Mechanics of Materials
CHAPTER 14 ALTERNATING STRESS
§ 14–1 SUMMARY
§ 14–2 SOME NOMENCLATURE IN THE ALTERNATING STRESS
§ 14–3 ENDURANCE LIMIT OF A MATERIALS AND ITS TEST
§ 14–4 ENDURANCE LIMIT OF A MEMBER AND ITS CALCULATION
§ 14–5 CALCULATION OF THE FATIGUE STRENGTH OF THE
MEMBER UNDER SYMMETRIC CYCLES
§ 14–6 SIMPLE INTRODUCTION OF MECHANICAL PROPERTIES OF
MATERIALS AT NON-CONSTANT TEMPERATURES AND STATIC
LOADS
第十四章 交变应力
§ 14–1 概述
§ 14–2 交变应力的几个名词术语
§ 14–3 材料持久限及其测定
§ 14–4 构件持久限及其计算
§ 14–5 对称循环下 构件的疲劳强度计算
§ 14–6 非常温静载下材料力学性能简介
§ 14–１ SUMMARY
1,Alternating stress,Stress at a point inside a member is changing
periodically with time,This kind of stress is called the alternating stress,
Fold iron
thread
P
P
P
P high-pressure oil Low-pressure oil
§ 14–１ 概 述
一、交变应力,构件内一点处的应力随时间作周期性变化，这
种应力称为交变应力。
折铁丝
P
P
P
P
2,Developing process of the fatigue failure,
1).Sub-structures and microscopic structures produce changes and the core of
the permanent damage is formed,
2).Microcosmic cracks are produced,
3).Microcosmic cracks expand,
merge and form the―leading‖ crack,
4).The macroscopic ―leading‖ crack
expand steadily,
5).The structure losts its stability or
ruptures completely,
Failure of materials under the action of an alternating stress,in custom,is
called the fatigue failure,
二、疲劳破坏的发展过程,
1.亚结构和显微结构发生变化，从而永久损伤形核。
2.产生微观裂纹。
3.微观裂纹长大并合并，
形成“主导”裂纹。
4.宏观主导裂纹稳定扩展。
5.结构失稳或完全断裂。
材料在交变应力下的破坏，习惯上称为疲劳破坏。
3,Characteristics of the fatigue failure,
,).1 jxw o rk ?? ??
2).The happening of rupture needs some numbers of cycling,
3).The failure is of the brittle
rupture,
4).The rupture area shows
obviously different zones,
（ Smooth zone and coarse zone）
三、疲劳破坏的特点,
2.断裂发生要经过一定的循环次数。
3.破坏均呈脆断。
4.,断口”分区明显。
（光滑区和粗糙区）
jx?? ??工作,1
。
§ 14–2 SOME NOMENCLATURE IN THE ALTERNATING STRESS
1,Cycle characteristics,
?
?
?
?
?
?
?
?
?
?
)(;
)(;
m i nm a x
m i n
m a x
m a xm i n
m a x
m i n
??
?
?
??
?
?
r
3,Amplitude of stress,
2
m i nm a x ??? ??
a
2,Mean stress,
2
m i nm a x ??? ??
m
?m
?min
?max
?a
T
t
?
o
§ 14–2 交变应力的几个名词术语
一、循环特征,
?
?
?
?
?
?
?
?
?
?
)(;
)(;
m i nm a x
m i n
m a x
m a xm i n
m a x
m i n
??
?
?
??
?
?
r
三、应力幅,
2
m i nm a x ??? ??
a
二、平均应力,
2
m i nm a x ??? ??
m
?m
?min
?max
?a
T
t
?
o
?m t
4,Several special alternating stresses,
1).Symmetric cycle,
1
m a x
m in ???
?
?r
m ax?? ?a
0?m??
min
?max
?a
T
?
o
?m t
四、几种特殊的交变应力,
1.对称循环,
1
m a x
m in ???
?
?r
m ax?? ?a
0?m??
min
?max
?a
T
?
o
t
?
2).Fluctuating cycle,
0
m ax
m in ??
?
?r
2
max?
??
?
? ma
3).Static cycle,
1
max
min ??
?
?r
0?a?
m a x?? ?m
5,Stable alternating stress,cycle characteristics
and the cycle period are not changed,
?min
?max
?a t
?
?m
?m
?min
?max
min??
o
o
t
?
2.脉动循环,
0
m ax
m in ??
?
?r
2
max?
??
?
? ma
3.静循环,
1
max
min ??
?
?r
0?a?
m ax?? ?m
五、稳定交变应力：循环特征及周期不变。
?min
?max
?a t
?
?m
?m
?min
?max
min??
o
o
M P a561
0 1 1 5.0
5 8 3 0 04
2
m a x
m a x ??
???
?
?
A
P
M P a2.537
0115.0
558004
2
m i n
m i n ??
???
?
?
A
P
M P a12
2
537561
2
m i nm a x ????? ???
a
M P a549
2
537561
2
m i nm a x ????? ???
m
957.0
561
537
m a x
m i n ???
?
?r
Example 1 A bolt connected the connecting with rod in a generator is
subjected to the maximum tensile force in work Pmax =58.3kN,the minimum
tensile force is Pmin =55.8kN,Inside diameter of the whorl is d=11.5mm,try to
determine ?a, ?m and r,
Solution,
M P a561
0 1 1 5.0
5 8 3 0 04
2
m a x
m a x ??
