1
Elasticity
3
Summarize
§ 1-1 Movement differential equations of elastic objects
§ 11-2 Without rotating wave and equal volume wave
§ 11-3 Transverse wave and vertical wave
§ 11-4 Spherical wave
Chapter 11 Elastic Wave
4
概述
§ 1-1 弹性体的运动微分方程
§ 11-2 无旋波与等容波
§ 11-3 横波与纵波
§ 11-4 球面波
第十一章 弹性波
5
Summarize,
When elastic object bears loads in static force equilibrium
conditions,not all the parts of object has displacement,
distortion and stress,At the beginning of the loads,the parts
which are more far from the loads have no impacts, After
then,the displacement,distortion and stress caused by loads
transmit to other places in a finite speed of wave,This wave is
called elastic wave,
This chapter will first give movement differential equations
of elastic objects,then introduce some conceptions of elastic
wave and simplify the equations according to different elastic
waves,at last give the speed transmitting formulas of wave in
infinite elastic objects,
6
概述
当静力平衡状态下的弹性体受到荷载作用时，并不是在
弹性体的所有各部分都立即引起位移、形变和应力。在作用
开始时，距荷载作用处较远的部分仍保持不受干扰。在作用
开始后，荷载所引起的位移、形变和应力，就以波动的形式
用有限大的速度向别处传播。这种波动就称为 弹性波 。
本章将首先给出描述弹性体运动的基本微分方程，然后
介绍弹性波的几个概念，针对不同的弹性波，对运动微分方
程进行简化，最后给出波在无限大弹性体中传播速度公式。
7
§ 11-1 Movement differential
equations of elastic objects
The two assumptions are equal to the basic assumptions when we discuss
static force questions,So the physic and geometry equations and elastic
equations where stress component is expressed by displacement
component,still are the same with movement equations at any
instantaneous time,The only difference is that the equilibrium differential
equations of static questions must be substituted by movement differential
equations,
This chapter we still adopt the assumptions,
（ 1） Elastic objects are ideal elastic objects,
（ 2） The displacement and distortion are tinny,
8
§ 11-1 弹性体的运动微分方程
上述两条假设，完全等同于讨论静力问题的基本假
设。因此，在静力问题中给出的物理方程和几何方程，
以及把应力分量用位移分量表示的弹性方程，仍然适用
于讨论动力问题的任一瞬时，所不同的仅仅在于，静力
问题中的平衡微分方程必须用运动微分方程来代替。
本章仍然采用如下假设,
（ 1） 弹性体为理想弹性体。
（ 2） 假定位移和形变都是微小的。
9
Toward any tiny object,when we apply d’Alembert
theory,we must consider stress,body force and the
inertia force of elastic objects caused by acceleration,
In space right-angle coordinate system,the x,y,z
directions component of inertia force of every unite
volume are,
Where ρ is the density of elastic objects,
2
2
t
u
?
?? ?
2
2
t?
?? ??
2
2
t
w
?
?? ?
10
对于任取的微元体，运用达朗伯尔原理，除了
考虑应力和体力以外，还须考虑弹性体由于具有加
速度而产生的惯性力。每单位体积上的惯性力在空
间直角坐标系的 x,y,z方向的分量分别为,
其中 ρ为弹性体的密度。
2
2
t
u
?
?? ?
2
2
t?
?? ??
2
2
t
w
?
?? ?
11
Because of the equilibrium relations,we simplify them and get,
The above formulas are called movement differential equations
of elastic objects,They and geometry equations and physic
equations are the basic equations of movement questions of
elasticity mechanics
02
2
????????????? t uXzyx zxyxx ????
02
2
????????????? tYxzy xyzyy ?????
02
2
????????????? t wZyxz yzxzz ????
12
由平衡关系，并简化后得,
上式称为 弹性体的运动微分方程 。它同几何方程和物理方程
一起构成弹性力学动力问题的基本方程。
02
2
????????????? t uXzyx zxyxx ????
02
2
????????????? tYxzy xyzyy ?????
02
2
????????????? t wZyxz yzxzz ????
13
Note 1,geometry equations,
x
u
x ?
???
yy ?
?? ??
z
w
z ?
???
zy
w
yz ?
??
?
?? ??
x
w
z
u
zx ?
??
?
???
y
u
xxy ?
??
?
?? ??
14
注 1：几何方程
x
u
x ?
???
yy ?
?? ??
z
w
z ?
???
zy
w
yz ?
??
?
?? ??
x
w
z
u
zx ?
??
?
???
y
u
xxy ?
??
?
?? ??
15
Note2,physic equations
)]([1 zyxx E ????? ???
)]([1 xzyy E ????? ???
)]([1 yxzz E ????? ???
yzyz E ?
?? )1(2 ??
zxzx E ?
?? )1(2 ??
xyxy E ?
?? )1(2 ??
16
注 2：物理方程
)]([1 zyxx E ????? ???
)]([1 xzyy E ????? ???
)]([1 yxzz E ????? ???
yzyz E ?
?? )1(2 ??
zxzx E ?
?? )1(2 ??
xyxy E ?
?? )1(2 ??
17
Because the displacement component is difficult to be expressed by stress
and its derivative,so movement equations of elasticity mechanics are
usually solved according to the displacement,Substitute the elasticity
equations where stress components are expressed by displacement
component into movement differential equations,and we let,
Then we get,zwyxue ????????? ?
0)21 1()1(2 2
2
2 ?
?
?????
?
?
