Mechanics of Materials
1
2
CHAPTER 8 STHENGTH THEORIES
§ 8–1 Concepts of strength theories
§ 8–2 Four strength theories and their equivalent stresses
§ 8–3 Mohr’s strength theory and its equivalent stress
§ 8-4 Applications of strength theories
3
第八章 强度理论
§ 8–1 强度理论的概念
§ 8–2 四个强度理论及其相当应力
§ 8–3 莫尔强度理论及其相当应力
§ 8-4 强度理论的应用
4
1,Opening words,
§ 8–１ CONCEPTS OF STRENGTH THEORIES
1),Investigation of the tensile,compressive and torsional tests of cast iron
and low-carbon steel
M
Low-
carbon steel
Cast iron
P
P Tension of
cast iron P
Compression of
cast iron
2),How will the member rupture in
Combined deformations？ M
P
5
一、引子,
§ 8–１ 强度理论的概念
1、铸铁与低碳钢的拉、压、扭试验现象是怎样产生的？
M
低碳钢
铸铁 P
P 铸铁拉伸
P
铸铁压缩
2、组合变形杆将怎样破坏？
M
P
6
2,Theories of strength,Some assumptions about the cause of the
strength failure of materials,
3,Types of failure of materials：⑴ yield； ⑵ rupture
1),Galileo planted the seed of the first strength theory；
2),Mariotto’s statement about the rupture of material due to
excessive deformation is the flush of the second strength theory；
3),C.Duguet proposed the theory of the maximum shearing stress；
4),Maxwell brought forward the theory of the maximum
distortional energy theory early,which was known after his letters
were published,
7
二、强度理论：是关于, 构件发生强度失效（ failure by lost
strength）起因, 的假说。
1、伽利略播下了第一强度理论的种子；
三、材料的破坏形式：⑴ 屈服； ⑵ 断裂 。
2、马里奥特关于变形过大引起破坏的论述，是第二强度理论的
萌芽；
3、杜奎特（ C.Duguet)提出了最大剪应力理论；
4、麦克斯威尔最早提出了最大畸变能理论（ maximum distortion
energy theory）；这是后来人们在他的书信出版后才知道的。
8
§ 8–2 FOUR STRENGTH THEORIES AND
THEIR EQUIVALENT STRESSES
1,Theory of the maximum tensile stress（ the first strength）,
This theory considers the main cause of rupture to be the maximum tensile
stress,The member will rupture as the maximum tensile stress reaches the strength
limit in axial tension,
1),Criterion of rupture,
0)(; 11 ?? ??? b
2),strength condition,? ? 0)( ;
11 ?? ???
3),Application range,May be applied to the members with the
failure form of brittle rupture,
9
§ 8–2 四个强度理论及其相当应力
一、最大拉应力（第一强度）理论,
认为构件的断裂是由最大拉应力引起的。当最大拉应力达到
单向拉伸的强度极限时，构件就断了。
1、破坏判据,0)(;
11 ?? ??? b
2、强度准则,? ? 0)( ;
11 ?? ???
3、实用范围：实用于破坏形式为脆断的构件。
10
2,Theory of the maximum tensile strain（ the second strength）,
This theory considers the main cause of rupture to be the maximum tensile strain,The
member will rupture as the maximum tensile strain reaches the limit strain in axial
tension
0)(; 11 ?? ??? b ? ?? ? EE b?????? ???? 3211 1
1),Criterion of failure,
2),Strength condition,
3),Application range,May be applied to the members with the
failure form of brittle rupture,
? ? b????? ??? 321
? ? ? ?????? ??? 321
11
二,最大伸长线应变（第二强度）理论,
认为构件的断裂是由最大伸长线应变引起的。当最大伸长线
应变达到单向拉伸试验下的极限应变时，构件就断了。
1、破坏判据,
0)(; 11 ?? ??? b
2、强度准则,
3、实用范围：实用于破坏形式为脆断的构件。
? ?? ? EE b?????? ???? 3211 1
? ? b????? ??? 321
? ? ? ?????? ??? 321
12
3,Theory of the maximum shearing stress（ the third strength）,
This theory considers the main cause of rupture to be the maximum shearing stress,
The member will rupture as the maximum shearing stress reaches the limit shearing
stress in axial tension,
s?? ?m ax ss ????? ???? 22 31m a x
1),Criterion of failure,
3),Application range,May be applied to the members with the
failure form of plastic yield,
s??? ?? 31
2),Strength condition,? ?
