1
Mechanics of Materials
2
CHAPTER 13 DYNAMIC LOADS
§ 13–1 BASIC CONCEPTS
§ 13–2 DYNAMIC RESPONSE ABOUT THE PROBLEM
OF ACCELERATIVE MOTIONS
§ 13–3 DYNAMIC RESPONSE ABOUT THE PROBLEM
OF IMPACT LOADS
第十三章 动荷载
§ 13–1 基本概念
§ 13–2 加速运动问题的动响应
§ 13–3 冲击荷载问题的动响应
1,Dynamic loads,
§ 13-1 BASIC CONCEPTS
2,Dynamic responses,
The loads don’t change with time(or change very stably and slowly) and
acceleration of each member is zero or may be neglected, this kind of the loads
are called static loads,
The loads change sharply with time and the velocity of the member changes
obviously（ the member produces the inertia force）,this kind of load is loads
are called dynamic loads,
Various responses (such as stress,strain,displacement and so on)of the
member under the action of dynamic loads are called dynamic responses,
Experiment prove that Hooke’law applied validly under static loads may be
applied to the case of dynamic loads with Estatic=Edynamic as long as the stress does
not exceed the proportional limit,
一、动载荷,
载荷不随时间变化（或变化极其平稳缓慢）且使构件各部件
加速度保持为零（或可忽略不计），此类载荷为 静载荷 。
载荷随时间急剧变化且使构件的速度有显著变化（系统产生
惯性力），此类载荷为 动载荷 。
§ 13-1 基本概念
二、动响应,
构件在动载荷作用下产生的各种响应（如应力、应变、位
移等），称为 动响应 。
实验表明：在静载荷下服从虎克定律的材料，只要应力不
超过比例极限,在动载荷下虎克定律仍成立且 E静 =E动 。
3,Dynamic load coefficient,
jdd K ?? ?
r e s p o n s e St a t ic
r e s p o n s e D y n a m ic?
dK
4,Classification of the dynamic stress,
1).Simple dynamic stress,The acceleration can be determined,We may
apply the“method of kineto static” to solve this kind of problems,
3).Alternating stress,Stress changes periodically with time,It belongs to
the fatigue problem,
4).Vibration problems,There are many methods to solve this kind of problems,
Dynamic load coefficient
2).Impact load,The velocity changes sharply in a very short time,In this
case the acceleration can not be determined,We must apply the,energy
method” to solve this kind of problems；
三、动荷系数,
jdd K ?? ?
静响应
动响应动荷系数 ?
dK
四、动应力分类,
1.简单动应力,加速度可以确定，采用, 动静法, 求解
。 2.冲击载荷,速度在极短暂的时间内有急剧改变，此时，加
速度不能确定，要采用, 能量法, 求解；
3.交变应力,应力随时间作周期性变化，属疲劳问题。
4.振动问题,求解方法很多。
§ 13-2 DYNAMIC RESPONSE OF THE PROBLEM OF
ACCELERATIVE MOTIONS
Principle of the method,D’Alembert’s principle
( Method of kineto statics ）
D’Alembert’s principle think there is inertial force on the body in un-
equilibrium,The direction of the inertial force is opposite to the acceleration of
the body and the magnitude of the inertial force is the product of the mass and
the acceleration of the body,After the inertial force is applied on the body the
dynamic problem may be dealt with the static problem in form,which is called
the method of kineto statics,
§ 13-2 加速运动问题的动响应
方法原理,D’Alembert’s principle
( 动静法 ）
达朗伯原理认为：处于不平衡状态的物体，存在惯性
力，惯性力的方向与加速度方向相反，惯性力的数值等于
加速度与质量的乘积。只要在物体上加上惯性力，就可以
把动力学问题在形式上作为静力学问题来处理，这就是动
静法。
a
g
Aq
G
??
)1()(
g
aAxxqqN
Gjd ???? ?
)1(
g
ax
A
N d
d ??? ??
Solution：① Analysis of the forces is
shown in the figure,
② Dynamic stress
1,Dynamic stress of the body in the straight-line motion
L
x
m n
a
x
a
Nd
qj
qG
Inertial force
Example 1 The effective area of the steel wire rope in a crane is A,[?] is
known, Weight per unit volume of the body is ? and the body moves up at the
acceleration a,Try to check the strength of the rope（ neglect the weight of the
rope）
a
g
A
q G
?
?惯性力：
)1()(
g
aAxxqqN
Gjd ???? ?
)1(
g
ax
A
N d
d ??? ??
[例 1 ] 起重机钢丝绳的有效横截面面积为 A,已知 [?],单位
体积重为 ?,以加速度 a上升，试校核钢丝绳的强度（不计绳
重）。 解：① 受力分析如图,
② 动应力
一、直线运动构件的动应力
L
x
m n
a
x
a
Nd
qj
qG
m a xm a x )1( jdd Kg
a
L ??? ???
g
a
K d ?? 1
Coefficient of the dynamic load,
? ???? ?? m a xm a x jdd K
Strength condition,
If, ? ?
,m a x ?? ?d
? ?,m a x ?? ?d
then the strength condition is satisfied,
then the strength condition is not satisfied,If,
m a xm a x )1( jdd Kg
a
L ??? ???
g
a
K d ?? 1
动荷系数,
? ???? ?? m a xm a x jdd K强度条件,
若,? ?
?? ?m a xd
? ??? ?m a xd
满足
不满足
)1)(( gaqLGN d ???
)
8.9
21)(605.251050(
109.2
1 3
4 ?????? ?
? ? M P a3 0 0M P a2 1 4 ??? ?
