Mechanics of Materials
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2
CHAPTER 1 AXIAL TENSION AND COMPRESSION
§ 1–1 Concepts and practical examples of axial tension and compression
§ 1–2 Internal force,method of section,axial force and its diagram
§ 1–3 Stresses on the section and strength conditions
§ 1-4 Deformation of the rod in axial tension and compression ? law of
elasticity
§ 1-5 Elastic strain energy of the rod in axial tension and compression
§ 1-6 Statically indeterminate problems and their treatment methods of axial
tension and compression
§ 1-7 Mechanical properties of materials in axial tension and compression
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4
§ 1–1 轴向拉压的概念及实例
§ 1–2 内力、截面法,轴力及轴力图
§ 1–3 截面上的应力及强度条件
第一章 轴向拉伸和压缩
§ 1-4 拉压杆的变形 ? 弹性定律
§ 1-5 拉压杆的弹性应变能
§ 1-6 拉压超静定问题及其处理方法
§ 1-7 材料在拉伸和压缩时的力学性能
§ 1–1 CONCEPTS AND PRACTICAL EXAMPLES OF
AXIAL TENSION AND COMPRESSION
Characteristic of the external force,The acting line of the resultant of
e x t e r n a l fo r c e s is co i n c i d ed with t h e a x i s of t h e ro d,
1,Concepts
Characteristic of the deformation,Deformation of the rod is mainly
elongation or contraction along the axis of the rod and companied with lateral
reduction or enlargement,
Axial tension, Deformation of the rod is axial elongation
and lateral shortening,
Axial compression,Deformation of the rod is axial
shortening and lateral enlargement,
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6
§ 1–1 轴向拉压的概念及实例
轴向拉压的外力特点,外力的合力作用线与杆的轴线重合 。
一、概念
轴向拉压的变形特点,杆的变形主要是轴向伸缩，伴随横向
缩扩。
轴向拉伸：杆的变形是轴向伸长，横向缩短。
轴向压缩：杆的变形是轴向缩短，横向变粗。
In axial compression,the corresponding force is called compressive force,
In axial tension,the corresponding force is called tensile
force,
Mechanical models are shown in the figures
PP
PP
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8
轴向压缩，对应的力称为压力。
轴向拉伸，对应的力称为拉力。
力学模型如图
PP
PP
Practical
examples in
engineering
2,
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10
工
程
实
例
二,
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1,Internal force
Internal force is the resultant of internal forces,which is acting mutually
between two neighbour parts inside the body,caused by the external forces,
§ 1–2 INTERNAL FORCE,METHOD OF SECTION,AXIAL
FORCE AND ITS DIAGRAM
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一、内力
指由外力作用所引起的、物体内相邻部分之间分布内
力系的合成（附加内力）。
§ 1–2 内力 · 截面法 · 轴力及轴力图
2,Method of section · axial force
Calculation of the internal forces is the foundation to analyze the
problems of strength,rigidity,stability etc,The general method to
determine internal forces is the method of section,
1),Basic steps of the method of section,
① Cut off,Assume to separate the rod into two distinct parts in the section in
which the internal forces are to be determined,
② Substitute,Take arbitrary part and substitute the action of another part on it by
the corresponding internal force in the cut-off section,
③ Equilibrium,Set up equilibrium equations for the remained part and
determine the unknown internal forces according to the external forces acted on it.
（ Here the internal forces in the cut-off section are the external forces for the
remained part） 15
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二、截面法 · 轴力
内力的计算是分析构件强度、刚度、稳定性等问题的
基础。求内力的一般方法是截面法。
1,截面法的基本步骤,
① 截开,在所求内力的截面处，假想地用截面将杆件一分为二。
②代替,任取一部分，其弃去部分对留下部分的作用，用作用
在截开面上相应的内力（力或力偶）代替。
③平衡,对留下的部分建立平衡方程，根据其上的已知外力来
计算杆在截开面上的未知内力（此时截开面上的内力
对所留部分而言是外力）。
2,Axial force—internal force of the rod in axial
tension or compression,designated by N,
Such as,Determine N by the method of section,
0?? X 0NP?? NP ?
A P P
P
A
N
Simple sketch
A P P Cut off:,
Substitute,
Equilibrium,
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2,轴力 ——轴向拉压杆的内力，用 N 表示。
例如,截面法求 N。
0?? X 0NP?? NP ?
A P P
简图
A P P 截开,
代替,
平衡,
P
A
N
x
① Reflected the variety relation between
the corresponding axial force and the
position of the section,
② Find out value of the maximum axial
force and the position of the section in which the maximum axial force act,That is
to determine the position of the critical section and supply the information for the
calculation of strength,
3,Diagram of the axial force— sketch
expression of N (x)
3),Sign conventions for the axial force,
axial force N (tensile force)is positive when its
direction point to the outward direction of the
normal line of the section,(compressive
force)negative inward
N > 0 N N
N<0 N N
x
N
P
+
meaning
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① 反映出轴力与横截面位置变化关系，较直观；
②确定出最大轴力的数值
及其所在横截面的位置,
即确定危险截面位置，为
强度计算提供依据。
三,轴力图 —— N (x) 的图象表示。
3,轴力的正负规定,
N 与外法线同向,为正轴力 (拉力 )
N与外法线反向,为负轴力 (压力 )
N > 0 N N
N<0 N N
N
x
P
+
意
义
Example 1 The forces with magnitudes 5P,8P,4P and P act respectively at
points A,B,C,D of the rod,Their directions are shown in the figure,Try to plot
the diagram of the axial force of the rod,
Solution,Determine the internal force N1 in segment OA,
Take the free body as shown in the figure,
A B C D
PA PB PC PD
O
A B C D
PA PB PC PD
N1
0?? X 0
1 ????? DCBA PPPPN
04851 ????? PPPPN PN 21 ?
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[例 1] 图示杆的 A,B,C,D点分别作用着大小为 5P,8P,4P,
P 的力，方向如图，试画出杆的轴力图。
解,求 OA段内力 N1：设置截面如图
A B C D
PA PB PC PD
O
0?? X
1 0A B C DN P P P P? ? ? ? ? ?
04851 ????? PPPPN PN 21 ?
A B C D
PA PB PC PD
N1
x
Similarly,we get the
internal forces in segment
AB,BC,CD, They are
respectively,
N2= –3P
N3= 5P
N4= P
The diagram
of the axial
force is shown
in the right
figure,
B C D
PB PC PD
N2
C D
PC PD
N3
D
PD
N4
N
x 2P
3P
5P
P + +
–
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同理，求得 AB、
BC,CD段内力分
别为,
N2= –3P
N3= 5P
N4= P
轴力图如右图
B C D
PB PC PD
N2
C D
PC PD
N3
D
PD
N4
N
x 2P
3P
5P
P + +
–
Simple method to plot the diagram of axial force,From the left to the right,
Characteristic of the diagram of the axial force,Value of sudden
change = concentrated load
If meeting the force P to the left ?,the increase of the axial force N is positive；
If meeting the force to the right ?, the increase of the axial force N is negative,
5kN 8kN 3kN
+
– 3kN
5kN
8kN
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轴力 (图 )的简便求法,自左向右,
轴力图的特点：突变值 = 集中载荷
遇到向左的 P?,轴力 N 增量为正；
遇到向右的 P?, 轴力 N 增量为负。
5kN 8kN 3kN
+
– 3kN
5kN
8kN
Solution,The free end of the rod is the
origin of the coordinate and coordinate x to the
right is positive,Take the segment of length x
on the left of point x,its internal force is q
K L
x O
2
0 2
1d)( kxxkxxN x ??? ??
2
m a x 2
1)( kLxN ??
Example 2 Length of the rod shown in the figure is L,Distributed force q = kx
is acted on it,direction of the force is shown in the figure,Try to plot the diagram of
axial force of the the rod,
L
q(x)
N(x)
x
q(x)
N x
O
–
2
2kL
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解,x 坐标向右为正，坐标原点在
自由端。
取左侧 x 段为对象，内力 N(x)为,
q
q L
x O
2
0 2
1d)( kxxkxxN x ??? ??
2
m a x 2
1)( kLxN ??
[例 2] 图示杆长为 L，受分布力 q = kx 作用，方向如图，试画出
杆的轴力图。
L
q(x)
N x
O
–
2
2kL
N(x)
x
q(x)
1,Concept of stress
§ 1–3 STRESSES ON THE SECTION AND STRENGTH CONDITIONS
Bring forward
the problem,
1),The magnitude of the internal force can not scale the strength of the
structure member,
2),Strength：① Intensity of the distributed internal forces in the
section? stress； ② The load-bearing capacity of the material,
1),Definition,Intensity of the internal force due to the external forces,
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P P
P P
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一、应力的概念
§ 1–3 截面上的应力及强度条件
问题提出,
1,内力大小不能衡量构件强度的大小。
2,强度：①内力在截面的分布集度 ? 应力；
②材料承受荷载的能力。
1,定义,由外力引起的内力 集度 。
P P
P P
Under most cases distribution of the internal force inside engineering members
is not uniform,Definition of intensity is neither accurate and important because
breakage or failure often begins from the point at which intensity of the internal
force is maximum,
?P
?A
M ① Average stress,
② Whole stress（ sum stress）,
A
Pp
M Δ
Δ?
Δ 0
Δ d
Δ dl i mA
PP
p
AA?
??
2),Expression of stress,
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工程构件，大多数情形下，内力并非均匀分布，集度的定
义不仅准确而且重要，因为“破坏”或“失效”往往从内力集
度最大处开始。
?P
?A
M ① 平均应力,
② 全应力（总应力）,
A
Pp
M Δ
Δ?
Δ 0
Δ d
Δ dl i mA
PP
p
AA?
??
2,应力的表示,
③ Whole stress may be decomposed into,
p ?
M ?
A
N
A
N
A d
d
Δ
Δlim
0Δ
??
?
?
A
T
A
T
A d
d
Δ
Δlim
0Δ
??
?
?
a,Stress perpendicular to the section is called―normal stress‖
b,Stress lying in the section is called―shearing stress‖
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③ 全应力分解为,
p ?
M ?
A
N
A
N
A d
d
Δ
Δlim
0Δ
??
?
?
A
T
A
T
A d
d
Δ
Δlim
0Δ
??
?
?
a.垂直于截面的应力称为,正应力” (Normal Stress)；
b.位于截面内的应力称为,剪应力” (Shearing Stress)。
Before
deformation
1),Experiment on the law of deformation and the hypothesis of plane
section,
Hypothesis of plane section,Cross sections remain planes before and
after deformations,
Deformations of longitudinal fibers are the same
a b
c d
After
loading
P P
d ′
a′
c′
b′
2,Stress in the cross section of the rod in tension or compression
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变形前
1,变形规律试验及平面假设,
平面假设,原为平面的横截面在变形后仍为平面。
纵向纤维变形相同。
a b
c d
受载后 P P d ′ a′ c′ b′
二、拉（压）杆横截面上的应力
The material is homogeneous and,its deformation is uniform,so the internal
force is distributed uniformly。
2,Tensile stress,
A
xN )( ??
Normal stress due to the axial forces — ?, distributes uniformly in the cross
section,
Critical section,The section in which internal force is maximum
and of which the dimension is smallest,
Critical point,The point at which the stress is maximum,
3,Critical section and the maximum working stress,
))( )(m a x ( m a x xA xN?? 37
? N(x) P
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均匀材料、均匀变形，内力当然均匀分布。
2,拉伸应力,
A
xN )( ??
轴力引起的正应力 —— ?, 在横截面上均布。
危险截面：内力最大的面，截面尺寸最小的面。
危险点：应力最大的点。
3,危险截面及最大工作应力,
))( )(m a x ( m a x xA xN??
? N(x) P
Straight rod,cross section of the rod is without sudden change,there is a
certain distance from the section to the point at which the load acts,
4),Application conditions of the formula,
6),Stress concentration,
Stress increases abruptly near the cross section with a sudden change in dimension
5),Saint-Venant principle,
Distribution and magnitude of the stress in the section at a certain distance from
the point at which the load is acted are not affected by the acting form of external
loads,
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直杆、杆的截面无突变、截面到载荷作用点有一定 的距离。
4,公式的应用条件,
6,应力集中（ Stress Concentration）,
在截面尺寸突变处，应力急剧变大。
5,Saint-Venant原理,
离开载荷作用处一定距离，应力分布与大小不受外载荷作
用方式的影响。
Sketch of Saint-Venant principle and stress concentrations
(Red real lines denote the line before deformation and red dashed lines denote the
shape after deformation.)
Sketch of deformation,a b c P P
Sketch of the stress
distribution,
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Saint-Venant原理与应力集中示意图
(红色实线为变形前的线，红色虚线为红色实线变形后的形状。 )
变形示意图,a b c P P
应力分布示意图,
7),Criterion of the strength design,
? ? ))( )(m a x ( m a x ?? ?? xA xN
where,[?]—allowable stress,?max– the maximum working stress
at the critical point,
② Design the dimension of the section,
][
m a x
m in ?
NA ?
? ? ; m a x ?AN ?
? ? m a x()P f N?
Three kinds of calculation of strength may be done
according to the criterion of strength,
That structure members are ensured not to be wrecked and have certain safe
degree,
? ? m a x ?? ?① Check the strength,
③ Determine the allowable load,
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7,强度设计准则（ Strength Design）,
? ? ))( )(m a x ( m a x ?? ?? xA xN
其中,[?]--许用应力,?max--危险点的最大工作应力。
② 设计截面尺寸,
][
m a x
m in ?
NA ?
? ? ; m a x ?AN ? ? ? )N(fP i?
依强度准则可进行三种强度计算,
保证构件不发生强度破坏并有一定安全余量的条件准则。
? ? m a x ?? ?
① 校核强度,
③ 许可载荷,
Example 3 A circular rod is subjected to a tensile force P =25 kN,Its diameter
is d =14mm and its allowable stress is [?]=170MPa,Try to check the strength of
the rod,
Solution,① Axial force,N = P =25kN
M P a1 6 20 1 40143 102544 2
3
2m a x ??
?????
..π d
P
A
N?② Stress,
③ Check the strength,
? ? 1 7 0 M P a1 6 2 M P am a x ??? ??
④ Conclusion,The strength of the rod satisfies request,
The rod can work normally,
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[例 3] 已知一圆杆受拉力 P =25 k N，直径 d =14mm，许用应力
[?]=170MPa，试校核此杆是否满足强度要求。
解,① 轴力,N = P =25kN
M P a1 6 20 1 40143 102544 2
3
2m a x ??
?????
..π d
P
A
N?
② 应力,
③ 强度校核,
? ? 1 7 0 M P a1 6 2 M P am a x ??? ??
④ 结论：此杆满足强度要求，能够正常工作。
Example 4 A three-pin house frame on which a vertical uniform load,with
the in density intensity is q =4.2kN/m is applied is shown in the figure,Diameter
of the steel tensile rod in the frame is d =16 mm and its allowable stress is
[?]=170MPa,Try to check the strength of the rod,
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Tie bar
8.5m
q
q
A
B
C
48
[例 4] 已知三铰屋架如图，承受竖向均布载荷，载荷的分布
集度为,q =4.2kN/m，屋架中的钢拉杆直径 d =16 mm，许用
应力 [?]=170M Pa。 试校核钢拉杆的强度。
钢拉杆
8.5m
q
q
A
B
C
① Determine the reactions first according to the global equilibrium
Solution,
kN519 0
0 0
.Rm
HX
AB
A
?? ?
?? ?
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Tie bar
8.5m
RA RB
HA
q
q
A
C
50
① 整体平衡求支反力 解,
kN519 0
0 0
.Rm
HX
AB
A
?? ?
?? ?
钢拉杆
8.5m
RA RB
HA
q
q
A
C
③ Stress,
④ Strength check
and conclusion,? ? M P a 1 7 0 M P a 1 3 1 m a x ??? ??
This rod satisfies the request of strength,It is safe,
M P a131
0160143
103264
d
4
2
3
2m a x
?
?
??
?
??
..
.
P
A
N
?
?
② Determine the axial force
according to the partial equilibrium
,
kN326 0,Nm C ?? ?