???
?
?
A
P
M P a2.537
0115.0
558004
2
m i n
m i n ??
???
?
?
A
P
M P a12
2
537561
2
m i nm a x ????? ???
a
M P a549
2
537561
2
m i nm a x ????? ???
m
957.0
561
537
m a x
m i n ???
?
?r
[例 1] 发动机连杆大头螺钉工作时最大拉力 Pmax =58.3kN,最小
拉力 Pmin =55.8kN,螺纹内径为 d=11.5mm,试求 ?a, ?m 和 r。
解,
§ 14–3 ENDURANCE LIMIT OF A MATERIAL AND ITS TEST
1,Endurance limit (fatigue limit)of a materials,
As long as the maximum stress of the cycle stress does not exceed a―the
maximum limit‖,a material may be subjected to millions of times cycling without
fatigue,This maximum stress is called the ―fatigue limit‖,It is designated by ?r,
2,? —N curve（ Stress—life curve）,
N0—cycle base
?r—Endurance limit of materials
?A—Nominal endurance limit ?
N(cycling
numbers)
NA
?A
?r
N0 o
§ 14–3 材料持久限及其测定
一、材料持久限 (疲劳极限 ),
循环应力中的最大只要不超过某个“最大限度”,构件就可
以经历无数次循环而不发生疲劳破坏，这个限度值称为“疲劳
极限”，用 ?r 表示 。
二,? —N 曲线（应力 —寿命曲线）,
N0—循环基数。
?r—材料持久限。
?A—名义持久限。 ?
N(循环
次数 )
NA
?A
?r
N0 o
§ 14–4 ENDURANCE LIMIT OF A MEMBER AND ITS CALCULATION
1,Endurance limit of a material—?r 0
The relation between ?r0 and ?r,
1),K? —Effective stress- concentration
coefficient,
2),?? —Size coefficient,
rr K ?
??
?
?
??0
r
r
?
? ?)(
?
Endurance limit of the smooth
specimen with larger dimensions
Endurance limit of the small smooth specimen ??
=
kr
dr
)(
)(
?
??
Endurance limit of the smooth
specimen without stress concentration
Endurance limit of the same-dimension
specimen with stress concentration
K? =
§ 14–4 构件持久限及其计算
一、构件持久限 —?r 0
?r0 与 ?r 的关系,
1,K? —有效应力集中系数,
kr
drK
)(
)(
?
?
? ??
件的持久限同尺寸有应力集中的试
的持久限无应力集中的光滑试件
2,?? —尺寸系数,
r
r
?
?
? ??
)(
??
光滑小试件的持久限
限大尺寸光滑试件的持久
rr K ?
??
?
?
??0
3),? —Surface quality coefficient,
If the cycle stress is the shearing stress,we can change the normal
stress in above formula by the shearing stress,
Under the case of a symmetric cycle,r = -1,All the above
coefficients can be obtained by looking up tables,
rr K ?
??
?
?
??0
dr
r
)(
)(
?
? ?
?Endurance limit of the smooth
specimen
Endurance limit of the member ? =
3,? —表面质量系数,
dr
r
)(
)(
?
?
? ???
光滑试件持久限
构件持久限
如果循环应力为剪应力，将上述公式中的正应力换为剪应力即可。
对称循环下, r= -1 。上述各系数均可查表而得。
rr K ?
??
?
?
??0
Example 2 A stepped shaft is shown in the figure,The material is the alloy steel of
chrominium and nickl.Knowing ?b=920MPa,?–1= 420MPa and ?–1= 250MPa,Try
to determine the effective stress-concentration coefficient and the dimension
coefficient of the shaft respectively in bending and torsion,
Solution,1,Coefficients of the effective stress-
concentration and the dimension in bending,
25.14050 ??dD 1 2 5.0405 ??dr
55.1,M P a1 0 0 0,?? ?? KAs b
Looking up the figure and table we can get the
effective coefficient of the stress concentration,
55.1,M P a9 0 0:s ?? ?? KA b
55.1,M P a9 2 0:s ?? ?? KA b
77.0???
Looking up the table we get the
dimension coefficient,
?50
?40
r=5
[例 2 ] 阶梯轴如图，材料为铬镍合金钢,?b=920MPa,?–1=
420MPa, ?–1= 250MPa,分别求出弯曲和扭转时的有效应力集中
系数和尺寸系数。
解,1.弯曲时的有效应力集中系数和尺寸系数
25.14050 ??dD 1 2 5.0
40
5 ??
d
r
55.1,M P a1 0 0 0,?? ?? Kb 时当
由图 表 查有效应力集中系数
55.1,M P a9 0 0,?? ?? Kb 时当
55.1,M P a9 2 0,?? ?? Kb 时当
77.0???
由表查尺寸系数
?50
?40
r=5
26.1)900920(900100 25.128.125.1 ???????K
81.0???
2,Coefficients of the effective stress concentration and the dimension in
torsion,
Looking up the figure and table we can get the effective coefficient of
stress concentration,
Looking up the table we get
the dimension coefficient
28.1,M P a1 0 0 0 ?? ?? KbAs
25.1 M P a900 ?? ?? Kb,
As
We use the rectilinear interpolation method and get,M P a920?
b?
