M e c h a n i c s o f M a t e r i a l s
1
2
CHAPTER 9 COMPOSITE DEFORMATIONS
§ 9–1 SUMMARY
§ 9–2 SKEW BENDING
§ 9–3 COMBINATION OF BENDING AND TORSION
§ 9-4 BENDING AND TENSION OR COMPRESSION ?
ECCENTRIC TENSION OR COMPRESSION ?
KERNEL OF THE SECTION
3
第九章 组合变形
§ 9–1 概述
§ 9–2 斜弯曲
§ 9–3 弯曲与扭转的组合
§ 9-4 拉 (压 )弯组合 ? 偏心拉（压） ? 截面核心
4
§ 9–1 SUMMARY
M
P
R z
x
y
P
P
1,Composite deformation,Structural members will produce several types
of simple deformations when subjected to complex external loads,The stress
corresponding to each simple deformation can not be neglected when the magnitude
of each stress has the same order,This kind of deformation is called the composite
deformation,
5
一、组合变形,在复杂外载作用下，构件的变形会包含几种简
单变形，当几种变形所对应的应力属同一量级时，不能忽略
之，这类构件的变形称为组合变形。
§ 9–1 概 述
M
P
R z
x
y
P
P
6
P
hg
7
P
hg
8
Dam
q
P
hg
9
水坝
q
P
hg
10
2,Methods to study composite deformation s—Principle of superposition
① Analysis of external forces,External forces are reduced along the
centroid of section and resolved along principal axes of inertia,
② Analysis of internal forces,Determine the internal-force equation
and its diagram corresponding to each external force component
and the critical section,
③ Analysis of stresses,Plot the distribution diagram of the stress
in the critical section,do the superposition of the stresses and
establish the strength condition of the critical point,
11
二、组合变形的研究方法 —— 叠加原理
① 外力分析,外力向形心 (或弯心 )简化并沿形心主惯性轴分解
② 内力分析,求每个外力分量对应的内力方程和内力图，确
定危险面。
③ 应力分析,画危险面应力分布图，叠加，建立危险点的强
度条件。
12
§ 9–2 SKEW BENDING
1,Skew bending,After bending deformation,the deflection curve and
the external forces（ transversal forces） of the rod are not in the same plane
2,Methods to study the skew bending,
1).Resolve,Resolve the external load along two centroid principal axes of inertia
of the cross section and get two perpendicular planar bending,
Pz
Py
y
z
P
j
x
y
z
P
Py Pz
13
§ 9–2 斜弯曲
一、斜弯曲, 杆件产生弯曲变形，但弯曲后，挠曲线与外力（横
向力）不共面。
二、斜弯曲的研究方法,
1.分解：将外载沿横截面的两个形心主轴分解，于是得到两个正交
的平面弯曲。
Py Pz
Pz
Py
y
z
P
j
14
x
y
z
P 13
2).Sum,Analyze bending in two perpendicular planes and sum the
results of the calculation,
x
y
z
Py Pz
P
Pz
Py
y
z
P
j
15
2.叠加：对两个平面弯曲进行研究；然后将计算结果叠加起来。
x
y
z
Py Pz
P
Pz
Py
y
z
P
j
jsinPPy? jc o sPPz?
Solution,1.Resolve the external force along the centroid principal
axis of inertia of the cross section
2.Study the bending in two planes,
jj s i ns i n)()( MxLPxLPM yz ?????
jco sMM y ?
① Internal forces
x
y
z
Py Pz
P
Pz
Py
y
z
P
j
L
m
m x
17
jsinPPy? jc o sPPz?
解,1.将外载沿横截面的形心主轴分解
2.研究两个平面弯曲
j
j
s in
s in)(
)(
M
xLP
xLPM yz
?
??
??
jco sMM y ?
①
内
力
18
Pz
Py
y
z
P
j
j? c o s
yy
y
I
M
I
zM z?????② Stresses
j? s in
zz
z
I
M
I
yM y??????
)s i nc o s( jj???
zy I
y
I
zM ????????
Stress due to My,
Stress due to M z,
Resultant stress,
L
Pz
Py
y
z
P
j x
y
z
Py Pz
P L
m
m x
19
j? c o s
yy
y
I
M
I
zM z?????② 应
力
j? s in
zz
z
I
M
I
yM y??????
)s i nc o s( jj???
zy I
y
I
zM ????????
My引起的应力,
M z引起的应力,
合应力,
20
L
Pz
Py
y
z
P
j x
y
z
Py Pz
P L
m
m x
④ Maximum normal stress
⑤ Calculation of the deformation
0)s i nc o s( 00 ???? jj?
zy I
y
I
zM
③ Equation of the neutral axis
j? c t gtg
0
0
y
z
I
I
z
y ??
It is obvious that only as Iy = Iz the neutral axis
is perpendicular to the external force,
2m a x DL ?? ? 1m a x Dy ?? ?
22 zy fff ??
z
y
f
f??tg
As j = ?,it is the planar bending,
Pz
Py
y
z
P
j
D1
D2
?
Neutral axis
f
fz
fy
?
The maximum normal stress of tension or
compression occurs in the points that lie in two sides
of the neutral axis and have the farthest distance to
the neutral axis,
21
④ 最大正应力
⑤ 变形计算
0)s i nc o s( 00 ???? jj?
zy I
y
I
zM
③ 中性轴方程
j? c t gtg
0
0
y
z
I
I
z
y ??
可见：只有当 Iy = Iz时，中性轴与外力才垂直。
在中性轴两侧，距中性轴最远的点为拉压最大正应力点。
22 zy fff ??
z
y
f
f??tg
当 j = ?时，即为平面弯曲。
D1
D2
?
中性轴
22
2m a x DL ?? ? 1m a x Dy ?? ?
Pz
Py
y
z
P
j
f
fz
fy
?
Example 1 Force P is through the center of section and makes an angle the j
with axis z in the beam as shown in the figure,Determine the maximum stress
and deflection of the beam,
Maximum normal stress Calculation of deformation
2
m a xm a x
1m a x D
y
y
z
z
DL W
M
W
M ??? ?????