51
RA
HA
RC
HC
N
Cq
A
52
③ 应力,
④ 强度校核与结论,? ? M P a 1 7 0 M P a 1 3 1
m a x ??? ??
此杆满足强度要求，是安全的。
M P a131
0160143
103264
d
4
2
3
2m a x
?
?
??
?
??
..
.
P
A
N
?
?
② 局部平衡求 轴力,
HC
kN326 0,Nm C ?? ?
RA
HA
RC
N
Cq
A
? ?/ ;
/ s i n
B D B D
BD
AN
Lh
?
?
?
? 。
Example 5 A simple crane is shown in the figure,AC is a rigid beam, sum
weight of the hoist and heavy body that is lifted is P,What should be the angle ?
so that the rod BD has the minimum weight？ The allowable stress of the rod [?]
is known,;BDBD LAV ?
Analysis,x L
h
? P
A B
C
D
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54
? ?/ ;
/ s i n
B D B D
BD
AN
Lh
?
?
?
? 。
[例 5] 简易起重机构如图,AC为刚性梁，吊车与吊起重物总重
为 P，为使 BD杆最轻，角 ? 应为何值？ 已知 BD 杆的 许用应力
为 [?]。;BDBD LAV ?
分析,x L
h
? P
A B
C
D
PxhNm BDA ??? ? )c t g() s i n(,0 ??
?c o sh
PLN
BD ?
? ??/NA BD?
? The cross-section area A of the rod BD,
Solution,? Internal force N((? ) of the rod BD,Take AC as
our study object as shown in the figure,
YA
XA
?
NBD
x
L
P
A B
C
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56
PxhNm BDA ??? ? )c t g() s i n(,0 ??
?c o sh
PLN
BD ?
? ??/NA BD?? BD杆横截面面积 A,
解,? BD杆 内力 N(? ),取 AC为研究对象，如图
YA
XA
?
NBD
x
L
P
A B
C
YA
XA
?
NBD
x
L
P
A B
C
③ Determine the minimum value of VBD,
2/ s i n ;
[ ] s i n 2BD
PLV A L A h ?
??
? ? ?
o
m i n
24 5,
[]
PLA s V?
?
? ? ?
67
58
YA
XA
?
NBD
x
L
P
A B
C
③ 求 VBD 的 最小值,2
/ s i n ;
[ ] s i n 2BD
PLV A L A h ?
??
? ? ?
][
2 45
m i n
o
??
PLV,??? 时
3,Stresses in the inclined section of the rod in tension or compression
P P
k
k a
a
a
a A
Pp ?
Aa:Area of the inclined section； Pa,Internal
force in the inclined section,
aa aa c o s c o s AAA A ???
,
a?a
a
a
a c o sc o s 0???? A
P
A
Pp a?a c o s0?p
P
k
k a
Pa
From geometric
relation
Substituting it into the above
formula we get
Solution,Adopt the method of
section,According to the equilibrium
equation,Pa=P then
Assume a straight rod is subjected
to a tensile force P,Determine the
stress in the inclined section k-k,
Whole stress in the
inclined ection,
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三、拉 (压 )杆斜截面上的应力
设有一等直杆受拉力 P作用。
求：斜截面 k-k上的应力。
P P
k
k a 解：采用截面法
由平衡方程,Pa=P
则,
a
a
a A
Pp ? Aa,斜截面面积； Pa：斜截面上内力。
由几何关系,
a
a a
a c o s
c o s AA
A
A ???
代入上式，得,
a?a
a
a
a c o sc o s 0???? A
P
A
Pp 斜截面上全应力,a?a c o s0?p
P
k
k a
Pa
a?a c o s0?p
Decomposition,
pa ? a?a? aa 20 c o sc o s ?? p
a?aa?a? aa 2s i n2s i nc o ss i n 00 ??? p
It indicates the change of stresses in different sections
through a point,
As a =90°,
0)( m in ?a?
As a = 0,90°, 0||
min ?a?
As a = 0°,
)( 0m a x ?? a ?
(The maximum normal stress exists
in the cross section)
As a = ± 45°,
2
|| 0m ax ?? a ?
(The maximum shearing stress exists
in the inclined section of 45 ° )
Whole stress in the
inclined section,
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P P
k
k a
P
k
k a
pa
?a
?a
a
62
斜截面上全应力,a?
a c o s0?p
分解,
pa ? a?a? aa 20 c o sc o s ?? p
a?aa?a? aa 2s i n2s i nc o ss i n 00 ??? p
反映：通过构件上一点不同截面上应力变化情况。
当 a = 90° 时,
0)( m in ?a?
当 a = 0,90° 时,0||
min ?a?
当 a = 0° 时,
)( 0m a x ?? a ?
(横截面上存在最大正应力 )
当 a = ± 45° 时,
2
|| 0m ax ?? a ?
(45 ° 斜截面上剪应力达到最大 )
P P
k
k a
P
k
k a
p a
a ?
a
?a
a
2,Element,?Element— delegate of a point inside the member,infinitesimal
geometric body which envelops the study point,The element in common use is just
hexahedron ?properties of an element—a,stress is distributed uniformly in an
arbitrary parallel arbitrary plane； b,stresses in the parallel plane opposite plane
are equal,
3,stress element at a point M in
the rod in tension or compression,
1,State of stress at a point,There are countless sections through a point,
Sum of stresses in the different section through a point is called the state of
stress at this point,
Complementary,
? P M
? ? ? ?
63
64
2、单元体,?单元体 —构件内的点的代表物，是包围被研究点的
无限小的几何体，常用的是正六面体。
?单元体的性质 —a、平行面上，应力均布；
b、平行面上，应力相等。
3、拉压杆内一点 M 的应力单元体,
1.一点的应力状态,过一点有无数的截面，这一点的各个截面
上的应力情况，称为这点的应力状态。
补充,
? P M
? ? ? ?
??
?
?
?
?
?
aa??
a??
a
a
c o ss in
c o s
0
2
0
Take a free body as shown in the Fig.3,a is
positive if it is along countclockwise； ? a
is positive if it makes the free body rotate
clockwise,From the equilibrium of the free
body we get,
?
?
?
??
?
?
?
??
a
?
?
a
?
?
a
a
2s i n
2
)2c o s(1
2
:r
0
0
o
4,Stress in the inclined section of the rod in tension or compression
? ? ? ?
a a
x
Fig.3
65
66
?
?
?
?
?
aa??
a??
a
a
c o ss i n
c o s
0
2
0
取分离体如图 3,a 逆时针为正；
? a 绕研究对象顺时针转为正；
由分离体平衡得,
?
?
?
??
?
?
?
??
a
?
?
a
?
?
a
a
2s i n
2
)2c o s(1
2
:
0
0
或
4、拉压杆斜截面上的应力
? ? ? ?
a a
x
图 3
M P a7.632/4.1272/0m a x ??? ??
M P a5.95)60c o s1(
2
4.127)2c o s1(
2
0 ????? a??
a
M P a2.5560s i n
2
4.1272s i n
2
0 ??? a??
a
M P a4.127 1014.3 1 0 0 0 04 20 ????? AP?
Example 6 A rod,which the diameter d =1 cm is subjected to a tensile force P
=10kN,Determine the maximum shearing stress,the normal stress and shearing
stress in the inclined section of an angle 30° about the cross section,
Solution,Stresses in the inclined section of the rod in tension or compression
can be determined directly by the formula,
67
68
M P a7.632/4.1272/0m a x ??? ??
00 1 2 7, 4( 1 c o s 2 ) ( 1 c o s 6 0 ) 9 5, 5 M P a
22a
??a? ? ? ? ?
00 1 2 7, 4s i n 2 s i n 6 0 5 5, 2 M P a
22a
??a? ? ?
M P a4.127 1014.3 1 0 0 0 04 20 ????? AP?
[例 6] 直径为 d =1 cm 杆受拉力 P =10 kN的作用，试求最大剪应
力，并求与横截面夹角 30° 的斜截面上的正应力和剪应力 。
解：拉压杆斜截面上的应力，直接由公式求之,
Example 7 A tensile rod as shown in the figure is made from two parts glued
mutually together along mn,It is subjected to the action of force P,Assume that the
allowable normal stress is [?]=100MPa and allowable shearing stress is [?]=50MPa for the
adhesive,Area of cross section of the rod is A= 4cm2,If strength of the rod is controlled by
the adhesive what is the angle a ( a,between 0 0~60 0) to get the largest tensile force?
kN50,6.26 ??? BB Pa
Combine (1),(2) and get,
P P
m
n a
Solution,
)1( ][c o s 2 ???a? a ??
A
P
)2( ][c o ss i n ???aa? a ??
A
P
69
?
?
P
a 60 0 30 0
B
0 0
70
[例 7]图示拉杆沿 mn由两部分胶合而成,受力 P，设胶合面的许用
拉应力为 [?]=100MPa ；许用剪应力为 [?]=50MPa,并设杆的
强度由胶合面控制,杆的横截面积为 A= 4cm2，试问,为使杆承受
最大拉力,a角值应为多大?(规定, a在 0~60度之间 )。
kN50,6.26 ??? BB Pa
联立 (1),(2)得,
P P
m
n a
解,
)1( ][c o s 2 ???a? a ??
A
P
)2( ][c o ss i n ???aa? a ??
A
P
?
?
P
a 60 0 30 0
B
0 0
? ?0 0 0 260 / ( c o s 6 0 s i n 6 0 ) 4 5 0 1 0 4 / 3 4 6, 2 k NPA ?? ? ? ? ? ?
kN50m a x ?? P
The curves of formula (1)and,(2) are shown in the Tig.(2),Obviously the strength of
the rod on the left of point B is controlled by the normal stress,that on the right of
point B is controlled by the shearing stress,As a=60°, from formula (2) we can get
? ?1 2,6 0 / ( c o s 6 0 s i n 6 0 ) 4 6 0 1 0 4 / 3 5 5, 4 4 k NBPA ?? ? ? ? ? ? ? ?
kN44.55m a x ?? P
Solution,At the point of intersection
of curves (1) and (2),
kN4.54;31 11 ??? BB Pa;M P a60][ m a x ?? P?Discussion,As
71
?
?
P
a 60 0 30 0
B1
0 0
72
? ?0 0 0 260 / ( c o s 6 0 s i n 6 0 ) 4 5 0 1 0 4 / 3 4 6, 2 k NPA ?? ? ? ? ? ?
kN50m a x ?? P
(1),(2)式的曲线如图 (2)，显然,B点左 侧由 正 应力控制杆的强
度,B点右侧由 剪 应力控制杆的强度,当 a=60° 时，由 (2)式得
kN44.55m a x ?? P
解 (1),(2)曲线交点处,
kN4.54;31 11 ??? BB Pa;M P a60][ m a x ?? P?讨论：若
?
?
P
a 60 0 30 0
B1
0 0
? ?1 2,6 0 / ( c o s 6 0 s i n 6 0 ) 4 6 0 1 0 4 / 3 5 5, 4 4 k NBPA ?? ? ? ? ? ? ? ?
1),The whole longitudinal
deformation of the rod,
3),Average stain,
1LLL
LL
? ????
2),Strain,linear deformation per unit length,
1,Deformation and strain of the rod in tension or compression
1L L L? ? ?
§ 1－ 4 DEFORMATION OF THE ROD IN AXIAL TENSION
AND COMPRESSION ? LAW OF ELASTICITY
73
a b
c d
x?
L
Cross section
74
1、杆的纵向总变形,
3、平均线应变,
1LLL
LL
? ????
2、线应变：单位长度的线变形。
一、拉压杆的变形及应变
1L L L? ? ?
§ 1－ 4 拉压杆的变形 ? 弹性定律
a b
c d
x?
L
4,Longitudinal strain at point x,
x
x
x ?
??
??
dlim
0
?
6,Lateral strain at point x,
5,Lateral deformation of the rod,
accaac ?????
ac
ac????
P P
d ′
a′
c′
b′
xx ??? d
L1
75
76
4,x点处的纵向线应变,
x
x
x ?
??
??
dlim
0
?
6,x点处的横向线应变,
5、杆的横向变形,
accaac ?????
ac
ac????
P P
d ′
a′
c′
b′
xx ??? d
L1
2,Elastic law of the rod in tension or compression
PLL
A?? EA
NL
EA
PLL ???
1),Case of equal internal forces
2),Case of variable internal forces
)(
d)()d(
xEA
xxNx ??
( ) d( d )
()LL
N x xLx
E A x? ? ? ???
?
?
??
n
i ii
ii
AE
LNL
1
When internal forces in n
segments are constant
※ ―EA‖is called the axial rigidity of the
rod in tension or compression,
P P
N （ x ）
x
d x
N(x)
dx
x
77
78
二、拉压杆的弹性定律
PLL
A??
P L N LL
E A E A? ? ?
1、等内力拉压杆的弹性定律
2、变内力拉压杆的弹性定律
)(
d)()d(
xEA
xxNx ??
( ) d( d )
()LL
N x xLx
E A x? ? ? ???
1
n
ii
i ii
NLL
EA?
?? ?
内力在 n段中分别为常量时
※,EA” 称为杆的抗拉压刚度。
P P
N （ x ）
x
d x
N(x)
dx
x
3),Elastic law in uniaxial stressed state
4),Possion,s ratio（ or coefficient of the lateral
deformation）
3,Who firstly proposed the Elastic Law
The Elastic Law is the important foundation of solid mechanics
such Mechanics of Materials,Generally it is considered to be
proposed firstly by the English scientist Hook (1635-1703),So the
Elastic Law is also called Hook,s Law,Actually,there was an early
record of the proportional ration between force and deformation,
which is 1500 years earlier than Hook,
??
?
??
? ? ?? ??
or
1)( )(1)d( ?? ExA xNEdx x ???? 1 ??
E
?
That is
79
80
1)( )(1)d( ?? ExA xNEdx x ????
3、单向应力状态下的弹性定律
1,??
E
?即
4、泊松比（或横向变形系数）
??
?
??
,? ? ?? ??或
三、是谁首先提出弹性定律
弹性定律是材料力学等固体力学一个非常重要的基础。一般
认为它是由英国科学家胡克 (1635一 1703)首先提出来的，所以通
常叫做胡克定律。其实，在胡克之前 1500年，我国早就有了关于
力和变形成正比关系的记载。
― ‖ 胡：请问,弛其弦，以绳缓 缓 之
是什么意思?
郑：这是讲测量弓力时，先将弓的弦
松开，另外用绳子松松地套住弓
的两端，然后加重物，测量。
胡：我明白了。这样弓体就没有初始应力，处于自 然状态。
东汉经学家郑玄 (127—200)对, 考工记 ·弓人, 中, 量其力，
有三均, 作了 这样的注释：, 假令弓力胜三石，引之中三尺，弛
其弦，以绳缓擐之，每加物一石，则张一尺。, (图 )
82
― ‖ 胡：请问,弛其弦，以绳缓 缓 之
是什么意思?