As
28.1,M P a1 0 0 0,?? ?? Kb 时当
25.1 M P a900,?? ?? Kb 时，当
，时当 M P a9 2 0,?b?
26.1)900920(900100 25.128.125.1 ???????K
81.0???
2.扭转时的有效应力集中系数和尺寸系数
由图 表 查有效应力集中系数
应用直线插值法
由表查尺寸系数
§ 14–5 CALCULATION OF THE FATIGUE STRENGTH OF
THE MEMBER UNDER SYMMETRIC CIRCLES
1,The permissible stress of fatigue under symmetric cycles,
? ? 1
0
1
1
1
?
?
? ?? ?
???
?
?
?
Knn
2,The strength condition of fatigue under symmetric cycles,
? ?1m a x ?? ??
§ 14–5 对称循环下 构件的疲劳强度计算
一, 对称循环的 疲劳许用应力,
? ? 1
0
1
1
1
?
?
? ?? ?
???
?
?
?
Knn
二, 对称循环的 疲劳强度条件,
? ?1m a x ?? ??
Solution,① Determine the stress at the
critical point and the cycle characteristic
m i nm a x ?? ??? W
M
M P a2.65
05.0
328 0 0
3 ?
??
?
1
m a x
m i n ???
?
?
r
It is the symmetric cycle,
?70
?50
r=7.5
M M
Example 3 A constant force couple M=0.8kN·m is acted on a rotating
carbon-steel shaft,The surface of the shaft is of finish turning,Knowing ?
b=600MPa,?–1= 250MPa and n=1.9,Try to check the strength of the shaft,
[例 3 ] 旋转碳钢轴上，作用一不变的力偶 M=0.8kN·m，轴表面
经过精车,? b=600MPa,?–1= 250MPa,规定 n=1.9,试校核轴
的强度。
解, ① 确定危险点应力及循环
特征
m i nm a x ?? ??? W
M
M P a2.65
05.0
328 0 0
3 ?
??
?
1
m a x
m i n ???
?
?
r
为对称循环
?70
?50
r=7.5
M M
③ Check strength
? ?1m a x ?? ??
② Look up the figure and table to determine each influence coefficient and
calculate the endurance limit of the member,
M P a600 ; 15.0 ; 4.1 ??? b
d
r
d
D ?
Determine K?,
Determine ??, From the figure we get
Looking up the figure we get
4.1??K
79.0???
Determine ?, the surface is of finish turning,? =0.94
? ? M P a8.69250
4.19.1
94.079.0 1
1
0
1
1 ???
????
?
?
? ?
????
?
?
Knn
Safe
③ 强度校核
? ?1m a x ?? ??
② 查图表求各影响系数, 计算构件持久限 。
M P a600 ; 15.0 ; 4.1 ??? b
d
r
d
D ?求 K?,
求 ??,查图得
查图得
4.1??K
79.0???
求 ?,表面精车,? =0.94
? ? M P a8.69250
4.19.1
94.079.0 1
1
0
1
1 ???
????
?
?
? ?
????
?
?
Knn
安全
§ 14–6 SIMPLE INTRODUCTION OF MECHANICAL PROPERTIES OF
MATERIALS AT NON-CONSTANT TEMPERATURES AND STATIC LOADS
1,Effect of the stress rate on mechanical properties of
materials,
dt
d?? ??
Low-carbon steel
O
?
?
1
Static
load
2 Dynamic load
Relation between the stress
rate and the yield limit
0 20 40 60 80 100
320
300
280
260
240
220
200
? s
(M
Pa
)
? (MPa/s)
·
3??? the load is called the dynamic load,
As
§ 14–6 非常温静载下材料力学性能简介
一、应力速率对材料力学性能的影响
dt
d?? ??
时称为动载 3???
低碳钢
O
?
?
1
静荷载
2 动荷载
应力速率与屈服极限的关系
0 20 40 60 80 100
320
300
280
260
240
220
200
? s
(MPa
)
? (MPa/s) ·
2,Effects of temperature on
mechanical properties of
materials
General tendency,
As temperature increases,E、
?S and?b descend ； ? and?
increase,
As temperature decreases,?b
increases； ? and? descend,
)( C?
)MPa(?
)GPa(E
0 100 200 300 400 500
216
177
137
700
600
500
400
300
200
100
100
90
80
70
60
50
40
30
20
10
(%),??
E
S?
?
b?
?
Effects of temperature on mechanical
properties of the low-carbon steel
二、温度对材料力学性能的影响
总趋势,
温度升高,E,?S, ?b下降；
?,? 增大。
温度下降,?b增大；
?,? 减小。
)( C?
)MPa(?
)GPa(E
0 100 200 300 400 500
216
177
137
700
600
500
400
300
200
100
100
90
80
70
60
50
40
30
20
10
(%),??
E
S?
?
b?
?
温度对低碳钢力学性能的影响
2000
1750
1500
1250
1000
750
500
250
0
-200 -100 0 100 200 300 400 500 600 700 800
)MPa(?
)( C?
?
b?
2.0?
80
70
60
50
40
30
20
10
0
(%)?
Effects of temperature on mechanical properties
of the chromium and manganese alloy
2000
1750
1500
1250
1000
750
500
250
0
-200 -100 0 100 200 300 400 500 600 700 800
)MPa(?