??? ?? 31
13
三、最大剪应力（第三强度）理论,
认为构件的屈服是由最大剪应力引起的。当最大剪应力达
到单向拉伸试验的极限剪应力时，构件就破坏了。
1、破坏判据,
s?? ?m ax
3、实用范围：实用于破坏形式为屈服的构件。
s
s ????? ????
22
31
m a x
s??? ?? 31
2、强度准则,? ???? ??
31
14
4,Theory of the maximum torsional (shearing)strain energy
(the fourth strength）,
This theory consider the main cause of yield to be the distortional strain
energy., The member will rupture as the distortional strain energy reaches the
distortional strain energy of yield in axial tension
xsx uu ?m a x ? ? ? ? ? ?? ?213232221
6
1 ??????? ???????
Eu x
1),Criterion of failure,
2),Strength condition,
3),Application range,May be applied to the members with the
failure form of plastic yield,
? ? ? ? ? ?? ? s??????? ?????? 21323222121
? ? ? ? ? ?? ? ? ???????? ?????? 21323222121
15
四、形状改变比能（第四强度）理论,
认为构件的屈服是由形状改变比能引起的。当形状改变比
能达到单向拉伸试验屈服时形状改变比能时，构件就破坏了。
1、破坏判据,
xsx uu ?m a x
2、强度准则
3、实用范围：实用于破坏形式为屈服的构件。
? ? ? ? ? ?? ?21323222161 ??????? ??????? Eu x
? ? ? ? ? ?? ? s??????? ?????? 21323222121
? ? ? ? ? ?? ? ? ???????? ?????? 21323222121
16
§ 8–3 MOHR’S STRENGTH THEORY AND
ITS EQUIVALENT STRESS
Mohr thought the maximum shearing
stress is the main cause of failure of
materials,but the friction force in the sliding
section is not neglected,(Law of Mohr
friction）,Combining the factors of the
maximum shearing stress and the maximum
normal stress Mohr obtained his strength
theory,°￠ íD ? ?a ?? ( O,M o hr),1 8 3 5 ?? 1 9 1 8
17
§ 8–3 莫尔强度理论及其相当应力
莫尔认为：最大剪应力
是使物体破坏的主要因素，
但滑移面上的摩擦力也不可
忽略（莫尔摩擦定律）。综
合最大剪应力及最大正应力
的因素，莫尔得出了他自己
的强度理论。 °￠ íD ? ?a ?? ( O,M o hr),1 8 3 5 ?? 1 9 1 8
18
Approximate envelope
Envelope of circles of limit stress
1,Two concepts,1),Circle of limit stress,
2),Limit curve,Envelope of circles of limit stress,
O
? s
Circle of limit stress
??
??
1s?
2s?3s
?
19
近似包络线
极限应力圆的包络线 O
? s
极限应力圆
一、两个概念,1、极限应力圆,
2、极限曲线：极限应力圆的包络线（ envelope）。
??
??
1s?
2s?3s
?
20
L j x
by
bL ??
?
?? ??
31
1),Criterion of failure,
2),Strength condition,
? ??????? ??? 31 ][ ][
y
L
rM
[? y]
? ?
?
o
?
[? L] O1 O2
Derivation of the critical condition of Mohr,s theory
O3 ? 1 ? 3
M
K L
P
N
2,Mohr’s strength theory,If the stress circle of an arbitrary point Contacts
with the limit curve,the material will yield or break due to shearing,
21
[? y]
? ?
?
o
?
[? L] O1 O2
莫尔理论危险条件的推导
L j x
by
bL ??
?
?? ??
31
2、强度准则,
1、破坏判据,
O3 ? 1 ? 3
M
K L
P
N
二、莫尔强度理论,任意一点的应力圆若与极限曲线相接触,
则材料即将屈服或剪断。
22
? ??????? ??? 31 ][ ][
y
L
rM
3,Equivalent stress：（ unified form of strength conditions）
? ? ?? ?r where,? r— equivalent stress ? ? ? ?
n
s????,,2.0b?
11 ?? ?r
? ?3212 ????? ???r
? ? ? ? ? ?? ?2132322214 21 ??????? ??????r
313 ??? ??r
?
?
?
?
?
?
?
?
?
31 ][
][ ?
?
???
y
L
rM ??