2
3
2
3
22 )
3()3( y
z
z
y
zy EI
LP
EI
LPfff ????
j? tgtg
z
y
z
y
I
I
f
f ??
As Iy = Iz the beam produce
planar bending,
Solution,Analysis of the critical point is shown in the figure
f
fz
fy
?
y
z
L
x
Py
Pz
P
h
b P
z
Py
y
z
P
j
D2
D1
?
Neutral axis
23
[例 1]结构如图,P过形心且与 z轴成 j角,求此梁的最大应力与挠
度。
2
3
2
3
22 )
3()3( y
z
z
y
zy EI
LP
EI
LPfff ????
j? tgtg
z
y
z
y
I
I
f
f ??当 Iy = Iz时，即发生平面弯曲。
解：危险点分析如图
24
2
m a xm a x
1m a x D
y
y
z
z
DL W
M
W
M ??? ?????
最大正应力 变形计算
中性轴
f
fz
fy
?
y
z
L
x
Py
Pz
P
h
b P
z
Py
y
z
P
j
D2
D1
?
Example 2 A wood purline is shown in the figure,Its span is L=3m and a
uniformly distributed load q=800N/m is acting on it,The permissible stress and
deflection of it is respectively [?]=12MPa and L/200, E=9GPa,Try to
determine the dimension of the section and check the rigidity of the beam,
N / m358447.0800s i n ???? ?qq y
Solution,① Analysis of external force—
resolve q
? ??? ???
y
y
z
z
W
M
W
M
m a x
N / m715894.0800co s ???? ?qq z
Nm4 0 38 33 5 88
22
m a x ?
??? LqM y
z
Nm8048 37158
22
m a x ?
??? LqM z
y
? ?26° 34′
y z q
q
L
A B
25
[例 2] 矩形截面木檩条如图，跨长 L=3m,受集度为 q=800N/m的
均布力作用,[?]=12MPa,许可挠度为,L/200, E=9GPa，试
选择截面尺寸并校核刚度。
N / m358447.0800s i n ???? ?qq y
解,① 外力分析 — 分解 q
? ??? ???
y
y
z
z
W
M
W
M
ma x
N / m715894.0800co s ???? ?qq z
Nm4 0 38 33 5 88
22
m a x ?
??? LqM y
z
Nm8048 37158
22
m a x ?
??? LqM z
y
? ?26° 34′
y z q
q
L
A B
26
§ 9–3 COMBINATION OF BENDING AND TORSION
80o P2 z
y
x
P1
150 200 100 A B C D
27
§ 9–3 弯曲与扭转的组合
80o P2 z
y
x
P1
150 200 100 A B C D
28
Solution,① Reduce
the external force to the
centroid of the section
and resolve it
Establish the strength
condition of the rod
shown in the figure,
Bending and torsion
80o P2 z
y
x
P1
150 200 100 A B C D
150 200 100 A B C D
P1
Mx
z
x
y
P2y
P2z
Mx
29
解, ① 外力向形心
简化并分解
建立图示杆件的强度条件
弯扭组合变形
80o P2 z
y
x
P1
150 200 100 A B C D
30
150 200 100 A B C D
P1
Mx
z
x
y
P2y
P2z
Mx
③ Sum the bending moments
and plot the diagram of the
resultant bending moment,
)()()( 22 xMxMxM zy ??
④ Determine the critical
section
)( ; )( ; )( xMxMxM nzy
M Z ( N m)
X
(Nm) Mz
x
M y ( N m)
X
My (Nm) x
（ Nm ）
x
M nM n
Mn (Nm)
x
M ( N m)
X
M max
M (Nm) Mmax
x
② Internal equations and
diagrams corresponding to
each external force component
31
② 每个外力分量对应
的内力方程和内力图
③ 叠加 弯矩, 并画图
)()()( 22 xMxMxM zy ??
④ 确定危险面
)( ; )( ; )( xMxMxM nzy
M Z ( N m)
X
(Nm) Mz
x
M y ( N m)
X
My (Nm) x
（ Nm ）
x
M nM n
Mn (Nm)
x
M ( N m)
X
M max
M (Nm) Mmax
x
32
⑤ Plot the diagram of stress distribution
in the critical section and determine
the critical point,
W
M
xB
m a x
1 ??
P
n
B W
M?
1
?
22
3
1 )
2(2 ?
??
?
? ???
?
?
?
⑥ Establish the strength condition
22313 4 ????? ????r
2
2
2
2
m a x 4
P
n
W
M
W
M ??
1xB?
1B?
1xB?
1B?
33
x B1
B2 My
Mz
Mn
M
x
1xB?
2xB?
M
1B?
⑤ 画危险面应力分布图，找危险点
W
M
xB
m a x
1 ??
P
n
B W
M?
1
?
22
3
1 )
2(2 ?
??
?
? ???
?
?
?
⑥ 建立强度条件
2
2
2
2
m a x 4
P
n
W
M
W
M ??
1xB?
1B?
1xB?
1B?
34
x B1
B2 My
Mz
Mn
M
x
1xB?
2xB?
M
1B?
22313 4 ????? ????r
? ? ? ? ? ?? ?2132322214 21 ??????? ??????r
1xB?
1B?
22 3?? ??
W
MM n22 75.0?
?
W
MMM nzy 222 75.0??
?
W
MMM nzy
r
222
4
75.0??
??
W
MMM nzy
r
222
3
??
??
22313 4 ????? ????r
2
2
2
2
m a x 4
P
n
W
M
W
M ??
W
MMM nzy 222 ??
?
35
36
22 3?? ??
W
MM n22 75.0?
?
W
MMM nzy 222 75.0??
?
W
MMM nzy
r
222
4
75.0??
??
W
MMM nzy
r
222
3
??
??
22313 4 ????? ????r
2
2
2
2
m a x 4
P
n
W
M
W
M ??
W
MMM nzy 222 ??
?
35
1xB?
1B?