Example 2 Length of the steel rope of a crane is 60m,its nominal diameter is
28cm,its effective cross-section area is A=2,9cm2,weight per unit length is q=25,
5N/m and [?] =300MPa,The acceleration when the rope lifts an object of 50kN is
a=2m/s2,Try to check the strength of the steel rope,
G(1+a/g)
Nd
L q(1+a/g)
)1)((1 gaqLGAAN dd ?????
Solution：① The free body
diagram is shown in the figure,
② Dynamic stress
)1)(( gaqLGN d ???
)
8.9
21)(605.251050(
109.2
1 3
4 ?????? ?
? ? M P a3 0 0M P a2 1 4 ??? ?
[例 2 ] 起重机钢丝绳长 60m，名义直径 28cm，有效横截面面积
A=2,9cm2,单位长度重量 q=25,5N/m,[?] =300MPa,以 a=2m/s2的
加速度提起重 50kN 的物体，试校核钢丝绳的强度。
G(1+a/g)
Nd
L q(1+a/g)
)1)((1 gaqLGAAN dd ?????
解：① 受力分析如图,
② 动应力
gLGRmmaG nG /22 ?? ???
? ??? ?? AG G /
? ? ? ?)(
2
?
?
? g
GLGA G ??
② Strength condition
Solution,① The free body diagram is
shown in the figure,
?
GG
L
O
2,Dynamic stress of the rotating member,
The inertia force is
Example 3 A ball of weight G is install on one end of the rotating arm
with length L,They are rotating about point O at an equal angular speed in the
smooth plane,Knowing permissible stress is [?],Determine the cross-section
areaof the rotating arm（ neglecting the weight of the rotating arm）,
gLGRmmaG nG /22 ?? ???
惯性力：
? ??? ?? AG G /
? ? ? ?)(
2
?
?
? g
GLGA G ??
[例 3 ] 重为 G的球装在长 L的转臂端部，以等角速度在光滑水
平面上绕 O点旋转,已知许用应力 [?],求转臂的截面面积
（不计转臂自重）。
② 强度条件
解：① 受力分析如图,
?
GG
L
O
二、转动构件的动应力,
Fig.1
qG
② Analysis of the internal force
is shown in Fig.2
g
AD
g
Aaq n
G 2
2???
??
02 ?? DqN Gd
2
2
42
??
g
ADDqN G
d ??
2
2?D
a n ?
Solution,① Analysis of the
inertia forceis shown in Fig.1
O
D
t
Fig.2
qG
NG NG
Example 4 Suppose the average diameter of a ring is D,its thickness is t
with t<<D.Its cross-section area is A and its weight per unit volume is ?,As
shown in the figure,the ring rotates at a constant angular speed ? about the axis
that is through the center of the circle and perpendicular to the plane where the
ring lies.Try to determine the dynamic stress of the ring and establish the
strength condition,
?
图 1
qG
[例 4] 设圆环的平均直径 D、厚度 t,且 t?D，环的横截面面积
为 A，单位体积重量为 ?,圆环绕过圆心且垂直于圆环平面的轴
以等角速度 ?旋转，如图所示,试确定圆环的动应力，并建立强
度条件。
② 内 力分析如图 2
g
AD
g
Aaq n
G 2
2???
??
02 ?? DqN Gd
2
2
42
??
g
ADDqN G
d ??
2
2?D
a n ?
解：① 惯性力分析，见图１
O
D
t
图 2
qG
NG NG
?
22
2
4
?????
gg
D
A
N d
d ???
? ????? ??
gd
2
?
?
?
g ][
??
③ Analysis of stress
④ Strength condition
Maximum linear speed,
?
?
?
g ][
m a x ?
22
2
4
?????
gg
D
A
N d
d ???
? ????? ??
gd
2
?
?
?
g ][
??
③ 应力分析
④ 强度条件
最大线速度,
?
?
?
g ][
m a x ?
§ 13-3 DYNAMIC RESPONSE ABOUT THE PROBLEM OF IMPACT LOADS
Principle of the method,Energy method
( Conservation of the mechanical energy ）
In the touching area between the impact and impacted members the
stressed state of each point is very complex and the lasting time of impact is
very short,therefore,it is difficult to analyze the variety of the touching force
with time accurately, In engineering this kind of impact problems are usually
solved by the energy method,That is to say,on the basis of some assumptions
we can do some conservative simplifications for the stress and deformation of
the impacted member according to the conservation law of energy,
§ 13-3 冲击荷载问题的动响应
方法原理：能量法
( 机械能守恒 ）
在冲击物与受冲构件的接触区域内，应力状态异常复杂，
且冲击持续时间非常短促，接触力随时间的变化难以准确分
析。工程中通常采用能量法来解决冲击问题，即在若干假设
的基础上，根据能量守恒定律对受冲击构件的应力与变形进
行偏于安全的简化计算。
2.Kinetic energy T, potential energy V, strain energy U,
The energy before and after impact is conservative,
Maximum impact effect,the kinetic energy after impact is zero,T2= 0
The strain energy for an impact force is U2=(1/2)PdΔd
1,Assumptions,
① The impact body is rigid；
② The impact body do not rebound；
③ The loss of energy of sound,light,heat etc.in the process of impact
is negleced （ energy is conservative）；
④ The process of impact shows linear-elastic deformations is linearly
elastic.(conservative calculation)
111 UVT ?? 222 UVT ???
(Before impact) (After impact)
① 冲击物为刚体；
②冲击物不反弹；
③ 不计冲击过程中的声、光、热等能量损耗（能量守恒）；
④ 冲击过程为线弹性变形过程。 (保守计算 )
111 )( UVT ??冲击前 )( 222 冲击后UVT ???
2.动能 T,势能 V,变形能 U，冲击前、后，能量守恒,
最大冲击效应：冲击后的动能为零,T2=0
一个冲击力的变形能为 U2=(1/2)PdΔ d
1.假设,
3,Coefficient of the dynamic load is Kd,
jdd
jdd
jdd
K
K
PKP
?? ?