郑：这是讲测量弓力时，先将弓的弦
松开，另外用绳子松松地套住弓
的两端，然后加重物，测量。
东汉经学家郑玄 (127—200)对, 考工记 ·弓人, 中, 量其力，
有三均, 作了 这样的注释：, 假令弓力胜三石，引之中三尺，弛
其弦，以绳缓擐之，每加物一石，则张一尺。, (图 )
胡：我明白了。这样弓体就没有初始应力，处于自 然状态。
郑：后来，到了唐代初期，贾公彦对我的注释又作 了注疏，他说,
郑又云假令弓力胜三石，引之 中三尺者，此即三石力弓也。
必知弓力三石者，当弛其弦以绳缓擐之者，谓不张之，别以
绳系两箭，乃加物一石张一尺、二石张二尺、三石张三
尺。 其中 ‖ ― 两萧 就是指弓的两端。
一条
―
胡：郑老先生讲, 每加物一石，则张一尺”。和我讲的完全是同一
个意思。您比我早 1500 中就记录下这种正比关系，的确了不起,
和推测, 一文中早就推崇过贵国的古代文化,目前我们还只
是刚刚走到这个知识领域的边缘，然而一旦对它有了充分的认
识，就将会在我们面 前展现出一个迄今为止只被人们神话般
地加以描述的知识王国”。
1686 年, 关于中国文字和语言的研究 真是令人佩服之至 』 我在
83
84
郑：后来，到了唐代初期，贾公彦对我的注释又作 了注疏，他说,
郑又云假令弓力胜三石，引之 中三尺者，此即三石力弓也。
必知弓力三石者，当弛其弦以绳缓擐之者，谓不张之，别以
绳系两箭，乃加物一石张一尺、二石张二尺、三石张三
尺。 其中 ‖ ― 两萧 就是指弓的两端。
一条
―
胡：郑老先生讲, 每加物一石，则张一尺”。和我讲的完全是同一
个意思。您比我早 1500 中就记录下这种正比关系，的确了不起,
和推测, 一文中早就推崇过贵国的古代文化,目前我们还只
是刚刚走到这个知识领域的边缘，然而一旦对它有了充分的认
识，就将会在我们面 前展现出一个迄今为止只被人们神话般
地加以描述的知识王国”。
1686 年, 关于中国文字和语言的研究 真是令人佩服之至 』 我在
C'
1,How to plot the enlargement sketch of the small deformation
?Accurate method to plot diagram
of deformation,the arc line as
shown in the figure；
?Determine deformation △ Li of
each rod as shown in Fig l,
?Approximate method to plot the
diagram of deformation; the tangent
of the arc line shown in the figure,
Example 8 Enlargement sketch of the small deformation the and method to
determine displacements
A B
C
L1 L
2
P 1L?2L?
C"
85
86
C'
1、怎样画小变形放大图？
?变形图严格画法，图中弧线；
?求各杆的变形量△ Li,如图 1；
?变形图近似画法，图中弧之切线。
[例 8] 小变形放大图与位移的求法。
A B
C
L1 L
2
P 1L?2L?
C"
2,Write the relation between the displacement of point B shown in Fig.2 and
deformations of two rods,
1Lu B ??
Solution,The diagram of deformation is shown in the Fig.2,Point B moves
to point B',From the diagram of displacement we may know,
aa s i nc t g
2
1
LLv
B
????
87
A B
C
L1
L2
a
1L?
2L?
Bu
Bv
B'
P
Fig, 2
88
2、写出图 2中 B点位移与两杆变形间的关系
1Lu B ??
解：变形图如图 2,B点位移至 B'点，由图知,
aa s i nc t g
2
1
LLv
B
????
A B
C
L1
L2
a
1L?
2L?
Bu
Bv
B'
P
图 2
? ????? 060s i n6.12.18.060s i n ooA TPTm
kN55.113/ ??? PT
M P a1511036.76 55.11 9 ???? AT?
Example 9 Suppose the crossbeam ABCD is rigid,A steel cable with the
cross-section area
76.36mm2 is around a pulley without friction,Knowing P=20kN,E =177GPa,
Determine the stress of the steel cable and the upright displacement of point C,
Solution,method 1,Enlargement sketch
method of the small deformation,
1)Determine the internal force of the steel
cable,Take ABD as our study object,
2) Stress and elongation of the steel cable
P
A B
C
D T T
YA
XA
89
800 400 400
D
C P
A B 60° 60°
90
? ????? 060s i n6.12.18.060s i n ooA TPTm
kN55.113/ ??? PT
M P a1511036.76 55.11 9 ???? AT?
[例 9] 设横梁 ABCD为刚梁，横截面面积为 76.36mm2 的钢索绕
过无摩擦的定滑轮。设 P=20kN，试求钢索内的应力和 C点的
垂直位移。设钢索的 E =177GPa。
解：方法 1：小变形放大图法
1）求钢索内力：以 ABCD为研究对象
2) 钢索的应力和伸长分别为,
P
A B
C
D T T
YA
XA
800 400 400
D
C P
A B 60° 60°
mm36.1m17736.76 6.155.11 ?????? EATLL
C P
A B 60° 60°
800 400 400
D
A B 60° 60° D
B' D'
1?
2?
C?
C 3） Deformation is shown in the figure,
Upright displacement of point C is,
0 0
12
2
s i n 6 0 s i n 6 0
2
C
B B D D
L
???
??
? ? ?
?
00
1, 3 6
2 s i n 6 0 2 s i n 6 0
0, 7 9 m m
L?
??
?
91
92
mm36.1m17736.76 6.155.11 ?????? EATLL
C P
A B 60° 60°
800 400 400
D
A B 60° 60° D
B' D'
1?
2?
C?
C 3）变形图如左图,
C点的垂直位移为,
0 0
12
2
s i n 6 0 s i n 6 0
2
C
B B D D
L
???
??
? ? ?
?
00
1, 3 6
2 s i n 6 0 2 s i n 6 0
0, 7 9 m m
L?
??
?
§ 1－ 5 ELASTIC STRAIN ENERGY OF THE ROD IN AXIAL TENSION
AND COMPRESSION
1,Elastic strain energy,The work done by the external forces
results in the increase of some energy associated with the
deformation of the rod,This energy is referred to as the strain
energy of the rod,It is written as U,
2,Calculation of the strain energy of the rod in tension and compression,
Neglecting the lost energy,the work done by external forces is equal to the
strain energy,
) d)(d ( xEA xNx ???
1d d ( ) ( d )
2U W N x x? ? ? ?
x
EA
xNU d
2
)(d 2?
?? L xEA xNU d2 )(
2 ?
?
?
n
i ii
ii
AE
LN
U
1
2
2
Internal force is constant
in a small segment
N （ x ）
x
d x
N(x)
dx
x
93
94
§ 1－ 5 拉压杆的弹性应变能
一, 弹性应变能,杆件发生弹性变形，外力功转变为变形能贮存
于杆内，这种能成为应变能（ Strain Energy）用,U” 表示。
二,拉压杆的应变能计算,
不计能量损耗时，外力功等于应变能 。
) d)(d ( xEA xNx ???
1d d ( ) ( d )
2U W N x x? ? ? ?
x
EA
xNU d
2
)(d 2?
?? L xEA xNU d2 )(
2
?
?
?
n
i ii
ii
AE
LN
U
1
2
2
内力为分
段常 量 时
N （ x ）
x
d x
N(x)
dx
x
3,Strain－ energy density u of the rod
in tension and compression Strain
energy per unit volume,
d 1 ( ) ( d ) 1
d 2 d 2
U N x xu
V A x
???? ? ?
N （ x ）
x
d x
N(x)
dx
x
dx
xx dd ??
N(x) N(x)
(d )x?
)(xN
95
96
三,拉压杆的比能 u,
单位体积内的应变能 。
d 1 ( ) ( d ) 1
d 2 d 2
U N x xu
V A x
???? ? ?
N （ x ）
x
d x
N(x)
dx
x
dx
xx dd ??
N(x) N(x)
(d )x?
)(xN
kN55.113/ ??? PT
Solution,Method 2,Energy method,
（ Work of external forces is equal to the
strain energy）
（ 1） Determine the internal force of the steel
cable,Take ABCD as our study object,
? ????? 060s i n6.12.18.060s i n ooA TPTm
Example 9 Suppose the crossbeam ABCD is rigid,A steel cable with the cross-
section area
76.36mm2 is around a pulley without friction,Knowing P=20kN,E =177GPa,
Determine the stress of the steel cable and the upright displacement of point C,
97
800 400 400
C P
A B 60° 60°
P
A B
C
D T T
YA
XA
98
kN55.113/ ??? PT
解：方法 2,能量法,
（外力功等于变形能）
（ 1）求钢索内力：以 ABCD为研究对象,
? ????? 060s i n6.12.18.060s i n ooA TPTm
[例 9] 设横梁 ABCD为刚梁，横截面面积为 76.36mm2 的钢索绕
过无摩擦的定滑轮。设 P=20kN，试求钢索内的应力和 C点的
垂直位移。设钢索的 E =177GPa。
800 400 400
C P
A B 60° 60°
P
A B
C
D T T
YA
XA
EA
LTP C
22
2
?? 2
2
1 1, 5 5 1, 6
2 0 1 7 7 7 6, 3 6
0,7 9 m m
C
TL
So
PEA
??
?
?
??
?
M P a1511036.76 55.11 9 ???? AT?
（ 2) Stress of the steel cable is,
（ 3) Displacement of point C is,
Energy method,The method by which the
problems relative to the elastic deformation of the
structure members are soloed according to the
concept of strain energy is called the energy method,
99
800 400 400
C P
A B 60° 60°
100
EA
LTP C
22
2
??
mm79.0
36.761 7 720
6.155.11
2
2
?
??
?
?
???
PEA
LT
C
M P a1511036.76 55.11 9 ???? AT?
（ 2) 钢索的应力为,
（ 3) C点位移为,
能量法,利用应变能的概念解决与结构物
或构件的弹性变形有关的问题，这种方法
称为能量法。
800 400 400
C P
A B 60° 60°
§ 1－ 6 STATICALLY INDETERMINATE PROBLEMS AND THEIR
TREATMENT METHODS OF AXIAL TENSION AND COMPRESSION
1),Statically indeterminate problems,The problems in
which unknown forces(external forces,internal forces,stresses) cannot
be wholly determined only by static equilibrium equations
1,Statically indeterminate problems and their treatment methods
2,Method to solve statically indeterminate problems,
Equilibrium equations,compatibility equation of
deformation and physical equations must be combined
together to solve to solve these problems,
101
102
§ 1－ 6 拉压超静定问题及其处理方法
1、超静定问题,单凭静力平衡方程不能确定出全部未知力
（外力、内力、应力）的问题。
一、超静定问题及其处理方法
2、超静定问题的处理方法,平衡方程、变形协调方程、
物理方程相结合，进行求解。
Example 10 Rods 1,2 and 3 are connected together with a pin as shown
in the figure,Knowing the length of each rod is,L1=L2,L3 =L； and the area
of each rod is A1=A2=A,A3； modulus of elasticity of each rod is,E1=E2=E、
E3,External force is along the upright direction,Determine the internal force of
each rod,
C
P
A
B D
aa
1 2
3
Solution,?Equilibrium equations,
? ??? 0s i ns i n 21 aa NNX
? ????? 0c o sc o s 321 PNNNY aa
P
A
aaN1
N3
N2
103
104
[例 10] 设 1,2,3三杆用铰链连接如图，已知：各杆长为：
L1=L2,L3 =L ；各杆面积为 A1=A2=A,A3 ；各杆弹性模量
为,E1=E2=E,E3。外力沿铅垂方向，求各杆的内力。
C
P
A
B D
aa
1 2
3
解,?、平衡方程,
? ??? 0s i ns i n 21 aa NNX
? ????? 0c o sc o s 321 PNNNY aa
P
A
aaN1
N3
N2
11
11
1 AE
LNL ??
33
33
3 AE
LNL ??
?Geometric equation—compatibility
equation of deformation,
?Physical equation—elastic law,
?Complementary equation,detaining from the
geometric equation and physical equation,
?Solving the equilibrium equations and complementary equation we get,
ac o s31 LL ???
ac o s
33
33
11
11
AE
LN
AE
LN ?
33
3
11
33
3
33
3
11
2
11
21 c o s2 ; c o s2
c o s
AEAE
PAEN
AEAE
PAENN
?
?
?
??
aa
a
C
A
B D
aa
1 2
3
A1
1L?
2L?
3L?
105
106
11
11
1 AE
LNL ??
33
33
3 AE
LNL ??
?几何方程 ——变形协调方程,
?物理方程 ——弹性定律,
?补充方程：由几何方程和物理方程得。
?解由平衡方程和补充方程组成的方程组，得,
ac o s31 LL ???
ac o s
33
33
11
11
AE
LN
AE
LN ?
33
3
11
33
3
33
3
11
2
11
21 c o s2 ; c o s2
c o s
AEAE
PAEN
AEAE
PAENN
?
?
?
??
aa
a
C
A
B D
aa
1 2
3
A1
1L?
2L?
3L?
?Equilibrium equations；
?Geometric equation—compatibility equation of
deformation；
?Physical equation—elastic law；
?Complementary equation,from geometric equation and
physical equation
?Solving equilibrium equations and the complementary
equation,
3,Method and steps for solving statically indeterminate problems,
107
108
?平衡方程；
?几何方程 ——变形协调方程；
?物理方程 ——弹性定律；
?补充方程：由几何方程和物理方程得；
?解由平衡方程和补充方程组成的方程组 。
3、超静定问题的处理方法步骤,
Example 11 Four angles of a wooden pole are reinforced with four equal leg
angle steel of 40?40?4,The allowable stresses of steel and wood are respectively
[?]1=160MPa and [?]2=12MPa,module of elasticity of them are E1=200GPa and
E2=10GPa,Respectively Determine the allowable permissible load P,
? ???? 04 21 PNNY
21 LL ???
2
22
22
11
11
1 LAE
LN
AE
LNL ?????
?Geometric equation
?Physical equation and
complementary equation,
Solution,?Equilibrium equations,
P P y
4N1
N2
109
110
[例 11] 木制短柱的四角用四个 40?40?4的等边角钢加固，角钢
和木材的许用应力分别为 [?]1=160M Pa和 [?]2=12MPa,弹性模
量分别为 E1=200GPa 和 E2 =10GPa； 求许可载荷 P。
? ???? 04 21 PNNY
21 LL ???
2
22
22
11
11
1 LAE
LN
AE
LNL ?????
?几何方程
?物理方程及 补充方程,
解,?平衡方程, P P
y
4N1
N2
P P
y
4N1
N2
? Solving equilibrium equations and
the complementary equation we get,
PNPN 72.0 ; 07.0 21 ??
? ?111 07.0 ?APN ??
?Determine the permissible load of
the structure,
Method 1,
Section area of the angle steel may be obtained
from the table of hot-rolled steel,A1=3.086cm2
? ?2 2 20, 7 2N P A ???
? ? ? ? 222 2 / 0, 7 2 2 5 0 1 2 / 0, 7 2 1 0 4 2 k Na n d P A ?? ? ? ?
? ? ? ?11 1 / 0, 0 7 3 0 8, 6 1 6 0 / 0, 0 7 7 0 5, 4 k NS o P A ?? ? ? ?
111
112
P P
y
4N1
N2
? 解平衡方程和补充方程，得,
PNPN 72.0 ; 07.0 21 ??
? ?1 1 10, 0 7N P A ???
?求结构的许可载荷,
方法 1,
角钢截面面积由型钢表查得,
A1=3.086cm2
? ?2 2 20, 7 2N P A ???
? ? ? ? 222 2 / 0, 7 2 2 5 0 1 2 / 0, 7 2 1 0 4 2 k NPA ?? ? ? ? ?
? ? ? ?11 1 / 0, 0 7 3 0 8, 6 1 6 0 / 0, 0 7 7 0 5, 4 k NPA ?? ? ? ? ?
? ? ? ?11 1 / 0, 8 m mLE?? ? ?
? ? ? ?22 2 / 1, 2 m mLE?? ? ?
In the case of △ 1=△ 2,the angle steel will reach the limit state
first,that is the maximum load is determined by the angle steel,
?Determine the permissible load of the structure,
? ? ? ?
07.0
07.0
111 ANP ??? kN4.7 0 5
07.0
6.3 0 81 6 0 ???
The maximum load of the structure is still controlled by the steel,
Method 2,
Additionally,What about would it the like if area of the angle
steel is increased as 5 times？
What about would it like if the area of the wood is
changed into 25mm2?
113
114
? ? ? ?11 1 / 0, 8 m mLE?? ? ?
? ? ? ?22 2 / 1, 2 m mLE?? ? ?
所以在 △ 1=△ 2 的前提下，角钢将先达到极限状态,
即角钢决定最大载荷。
?求结构的许可载荷,
? ? ? ?
07.0
07.0
111 ANP ??? kN4.7 0 5
07.0
6.3 0 81 6 0 ???
另外：若将钢的面积增大 5倍，怎样？
若将木的面积变为 25mm2,又 怎样？
结构的最大载荷永远由钢控制着 。
方法 2,
?Geometric equation
Solution,?Equilibrium equations,
? ??? 0s i ns i n 21 aa NNX
? ???? 0c o sc o s 321 NNNY aa
13 c o s)( LL ???? a?