? ? ? ? ? ?? ?2132322214 21 ??????? ??????r
① Analysis of external forces,Reduce the external forces to the
centroid of section and resolve them,
③ Analysis of stresses, Establish strength conditions,
Steps of solving the problem of composite deformation of
bending and torsion,
② Analysis of internal forces, Determine the internal equation
and its diagram corresponding toeach external force component
and critical section,
37
W
MMM nzy
r
222
3
??
??
W
MMM nzy
r
222
4
75.0??
??
① 外力分析,外力向形心简化并 分解。
② 内力分析,每个外力分量对应的内力方程和内力图,确定危
险面。
③ 应力分析,建立强度条件。
W
MMM nzy
r
222
3
??
??
W
MMM nzy
r
222
4
75.0??
??
弯扭组合问题的求解步骤,
38
Example 3 A hollow
circular shaft is shown in the
figure.Its inside diameter is
d=24mm and its outside
diameter is
The diameters of pulley B and
D are respectively D=30mm
D1 =400mm and D2 =600mm，
P1=600N,[?]=100MPa,Try
to check the strength of the
shaft with the third strength,
① Analysis of
external forces,
Bending and torsion
80o P2 z
y
x
P1
150 200 100 A B C D
150 200 100 A B C D
P1
Mx
z
x
y
P2y
P2z
Mx
Solution,
39
[例 3] 图示空心圆轴，
内径 d=24mm，外
径 D=30mm,B 轮
直径 D 1 ＝ 400mm，
D轮直径 D 2＝
600mm,P1=600N，
[?]=100MPa，试用
第三强度理论校核
此轴的强度。
① 外力分析,
弯扭组合变形
80o P2 z
y
x
P1
150 200 100 A B C D
150 200 100 A B C D
P1
Mx
z
x
y
P2y
P2z
Mx
解,
40
② Analysis of internal
forces,Internal forces
in the critical section are,
③ Analysis of stress,
W
MM n
r
22
m a x
3
???
Nm3.71m a x ?M
Nm1 2 0?nM
)8.01(03.014.3
1203.7132
43
22
??
??
? ???? M P a5.97
It is safe
M Z ( N m)
X
(Nm) Mz
x
M y ( N m)
X
My (Nm) x
（ Nm ）
x
M nM n
Mn (Nm)
x
M ( N m)
X
M max
M (Nm) 71.3
x
71.25
40
7.05
120
5.5
40.6
41
② 内力分析,危
险面内力为,
③ 应力分析,
Nm3.71m a x ?M
Nm1 2 0?nM
)8.01(03.014.3
1203.7132
43
22
??
??
? ???? M P a5.97
安全
M Z ( N m)
X
(Nm) Mz
x
M y ( N m)
X
My (Nm) x
（ Nm ）
x
M nM n
Mn (Nm)
x
M ( N m)
X
M max
M (Nm) 71.3
x
71.25
40
7.05
120
5.5
40.6
42
W
MM n
r
22
m a x
3
???
§ 9–4 BENDING AND TENSION OR COMPRESSION ? ECCENTRIC
TENSION OR COMPRESSION ? KERNEL OF THE SECTION
P
R
P
x
y z P
My
x
y z P
My
Mz
1,Composite deformation of bending and tension or compression：
Deformation of the rod due to simultaneous action of transversal and axial
forces,
43
§ 9–4 拉 (压 )弯组合 ? 偏心拉（压） ? 截面核心
一、拉 (压 )弯组合变形,杆件同时受横向力和轴向力的作用而产
生的变形。
P
R
44
P
x
y z P
My
x
y z P
My
Mz
A
P
xP ??
z
z
xM I
yM
z
???
y
y
xM I
zM
y
??
y
y
z
z
x I
zM
I
yM
A
P ????
2,Analysis of stress,
45
P
My
Mz
P MZ My x
y z z y
A
P
xP ??
z
z
xM I
yM
z
???
y
y
xM I
zM
y
??
y
y
z
z
x I
zM
I
yM
A
P ????
二、应力分析,
46
P
My
Mz
P MZ My x
y z z y
000 ????
y
y
z
z
x I
zM
I
yM
A
P?
4,Critical point
（ Farthest point from the neutral axis）
3,Equation of the neutral
axis
For the problem of eccentric
tension or compression
0)1( 2 02 02 02 0 ??????
y
P
z
P
y
P
z
P
i
zz
i
yy
A
P
Ai
zPz
Ai
yPy
A
P
y
y
z
z
L W
M
W
M
A
P ???
ma x?
y
y
z
z
y W
M
W
M
A
P ???
m a x?
01 2 02 0 ???
y
P
z
P
i
zz
i
yy
47
P （ z P, y P ）
y
z
y
z
),( PP yzP
Neutral axis
000 ????
y
y
z
z
x I
zM
I
yM
A
P?
四、危险点
（距中性轴最远的点）
三、中性轴方程
对于偏心拉压问题
0)1( 2 02 02 02 0 ??????
y
P
z
P
y
P
z
P
i
zz
i
yy
A
P
Ai
zPz
Ai
yPy
A
P
01 2 02 0 ???
y
P
z
P
i
zz
i
yy
中性轴
48
y
y
z
z
L W
M
W
M
A
P ???
ma x?
y
y
z
z
y W
M
W
M
A
P ???
m a x?
P （ z P, y P ）
y
z
y
z
),( PP yzP
y
z
5,Kernel of section in the problem of the eccentric tension、
compression,
ay
az
01 2 ??
z
yP
i
ay
01 2 ??
y
zP
i
az
After knowing ay and az,
The action range of the compressive force,As the compressive force is acted in
this range there are no tensile stresses in the section,
May determine an action point
of the force P,),( PP yz
01 2 02 0 ???
y
P
z
P
i
zz
i
yy
Neutral axis ),( PP yzP
Kernel of section
49
y
z
五、（偏心拉、压问题的）截面核心,
ay
az
01 2 ??
z
yP
i
ay
01 2 ??
y
zP
i
az
已知 ay,az 后,
压力作用区域。
当压力作用在此区域内时，横截面上无拉应力。
可求 P 力的一个作用点 ),(
PP yz
01 2 02 0 ???
y
P
z
P
i
zz
i
yy
中性轴 ),( PP yzP
截面核心
50
M P a75.82.02.03 5 0 0 0 0 m a x2 ???? AP?