???
?
3.动荷系数为 Kd,
jdd
jdd
jdd
K
K
PKP
?? ?
???
?
The energy before and after impact is conservative and
j
2
jd
2
2
)(
2
1 ?????
dK
mgKhmgm ?
j
d
hg
K
?
?
???
2/
11
2?
1,Impact problem about the free falling body along the axis
jddjjdd KmgPPKP ????? ),(
△ j:Static displacement at the falling
point of the impact body,
?d mg
v mg
h
0
2/
1
1
2
1
?
?
?
U
m g hV
mvT
Before impact,
2/
0
2
2
2
dd
d
PU
mgV
T
??
???
?
After impact,
Kinetic energy
Potential energy
Strain energy
0
2/
1
1
2
1
?
?
?
U
m g hV
mvT
变形能
势能
动能
冲击前后能量守恒，且
j
2
jd
2
2
)(
2
1 ?????
dK
mgKhmgm ?
j
d
hg
K
?
?
???
2/
11
2?
一、轴向自由落体冲击问题
冲击前,
2/
0
2
2
2
dd
d
PU
mgV
T
??
???
?
变形能
势能
动能
冲击后,
jdd
jjdd
K
mgPPKP
???
?? )(
△ j:冲击物落点的静位移。
?d mg
v mg
h
Discussions,
j
h
dK ?????
2
11:,0)1( ?
2,0 ?? dKh
(2)Sudden loads
讨论,
j
h
dK ?????
2
11:,0)1( ?
2,0 ?? dKh(2)突加荷载
jdK
mgmv ?? 22
22
1
2,Axial impact of neglecting weight,
0
0
2/
1
1
2
1
?
?
?
U
V
mvT
Before impact,
2/
0
0
2
2
2
ddPU
V
T
??
?
?
After impact,
The energy before and after
impact is conservative and
jdd
jjdd
K
mgPPKP
???
?? )(
Coefficient of
The dynamic load
j
d gK ??
2?
Kinetic energy
Potential energy
Strain energy
Kinetic energy
Potential energy
Strain energy
mg
?
jdK
mgmv ?? 22
22
1
二、不计重力的轴向冲击,
0
0
2/
1
1
2
1
?
?
?
U
V
mvT
变形能
势能
动能
冲击前,
2/
0
0
2
2
2
ddPU
V
T
??
?
?
变形能
势能
动能
冲击后,冲击前后能量守恒，且
jdd
jjdd
K
mgPPKP
???
?? )(
动荷系数
j
d gK ??
2?
mg
?
3,Calculation of the impact response
② Coefficient of the dynamic
load
③ Determine the dynamic stress
Solution：① Determine
the static deformation
9.217
425
1000211211 ?????
?
???
j
h
dK
mm425???? EAWLEA LP jj
M P a41.15?? jdd K ??
Static stress,M P a0 7 0 7 4.0/ ?? AW
j?
Dynamic stress,
h=1m
v W
f 6m
The impact response is equal to the product of the static response and the
coefficient of the dynamic load,
Example 5 The wooden stake with the diameter of 0.3m is
impacted by a free falling hammer,Weight of the falling
hammer is 5kN,Determine the maximum dynamic stress of
the wooden stake,E=10GPa
三、冲击响应计算
② 动荷系数
③ 求动应力
解：① 求静变形
9.217
425
1000211211 ?????
?
???
j
h
dK
mm425????
EA
WL
EA
LP j
j
M P a41.15?? jdd K ??
等于静响应 与 动荷系数之积,
[例 5 ] 直径 0.3m的木桩受自由落锤冲击，落锤重 5kN,
求：桩的最大动应力。 E=10GPa
静应力,
M P a0 7 0 7 4.0/ ?? AWj?
动应力,
h=1m
v W
f 6m
4,Impact problem of the beam
0)(
2
1 2
111
???
???
dfhmgmv
UVT
mg
L
h A B C
A B C x
f
fd
1).Assumptions,
① Impact bodies are rigid ones；
② The potential energy of weight and kinetic
energy of the impacted bodies is neglected ；
③ Impact bodies do not rebound；
④ The loss of energy of sound,light,heat
and so on(energy is conservative)is neglected,
Before impact
四,梁的冲击问题 1.假设,
① 冲击物为刚体；
② 不计被冲击物的重力势能和动能；
③ 冲击物不反弹；
④ 不计声、光、热等能量损耗（能
量守恒）。
0)(
2
1
冲 击击
2
111
???
???
dfhmgm
UVT
?
mg
L
h A B C
A B C x
f
fd
A B C x
f
fd
The energy before and after impact
is conservative,therefore,
22 )(
2
)(
2
1
d
j
d ff
mgfhmgmv ???
jdj
j
fKf
f
hgf ????? )2)(11( 2
d
?
jj
d
d f
hg
f
f
K
2)2(
11
?
????
?
jf
h
dK
211 ??? 2?dK
Coefficient of
the dynamic
load
(2)Sudden load,(1)Free falling body
22
)(
2
1
)(
2
1
2
1
00
d
j
d
j
j
dd
f
f
mg
f
f
P
fP
??
???
After impact
222 UVT ??
22
222
)(
2
1
)(
2
1
2
1
00
冲 击击
d
j
d
j
j
dd
f
f
mg
f
f
P
fP
UVT
??
???
??
冲击前、后，能量守恒，所以,
A B C x
f
fd
22 )(
2
)(
2
1
d
j
d ff
mgfhmgmv ???
jdj
j
fKf
f
hgf ????? )2)(11( 2
d
?
jj
d
d f
hg
f
f
K
2)2(
11:
?
????
?
