手册上有如下定义,
hbIGI T t
t
3,,?? ?? 其中
m a x1 ??? ?
3
1 ; ) 10
, ( ??? ??bh即对于狭长矩形
查表求 ? 和 ? 时一定要 注意，表中 ? 和 ? 与那套公式对应。
h ?
b
h
? 1 T
? max
注意！ b
115
Example 7 For a straight steel rod with a rectangular section,its
dimensions of the cross section are h = 100 mm,b=50mm,its length is L=2m，
External torques acting on two ends of the rod are T=4000N·m,G =80GPa,
[?]=100MPa,[?]=1o/m, Try to check the strength and rigidity of the rod,
Solution,① To reference to the table to determine ?, ?
② Check the strength
0, 4 9 3 ; 0, 4 5 7 ; 250100 ????? ??bh
m106610504930 3633 ??????,..btW ?
116
[例 8] 一矩形截面等直钢杆，其横截面尺寸为,h = 100 mm,
b=50mm，长度 L=2m，杆的两端受扭转力偶 T=4000N·m 的
作用,钢的 G =80GPa, [?]=100M Pa,[?]=1o/m,试校核
此杆的强度和刚度。
解,①查表求 ?, ?
② 校核强度
0, 4 9 3 ; 0, 4 5 7 ; 250100 ????? ??bh
m106610504930 3633 ??????,..btW ?
117
③ Check the rigidity
? ??? ??
?
?? ? M P a65
10661
4000
6
m a x
m a x,W
T
t
4844 m102840504570 ??????,.bI
t ?
? ??? ???
???
?? ? /m1r a d / m017450
102861080
4000 o
89,GI
T
t
To sum up,the rod satisfies the requests in strength and rigidity,
118
③ 校核刚度
? ??? ??
?
?? ? M P a65
10661
4000
6
m a x
m a x,W
T
t
4844 m102840504570 ??????,.bI
t ?
? ??? ???
???
?? ? /m1r a d / m017450
102861080
4000 o
89,GI
T
t
综上,此杆满足强度和刚度要求。
119
1,The direction of the shearing stress current
coincides with that of the internal torque,
2 Distribution of the
shearing stress of the
opened thin-walled tubes
in free torsion is shown in
the figure（ a）,The
shearing stress at each
middle point of thickness
is zero,
§ 3–9 STRESSES OF OPENED AND CLOSED RODS WITH
THIN-WALLED SECTIONS IN FREE TORSION
Fig,
120
一、剪应力流的方向与扭矩的方向一致。
二、开口薄壁截面杆在自由扭转时的剪应力分布如图（ a），厚
度中点处，应力为零。
§ 3–9 开口和闭口薄壁截面杆在自由扭转时的应力
121
3,Distribution of the shearing stress of the closed thin-walled
tubes in free torsion is shown in the figure（ b）,Shearing
stresses distribute uniformly in the same thickness,
Fig,
122
三、闭口薄壁截面杆在自由扭转时的剪应力分布如图（ b），同
一厚度处，应力均匀分布。
123
4,Calculation of the shearing stress of the closed thin-walled
tubes in free torsion,Take an element in the figure （ c） as
shown in the figure（ d）。
Fig.（ c）
? 1
? 2
Fig.（ d）
2211 d d ; 0 ???? ????? xxX
co n s t a n t 2211 ?? ????
124
四、闭口薄壁截面杆自由扭转时的剪应力计算，在（ c）图上取
单元体如图（ d）。
图（ c）
? 1
? 2
图（ d）
2211 d d ; 0 ???? ??? ??? xxX
常量??? 2211 ????
125
??
?
m i n
m a x 2
T?
? ? ???????
?
22 ????? d)ds(T
2
1
?? ??
?
??? dds
?
is the area of the circle at average radius
126
??
?
m i n
m a x 2
T?
? ? ???????
?
22 ????? d)ds(T
积。为厚度中线所包围的面
2
1
????
?
??? dds
?
127
Example 8 The elliptical cross-section of a closed thin-walled tube is shown
in the following figure,Its dimensions are a=50 mm,b=75mm,the thickness
is t =5mm,the external torque acting on the two ends of the tube is T=5000N·m,
Try to determine the maximum shearing stress,
Solution,The maximum shearing stress of
the closed thin-walled tube in free torsion,
b
a t
M P a42
10755052
5 0 0 0
22
9
m i n
m a x
?
????
?
??
?
?
???
?
abt
TT
128
[例 8]下图示椭圆形薄壁截面杆，横截面尺寸为,a=50 mm，
b=75mm，厚度 t =5mm，杆两端受扭转力偶 T=5000N·m,试
求此杆的最大剪应力。
解,闭口薄壁 杆自由扭转时的最大剪应力,
b
a t
M P a42
10755052
5000
22
9
m i n
m a x
?
????
?
??
?
?
???
?
abt
TT
Chapter 3 Exercises
1,In a gear-box,why is the shaft with a slow speed thicker than that with
a height speed？
2,When the shearing stress and the normal stress coexist in an element,is
the theorem of the conjugate shearing stress amenable? Why?
3,For a hollow circular tube of Al,the outside diameter is D＝ 100mm，
the inside diameter is d＝ 80mm,and the length is L＝ 2.5m,The Shearing
elastic modulus is G＝ 28GPa,
① If a force couple is acted on both ends of the circular tube and results
in the pure torsion,try to determine the torsional angle when the maximum
shearing stress is 50MPa,
② For the solid shaft of Al subjected to the same force couple and having
the same maximum shearing stress,what is the diameter D2?
③ Determine the ratio of the weight of the hollow tube to that of the solid
shaft,
130
第三章 练习题
一、在变速箱中，为什么低速轴比高速轴粗？
二、当单元体上同时存在剪应力和正应力时，剪
应力互等定理是否成立？为什么？
三、铝制空心圆管，外径 D＝ 100mm，内径 d＝
80mm，长度 L＝ 2.5m。铝的剪切弹性模量 G＝ 28GPa。
①若圆管两端受力偶矩作用产生纯扭转，试求当
最大剪应力为 50MPa时的扭转角。
②对于承受相同力偶矩并产生相同最大剪应力的
铝制实心轴，其直径 应为多大？
③求空心管与实心轴的重量之比。
2D
131
Solution,①
②
③ As the two have the same material and the same length,
the ratio of weights is equal to the ratio of cross-section areas,
??
?
?????? 12.5
1 0 028
5.25022 m a xm a x
GD
L
GI
LW
GI
TL
p
t
p
???
tt W
T
W
T ?
1
)1(1616 4
33
1 ??? ?? DD?
mmDD 9.838.011 0 01 3 43 41 ????? ?
51.0
9.83
801 0 0
4
/)(
4 2
22
2
1
22 ???? DdD ??
132
解,①
②
③ 两者材料、长度相同，重量之比等于横
截面面积之比,
??
?
?????? 12.5
1 0 028
5.25022 m a xm a x
GD
L
GI
LW
GI
TL
p
t
p
???
tt W
T
W
T ?
1
)1(1616 4
33
1 ??? ?? DD?
mmDD 9.838.011 0 01 3 43 41 ????? ?