???
11
m a x1
zW
M
A
P?
M P a7.11
3.02.0
6503 5 0
3.02.0
3 5 0 0 0 0
2
?
?
???
?
Solution,The stresses
in the cross sections of the
two poles are both compressive
ones,
Example 4 Two poles subjected to the force P=350kN are shown in
the figure,The section of one pole is unequal and the section of the other pole is
equal,Try to determine the normal stress with maximum absolute value in the
poles,
M P
P d
51
Fig, Fig,
P
300
200
P
200
M P a75.8
2.02.0
3 5 0 0 0 0
m a x2
?
?
?
?
A
P
?
???
11
m a x1
zW
M
A
P?
M P a7.11
3.02.0
6503 5 0
3.02.0
3 5 0 0 0 0
2
?
?
???
?
解,两柱横截面上的最大正
应力均为 压应力
[例 4] 图示不等截面与等截面柱，受力 P=350kN，试分别求出两
柱内的绝对值最大正应力。
图（ 1） 图（ 2）
M P
P d
52
.,
P
300
200
P
200
mm5102010100 201020 ???? ???Cz
2
3
5100101210010 ?????CyI
45
2
3
mm1027.7
]252010
12
2010[
??
?????
Solution,Analysis of the internal force is
shown in the figure,Centroid of the slot in
the coordinates shown in the figure
Nm5 0 0105 3 ??? ?PM
P
P
Slot
Example 5 A steel plate shown in the figure is subjected to forces
P=100kN,Try to determine the maximum normal stress； If the slot is moved to the
middle of the plate and the maximum normal stress is kept constant,how much
should the width of the slot be?
53
P
P
M
N
20 100 20
y
z
yC
10
mm5102010100 201020 ???? ???Cz
2
3
5100101210010 ?????CyI
45
2
3
mm1027.7
]252010
12
2010[
??
?????
例 5 图示钢板受力 P=100kN,试求最大正应力；若将缺口移至板
宽的中央，且使最大正应力保持不变，则挖空宽度为多少？
解,内力分析 如图
坐标如图，挖孔处的形心
Nm5 0 0105 3 ??? ?PM
P
P
54
P
P
M
N
20 100 20
y
z
yC
10
P
P
M
N
ycI
zM
A
N m a x
m a x ???
M P a8.1628.37125 ???
Analysis of stress is
shown in the figure,
7
3
6
3
1027.7
1055500
10800
10100
?
?
? ?
???
?
??
As the hole is moved to
the middle of the plate
)100(10mm9.631108.162 10100 26
3
m a x
xNA ??????? ?
mm8.36 so ?x
20 100 20
y
z
yC
55
＋
P
P
M
N
M P a8.1628.37125 ???
应力分析 如图
7
3
6
3
1027.7
1055500
10800
10100
?
?
? ?
???
?
??
孔移至板中间时
)100(10mm9.631108.162 10100 26
3
m a x
xNA ??????? ?
mm8.36 ?? x
20 100 20
y
z
yC
56
ycI
zM
A
N m a x
m a x ???
＋
M P a7.351.07 0 0 016 3 ????? ??
nW
T
M P a37.6101.0504 32 ?????? ?? AP
Solution,For the composite
deformation of tension and
torsion,the stressed state at the
critical point is shown in the figure,
Example 6 A circular rod which the diameter d =0.1m is subjected to forces
T=7kNm and P=50kN as shown in the figure,[?]=100MPa,Try to check the
strength of the rod with the third strength theory,
Therefore,the rod is safe,
223 4??? ??r
? ????
???
M P a7.71
7.35437.6 22
?
?
A
A P
P
T
T
57
M P a7.351.07 0 0 016 3 ????? ??
nW
T
M P a37.6101.0504 32 ?????? ?? AP
解,拉扭组合,危险点 应力状态如图
[例 6] 直径为 d=0.1m的圆杆受力如图,T=7kNm,P=50kN,
[?]=100MPa,试按 第三强度理论校核此杆的强度 。
故，安全。 ? ???? ??? M P a7.71 7.35437.6 22
?
?
A
A P
P
T
T
58
223 4??? ??r
59
Chapter 9 Exercises
1,A circular shaft of steel is deformed under tension and torsion,
Try to write out the strength conditions,If it shows the tension-torsion-
bending composite deformation,try to write out the strength conditions,
2,The cross-section area of the square-section rod is reduced half at
the section mn,Try to determine the maximum tensile stress at the
section mn induced by the axial force P,
Solution,
3,A rectangular-section beam is shown in the figure,Knowing b =
50mm and h =75mm,Try to determine the maximum normal stress of
the beam,If the beam is changed to have circular sections with the
diameter d = 65mm, what is the maximum normal stress?
W
M
A
N ??
m a x?
2
22 8
64/42/ a
PaaaPaP ?
?
??
?
??
?
? ???
60
第九章 练习题
一、钢圆轴为拉伸与扭转的组合变形，试写出
其强度条件。若为拉伸、扭转和弯曲的组合变形，
试写出其强度条件。
二、方形截面杆的横截面面积在 mn 处减少一
半，试求由轴向载荷 P 引起的 mn 截面上的最大拉
应力。 解,
三、矩形截面梁如图。已知 b = 50mm,h =75
mm,求梁内的最大正应力。如改为 d = 65mm 的
圆截面，最大正应力为多少？
W
M
A
N ??
m a x?
2
22 8
64/42/ a
PaaaPaP ?
?
??
?
??
?
? ???
61
Soluion,
If the beam is changed to have circular sections,
y
y
z
z
W
M
W
M m a xm a x
m a x ???
M P a9605.0075.0 61021075.005.0 6105.1 2323 ?? ????? ???
M P aoW MM yz 8.9265,25.132 3 22
2
m a x
2
m a x
m a x ??
????
??
62
解,
如改为圆截面
y
y
z
z
W
M
W
M m a xm a x
m a x ???
M P aoW MM yz 8.9265,25.132 3 22
2
m a x
2
m a x
m a x ??
????
??
M P a9605.0075.0 61021075.005.0 6105.1 2323 ?? ????? ???