动荷系数
jf
h
dK
211:)1( ???自由落体 2:)2( ?
dK突加荷载
h
B A C
mg E =P
5,Calculation of the dynamic response,
Solution：① Determine
the static deflection of point
C
21
1
2 CC
AAf
Cj ??
The dynamic response is equal to the product of the static response and the
coefficient of the dynamic load,
ABDE
A
EI
PL
EI
LR
4896
33
?? EI
PL
192
5 3?EIEIEI
DEAB ??
D
C2
C1
A1
L
Example 6 A structure is
shown in the figure,AB=DE=L，
points A and C are separately the
middle points of AB and DE,
Determine the dynamic stress of
section C under the impact of
weight mg,
h
B A C
mg E =P
五、动响应计算,
解：① 求 C点静挠度
21
1
2 CC
AAf
Cj ??
动响应计算等于静响应计算与动荷系数之积,
[例 6 ] 结构如图,AB=DE=L,A,C 分别为 AB 和 DE 的中
点，求梁在重物 mg 的冲击下,C 面的动应力。
ABDE
A
EI
PL
EI
LR
4896
33
??
EI
PL
192
5 3?
EIEIEI DEAB ??
D
C2
C1
A1
L
② Coefficient of the dynamic load
3
3
5
384
11
192
5
2
11
PL
E I h
EI
PL
h
d
K
???
???
③ Determine the dynamic stress of section C
zz
C
dCjdCd W
PL
PL
E I h
W
M
KK
4
)
5
384
11( 3m a xm a x ????? ??
h
B A C
mg E =P
C1
A1
D
EIEIEI DEAB ??
L
C2
② 动荷系数
3
3
5
384
11
192
5
2
11
PL
E I h
EI
PL
h
d
K
???
???
③ 求 C面的动应力
zz
C
dCjdCd W
PL
PL
E I h
W
M
KK
4
)
5
384
11( 3m a xm a x ????? ??
h
B A C
mg E =P
C1
A1
D
EIEIEI DEAB ??
L
C2
45
Chapter 13 Exercises
1,A steel disc is installed on a shaft,There is a circular
hole in the disc,If the shaft and the disc rotate with a
uniform angular speed,try to determine the
maximum normal stress of the shaft,
Solution,The unsymmetry of the disc results in the normal
stress of bending in the shaft,The inertia force resulting in
bending of the shaft is
)/1(40 s??
KNP d 58.10404.08.9 104.7603.03.04 2
3
2 ???????? ?
mKNM d ????? 12.24.058.1021m a x
M P aWMd d 5.123m a x 12.0 2 1 2 032m a x ??? ????
46
第十三章 练习题
一、轴上装一钢制圆盘，盘上有一圆孔。若轴与
圆盘以匀角速度 旋转。试求轴内的最大正
应力。
解：圆盘结构上的不对称性是引起轴内弯曲正应
力的原因。引起轴弯曲的惯性力,
)/1(40 s??
KNP d 58.10404.0
8.9
104.7603.03.0
4
2
3
2 ???????? ?
mKNM d ????? 12.24.058.1021m a x
M P aWMd d 5.123m a x 12.0 2 1 2 032m a x ??? ????
47
2,The end of Rod AB at the bottom is fixed,Point C of the rod is
impacted by the weight Q that is moving at a uniform speed along
the horizontal direction,Assume E,I and W of Rod AB are known,
Try to determine the maximum impact stress of the rod,
Solution,There is no change in the strain energy under the action of
the horizontal impact,
?
ddPvg
Q ??
2
1
2
1 2
QKP dd ? jdd K ???
EIQaj 3/3??
3
22 3
gQ a
E I vg v
d jK ?? ?
2
23
m a xm a x ga W
QE I v
jdd K ?? ??
W
Qa
j ?m a x?
48
二、杆 AB下端固定，在 C点受到以匀速 沿
水平运动的重物 Q冲击。设 AB杆的 E,I及 W均为已
知。试求杆内的最大冲击应力。
解：水平冲击无势能变化
,
,
?
ddPvg
Q ??
2
1
2
1 2
QKP dd ? jdd K ???
EIQaj 3/3??
3
22 3
gQ a
E I vg v
d jK ?? ?
2
23
m a xm a x ga W
QE I v
jdd K ?? ??
W
Qa
j ?m a x?
49
3,For a round wooden stake,its diameter is d=30cm and its length is
L=1m,The end of the wooden stake at the bottom is fixed,E=10GPa,A
heavy hammer with the weight Q=5KN falls freely from the height h=1m
above the wooden stake,Try to determine the dynamic load coefficients in
the following two cases,
① Put a rubber cushion with the diameter and E=8MPa on
the top of the wooden stake,
② No rubber cushion on the top of the wooden stake,
Solution:①
②
cmd 151 ?
EAQLAEQhj // 111 ???
2623 3.01010
415
15.0108
402.05
??
??
??
?? ??
m5105.71 ???
? ? 4.5311
5105.71
02.012 ????
??
??dK
26 3.01010 415/ ??? ????? ?EAQLj
m5107 0 7.0 ???
53311 5107 0 7.0 12 ???? ???dK
50
三、直径 d=30cm，长度 L=1m的圆木桩，下端固
定，材料 E=10GPa。重为 Q=5KN的重锤从离木桩顶
为 h=1m的高度自由落下。求下列两种情况下的动荷
系数,① 木桩顶放置直径,厚度
的橡皮垫，橡皮 E=8MPa.② 无橡皮垫。
解：①
②
cmd 151 ?
EAQLAEQhj // 111 ???
2623 3.01010
415
15.0108
402.05
??
??
??
?? ??
m5105.71 ???
? ? 4.5311
5105.71
02.012 ????
??
??dK
26 3.01010 415/ ??? ????? ?EAQLj
m5107 0 7.0 ???