)( C?
?
b?
2.0?
80
70
60
50
40
30
20
10
0
(%)?
温度对铬锰合金力学性能的影响
As temperature decreases,the plasticity
decreases and the strength limit increases,
-
-
-
- - -
0 5 10 15
30
20
10
0
C?20?
C?196?
C?253?
P(KN)
Dl(mm)
mid-carbon
steel
P(KN)
-
-
-
- - -
0 5 10 15
30
20
10
0
C?20?
C?196?
C?253?
Dl(mm)
Pure
iron
-
-
-
P(kN)
-
-
-
- - -
0 5 10 15
30
20
10
0
C?20?
C?196?
C?253?
Dl(mm)
- - -
0 5 10 15
30
20
10
0
C?20?
C?196?
C?253?
P(kN)
Dl(mm)
温度降低，塑性降低，强度极限提高
纯铁 中碳钢
As temperature decreases,?b increases,why does the structure show the
brittleness rupture at low-temperature?
The Albert bridge of Belgium ruptured at low temperatures on March 4,1938,
温度降低,?b增大，为什么结构会发生低温脆断？
The working segment of the
member can not go beyond the
stable range！ ?
t
O
A B
C
D
E
Unstable
range
Stable range
Accelerative
range
Destroy
range
? 0
Creep curve of the material
构件的工作阶段不能超过稳定阶段
！
?
t
O
A B
C
D
E
不稳定
阶段
稳定阶段
加速阶段
破坏
阶段
? 0
材料的蠕变曲线
Stress is not
changed 4321 TTTT ???
The higher the temperature is,
the faster the creep is,
T1
T2
T3
T4
?1
?2
?3
?4
Temperature is
not changed 1234 ???? ???
The larger the stress is,the
faster the creep is,
应力不变
4321 TTTT ???
温度越高蠕变越快
T1
T2
T3
T4
?1
?2
?3
?4
温度不变
1234 ???? ???
应力越高蠕变越快
3,Stress relaxation,
At certain high temperature the general deformation of a member
is not changed,while the elastic deformation will change into the
plastic one with time,which results in the decreasing of stress in the
member,This phenomenon is called stress relaxation,
三、应力松弛,
在一定的高温下，构件上的总变形量不变时，弹性变形
会随时间的增长而转变为塑性变形，从而使构件内的应力变
小。这种现象称为应力松弛。
Temperature is not changed
123 ??? ??
?2 ?
1
?3
The larger the initial stress is,the larger
the initial rate of relaxation is,
Initial linear strain is not changed
321 TTT ??
T1
T3
T2
The higher the temperature is,the
larger the initial rate of relaxation is,
温度不变
123 ??? ??
?2 ?
1
?3
初应力越大，松弛的初速率越大
初始弹性应变不变
321 TTT ??
T1
T3
T2
温度越高，松弛的初速率越大
4,Mechanical properties of materials under the impact load ·
impact ductility · change temperature
With the decreasing of temperature,?b increases,but the structure
ruptures easily in brittleness at low temperatures,why?
As temperature decreases,?b increases,but the impact
ductility of the material decreases and the capacity to resist rupture
is almost not change.So the structure ruptures easily in brittleness
at low temperatures,
四、冲击荷载下材料力学性能 ·冲击韧度 ·转变温度
温度降低,?b增大，结构反而还发生低温脆断，原因何在？
温度降低,?b增大，但材料的冲击韧性下降，且抗断裂能
力基本不变，所以，结构易发生低温脆断。
1,Specimen of the impact test
40
55
40
55
10
10
10
10
?45 2
R 0.5
2 R1
The specimen with V-cut
The specimen with U-cut
Specimen
1.冲击试验试件
40
55
40
55
10
10
10
10
?45 2
R 0.5
2 R1
V型切口试样
U型切口试样
试件
2,Impact test
Specimen
2.冲击试验
试件
① Impact ductility of the specimen with―U‖cut,
② Impact ductility of the specimen with―V‖ cut,
③ Cold-short,the phenomenon that the impact ductility
decreases with the decreasing of temperature,
?k? Work of the impact force = W
?k?
Work of the impact force
Area of the cut =
W
A
① ―U‖型口试件的冲击韧性,
A
W
k ?? 断口面积
冲击力功?
② ―V‖型口试件的冲击韧性,
Wk ?? 冲击力功?
③ 冷脆：温度降低，冲击韧性下降的现象称为冷脆。
Complete failure of an oil-tanker at
low temperature
Chapter 14 Exercises
1,Try to explain the differences among the maximum stress,the
endurance limit of a material and the endurance limit of a member of
the alternating stresses,
2,The stress-time curve of the alternating stress is shown in the figure,
Determine the cycle characteristic,the mean stress and the stress
amplitude,
Solution,
311 2 040 ??? ?r
( ) M P a
m 402
1 2 040 ??? ???
( ) MP a80
2
1 2 040 ?? ??
??
第十四章 练习题
一、试述交变应力中的最大应力、材料持久极限
和构件持久极限之间的区别。
二、交变应力的应力 ─ 时间曲线如图，求其循环
特征、平均应力和应力幅。
解,
311 2 040 ??? ?r
( ) M P a
m 402
1 2 040 ??? ???
( ) MP a80
2
1 2 040 ?? ??