?? t
uXu
x
eE ?
??
0)21 1()1(2 2
2
2 ?
?
?????
?
?
?? tYy
eE ???
??
0)21 1()1(2 2
2
2 ?
?
?????
?
?
?? t
wZw
z
eE ?
??
18
由于位移分量很难用应力及其导数来表示，所以弹
性力学动力问题通常要按位移求解。将应力分量用位移
分量表示的弹性方程代入运动微分方程，并令,
得,z
w
yx
ue
?
??
?
??
?
?? ?
0)21 1()1(2 2
2
2 ?
?
?????
?
?
?? t
uXu
x
eE ?
??
0)21 1()1(2 2
2
2 ?
?
?????
?
?
?? tYy
eE ???
??
0)21 1()1(2 2
2
2 ?
?
?????
?
?
?? t
wZw
z
eE ?
??
19
These are the basic differential equations of movement equations
solved by displacement,They are also called Lame equations,
We need boundary conditions to solve Lame equations,Besides these we
still need original conditions,because displacement components are the
function of time variable,
In order to simplify calculation,usually we neglect body force,Now the
movement differential equations of elastic objects can be simplified as,
)21 1()1(2 22
2
uxeEt u ????????? ???
)21 1()1(2 22
2
????? ????????? yeEt
)21 1()1(2 22
2
wzeEt ????????? ????
20
这就是按位移求解动力问题的基本微分方程，也称
为 拉密（ Lame)方程 。
要求解拉密方程，显然需要边界条件。除此之外，
由于位移分量还是时间变量的函数，因此求解动力问题
还要给出初始条件。
为求解上的简便，通常不计体力，此时弹性体的运
动微分方程简化为,
)21 1()1(2 22
2
uxeEt u ????????? ???
)21 1()1(2 22
2
????? ????????? yeEt
)21 1()1(2 22
2
wzeEt ????????? ????
21
§ 11-2 Without rotating
Wave and equal volume wave
1,Without rotating waves
Without rotating wave means that in elastic objects,the distortion
caused by waves is not rotating, That means rotating values of
three vertical coordinates at any point in the elastic object are zero,
xu ?
?? ?
y?
?? ?? zw ??? ?
Where is potential function of displacement,
This displacement is called without rotating displacement,and the
elastic wave corresponds to the displacement are called without
rotating wave,
),,,( tzyx?? ?
Suppose the displacement of elastic objects can be
expressed by,
u,v,w
22
§ 11-2 无旋波与等容波
一、无旋波
所谓 无旋波 是指 在弹性体中，波动所产生的变形不存在旋
转，即弹性体在任一点对三个垂直坐标轴的旋转量皆为零。
假定弹性体的位移 u,v,w可以表示成为,
xu ?
?? ?
y?
?? ?? zw ??? ?
其中 是位移的势函数。这种位移称为无旋位
移。而相应于这种位移状态的弹性波就称 无旋波 。
),,,( tzyx?? ?
23
[proving],at any point of elastic object,the rotating value of z
axis is,
So the rotating values of the three coordinates at any point of the
elastic object are zero,
y
u
xz ?
??
?
?? ??
substitute into the formula,we can
get,xu ??? ? y??? ??
0?z?
Similarly,
0?x? 0?y?
24
[证 ],在弹性体的任一点处，该点对 z 轴的旋转量
即弹性体的任一点对三个坐标的旋转量都等于零。
y
u
xz ?
??
?
?? ??
将 代入，可得,
xu ?
?? ?
y?
?? ??
0?z?
同理
0?x? 0?y?
25
In the condition of without rotating displacement,
?? 2??????????? zwyxue
so
uxxxe 222 ???????????? ??
similarly,
?2????ye
wze 2????
Substitute the above three formulas into movement differential
equations without including body force,simplify it we can get
wave movement equations of without rotating wave,
26
在无旋位移状态下
?? 2??????????? zwyxue
从而
uxxxe 222 ???????????? ??
同理
?2????ye
wze 2????
将上三式代入不计体力的运动微分方程，并简化后
得无旋波的波动方程
27
uct u 2212
2
????
?? 2212
2
???? ct
wct w 2212
2
????
???
?
)21)(1(
)1(
1 ??
?? EcWhere,
1c
is the transmitting speed of without rotating
wave in infinite elastic objects,
28
uct u 2212
2
????
?? 2212
2
???? ct
wct w 2212
2
????
???
?
)21)(1(
)1(
1 ??
?? Ec其中
1c
就是无旋波在无限大弹性体中的传播速度
29
Equal volume wave means that in the distortion caused by waves in
elastic objects,the volume strain is zero,That means the volume of
elastic object remains unchanged,
2,Equal volume wave
Suppose the displacement u,v,w of elastic objects satisfy the
condition that the volume strain is zero,
0?????????? zwyxue ?
This displacement is called equal volume displacement and the
elastic wave corresponds to this displacement is called equal
volume wave,
30
所谓 等容波 是指 在弹性体内，波动所产生的变形中体积应
变为零 。即弹性体中任一部分的容积（即体积）保持不变 。
二、等容波
假定弹性体的位移 u,v,w满足体积应变为零的条件，即,
0?????????? zwyxue ?
这种位移称为 等容位移 。而相应于这种位移状态的弹性
波就是等容波。
31
Because, so simplify movement differential equations
without including body force we can get wave movement equations
of equal volume wave,
0?e
uct u 2222
2
????
?? 2222
2
???? ct
wct w 2222
2
????
where
?? )1(22 ??