2,Assemble stresses—initial stresses
1),There is no the assemble stress in
statically determinate structure,
2),There is the assemble
stress in statically indeterminate structure,
The dimension error of rod 3 is ? as shown in the
figure,Determine the assemble internal force of each
rod,
A
B C
1 2
A
B C
1 2
D
A1
3
a a
?
115
116
?、几何方程
解,?、平衡方程,
2,静不定结构存在装配应力 。
? ??? 0s i ns i n 21 aa NNX
? ???? 0c o sc o s 321 NNNY aa
13 c o s)( LL ???? a?
二、装配应力 ——预应力
1、静定结构无装配应力。
如图,3号杆的尺寸误差为 ?，求各杆
的装配内力 。 A
B C
1 2
A
B C
1 2
D
A1
3
a a
?
a? c o s)(
33
33
11
11
AE
LN
AE
LN
??
?Physical equation and complementary equation,
? Solving equilibrium equations and
the complementary equation we get,
/ c o s21
c o s
3311
3
2
11
3
21 AEAE
AE
L
NN
a
a?
?
???
/ c o s21
c o s2
3311
3
3
11
3
3 AEAE
AE
L
N
a
a?
?
??
?
A1
a aN1 N2
N3
A
A1 3L?
2L?1L?
117
118
a? c o s)(
33
33
11
11
AE
LN
AE
LN
??
?、物理方程及 补充方程,
?,解平衡方程和补充方程，得,
/ c o s21
c o s
3311
3
2
11
3
21 AEAE
AE
L
NN
a
a?
?
???
/ c o s21
c o s2
3311
3
3
11
3
3 AEAE
AE
L
N
a
a?
?
??
?
A1
a aN1 N2
N3
A
A1 3L?
2L?1L?
1),There is no the temperature stress in
statically determinate structure,
2),There is the temperature
stress in statically indeterminate structure,
3,Temperature stress
Materials and dimensions of rod 1and2 are all
the same as shown in the figure,If temperature of
the structure changes from T1 to T2,determine the
temperature internal forces of each rod.（ linear
thermal expansion coefficient of each rod ai ; △ T=
T2 -T1)
A
B C
1 2
C
A
B D
??
1 2
3
A1
1L?
2L?
3L?
119
120
1、静定结构无温度应力。
三,应力温度
如图,1,2号杆的尺寸及材料都相
同，当结构温度由 T1变到 T2时,求各杆
的温度内力。（各杆的线膨胀系数分别
为 ai ; △ T= T2 -T1)
A
B C
1 2
C
A
B D
??
1 2
3
A1
1L?
2L?
3L?
2、静不定结构存在温度应力。
C
A
B D
??
1 2
3
A1
1L?
2L?
3L?
?,Geometric equation
Solution,?,Equilibrium equations,
12s i n s i n 0X N N??? ? ??
? ???? 0c o sc o s 321 NNNY ??
?c o s31 LL ???
ii
ii
ii
i LTAE
LNL a????
?,Physical equation,
A
??N1
N3
N2
121
122
C
A
B D
??
1 2
3
A1
1L?
2L?
3L?
?、几何方程
解,?、平衡方程,
?c o s31 LL ???
ii
ii
ii
i LTAE
LNL a????
?、物理方程,
A
??N1
N3
N2
12s i n s i n 0X N N??? ? ??
? ???? 0c o sc o s 321 NNNY ??
C
A
B D
??
1 2
3
A1
1L?
2L?
3L?
?,Complementary equation,
?aa c o s)( 33
33
33
11
11
11 LT
AE
LNLT
AE
LN ?????
Solving equilibrium equations and the
complementary equation we get,
/ c o s21
)c o s(
3311
3
2
3111
21 AEAE
TAE
NN
?
?aa
?
??
???
/ c o s21
c o s)c o s(2
3311
3
2
3111
3 AEAE
TAE
N
?
??aa
?
??
?
123
124
C
A
B D
??
1 2
3
A1
1L?
2L?
3L?
?,补充方程
?aa c o s)( 33
33
33
11
11
11 LT
AE
LNLT
AE
LN ?????
解平衡方程和补充方程，得,
/ c o s21
)c o s(
3311
3
2
3111
21 AEAE
TAE
NN
?
?aa
?
??
???
/ c o s21
c o s)c o s(2
3311
3
2
3111
3 AEAE
TAE
N
?
??aa
?
??
?
a
a
a
a
N1
N2
Example 12 Theupper and lower ends of a ladder-like steel
shaft are fixed at temperature T1=5℃ as shown in the figure,
Areas of the upper and lower segments are respectively ?1=?cm2
and ?2=??cm2,When its temperature reaches T2=25℃,determine
the temperature stress of each rod.(Linear thermal expansion
coefficient a? ?2.5× 10 -6 1/0C； modulus of elasticity
E=200GPa)
?,Geometric equation,
Solution,?,Equilibrium equation,
? ??? 021 NNY
0?????? NT LLL 125
126
a
a
a
a
N1
N2
[例 12] 如图，阶梯钢杆的上下两端在 T1=5℃
时被固定,杆的上下两段的面积分别
??=?cm2, ??=??cm2，当温度升至 T2
=25℃ 时,求各杆的温度应力。
(线膨胀系数 a =12.5× ；
弹性模量 E=200GPa)
C?110 6?
?、几何方程,
解,?、平衡方程,
? ??? 021 NNY
0?????? NT LLL
?,Physical equation,
Solving equilibrium equations and
the complementary equation we get,k N 3.33
21 ?? NN
?,Complementary equation,
2
2
1
1 ; 2
EA
aN
EA
aNLTaL
NT ?????? a
2
2
1
12
EA
N
EA
N
T ??? a
?,Temperature stresses,
M P a 7.66
1
1
1 ?? A
N? M P a 3.33
2
2
2 ?? A
N?
127
128
?、物理方程
解平衡方程和补充方程，得,
k N 3.3321 ?? NN
?,补充方程
2
2
1
1 ; 2
EA
aN
EA
aNLTaL
NT ?????? a
2
2
1
12
EA
N
EA
N
T ??? a
?、温度应力
M P a 7.66
1
1
1 ?? A
N? M P a 3.33
2
2
2 ?? A
N?
§ 1－ 7 MECHANICAL PROPERTIES OF MATERIALS IN AXIAL TENSION
AND COMPRESSION
1,Testing conditions and instruments
d
h
Mechanical properties,Some properties about strength and deformation shown by
materials under the action of external loads,
1),Testing conditions,normal temperature(20℃ )；
static load（ loaded gradually)
Standard specimen
129
130
§ 1－ 7 材料在拉伸和压缩时的力学性能
一、试验条件及试验仪器
1、试验条件：常温 (20℃) ；静载（极其缓慢地加载）；
标准试件。
d
h
力学性能：材料在外力作用下表现的有关强度、变形方面的特性。
2,Experimental implement,Universal testing machine ； Extensometer
131
132
2、试验仪器：万能材料试验机；变形仪（常用引伸仪）。
meter-
pedestal
plate
centesi
mal
meter
meter
pedestal
bolt for
installing
the meter
standard
specimen
spring
EEA
P
L
L ?? ????
2,Tensile diagram of the low carbon steel specimen(P--?L diagram)
3,Stress-strain curve of the low carbon steel specimen (? --? diagram)
EA
PLL ??
133
Typical points in the curve of the
low-carbon steel
???
134
EEA
P
L
L ?? ????
二、低碳钢试件的拉伸图 (P-- ?L图 )
三、低碳钢试件的应力 --应变曲线 (? --? 图 )
EA
PLL ??
(1) Elastic range of the low-carbon steel in tension (oe)
1,op – Proportional segment,
?p – Proportional limit
E
?? ?
atg?E
2,pe – Curved segment,
?e – Elastic limit
)( nf ?? ?
135
Stress-strain relation ship
in the elastic region
136
(一 ) 低碳钢拉伸的弹性阶段 (oe段 )
1,op -- 比例段,
?p -- 比例极限
E
?? ?
atg?E
2,pe --曲线段,
?e -- 弹性极限
)( nf ?? ?
(2) Yielding(flowing) range of the low carbon steel in
tension(es)
e s –Yielding range,?s ---Yielding limit
Sliding lines,
Failure stress of plastic material,?s 。
137
Single-crystal Cu-A1 specimen after
tension
138
(二 ) 低碳钢拉伸的屈服 (流动）阶段 (es 段 )
e s --屈服 段, ?s ---屈服极限
滑移线,
塑性材料的失效应力,?s 。
２,Unloaded law,
１,?ｂ --Strength limit
３,Cold hardening,
４,hard-drawn
time effect,
(3),Hardening range of the low carbon steel(ｓｂ )
139
Typical points in the curve of the low-carbon
steel ???
140
２、卸载定律,
１,?ｂ ---强度 极限
３、冷作硬化,
４、冷拉时效,
(三 )、低碳钢拉伸的强化阶段 (ｓｂ 段 )
1,Percent elongation
?
10 0
0
0
100LL
L
? ???
2,Percent reduction
in area ?
01 0
0
0
100AA
A
? ???
3,Brittleness,ductility、
relativity
0 05? ? the division line
(4),Necking(crack) stage (b f ) of the low carbon steel in tension
141
Typical points in the curve of the low-carbon
steel ???
eulargeneut
142
1、延伸率,? 2、面缩率,? 3、脆性、塑性及相对性
为界以 005??
(四 )、低碳钢拉伸的颈缩（断裂）阶段 (b f 段 )
10 0
0
0
100LL
L
? ??? 01 0 0
0
100AA
A
? ???
? ?
?
4,Plastic material without obvious yield
?.?
? 0.2
Nominal yield stress,
? 0.2 is the failure stress of this type of
material,
5,Mechanical properties of the cast
iron in tension
?ｂ L --- Tensile strength limit of cast iron(fail
stress）; tg a?E
?
?
bL?
143
0
0
144
? ?
?
四、无明显屈服现象的塑性材料
?.?
? 0.2
名义屈服应力,
? 0.2,即此类材料的失效应力。
五、铸铁拉伸时的机械性能
?ｂ L ---铸铁拉伸强度 极限（失效应力）
割线斜率 ; tg a?E
?
?
bL?
0
0
6,Mechanical properties of
the material in compression
?ｂ y – Compressive strength
limit of cast iron；
?ｂ y ?（ 4 ～ 6） ?ｂ L
145
compression
tension
comparison of tension and compression for the low-
carbon steel
Compression of the cast iron
146
六、材料压缩时的机械性能
?ｂ y ---铸铁压缩强度 极限；
?ｂ y ?（ 4 ～ 6） ?ｂ L
7,Safety factor,permissible stress,limit stress
? ?
n
jx?? ?
? ?bs
jx
????,,2.0?
N>1
1),Permissible stress,
2),Limit stress,
3),Safety factor,
147
148
七、安全系数、许用应力、极限应力
? ?
n
jx?? ?
? ?bs
jx
????,,2.0?
N>1
1,许用应力,
2、极限应力,
3、安全系数,
006500/30 ???
N5024/160214.3 2 ?????? ?AP
Solution,Deformation may exceed the range of
linear elasticity,therefore the elastic law is not
applied here,Calculation should be done in the
following,
M P a1 6 0??
Example 13 Diameter of a copper wire is d=2mm and its length is L=500mm,
Tensile curve of copper is shown in the figure,To make elongation of the copper wire
is 30mm whet is the force P that we must act？
0 5 1 0 1 5 2 0 （ ?? ）
1
0
0
2
0
0
3
0
0
? （ M Pa ）
From the tensile curve,
? (MPa)
? (%)
149
150
006500/30 ???
N5024/160214.3 2 ?????? ?AP
解：变形量可能已超出了“线弹性”
范围，故，不可 再 应用 ―弹性 定律,
。 应如下计算,
M P a1 6 0??
[例 13] 铜丝直径 d=2mm，长 L=500mm,材料的 拉伸 曲线如 图
所示。如欲使铜丝的伸长量为 30mm,则大约需加多大的力 P？
0 5 1 0 1 5 2 0 （ ?? ）
1
0
0
2
0
0
3
0
0
? （ M Pa ）
由拉伸图知,
? (MPa)
? (%)
151
1,Knowing the elastic modulus of steel is E＝
200GPa,the elastic modulus of Aluminum is E＝
71GPa.Try to compare:① which material produces a
larger strain when they are subjected to the same stress？
② which material corresponds to a large stress when
they have the same strain？
Chapter 1 Exercises
152
一、钢的弹性模量 E＝ 200GPa，铝的弹性模
量 E＝ 71GPa。试比较在同一应力作用下，那种
材料的应变大？在产生同一应变的情况下，那
种材料的应力大？
第一章 练习题
153
2,For the different members made of a same material,do
they have the same allowable stress? In general cases,the safety
coefficient of the brittle material is selected to be larger than that of
the plastic material,why?
3,As shown in the figure,the Al-alloy circular bar is
subjected to the axial tensile force P,Knowing E＝ 73GPa,and
μ=1/3,When the elongation of the bar is L=7mm,try to determine:
① the change in its diameter Δd； ② the axial force P,
154
二、由同一材料制成的不同构件，其许用
应力是否相同?一般情况下脆性材料的安全系数
要比塑性材料的安全系数选得大些，为什么？
三、图示铝合金圆杆受轴向拉力 P。已知材
料的弹性模量 E＝ 73GPa，泊松比 μ ＝ 。试求
当杆伸长量 ＝ 7mm时，①直径的减少量 ；
② P力的大小。
3
1
d?
L?
155
Solution,
KN
L
AEL
N
mm
L
Ld
d
4.120
43
30737
0 2 3 3.0
3 0 0 03
7301
2
?
?
???
?
???
?
?
?
??
?
???
??
?
?
②
①
4,A bracket is shown in the figure,The cross-section area of the steel bar
AB is A1=6cm2； the cross-section area of the wooden bar BC is A2=300cm2，
Knowing the allowable stress of steel is,the allowable tensile
stress of wood is,and the allowable compressive stress of wood
is, Try determine the allowable load P of the bracket,
? ? M P a1 4 0??
? ? M P aL 8??
? ? M P aY 4??
156
解,
KN
L
AEL
N
mm
L
Ld
d
4.120
43
30737
0 2 3 3.0
3 0 0 03
7301
2
?
?
???
?
???
?
?
?
??
?
???
??
?
?
②
①
? ? M P aL 8??
四、图示支架,AB为钢杆，横截面面积 ；
BC为木杆，横截面面积 。钢的许用应力
,木材的许用拉应力,许用压应
力 。试求支架的许可载荷 。
21 6A cm?
22 300A cm?
? ? M P a1 4 0??
? ? M P aY 4?? ??P
157
Solution,From the equilibration condition we get,
The allowable axial force of bar AB and bar BC is respectively,
The bar AB should satisfy the strength condition,
The bar BC should satisfy the strength condition,
As the strength of the two bars should both be satisfied,the allowable
load of the bracket should be taken as,
[P] =101 KN
21 61.2
2.2,
4.1
2.2 NPNP ??
? ? ? ? KNAN 84101 4 06 211 ????? ?
? ? ? ? KNAN Y 1 2 01043 0 0 222 ????? ?
? ? ? ? KNNP 1 3 24.1 2.2 1 ??
? ? ? ? KNNP 1 0 161.2 2.2 2 ??
158
解：由平衡条件求得,
AB杆与 BC杆的许可轴力分别为,
AB杆要满足强度条件,
BC杆要满足强度条件,
两杆的强度都应得到满足，故支架的许可载荷
应取
[P] =101 KN
21 61.2
2.2,
4.1
2.2 NPNP ??
? ? ? ? KNAN 84101 4 06 211 ????? ?
? ? ? ? KNAN Y 1 2 01043 0 0 222 ????? ?
? ? ? ? KNNP 1 3 24.1 2.2 1 ??
? ? ? ? KNNP 1 0 161.2 2.2 2 ??