53311 5107 0 7.0 12 ???? ???dK
51
52
Mechanics of Materials
2
CHAPTER 13 DYNAMIC LOADS
§ 13–1 BASIC CONCEPTS
§ 13–2 DYNAMIC RESPONSE ABOUT THE PROBLEM
OF ACCELERATIVE MOTIONS
§ 13–3 DYNAMIC RESPONSE ABOUT THE PROBLEM
OF IMPACT LOADS
第十三章 动荷载
§ 13–1 基本概念
§ 13–2 加速运动问题的动响应
§ 13–3 冲击荷载问题的动响应
1,Dynamic loads,
§ 13-1 BASIC CONCEPTS
2,Dynamic responses,
The loads don’t change with time(or change very stably and slowly) and
acceleration of each member is zero or may be neglected, this kind of the loads
are called static loads,
The loads change sharply with time and the velocity of the member changes
obviously（ the member produces the inertia force）,this kind of load is loads
are called dynamic loads,
Various responses (such as stress,strain,displacement and so on)of the
member under the action of dynamic loads are called dynamic responses,
Experiment prove that Hooke’law applied validly under static loads may be
applied to the case of dynamic loads with Estatic=Edynamic as long as the stress does
not exceed the proportional limit,
一、动载荷,
载荷不随时间变化（或变化极其平稳缓慢）且使构件各部件
加速度保持为零（或可忽略不计），此类载荷为 静载荷 。
载荷随时间急剧变化且使构件的速度有显著变化（系统产生
惯性力），此类载荷为 动载荷 。
§ 13-1 基本概念
二、动响应,
构件在动载荷作用下产生的各种响应（如应力、应变、位
移等），称为 动响应 。
实验表明：在静载荷下服从虎克定律的材料，只要应力不
超过比例极限,在动载荷下虎克定律仍成立且 E静 =E动 。
3,Dynamic load coefficient,
jdd K ?? ?
r e s p o n s e St a t ic
r e s p o n s e D y n a m ic?
dK
4,Classification of the dynamic stress,
1).Simple dynamic stress,The acceleration can be determined,We may
apply the“method of kineto static” to solve this kind of problems,
3).Alternating stress,Stress changes periodically with time,It belongs to
the fatigue problem,
4).Vibration problems,There are many methods to solve this kind of problems,
Dynamic load coefficient
2).Impact load,The velocity changes sharply in a very short time,In this
case the acceleration can not be determined,We must apply the,energy
method” to solve this kind of problems；
三、动荷系数,
jdd K ?? ?
静响应
动响应动荷系数 ?
dK
四、动应力分类,
1.简单动应力,加速度可以确定，采用, 动静法, 求解
。 2.冲击载荷,速度在极短暂的时间内有急剧改变，此时，加
速度不能确定，要采用, 能量法, 求解；
3.交变应力,应力随时间作周期性变化，属疲劳问题。
4.振动问题,求解方法很多。
§ 13-2 DYNAMIC RESPONSE OF THE PROBLEM OF
ACCELERATIVE MOTIONS
Principle of the method,D’Alembert’s principle
( Method of kineto statics ）
D’Alembert’s principle think there is inertial force on the body in un-
equilibrium,The direction of the inertial force is opposite to the acceleration of
the body and the magnitude of the inertial force is the product of the mass and
the acceleration of the body,After the inertial force is applied on the body the
dynamic problem may be dealt with the static problem in form,which is called
the method of kineto statics,
§ 13-2 加速运动问题的动响应
方法原理,D’Alembert’s principle
( 动静法 ）
达朗伯原理认为：处于不平衡状态的物体，存在惯性
力，惯性力的方向与加速度方向相反，惯性力的数值等于
加速度与质量的乘积。只要在物体上加上惯性力，就可以
把动力学问题在形式上作为静力学问题来处理，这就是动
静法。
a
g
Aq
G
??
)1()(
g
aAxxqqN
Gjd ???? ?
)1(
g
ax
A
N d
d ??? ??
Solution：① Analysis of the forces is
shown in the figure,
② Dynamic stress
1,Dynamic stress of the body in the straight-line motion
L
x
m n
a
x
a
Nd
qj
qG
Inertial force
Example 1 The effective area of the steel wire rope in a crane is A,[?] is
known, Weight per unit volume of the body is ? and the body moves up at the
acceleration a,Try to check the strength of the rope（ neglect the weight of the
rope）
a
g
A
q G
?
?惯性力：
)1()(
g
aAxxqqN
Gjd ???? ?
)1(
g
ax
A
N d
d ??? ??
[例 1 ] 起重机钢丝绳的有效横截面面积为 A,已知 [?],单位
体积重为 ?,以加速度 a上升，试校核钢丝绳的强度（不计绳
重）。 解：① 受力分析如图,
② 动应力
一、直线运动构件的动应力
L
x
m n
a
x
a
Nd
qj
qG
m a xm a x )1( jdd Kg
a
L ??? ???
g
a
K d ?? 1
Coefficient of the dynamic load,
? ???? ?? m a xm a x jdd K
Strength condition,
If, ? ?
,m a x ?? ?d
? ?,m a x ?? ?d
then the strength condition is satisfied,
then the strength condition is not satisfied,If,
m a xm a x )1( jdd Kg
a
L ??? ???
g
a
K d ?? 1
动荷系数,
? ???? ?? m a xm a x jdd K强度条件,
若,? ?
?? ?m a xd
? ??? ?m a xd
满足
不满足
)1)(( gaqLGN d ???
)
8.9
21)(605.251050(
109.2
1 3
4 ?????? ?
? ? M P a3 0 0M P a2 1 4 ??? ?