Ec
is the transmitting speed of equal volume waves
in infinite elastic objects,
2c
32
由于,故不计体力的运动微分方程，简化后得等
容波的波动方程,
0?e
uct u 2222
2
????
?? 2222
2
???? ct
wct w 2222
2
????
其中
?? )1(22 ??
Ec
就是等容波在无限大弹性体中的传播速度。
2c
33
According to without rotating wave and equal volume
wave,we give the conclusion without proving that,in elastic
objects,stress,strain and speed of particle transmit in the
same way and the same speed as displacement,
34
对于无旋波和等容波，我们不加证明地给出如下 结论,
在弹性体中，形变、应力以及质点速度，都将和位移以相
同的方式与速度进行传播。
35
1,vertical wave
[definition] The particle movement direction of elastic objects is
parallel to the transmitting direction of elastic wave,（ seen in the
Fig.）
Transmitting format of
vertical wave
§ 11-3 Vertical wave and transverse wave
36
一、纵波
[定义 ] 弹性体的质点运动方向平行弹性波的传播方向 （图示）
纵波的传播形式
§ 11-3 纵波与横波
37
Let x axis be the transmitting direction of waves,then the
displacement components of any point in elastic objects are,
),( txuu ? 0?? 0?w
So,
x
ue
?
??
And,
2
2
x
u
x
e
?
??
?
?
0???ye 0???ze
2
2
2
x
uu
?
??? 02 ??? 02 ?? w
38
将 x轴取为波的传播方向，则弹性体内任取一点的位移
分量都有,
),( txuu ? 0?? 0?w
从而
x
ue
?
??
而
2
2
x
u
x
e
?
??
?
?
0???ye 0???ze
2
2
2
x
uu
?
??? 02 ??? 02 ?? w
39
Substitute them into movement differential equations without
including body force,we can find that the second and third
formulas are identical equations,So the first formula can be
simplified as,
2
2
2
12
2
x
uc
t
u
?
??
?
?
where
???
?
)21)(1(
)1(
1 ??
?? Ec
is transmitting speed of vertical wave in elastic objects,
1c
It is obvious that transmitting speed of vertical waves is the same
as without rotating waves’, In fact vertical wave is a kind of
without rotating wave,
40
代入不计体力的运动微分方程，可见其第二、第三式成为
恒等式，而第一式简化为,
2
2
2
12
2
x
uc
t
u
?
??
?
?
其中
???
?
)21)(1(
)1(
1 ??
?? Ec
为纵波在弹性体中的传播速度。
1c
显然纵波的传播速度与无旋波相同。事实上，纵波就是
一种无旋波。
41
The general solution of wave movement equations of vertical
wave is,)()(),(
1211 tcxftcxftxu ????
The physic meaning of the general solution is,let us take the
first item for example,function at a fixed time is
function of x,and can be expressed by curve abc in Fig.(a)
(suppose the form is like this ),After,the function
becomes,
)( 11 tcxf ?
t?
)( 111 tctcxf ???
If let, then function becomes,
Its format is similar to the original function,
From the Fig we can see the only difference is transverse
coordinate moves in level, So expresses
the wave whose speed is along x positive direction,
tcxx ??? 11 )( 111 tcxf ?
)( 11 tcxf ?
tc?1 )( 11 tcxf ?
1c
42
纵波波动方程的通解是,
)()(),( 1211 tcxftcxftxu ????
该通解的 物理意义,以其第一项为例，函数 在某
一个固定时刻将是 x的函数，可以用图 (a)中的曲线 abc表示
（假设是这种形状），在 时间之后，函数变为,
)( 11 tcxf ?
t?
)( 111 tctcxf ???
如果令,则函数可写为,其形式
同原函数 完全类同，只是横坐标发生平移
tcxx ??? 11 )( 111 tcxf ?
)( 11 tcxf ? tc?1
见图。因此 表示以速度 向 x轴正向传播的波。 )(
11 tcxf ? 1c
43
Similarly expresses the wave whose speed is
along x negative direction, The general solution expresses two
waves transmitting in opposite directions (in Fig.b),And the
speed is the modulus of wave movement equations,
1c)( 12 tcxf ?
1c
1f
c a
b
x
(a)
(b)
tc?1
tc?1 tc?1
44
同理,表示以同样速度 向 x轴负向传播的波。
整个通解表示朝相反两个方向传播的两个波（如图 b），其
传播速度为波动方程的系数 。
1c)( 12 tcxf ?
1c
1f
c a
b
x
(a)
(b)
tc?1
tc?1 tc?1
45
2,Transverse wave
[definition] The partial movement direction of elastic object is
vertical to the transmitting direction of elastic wave,
Transmitting format of
transverse wave
46
二、横波
[定义 ] 弹性体的质点运动方向垂直于弹性波的传播方向。
横波的传播形式
47
Still we let x axis be the transmitting direction of wave and y axis
be displacement direction of partial,Then the displacement
components of any point in the elastic object have,
0?u ),( tx?? ? 0?w
so 0?e
and 02 ?? u
2
2
2
x?
??? ?? 02 ?? w
Substitute them into movement differential equations without
including body force,we can find that the first and third formulas
are identical equations,So the second formula can be simplified
as,
2
2
2
22
2
xct ?
??
?
? ??
?? )1(22 ??
Ec
2c
is transmitting speed of transverse wave in elastic object,
Because volume strain of transverse wave
48
仍然将 x轴放在波的传播方向,y轴为质点位移方向，则
弹性体内任取一点的位移分量都有
0?u ),( tx?? ? 0?w
从而 0?e
而 02 ?? u
2
2
2
x?