51.0
9.83
801 0 0
4
/)(
4 2
22
2
1
22 ???? DdD ??
133
134
Mechanics of Materials
2
3
§ 3–1 SUMMARY
§ 3–2 EXTERNAL TORQUE OF A TRANSMISSION SHAFT INTERNAL
TORQUE AND DIAGRAMS
§ 3–3 TORSION OF THIN-WALLED HOLLOW SHAFTS
§ 3–4 STRESSES OF CIRCULAR SHAFTS IN TORSION · STRENGTH
CONDITIONS
§ 3–5 DEFORMATIONS OF A CIRCULAR SHAFT IN TORSION · RIGIDITY
CONDITIONS
§ 3–6 STATICALLY INDETERMINATE PROBLEMS OF CIRCULAR
SHAFTS IN TORSION
§ 3–7 STAIN ENERGY OF CIRCULAR SHAFTS IN TORSION
§ 3–8 STRESSES AND DEFORMATIONS OF NONCIRCULAR SHAFTS IN
FREE TORSION
§ 3–9 STRESSES OF OPENED AND CLOSED RODS WITH THIN WALLED
SECTIONS IN FREE TORSION
CHAPER 3 TORSION
4
§ 3–1 概述
§ 3–2 传动轴的外力偶矩 · 扭矩及扭矩图
§ 3–3 薄壁圆筒的扭转
§ 3–4 等直圆杆在扭转时的应力 · 强度条件
§ 3–5 等直圆杆在扭转时的变形 · 刚度条件
§ 3–6 等直圆杆的扭转超静定问题
§ 3–7 等直圆杆在扭转时的应变能
§ 3–8 非圆截面等直杆在自由扭转时的应力和变形
§ 3–9 开口和闭口薄壁截面杆在自由扭转时的应力
第三章 扭 转
5
§ 3–1 SUMMARY
Shaft,In engineering the members of which deformations are
mainly torsion,Such as transmission shafts in machines,drill rods
in oil-drilling rigs etc,
Torsion,Resultant of the external forces is a force couple and
its acting plane is perpendicular to the axis of the shaft,Under
this case the deformation of the rod is torsion,
A B O
m m
? O
B
A ?
6
§ 3–1 概 述
轴,工程中以扭转为主要变形的构件。如：机器中的传动轴,
石油钻机中的钻杆等。
扭转,外力的合力为一力偶，且力偶的作用面与直杆的轴线
垂直，杆发生的变形为扭转变形。
A B O
m m
? O
B
A ?
7
The angle of twist (?),The angle of rotation of one section
with respect to another,
Shearing strain (? ),The change of a right angle
between two straight lines,
m m
? O
B
A ?
8
扭转角（ ?）,任意两截面绕轴线转动而发生的角位移。
剪应变（ ?）,直角的改变量。
m m
? O
B
A ?
9
Practical examples in engineering
Transmission shaft
10
工
程
实
例
11
§ 3–2 EXTERNAL TORQUE OF A TRANSMISSION SHAFT
INTERNAL TORQUE AND ITS DIAGRAM
1,External torque of a transmission shaft The relation
between the transmission power,revolution and external torque
of the transmission shaft,
m)( k N559 ?? nP.m
m)( k N0 2 47 ?? nP.m
m)( k N1217 ?? nP.m
where,P - power,unit,kilowatt (kW)
n – rotational speed,unit,r/min or（ rpm）
where,P - power,unit,horsepower (HP)
n – rotational speed,unit,r/min or（ rpm）
where,P - power,unit,kilowatt (kW)
n – rotational speed,unit,r/min or（ rpm）
1PS=735.5N·m/s,1HP=745.7N·m/s,1kW=1.36PS
12
§ 3–2 传动轴的外力偶矩 · 扭矩及扭矩图
一、传动轴的外力偶矩
传递轴的传递功率、转数与外力偶矩的关系,
m)( k N559 ?? nP.m
m)( k N0 2 47 ?? nP.m
m)( k N1217 ?? nP.m
其中,P — 功率，千瓦（ kW）
n — 转速，转 /分（ rpm）
其中,P — 功率，马力（ PS）
n — 转速，转 /分（ rpm）
其中,P — 功率，马力（ HP）
n — 转速，转 /分（ rpm）
1PS=735.5N·m/s,1HP=745.7N·m/s,1kW=1.36PS
13
3) Sign conventions for internal torque,
, T”will be considered to be positive when the relation
between its turning direction and the out normal line obeys the
right hand rule,otherwise it will be considered to be negative,
2,Internal torque and its diagram
1) Internal torque,The moment of internal forces acting in arbitrary section
of the member in torsion,Designated by―T‖,
2) Determine the internal torque
by the method of section,
m m
m T
mT
mTm
x
?
???? 0,0
x
14
3 扭矩的符号规定,
, T”的转向与截面外法线方向满足右手螺旋规则为正，
反之为负。
二、扭矩及扭矩图
1 扭矩,构件受扭时，横截面上的内力偶矩，记作,T‖。
2 截面法求扭矩
m m
m T
mT
mT
m
x
?
??
??
0
0
x
15
4) Internal torque diagram,Sketch that expresses the law of
change of the torque in each cross section along the axis,
Purpose
① the law of change of the torque ；
② Value of |T|max and the position of its section
strength calculation（ critical section）
x
T
?
16
4 扭矩 图,表示沿杆件轴线各横截面上扭矩变化规律的图线。
目
的
① 扭矩变化规律；
② |T|max值及其截面位置 强度计算（危险截面）。
x
T
?
17
Example 1 A transmission shaft is rotating at n =300r/min,Knowing its
input power is P1=500kW,and its output are P2=150kW,P3=150kW，
P4=200kW,Try to plot the internal torque diagram,
n
A B C D
m2 m3 m1 m4 Solution：① Calculate
the external torque
m)15.9( kN
300
500
9.55
55.9
1
1
????
?
n
P
m
m)( k N 7843001509, 5 5559 232 ??????,nP.mm
m)( k N 3763002009, 5 5559 44 ?????,nP.m
18
[例 1]已知：一传动轴,n =300r/min，主动轮输入 P1=500kW，
从动轮输出 P2=150kW,P3=150kW,P4=200kW，试绘制扭矩
图。
n
A B C D
m2 m3 m1 m4
解：①计算外力偶矩
m)15,9( kN
30 0
50 0
9,55559 1
1
??
???
n
P
.m
m)( k N 7843001509, 5 5559 232 ??????,nP.mm
m)( k N 3763002009, 5 5559 44 ?????,nP.m
19
n
A B C D
m2 m3 m1 m4 1
1
2
2 3
3
② Determine the internal torque（ suppose it is positive）
mkN78.4
0,0
21
21
?????
????
mT
mTm
x
mkN569784784(
,0
322
322
?????????
???
.)..mmT
mmT
mkN37.6
,0
42
43
???
???
mT
mT
x
20
n
A B C D
m2 m3 m1 m4 1
1
2
2 3
3
② 求扭矩（扭矩按正方向设）
mkN78.4
0,0
21
21
?????
????
mT
mTm
x
mkN569784784(
,0
322
322
?????????
???
.)..mmT
mmT
mkN37.6
,0-
42
43
???