151
160
1
2
CHAPTER 1 AXIAL TENSION AND COMPRESSION
§ 1–1 Concepts and practical examples of axial tension and compression
§ 1–2 Internal force,method of section,axial force and its diagram
§ 1–3 Stresses on the section and strength conditions
§ 1-4 Deformation of the rod in axial tension and compression ? law of
elasticity
§ 1-5 Elastic strain energy of the rod in axial tension and compression
§ 1-6 Statically indeterminate problems and their treatment methods of axial
tension and compression
§ 1-7 Mechanical properties of materials in axial tension and compression
3
4
§ 1–1 轴向拉压的概念及实例
§ 1–2 内力、截面法,轴力及轴力图
§ 1–3 截面上的应力及强度条件
第一章 轴向拉伸和压缩
§ 1-4 拉压杆的变形 ? 弹性定律
§ 1-5 拉压杆的弹性应变能
§ 1-6 拉压超静定问题及其处理方法
§ 1-7 材料在拉伸和压缩时的力学性能
§ 1–1 CONCEPTS AND PRACTICAL EXAMPLES OF
AXIAL TENSION AND COMPRESSION
Characteristic of the external force,The acting line of the resultant of
e x t e r n a l fo r c e s is co i n c i d ed with t h e a x i s of t h e ro d,
1,Concepts
Characteristic of the deformation,Deformation of the rod is mainly
elongation or contraction along the axis of the rod and companied with lateral
reduction or enlargement,
Axial tension, Deformation of the rod is axial elongation
and lateral shortening,
Axial compression,Deformation of the rod is axial
shortening and lateral enlargement,
5
6
§ 1–1 轴向拉压的概念及实例
轴向拉压的外力特点,外力的合力作用线与杆的轴线重合 。
一、概念
轴向拉压的变形特点,杆的变形主要是轴向伸缩，伴随横向
缩扩。
轴向拉伸：杆的变形是轴向伸长，横向缩短。
轴向压缩：杆的变形是轴向缩短，横向变粗。
In axial compression,the corresponding force is called compressive force,
In axial tension,the corresponding force is called tensile
force,
Mechanical models are shown in the figures
PP
PP
7
8
轴向压缩，对应的力称为压力。
轴向拉伸，对应的力称为拉力。
力学模型如图
PP
PP
Practical
examples in
engineering
2,
9
10
工
程
实
例
二,
11
12
1,Internal force
Internal force is the resultant of internal forces,which is acting mutually
between two neighbour parts inside the body,caused by the external forces,
§ 1–2 INTERNAL FORCE,METHOD OF SECTION,AXIAL
FORCE AND ITS DIAGRAM
13
14
一、内力
指由外力作用所引起的、物体内相邻部分之间分布内
力系的合成（附加内力）。
§ 1–2 内力 · 截面法 · 轴力及轴力图
2,Method of section · axial force
Calculation of the internal forces is the foundation to analyze the
problems of strength,rigidity,stability etc,The general method to
determine internal forces is the method of section,
1),Basic steps of the method of section,
① Cut off,Assume to separate the rod into two distinct parts in the section in
which the internal forces are to be determined,
② Substitute,Take arbitrary part and substitute the action of another part on it by
the corresponding internal force in the cut-off section,
③ Equilibrium,Set up equilibrium equations for the remained part and
determine the unknown internal forces according to the external forces acted on it.
（ Here the internal forces in the cut-off section are the external forces for the
remained part） 15
16
二、截面法 · 轴力
内力的计算是分析构件强度、刚度、稳定性等问题的
基础。求内力的一般方法是截面法。
1,截面法的基本步骤,
① 截开,在所求内力的截面处，假想地用截面将杆件一分为二。
②代替,任取一部分，其弃去部分对留下部分的作用，用作用
在截开面上相应的内力（力或力偶）代替。
③平衡,对留下的部分建立平衡方程，根据其上的已知外力来
计算杆在截开面上的未知内力（此时截开面上的内力
对所留部分而言是外力）。
2,Axial force—internal force of the rod in axial
tension or compression,designated by N,
Such as,Determine N by the method of section,
0?? X 0NP?? NP ?
A P P
P
A
N
Simple sketch
A P P Cut off:,
Substitute,
Equilibrium,
17
18
2,轴力 ——轴向拉压杆的内力，用 N 表示。
例如,截面法求 N。
0?? X 0NP?? NP ?
A P P
简图
A P P 截开,
代替,
平衡,
P
A
N
x
① Reflected the variety relation between
the corresponding axial force and the
position of the section,
② Find out value of the maximum axial
force and the position of the section in which the maximum axial force act,That is
to determine the position of the critical section and supply the information for the
calculation of strength,
3,Diagram of the axial force— sketch
expression of N (x)
3),Sign conventions for the axial force,
axial force N (tensile force)is positive when its
direction point to the outward direction of the
normal line of the section,(compressive
force)negative inward
N > 0 N N
N<0 N N
x
N
P
+
meaning
19
20
① 反映出轴力与横截面位置变化关系，较直观；
②确定出最大轴力的数值
及其所在横截面的位置,
即确定危险截面位置，为
强度计算提供依据。
三,轴力图 —— N (x) 的图象表示。
3,轴力的正负规定,
N 与外法线同向,为正轴力 (拉力 )
N与外法线反向,为负轴力 (压力 )
N > 0 N N
N<0 N N
N
x
P
+
意
义
Example 1 The forces with magnitudes 5P,8P,4P and P act respectively at
points A,B,C,D of the rod,Their directions are shown in the figure,Try to plot
the diagram of the axial force of the rod,
Solution,Determine the internal force N1 in segment OA,
Take the free body as shown in the figure,
A B C D
PA PB PC PD
O
A B C D
PA PB PC PD
N1
0?? X 0
1 ????? DCBA PPPPN
04851 ????? PPPPN PN 21 ?
21
22
[例 1] 图示杆的 A,B,C,D点分别作用着大小为 5P,8P,4P,
P 的力，方向如图，试画出杆的轴力图。
解,求 OA段内力 N1：设置截面如图
A B C D
PA PB PC PD
O
0?? X
1 0A B C DN P P P P? ? ? ? ? ?
04851 ????? PPPPN PN 21 ?
A B C D
PA PB PC PD
N1
x
Similarly,we get the
internal forces in segment
AB,BC,CD, They are
respectively,
N2= –3P
N3= 5P
N4= P
The diagram
of the axial
force is shown
in the right
figure,
B C D
PB PC PD
N2
C D
PC PD
N3
D
PD
N4
N
x 2P
3P
5P
P + +
–
23
24
同理，求得 AB、
BC,CD段内力分
别为,
N2= –3P
N3= 5P
N4= P
轴力图如右图
B C D
PB PC PD
N2
C D
PC PD
N3
D
PD
N4
N
x 2P
3P
5P
P + +
–
Simple method to plot the diagram of axial force,From the left to the right,
Characteristic of the diagram of the axial force,Value of sudden
change = concentrated load
If meeting the force P to the left ?,the increase of the axial force N is positive；
If meeting the force to the right ?, the increase of the axial force N is negative,
5kN 8kN 3kN
+
– 3kN
5kN
8kN
25
26
轴力 (图 )的简便求法,自左向右,
轴力图的特点：突变值 = 集中载荷
遇到向左的 P?,轴力 N 增量为正；
遇到向右的 P?, 轴力 N 增量为负。
5kN 8kN 3kN
+
– 3kN
5kN
8kN
Solution,The free end of the rod is the
origin of the coordinate and coordinate x to the
right is positive,Take the segment of length x
on the left of point x,its internal force is q
K L
x O
2
0 2
1d)( kxxkxxN x ??? ??
2
m a x 2
1)( kLxN ??
Example 2 Length of the rod shown in the figure is L,Distributed force q = kx
is acted on it,direction of the force is shown in the figure,Try to plot the diagram of
axial force of the the rod,
L
q(x)
N(x)
x
q(x)
N x
O
–
2
2kL
27
28
解,x 坐标向右为正，坐标原点在
自由端。
取左侧 x 段为对象，内力 N(x)为,
q
q L
x O
2
0 2
1d)( kxxkxxN x ??? ??
2
m a x 2
1)( kLxN ??
[例 2] 图示杆长为 L，受分布力 q = kx 作用，方向如图，试画出
杆的轴力图。
L
q(x)
N x
O
–
2
2kL
N(x)
x
q(x)
1,Concept of stress
§ 1–3 STRESSES ON THE SECTION AND STRENGTH CONDITIONS
Bring forward
the problem,
1),The magnitude of the internal force can not scale the strength of the
structure member,
2),Strength：① Intensity of the distributed internal forces in the
section? stress； ② The load-bearing capacity of the material,
1),Definition,Intensity of the internal force due to the external forces,
29
P P
P P
30
一、应力的概念
§ 1–3 截面上的应力及强度条件
问题提出,
1,内力大小不能衡量构件强度的大小。
2,强度：①内力在截面的分布集度 ? 应力；
②材料承受荷载的能力。
1,定义,由外力引起的内力 集度 。
P P
P P
Under most cases distribution of the internal force inside engineering members
is not uniform,Definition of intensity is neither accurate and important because
breakage or failure often begins from the point at which intensity of the internal
force is maximum,
?P
?A
M ① Average stress,
② Whole stress（ sum stress）,
A
Pp
M Δ
Δ?
Δ 0
Δ d
Δ dl i mA
PP
p
AA?
??
2),Expression of stress,
31
32
工程构件，大多数情形下，内力并非均匀分布，集度的定
义不仅准确而且重要，因为“破坏”或“失效”往往从内力集
度最大处开始。
?P
?A
M ① 平均应力,
② 全应力（总应力）,
A
Pp
M Δ
Δ?
Δ 0
Δ d
Δ dl i mA
PP
p
AA?
??
2,应力的表示,
③ Whole stress may be decomposed into,
p ?
M ?
A
N
A
N
A d
d
Δ
Δlim
0Δ
??
?
?
A
T
A
T
A d
d
Δ
Δlim
0Δ
??
?
?
a,Stress perpendicular to the section is called―normal stress‖
b,Stress lying in the section is called―shearing stress‖
33
34
③ 全应力分解为,
p ?
M ?
A
N
A
N
A d
d
Δ
Δlim
0Δ
??
?
?
A
T
A
T
A d
d
Δ
Δlim
0Δ
??
?
?
a.垂直于截面的应力称为,正应力” (Normal Stress)；
b.位于截面内的应力称为,剪应力” (Shearing Stress)。
Before
deformation
1),Experiment on the law of deformation and the hypothesis of plane
section,
Hypothesis of plane section,Cross sections remain planes before and
after deformations,
Deformations of longitudinal fibers are the same
a b
c d
After
loading
P P
d ′
a′
c′
b′
2,Stress in the cross section of the rod in tension or compression
35
36
变形前
1,变形规律试验及平面假设,
平面假设,原为平面的横截面在变形后仍为平面。
纵向纤维变形相同。
a b
c d
受载后 P P d ′ a′ c′ b′
二、拉（压）杆横截面上的应力
The material is homogeneous and,its deformation is uniform,so the internal
force is distributed uniformly。
2,Tensile stress,
A
xN )( ??
Normal stress due to the axial forces — ?, distributes uniformly in the cross
section,
Critical section,The section in which internal force is maximum
and of which the dimension is smallest,
Critical point,The point at which the stress is maximum,
3,Critical section and the maximum working stress,
))( )(m a x ( m a x xA xN?? 37
? N(x) P
38
均匀材料、均匀变形，内力当然均匀分布。
2,拉伸应力,
A
xN )( ??
轴力引起的正应力 —— ?, 在横截面上均布。
危险截面：内力最大的面，截面尺寸最小的面。
危险点：应力最大的点。
3,危险截面及最大工作应力,
))( )(m a x ( m a x xA xN??
? N(x) P
Straight rod,cross section of the rod is without sudden change,there is a
certain distance from the section to the point at which the load acts,
4),Application conditions of the formula,
6),Stress concentration,
Stress increases abruptly near the cross section with a sudden change in dimension
5),Saint-Venant principle,
Distribution and magnitude of the stress in the section at a certain distance from
the point at which the load is acted are not affected by the acting form of external
loads,
39
40
直杆、杆的截面无突变、截面到载荷作用点有一定 的距离。
4,公式的应用条件,
6,应力集中（ Stress Concentration）,
在截面尺寸突变处，应力急剧变大。
5,Saint-Venant原理,
离开载荷作用处一定距离，应力分布与大小不受外载荷作
用方式的影响。
Sketch of Saint-Venant principle and stress concentrations
(Red real lines denote the line before deformation and red dashed lines denote the
shape after deformation.)
Sketch of deformation,a b c P P
Sketch of the stress
distribution,
41
42
Saint-Venant原理与应力集中示意图
(红色实线为变形前的线，红色虚线为红色实线变形后的形状。 )
变形示意图,a b c P P
应力分布示意图,
7),Criterion of the strength design,
? ? ))( )(m a x ( m a x ?? ?? xA xN
where,[?]—allowable stress,?max– the maximum working stress
at the critical point,
② Design the dimension of the section,
][
m a x
m in ?
NA ?
? ? ; m a x ?AN ?
? ? m a x()P f N?
Three kinds of calculation of strength may be done
according to the criterion of strength,
That structure members are ensured not to be wrecked and have certain safe
degree,
? ? m a x ?? ?① Check the strength,
③ Determine the allowable load,
43
44
7,强度设计准则（ Strength Design）,
? ? ))( )(m a x ( m a x ?? ?? xA xN
其中,[?]--许用应力,?max--危险点的最大工作应力。
② 设计截面尺寸,
][
m a x
m in ?
NA ?
? ? ; m a x ?AN ? ? ? )N(fP i?
依强度准则可进行三种强度计算,
保证构件不发生强度破坏并有一定安全余量的条件准则。
? ? m a x ?? ?
① 校核强度,
③ 许可载荷,
Example 3 A circular rod is subjected to a tensile force P =25 kN,Its diameter
is d =14mm and its allowable stress is [?]=170MPa,Try to check the strength of
the rod,
Solution,① Axial force,N = P =25kN
M P a1 6 20 1 40143 102544 2
3
2m a x ??
?????
..π d
P
A
N?② Stress,
③ Check the strength,
? ? 1 7 0 M P a1 6 2 M P am a x ??? ??
④ Conclusion,The strength of the rod satisfies request,
The rod can work normally,
45
46
[例 3] 已知一圆杆受拉力 P =25 k N，直径 d =14mm，许用应力
[?]=170MPa，试校核此杆是否满足强度要求。
解,① 轴力,N = P =25kN
M P a1 6 20 1 40143 102544 2
3
2m a x ??
?????
..π d
P
A
N?
② 应力,
③ 强度校核,
? ? 1 7 0 M P a1 6 2 M P am a x ??? ??
④ 结论：此杆满足强度要求，能够正常工作。
Example 4 A three-pin house frame on which a vertical uniform load,with
the in density intensity is q =4.2kN/m is applied is shown in the figure,Diameter
of the steel tensile rod in the frame is d =16 mm and its allowable stress is
[?]=170MPa,Try to check the strength of the rod,
47
Tie bar
8.5m
q
q
A
B
C
48
[例 4] 已知三铰屋架如图，承受竖向均布载荷，载荷的分布
集度为,q =4.2kN/m，屋架中的钢拉杆直径 d =16 mm，许用
应力 [?]=170M Pa。 试校核钢拉杆的强度。
钢拉杆
8.5m
q
q
A
B
C
① Determine the reactions first according to the global equilibrium
Solution,
kN519 0
0 0
.Rm
HX
AB
A
?? ?
?? ?
49
Tie bar
8.5m
RA RB
HA
q
q
A
C
50
① 整体平衡求支反力 解,
kN519 0
0 0
.Rm
HX
AB
A
?? ?
?? ?
钢拉杆
8.5m
RA RB
HA
q
q
A
C
③ Stress,
④ Strength check
and conclusion,? ? M P a 1 7 0 M P a 1 3 1 m a x ??? ??
This rod satisfies the request of strength,It is safe,
M P a131
0160143
103264
d
4
2
3
2m a x
?
?
??
?
??
..
.
P
A
N
?
?