Example 2 Length of the steel rope of a crane is 60m,its nominal diameter is
28cm,its effective cross-section area is A=2,9cm2,weight per unit length is q=25,
5N/m and [?] =300MPa,The acceleration when the rope lifts an object of 50kN is
a=2m/s2,Try to check the strength of the steel rope,
G(1+a/g)
Nd
L q(1+a/g)
)1)((1 gaqLGAAN dd ?????
Solution：① The free body
diagram is shown in the figure,
② Dynamic stress
)1)(( gaqLGN d ???
)
8.9
21)(605.251050(
109.2
1 3
4 ?????? ?
? ? M P a3 0 0M P a2 1 4 ??? ?
[例 2 ] 起重机钢丝绳长 60m，名义直径 28cm，有效横截面面积
A=2,9cm2,单位长度重量 q=25,5N/m,[?] =300MPa,以 a=2m/s2的
加速度提起重 50kN 的物体，试校核钢丝绳的强度。
G(1+a/g)
Nd
L q(1+a/g)
)1)((1 gaqLGAAN dd ?????
解：① 受力分析如图,
② 动应力
gLGRmmaG nG /22 ?? ???
? ??? ?? AG G /
? ? ? ?)(
2
?
?
? g
GLGA G ??
② Strength condition
Solution,① The free body diagram is
shown in the figure,
?
GG
L
O
2,Dynamic stress of the rotating member,
The inertia force is
Example 3 A ball of weight G is install on one end of the rotating arm
with length L,They are rotating about point O at an equal angular speed in the
smooth plane,Knowing permissible stress is [?],Determine the cross-section
areaof the rotating arm（ neglecting the weight of the rotating arm）,
gLGRmmaG nG /22 ?? ???
惯性力：
? ??? ?? AG G /
? ? ? ?)(
2
?
?
? g
GLGA G ??
[例 3 ] 重为 G的球装在长 L的转臂端部，以等角速度在光滑水
平面上绕 O点旋转,已知许用应力 [?],求转臂的截面面积
（不计转臂自重）。
② 强度条件
解：① 受力分析如图,
?
GG
L
O
二、转动构件的动应力,
Fig.1
qG
② Analysis of the internal force
is shown in Fig.2
g
AD
g
Aaq n
G 2
2???
??
02 ?? DqN Gd
2
2
42
??
g
ADDqN G
d ??
2
2?D
a n ?
Solution,① Analysis of the
inertia forceis shown in Fig.1
O
D
t
Fig.2
qG
NG NG
Example 4 Suppose the average diameter of a ring is D,its thickness is t
with t<<D.Its cross-section area is A and its weight per unit volume is ?,As
shown in the figure,the ring rotates at a constant angular speed ? about the axis
that is through the center of the circle and perpendicular to the plane where the
ring lies.Try to determine the dynamic stress of the ring and establish the
strength condition,
?
图 1
qG
[例 4] 设圆环的平均直径 D、厚度 t,且 t?D，环的横截面面积
为 A，单位体积重量为 ?,圆环绕过圆心且垂直于圆环平面的轴
以等角速度 ?旋转，如图所示,试确定圆环的动应力，并建立强
度条件。
② 内 力分析如图 2
g
AD
g
Aaq n
G 2
2???
??
02 ?? DqN Gd
2
2
42
??
g
ADDqN G
d ??
2
2?D
a n ?
解：① 惯性力分析，见图１
O
D
t
图 2
qG
NG NG
?
22
2
4
?????
gg
D
A
N d
d ???
? ????? ??
gd
2
?
?
?
g ][
??
③ Analysis of stress
④ Strength condition
Maximum linear speed,
?
?
?
g ][
m a x ?
22
2
4
?????
gg
D
A
N d
d ???
? ????? ??
gd
2
?
?
?
g ][
??
③ 应力分析
④ 强度条件
最大线速度,
?
?
?
g ][
m a x ?
§ 13-3 DYNAMIC RESPONSE ABOUT THE PROBLEM OF IMPACT LOADS
Principle of the method,Energy method
( Conservation of the mechanical energy ）
In the touching area between the impact and impacted members the
stressed state of each point is very complex and the lasting time of impact is
very short,therefore,it is difficult to analyze the variety of the touching force
with time accurately, In engineering this kind of impact problems are usually
solved by the energy method,That is to say,on the basis of some assumptions
we can do some conservative simplifications for the stress and deformation of
the impacted member according to the conservation law of energy,
§ 13-3 冲击荷载问题的动响应
方法原理：能量法
( 机械能守恒 ）
在冲击物与受冲构件的接触区域内，应力状态异常复杂，
且冲击持续时间非常短促，接触力随时间的变化难以准确分
析。工程中通常采用能量法来解决冲击问题，即在若干假设
的基础上，根据能量守恒定律对受冲击构件的应力与变形进
行偏于安全的简化计算。
2.Kinetic energy T, potential energy V, strain energy U,
The energy before and after impact is conservative,
Maximum impact effect,the kinetic energy after impact is zero,T2= 0
The strain energy for an impact force is U2=(1/2)PdΔd
1,Assumptions,
① The impact body is rigid；
② The impact body do not rebound；
③ The loss of energy of sound,light,heat etc.in the process of impact
is negleced （ energy is conservative）；
④ The process of impact shows linear-elastic deformations is linearly
elastic.(conservative calculation)
111 UVT ?? 222 UVT ???
(Before impact) (After impact)
① 冲击物为刚体；
②冲击物不反弹；
③ 不计冲击过程中的声、光、热等能量损耗（能量守恒）；
④ 冲击过程为线弹性变形过程。 (保守计算 )
111 )( UVT ??冲击前 )( 222 冲击后UVT ???
2.动能 T,势能 V,变形能 U，冲击前、后，能量守恒,
最大冲击效应：冲击后的动能为零,T2=0
一个冲击力的变形能为 U2=(1/2)PdΔ d
1.假设,
3,Coefficient of the dynamic load is Kd,
jdd
jdd
jdd
K
K
PKP
?? ?