??
CHAPTER 14 ALTERNATING STRESS
§ 14–1 SUMMARY
§ 14–2 SOME NOMENCLATURE IN THE ALTERNATING STRESS
§ 14–3 ENDURANCE LIMIT OF A MATERIALS AND ITS TEST
§ 14–4 ENDURANCE LIMIT OF A MEMBER AND ITS CALCULATION
§ 14–5 CALCULATION OF THE FATIGUE STRENGTH OF THE
MEMBER UNDER SYMMETRIC CYCLES
§ 14–6 SIMPLE INTRODUCTION OF MECHANICAL PROPERTIES OF
MATERIALS AT NON-CONSTANT TEMPERATURES AND STATIC
LOADS
第十四章 交变应力
§ 14–1 概述
§ 14–2 交变应力的几个名词术语
§ 14–3 材料持久限及其测定
§ 14–4 构件持久限及其计算
§ 14–5 对称循环下 构件的疲劳强度计算
§ 14–6 非常温静载下材料力学性能简介
§ 14–１ SUMMARY
1,Alternating stress,Stress at a point inside a member is changing
periodically with time,This kind of stress is called the alternating stress,
Fold iron
thread
P
P
P
P high-pressure oil Low-pressure oil
§ 14–１ 概 述
一、交变应力,构件内一点处的应力随时间作周期性变化，这
种应力称为交变应力。
折铁丝
P
P
P
P
2,Developing process of the fatigue failure,
1).Sub-structures and microscopic structures produce changes and the core of
the permanent damage is formed,
2).Microcosmic cracks are produced,
3).Microcosmic cracks expand,
merge and form the―leading‖ crack,
4).The macroscopic ―leading‖ crack
expand steadily,
5).The structure losts its stability or
ruptures completely,
Failure of materials under the action of an alternating stress,in custom,is
called the fatigue failure,
二、疲劳破坏的发展过程,
1.亚结构和显微结构发生变化，从而永久损伤形核。
2.产生微观裂纹。
3.微观裂纹长大并合并，
形成“主导”裂纹。
4.宏观主导裂纹稳定扩展。
5.结构失稳或完全断裂。
材料在交变应力下的破坏，习惯上称为疲劳破坏。
3,Characteristics of the fatigue failure,
,).1 jxw o rk ?? ??
2).The happening of rupture needs some numbers of cycling,
3).The failure is of the brittle
rupture,
4).The rupture area shows
obviously different zones,
（ Smooth zone and coarse zone）
三、疲劳破坏的特点,
2.断裂发生要经过一定的循环次数。
3.破坏均呈脆断。
4.,断口”分区明显。
（光滑区和粗糙区）
jx?? ??工作,1
。
§ 14–2 SOME NOMENCLATURE IN THE ALTERNATING STRESS
1,Cycle characteristics,
?
?
?
?
?
?
?
?
?
?
)(;
)(;
m i nm a x
m i n
m a x
m a xm i n
m a x
m i n
??
?
?
??
?
?
r
3,Amplitude of stress,
2
m i nm a x ??? ??
a
2,Mean stress,
2
m i nm a x ??? ??
m
?m
?min
?max
?a
T
t
?
o
§ 14–2 交变应力的几个名词术语
一、循环特征,
?
?
?
?
?
?
?
?
?
?
)(;
)(;
m i nm a x
m i n
m a x
m a xm i n
m a x
m i n
??
?
?
??
?
?
r
三、应力幅,
2
m i nm a x ??? ??
a
二、平均应力,
2
m i nm a x ??? ??
m
?m
?min
?max
?a
T
t
?
o
?m t
4,Several special alternating stresses,
1).Symmetric cycle,
1
m a x
m in ???
?
?r
m ax?? ?a
0?m??
min
?max
?a
T
?
o
?m t
四、几种特殊的交变应力,
1.对称循环,
1
m a x
m in ???
?
?r
m ax?? ?a
0?m??
min
?max
?a
T
?
o
t
?
2).Fluctuating cycle,
0
m ax
m in ??
?
?r
2
max?
??
?
? ma
3).Static cycle,
1
max
min ??
?
?r
0?a?
m a x?? ?m
5,Stable alternating stress,cycle characteristics
and the cycle period are not changed,
?min
?max
?a t
?
?m
?m
?min
?max
min??
o
o
t
?
2.脉动循环,
0
m ax
m in ??
?
?r
2
max?
??
?
? ma
3.静循环,
1
max
min ??
?
?r
0?a?
m ax?? ?m
五、稳定交变应力：循环特征及周期不变。
?min
?max
?a t
?
?m
?m
?min
?max
min??
o
o
M P a561
0 1 1 5.0
5 8 3 0 04
2
m a x
m a x ??
???
?
?
A
P
M P a2.537
0115.0
558004
2
m i n
m i n ??
???
?
?
A
P
M P a12
2
537561
2
m i nm a x ????? ???
a
M P a549
2
537561
2
m i nm a x ????? ???
m
957.0
561
537
m a x
m i n ???
?
?r
Example 1 A bolt connected the connecting with rod in a generator is
subjected to the maximum tensile force in work Pmax =58.3kN,the minimum
tensile force is Pmin =55.8kN,Inside diameter of the whorl is d=11.5mm,try to
determine ?a, ?m and r,
Solution,
M P a561
0 1 1 5.0
5 8 3 0 04
2
m a x
m a x ??