??? ?? 02 ?? w
代入不计体力的运动微分方程，可见其第一、第三式成为恒
等式，第二式简化为,
2
2
2
22
2
xct ?
??
?
? ??
?? )1(22 ??
Ec
2c
为横波在弹性体中的传播速度。由于横波的体积应变
49
General solution to wave movement equations of
transverse wave is,
, so transverse wave is
equal volume wave 0?e
)()(),( 2221 tcxftcxftx ?????
It is obvious that the general solution expresses two waves
transmitting in opposite directions, Its displacement is along y
direction and the transmitting direction is along x direction, The
transmitting speed is a constant,
2c
50
横波的波动方程的通解为,
,故横波为等容波。 0?e
)()(),( 2221 tcxftcxftx ?????
显然，整个通解表示朝相反两个方向传播的两个波，它的
位移沿着 y方向，而传播方向是沿着 x方向，传播速度等于常
量 。
2c
51
§ 11-4 Spherical wave
If elastic objects have spherical hole or spherical out surface
,when the spherical hole or spherical out surface bears spherical
force,the elastic wave transmitting from the hole to the outside or
form the surface to the inside is called spherical wave,
Spherical wave is symmetrical of sphere,Apply basic
differential equations of sphere symmetry,we get,
0)22()21)(1( )1( 222
2
?????? ? rrrr kurdrdurdr udE ?? ?
Now, If we ignore body force,and use radial
inertia force to substitute
),( truu rr ?
2
2
t
u r
?
?? ? rk
52
§ 11-4 球面波
如果弹性体具有圆球形的孔洞或具有圆球形的外表面，
则在圆球形孔洞或圆球形外表面上受到球对称的动力作用时
，由孔洞向外传播或由外表面向内传播的弹性波，称为 球面
波 。
球面波是球对称的。利用球对称的基本微分方程,
0)22()21)(1( )1( 222
2
?????? ? rrrr kurdrdurdr udE ?? ?
此时，,而不计体力时，用径向惯性力 ),( truu
rr ?
2
2
t
u r
?
?? ? 代替,
rk
53
Then the above formula can be simplified as,
We get,
0)22(
)21)(1(
)1(
2
2
22
2
?
?
???
?
??
?
?
??
?
t
u
r
u
r
u
rr
uE rrrr ?
??
?
let,
???
?
)21)(1(
)1(
1 ??
?? Ec
Suppose
ru r ?
?? ?
then is potential function of displacement,
Substitute it into formula (a),we get,
),( tr?? ?
0122 2
2
2
1
2
2
?????????? tucr ururru rr rrr
（ a）
54
则上式简写成
即得,
0)22(
)21)(1(
)1(
2
2
22
2
?
?
???
?
??
?
?
??
?
t
u
r
u
r
u
rr
uE rrrr ?
??
?
令,
???
?
)21)(1(
)1(
1 ??
?? Ec
假定
ru r ?
?? ?
则 是位移的势函数。代入（ a）式得 ),( tr?? ?
0122 2
2
2
1
2
2
?????????? tucr ururru rr rrr
（ a）
55
So formula (b) can be written as
Because ? ?
rrrrrrrrr ?
??
?
??
?
??
??
?
??
?
?
?
?
? ???? 221
2
2
3
3
2
2
???
?
???
?
?
?
?
???
?
??
?
?
?
?
?
?
2
2
2
2
trrt
??
0122 2
2
2
1
22
2
3
3
??
?
??
?
?
?
?
?
??
?
??
?
??
?
?
rtcrrrrr
???? （ b）
? ? 011 2
2
2
1
2
2
???
?
?
???
?
?
?
?
??
?
?
?
?
?
?
?
?
?
?
trc
r
rrr
??
56
所以（ b）式可写成
由于 ? ?
rrrrrrrrr ?
??
?
??
?
??
??
?
??
?
?
?
?
? ???? 221
2
2
3
3
2
2
???
?
???
?
?
?
?
???
?
??
?
?
?
?
?
?
2
2
2
2
trrt
??
0122 2
2
2
1
22
2
3
3
??
?
??
?
?
?
?
?
??
?
??
?
??
?
?
rtcrrrrr
???? （ b）
? ? 011 2
2
2
1
2
2
???
?
?
???
?
?
?
?
??
?
?
?
?
?
?
?
?
?
?
trc
r
rrr
??
57
Its general solution is,
Integrate r once,we get,
? ? ? ?tFtcrrr ?????? 222
1
2
2 11 ?
?
Because we let F(t)=0,so it doesn’t affect displacement and
above formula can be simplified as,
ru
? ? ? ??? rrcrt 222122 ?????
? ? ? ?tcrftcrfr 1211 ?????
It is obvious that transmitting speed of spherical wave is
(spherical wave is without rotating wave),expresses
spherical wave transmitting form inside to outside,and
expresses spherical wave transmitting from outside to inside,
1c
1f
2f
58
它的通解是,
对 r积分一次，得,
? ? ? ?tFtcrrr ?????? 222
1
2
2 11 ?
?
由于令 F(t)=0，并不会影响位移,因此上式可简写成为,
ru
? ? ? ??? rrcrt 222122 ?????
? ? ? ?tcrftcrfr 1211 ?????