???
?
3.动荷系数为 Kd,
jdd
jdd
jdd
K
K
PKP
?? ?
???
?
The energy before and after impact is conservative and
j
2
jd
2
2
)(
2
1 ?????
dK
mgKhmgm ?
j
d
hg
K
?
?
???
2/
11
2?
1,Impact problem about the free falling body along the axis
jddjjdd KmgPPKP ????? ),(
△ j:Static displacement at the falling
point of the impact body,
?d mg
v mg
h
0
2/
1
1
2
1
?
?
?
U
m g hV
mvT
Before impact,
2/
0
2
2
2
dd
d
PU
mgV
T
??
???
?
After impact,
Kinetic energy
Potential energy
Strain energy
0
2/
1
1
2
1
?
?
?
U
m g hV
mvT
变形能
势能
动能
冲击前后能量守恒，且
j
2
jd
2
2
)(
2
1 ?????
dK
mgKhmgm ?
j
d
hg
K
?
?
???
2/
11
2?
一、轴向自由落体冲击问题
冲击前,
2/
0
2
2
2
dd
d
PU
mgV
T
??
???
?
变形能
势能
动能
冲击后,
jdd
jjdd
K
mgPPKP
???
?? )(
△ j:冲击物落点的静位移。
?d mg
v mg
h
Discussions,
j
h
dK ?????
2
11:,0)1( ?
2,0 ?? dKh
(2)Sudden loads
讨论,
j
h
dK ?????
2
11:,0)1( ?
2,0 ?? dKh(2)突加荷载
jdK
mgmv ?? 22
22
1
2,Axial impact of neglecting weight,
0
0
2/
1
1
2
1
?
?
?
U
V
mvT
Before impact,
2/
0
0
2
2
2
ddPU
V
T
??
?
?
After impact,
The energy before and after
impact is conservative and
jdd
jjdd
K
mgPPKP
???
?? )(
Coefficient of
The dynamic load
j
d gK ??
2?
Kinetic energy
Potential energy
Strain energy
Kinetic energy
Potential energy
Strain energy
mg
?
jdK
mgmv ?? 22
22
1
二、不计重力的轴向冲击,
0
0
2/
1
1
2
1
?
?
?
U
V
mvT
变形能
势能
动能
冲击前,
2/
0
0
2
2
2
ddPU
V
T
??
?
?
变形能
势能
动能
冲击后,冲击前后能量守恒，且
jdd
jjdd
K
mgPPKP
???
?? )(
动荷系数
j
d gK ??
2?
mg
?
3,Calculation of the impact response
② Coefficient of the dynamic
load
③ Determine the dynamic stress
Solution：① Determine
the static deformation
9.217
425
1000211211 ?????
?
???
j
h
dK
mm425???? EAWLEA LP jj
M P a41.15?? jdd K ??
Static stress,M P a0 7 0 7 4.0/ ?? AW
j?
Dynamic stress,
h=1m
v W
f 6m
The impact response is equal to the product of the static response and the
coefficient of the dynamic load,
Example 5 The wooden stake with the diameter of 0.3m is
impacted by a free falling hammer,Weight of the falling
hammer is 5kN,Determine the maximum dynamic stress of
the wooden stake,E=10GPa
三、冲击响应计算
② 动荷系数
③ 求动应力
解：① 求静变形
9.217
425
1000211211 ?????
?
???
j
h
dK
mm425????
EA
WL
EA
LP j
j
M P a41.15?? jdd K ??
等于静响应 与 动荷系数之积,
[例 5 ] 直径 0.3m的木桩受自由落锤冲击，落锤重 5kN,
求：桩的最大动应力。 E=10GPa
静应力,
M P a0 7 0 7 4.0/ ?? AWj?
动应力,
h=1m
v W
f 6m
4,Impact problem of the beam
0)(
2
1 2
111
???
???
dfhmgmv
UVT
mg
L
h A B C
A B C x
f
fd
1).Assumptions,
① Impact bodies are rigid ones；
② The potential energy of weight and kinetic
energy of the impacted bodies is neglected ；
③ Impact bodies do not rebound；
④ The loss of energy of sound,light,heat
and so on(energy is conservative)is neglected,
Before impact
四,梁的冲击问题 1.假设,
① 冲击物为刚体；
② 不计被冲击物的重力势能和动能；
③ 冲击物不反弹；
④ 不计声、光、热等能量损耗（能
量守恒）。
0)(
2
1
冲 击击
2
111
???
???
dfhmgm
UVT
?
mg
L
h A B C
A B C x
f
fd
A B C x
f
fd
The energy before and after impact
is conservative,therefore,
22 )(
2
)(
2
1
d
j
d ff
mgfhmgmv ???
jdj
j
fKf
f
hgf ????? )2)(11( 2
d
?
jj
d
d f
hg
f
f
K
2)2(
11
?
????
?
jf
h
dK
211 ??? 2?dK
Coefficient of
the dynamic
load
(2)Sudden load,(1)Free falling body
22
)(
2
1
)(
2
1
2
1
00
d
j
d
j
j
dd
f
f
mg
f
f
P
fP
??
???
After impact
222 UVT ??
22
222
)(
2
1
)(
2
1
2
1
00
冲 击击
d
j
d
j
j
dd
f
f
mg
f
f
P
fP
UVT
??
???
??
冲击前、后，能量守恒，所以,
A B C x
f
fd
22 )(
2
)(
2
1
d
j
d ff
mgfhmgmv ???
jdj
j
fKf
f
hgf ????? )2)(11( 2
d
?
jj
d
d f
hg
f
f
K
2)2(
11:
?
????
?