3),Application range,May be applied to the members with the failure form of
plastic yield and the brittle members （ rock,concrete and so on.） with different
limit strength of tension and compression in complex stressed state,
23
三、相当应力：（强度准则的统一形式）。
其中, ? r— 相当应力 。 ? ? ? ?
n
s????,,2.0b?
3、实用范围：实用于破坏形式为屈服的构件及其拉压极限强度不
等的处于复杂应力状态的脆性材料的破坏（岩石、混凝土等）。
24
? ? ?? ?r
11 ?? ?r
? ?3212 ????? ???r
? ? ? ? ? ?? ?2132322214 21 ??????? ??????r
313 ??? ??r
?
?
?
?
?
?
?
?
?
31 ][
][ ?
?
???
y
L
rM ??
§ 8–4 APPLICATIONS OF STRENGTH THEORIES
1,Steps of calculation of strength,
1),Analysis of external forces,Determine the value of the request external forces
2),Analysis of internal forces,Plot the diagram of the internal forces and
determine the critical section,
3),Analysis of stresses,Plot the diagram of stress distribution in the critical
section,determine the critical point,plot the element and determine the principal
stresses,
4),Analysis of strength,Select the proper theory of strength,calculate the
equivalent stress and then do the calculation of strength,
25
§ 8–4 强度理论的应用
一、强度计算的步骤,
1、外力分析：确定所需的外力值。
2、内力分析：画内力图，确定可能的危险面。
3、应力分析：画危面应力分布图，确定危险点并画出单元体,
求主应力。
4、强度分析：选择适当的强度理论，计算相当应力，然后进行
强度计算。
26
2,Rules to select strength theories,decide according to the
form of failure,
1),Brittle materials,As the minimum principal stress is larger than or
equal to zero the first theory may be applied； As the minimum principal stress is
smaller than zero and the maximum principal stress is larger than zero Mohr’s
theory may be applied； As the maximum principal stress is smaller than or equal to
zero the third or fourth theory may be applied；
2),Ductile material,As the minimum principal stress is larger than or equal to
zero the first theory may be applied； For the other stressed state the third or fourth
theory may be applied,
4),Forms of failure are relative with temperature,speed of deformation etc,
3),In simple deformation,Apply the strength condition corresponding to the
simple deformation,Such as in torsion we may apply,
? ??? ?m ax
27
二、强度理论的选用原则：依破坏形式而定。
1、脆性材料：当最小主应力大于等于零时，使用第一理论；
3、简单变形时：一律用与其对应的强度准则。如扭转，都用,
2、塑性材料：当最小主应力大于等于零时，使用第一理论；
? ??? ?m ax
4、破坏形式还与温度、变形速度等有关！
当最小主应力小于零而最大主应力大于零时，使
用莫尔理论。
当最大主应力小于等于零时，使用第三或第四理论。
其它应力状态时，使用第三或第四理论。
28
M P a7.351.07 0 0 016 3 ????? ??
nW
T
M P a37.6101.0504 32 ?????? ?? AP
Solution,Stressed state at the critical
point A is shown in the figure,
Example 1 A circular rod made of cast iron is subjected to the loads
T=7kNm,P=50kN as shown in the figure,Its diameter is d=0.1m,[?]=40MPa,Try
to check the strength of the rod according to the theory of the first strength,
22
2
1 )2(2 ?
??? ??? M P a7.327.35)
2
37.6(
2
37.6 22 ?????
M P a32,0,M P a39 321 ????? ??? ? ??? ?1 Safe
P P
T
T
A
A
?
?
A ?
?
29
M P a7.351.07 0 0 016 3 ????? ??
nW
T
M P a37.6101.0504 32 ?????? ?? AP
解,危险点 A的应力状态如图,
[例 1] 直径为 d=0.1m的圆杆受力如图,T=7kNm,P=50kN,为 铸铁构
件,[?]=40MPa,试 用第一强度理论校核 杆的 强度。
故，安全。
P P
T
T
A
A
?
?
A ?
?
30
22
2
1 )2(2 ?
??? ??? M P a7.327.35)
2
37.6(
2
37.6 22 ?????
M P a32,0,M P a39 321 ????? ??? ? ??? ?1
A ? x
? y
x
y A
Solution,From generalized Hooke’s
law we get,
???? )(1 2 yxx E ????? M P a4.9410)37.73.088.1(3.01 1.2 72 ?????