???
?
?
A
P
M P a2.537
0115.0
558004
2
m i n
m i n ??
???
?
?
A
P
M P a12
2
537561
2
m i nm a x ????? ???
a
M P a549
2
537561
2
m i nm a x ????? ???
m
957.0
561
537
m a x
m i n ???
?
?r
[例 1] 发动机连杆大头螺钉工作时最大拉力 Pmax =58.3kN,最小
拉力 Pmin =55.8kN,螺纹内径为 d=11.5mm,试求 ?a, ?m 和 r。
解,
§ 14–3 ENDURANCE LIMIT OF A MATERIAL AND ITS TEST
1,Endurance limit (fatigue limit)of a materials,
As long as the maximum stress of the cycle stress does not exceed a―the
maximum limit‖,a material may be subjected to millions of times cycling without
fatigue,This maximum stress is called the ―fatigue limit‖,It is designated by ?r,
2,? —N curve（ Stress—life curve）,
N0—cycle base
?r—Endurance limit of materials
?A—Nominal endurance limit ?
N(cycling
numbers)
NA
?A
?r
N0 o
§ 14–3 材料持久限及其测定
一、材料持久限 (疲劳极限 ),
循环应力中的最大只要不超过某个“最大限度”,构件就可
以经历无数次循环而不发生疲劳破坏，这个限度值称为“疲劳
极限”，用 ?r 表示 。
二,? —N 曲线（应力 —寿命曲线）,
N0—循环基数。
?r—材料持久限。
?A—名义持久限。 ?
N(循环
次数 )
NA
?A
?r
N0 o
§ 14–4 ENDURANCE LIMIT OF A MEMBER AND ITS CALCULATION
1,Endurance limit of a material—?r 0
The relation between ?r0 and ?r,
1),K? —Effective stress- concentration
coefficient,
2),?? —Size coefficient,
rr K ?
??
?
?
??0
r
r
?
? ?)(
?
Endurance limit of the smooth
specimen with larger dimensions
Endurance limit of the small smooth specimen ??
=
kr
dr
)(
)(
?
??
Endurance limit of the smooth
specimen without stress concentration
Endurance limit of the same-dimension
specimen with stress concentration
K? =
§ 14–4 构件持久限及其计算
一、构件持久限 —?r 0
?r0 与 ?r 的关系,
1,K? —有效应力集中系数,
kr
drK
)(
)(
?
?
? ??
件的持久限同尺寸有应力集中的试
的持久限无应力集中的光滑试件
2,?? —尺寸系数,
r
r
?
?
? ??
)(
??
光滑小试件的持久限
限大尺寸光滑试件的持久
rr K ?
??
?
?
??0
3),? —Surface quality coefficient,
If the cycle stress is the shearing stress,we can change the normal
stress in above formula by the shearing stress,
Under the case of a symmetric cycle,r = -1,All the above
coefficients can be obtained by looking up tables,
rr K ?
??
?
?
??0
dr
r
)(
)(
?
? ?
?Endurance limit of the smooth
specimen
Endurance limit of the member ? =
3,? —表面质量系数,
dr
r
)(
)(
?
?
? ???
光滑试件持久限
构件持久限
如果循环应力为剪应力，将上述公式中的正应力换为剪应力即可。
对称循环下, r= -1 。上述各系数均可查表而得。
rr K ?
??
?
?
??0
Example 2 A stepped shaft is shown in the figure,The material is the alloy steel of
chrominium and nickl.Knowing ?b=920MPa,?–1= 420MPa and ?–1= 250MPa,Try
to determine the effective stress-concentration coefficient and the dimension
coefficient of the shaft respectively in bending and torsion,
Solution,1,Coefficients of the effective stress-
concentration and the dimension in bending,
25.14050 ??dD 1 2 5.0405 ??dr
55.1,M P a1 0 0 0,?? ?? KAs b
Looking up the figure and table we can get the
effective coefficient of the stress concentration,
55.1,M P a9 0 0:s ?? ?? KA b
55.1,M P a9 2 0:s ?? ?? KA b
77.0???
Looking up the table we get the
dimension coefficient,
?50
?40
r=5
[例 2 ] 阶梯轴如图，材料为铬镍合金钢,?b=920MPa,?–1=
420MPa, ?–1= 250MPa,分别求出弯曲和扭转时的有效应力集中
系数和尺寸系数。
解,1.弯曲时的有效应力集中系数和尺寸系数
25.14050 ??dD 1 2 5.0
40
5 ??
d
r
55.1,M P a1 0 0 0,?? ?? Kb 时当
由图 表 查有效应力集中系数
55.1,M P a9 0 0,?? ?? Kb 时当
55.1,M P a9 2 0,?? ?? Kb 时当
77.0???
由表查尺寸系数
?50
?40
r=5
26.1)900920(900100 25.128.125.1 ???????K
81.0???
2,Coefficients of the effective stress concentration and the dimension in
torsion,
Looking up the figure and table we can get the effective coefficient of
stress concentration,
Looking up the table we get
the dimension coefficient
28.1,M P a1 0 0 0 ?? ?? KbAs
25.1 M P a900 ?? ?? Kb,
As
We use the rectilinear interpolation method and get,M P a920?
b?