??
mT
mT
x
21
③ Plot the internal torque diagram
mkN 569m a x ??,T
Each section in segment BC is a critical section,
x
T
n
A B C D
m2 m3 m1 m4
4.78
9.56
6.37
?
–
–
22
③ 绘制扭矩图
mkN 569m a x ??,T
BC段为危险截面。
x
T
n
A B C D
m2 m3 m1 m4
4.78
9.56
6.37
?
–
–
23
§ 3–3 TORSION OF
THE THIN-WALLED HOLLOW SHAFTS
1,Experiment,
1),Preparation,
① Plot the longitudinal lines
and circumference lines；
② To act a pair of external
torque m。
Hollow round shaft with thin wall,
010
1 rt ? （ r0,average radius） Thickness of the wall
24
§ 3–3 薄壁圆筒的扭转
薄壁圆筒,壁厚
010
1 rt ?
（ r0,为平均半径）
一、实验,
1.实验前,
① 绘纵向线，圆周线；
②施加一对外力偶 m。
25
2,After deformation,
① The circumference lines
do not change；
② The longitudinal lines
are changed into slants 。
3.Conclusions,
① Shape,size and distance of the circumference lines on the shaft surface
do not change, while rotating with respect to one another along the axis of
the shaft。
② All the longitudinal lines revolve through a same angle ? 。
③ All squares drawn on the shaft surface warp into rhombs with same sizes。
26
2.实验后,
① 圆周线不变；
②纵向线变成斜直线。
3.结论：① 圆筒表面的各圆周线的形状、大小和间距均未改
变，只是绕轴线作了相对转动。
② 各纵向线均倾斜了同一微小角度 ? 。
③ 所有矩形网格均歪斜成同样大小的平行四边形。
27
? ?
a
c d
dx
b
dy
?′
?′
? ① No normal stress
② There are only the shearing stress
? perpendicular to the radius at each
point on the cross section,Magnitude
of the shearing stresses are the same on
the same section,directions of them
coincide with that of the internal
torque,
4.) Relation between ? and ?, ? ?
L
RS
RL
??
???
??
??
o
A small cubic element is shown in the figure,
28
? ?
a
c d
dx
b
dy
?′
?′
?
① 无正应力
②横截面上各点处，只产
生垂直于半径的均匀分布的剪
应力 ?,沿周向大小不变，方
向与该截面的扭矩方向一致。
4,? 与 ? 的关系,
?
?
L
R
RL
???
???
??
??
微小矩形单元体如图所示,
29
2,Magnitude of the shearing stress ? in the hollow shaft with thin wall,
?
?
tA
T
tr
T
TtrrArS
TrA
A
A
2 2
2d o
d
0
2
0
000
0
??
???????
???
?
?
?
?
???
?
A0,Area of the circle with an average radius
30
二、薄壁圆筒剪应力 ? 大小,
?
?
tA
T
tr
T
TtrrAr
TrA
A
A
2 2
2d
d
0
2
0
000
0
???
?????????
????
?
?
???
?
A0：平均半径所作圆的面积。
31
3,Theorem of conjugate shearing
stresses,
)( h u s,
0
aT
d x d ytd x d yt
m
z
??
??
??
??????
??
Formula (a) is called theorem of conjugate shearing stresses,
It indicates shearing stresses always exist on mutually
perpendicular plane and occur in equal and opposite pairs
and point,perpendicularly,either toward or away from the
intersection line of the planes,
a
c d
dx
b
? ? dy
?′
?′
t
z
?
32
三、剪应力互等定理,
??
??
??
??????
??
0
故
d x d ytd x d yt
m
z
上式称 为剪应力互等定理 。
该定理表明,在单元体相互垂直的两个平面上，剪应
力必然成对出现，且数值相等，两者都垂直于两平面的交
线，其方向则共同指向或共同背离该交线。
a
c d
dx
b
? ? dy
?′
?′
t
z
?
33
4,Hooke’s law of shear,
There are only shearing stresses and no normal
stresses on the four side planes of the element,which is
called pure-shear stressed state。
l
34
四、剪切虎克定律,
单元体的四个侧面上只有剪应力而无正应力作用，这
种应力状态称为 纯剪切应力状态。
l
35
?
?
T=m
?
?? ?
)( ) 2(
0 R
LtA
T
??
?
??
?
Hooke’s law in shear,The shearing stress is directly
proportional to the shearing strain if the shearing stress does
not exceed smaller than the proportional sheer limit of the
material(τ ≤τp),
36
?
?
T=m
?
?? ?
)( ) 2(
0 R
LtA
T
??
?
??
?
剪切虎克定律,当剪应力不超过材料的剪切比例极限
时（ τ ≤τp)，剪应力与剪应变成正比关系。
37
?? ?? G
In above formula G is a elastic constant of material,It is called modulus of
elasticity in shearing,as ? has no dimension,G has the same dimension with ?,
Values of G for different materials can be determined by experiments,Value of G for
steel is about 80GPa,
Modulus of elasticity in shearing,Young’s modulus and Possion’s ratio are
three elastic constants,They indicate elastic properties of materials,For the
isotropic materials there is the following relation between the three constants.（ See
later chapters about deduction of the formula）,
It is obvious that we can find out the third if any two of the three
elastic constants are known,
)1(2 ??
?
E
G
38
?? ?? G
式中,G是材料的一个弹性常数，称为剪切弹性模量，因 ? 无
量纲，故 G的量纲与 ? 相同，不同材料的 G值可通过实验确定，钢
材的 G值约为 80GPa。
剪切弹性模量、弹性模量和泊松比是表明材料弹性性质的三
个常数。对各向同性材料，这三个弹性常数之间存在下列关系
（推导详见后面章节）,
可见，在三个弹性常数中，只要知道任意两个，第三个量
就可以推算出来。
)1(2 ???
EG
39
§ 3–4 STRESSES OF CIRCULAR SHAFTS IN
TORSION · STRENGTH CONDITIONS
Stress on the cross
section of the round shaft
① Geometric deformations
② Physical relations
③ Statics
1),After deformation the cross
sections still remain planes;
2),No elongation or contraction
along the axis；
3),Longitudinal lines are still
parallel to each other,
1,Observation of the
torsional test of the
straight circular rod,
40
§ 3–4 等直圆杆在扭转时的应力 · 强度条件
等直圆杆横截面应力
① 变形几何方面
②物理关系方面
③静力学方面
1,横截面变形后
仍为平面；
2,轴向无伸缩；
3,纵向线变形后仍为平行。
一、等直圆杆扭转实验观察,
41
2,Stresses on the cross section of the round shaft in torsion,
1),Geometric deformation
relation,
xx
GG
d
d
dtg
1 ????
??
?????
xd
d ???
? ?
The shearing strain ?? at an arbitrary point is
directly proportional to the distance ? from the
center of the section to the point,
xd
d? —— of the variation rate twisting angle along the length,
42
二、等直圆杆扭转时横截面上的应力,
1,变形几何关系,
xx
GG
d
d
dtg
1 ????
??
?????
xd
d ???
? ?
距圆心为 ? 任一点处的 ??与该点到圆心的距离 ?成正比。
xd
d?