63
64
1
2
CHAPTER 9 COMPOSITE DEFORMATIONS
§ 9–1 SUMMARY
§ 9–2 SKEW BENDING
§ 9–3 COMBINATION OF BENDING AND TORSION
§ 9-4 BENDING AND TENSION OR COMPRESSION ?
ECCENTRIC TENSION OR COMPRESSION ?
KERNEL OF THE SECTION
3
第九章 组合变形
§ 9–1 概述
§ 9–2 斜弯曲
§ 9–3 弯曲与扭转的组合
§ 9-4 拉 (压 )弯组合 ? 偏心拉（压） ? 截面核心
4
§ 9–1 SUMMARY
M
P
R z
x
y
P
P
1,Composite deformation,Structural members will produce several types
of simple deformations when subjected to complex external loads,The stress
corresponding to each simple deformation can not be neglected when the magnitude
of each stress has the same order,This kind of deformation is called the composite
deformation,
5
一、组合变形,在复杂外载作用下，构件的变形会包含几种简
单变形，当几种变形所对应的应力属同一量级时，不能忽略
之，这类构件的变形称为组合变形。
§ 9–1 概 述
M
P
R z
x
y
P
P
6
P
hg
7
P
hg
8
Dam
q
P
hg
9
水坝
q
P
hg
10
2,Methods to study composite deformation s—Principle of superposition
① Analysis of external forces,External forces are reduced along the
centroid of section and resolved along principal axes of inertia,
② Analysis of internal forces,Determine the internal-force equation
and its diagram corresponding to each external force component
and the critical section,
③ Analysis of stresses,Plot the distribution diagram of the stress
in the critical section,do the superposition of the stresses and
establish the strength condition of the critical point,
11
二、组合变形的研究方法 —— 叠加原理
① 外力分析,外力向形心 (或弯心 )简化并沿形心主惯性轴分解
② 内力分析,求每个外力分量对应的内力方程和内力图，确
定危险面。
③ 应力分析,画危险面应力分布图，叠加，建立危险点的强
度条件。
12
§ 9–2 SKEW BENDING
1,Skew bending,After bending deformation,the deflection curve and
the external forces（ transversal forces） of the rod are not in the same plane
2,Methods to study the skew bending,
1).Resolve,Resolve the external load along two centroid principal axes of inertia
of the cross section and get two perpendicular planar bending,
Pz
Py
y
z
P
j
x
y
z
P
Py Pz
13
§ 9–2 斜弯曲
一、斜弯曲, 杆件产生弯曲变形，但弯曲后，挠曲线与外力（横
向力）不共面。
二、斜弯曲的研究方法,
1.分解：将外载沿横截面的两个形心主轴分解，于是得到两个正交
的平面弯曲。
Py Pz
Pz
Py
y
z
P
j
14
x
y
z
P 13
2).Sum,Analyze bending in two perpendicular planes and sum the
results of the calculation,
x
y
z
Py Pz
P
Pz
Py
y
z
P
j
15
2.叠加：对两个平面弯曲进行研究；然后将计算结果叠加起来。
x
y
z
Py Pz
P
Pz
Py
y
z
P
j
jsinPPy? jc o sPPz?
Solution,1.Resolve the external force along the centroid principal
axis of inertia of the cross section
2.Study the bending in two planes,
jj s i ns i n)()( MxLPxLPM yz ?????
jco sMM y ?
① Internal forces
x
y
z
Py Pz
P
Pz
Py
y
z
P
j
L
m
m x
17
jsinPPy? jc o sPPz?
解,1.将外载沿横截面的形心主轴分解
2.研究两个平面弯曲
j
j
s in
s in)(
)(
M
xLP
xLPM yz
?
??
??
jco sMM y ?
①
内
力
18
Pz
Py
y
z
P
j
j? c o s
yy
y
I
M
I
zM z?????② Stresses
j? s in
zz
z
I
M
I
yM y??????
)s i nc o s( jj???
zy I
y
I
zM ????????
Stress due to My,
Stress due to M z,
Resultant stress,
L
Pz
Py
y
z
P
j x
y
z
Py Pz
P L
m
m x
19
j? c o s
yy
y
I
M
I
zM z?????② 应
力
j? s in
zz
z
I
M
I
yM y??????
)s i nc o s( jj???
zy I
y
I
zM ????????
My引起的应力,
M z引起的应力,
合应力,
20
L
Pz
Py
y
z
P
j x
y
z
Py Pz
P L
m
m x
④ Maximum normal stress
⑤ Calculation of the deformation
0)s i nc o s( 00 ???? jj?
zy I
y
I
zM
③ Equation of the neutral axis
j? c t gtg
0
0
y
z
I
I
z
y ??
It is obvious that only as Iy = Iz the neutral axis
is perpendicular to the external force,
2m a x DL ?? ? 1m a x Dy ?? ?
22 zy fff ??
z
y
f
f??tg
As j = ?,it is the planar bending,
Pz
Py
y
z
P
j
D1
D2
?
Neutral axis
f
fz
fy
?
The maximum normal stress of tension or
compression occurs in the points that lie in two sides
of the neutral axis and have the farthest distance to
the neutral axis,
21
④ 最大正应力
⑤ 变形计算
0)s i nc o s( 00 ???? jj?
zy I
y
I
zM
③ 中性轴方程
j? c t gtg
0
0
y
z
I
I
z
y ??
可见：只有当 Iy = Iz时，中性轴与外力才垂直。
在中性轴两侧，距中性轴最远的点为拉压最大正应力点。
22 zy fff ??
z
y
f
f??tg
当 j = ?时，即为平面弯曲。
D1
D2
?
中性轴
22
2m a x DL ?? ? 1m a x Dy ?? ?
Pz
Py
y
z
P
j
f
fz
fy
?
Example 1 Force P is through the center of section and makes an angle the j
with axis z in the beam as shown in the figure,Determine the maximum stress
and deflection of the beam,
Maximum normal stress Calculation of deformation
2
m a xm a x
1m a x D
y
y
z
z
DL W
M
W
M ??? ?????