???? )(1 2 xyy E ????? M P a1.1 8 310)88.13.037.7(3.01 1.2 72 ?????
0,M P a4.94,M P a1.183 321 ???? ???
? ????? ???? MP a 1.183313 ? ?? ? 003 7.7
170
1701.183 ????
?
?? r
This container does not satisfy the third strength theory and is not safe,
Example 2 As a circular thin-walled tube is subjected to the maximum internal
pressure,tested ?x=1.88?10-4,?y=7.37?10-4,Knowing for steel E=210GPa，
[?]=170MPa,Possion’s ratio is ?=0.3,Check its
strength according to the third strength theory,
31
[例 2] 薄壁圆筒受最大内压时,测得 ?x=1.88?10-4,?y=7.37?10-4,已知
钢的 E=210GPa,[?]=170MPa,泊松比 ?=0.3，试用第三强度理论 校
核 其 强度。
???? )(1 2 yxx E ????? M P a4.9410)37.73.088.1(3.01 1.2 72 ?????
???? )(1 2 xyy E ????? M P a1.1 8 310)88.13.037.7(3.01 1.2 72 ?????
解：由广义虎克定律得,
A ? x
? y
x
y A
0,M P a4.94,M P a1.183 321 ???? ???
? ?
? ? 003 7.7170
1701.183 ????
?
?? r?
所以，此容器不满足第三强度理论。不安全 。 32
? ????? ???? MP a 1.183313
21
12s in
OO
LOMO ???
bLby
bLby
??
??
?
??
5.04 0 01 2 0 0 4 0 01 2 0 0 ????
Solution,Plot the diagram
of Mohr’theory
[? y]
? ?
?
o
?
[? L] O1 O2
Critical diagram of Mohr,s theory
O3 ? 1 ? 3
M
K L
P
N
?
Example 3 For a structural member made of cast iron,?bL= 400MPa,
?by= 1200MPa,As a point with with the of plane stress yields according to
Mohr’s strength theory,the maximum shearing stress is 450MPa,Try to
determine the principal stresses at the point,
33
破坏判据,
[例 3] 一铸铁构件 ?bL= 400MPa,?by= 1200MPa，一平面应力
状态点按 莫尔强度理论屈服时，最大剪应力为 450MPa，试求该
点的主应力值。
21
12s in
OO
LOMO ???
???? s i n)/2(s i n)/( m a x1331 bLLOKOOO ????
bLby
bLby
??
??
?
??
5.0
4001200
4001200
?
?
??
解,做莫尔理论分析图
[? y]
? ?
?
o
?
[? L] O1 O2
莫尔理论危险图
O3 ? 1 ? 3
M
K L
P
N ?
34
M P a750 ; M P a150 31 ??? ??
Sowing the above simultaneous equations,we get,
?????? s i n/)2(22 m a x31 bLbL ????
4 5 02/)( m a x31 ??? ???
3005.0/)200450(2002 31 ?????? ??
That is
???? s i n/)2(s i n/)( m a x1331 bLLOKOOO ????
Criterion of failure,
35
?????? s i n/)2(22, m a x31 bLbL ????即
4502)/( m a x31 ??? ???
M P a750 ; M P a150 31 ??? ??
3 0 05.0)/2 0 04 5 0(2 0 02 31 ??????? ??
解 上述联立方程得,
36
37
Chapter 8 Exercises
1,When establishing the strength conditions in the complex stress
state,why can we determine the allowable stress of a material according
to the results of the simple-tension test?
2,The first and the second strength theories are only suitable for
brittle materials and the third and the fourth strength theories are only
suitable for plastic materials,Can we say like this? Why?
3,Try to write out the expressions of the equivalent stress of a circular
shaft in torsion according to the four strength theories,
Solution,
t
W
T?? ?? ?
1 02 ?? ?? ??3?
?? ?1r ? ???? ?? 12r ?? 23 ?r ?? 34 ?r
38
第八章 练习题
一、在建立复杂应力状态下的强度条件时，为
什么可以根据简单拉伸的试验结果来确定材料的许
用应力？
二、第一和第二强度理论只适用于脆性材料；
第三和第四强度理论只适用于塑性材料。这种说法
是否正确，为什么？
三、试按四个强度理论写出圆轴扭转时的相当
应力表达式。
解,
tW
T?? ?? ?
1 02 ?? ?? ??3?