—— 扭转角沿长度方向变化率。
43
2),Physical relation,
Hooke’s law,
Substituting it into the preceding formula we get,
?? ?? G
xGxGG d
d
d
d ??????
?? ??????
xG d
d ???
? ?
44
2,物理关系,
虎克定律,
代入上式得,
?? ?? G
xGxGG d
d
d
d ??????
?? ??????
xG d
d ???
? ?
45
3),Static relation,
A
x
GA
x
G
AT
AA
A
d
d
d
d
d
d
d
22
?
??
?
?? ?
????
????
AI Ap d2???
Let
xGI T p d
d ??
pGI
T
x ? d
d ?
After substituting into the physical relation
xG d
d ???
? ?
pI
T ??
?
??we get
O
τp ?
dA
46
3,静力学关系,
A
x
G
A
x
G
AT
A
A
A
d
d
d
d
d
d
d
2
2
?
?
?
?
??
?
??
??
????
AI Ap d2???
令
xGI T p d
d ??
pGI
T
x ? d
d ?
代入物理关系式 得,
xG d
d ???
? ?
pI
T ??
?
??
O
τp ?
dA
47
pI
T ??
?
?? —Formula to calculate the shearing stress of any point in the
section with distance ? to the center,
4),Discussions on the formula,
① It can be only applied to the isotropic,linear-elastic straight-
circular rods with small deformations,
② In the above formula,T—Torque in the cross section,It is
determined from the external torques by the method of section,
? — Distance of the point to the center of the circle,
Ip—Polar moment of inertia of section,It is a purely
geometric quantity without any physical meaning,
48
pI
T ??
?
?? —横截面上距圆心为 ?处任一点剪应力计算公式。
4,公式讨论,
① 仅适用于各向同性、线弹性材料，在小变形时的等圆截面
直杆。
② 式中,T—横截面上的扭矩，由截面法通过外力偶矩求得。
? —该点到圆心的距离。
Ip—截面极惯性矩，纯几何量，无物理意义。
49
unit,mm4,m4。
AI Ap d2???
③ The formula is deduced from the solid circular shaft,but it
can also be applied to the hollow circular shaft,Values of their Ip
are different,
4
4
2
0
2
2
10
32
d2
d
D.
D
AI
D
Ap
??
? ????
??
?
????
?
a,For a solid circular section,
D ?
d?
O
50
单位,mm4,m4。
AI Ap d2???
③ 尽管由实心圆截面杆推出，但同样适用于空心圆截面杆,
只是 Ip值不同。
4
4
2
0
2
2
10
32
d2
d
D.
D
AI
D
Ap
??
? ????
??
?
????
?
a,对于实心圆截面,
D ?
d?
O
51
b,For a hollow circular section,
)1(10)1(
32
)(
32
d2
d
444
4
44
2
2
2
2
??
?
?
????
?
????
??
?
????
??
D.
D
dD
AI
D
d
Ap
)( Dd??
d D O
?
d?
52
b,对于空心圆截面,
)1(10)1(
32
)(
32
d2
d
444
4
44
2
2
2
2
??
?
?
????
?
????
??
?
????
??
D.
D
dD
AI
D
d
Ap
)( Dd??
d D O
?
d?
53
④ Distribution of the shearing stresses
（ solid section） （ hollow section）
In engineering the members with hollow section are widely
used to increase the strength,save materials and decrease the
weight of structures,
54
④ 应力分布
（实心截面） （空心截面）
工程上采用空心截面构件：提高强度，节约材料，重量轻,
结构轻便，应用广泛。
55
⑤ Determine the maximum shearing stress,
pI
T ??
?
??From
when
m a x,2 ??? ? ???
dR
)
2
e t (
2
2
h e n
m a x
d
IWl
W
T
d
I
T
I
d
T
T
p
t
p
p
???
?
??
tW
T?
m a x?
Wt — section factor in torsion（ section
factor modulus in torsion ),geometric
quantity,unit,mm3 or m3
a,For the solid circular section,
33 2016 D.DRIW
pt ??? ?
b,For the hollow circular section,
)-(12016)1( 4343 ??? D.DRIW pt ????
56
⑤ 确定最大剪应力,
pI
T ??
?
??
由 知：当
m a x,2 ??? ? ???
dR
)
2
(
2
2
m a x
d
IW
W
T
d
I
T
I
d
T
p
t
p
p
???
?
?? 令?
tW
T?
m a x?
Wt — 抗扭截面系数（抗扭截面模量）,
几何量，单位,mm3或 m3。
对于实心圆截面,
33 2016 D.DRIW
pt ??? ?
对于空心圆截面,
)-(12016)1( 4343 ??? D.DRIW pt ????
57
3,Stress on the inclined plane of the round shaft in torsion
Lower-carbon steel
specimen,
Rupture along the cross
section,
Cast-iron specimen,
Rupture along the helix
direction with an
approximate inclined
angle 45?,
So the stresses on the inclined plane must be analyzed,
58
三、等直圆杆扭转时斜截面上的应力
低碳钢试件,
沿横截面断开。
铸铁试件,
沿与轴线约成 45?的
螺旋线断开。
因此还需要研究斜截面上的应力。
59
1,Stressed element at point M
is shown in Fig.(b),
(a)
M
(b)
′ ′
(c)
2,Stresses on the inclined
section； Take a free body as
shown in Fig.(d),
(d)
?
′
?
x
60
1,点 M的应力单元体如图 (b),
(a)
M
(b)
′ ′
(c)
2,斜截面上的应力；
取分离体如图 (d),
(d)
?
′
?
x
61
(d)
?
′
?
x
n
t
Conventions for the angle of rotation,
Positive direction of the axis
turns to the out normal line of section
countclockwise：,+‖
clockwise：,–‖
From the equilibrium equations,
0) c o ss i nd() s i nc o sd(d ; 0 ?????? ??????? ? AAAF n
0) s i ns i nd() c o sc o sd(d ; 0 ?????? ??????? ? AAAF t
?? ??
We have,
?????? ?? 2c o s ; 2s i n ???
62
(d)
?
′
?
x
n
t
转角 α规定,
x轴正向转至截面外法线 逆时针：为,+‖
顺时针：为,–‖
由平衡方程,
0) c o ss i nd() s i nc o sd(d ; 0 ?????? ??????? ? AAAF n
0) s i ns i nd() c o sc o sd(d ; 0 ?????? ??????? ? AAAF t
?? ??
解得,
?????? ?? 2c o s ; 2s i n ???
63
?????? ?? 2c o s ; 2s i n ???
Analysis,As ? = 0°,
???? ??? ?? m a x00,0
As ? = 45°,
0,45m i n45 ???? ?? ????
As ? = – 45°,
0,45m a x45 ??? ???? ????
As ? = 90°,
???? ????? ?? m a x9090,0
′
45°
It is found that for a circular shaft in
torsion the maximum shearing stress occurs on
transverse and longitudinal sections,and that the
maximum tensile and compressive normal
stresses occur on the inclined sections oriented at
? = ? 45?,From these conclusions we can
explain the forementioned rupture phenomenon,
64
?????? ?? 2c o s ; 2s i n ???