2
3
2
3
22 )
3()3( y
z
z
y
zy EI
LP
EI
LPfff ????
j? tgtg
z
y
z
y
I
I
f
f ??
As Iy = Iz the beam produce
planar bending,
Solution,Analysis of the critical point is shown in the figure
f
fz
fy
?
y
z
L
x
Py
Pz
P
h
b P
z
Py
y
z
P
j
D2
D1
?
Neutral axis
23
[例 1]结构如图,P过形心且与 z轴成 j角,求此梁的最大应力与挠
度。
2
3
2
3
22 )
3()3( y
z
z
y
zy EI
LP
EI
LPfff ????
j? tgtg
z
y
z
y
I
I
f
f ??当 Iy = Iz时，即发生平面弯曲。
解：危险点分析如图
24
2
m a xm a x
1m a x D
y
y
z
z
DL W
M
W
M ??? ?????
最大正应力 变形计算
中性轴
f
fz
fy
?
y
z
L
x
Py
Pz
P
h
b P
z
Py
y
z
P
j
D2
D1
?
Example 2 A wood purline is shown in the figure,Its span is L=3m and a
uniformly distributed load q=800N/m is acting on it,The permissible stress and
deflection of it is respectively [?]=12MPa and L/200, E=9GPa,Try to
determine the dimension of the section and check the rigidity of the beam,
N / m358447.0800s i n ???? ?qq y
Solution,① Analysis of external force—
resolve q
? ??? ???
y
y
z
z
W
M
W
M
m a x
N / m715894.0800co s ???? ?qq z
Nm4 0 38 33 5 88
22
m a x ?
??? LqM y
z
Nm8048 37158
22
m a x ?
??? LqM z
y
? ?26° 34′
y z q
q
L
A B
25
[例 2] 矩形截面木檩条如图，跨长 L=3m,受集度为 q=800N/m的
均布力作用,[?]=12MPa,许可挠度为,L/200, E=9GPa，试
选择截面尺寸并校核刚度。
N / m358447.0800s i n ???? ?qq y
解,① 外力分析 — 分解 q
? ??? ???
y
y
z
z
W
M
W
M
ma x
N / m715894.0800co s ???? ?qq z
Nm4 0 38 33 5 88
22
m a x ?
??? LqM y
z
Nm8048 37158
22
m a x ?
??? LqM z
y
? ?26° 34′
y z q
q
L
A B
26
§ 9–3 COMBINATION OF BENDING AND TORSION
80o P2 z
y
x
P1
150 200 100 A B C D
27
§ 9–3 弯曲与扭转的组合
80o P2 z
y
x
P1
150 200 100 A B C D
28
Solution,① Reduce
the external force to the
centroid of the section
and resolve it
Establish the strength
condition of the rod
shown in the figure,
Bending and torsion
80o P2 z
y
x
P1
150 200 100 A B C D
150 200 100 A B C D
P1
Mx
z
x
y
P2y
P2z
Mx
29
解, ① 外力向形心
简化并分解
建立图示杆件的强度条件
弯扭组合变形
80o P2 z
y
x
P1
150 200 100 A B C D
30
150 200 100 A B C D
P1
Mx
z
x
y
P2y
P2z
Mx
③ Sum the bending moments
and plot the diagram of the
resultant bending moment,
)()()( 22 xMxMxM zy ??
④ Determine the critical
section
)( ; )( ; )( xMxMxM nzy
M Z ( N m)
X
(Nm) Mz
x
M y ( N m)
X
My (Nm) x
（ Nm ）
x
M nM n
Mn (Nm)
x
M ( N m)
X
M max
M (Nm) Mmax
x
② Internal equations and
diagrams corresponding to
each external force component
31
② 每个外力分量对应
的内力方程和内力图
③ 叠加 弯矩, 并画图
)()()( 22 xMxMxM zy ??
④ 确定危险面
)( ; )( ; )( xMxMxM nzy
M Z ( N m)
X
(Nm) Mz
x
M y ( N m)
X
My (Nm) x
（ Nm ）
x
M nM n
Mn (Nm)
x
M ( N m)
X
M max
M (Nm) Mmax
x
32
⑤ Plot the diagram of stress distribution
in the critical section and determine
the critical point,
W
M
xB
m a x
1 ??
P
n
B W
M?
1
?
22
3
1 )
2(2 ?
??
?
? ???
?
?
?
⑥ Establish the strength condition
22313 4 ????? ????r
2
2
2
2
m a x 4
P
n
W
M
W
M ??
1xB?
1B?
1xB?
1B?
33
x B1
B2 My
Mz
Mn
M
x
1xB?
2xB?
M
1B?
⑤ 画危险面应力分布图，找危险点
W
M
xB
m a x
1 ??
P
n
B W
M?
1
?
22
3
1 )
2(2 ?
??
?
? ???
?
?
?
⑥ 建立强度条件
2
2
2
2
m a x 4
P
n
W
M
W
M ??
1xB?
1B?
1xB?
1B?
34
x B1
B2 My
Mz
Mn
M
x
1xB?
2xB?
M
1B?
22313 4 ????? ????r
? ? ? ? ? ?? ?2132322214 21 ??????? ??????r
1xB?
1B?
22 3?? ??
W
MM n22 75.0?
?
W
MMM nzy 222 75.0??
?
W
MMM nzy
r
222
4
75.0??
??
W
MMM nzy
r
222
3
??
??
22313 4 ????? ????r
2
2
2
2
m a x 4
P
n
W
M
W
M ??
W
MMM nzy 222 ??
?
35
36
22 3?? ??
W
MM n22 75.0?
?
W
MMM nzy 222 75.0??
?
W
MMM nzy
r
222
4
75.0??
??
W
MMM nzy
r
222
3
??
??
22313 4 ????? ????r
2
2
2
2
m a x 4
P
n
W
M
W
M ??
W
MMM nzy 222 ??
?
35
1xB?
1B?
? ? ? ? ? ?? ?2132322214 21 ??????? ??????r
① Analysis of external forces,Reduce the external forces to the
centroid of section and resolve them,
③ Analysis of stresses, Establish strength conditions,
Steps of solving the problem of composite deformation of
bending and torsion,
② Analysis of internal forces, Determine the internal equation
and its diagram corresponding toeach external force component
and critical section,
37
W
MMM nzy
r
222
3
??