动荷系数
jf
h
dK
211:)1( ???自由落体 2:)2( ?
dK突加荷载
h
B A C
mg E =P
5,Calculation of the dynamic response,
Solution：① Determine
the static deflection of point
C
21
1
2 CC
AAf
Cj ??
The dynamic response is equal to the product of the static response and the
coefficient of the dynamic load,
ABDE
A
EI
PL
EI
LR
4896
33
?? EI
PL
192
5 3?EIEIEI
DEAB ??
D
C2
C1
A1
L
Example 6 A structure is
shown in the figure,AB=DE=L，
points A and C are separately the
middle points of AB and DE,
Determine the dynamic stress of
section C under the impact of
weight mg,
h
B A C
mg E =P
五、动响应计算,
解：① 求 C点静挠度
21
1
2 CC
AAf
Cj ??
动响应计算等于静响应计算与动荷系数之积,
[例 6 ] 结构如图,AB=DE=L,A,C 分别为 AB 和 DE 的中
点，求梁在重物 mg 的冲击下,C 面的动应力。
ABDE
A
EI
PL
EI
LR
4896
33
??
EI
PL
192
5 3?
EIEIEI DEAB ??
D
C2
C1
A1
L
② Coefficient of the dynamic load
3
3
5
384
11
192
5
2
11
PL
E I h
EI
PL
h
d
K
???
???
③ Determine the dynamic stress of section C
zz
C
dCjdCd W
PL
PL
E I h
W
M
KK
4
)
5
384
11( 3m a xm a x ????? ??
h
B A C
mg E =P
C1
A1
D
EIEIEI DEAB ??
L
C2
② 动荷系数
3
3
5
384
11
192
5
2
11
PL
E I h
EI
PL
h
d
K
???
???
③ 求 C面的动应力
zz
C
dCjdCd W
PL
PL
E I h
W
M
KK
4
)
5
384
11( 3m a xm a x ????? ??
h
B A C
mg E =P
C1
A1
D
EIEIEI DEAB ??
L
C2
45
Chapter 13 Exercises
1,A steel disc is installed on a shaft,There is a circular
hole in the disc,If the shaft and the disc rotate with a
uniform angular speed,try to determine the
maximum normal stress of the shaft,
Solution,The unsymmetry of the disc results in the normal
stress of bending in the shaft,The inertia force resulting in
bending of the shaft is
)/1(40 s??
KNP d 58.10404.08.9 104.7603.03.04 2
3
2 ???????? ?
mKNM d ????? 12.24.058.1021m a x
M P aWMd d 5.123m a x 12.0 2 1 2 032m a x ??? ????
46
第十三章 练习题
一、轴上装一钢制圆盘，盘上有一圆孔。若轴与
圆盘以匀角速度 旋转。试求轴内的最大正
应力。
解：圆盘结构上的不对称性是引起轴内弯曲正应
力的原因。引起轴弯曲的惯性力,
)/1(40 s??
KNP d 58.10404.0
8.9
104.7603.03.0
4
2
3
2 ???????? ?
mKNM d ????? 12.24.058.1021m a x
M P aWMd d 5.123m a x 12.0 2 1 2 032m a x ??? ????
47
2,The end of Rod AB at the bottom is fixed,Point C of the rod is
impacted by the weight Q that is moving at a uniform speed along
the horizontal direction,Assume E,I and W of Rod AB are known,
Try to determine the maximum impact stress of the rod,
Solution,There is no change in the strain energy under the action of
the horizontal impact,
?
ddPvg
Q ??
2
1
2
1 2
QKP dd ? jdd K ???
EIQaj 3/3??
3
22 3
gQ a
E I vg v
d jK ?? ?
2
23
m a xm a x ga W
QE I v
jdd K ?? ??
W
Qa
j ?m a x?
48
二、杆 AB下端固定，在 C点受到以匀速 沿
水平运动的重物 Q冲击。设 AB杆的 E,I及 W均为已
知。试求杆内的最大冲击应力。
解：水平冲击无势能变化
,
,
?
ddPvg
Q ??
2
1
2
1 2
QKP dd ? jdd K ???
EIQaj 3/3??
3
22 3
gQ a
E I vg v
d jK ?? ?
2
23
m a xm a x ga W
QE I v
jdd K ?? ??
W
Qa
j ?m a x?
49
3,For a round wooden stake,its diameter is d=30cm and its length is
L=1m,The end of the wooden stake at the bottom is fixed,E=10GPa,A
heavy hammer with the weight Q=5KN falls freely from the height h=1m
above the wooden stake,Try to determine the dynamic load coefficients in
the following two cases,
① Put a rubber cushion with the diameter and E=8MPa on
the top of the wooden stake,
② No rubber cushion on the top of the wooden stake,
Solution:①
②
cmd 151 ?
EAQLAEQhj // 111 ???
2623 3.01010
415
15.0108
402.05
??
??
??
?? ??
m5105.71 ???
? ? 4.5311
5105.71
02.012 ????
??
??dK
26 3.01010 415/ ??? ????? ?EAQLj
m5107 0 7.0 ???
53311 5107 0 7.0 12 ???? ???dK
50
三、直径 d=30cm，长度 L=1m的圆木桩，下端固
定，材料 E=10GPa。重为 Q=5KN的重锤从离木桩顶
为 h=1m的高度自由落下。求下列两种情况下的动荷
系数,① 木桩顶放置直径,厚度
的橡皮垫，橡皮 E=8MPa.② 无橡皮垫。
解：①
②
cmd 151 ?
EAQLAEQhj // 111 ???
2623 3.01010
415
15.0108
402.05
??
??
??
?? ??
m5105.71 ???
? ? 4.5311
5105.71
02.012 ????
??
??dK
26 3.01010 415/ ??? ????? ?EAQLj
m5107 0 7.0 ???
53311 5107 0 7.0 12 ???? ???dK
51
52