分析,当 ? = 0° 时,
???? ??? ?? m a x00,0
当 ? = 45° 时,
0,45m i n45 ???? ?? ????
当 ? = – 45° 时,
0,45m a x45 ??? ???? ????
当 ? = 90° 时,
???? ????? ?? m a x9090,0
′
45°
由此可见：圆轴扭转时，在横截
面和纵截面上的剪应力为最大值；在
方向角 ? = ? 45?的斜截面上作用有最
大压应力和最大拉应力。根据这一结
论，就可解释前述的破坏现象。
65
4,Strength calculation of the round shaft in torsion
Strength condition,
For the round shaft
with equal sections,
][m a x ?? ?
][m a x ??
tW
T
([?] is called permissible
shearing stress.)
Three aspects of strength calculations,
① Check the strength,
② Design the dimension
of the section,
③ Calculate the permissible load,
][m a xm a x ?? ??
tW
T
][
m a x
?
TW
t ?
][m a x ?tWT ?
?
?
?
?
?
?
?
?
?
?
? ）（
4
3
3
1
16
16
?
?
?
D
h o l l o w
Ds o l i d
W
t
66
四、圆轴扭转时的强度计算
强度条件,
对于等截面圆轴,
][m a x ?? ?
][m a x ??
tW
T
([?] 称为许用剪应力。 )
强度计算三方面,
① 校核强度,
② 设计截面尺寸,
③ 计算许可载荷,
][m a xm a x ?? ??
tW
T
][
m a x
?
TW
t ?
][m a x ?tWT ?
??
?
?
?
??
?
?
?
? ）（空：
实：
4
3
3
1
16
16
?
?
?
D
D
W
t
67
Example 2 Power is 150kW,speed of rotation is 15.4r/s for the rotator of
a motor as shown in the figure.the Permissible shearing stress is [?]=30MPa,Try
to check its strength,
n
NmT
BC ?2
10 3???
m)( k N551
m)(N
4151432
10150
3
??
?
??
?
?
.
..
T
m
Solution,① Determine its
torques and plot the torque diagram
② Calculate and check the
shearing stress strength
③ Strength of this shaft satisfies request
D3 =135 D2=75 D1=70
A B C
m m
x
][M P a23
16070
10551
3
3
m a x ??? ???
???
.
.
W
T
t
68
[例 2] 功率为 150kW，转速为 15.4转 /秒的电动机转子轴如图,
许用剪应力 [?]=30M Pa,试校核其强度。
n
NmT
BC ?2
10 3???
m)( k N551
m)(N
4151432
10150
3
??
?
??
?
?
.
..
T
m
解：①求扭矩及扭矩图
② 计算并校核剪应力强度
③ 此轴满足强度要求。
D3 =135 D2=75 D1=70
A B C
m m
x
][M P a23
16070
10551
3
3
m a x ??? ???
???
.
.
W
T
t
69
§ 3–5 DEFORMATION OF A CIRCULAR SHAFT IN TORSION
RIGIDITY CONDITIONS
1,Deformation in torsion
From the formula
pGI
T
x ? d
d ?
We can determine the relative angle ? of twist
between two end sections of the rod with the length l,
）c o n s t a n t is ( i f d d
0
T
GI
Tl
x
GI
T
p
l
p
???? ???
70
§ 3–5 等直圆杆在扭转时的变形 · 刚度条件
一、扭转时的变形
由公式
pGI
T
x ? d
d ?
知,长为 l一段杆两截面间相对扭转角 ? 为
值不变）若 (
d d
0
T
GI
Tl
x
GI
T
p
l
p
?
???? ??
71
2,The angle of twist per unit length?,
( r a d / m )
d
d
pGI
T
x
??
?
? / m )(
1 8 0
d
d ????
?
??
pGI
T
x
or
3,Rigidity condition
? ? ( r a d / m ) m a xm a x ?? ??
pGI
T
? ? / m )(
1 8 0
m a xm a x ???? ?
?
?
pGI
T
or
G Ip indicates the capacity that the dimension and material
properties of the section resist torsional deformations,It is called
the torsional rigidity of the circular shaft,
[? ] is called the permissible angle of twist per unit length of the shaft,
72
二、单位长度扭转角 ?,
( r a d / m ) dd
pGI
T
x ??
??
/ m )( 1 8 0 dd ???? ???
pGI
T
x
或
三、刚度条件
或
GIp反映了截面尺寸和材料性能抵抗扭转变形的能力，称为
圆轴 的抗扭刚度 。
[? ]称为许用单位长度扭转角。
? ? ( r a d / m ) m a xm a x ?? ??
pGI
T
? ? / m )( 1 8 0 m a xm a x ???? ?
?
?
pGI
T
73
Three aspects of rigidity calculations,
① Check the rigidity,
② Design the dimension
of the section,
③ Calculate the
permissible load,
? ? m a x ?? ?
] [
m a x
?G
T
I p ?
] [ m a x ?pGIT ?
Sometimes,these conditions may be considered to select materials.,
74
刚度计算的三方面,
① 校核刚度,
② 设计截面尺寸,
③ 计算许可载荷,
? ? m a x ?? ?
] [
m a x
?G
T
I p ?
] [ m a x ?pGIT ?
有时，还可依据此条件进行选材。
75
Example 3 Uniformly distributed couple m=20Nm/m acted a circular shaft
with length L=2m is shown in the figure,If the ratio of the inside diameter and the
outside diameter of the shaft is ? =0.8,G=80GPa,the permissible shearing
stress is [?]=30MPa,Try to design the outside diameter of the shaft； As
[?]=2o/m,try to check the rigidity of the shaft and determine the angle of rotation
of the right-end section,
Solution,① Design the outside diameter of the rod
][
m a x
?
TW
t ?
116D 4
3
）（ ?? ??tW
3
1
4
m a x
][ 1
16
?
?
?
?
?
?
?
?
??? ）（
T
D
76
[例 3]长为 L=2m 的圆杆受均布力偶 m=20Nm/m 的作用，如图，
若杆的内外径之比为 ? =0.8, G=80GPa,许用剪应力
[?]=30MPa，试设计杆的外径；若 [?]=2o/m,试校核此杆的刚
度，并求右端面转角。
解,①设计杆的外径
][
m a x
?
TW
t ?
116D 4
3
）（ ?? ??tW
3
1
4
m a x
][ 1
16
?
?
?
?
?
?
?
?
??? ）（
T
D
77
3
1
4
m a x
][ 1
16
?
?
?
?
?
?
?
?
??? ）（
T
D
40Nm
x
T
Substituting the values we
get,
D ? 0.0226m。
② Check the rigidity according
to the rigidity condition in
torsion
??
1 8 0m a x
m a x ??
PGI
T
78
3
1
4
m a x
][ 1
16
?
?
?
?
?
?
?
?
??? ）（
T
D
40Nm
x
T
代入数值得,
D ? 0.0226m。
② 由扭转刚度条件校核刚度
??
1 8 0m a x
m a x ??
PGI
T
79
40Nm
x
T
??
1 8 0m a x
m a x ??
PGI
T ? ?
?
??
??
???