??
W
MMM nzy
r
222
4
75.0??
??
① 外力分析,外力向形心简化并 分解。
② 内力分析,每个外力分量对应的内力方程和内力图,确定危
险面。
③ 应力分析,建立强度条件。
W
MMM nzy
r
222
3
??
??
W
MMM nzy
r
222
4
75.0??
??
弯扭组合问题的求解步骤,
38
Example 3 A hollow
circular shaft is shown in the
figure.Its inside diameter is
d=24mm and its outside
diameter is
The diameters of pulley B and
D are respectively D=30mm
D1 =400mm and D2 =600mm，
P1=600N,[?]=100MPa,Try
to check the strength of the
shaft with the third strength,
① Analysis of
external forces,
Bending and torsion
80o P2 z
y
x
P1
150 200 100 A B C D
150 200 100 A B C D
P1
Mx
z
x
y
P2y
P2z
Mx
Solution,
39
[例 3] 图示空心圆轴，
内径 d=24mm，外
径 D=30mm,B 轮
直径 D 1 ＝ 400mm，
D轮直径 D 2＝
600mm,P1=600N，
[?]=100MPa，试用
第三强度理论校核
此轴的强度。
① 外力分析,
弯扭组合变形
80o P2 z
y
x
P1
150 200 100 A B C D
150 200 100 A B C D
P1
Mx
z
x
y
P2y
P2z
Mx
解,
40
② Analysis of internal
forces,Internal forces
in the critical section are,
③ Analysis of stress,
W
MM n
r
22
m a x
3
???
Nm3.71m a x ?M
Nm1 2 0?nM
)8.01(03.014.3
1203.7132
43
22
??
??
? ???? M P a5.97
It is safe
M Z ( N m)
X
(Nm) Mz
x
M y ( N m)
X
My (Nm) x
（ Nm ）
x
M nM n
Mn (Nm)
x
M ( N m)
X
M max
M (Nm) 71.3
x
71.25
40
7.05
120
5.5
40.6
41
② 内力分析,危
险面内力为,
③ 应力分析,
Nm3.71m a x ?M
Nm1 2 0?nM
)8.01(03.014.3
1203.7132
43
22
??
??
? ???? M P a5.97
安全
M Z ( N m)
X
(Nm) Mz
x
M y ( N m)
X
My (Nm) x
（ Nm ）
x
M nM n
Mn (Nm)
x
M ( N m)
X
M max
M (Nm) 71.3
x
71.25
40
7.05
120
5.5
40.6
42
W
MM n
r
22
m a x
3
???
§ 9–4 BENDING AND TENSION OR COMPRESSION ? ECCENTRIC
TENSION OR COMPRESSION ? KERNEL OF THE SECTION
P
R
P
x
y z P
My
x
y z P
My
Mz
1,Composite deformation of bending and tension or compression：
Deformation of the rod due to simultaneous action of transversal and axial
forces,
43
§ 9–4 拉 (压 )弯组合 ? 偏心拉（压） ? 截面核心
一、拉 (压 )弯组合变形,杆件同时受横向力和轴向力的作用而产
生的变形。
P
R
44
P
x
y z P
My
x
y z P
My
Mz
A
P
xP ??
z
z
xM I
yM
z
???
y
y
xM I
zM
y
??
y
y
z
z
x I
zM
I
yM
A
P ????
2,Analysis of stress,
45
P
My
Mz
P MZ My x
y z z y
A
P
xP ??
z
z
xM I
yM
z
???
y
y
xM I
zM
y
??
y
y
z
z
x I
zM
I
yM
A
P ????
二、应力分析,
46
P
My
Mz
P MZ My x
y z z y
000 ????
y
y
z
z
x I
zM
I
yM
A
P?
4,Critical point
（ Farthest point from the neutral axis）
3,Equation of the neutral
axis
For the problem of eccentric
tension or compression
0)1( 2 02 02 02 0 ??????
y
P
z
P
y
P
z
P
i
zz
i
yy
A
P
Ai
zPz
Ai
yPy
A
P
y
y
z
z
L W
M
W
M
A
P ???
ma x?
y
y
z
z
y W
M
W
M
A
P ???
m a x?
01 2 02 0 ???
y
P
z
P
i
zz
i
yy
47
P （ z P, y P ）
y
z
y
z
),( PP yzP
Neutral axis
000 ????
y
y
z
z
x I
zM
I
yM
A
P?
四、危险点
（距中性轴最远的点）
三、中性轴方程
对于偏心拉压问题
0)1( 2 02 02 02 0 ??????
y
P
z
P
y
P
z
P
i
zz
i
yy
A
P
Ai
zPz
Ai
yPy
A
P
01 2 02 0 ???
y
P
z
P
i
zz
i
yy
中性轴
48
y
y
z
z
L W
M
W
M
A
P ???
ma x?
y
y
z
z
y W
M
W
M
A
P ???
m a x?
P （ z P, y P ）
y
z
y
z
),( PP yzP
y
z
5,Kernel of section in the problem of the eccentric tension、
compression,
ay
az
01 2 ??
z
yP
i
ay
01 2 ??
y
zP
i
az
After knowing ay and az,
The action range of the compressive force,As the compressive force is acted in
this range there are no tensile stresses in the section,
May determine an action point
of the force P,),( PP yz
01 2 02 0 ???
y
P
z
P
i
zz
i
yy
Neutral axis ),( PP yzP
Kernel of section
49
y
z
五、（偏心拉、压问题的）截面核心,
ay
az
01 2 ??
z
yP
i
ay
01 2 ??
y
zP
i
az
已知 ay,az 后,
压力作用区域。
当压力作用在此区域内时，横截面上无拉应力。
可求 P 力的一个作用点 ),(
PP yz
01 2 02 0 ???
y
P
z
P
i
zz
i
yy
中性轴 ),( PP yzP
截面核心
50
M P a75.82.02.03 5 0 0 0 0 m a x2 ???? AP?