?? ?1r ? ???? ?? 12r ?? 23 ?r ?? 34 ?r
39
40
1
2
CHAPTER 8 STHENGTH THEORIES
§ 8–1 Concepts of strength theories
§ 8–2 Four strength theories and their equivalent stresses
§ 8–3 Mohr’s strength theory and its equivalent stress
§ 8-4 Applications of strength theories
3
第八章 强度理论
§ 8–1 强度理论的概念
§ 8–2 四个强度理论及其相当应力
§ 8–3 莫尔强度理论及其相当应力
§ 8-4 强度理论的应用
4
1,Opening words,
§ 8–１ CONCEPTS OF STRENGTH THEORIES
1),Investigation of the tensile,compressive and torsional tests of cast iron
and low-carbon steel
M
Low-
carbon steel
Cast iron
P
P Tension of
cast iron P
Compression of
cast iron
2),How will the member rupture in
Combined deformations？ M
P
5
一、引子,
§ 8–１ 强度理论的概念
1、铸铁与低碳钢的拉、压、扭试验现象是怎样产生的？
M
低碳钢
铸铁 P
P 铸铁拉伸
P
铸铁压缩
2、组合变形杆将怎样破坏？
M
P
6
2,Theories of strength,Some assumptions about the cause of the
strength failure of materials,
3,Types of failure of materials：⑴ yield； ⑵ rupture
1),Galileo planted the seed of the first strength theory；
2),Mariotto’s statement about the rupture of material due to
excessive deformation is the flush of the second strength theory；
3),C.Duguet proposed the theory of the maximum shearing stress；
4),Maxwell brought forward the theory of the maximum
distortional energy theory early,which was known after his letters
were published,
7
二、强度理论：是关于, 构件发生强度失效（ failure by lost
strength）起因, 的假说。
1、伽利略播下了第一强度理论的种子；
三、材料的破坏形式：⑴ 屈服； ⑵ 断裂 。
2、马里奥特关于变形过大引起破坏的论述，是第二强度理论的
萌芽；
3、杜奎特（ C.Duguet)提出了最大剪应力理论；
4、麦克斯威尔最早提出了最大畸变能理论（ maximum distortion
energy theory）；这是后来人们在他的书信出版后才知道的。
8
§ 8–2 FOUR STRENGTH THEORIES AND
THEIR EQUIVALENT STRESSES
1,Theory of the maximum tensile stress（ the first strength）,
This theory considers the main cause of rupture to be the maximum tensile
stress,The member will rupture as the maximum tensile stress reaches the strength
limit in axial tension,
1),Criterion of rupture,
0)(; 11 ?? ??? b
2),strength condition,? ? 0)( ;
11 ?? ???
3),Application range,May be applied to the members with the
failure form of brittle rupture,
9
§ 8–2 四个强度理论及其相当应力
一、最大拉应力（第一强度）理论,
认为构件的断裂是由最大拉应力引起的。当最大拉应力达到
单向拉伸的强度极限时，构件就断了。
1、破坏判据,0)(;
11 ?? ??? b
2、强度准则,? ? 0)( ;
11 ?? ???
3、实用范围：实用于破坏形式为脆断的构件。
10
2,Theory of the maximum tensile strain（ the second strength）,
This theory considers the main cause of rupture to be the maximum tensile strain,The
member will rupture as the maximum tensile strain reaches the limit strain in axial
tension
0)(; 11 ?? ??? b ? ?? ? EE b?????? ???? 3211 1
1),Criterion of failure,
2),Strength condition,
3),Application range,May be applied to the members with the
failure form of brittle rupture,
? ? b????? ??? 321
? ? ? ?????? ??? 321
11
二,最大伸长线应变（第二强度）理论,
认为构件的断裂是由最大伸长线应变引起的。当最大伸长线
应变达到单向拉伸试验下的极限应变时，构件就断了。
1、破坏判据,
0)(; 11 ?? ??? b
2、强度准则,
3、实用范围：实用于破坏形式为脆断的构件。
? ?? ? EE b?????? ???? 3211 1
? ? b????? ??? 321
? ? ? ?????? ??? 321
12
3,Theory of the maximum shearing stress（ the third strength）,
This theory considers the main cause of rupture to be the maximum shearing stress,
The member will rupture as the maximum shearing stress reaches the limit shearing
stress in axial tension,
s?? ?m ax ss ????? ???? 22 31m a x
1),Criterion of failure,
3),Application range,May be applied to the members with the
failure form of plastic yield,
s??? ?? 31
2),Strength condition,? ?