② Determine the axial force
according to the partial equilibrium
,
kN326 0,Nm C ?? ?
51
RA
HA
RC
HC
N
Cq
A
52
③ 应力,
④ 强度校核与结论,? ? M P a 1 7 0 M P a 1 3 1
m a x ??? ??
此杆满足强度要求，是安全的。
M P a131
0160143
103264
d
4
2
3
2m a x
?
?
??
?
??
..
.
P
A
N
?
?
② 局部平衡求 轴力,
HC
kN326 0,Nm C ?? ?
RA
HA
RC
N
Cq
A
? ?/ ;
/ s i n
B D B D
BD
AN
Lh
?
?
?
? 。
Example 5 A simple crane is shown in the figure,AC is a rigid beam, sum
weight of the hoist and heavy body that is lifted is P,What should be the angle ?
so that the rod BD has the minimum weight？ The allowable stress of the rod [?]
is known,;BDBD LAV ?
Analysis,x L
h
? P
A B
C
D
53
54
? ?/ ;
/ s i n
B D B D
BD
AN
Lh
?
?
?
? 。
[例 5] 简易起重机构如图,AC为刚性梁，吊车与吊起重物总重
为 P，为使 BD杆最轻，角 ? 应为何值？ 已知 BD 杆的 许用应力
为 [?]。;BDBD LAV ?
分析,x L
h
? P
A B
C
D
PxhNm BDA ??? ? )c t g() s i n(,0 ??
?c o sh
PLN
BD ?
? ??/NA BD?
? The cross-section area A of the rod BD,
Solution,? Internal force N((? ) of the rod BD,Take AC as
our study object as shown in the figure,
YA
XA
?
NBD
x
L
P
A B
C
55
56
PxhNm BDA ??? ? )c t g() s i n(,0 ??
?c o sh
PLN
BD ?
? ??/NA BD?? BD杆横截面面积 A,
解,? BD杆 内力 N(? ),取 AC为研究对象，如图
YA
XA
?
NBD
x
L
P
A B
C
YA
XA
?
NBD
x
L
P
A B
C
③ Determine the minimum value of VBD,
2/ s i n ;
[ ] s i n 2BD
PLV A L A h ?
??
? ? ?
o
m i n
24 5,
[]
PLA s V?
?
? ? ?
67
58
YA
XA
?
NBD
x
L
P
A B
C
③ 求 VBD 的 最小值,2
/ s i n ;
[ ] s i n 2BD
PLV A L A h ?
??
? ? ?
][
2 45
m i n
o
??
PLV,??? 时
3,Stresses in the inclined section of the rod in tension or compression
P P
k
k a
a
a
a A
Pp ?
Aa:Area of the inclined section； Pa,Internal
force in the inclined section,
aa aa c o s c o s AAA A ???
,
a?a
a
a
a c o sc o s 0???? A
P
A
Pp a?a c o s0?p
P
k
k a
Pa
From geometric
relation
Substituting it into the above
formula we get
Solution,Adopt the method of
section,According to the equilibrium
equation,Pa=P then
Assume a straight rod is subjected
to a tensile force P,Determine the
stress in the inclined section k-k,
Whole stress in the
inclined ection,
59
60
三、拉 (压 )杆斜截面上的应力
设有一等直杆受拉力 P作用。
求：斜截面 k-k上的应力。
P P
k
k a 解：采用截面法
由平衡方程,Pa=P
则,
a
a
a A
Pp ? Aa,斜截面面积； Pa：斜截面上内力。
由几何关系,
a
a a
a c o s
c o s AA
A
A ???
代入上式，得,
a?a
a
a
a c o sc o s 0???? A
P
A
Pp 斜截面上全应力,a?a c o s0?p
P
k
k a
Pa
a?a c o s0?p
Decomposition,
pa ? a?a? aa 20 c o sc o s ?? p
a?aa?a? aa 2s i n2s i nc o ss i n 00 ??? p
It indicates the change of stresses in different sections
through a point,
As a =90°,
0)( m in ?a?
As a = 0,90°, 0||
min ?a?
As a = 0°,
)( 0m a x ?? a ?
(The maximum normal stress exists
in the cross section)
As a = ± 45°,
2
|| 0m ax ?? a ?
(The maximum shearing stress exists
in the inclined section of 45 ° )
Whole stress in the
inclined section,
61
P P
k
k a
P
k
k a
pa
?a
?a
a
62
斜截面上全应力,a?
a c o s0?p
分解,
pa ? a?a? aa 20 c o sc o s ?? p
a?aa?a? aa 2s i n2s i nc o ss i n 00 ??? p
反映：通过构件上一点不同截面上应力变化情况。
当 a = 90° 时,
0)( m in ?a?
当 a = 0,90° 时,0||
min ?a?
当 a = 0° 时,
)( 0m a x ?? a ?
(横截面上存在最大正应力 )
当 a = ± 45° 时,
2
|| 0m ax ?? a ?
(45 ° 斜截面上剪应力达到最大 )
P P
k
k a
P
k
k a
p a
a ?
a
?a
a
2,Element,?Element— delegate of a point inside the member,infinitesimal
geometric body which envelops the study point,The element in common use is just
hexahedron ?properties of an element—a,stress is distributed uniformly in an
arbitrary parallel arbitrary plane； b,stresses in the parallel plane opposite plane
are equal,
3,stress element at a point M in
the rod in tension or compression,
1,State of stress at a point,There are countless sections through a point,
Sum of stresses in the different section through a point is called the state of
stress at this point,
Complementary,
? P M
? ? ? ?
63
64
2、单元体,?单元体 —构件内的点的代表物，是包围被研究点的
无限小的几何体，常用的是正六面体。
?单元体的性质 —a、平行面上，应力均布；
b、平行面上，应力相等。
3、拉压杆内一点 M 的应力单元体,
1.一点的应力状态,过一点有无数的截面，这一点的各个截面
上的应力情况，称为这点的应力状态。
补充,
? P M
? ? ? ?
??
?
?
?
?
?
aa??
a??
a
a
c o ss in
c o s
0
2
0
Take a free body as shown in the Fig.3,a is
positive if it is along countclockwise； ? a
is positive if it makes the free body rotate
clockwise,From the equilibrium of the free
body we get,
?
?
?
??
?
?
?
??
a
?
?
a
?
?
a
a
2s i n
2
)2c o s(1
2
:r
0
0
o
4,Stress in the inclined section of the rod in tension or compression
? ? ? ?
a a
x
Fig.3
65
66
?
?
?
?
?
aa??
a??
a
a
c o ss i n
c o s
0
2
0
取分离体如图 3,a 逆时针为正；
? a 绕研究对象顺时针转为正；
由分离体平衡得,
?
?
?
??
?
?
?
??
a
?
?
a
?
?
a
a
2s i n
2
)2c o s(1
2
:
0
0
或
4、拉压杆斜截面上的应力
? ? ? ?
a a
x
图 3
M P a7.632/4.1272/0m a x ??? ??
M P a5.95)60c o s1(
2
4.127)2c o s1(
2
0 ????? a??
a
M P a2.5560s i n
2
4.1272s i n
2
0 ??? a??
a
M P a4.127 1014.3 1 0 0 0 04 20 ????? AP?
Example 6 A rod,which the diameter d =1 cm is subjected to a tensile force P
=10kN,Determine the maximum shearing stress,the normal stress and shearing
stress in the inclined section of an angle 30° about the cross section,
Solution,Stresses in the inclined section of the rod in tension or compression
can be determined directly by the formula,
67
68
M P a7.632/4.1272/0m a x ??? ??
00 1 2 7, 4( 1 c o s 2 ) ( 1 c o s 6 0 ) 9 5, 5 M P a
22a
??a? ? ? ? ?
00 1 2 7, 4s i n 2 s i n 6 0 5 5, 2 M P a
22a
??a? ? ?
M P a4.127 1014.3 1 0 0 0 04 20 ????? AP?
[例 6] 直径为 d =1 cm 杆受拉力 P =10 kN的作用，试求最大剪应
力，并求与横截面夹角 30° 的斜截面上的正应力和剪应力 。
解：拉压杆斜截面上的应力，直接由公式求之,
Example 7 A tensile rod as shown in the figure is made from two parts glued
mutually together along mn,It is subjected to the action of force P,Assume that the
allowable normal stress is [?]=100MPa and allowable shearing stress is [?]=50MPa for the
adhesive,Area of cross section of the rod is A= 4cm2,If strength of the rod is controlled by
the adhesive what is the angle a ( a,between 0 0~60 0) to get the largest tensile force?
kN50,6.26 ??? BB Pa
Combine (1),(2) and get,
P P
m
n a
Solution,
)1( ][c o s 2 ???a? a ??
A
P
)2( ][c o ss i n ???aa? a ??
A
P
69
?
?
P
a 60 0 30 0
B
0 0
70
[例 7]图示拉杆沿 mn由两部分胶合而成,受力 P，设胶合面的许用
拉应力为 [?]=100MPa ；许用剪应力为 [?]=50MPa,并设杆的
强度由胶合面控制,杆的横截面积为 A= 4cm2，试问,为使杆承受
最大拉力,a角值应为多大?(规定, a在 0~60度之间 )。
kN50,6.26 ??? BB Pa
联立 (1),(2)得,
P P
m
n a
解,
)1( ][c o s 2 ???a? a ??
A
P
)2( ][c o ss i n ???aa? a ??
A
P
?
?
P
a 60 0 30 0
B
0 0
? ?0 0 0 260 / ( c o s 6 0 s i n 6 0 ) 4 5 0 1 0 4 / 3 4 6, 2 k NPA ?? ? ? ? ? ?
kN50m a x ?? P
The curves of formula (1)and,(2) are shown in the Tig.(2),Obviously the strength of
the rod on the left of point B is controlled by the normal stress,that on the right of
point B is controlled by the shearing stress,As a=60°, from formula (2) we can get
? ?1 2,6 0 / ( c o s 6 0 s i n 6 0 ) 4 6 0 1 0 4 / 3 5 5, 4 4 k NBPA ?? ? ? ? ? ? ? ?
kN44.55m a x ?? P
Solution,At the point of intersection
of curves (1) and (2),
kN4.54;31 11 ??? BB Pa;M P a60][ m a x ?? P?Discussion,As
71
?
?
P
a 60 0 30 0
B1
0 0
72
? ?0 0 0 260 / ( c o s 6 0 s i n 6 0 ) 4 5 0 1 0 4 / 3 4 6, 2 k NPA ?? ? ? ? ? ?
kN50m a x ?? P
(1),(2)式的曲线如图 (2)，显然,B点左 侧由 正 应力控制杆的强
度,B点右侧由 剪 应力控制杆的强度,当 a=60° 时，由 (2)式得
kN44.55m a x ?? P
解 (1),(2)曲线交点处,
kN4.54;31 11 ??? BB Pa;M P a60][ m a x ?? P?讨论：若
?
?
P
a 60 0 30 0
B1
0 0
? ?1 2,6 0 / ( c o s 6 0 s i n 6 0 ) 4 6 0 1 0 4 / 3 5 5, 4 4 k NBPA ?? ? ? ? ? ? ? ?
1),The whole longitudinal
deformation of the rod,
3),Average stain,
1LLL
LL
? ????
2),Strain,linear deformation per unit length,
1,Deformation and strain of the rod in tension or compression
1L L L? ? ?
§ 1－ 4 DEFORMATION OF THE ROD IN AXIAL TENSION
AND COMPRESSION ? LAW OF ELASTICITY
73
a b
c d
x?
L
Cross section
74
1、杆的纵向总变形,
3、平均线应变,
1LLL
LL
? ????
2、线应变：单位长度的线变形。
一、拉压杆的变形及应变
1L L L? ? ?
§ 1－ 4 拉压杆的变形 ? 弹性定律
a b
c d
x?
L
4,Longitudinal strain at point x,
x
x
x ?
??
??
dlim
0
?
6,Lateral strain at point x,
5,Lateral deformation of the rod,
accaac ?????
ac
ac????
P P
d ′
a′
c′
b′
xx ??? d
L1
75
76
4,x点处的纵向线应变,
x
x
x ?
??
??
dlim
0
?
6,x点处的横向线应变,
5、杆的横向变形,
accaac ?????
ac
ac????
P P
d ′
a′
c′
b′
xx ??? d
L1
2,Elastic law of the rod in tension or compression
PLL
A?? EA
NL
EA
PLL ???
1),Case of equal internal forces
2),Case of variable internal forces
)(
d)()d(
xEA
xxNx ??
( ) d( d )
()LL
N x xLx
E A x? ? ? ???
?
?
??
n
i ii
ii
AE
LNL
1
When internal forces in n
segments are constant
※ ―EA‖is called the axial rigidity of the
rod in tension or compression,
P P
N （ x ）
x
d x
N(x)
dx
x
77
78
二、拉压杆的弹性定律
PLL
A??
P L N LL
E A E A? ? ?
1、等内力拉压杆的弹性定律
2、变内力拉压杆的弹性定律
)(
d)()d(
xEA
xxNx ??
( ) d( d )
()LL
N x xLx
E A x? ? ? ???
1
n
ii
i ii
NLL
EA?
?? ?
内力在 n段中分别为常量时
※,EA” 称为杆的抗拉压刚度。
P P
N （ x ）
x
d x
N(x)
dx
x
3),Elastic law in uniaxial stressed state
4),Possion,s ratio（ or coefficient of the lateral
deformation）
3,Who firstly proposed the Elastic Law
The Elastic Law is the important foundation of solid mechanics
such Mechanics of Materials,Generally it is considered to be
proposed firstly by the English scientist Hook (1635-1703),So the
Elastic Law is also called Hook,s Law,Actually,there was an early
record of the proportional ration between force and deformation,
which is 1500 years earlier than Hook,
??
?
??
? ? ?? ??
or
1)( )(1)d( ?? ExA xNEdx x ???? 1 ??
E
?
That is
79
80
1)( )(1)d( ?? ExA xNEdx x ????
3、单向应力状态下的弹性定律
1,??
E
?即
4、泊松比（或横向变形系数）
??
?
??
,? ? ?? ??或
三、是谁首先提出弹性定律
弹性定律是材料力学等固体力学一个非常重要的基础。一般
认为它是由英国科学家胡克 (1635一 1703)首先提出来的，所以通
常叫做胡克定律。其实，在胡克之前 1500年，我国早就有了关于
力和变形成正比关系的记载。
― ‖ 胡：请问,弛其弦，以绳缓 缓 之
是什么意思?
郑：这是讲测量弓力时，先将弓的弦
松开，另外用绳子松松地套住弓
的两端，然后加重物，测量。
胡：我明白了。这样弓体就没有初始应力，处于自 然状态。
东汉经学家郑玄 (127—200)对, 考工记 ·弓人, 中, 量其力，
有三均, 作了 这样的注释：, 假令弓力胜三石，引之中三尺，弛
其弦，以绳缓擐之，每加物一石，则张一尺。, (图 )
82
― ‖ 胡：请问,弛其弦，以绳缓 缓 之
是什么意思?