显然，球面波的传播速度等于 (球面波是无旋波）。
表示由内向外传播的球面波,表示由外向内传播的球面
波。
1c 1f
2f
59
Exercise 11.1 What is elastic wave? What is
the meaning to study elastic wave?
Solution,(omitted)
Exercise11.2 Given,elasticity module of steel E=210GPa，
density ?=7950kg/m3,elasticity module of concrete E=30GPa，
density ?=2400kg/m3,calculate the transmitting speed of
vertical wave in the two material,
Solution, transmitting speed formula of vertical wave in one
dimension strait pole is,
?Ev ?
so s m v s m v / 3500,/ 5130 ? ? concrete steel
60
练习 11.1 什么是弹性波？研究弹性波有何意义？
答：（略）
练习 11.2 已知钢的弹性模量 E=210GPa，密度 ?=7950kg/m3,
混凝土的弹性模量 E=30GPa,密度 ?=2400kg/m3,问在此两
种材料杆中纵波的传播速度。
解,由纵波在一维直杆中的传播速度公式
smvsmv /3 5 0 0,/5 1 3 0 ?? 混凝土钢
?Ev ?
得
61
62
Elasticity
3
Summarize
§ 1-1 Movement differential equations of elastic objects
§ 11-2 Without rotating wave and equal volume wave
§ 11-3 Transverse wave and vertical wave
§ 11-4 Spherical wave
Chapter 11 Elastic Wave
4
概述
§ 1-1 弹性体的运动微分方程
§ 11-2 无旋波与等容波
§ 11-3 横波与纵波
§ 11-4 球面波
第十一章 弹性波
5
Summarize,
When elastic object bears loads in static force equilibrium
conditions,not all the parts of object has displacement,
distortion and stress,At the beginning of the loads,the parts
which are more far from the loads have no impacts, After
then,the displacement,distortion and stress caused by loads
transmit to other places in a finite speed of wave,This wave is
called elastic wave,
This chapter will first give movement differential equations
of elastic objects,then introduce some conceptions of elastic
wave and simplify the equations according to different elastic
waves,at last give the speed transmitting formulas of wave in
infinite elastic objects,
6
概述
当静力平衡状态下的弹性体受到荷载作用时，并不是在
弹性体的所有各部分都立即引起位移、形变和应力。在作用
开始时，距荷载作用处较远的部分仍保持不受干扰。在作用
开始后，荷载所引起的位移、形变和应力，就以波动的形式
用有限大的速度向别处传播。这种波动就称为 弹性波 。
本章将首先给出描述弹性体运动的基本微分方程，然后
介绍弹性波的几个概念，针对不同的弹性波，对运动微分方
程进行简化，最后给出波在无限大弹性体中传播速度公式。
7
§ 11-1 Movement differential
equations of elastic objects
The two assumptions are equal to the basic assumptions when we discuss
static force questions,So the physic and geometry equations and elastic
equations where stress component is expressed by displacement
component,still are the same with movement equations at any
instantaneous time,The only difference is that the equilibrium differential
equations of static questions must be substituted by movement differential
equations,
This chapter we still adopt the assumptions,
（ 1） Elastic objects are ideal elastic objects,
（ 2） The displacement and distortion are tinny,
8
§ 11-1 弹性体的运动微分方程
上述两条假设，完全等同于讨论静力问题的基本假
设。因此，在静力问题中给出的物理方程和几何方程，
以及把应力分量用位移分量表示的弹性方程，仍然适用
于讨论动力问题的任一瞬时，所不同的仅仅在于，静力
问题中的平衡微分方程必须用运动微分方程来代替。
本章仍然采用如下假设,
（ 1） 弹性体为理想弹性体。
（ 2） 假定位移和形变都是微小的。
9
Toward any tiny object,when we apply d’Alembert
theory,we must consider stress,body force and the
inertia force of elastic objects caused by acceleration,
In space right-angle coordinate system,the x,y,z
directions component of inertia force of every unite
volume are,
Where ρ is the density of elastic objects,
2
2
t
u
?
?? ?
2
2
t?
?? ??
2
2
t
w
?
?? ?
10
对于任取的微元体，运用达朗伯尔原理，除了
考虑应力和体力以外，还须考虑弹性体由于具有加
速度而产生的惯性力。每单位体积上的惯性力在空
间直角坐标系的 x,y,z方向的分量分别为,
其中 ρ为弹性体的密度。
2
2
t
u
?
?? ?
2
2
t?
?? ??
2
2
t
w
?
?? ?
11
Because of the equilibrium relations,we simplify them and get,
The above formulas are called movement differential equations
of elastic objects,They and geometry equations and physic
equations are the basic equations of movement questions of
elasticity mechanics
02
2
????????????? t uXzyx zxyxx ????
02
2
????????????? tYxzy xyzyy ?????
02
2
????????????? t wZyxz yzxzz ????
12
由平衡关系，并简化后得,
上式称为 弹性体的运动微分方程 。它同几何方程和物理方程
一起构成弹性力学动力问题的基本方程。
02
2
????????????? t uXzyx zxyxx ????
02
2
????????????? tYxzy xyzyy ?????
02
2
????????????? t wZyxz yzxzz ????
13
Note 1,geometry equations,
x
u
x ?
???
yy ?
?? ??
z
w
z ?
???
zy
w
yz ?
??
?
?? ??
x
w
z
u
zx ?
??
?
???
y
u
xxy ?
??
?
?? ??
14
注 1：几何方程
x
u
x ?
???
yy ?
?? ??
z
w
z ?
???
zy
w
yz ?