??? ?? 31
13
三、最大剪应力（第三强度）理论,
认为构件的屈服是由最大剪应力引起的。当最大剪应力达
到单向拉伸试验的极限剪应力时，构件就破坏了。
1、破坏判据,
s?? ?m ax
3、实用范围：实用于破坏形式为屈服的构件。
s
s ????? ????
22
31
m a x
s??? ?? 31
2、强度准则,? ???? ??
31
14
4,Theory of the maximum torsional (shearing)strain energy
(the fourth strength）,
This theory consider the main cause of yield to be the distortional strain
energy., The member will rupture as the distortional strain energy reaches the
distortional strain energy of yield in axial tension
xsx uu ?m a x ? ? ? ? ? ?? ?213232221
6
1 ??????? ???????
Eu x
1),Criterion of failure,
2),Strength condition,
3),Application range,May be applied to the members with the
failure form of plastic yield,
? ? ? ? ? ?? ? s??????? ?????? 21323222121
? ? ? ? ? ?? ? ? ???????? ?????? 21323222121
15
四、形状改变比能（第四强度）理论,
认为构件的屈服是由形状改变比能引起的。当形状改变比
能达到单向拉伸试验屈服时形状改变比能时，构件就破坏了。
1、破坏判据,
xsx uu ?m a x
2、强度准则
3、实用范围：实用于破坏形式为屈服的构件。
? ? ? ? ? ?? ?21323222161 ??????? ??????? Eu x
? ? ? ? ? ?? ? s??????? ?????? 21323222121
? ? ? ? ? ?? ? ? ???????? ?????? 21323222121
16
§ 8–3 MOHR’S STRENGTH THEORY AND
ITS EQUIVALENT STRESS
Mohr thought the maximum shearing
stress is the main cause of failure of
materials,but the friction force in the sliding
section is not neglected,(Law of Mohr
friction）,Combining the factors of the
maximum shearing stress and the maximum
normal stress Mohr obtained his strength
theory,°￠ íD ? ?a ?? ( O,M o hr),1 8 3 5 ?? 1 9 1 8
17
§ 8–3 莫尔强度理论及其相当应力
莫尔认为：最大剪应力
是使物体破坏的主要因素，
但滑移面上的摩擦力也不可
忽略（莫尔摩擦定律）。综
合最大剪应力及最大正应力
的因素，莫尔得出了他自己
的强度理论。 °￠ íD ? ?a ?? ( O,M o hr),1 8 3 5 ?? 1 9 1 8
18
Approximate envelope
Envelope of circles of limit stress
1,Two concepts,1),Circle of limit stress,
2),Limit curve,Envelope of circles of limit stress,
O
? s
Circle of limit stress
??
??
1s?
2s?3s
?
19
近似包络线
极限应力圆的包络线 O
? s
极限应力圆
一、两个概念,1、极限应力圆,
2、极限曲线：极限应力圆的包络线（ envelope）。
??
??
1s?
2s?3s
?
20
L j x
by
bL ??
?
?? ??
31
1),Criterion of failure,
2),Strength condition,
? ??????? ??? 31 ][ ][
y
L
rM
[? y]
? ?
?
o
?
[? L] O1 O2
Derivation of the critical condition of Mohr,s theory
O3 ? 1 ? 3
M
K L
P
N
2,Mohr’s strength theory,If the stress circle of an arbitrary point Contacts
with the limit curve,the material will yield or break due to shearing,
21
[? y]
? ?
?
o
?
[? L] O1 O2
莫尔理论危险条件的推导
L j x
by
bL ??
?
?? ??
31
2、强度准则,
1、破坏判据,
O3 ? 1 ? 3
M
K L
P
N
二、莫尔强度理论,任意一点的应力圆若与极限曲线相接触,
则材料即将屈服或剪断。
22
? ??????? ??? 31 ][ ][
y
L
rM
3,Equivalent stress：（ unified form of strength conditions）
? ? ?? ?r where,? r— equivalent stress ? ? ? ?
n
s????,,2.0b?
11 ?? ?r
? ?3212 ????? ???r
? ? ? ? ? ?? ?2132322214 21 ??????? ??????r
313 ??? ??r
?
?
?
?
?
?
?
?
?
31 ][
][ ?
?
???
y
L
rM ??