As
28.1,M P a1 0 0 0,?? ?? Kb 时当
25.1 M P a900,?? ?? Kb 时，当
，时当 M P a9 2 0,?b?
26.1)900920(900100 25.128.125.1 ???????K
81.0???
2.扭转时的有效应力集中系数和尺寸系数
由图 表 查有效应力集中系数
应用直线插值法
由表查尺寸系数
§ 14–5 CALCULATION OF THE FATIGUE STRENGTH OF
THE MEMBER UNDER SYMMETRIC CIRCLES
1,The permissible stress of fatigue under symmetric cycles,
? ? 1
0
1
1
1
?
?
? ?? ?
???
?
?
?
Knn
2,The strength condition of fatigue under symmetric cycles,
? ?1m a x ?? ??
§ 14–5 对称循环下 构件的疲劳强度计算
一, 对称循环的 疲劳许用应力,
? ? 1
0
1
1
1
?
?
? ?? ?
???
?
?
?
Knn
二, 对称循环的 疲劳强度条件,
? ?1m a x ?? ??
Solution,① Determine the stress at the
critical point and the cycle characteristic
m i nm a x ?? ??? W
M
M P a2.65
05.0
328 0 0
3 ?
??
?
1
m a x
m i n ???
?
?
r
It is the symmetric cycle,
?70
?50
r=7.5
M M
Example 3 A constant force couple M=0.8kN·m is acted on a rotating
carbon-steel shaft,The surface of the shaft is of finish turning,Knowing ?
b=600MPa,?–1= 250MPa and n=1.9,Try to check the strength of the shaft,
[例 3 ] 旋转碳钢轴上，作用一不变的力偶 M=0.8kN·m，轴表面
经过精车,? b=600MPa,?–1= 250MPa,规定 n=1.9,试校核轴
的强度。
解, ① 确定危险点应力及循环
特征
m i nm a x ?? ??? W
M
M P a2.65
05.0
328 0 0
3 ?
??
?
1
m a x
m i n ???
?
?
r
为对称循环
?70
?50
r=7.5
M M
③ Check strength
? ?1m a x ?? ??
② Look up the figure and table to determine each influence coefficient and
calculate the endurance limit of the member,
M P a600 ; 15.0 ; 4.1 ??? b
d
r
d
D ?
Determine K?,
Determine ??, From the figure we get
Looking up the figure we get
4.1??K
79.0???
Determine ?, the surface is of finish turning,? =0.94
? ? M P a8.69250
4.19.1
94.079.0 1
1
0
1
1 ???
????
?
?
? ?
????
?
?
Knn
Safe
③ 强度校核
? ?1m a x ?? ??
② 查图表求各影响系数, 计算构件持久限 。
M P a600 ; 15.0 ; 4.1 ??? b
d
r
d
D ?求 K?,
求 ??,查图得
查图得
4.1??K
79.0???
求 ?,表面精车,? =0.94
? ? M P a8.69250
4.19.1
94.079.0 1
1
0
1
1 ???
????
?
?
? ?
????
?
?
Knn
安全
§ 14–6 SIMPLE INTRODUCTION OF MECHANICAL PROPERTIES OF
MATERIALS AT NON-CONSTANT TEMPERATURES AND STATIC LOADS
1,Effect of the stress rate on mechanical properties of
materials,
dt
d?? ??
Low-carbon steel
O
?
?
1
Static
load
2 Dynamic load
Relation between the stress
rate and the yield limit
0 20 40 60 80 100
320
300
280
260
240
220
200
? s
(M
Pa
)
? (MPa/s)
·
3??? the load is called the dynamic load,
As
§ 14–6 非常温静载下材料力学性能简介
一、应力速率对材料力学性能的影响
dt
d?? ??
时称为动载 3???
低碳钢
O
?
?
1
静荷载
2 动荷载
应力速率与屈服极限的关系
0 20 40 60 80 100
320
300
280
260
240
220
200
? s
(MPa
)
? (MPa/s) ·
2,Effects of temperature on
mechanical properties of
materials
General tendency,
As temperature increases,E、
?S and?b descend ； ? and?
increase,
As temperature decreases,?b
increases； ? and? descend,
)( C?
)MPa(?
)GPa(E
0 100 200 300 400 500
216
177
137
700
600
500
400
300
200
100
100
90
80
70
60
50
40
30
20
10
(%),??
E
S?
?
b?
?
Effects of temperature on mechanical
properties of the low-carbon steel
二、温度对材料力学性能的影响
总趋势,
温度升高,E,?S, ?b下降；
?,? 增大。
温度下降,?b增大；
?,? 减小。
)( C?
)MPa(?
)GPa(E
0 100 200 300 400 500
216
177
137
700
600
500
400
300
200
100
100
90
80
70
60
50
40
30
20
10
(%),??
E
S?
?
b?
?
温度对低碳钢力学性能的影响
2000
1750
1500
1250
1000
750
500
250
0
-200 -100 0 100 200 300 400 500 600 700 800
)MPa(?
)( C?
?
b?
2.0?
80
70
60
50
40
30
20
10
0
(%)?
Effects of temperature on mechanical properties
of the chromium and manganese alloy
2000
1750
1500
1250
1000
750
500
250
0
-200 -100 0 100 200 300 400 500 600 700 800
)MPa(?