??? m
D
/89.1
)1(1080
1804032
4429
?
③ Angle of twist of the right-end section,
）ad( 033.0
)4(
102040 2
0
2
2
00
r
xx
GI
dx
GI
x
dx
GI
T
PP
L
P
?
??
?
?? ???
80
40Nm
x
T
??
1 8 0m a x
m a x ??
PGI
T ? ??
??
??
???
??? ?891
11080
1804032
4429,)(D
③ 右端面转角 为,
弧度）( 0330
4
102040 2
0
22
00
.
)xx(
GI
dx
GI
x
dx
GI
T
PP
L
P
?
???
?
????
81
Example 4 It is Required that the speed of rotation of the shaft is n = 500 r /
min,input power is N1 = 500 HP,output power is respectively N2 = 200HP
and N3 = 300HP in the design of a transmission shaft,Knowing,G=80GPa,
[? ]=70MPa,[? ]=1o/m,Try to determine,① Diameter d1 of segment AB and
diameter d2 of segment BC？ ② If select the same diameter for the whole shaft,
how much is the diameter？ ③ How to
arrange the driver and the driven pulleys
is reasonable？
Solution,① Torque diagram is
shown in the lower figure of the
shaft for the structure shown in the
upper figure,From the strength
condition we get,
500 400
N1 N3 N2
A C B
T x
–7.024 – 4.21
(kNm)
m)( k N0247 ?? nN.m
82
[例 4] 某传动轴设计要求转速 n = 500 r / min，输入功率 N1 = 500
马力,输出功率分别 N2 = 200马力及 N3 = 300马力，已知：
G=80GPa, [? ]=70M Pa,[? ]=1o/m,试确定,
① AB 段直径 d1和 BC 段直径 d2？
② 若全轴选同一直径，应为多少？
③ 主动轮与从动轮如何安排合理？
解,① 图示状态下,扭矩如
图,由强度条件得,
500 400
N1 N3 N2
A C B
T x
–7.024 – 4.21
(kNm)
m)( k N0247 ?? nN.m
83
][16
31
?
? TdW
t ??
? ? mm4671070143
42101616
3
632,.
Td ?
??
?????
??
] [ 32
4
?
?
G
TdI
p ??
? ? mm801070143
7 0 2 41616
3
6
31 ?
??
?????
.
Td
??
From the rigidity condition,
500 400
N1 N3 N2
A C B
T
x
–7.024 –4.21
(kNm)
84
][16
31
?
? TdW
t ??
? ? mm4671070143
42101616
3
632,.
Td ?
??
?????
??
] [ 32
4
?
?
G
TdI
p ??
? ? mm801070143
7 0 2 41616
3
6
31 ?
??
?????
.
Td
??
由刚度条件得,
500 400
N1 N3 N2
A C B
T
x
–7.024 –4.21
(kNm)
85
mm474
11080143
180421032
] [
32
4
9242,.G
Td ?
???
??????
??
mm84
11080143
180702432
] [
32
4
9241 ????
???????
.G
Td
??
? ? ? ? mm75 mm85 21 ?? d,dSum up,
② If select the same diameter for
the whole shaft ? ? ? ? mm851 ?? dd
86
mm474
11080143
180421032
] [
32
4
9242,.G
Td ?
???
??????
??
mm84
11080143
180702432
] [
32
4
9241 ????
???????
.G
Td
??
? ? ? ? mm75 mm85 21 ?? d,d综上,
② 全轴选同一直径时 ? ? ? ?
mm851 ?? dd
87
③ the smaller the torque with the maximum absolute value in the shaft is,the more
reasonable it is,So wheel 1 must be replaced with wheel 2,After transposition,
torque diagram of the shaft is shown in the figure,Under this case,the maximum
diameter of the shaft is 75mm。
T
x
– 4.21
(kNm)
2.814
88
③ 轴上的 绝对值 最大 的 扭矩 越小越 合理，所以,1轮和 2轮应
该 换 位。 换位后,轴的扭矩如图所示,此时,轴的最大直径才
为 75mm。
T
x
– 4.21
(kNm)
2.814
89
§ 3–6 STATICALLY INDETERMINATE PROBLEMS OF
ROUND SHAFT IN TORSION
Method and Steps to solve the statically indeterminate
problems in torsion,
Write out the equilibrium equations；
Write out the geometric equation—compatibility equation of
deformation；
Find out the complementary equations when combined
geometric equations and physical equations；
Write out the physical equations；
Solve the combined equations including equilibrium equations
and complementary equations,
90
§ 3–6 等直圆杆的扭转超静定问题
解决扭转超静定问题的方法步骤,
平衡方程；
几何方程 ——变形协调方程；
补充方程：由几何方程和物理方程得；
物理方程；
解由平衡方程和补充方程组成的方程组。
91
Example 5 A circular rod is subjected to uniformly distributed a force couple as
shown in the figure,Length of the rod is L=2m,If the ratio between the inside
and outside diameter is ? =0.8,and outside diameter of the rod is D=0.0226m，
G =80GPa,Try to determine the constraint couple in the fixed ends,
Solution,① Free body
diagram of the rod is shown
in the figure,It is a one-time
statically indeterminate
problem,Its equilibrium
equation is,
02 ??? BA mmm
A B
92
[例 5]长为 L=2m 的圆杆受均布力偶 m=20Nm/m 的作用，如图，
若杆的内外径之比为 ? =0.8,外径 D=0.0226m, G=80GPa，
试求固定端反力偶。
解, ①杆的受力图如图示,
这是一次超静定问题。
平衡方程为,
02 ??? BA mmm
A B
93
② Geometric equation—compatibility equation of deformation
0?BA?
③ Summing up the physical equation and the geometric equation we can get
the complementary equation,
0
402202
00
?
?
?
?
?? ??
P
A
P
A
L
P
BA GI
m
dx
GI
xm
dx
GI
T?
mN 20 ??? Am
④ From the equilibrium equation and the complementary equation we get,
Additionally,this problem can be directly solved by using symmetry,
mN 20 ??Bm
94
② 几何方程 ——变形协调方程
0?BA?
③ 综合物理方程与几何方程，得补充方程,
040220200 ???? ????
P
A
P
AL
P
BA GI
mdx
GI
xmdx
GI
T?
mN 20 ??? Am
④ 由平衡方程和补充方程得,
另,此题可由对称性直接求得结果。
mN 20 ??Bm
95
§ 3–7 STAIN ENERGY OF ROUND SHAFTS IN TORSION
?? G?
dV)dx(d z d y )(dW ???? 2121 ??
2
2
1
2
1
d
d
d
d ??? G
V
W
V
Uu ????
1,Strain energy and its density
a
c d
dx
b
? ? dy
?′
?′
dz
z
?
x
y Infinitesimal work of the element,
Strain energy per unit volume,
96
§ 3–7 等直圆杆在扭转时的应变能
?? G?
dV)dx(d z d y )(dW ???? 2121 ??
2
2
1
2
1
d
d
d
d ??? G
V
W
V
Uu ????
一,应变能与能密度
a
c d
dx
b
? ? dy
?′
?′
dz
z
?
x
y 单元体微功,
应变比能,
97
2,Calculation of the stress in cylindric close-coiled helical
springs
1),Calculation of the stress
= +
? Q ?