3),Application range,May be applied to the members with the failure form of
plastic yield and the brittle members （ rock,concrete and so on.） with different
limit strength of tension and compression in complex stressed state,
23
三、相当应力：（强度准则的统一形式）。
其中, ? r— 相当应力 。 ? ? ? ?
n
s????,,2.0b?
3、实用范围：实用于破坏形式为屈服的构件及其拉压极限强度不
等的处于复杂应力状态的脆性材料的破坏（岩石、混凝土等）。
24
? ? ?? ?r
11 ?? ?r
? ?3212 ????? ???r
? ? ? ? ? ?? ?2132322214 21 ??????? ??????r
313 ??? ??r
?
?
?
?
?
?
?
?
?
31 ][
][ ?
?
???
y
L
rM ??
§ 8–4 APPLICATIONS OF STRENGTH THEORIES
1,Steps of calculation of strength,
1),Analysis of external forces,Determine the value of the request external forces
2),Analysis of internal forces,Plot the diagram of the internal forces and
determine the critical section,
3),Analysis of stresses,Plot the diagram of stress distribution in the critical
section,determine the critical point,plot the element and determine the principal
stresses,
4),Analysis of strength,Select the proper theory of strength,calculate the
equivalent stress and then do the calculation of strength,
25
§ 8–4 强度理论的应用
一、强度计算的步骤,
1、外力分析：确定所需的外力值。
2、内力分析：画内力图，确定可能的危险面。
3、应力分析：画危面应力分布图，确定危险点并画出单元体,
求主应力。
4、强度分析：选择适当的强度理论，计算相当应力，然后进行
强度计算。
26
2,Rules to select strength theories,decide according to the
form of failure,
1),Brittle materials,As the minimum principal stress is larger than or
equal to zero the first theory may be applied； As the minimum principal stress is
smaller than zero and the maximum principal stress is larger than zero Mohr’s
theory may be applied； As the maximum principal stress is smaller than or equal to
zero the third or fourth theory may be applied；
2),Ductile material,As the minimum principal stress is larger than or equal to
zero the first theory may be applied； For the other stressed state the third or fourth
theory may be applied,
4),Forms of failure are relative with temperature,speed of deformation etc,
3),In simple deformation,Apply the strength condition corresponding to the
simple deformation,Such as in torsion we may apply,
? ??? ?m ax
27
二、强度理论的选用原则：依破坏形式而定。
1、脆性材料：当最小主应力大于等于零时，使用第一理论；
3、简单变形时：一律用与其对应的强度准则。如扭转，都用,
2、塑性材料：当最小主应力大于等于零时，使用第一理论；
? ??? ?m ax
4、破坏形式还与温度、变形速度等有关！
当最小主应力小于零而最大主应力大于零时，使
用莫尔理论。
当最大主应力小于等于零时，使用第三或第四理论。
其它应力状态时，使用第三或第四理论。
28
M P a7.351.07 0 0 016 3 ????? ??
nW
T
M P a37.6101.0504 32 ?????? ?? AP
Solution,Stressed state at the critical
point A is shown in the figure,
Example 1 A circular rod made of cast iron is subjected to the loads
T=7kNm,P=50kN as shown in the figure,Its diameter is d=0.1m,[?]=40MPa,Try
to check the strength of the rod according to the theory of the first strength,
22
2
1 )2(2 ?
??? ??? M P a7.327.35)
2
37.6(
2
37.6 22 ?????
M P a32,0,M P a39 321 ????? ??? ? ??? ?1 Safe
P P
T
T
A
A
?
?
A ?
?
29
M P a7.351.07 0 0 016 3 ????? ??
nW
T
M P a37.6101.0504 32 ?????? ?? AP
解,危险点 A的应力状态如图,
[例 1] 直径为 d=0.1m的圆杆受力如图,T=7kNm,P=50kN,为 铸铁构
件,[?]=40MPa,试 用第一强度理论校核 杆的 强度。
故，安全。
P P
T
T
A
A
?
?
A ?
?
30
22
2
1 )2(2 ?
??? ??? M P a7.327.35)
2
37.6(
2
37.6 22 ?????
M P a32,0,M P a39 321 ????? ??? ? ??? ?1
A ? x
? y
x
y A
Solution,From generalized Hooke’s
law we get,
???? )(1 2 yxx E ????? M P a4.9410)37.73.088.1(3.01 1.2 72 ?????
???? )(1 2 xyy E ????? M P a1.1 8 310)88.13.037.7(3.01 1.2 72 ?????