郑：这是讲测量弓力时，先将弓的弦
松开，另外用绳子松松地套住弓
的两端，然后加重物，测量。
东汉经学家郑玄 (127—200)对, 考工记 ·弓人, 中, 量其力，
有三均, 作了 这样的注释：, 假令弓力胜三石，引之中三尺，弛
其弦，以绳缓擐之，每加物一石，则张一尺。, (图 )
胡：我明白了。这样弓体就没有初始应力，处于自 然状态。
郑：后来，到了唐代初期，贾公彦对我的注释又作 了注疏，他说,
郑又云假令弓力胜三石，引之 中三尺者，此即三石力弓也。
必知弓力三石者，当弛其弦以绳缓擐之者，谓不张之，别以
绳系两箭，乃加物一石张一尺、二石张二尺、三石张三
尺。 其中 ‖ ― 两萧 就是指弓的两端。
一条
―
胡：郑老先生讲, 每加物一石，则张一尺”。和我讲的完全是同一
个意思。您比我早 1500 中就记录下这种正比关系，的确了不起,
和推测, 一文中早就推崇过贵国的古代文化,目前我们还只
是刚刚走到这个知识领域的边缘，然而一旦对它有了充分的认
识，就将会在我们面 前展现出一个迄今为止只被人们神话般
地加以描述的知识王国”。
1686 年, 关于中国文字和语言的研究 真是令人佩服之至 』 我在
83
84
郑：后来，到了唐代初期，贾公彦对我的注释又作 了注疏，他说,
郑又云假令弓力胜三石，引之 中三尺者，此即三石力弓也。
必知弓力三石者，当弛其弦以绳缓擐之者，谓不张之，别以
绳系两箭，乃加物一石张一尺、二石张二尺、三石张三
尺。 其中 ‖ ― 两萧 就是指弓的两端。
一条
―
胡：郑老先生讲, 每加物一石，则张一尺”。和我讲的完全是同一
个意思。您比我早 1500 中就记录下这种正比关系，的确了不起,
和推测, 一文中早就推崇过贵国的古代文化,目前我们还只
是刚刚走到这个知识领域的边缘，然而一旦对它有了充分的认
识，就将会在我们面 前展现出一个迄今为止只被人们神话般
地加以描述的知识王国”。
1686 年, 关于中国文字和语言的研究 真是令人佩服之至 』 我在
C'
1,How to plot the enlargement sketch of the small deformation
?Accurate method to plot diagram
of deformation,the arc line as
shown in the figure；
?Determine deformation △ Li of
each rod as shown in Fig l,
?Approximate method to plot the
diagram of deformation; the tangent
of the arc line shown in the figure,
Example 8 Enlargement sketch of the small deformation the and method to
determine displacements
A B
C
L1 L
2
P 1L?2L?
C"
85
86
C'
1、怎样画小变形放大图？
?变形图严格画法，图中弧线；
?求各杆的变形量△ Li,如图 1；
?变形图近似画法，图中弧之切线。
[例 8] 小变形放大图与位移的求法。
A B
C
L1 L
2
P 1L?2L?
C"
2,Write the relation between the displacement of point B shown in Fig.2 and
deformations of two rods,
1Lu B ??
Solution,The diagram of deformation is shown in the Fig.2,Point B moves
to point B',From the diagram of displacement we may know,
aa s i nc t g
2
1
LLv
B
????
87
A B
C
L1
L2
a
1L?
2L?
Bu
Bv
B'
P
Fig, 2
88
2、写出图 2中 B点位移与两杆变形间的关系
1Lu B ??
解：变形图如图 2,B点位移至 B'点，由图知,
aa s i nc t g
2
1
LLv
B
????
A B
C
L1
L2
a
1L?
2L?
Bu
Bv
B'
P
图 2
? ????? 060s i n6.12.18.060s i n ooA TPTm
kN55.113/ ??? PT
M P a1511036.76 55.11 9 ???? AT?
Example 9 Suppose the crossbeam ABCD is rigid,A steel cable with the
cross-section area
76.36mm2 is around a pulley without friction,Knowing P=20kN,E =177GPa,
Determine the stress of the steel cable and the upright displacement of point C,
Solution,method 1,Enlargement sketch
method of the small deformation,
1)Determine the internal force of the steel
cable,Take ABD as our study object,
2) Stress and elongation of the steel cable
P
A B
C
D T T
YA
XA
89
800 400 400
D
C P
A B 60° 60°
90
? ????? 060s i n6.12.18.060s i n ooA TPTm
kN55.113/ ??? PT
M P a1511036.76 55.11 9 ???? AT?
[例 9] 设横梁 ABCD为刚梁，横截面面积为 76.36mm2 的钢索绕
过无摩擦的定滑轮。设 P=20kN，试求钢索内的应力和 C点的
垂直位移。设钢索的 E =177GPa。
解：方法 1：小变形放大图法
1）求钢索内力：以 ABCD为研究对象
2) 钢索的应力和伸长分别为,
P
A B
C
D T T
YA
XA
800 400 400
D
C P
A B 60° 60°
mm36.1m17736.76 6.155.11 ?????? EATLL
C P
A B 60° 60°
800 400 400
D
A B 60° 60° D
B' D'
1?
2?
C?
C 3） Deformation is shown in the figure,
Upright displacement of point C is,
0 0
12
2
s i n 6 0 s i n 6 0
2
C
B B D D
L
???
??
? ? ?
?
00
1, 3 6
2 s i n 6 0 2 s i n 6 0
0, 7 9 m m
L?
??
?
91
92
mm36.1m17736.76 6.155.11 ?????? EATLL
C P
A B 60° 60°
800 400 400
D
A B 60° 60° D
B' D'
1?
2?
C?
C 3）变形图如左图,
C点的垂直位移为,
0 0
12
2
s i n 6 0 s i n 6 0
2
C
B B D D
L
???
??
? ? ?
?
00
1, 3 6
2 s i n 6 0 2 s i n 6 0
0, 7 9 m m
L?
??
?
§ 1－ 5 ELASTIC STRAIN ENERGY OF THE ROD IN AXIAL TENSION
AND COMPRESSION
1,Elastic strain energy,The work done by the external forces
results in the increase of some energy associated with the
deformation of the rod,This energy is referred to as the strain
energy of the rod,It is written as U,
2,Calculation of the strain energy of the rod in tension and compression,
Neglecting the lost energy,the work done by external forces is equal to the
strain energy,
) d)(d ( xEA xNx ???
1d d ( ) ( d )
2U W N x x? ? ? ?
x
EA
xNU d
2
)(d 2?
?? L xEA xNU d2 )(
2 ?
?
?
n
i ii
ii
AE
LN
U
1
2
2
Internal force is constant
in a small segment
N （ x ）
x
d x
N(x)
dx
x
93
94
§ 1－ 5 拉压杆的弹性应变能
一, 弹性应变能,杆件发生弹性变形，外力功转变为变形能贮存
于杆内，这种能成为应变能（ Strain Energy）用,U” 表示。
二,拉压杆的应变能计算,
不计能量损耗时，外力功等于应变能 。
) d)(d ( xEA xNx ???
1d d ( ) ( d )
2U W N x x? ? ? ?
x
EA
xNU d
2
)(d 2?
?? L xEA xNU d2 )(
2
?
?
?
n
i ii
ii
AE
LN
U
1
2
2
内力为分
段常 量 时
N （ x ）
x
d x
N(x)
dx
x
3,Strain－ energy density u of the rod
in tension and compression Strain
energy per unit volume,
d 1 ( ) ( d ) 1
d 2 d 2
U N x xu
V A x
???? ? ?
N （ x ）
x
d x
N(x)
dx
x
dx
xx dd ??
N(x) N(x)
(d )x?
)(xN
95
96
三,拉压杆的比能 u,
单位体积内的应变能 。
d 1 ( ) ( d ) 1
d 2 d 2
U N x xu
V A x
???? ? ?
N （ x ）
x
d x
N(x)
dx
x
dx
xx dd ??
N(x) N(x)
(d )x?
)(xN
kN55.113/ ??? PT
Solution,Method 2,Energy method,
（ Work of external forces is equal to the
strain energy）
（ 1） Determine the internal force of the steel
cable,Take ABCD as our study object,
? ????? 060s i n6.12.18.060s i n ooA TPTm
Example 9 Suppose the crossbeam ABCD is rigid,A steel cable with the cross-
section area
76.36mm2 is around a pulley without friction,Knowing P=20kN,E =177GPa,
Determine the stress of the steel cable and the upright displacement of point C,
97
800 400 400
C P
A B 60° 60°
P
A B
C
D T T
YA
XA
98
kN55.113/ ??? PT
解：方法 2,能量法,
（外力功等于变形能）
（ 1）求钢索内力：以 ABCD为研究对象,
? ????? 060s i n6.12.18.060s i n ooA TPTm
[例 9] 设横梁 ABCD为刚梁，横截面面积为 76.36mm2 的钢索绕
过无摩擦的定滑轮。设 P=20kN，试求钢索内的应力和 C点的
垂直位移。设钢索的 E =177GPa。
800 400 400
C P
A B 60° 60°
P
A B
C
D T T
YA
XA
EA
LTP C
22
2
?? 2
2
1 1, 5 5 1, 6
2 0 1 7 7 7 6, 3 6
0,7 9 m m
C
TL
So
PEA
??
?
?
??
?
M P a1511036.76 55.11 9 ???? AT?
（ 2) Stress of the steel cable is,
（ 3) Displacement of point C is,
Energy method,The method by which the
problems relative to the elastic deformation of the
structure members are soloed according to the
concept of strain energy is called the energy method,
99
800 400 400
C P
A B 60° 60°
100
EA
LTP C
22
2
??
mm79.0
36.761 7 720
6.155.11
2
2
?
??
?
?
???
PEA
LT
C
M P a1511036.76 55.11 9 ???? AT?
（ 2) 钢索的应力为,
（ 3) C点位移为,
能量法,利用应变能的概念解决与结构物
或构件的弹性变形有关的问题，这种方法
称为能量法。
800 400 400
C P
A B 60° 60°
§ 1－ 6 STATICALLY INDETERMINATE PROBLEMS AND THEIR
TREATMENT METHODS OF AXIAL TENSION AND COMPRESSION
1),Statically indeterminate problems,The problems in
which unknown forces(external forces,internal forces,stresses) cannot
be wholly determined only by static equilibrium equations
1,Statically indeterminate problems and their treatment methods
2,Method to solve statically indeterminate problems,
Equilibrium equations,compatibility equation of
deformation and physical equations must be combined
together to solve to solve these problems,
101
102
§ 1－ 6 拉压超静定问题及其处理方法
1、超静定问题,单凭静力平衡方程不能确定出全部未知力
（外力、内力、应力）的问题。
一、超静定问题及其处理方法
2、超静定问题的处理方法,平衡方程、变形协调方程、
物理方程相结合，进行求解。
Example 10 Rods 1,2 and 3 are connected together with a pin as shown
in the figure,Knowing the length of each rod is,L1=L2,L3 =L； and the area
of each rod is A1=A2=A,A3； modulus of elasticity of each rod is,E1=E2=E、
E3,External force is along the upright direction,Determine the internal force of
each rod,
C
P
A
B D
aa
1 2
3
Solution,?Equilibrium equations,
? ??? 0s i ns i n 21 aa NNX
? ????? 0c o sc o s 321 PNNNY aa
P
A
aaN1
N3
N2
103
104
[例 10] 设 1,2,3三杆用铰链连接如图，已知：各杆长为：
L1=L2,L3 =L ；各杆面积为 A1=A2=A,A3 ；各杆弹性模量
为,E1=E2=E,E3。外力沿铅垂方向，求各杆的内力。
C
P
A
B D
aa
1 2
3
解,?、平衡方程,
? ??? 0s i ns i n 21 aa NNX
? ????? 0c o sc o s 321 PNNNY aa
P
A
aaN1
N3
N2
11
11
1 AE
LNL ??
33
33
3 AE
LNL ??
?Geometric equation—compatibility
equation of deformation,
?Physical equation—elastic law,
?Complementary equation,detaining from the
geometric equation and physical equation,
?Solving the equilibrium equations and complementary equation we get,
ac o s31 LL ???
ac o s
33
33
11
11
AE
LN
AE
LN ?
33
3
11
33
3
33
3
11
2
11
21 c o s2 ; c o s2
c o s
AEAE
PAEN
AEAE
PAENN
?
?
?
??
aa
a
C
A
B D
aa
1 2
3
A1
1L?
2L?
3L?
105
106
11
11
1 AE
LNL ??
33
33
3 AE
LNL ??
?几何方程 ——变形协调方程,
?物理方程 ——弹性定律,
?补充方程：由几何方程和物理方程得。
?解由平衡方程和补充方程组成的方程组，得,
ac o s31 LL ???
ac o s
33
33
11
11
AE
LN
AE
LN ?
33
3
11
33
3
33
3
11
2
11
21 c o s2 ; c o s2
c o s
AEAE
PAEN
AEAE
PAENN
?
?
?
??
aa
a
C
A
B D
aa
1 2
3
A1
1L?
2L?
3L?
?Equilibrium equations；
?Geometric equation—compatibility equation of
deformation；
?Physical equation—elastic law；
?Complementary equation,from geometric equation and
physical equation
?Solving equilibrium equations and the complementary
equation,
3,Method and steps for solving statically indeterminate problems,
107
108
?平衡方程；
?几何方程 ——变形协调方程；
?物理方程 ——弹性定律；
?补充方程：由几何方程和物理方程得；
?解由平衡方程和补充方程组成的方程组 。
3、超静定问题的处理方法步骤,
Example 11 Four angles of a wooden pole are reinforced with four equal leg
angle steel of 40?40?4,The allowable stresses of steel and wood are respectively
[?]1=160MPa and [?]2=12MPa,module of elasticity of them are E1=200GPa and
E2=10GPa,Respectively Determine the allowable permissible load P,
? ???? 04 21 PNNY
21 LL ???
2
22
22
11
11
1 LAE
LN
AE
LNL ?????
?Geometric equation
?Physical equation and
complementary equation,
Solution,?Equilibrium equations,
P P y
4N1
N2
109
110
[例 11] 木制短柱的四角用四个 40?40?4的等边角钢加固，角钢
和木材的许用应力分别为 [?]1=160M Pa和 [?]2=12MPa,弹性模
量分别为 E1=200GPa 和 E2 =10GPa； 求许可载荷 P。
? ???? 04 21 PNNY
21 LL ???
2
22
22
11
11
1 LAE
LN
AE
LNL ?????
?几何方程
?物理方程及 补充方程,
解,?平衡方程, P P
y
4N1
N2
P P
y
4N1
N2
? Solving equilibrium equations and
the complementary equation we get,
PNPN 72.0 ; 07.0 21 ??
? ?111 07.0 ?APN ??
?Determine the permissible load of
the structure,
Method 1,
Section area of the angle steel may be obtained
from the table of hot-rolled steel,A1=3.086cm2
? ?2 2 20, 7 2N P A ???
? ? ? ? 222 2 / 0, 7 2 2 5 0 1 2 / 0, 7 2 1 0 4 2 k Na n d P A ?? ? ? ?
? ? ? ?11 1 / 0, 0 7 3 0 8, 6 1 6 0 / 0, 0 7 7 0 5, 4 k NS o P A ?? ? ? ?
111
112
P P
y
4N1
N2
? 解平衡方程和补充方程，得,
PNPN 72.0 ; 07.0 21 ??
? ?1 1 10, 0 7N P A ???
?求结构的许可载荷,
方法 1,
角钢截面面积由型钢表查得,
A1=3.086cm2
? ?2 2 20, 7 2N P A ???
? ? ? ? 222 2 / 0, 7 2 2 5 0 1 2 / 0, 7 2 1 0 4 2 k NPA ?? ? ? ? ?
? ? ? ?11 1 / 0, 0 7 3 0 8, 6 1 6 0 / 0, 0 7 7 0 5, 4 k NPA ?? ? ? ? ?
? ? ? ?11 1 / 0, 8 m mLE?? ? ?
? ? ? ?22 2 / 1, 2 m mLE?? ? ?
In the case of △ 1=△ 2,the angle steel will reach the limit state
first,that is the maximum load is determined by the angle steel,
?Determine the permissible load of the structure,
? ? ? ?
07.0
07.0
111 ANP ??? kN4.7 0 5
07.0
6.3 0 81 6 0 ???
The maximum load of the structure is still controlled by the steel,
Method 2,
Additionally,What about would it the like if area of the angle
steel is increased as 5 times？
What about would it like if the area of the wood is
changed into 25mm2?
113
114
? ? ? ?11 1 / 0, 8 m mLE?? ? ?
? ? ? ?22 2 / 1, 2 m mLE?? ? ?
所以在 △ 1=△ 2 的前提下，角钢将先达到极限状态,
即角钢决定最大载荷。
?求结构的许可载荷,
? ? ? ?
07.0
07.0
111 ANP ??? kN4.7 0 5
07.0
6.3 0 81 6 0 ???