T
Q
T
t
TQ W
T
A
Q ???? ???
m a x
Approximate value,
323
8
1
2
416
2
d
DP
D
d
d
P
d
PD
???
?
?
?
?
?
?
????
P
Q
T
98
二、圆柱形密圈螺旋弹簧的计算
1,应力的计算
= +
? Q ?
T
Q
T
t
TQ W
T
A
Q ???? ???
m a x
近似值,
323
8
1
2
416
2
d
DP
D
d
d
P
d
PD
???
?
?
?
?
?
?
????
P
Q
T
99
2) Strength condition of the spring wire,
? ??
?
? ?? 8 3
d
DPK
m a x
Exact value：（ Revised formula is to considered the effect of the curvature
and shearing stress of the spring）
33m a x
886150
44
14
d
DPK
d
DP
C
.
C
C
??
? ??
?
??
?
? ?
?
??
where,
d
DC ?
C
.
C
CK 6 1 50
44
14 ?
?
??
Is called the spring index,
is called the coefficient of
curvature,
100
2,弹簧丝的强度条件,
? ??
?
? ?? 8 3
d
DPK
m a x
精确值：（修正公式，考虑弹簧曲率及剪力的影响）
33m a x
886150
44
14
d
DPK
d
DP
C
.
C
C
??
? ??
?
??
?
? ?
?
??
其中,
d
DC ?
C
.
C
CK 6 1 50
44
14 ?
?
??
称为弹簧指数。
称为曲度系数。
101
3).Calculation of the displacement(energy method）
。c o n s t a n t s p r i n g t h es
64;
64
;
3
4
4
3
i
nR
Gd
K
K
P
Gd
nPR
UW
?????
?
?PW 21 ?
Work of the external force,
Strain energy,
AL
I
T
G
VUU
pVVV
d
2
1
d
2
1
d
2
?
?
?
?
?
?
??????
?
??
? ?
pp GI
PRRn
L
I
T
G 2
2
2
1 22 ???
102
3.位移的计算 (能量法）
为弹簧常数。
64; 64 ; 3
4
4
3
nR
GdK
K
P
Gd
nPRUW ????? ?
?PW 21 ?
外力功,
变形能,
AL
I
T
G
VUU
pVVV
d
2
1
d
2
1
d
2
?
?
?
?
?
?
??????
?
??
? ?
pp GI
PRRn
L
I
T
G 2
2
2
1 22 ???
103
Example 6 Mean diameter of a close-coiled helical spring is
D=125mm,wire diameter is d =18mm,It is subjected to a couple
of tensile forces with magnitude P=500N,Try to determine the
exact value and approximate value of the maximum shearing stress；
If G =82GPa,in order to make the deformation of the spring be
6mm,how many turns should the spring have at least？
Solution,① Approximate value of the maximum shearing stress,
M P a329
0180
50012508
1
1252
18
8
1
2
3
3m a x
.
.
.
)(
d
DP
)
D
d
(
?
?
??
?
?
?
??
?
?
?
104
[例 6] 圆柱形密圈螺旋弹簧的平均直径为,D=125mm，簧丝直
径为,d =18mm，受拉力 P=500N 的作用,试求最大剪应力
的近似值和精确值；若 G =82GPa，欲使弹簧变形等于 6mm,
问：弹簧至少应有几圈？
解,① 最大剪应力的近似值,
M P a329
0180
50012508
1
1252
18
8
1
2
3
3m a x
.
.
.
)(
d
DP
)
D
d
(
?
?
??
?
?
?
??
?
?
?
105
② Exact value of the maximum shearing stress,
091615044 14 ; 631518125,C.CCK.dDC ??????? ?
M P a2330180 500125080918 33m a x,...dDPK ?? ????? ???
③ The number of turns of the spring,
66
1 2 505 0 064
1018826
64 3
64
3
4
.
.PR
Gdn ?
??
????? ??
（ turns）
106
② 最大剪应力的精确值,
091615044 14 ; 631518125,C.CCK.dDC ??????? ?
M P a2330180 500125080918 33m a x,...dDPK ?? ????? ???
③ 弹簧圈数,
66
1 2 505 0 064
1018826
64 3
64
3
4
.
.PR
Gdn ?
??
????? ??
（圈）
107
§ 3–8 STRESSES AND DEFORMATIONS OF NONCIRCULAR
SHAFTS IN FREE TORSION
Noncircular straight shaft,Here the assumptions of plane can not be
applied,That is,cross-sections of the rod do not remain plane due to warping,
Therefore the preceding formulas about stress and deformation for the straight
circular shaft can not be applied either,This kind of problem must be solved by
the method in elastic mechanics,
108
§ 3–8 非圆截面等直杆在自由扭转时的应力和变形
非圆截面等直杆,平面假设不成立。即各截面发生翘曲不保
持平面。因此，由等直圆杆扭转时推出的应力、变形公式不
适用，须由弹性力学方法求解。
109
1,Free torsion,Warping of each cross section of the rod in torsion
is not restricted and the magnitude is completely same.,
2,Restricted torsion,Warping of each cross section of the rod in
torsion is restricted and the magnitude is different.,
3,shearing stress on the section of the
rectangular rod,
h ?
b
h
? 1 T
? max
Attention！ b
1),Distribution of the shearing
stress is shown in the figure,
（ Corner point,center of the section,
middle points of the long and short sides）
110
一,自由扭转,杆件扭转时，横截面的翘曲不受限制，任意两相
邻截面的翘曲程度完全相同。
二,约束扭转,杆件扭转时，横截面的翘曲受到限制，相邻截面
的翘曲程度不同。
三,矩形杆横截面上的剪应力,
h ?
b
h
? 1 T
? max
注意！ b
1,剪应力分布如图,
（角点、形心、长短边中点）
111
2),Maximum shearing stress and angle of twist per unit length
m a x1 ??? ?
h ?
b
h
? 1 T
? max
attention！ b
m a xm a x
tW
T
??
3 b
tW ??
where,
4 bI
t ??
,
tGI
T??
where,It—equivalent polar
moment of inertia,
112
2,最大剪应力及单位扭转角
m a x1 ??? ?
h ?
b
h
? 1 T
? max
注意！ b
m a xm a x
tW
T
?? 3 btW ??
其中,
4 bI
t ??
,
tGI
T??
其中,It—相当极惯性矩。
113
hbtWw
W
T
t
2m a x
m a x,h e r e ?? ??
Attention！ There are the following definitions for W t and It in most
teaching books and engineering handbooks,
hbIw
GI
T
t
t
3,h e r e,?? ??
m a x1 ??? ?
3
1; ) 10 ( ??? ??
b
h
In reference to the table about ? and ? we must pay attention to
which expression the ? and ? are corresponding to,
h ?
b
h
? 1 T
? max
Attention！ b For the narrow rectangular sections
114
hb
t
W
W
T
t
2m a x
m a x, ?? ?? 其中
注意！ 对于 W t 和 It, 多数教材与手册上有如下定义,
hbIGI T t
t
3,,?? ?? 其中
m a x1 ??? ?