???
11
m a x1
zW
M
A
P?
M P a7.11
3.02.0
6503 5 0
3.02.0
3 5 0 0 0 0
2
?
?
???
?
Solution,The stresses
in the cross sections of the
two poles are both compressive
ones,
Example 4 Two poles subjected to the force P=350kN are shown in
the figure,The section of one pole is unequal and the section of the other pole is
equal,Try to determine the normal stress with maximum absolute value in the
poles,
M P
P d
51
Fig, Fig,
P
300
200
P
200
M P a75.8
2.02.0
3 5 0 0 0 0
m a x2
?
?
?
?
A
P
?
???
11
m a x1
zW
M
A
P?
M P a7.11
3.02.0
6503 5 0
3.02.0
3 5 0 0 0 0
2
?
?
???
?
解,两柱横截面上的最大正
应力均为 压应力
[例 4] 图示不等截面与等截面柱，受力 P=350kN，试分别求出两
柱内的绝对值最大正应力。
图（ 1） 图（ 2）
M P
P d
52
.,
P
300
200
P
200
mm5102010100 201020 ???? ???Cz
2
3
5100101210010 ?????CyI
45
2
3
mm1027.7
]252010
12
2010[
??
?????
Solution,Analysis of the internal force is
shown in the figure,Centroid of the slot in
the coordinates shown in the figure
Nm5 0 0105 3 ??? ?PM
P
P
Slot
Example 5 A steel plate shown in the figure is subjected to forces
P=100kN,Try to determine the maximum normal stress； If the slot is moved to the
middle of the plate and the maximum normal stress is kept constant,how much
should the width of the slot be?
53
P
P
M
N
20 100 20
y
z
yC
10
mm5102010100 201020 ???? ???Cz
2
3
5100101210010 ?????CyI
45
2
3
mm1027.7
]252010
12
2010[
??
?????
例 5 图示钢板受力 P=100kN,试求最大正应力；若将缺口移至板
宽的中央，且使最大正应力保持不变，则挖空宽度为多少？
解,内力分析 如图
坐标如图，挖孔处的形心
Nm5 0 0105 3 ??? ?PM
P
P
54
P
P
M
N
20 100 20
y
z
yC
10
P
P
M
N
ycI
zM
A
N m a x
m a x ???
M P a8.1628.37125 ???
Analysis of stress is
shown in the figure,
7
3
6
3
1027.7
1055500
10800
10100
?
?
? ?
???
?
??
As the hole is moved to
the middle of the plate
)100(10mm9.631108.162 10100 26
3
m a x
xNA ??????? ?
mm8.36 so ?x
20 100 20
y
z
yC
55
＋
P
P
M
N
M P a8.1628.37125 ???
应力分析 如图
7
3
6
3
1027.7
1055500
10800
10100
?
?
? ?
???
?
??
孔移至板中间时
)100(10mm9.631108.162 10100 26
3
m a x
xNA ??????? ?
mm8.36 ?? x
20 100 20
y
z
yC
56
ycI
zM
A
N m a x
m a x ???
＋
M P a7.351.07 0 0 016 3 ????? ??
nW
T
M P a37.6101.0504 32 ?????? ?? AP
Solution,For the composite
deformation of tension and
torsion,the stressed state at the
critical point is shown in the figure,
Example 6 A circular rod which the diameter d =0.1m is subjected to forces
T=7kNm and P=50kN as shown in the figure,[?]=100MPa,Try to check the
strength of the rod with the third strength theory,
Therefore,the rod is safe,
223 4??? ??r
? ????
???
M P a7.71
7.35437.6 22
?
?
A
A P
P
T
T
57
M P a7.351.07 0 0 016 3 ????? ??
nW
T
M P a37.6101.0504 32 ?????? ?? AP
解,拉扭组合,危险点 应力状态如图
[例 6] 直径为 d=0.1m的圆杆受力如图,T=7kNm,P=50kN,
[?]=100MPa,试按 第三强度理论校核此杆的强度 。
故，安全。 ? ???? ??? M P a7.71 7.35437.6 22
?
?
A
A P
P
T
T
58
223 4??? ??r
59
Chapter 9 Exercises
1,A circular shaft of steel is deformed under tension and torsion,
Try to write out the strength conditions,If it shows the tension-torsion-
bending composite deformation,try to write out the strength conditions,
2,The cross-section area of the square-section rod is reduced half at
the section mn,Try to determine the maximum tensile stress at the
section mn induced by the axial force P,
Solution,
3,A rectangular-section beam is shown in the figure,Knowing b =
50mm and h =75mm,Try to determine the maximum normal stress of
the beam,If the beam is changed to have circular sections with the
diameter d = 65mm, what is the maximum normal stress?
W
M
A
N ??
m a x?
2
22 8
64/42/ a
PaaaPaP ?
?
??
?
??
?
? ???
60
第九章 练习题
一、钢圆轴为拉伸与扭转的组合变形，试写出
其强度条件。若为拉伸、扭转和弯曲的组合变形，
试写出其强度条件。
二、方形截面杆的横截面面积在 mn 处减少一
半，试求由轴向载荷 P 引起的 mn 截面上的最大拉
应力。 解,
三、矩形截面梁如图。已知 b = 50mm,h =75
mm,求梁内的最大正应力。如改为 d = 65mm 的
圆截面，最大正应力为多少？
W
M
A
N ??
m a x?
2
22 8
64/42/ a
PaaaPaP ?
?
??
?
??
?
? ???
61
Soluion,
If the beam is changed to have circular sections,
y
y
z
z
W
M
W
M m a xm a x
m a x ???
M P a9605.0075.0 61021075.005.0 6105.1 2323 ?? ????? ???
M P aoW MM yz 8.9265,25.132 3 22
2
m a x
2
m a x
m a x ??
????
??
62
解,
如改为圆截面
y
y
z
z
W
M
W
M m a xm a x
m a x ???
M P aoW MM yz 8.9265,25.132 3 22
2
m a x
2
m a x
m a x ??
????
??
M P a9605.0075.0 61021075.005.0 6105.1 2323 ?? ????? ???
63
64