??
?
?? ??
x
w
z
u
zx ?
??
?
???
y
u
xxy ?
??
?
?? ??
15
Note2,physic equations
)]([1 zyxx E ????? ???
)]([1 xzyy E ????? ???
)]([1 yxzz E ????? ???
yzyz E ?
?? )1(2 ??
zxzx E ?
?? )1(2 ??
xyxy E ?
?? )1(2 ??
16
注 2：物理方程
)]([1 zyxx E ????? ???
)]([1 xzyy E ????? ???
)]([1 yxzz E ????? ???
yzyz E ?
?? )1(2 ??
zxzx E ?
?? )1(2 ??
xyxy E ?
?? )1(2 ??
17
Because the displacement component is difficult to be expressed by stress
and its derivative,so movement equations of elasticity mechanics are
usually solved according to the displacement,Substitute the elasticity
equations where stress components are expressed by displacement
component into movement differential equations,and we let,
Then we get,zwyxue ????????? ?
0)21 1()1(2 2
2
2 ?
?
?????
?
?
?? t
uXu
x
eE ?
??
0)21 1()1(2 2
2
2 ?
?
?????
?
?
?? tYy
eE ???
??
0)21 1()1(2 2
2
2 ?
?
?????
?
?
?? t
wZw
z
eE ?
??
18
由于位移分量很难用应力及其导数来表示，所以弹
性力学动力问题通常要按位移求解。将应力分量用位移
分量表示的弹性方程代入运动微分方程，并令,
得,z
w
yx
ue
?
??
?
??
?
?? ?
0)21 1()1(2 2
2
2 ?
?
?????
?
?
?? t
uXu
x
eE ?
??
0)21 1()1(2 2
2
2 ?
?
?????
?
?
?? tYy
eE ???
??
0)21 1()1(2 2
2
2 ?
?
?????
?
?
?? t
wZw
z
eE ?
??
19
These are the basic differential equations of movement equations
solved by displacement,They are also called Lame equations,
We need boundary conditions to solve Lame equations,Besides these we
still need original conditions,because displacement components are the
function of time variable,
In order to simplify calculation,usually we neglect body force,Now the
movement differential equations of elastic objects can be simplified as,
)21 1()1(2 22
2
uxeEt u ????????? ???
)21 1()1(2 22
2
????? ????????? yeEt
)21 1()1(2 22
2
wzeEt ????????? ????
20
这就是按位移求解动力问题的基本微分方程，也称
为 拉密（ Lame)方程 。
要求解拉密方程，显然需要边界条件。除此之外，
由于位移分量还是时间变量的函数，因此求解动力问题
还要给出初始条件。
为求解上的简便，通常不计体力，此时弹性体的运
动微分方程简化为,
)21 1()1(2 22
2
uxeEt u ????????? ???
)21 1()1(2 22
2
????? ????????? yeEt
)21 1()1(2 22
2
wzeEt ????????? ????
21
§ 11-2 Without rotating
Wave and equal volume wave
1,Without rotating waves
Without rotating wave means that in elastic objects,the distortion
caused by waves is not rotating, That means rotating values of
three vertical coordinates at any point in the elastic object are zero,
xu ?
?? ?
y?
?? ?? zw ??? ?
Where is potential function of displacement,
This displacement is called without rotating displacement,and the
elastic wave corresponds to the displacement are called without
rotating wave,
),,,( tzyx?? ?
Suppose the displacement of elastic objects can be
expressed by,
u,v,w
22
§ 11-2 无旋波与等容波
一、无旋波
所谓 无旋波 是指 在弹性体中，波动所产生的变形不存在旋
转，即弹性体在任一点对三个垂直坐标轴的旋转量皆为零。
假定弹性体的位移 u,v,w可以表示成为,
xu ?
?? ?
y?
?? ?? zw ??? ?
其中 是位移的势函数。这种位移称为无旋位
移。而相应于这种位移状态的弹性波就称 无旋波 。
),,,( tzyx?? ?
23
[proving],at any point of elastic object,the rotating value of z
axis is,
So the rotating values of the three coordinates at any point of the
elastic object are zero,
y
u
xz ?
??
?
?? ??
substitute into the formula,we can
get,xu ??? ? y??? ??
0?z?
Similarly,
0?x? 0?y?
24
[证 ],在弹性体的任一点处，该点对 z 轴的旋转量
即弹性体的任一点对三个坐标的旋转量都等于零。
y
u
xz ?
??
?
?? ??
将 代入，可得,
xu ?
?? ?
y?
?? ??
0?z?
同理
0?x? 0?y?
25
In the condition of without rotating displacement,
?? 2??????????? zwyxue
so
uxxxe 222 ???????????? ??
similarly,
?2????ye
wze 2????
Substitute the above three formulas into movement differential
equations without including body force,simplify it we can get
wave movement equations of without rotating wave,
26
在无旋位移状态下
?? 2??????????? zwyxue
从而
uxxxe 222 ???????????? ??
同理
?2????ye
wze 2????
将上三式代入不计体力的运动微分方程，并简化后
得无旋波的波动方程
27
uct u 2212
2
????
?? 2212
2
???? ct
wct w 2212
2
????
???
?
)21)(1(
)1(
1 ??
?? EcWhere,
1c
is the transmitting speed of without rotating
wave in infinite elastic objects,
28
uct u 2212
2
????
?? 2212
2
???? ct
wct w 2212
2
????
???
?
)21)(1(
)1(
1 ??