3
1 ; ) 10
, ( ??? ??bh即对于狭长矩形
查表求 ? 和 ? 时一定要 注意，表中 ? 和 ? 与那套公式对应。
h ?
b
h
? 1 T
? max
注意！ b
115
Example 7 For a straight steel rod with a rectangular section,its
dimensions of the cross section are h = 100 mm,b=50mm,its length is L=2m，
External torques acting on two ends of the rod are T=4000N·m,G =80GPa,
[?]=100MPa,[?]=1o/m, Try to check the strength and rigidity of the rod,
Solution,① To reference to the table to determine ?, ?
② Check the strength
0, 4 9 3 ; 0, 4 5 7 ; 250100 ????? ??bh
m106610504930 3633 ??????,..btW ?
116
[例 8] 一矩形截面等直钢杆，其横截面尺寸为,h = 100 mm,
b=50mm，长度 L=2m，杆的两端受扭转力偶 T=4000N·m 的
作用,钢的 G =80GPa, [?]=100M Pa,[?]=1o/m,试校核
此杆的强度和刚度。
解,①查表求 ?, ?
② 校核强度
0, 4 9 3 ; 0, 4 5 7 ; 250100 ????? ??bh
m106610504930 3633 ??????,..btW ?
117
③ Check the rigidity
? ??? ??
?
?? ? M P a65
10661
4000
6
m a x
m a x,W
T
t
4844 m102840504570 ??????,.bI
t ?
? ??? ???
???
?? ? /m1r a d / m017450
102861080
4000 o
89,GI
T
t
To sum up,the rod satisfies the requests in strength and rigidity,
118
③ 校核刚度
? ??? ??
?
?? ? M P a65
10661
4000
6
m a x
m a x,W
T
t
4844 m102840504570 ??????,.bI
t ?
? ??? ???
???
?? ? /m1r a d / m017450
102861080
4000 o
89,GI
T
t
综上,此杆满足强度和刚度要求。
119
1,The direction of the shearing stress current
coincides with that of the internal torque,
2 Distribution of the
shearing stress of the
opened thin-walled tubes
in free torsion is shown in
the figure（ a）,The
shearing stress at each
middle point of thickness
is zero,
§ 3–9 STRESSES OF OPENED AND CLOSED RODS WITH
THIN-WALLED SECTIONS IN FREE TORSION
Fig,
120
一、剪应力流的方向与扭矩的方向一致。
二、开口薄壁截面杆在自由扭转时的剪应力分布如图（ a），厚
度中点处，应力为零。
§ 3–9 开口和闭口薄壁截面杆在自由扭转时的应力
121
3,Distribution of the shearing stress of the closed thin-walled
tubes in free torsion is shown in the figure（ b）,Shearing
stresses distribute uniformly in the same thickness,
Fig,
122
三、闭口薄壁截面杆在自由扭转时的剪应力分布如图（ b），同
一厚度处，应力均匀分布。
123
4,Calculation of the shearing stress of the closed thin-walled
tubes in free torsion,Take an element in the figure （ c） as
shown in the figure（ d）。
Fig.（ c）
? 1
? 2
Fig.（ d）
2211 d d ; 0 ???? ????? xxX
co n s t a n t 2211 ?? ????
124
四、闭口薄壁截面杆自由扭转时的剪应力计算，在（ c）图上取
单元体如图（ d）。
图（ c）
? 1
? 2
图（ d）
2211 d d ; 0 ???? ??? ??? xxX
常量??? 2211 ????
125
??
?
m i n
m a x 2
T?
? ? ???????
?
22 ????? d)ds(T
2
1
?? ??
?
??? dds
?
is the area of the circle at average radius
126
??
?
m i n
m a x 2
T?
? ? ???????
?
22 ????? d)ds(T
积。为厚度中线所包围的面
2
1
????
?
??? dds
?
127
Example 8 The elliptical cross-section of a closed thin-walled tube is shown
in the following figure,Its dimensions are a=50 mm,b=75mm,the thickness
is t =5mm,the external torque acting on the two ends of the tube is T=5000N·m,
Try to determine the maximum shearing stress,
Solution,The maximum shearing stress of
the closed thin-walled tube in free torsion,
b
a t
M P a42
10755052
5 0 0 0
22
9
m i n
m a x
?
????
?
??
?
?
???
?
abt
TT
128
[例 8]下图示椭圆形薄壁截面杆，横截面尺寸为,a=50 mm，
b=75mm，厚度 t =5mm，杆两端受扭转力偶 T=5000N·m,试
求此杆的最大剪应力。
解,闭口薄壁 杆自由扭转时的最大剪应力,
b
a t
M P a42
10755052
5000
22
9
m i n
m a x
?
????
?
??
?
?
???
?
abt
TT
Chapter 3 Exercises
1,In a gear-box,why is the shaft with a slow speed thicker than that with
a height speed？
2,When the shearing stress and the normal stress coexist in an element,is
the theorem of the conjugate shearing stress amenable? Why?
3,For a hollow circular tube of Al,the outside diameter is D＝ 100mm，
the inside diameter is d＝ 80mm,and the length is L＝ 2.5m,The Shearing
elastic modulus is G＝ 28GPa,
① If a force couple is acted on both ends of the circular tube and results
in the pure torsion,try to determine the torsional angle when the maximum
shearing stress is 50MPa,
② For the solid shaft of Al subjected to the same force couple and having
the same maximum shearing stress,what is the diameter D2?
③ Determine the ratio of the weight of the hollow tube to that of the solid
shaft,
130
第三章 练习题
一、在变速箱中，为什么低速轴比高速轴粗？
二、当单元体上同时存在剪应力和正应力时，剪
应力互等定理是否成立？为什么？
三、铝制空心圆管，外径 D＝ 100mm，内径 d＝
80mm，长度 L＝ 2.5m。铝的剪切弹性模量 G＝ 28GPa。
①若圆管两端受力偶矩作用产生纯扭转，试求当
最大剪应力为 50MPa时的扭转角。
②对于承受相同力偶矩并产生相同最大剪应力的
铝制实心轴，其直径 应为多大？
③求空心管与实心轴的重量之比。
2D
131
Solution,①
②
③ As the two have the same material and the same length,
the ratio of weights is equal to the ratio of cross-section areas,
??
?
?????? 12.5
1 0 028
5.25022 m a xm a x
GD
L
GI
LW
GI
TL
p
t
p
???
tt W
T
W
T ?
1
)1(1616 4
33
1 ??? ?? DD?
mmDD 9.838.011 0 01 3 43 41 ????? ?
51.0
9.83
801 0 0
4
/)(
4 2
22
2
1
22 ???? DdD ??
132
解,①
②
③ 两者材料、长度相同，重量之比等于横
截面面积之比,
??
?
?????? 12.5
1 0 028
5.25022 m a xm a x
GD
L
GI
LW
GI
TL
p
t
p
???
tt W
T
W
T ?
1
)1(1616 4
33
1 ??? ?? DD?
mmDD 9.838.011 0 01 3 43 41 ????? ?
51.0
9.83
801 0 0
4
/)(
4 2
22
2
1
22 ???? DdD ??
133
134