)( C?
?
b?
2.0?
80
70
60
50
40
30
20
10
0
(%)?
温度对铬锰合金力学性能的影响
As temperature decreases,the plasticity
decreases and the strength limit increases,
-
-
-
- - -
0 5 10 15
30
20
10
0
C?20?
C?196?
C?253?
P(KN)
Dl(mm)
mid-carbon
steel
P(KN)
-
-
-
- - -
0 5 10 15
30
20
10
0
C?20?
C?196?
C?253?
Dl(mm)
Pure
iron
-
-
-
P(kN)
-
-
-
- - -
0 5 10 15
30
20
10
0
C?20?
C?196?
C?253?
Dl(mm)
- - -
0 5 10 15
30
20
10
0
C?20?
C?196?
C?253?
P(kN)
Dl(mm)
温度降低，塑性降低，强度极限提高
纯铁 中碳钢
As temperature decreases,?b increases,why does the structure show the
brittleness rupture at low-temperature?
The Albert bridge of Belgium ruptured at low temperatures on March 4,1938,
温度降低,?b增大，为什么结构会发生低温脆断？
The working segment of the
member can not go beyond the
stable range！ ?
t
O
A B
C
D
E
Unstable
range
Stable range
Accelerative
range
Destroy
range
? 0
Creep curve of the material
构件的工作阶段不能超过稳定阶段
！
?
t
O
A B
C
D
E
不稳定
阶段
稳定阶段
加速阶段
破坏
阶段
? 0
材料的蠕变曲线
Stress is not
changed 4321 TTTT ???
The higher the temperature is,
the faster the creep is,
T1
T2
T3
T4
?1
?2
?3
?4
Temperature is
not changed 1234 ???? ???
The larger the stress is,the
faster the creep is,
应力不变
4321 TTTT ???
温度越高蠕变越快
T1
T2
T3
T4
?1
?2
?3
?4
温度不变
1234 ???? ???
应力越高蠕变越快
3,Stress relaxation,
At certain high temperature the general deformation of a member
is not changed,while the elastic deformation will change into the
plastic one with time,which results in the decreasing of stress in the
member,This phenomenon is called stress relaxation,
三、应力松弛,
在一定的高温下，构件上的总变形量不变时，弹性变形
会随时间的增长而转变为塑性变形，从而使构件内的应力变
小。这种现象称为应力松弛。
Temperature is not changed
123 ??? ??
?2 ?
1
?3
The larger the initial stress is,the larger
the initial rate of relaxation is,
Initial linear strain is not changed
321 TTT ??
T1
T3
T2
The higher the temperature is,the
larger the initial rate of relaxation is,
温度不变
123 ??? ??
?2 ?
1
?3
初应力越大，松弛的初速率越大
初始弹性应变不变
321 TTT ??
T1
T3
T2
温度越高，松弛的初速率越大
4,Mechanical properties of materials under the impact load ·
impact ductility · change temperature
With the decreasing of temperature,?b increases,but the structure
ruptures easily in brittleness at low temperatures,why?
As temperature decreases,?b increases,but the impact
ductility of the material decreases and the capacity to resist rupture
is almost not change.So the structure ruptures easily in brittleness
at low temperatures,
四、冲击荷载下材料力学性能 ·冲击韧度 ·转变温度
温度降低,?b增大，结构反而还发生低温脆断，原因何在？
温度降低,?b增大，但材料的冲击韧性下降，且抗断裂能
力基本不变，所以，结构易发生低温脆断。
1,Specimen of the impact test
40
55
40
55
10
10
10
10
?45 2
R 0.5
2 R1
The specimen with V-cut
The specimen with U-cut
Specimen
1.冲击试验试件
40
55
40
55
10
10
10
10
?45 2
R 0.5
2 R1
V型切口试样
U型切口试样
试件
2,Impact test
Specimen
2.冲击试验
试件
① Impact ductility of the specimen with―U‖cut,
② Impact ductility of the specimen with―V‖ cut,
③ Cold-short,the phenomenon that the impact ductility
decreases with the decreasing of temperature,
?k? Work of the impact force = W
?k?
Work of the impact force
Area of the cut =
W
A
① ―U‖型口试件的冲击韧性,
A
W
k ?? 断口面积
冲击力功?
② ―V‖型口试件的冲击韧性,
Wk ?? 冲击力功?
③ 冷脆：温度降低，冲击韧性下降的现象称为冷脆。
Complete failure of an oil-tanker at
low temperature
Chapter 14 Exercises
1,Try to explain the differences among the maximum stress,the
endurance limit of a material and the endurance limit of a member of
the alternating stresses,
2,The stress-time curve of the alternating stress is shown in the figure,
Determine the cycle characteristic,the mean stress and the stress
amplitude,
Solution,
311 2 040 ??? ?r
( ) M P a
m 402
1 2 040 ??? ???
( ) MP a80
2
1 2 040 ?? ??
??
第十四章 练习题
一、试述交变应力中的最大应力、材料持久极限
和构件持久极限之间的区别。
二、交变应力的应力 ─ 时间曲线如图，求其循环
特征、平均应力和应力幅。
解,
311 2 040 ??? ?r
( ) M P a
m 402
1 2 040 ??? ???
( ) MP a80
2
1 2 040 ?? ??
??