?? Ec其中
1c
就是无旋波在无限大弹性体中的传播速度
29
Equal volume wave means that in the distortion caused by waves in
elastic objects,the volume strain is zero,That means the volume of
elastic object remains unchanged,
2,Equal volume wave
Suppose the displacement u,v,w of elastic objects satisfy the
condition that the volume strain is zero,
0?????????? zwyxue ?
This displacement is called equal volume displacement and the
elastic wave corresponds to this displacement is called equal
volume wave,
30
所谓 等容波 是指 在弹性体内，波动所产生的变形中体积应
变为零 。即弹性体中任一部分的容积（即体积）保持不变 。
二、等容波
假定弹性体的位移 u,v,w满足体积应变为零的条件，即,
0?????????? zwyxue ?
这种位移称为 等容位移 。而相应于这种位移状态的弹性
波就是等容波。
31
Because, so simplify movement differential equations
without including body force we can get wave movement equations
of equal volume wave,
0?e
uct u 2222
2
????
?? 2222
2
???? ct
wct w 2222
2
????
where
?? )1(22 ??
Ec
is the transmitting speed of equal volume waves
in infinite elastic objects,
2c
32
由于,故不计体力的运动微分方程，简化后得等
容波的波动方程,
0?e
uct u 2222
2
????
?? 2222
2
???? ct
wct w 2222
2
????
其中
?? )1(22 ??
Ec
就是等容波在无限大弹性体中的传播速度。
2c
33
According to without rotating wave and equal volume
wave,we give the conclusion without proving that,in elastic
objects,stress,strain and speed of particle transmit in the
same way and the same speed as displacement,
34
对于无旋波和等容波，我们不加证明地给出如下 结论,
在弹性体中，形变、应力以及质点速度，都将和位移以相
同的方式与速度进行传播。
35
1,vertical wave
[definition] The particle movement direction of elastic objects is
parallel to the transmitting direction of elastic wave,（ seen in the
Fig.）
Transmitting format of
vertical wave
§ 11-3 Vertical wave and transverse wave
36
一、纵波
[定义 ] 弹性体的质点运动方向平行弹性波的传播方向 （图示）
纵波的传播形式
§ 11-3 纵波与横波
37
Let x axis be the transmitting direction of waves,then the
displacement components of any point in elastic objects are,
),( txuu ? 0?? 0?w
So,
x
ue
?
??
And,
2
2
x
u
x
e
?
??
?
?
0???ye 0???ze
2
2
2
x
uu
?
??? 02 ??? 02 ?? w
38
将 x轴取为波的传播方向，则弹性体内任取一点的位移
分量都有,
),( txuu ? 0?? 0?w
从而
x
ue
?
??
而
2
2
x
u
x
e
?
??
?
?
0???ye 0???ze
2
2
2
x
uu
?
??? 02 ??? 02 ?? w
39
Substitute them into movement differential equations without
including body force,we can find that the second and third
formulas are identical equations,So the first formula can be
simplified as,
2
2
2
12
2
x
uc
t
u
?
??
?
?
where
???
?
)21)(1(
)1(
1 ??
?? Ec
is transmitting speed of vertical wave in elastic objects,
1c
It is obvious that transmitting speed of vertical waves is the same
as without rotating waves’, In fact vertical wave is a kind of
without rotating wave,
40
代入不计体力的运动微分方程，可见其第二、第三式成为
恒等式，而第一式简化为,
2
2
2
12
2
x
uc
t
u
?
??
?
?
其中
???
?
)21)(1(
)1(
1 ??
?? Ec
为纵波在弹性体中的传播速度。
1c
显然纵波的传播速度与无旋波相同。事实上，纵波就是
一种无旋波。
41
The general solution of wave movement equations of vertical
wave is,)()(),(
1211 tcxftcxftxu ????
The physic meaning of the general solution is,let us take the
first item for example,function at a fixed time is
function of x,and can be expressed by curve abc in Fig.(a)
(suppose the form is like this ),After,the function
becomes,
)( 11 tcxf ?
t?
)( 111 tctcxf ???
If let, then function becomes,
Its format is similar to the original function,
From the Fig we can see the only difference is transverse
coordinate moves in level, So expresses
the wave whose speed is along x positive direction,
tcxx ??? 11 )( 111 tcxf ?
)( 11 tcxf ?
tc?1 )( 11 tcxf ?
1c
42
纵波波动方程的通解是,
)()(),( 1211 tcxftcxftxu ????
该通解的 物理意义,以其第一项为例，函数 在某
一个固定时刻将是 x的函数，可以用图 (a)中的曲线 abc表示
（假设是这种形状），在 时间之后，函数变为,
)( 11 tcxf ?
t?
)( 111 tctcxf ???
如果令,则函数可写为,其形式
同原函数 完全类同，只是横坐标发生平移
tcxx ??? 11 )( 111 tcxf ?
)( 11 tcxf ? tc?1
见图。因此 表示以速度 向 x轴正向传播的波。 )(
11 tcxf ? 1c
43
Similarly expresses the wave whose speed is
along x negative direction, The general solution expresses two
waves transmitting in opposite directions (in Fig.b),And the
speed is the modulus of wave movement equations,
1c)( 12 tcxf ?
1c
1f
c a
b
x
(a)
(b)
tc?1
tc?1 tc?1
44
同理,表示以同样速度 向 x轴负向传播的波。
整个通解表示朝相反两个方向传播的两个波（如图 b），其
传播速度为波动方程的系数 。
1c)( 12 tcxf ?