0,M P a4.94,M P a1.183 321 ???? ???
? ????? ???? MP a 1.183313 ? ?? ? 003 7.7
170
1701.183 ????
?
?? r
This container does not satisfy the third strength theory and is not safe,
Example 2 As a circular thin-walled tube is subjected to the maximum internal
pressure,tested ?x=1.88?10-4,?y=7.37?10-4,Knowing for steel E=210GPa，
[?]=170MPa,Possion’s ratio is ?=0.3,Check its
strength according to the third strength theory,
31
[例 2] 薄壁圆筒受最大内压时,测得 ?x=1.88?10-4,?y=7.37?10-4,已知
钢的 E=210GPa,[?]=170MPa,泊松比 ?=0.3，试用第三强度理论 校
核 其 强度。
???? )(1 2 yxx E ????? M P a4.9410)37.73.088.1(3.01 1.2 72 ?????
???? )(1 2 xyy E ????? M P a1.1 8 310)88.13.037.7(3.01 1.2 72 ?????
解：由广义虎克定律得,
A ? x
? y
x
y A
0,M P a4.94,M P a1.183 321 ???? ???
? ?
? ? 003 7.7170
1701.183 ????
?
?? r?
所以，此容器不满足第三强度理论。不安全 。 32
? ????? ???? MP a 1.183313
21
12s in
OO
LOMO ???
bLby
bLby
??
??
?
??
5.04 0 01 2 0 0 4 0 01 2 0 0 ????
Solution,Plot the diagram
of Mohr’theory
[? y]
? ?
?
o
?
[? L] O1 O2
Critical diagram of Mohr,s theory
O3 ? 1 ? 3
M
K L
P
N
?
Example 3 For a structural member made of cast iron,?bL= 400MPa,
?by= 1200MPa,As a point with with the of plane stress yields according to
Mohr’s strength theory,the maximum shearing stress is 450MPa,Try to
determine the principal stresses at the point,
33
破坏判据,
[例 3] 一铸铁构件 ?bL= 400MPa,?by= 1200MPa，一平面应力
状态点按 莫尔强度理论屈服时，最大剪应力为 450MPa，试求该
点的主应力值。
21
12s in
OO
LOMO ???
???? s i n)/2(s i n)/( m a x1331 bLLOKOOO ????
bLby
bLby
??
??
?
??
5.0
4001200
4001200
?
?
??
解,做莫尔理论分析图
[? y]
? ?
?
o
?
[? L] O1 O2
莫尔理论危险图
O3 ? 1 ? 3
M
K L
P
N ?
34
M P a750 ; M P a150 31 ??? ??
Sowing the above simultaneous equations,we get,
?????? s i n/)2(22 m a x31 bLbL ????
4 5 02/)( m a x31 ??? ???
3005.0/)200450(2002 31 ?????? ??
That is
???? s i n/)2(s i n/)( m a x1331 bLLOKOOO ????
Criterion of failure,
35
?????? s i n/)2(22, m a x31 bLbL ????即
4502)/( m a x31 ??? ???
M P a750 ; M P a150 31 ??? ??
3 0 05.0)/2 0 04 5 0(2 0 02 31 ??????? ??
解 上述联立方程得,
36
37
Chapter 8 Exercises
1,When establishing the strength conditions in the complex stress
state,why can we determine the allowable stress of a material according
to the results of the simple-tension test?
2,The first and the second strength theories are only suitable for
brittle materials and the third and the fourth strength theories are only
suitable for plastic materials,Can we say like this? Why?
3,Try to write out the expressions of the equivalent stress of a circular
shaft in torsion according to the four strength theories,
Solution,
t
W
T?? ?? ?
1 02 ?? ?? ??3?
?? ?1r ? ???? ?? 12r ?? 23 ?r ?? 34 ?r
38
第八章 练习题
一、在建立复杂应力状态下的强度条件时，为
什么可以根据简单拉伸的试验结果来确定材料的许
用应力？
二、第一和第二强度理论只适用于脆性材料；
第三和第四强度理论只适用于塑性材料。这种说法
是否正确，为什么？
三、试按四个强度理论写出圆轴扭转时的相当
应力表达式。
解,
tW
T?? ?? ?
1 02 ?? ?? ??3?
?? ?1r ? ???? ?? 12r ?? 23 ?r ?? 34 ?r
39
40