另外：若将钢的面积增大 5倍，怎样？
若将木的面积变为 25mm2,又 怎样？
结构的最大载荷永远由钢控制着 。
方法 2,
?Geometric equation
Solution,?Equilibrium equations,
? ??? 0s i ns i n 21 aa NNX
? ???? 0c o sc o s 321 NNNY aa
13 c o s)( LL ???? a?
2,Assemble stresses—initial stresses
1),There is no the assemble stress in
statically determinate structure,
2),There is the assemble
stress in statically indeterminate structure,
The dimension error of rod 3 is ? as shown in the
figure,Determine the assemble internal force of each
rod,
A
B C
1 2
A
B C
1 2
D
A1
3
a a
?
115
116
?、几何方程
解,?、平衡方程,
2,静不定结构存在装配应力 。
? ??? 0s i ns i n 21 aa NNX
? ???? 0c o sc o s 321 NNNY aa
13 c o s)( LL ???? a?
二、装配应力 ——预应力
1、静定结构无装配应力。
如图,3号杆的尺寸误差为 ?，求各杆
的装配内力 。 A
B C
1 2
A
B C
1 2
D
A1
3
a a
?
a? c o s)(
33
33
11
11
AE
LN
AE
LN
??
?Physical equation and complementary equation,
? Solving equilibrium equations and
the complementary equation we get,
/ c o s21
c o s
3311
3
2
11
3
21 AEAE
AE
L
NN
a
a?
?
???
/ c o s21
c o s2
3311
3
3
11
3
3 AEAE
AE
L
N
a
a?
?
??
?
A1
a aN1 N2
N3
A
A1 3L?
2L?1L?
117
118
a? c o s)(
33
33
11
11
AE
LN
AE
LN
??
?、物理方程及 补充方程,
?,解平衡方程和补充方程，得,
/ c o s21
c o s
3311
3
2
11
3
21 AEAE
AE
L
NN
a
a?
?
???
/ c o s21
c o s2
3311
3
3
11
3
3 AEAE
AE
L
N
a
a?
?
??
?
A1
a aN1 N2
N3
A
A1 3L?
2L?1L?
1),There is no the temperature stress in
statically determinate structure,
2),There is the temperature
stress in statically indeterminate structure,
3,Temperature stress
Materials and dimensions of rod 1and2 are all
the same as shown in the figure,If temperature of
the structure changes from T1 to T2,determine the
temperature internal forces of each rod.（ linear
thermal expansion coefficient of each rod ai ; △ T=
T2 -T1)
A
B C
1 2
C
A
B D
??
1 2
3
A1
1L?
2L?
3L?
119
120
1、静定结构无温度应力。
三,应力温度
如图,1,2号杆的尺寸及材料都相
同，当结构温度由 T1变到 T2时,求各杆
的温度内力。（各杆的线膨胀系数分别
为 ai ; △ T= T2 -T1)
A
B C
1 2
C
A
B D
??
1 2
3
A1
1L?
2L?
3L?
2、静不定结构存在温度应力。
C
A
B D
??
1 2
3
A1
1L?
2L?
3L?
?,Geometric equation
Solution,?,Equilibrium equations,
12s i n s i n 0X N N??? ? ??
? ???? 0c o sc o s 321 NNNY ??
?c o s31 LL ???
ii
ii
ii
i LTAE
LNL a????
?,Physical equation,
A
??N1
N3
N2
121
122
C
A
B D
??
1 2
3
A1
1L?
2L?
3L?
?、几何方程
解,?、平衡方程,
?c o s31 LL ???
ii
ii
ii
i LTAE
LNL a????
?、物理方程,
A
??N1
N3
N2
12s i n s i n 0X N N??? ? ??
? ???? 0c o sc o s 321 NNNY ??
C
A
B D
??
1 2
3
A1
1L?
2L?
3L?
?,Complementary equation,
?aa c o s)( 33
33
33
11
11
11 LT
AE
LNLT
AE
LN ?????
Solving equilibrium equations and the
complementary equation we get,
/ c o s21
)c o s(
3311
3
2
3111
21 AEAE
TAE
NN
?
?aa
?
??
???
/ c o s21
c o s)c o s(2
3311
3
2
3111
3 AEAE
TAE
N
?
??aa
?
??
?
123
124
C
A
B D
??
1 2
3
A1
1L?
2L?
3L?
?,补充方程
?aa c o s)( 33
33
33
11
11
11 LT
AE
LNLT
AE
LN ?????
解平衡方程和补充方程，得,
/ c o s21
)c o s(
3311
3
2
3111
21 AEAE
TAE
NN
?
?aa
?
??
???
/ c o s21
c o s)c o s(2
3311
3
2
3111
3 AEAE
TAE
N
?
??aa
?
??
?
a
a
a
a
N1
N2
Example 12 Theupper and lower ends of a ladder-like steel
shaft are fixed at temperature T1=5℃ as shown in the figure,
Areas of the upper and lower segments are respectively ?1=?cm2
and ?2=??cm2,When its temperature reaches T2=25℃,determine
the temperature stress of each rod.(Linear thermal expansion
coefficient a? ?2.5× 10 -6 1/0C； modulus of elasticity
E=200GPa)
?,Geometric equation,
Solution,?,Equilibrium equation,
? ??? 021 NNY
0?????? NT LLL 125
126
a
a
a
a
N1
N2
[例 12] 如图，阶梯钢杆的上下两端在 T1=5℃
时被固定,杆的上下两段的面积分别
??=?cm2, ??=??cm2，当温度升至 T2
=25℃ 时,求各杆的温度应力。
(线膨胀系数 a =12.5× ；
弹性模量 E=200GPa)
C?110 6?
?、几何方程,
解,?、平衡方程,
? ??? 021 NNY
0?????? NT LLL
?,Physical equation,
Solving equilibrium equations and
the complementary equation we get,k N 3.33
21 ?? NN
?,Complementary equation,
2
2
1
1 ; 2
EA
aN
EA
aNLTaL
NT ?????? a
2
2
1
12
EA
N
EA
N
T ??? a
?,Temperature stresses,
M P a 7.66
1
1
1 ?? A
N? M P a 3.33
2
2
2 ?? A
N?
127
128
?、物理方程
解平衡方程和补充方程，得,
k N 3.3321 ?? NN
?,补充方程
2
2
1
1 ; 2
EA
aN
EA
aNLTaL
NT ?????? a
2
2
1
12
EA
N
EA
N
T ??? a
?、温度应力
M P a 7.66
1
1
1 ?? A
N? M P a 3.33
2
2
2 ?? A
N?
§ 1－ 7 MECHANICAL PROPERTIES OF MATERIALS IN AXIAL TENSION
AND COMPRESSION
1,Testing conditions and instruments
d
h
Mechanical properties,Some properties about strength and deformation shown by
materials under the action of external loads,
1),Testing conditions,normal temperature(20℃ )；
static load（ loaded gradually)
Standard specimen
129
130
§ 1－ 7 材料在拉伸和压缩时的力学性能
一、试验条件及试验仪器
1、试验条件：常温 (20℃) ；静载（极其缓慢地加载）；
标准试件。
d
h
力学性能：材料在外力作用下表现的有关强度、变形方面的特性。
2,Experimental implement,Universal testing machine ； Extensometer
131
132
2、试验仪器：万能材料试验机；变形仪（常用引伸仪）。
meter-
pedestal
plate
centesi
mal
meter
meter
pedestal
bolt for
installing
the meter
standard
specimen
spring
EEA
P
L
L ?? ????
2,Tensile diagram of the low carbon steel specimen(P--?L diagram)
3,Stress-strain curve of the low carbon steel specimen (? --? diagram)
EA
PLL ??
133
Typical points in the curve of the
low-carbon steel
???
134
EEA
P
L
L ?? ????
二、低碳钢试件的拉伸图 (P-- ?L图 )
三、低碳钢试件的应力 --应变曲线 (? --? 图 )
EA
PLL ??
(1) Elastic range of the low-carbon steel in tension (oe)
1,op – Proportional segment,
?p – Proportional limit
E
?? ?
atg?E
2,pe – Curved segment,
?e – Elastic limit
)( nf ?? ?
135
Stress-strain relation ship
in the elastic region
136
(一 ) 低碳钢拉伸的弹性阶段 (oe段 )
1,op -- 比例段,
?p -- 比例极限
E
?? ?
atg?E
2,pe --曲线段,
?e -- 弹性极限
)( nf ?? ?
(2) Yielding(flowing) range of the low carbon steel in
tension(es)
e s –Yielding range,?s ---Yielding limit
Sliding lines,
Failure stress of plastic material,?s 。
137
Single-crystal Cu-A1 specimen after
tension
138
(二 ) 低碳钢拉伸的屈服 (流动）阶段 (es 段 )
e s --屈服 段, ?s ---屈服极限
滑移线,
塑性材料的失效应力,?s 。
２,Unloaded law,
１,?ｂ --Strength limit
３,Cold hardening,
４,hard-drawn
time effect,
(3),Hardening range of the low carbon steel(ｓｂ )
139
Typical points in the curve of the low-carbon
steel ???
140
２、卸载定律,
１,?ｂ ---强度 极限
３、冷作硬化,
４、冷拉时效,
(三 )、低碳钢拉伸的强化阶段 (ｓｂ 段 )
1,Percent elongation
?
10 0
0
0
100LL
L
? ???
2,Percent reduction
in area ?
01 0
0
0
100AA
A
? ???
3,Brittleness,ductility、
relativity
0 05? ? the division line
(4),Necking(crack) stage (b f ) of the low carbon steel in tension
141
Typical points in the curve of the low-carbon
steel ???
eulargeneut
142
1、延伸率,? 2、面缩率,? 3、脆性、塑性及相对性
为界以 005??
(四 )、低碳钢拉伸的颈缩（断裂）阶段 (b f 段 )
10 0
0
0
100LL
L
? ??? 01 0 0
0
100AA
A
? ???
? ?
?
4,Plastic material without obvious yield
?.?
? 0.2
Nominal yield stress,
? 0.2 is the failure stress of this type of
material,
5,Mechanical properties of the cast
iron in tension
?ｂ L --- Tensile strength limit of cast iron(fail
stress）; tg a?E
?
?
bL?
143
0
0
144
? ?
?
四、无明显屈服现象的塑性材料
?.?
? 0.2
名义屈服应力,
? 0.2,即此类材料的失效应力。
五、铸铁拉伸时的机械性能
?ｂ L ---铸铁拉伸强度 极限（失效应力）
割线斜率 ; tg a?E
?
?
bL?
0
0
6,Mechanical properties of
the material in compression
?ｂ y – Compressive strength
limit of cast iron；
?ｂ y ?（ 4 ～ 6） ?ｂ L
145
compression
tension
comparison of tension and compression for the low-
carbon steel
Compression of the cast iron
146
六、材料压缩时的机械性能
?ｂ y ---铸铁压缩强度 极限；
?ｂ y ?（ 4 ～ 6） ?ｂ L
7,Safety factor,permissible stress,limit stress
? ?
n
jx?? ?
? ?bs
jx
????,,2.0?
N>1
1),Permissible stress,
2),Limit stress,
3),Safety factor,
147
148
七、安全系数、许用应力、极限应力
? ?
n
jx?? ?
? ?bs
jx
????,,2.0?
N>1
1,许用应力,
2、极限应力,
3、安全系数,
006500/30 ???
N5024/160214.3 2 ?????? ?AP
Solution,Deformation may exceed the range of
linear elasticity,therefore the elastic law is not
applied here,Calculation should be done in the
following,
M P a1 6 0??
Example 13 Diameter of a copper wire is d=2mm and its length is L=500mm,
Tensile curve of copper is shown in the figure,To make elongation of the copper wire
is 30mm whet is the force P that we must act？
0 5 1 0 1 5 2 0 （ ?? ）
1
0
0
2
0
0
3
0
0
? （ M Pa ）
From the tensile curve,
? (MPa)
? (%)
149
150
006500/30 ???
N5024/160214.3 2 ?????? ?AP
解：变形量可能已超出了“线弹性”
范围，故，不可 再 应用 ―弹性 定律,
。 应如下计算,
M P a1 6 0??
[例 13] 铜丝直径 d=2mm，长 L=500mm,材料的 拉伸 曲线如 图
所示。如欲使铜丝的伸长量为 30mm,则大约需加多大的力 P？
0 5 1 0 1 5 2 0 （ ?? ）
1
0
0
2
0
0
3
0
0
? （ M Pa ）
由拉伸图知,
? (MPa)
? (%)
151
1,Knowing the elastic modulus of steel is E＝
200GPa,the elastic modulus of Aluminum is E＝
71GPa.Try to compare:① which material produces a
larger strain when they are subjected to the same stress？
② which material corresponds to a large stress when
they have the same strain？
Chapter 1 Exercises
152
一、钢的弹性模量 E＝ 200GPa，铝的弹性模
量 E＝ 71GPa。试比较在同一应力作用下，那种
材料的应变大？在产生同一应变的情况下，那
种材料的应力大？
第一章 练习题
153
2,For the different members made of a same material,do
they have the same allowable stress? In general cases,the safety
coefficient of the brittle material is selected to be larger than that of
the plastic material,why?
3,As shown in the figure,the Al-alloy circular bar is
subjected to the axial tensile force P,Knowing E＝ 73GPa,and
μ=1/3,When the elongation of the bar is L=7mm,try to determine:
① the change in its diameter Δd； ② the axial force P,
154
二、由同一材料制成的不同构件，其许用
应力是否相同?一般情况下脆性材料的安全系数
要比塑性材料的安全系数选得大些，为什么？
三、图示铝合金圆杆受轴向拉力 P。已知材
料的弹性模量 E＝ 73GPa，泊松比 μ ＝ 。试求
当杆伸长量 ＝ 7mm时，①直径的减少量 ；
② P力的大小。
3
1
d?
L?
155
Solution,
KN
L
AEL
N
mm
L
Ld
d
4.120
43
30737
0 2 3 3.0
3 0 0 03
7301
2
?
?
???
?
???
?
?
?
??
?
???
??
?
?
②
①
4,A bracket is shown in the figure,The cross-section area of the steel bar
AB is A1=6cm2； the cross-section area of the wooden bar BC is A2=300cm2，
Knowing the allowable stress of steel is,the allowable tensile
stress of wood is,and the allowable compressive stress of wood
is, Try determine the allowable load P of the bracket,
? ? M P a1 4 0??
? ? M P aL 8??
? ? M P aY 4??
156
解,
KN
L
AEL
N
mm
L
Ld
d
4.120
43
30737
0 2 3 3.0
3 0 0 03
7301
2
?
?
???
?
???
?
?
?
??
?
???
??
?
?
②
①
? ? M P aL 8??
四、图示支架,AB为钢杆，横截面面积 ；
BC为木杆，横截面面积 。钢的许用应力
,木材的许用拉应力,许用压应
力 。试求支架的许可载荷 。
21 6A cm?
22 300A cm?
? ? M P a1 4 0??
? ? M P aY 4?? ??P
157
Solution,From the equilibration condition we get,
The allowable axial force of bar AB and bar BC is respectively,
The bar AB should satisfy the strength condition,
The bar BC should satisfy the strength condition,
As the strength of the two bars should both be satisfied,the allowable
load of the bracket should be taken as,
[P] =101 KN
21 61.2
2.2,
4.1
2.2 NPNP ??
? ? ? ? KNAN 84101 4 06 211 ????? ?
? ? ? ? KNAN Y 1 2 01043 0 0 222 ????? ?
? ? ? ? KNNP 1 3 24.1 2.2 1 ??
? ? ? ? KNNP 1 0 161.2 2.2 2 ??
158
解：由平衡条件求得,
AB杆与 BC杆的许可轴力分别为,
AB杆要满足强度条件,
BC杆要满足强度条件,
两杆的强度都应得到满足，故支架的许可载荷
应取
[P] =101 KN
21 61.2
2.2,
4.1
2.2 NPNP ??
? ? ? ? KNAN 84101 4 06 211 ????? ?
? ? ? ? KNAN Y 1 2 01043 0 0 222 ????? ?
? ? ? ? KNNP 1 3 24.1 2.2 1 ??
? ? ? ? KNNP 1 0 161.2 2.2 2 ??
151
160