1c
1f
c a
b
x
(a)
(b)
tc?1
tc?1 tc?1
45
2,Transverse wave
[definition] The partial movement direction of elastic object is
vertical to the transmitting direction of elastic wave,
Transmitting format of
transverse wave
46
二、横波
[定义 ] 弹性体的质点运动方向垂直于弹性波的传播方向。
横波的传播形式
47
Still we let x axis be the transmitting direction of wave and y axis
be displacement direction of partial,Then the displacement
components of any point in the elastic object have,
0?u ),( tx?? ? 0?w
so 0?e
and 02 ?? u
2
2
2
x?
??? ?? 02 ?? w
Substitute them into movement differential equations without
including body force,we can find that the first and third formulas
are identical equations,So the second formula can be simplified
as,
2
2
2
22
2
xct ?
??
?
? ??
?? )1(22 ??
Ec
2c
is transmitting speed of transverse wave in elastic object,
Because volume strain of transverse wave
48
仍然将 x轴放在波的传播方向,y轴为质点位移方向，则
弹性体内任取一点的位移分量都有
0?u ),( tx?? ? 0?w
从而 0?e
而 02 ?? u
2
2
2
x?
??? ?? 02 ?? w
代入不计体力的运动微分方程，可见其第一、第三式成为恒
等式，第二式简化为,
2
2
2
22
2
xct ?
??
?
? ??
?? )1(22 ??
Ec
2c
为横波在弹性体中的传播速度。由于横波的体积应变
49
General solution to wave movement equations of
transverse wave is,
, so transverse wave is
equal volume wave 0?e
)()(),( 2221 tcxftcxftx ?????
It is obvious that the general solution expresses two waves
transmitting in opposite directions, Its displacement is along y
direction and the transmitting direction is along x direction, The
transmitting speed is a constant,
2c
50
横波的波动方程的通解为,
,故横波为等容波。 0?e
)()(),( 2221 tcxftcxftx ?????
显然，整个通解表示朝相反两个方向传播的两个波，它的
位移沿着 y方向，而传播方向是沿着 x方向，传播速度等于常
量 。
2c
51
§ 11-4 Spherical wave
If elastic objects have spherical hole or spherical out surface
,when the spherical hole or spherical out surface bears spherical
force,the elastic wave transmitting from the hole to the outside or
form the surface to the inside is called spherical wave,
Spherical wave is symmetrical of sphere,Apply basic
differential equations of sphere symmetry,we get,
0)22()21)(1( )1( 222
2
?????? ? rrrr kurdrdurdr udE ?? ?
Now, If we ignore body force,and use radial
inertia force to substitute
),( truu rr ?
2
2
t
u r
?
?? ? rk
52
§ 11-4 球面波
如果弹性体具有圆球形的孔洞或具有圆球形的外表面，
则在圆球形孔洞或圆球形外表面上受到球对称的动力作用时
，由孔洞向外传播或由外表面向内传播的弹性波，称为 球面
波 。
球面波是球对称的。利用球对称的基本微分方程,
0)22()21)(1( )1( 222
2
?????? ? rrrr kurdrdurdr udE ?? ?
此时，,而不计体力时，用径向惯性力 ),( truu
rr ?
2
2
t
u r
?
?? ? 代替,
rk
53
Then the above formula can be simplified as,
We get,
0)22(
)21)(1(
)1(
2
2
22
2
?
?
???
?
??
?
?
??
?
t
u
r
u
r
u
rr
uE rrrr ?
??
?
let,
???
?
)21)(1(
)1(
1 ??
?? Ec
Suppose
ru r ?
?? ?
then is potential function of displacement,
Substitute it into formula (a),we get,
),( tr?? ?
0122 2
2
2
1
2
2
?????????? tucr ururru rr rrr
（ a）
54
则上式简写成
即得,
0)22(
)21)(1(
)1(
2
2
22
2
?
?
???
?
??
?
?
??
?
t
u
r
u
r
u
rr
uE rrrr ?
??
?
令,
???
?
)21)(1(
)1(
1 ??
?? Ec
假定
ru r ?
?? ?
则 是位移的势函数。代入（ a）式得 ),( tr?? ?
0122 2
2
2
1
2
2
?????????? tucr ururru rr rrr
（ a）
55
So formula (b) can be written as
Because ? ?
rrrrrrrrr ?
??
?
??
?
??
??
?
??
?
?
?
?
? ???? 221
2
2
3
3
2
2
???
?
???
?
?
?
?
???
?
??
?
?
?
?
?
?
2
2
2
2
trrt
??
0122 2
2
2
1
22
2
3
3
??
?
??
?
?
?
?
?
??
?
??
?
??
?
?
rtcrrrrr
???? （ b）
? ? 011 2
2
2
1
2
2
???
?
?
???
?
?
?
?
??
?
?
?
?
?
?
?
?
?
?
trc
r
rrr
??
56
所以（ b）式可写成
由于 ? ?
rrrrrrrrr ?
??
?
??
?
??
??
?
??
?
?
?
?
? ???? 221
2
2
3
3
2
2
???
?
???
?
?
?
?
???
?
??
?
?
?
?
?
?
2
2
2
2
trrt
??
0122 2
2
2
1
22
2
3
3
??
?
??
?
?
?
?
?
??
?
??
?
??
?
?
rtcrrrrr
???? （ b）
? ? 011 2
2
2
1
2
2
???
?
?
???
?
?
?
?
??
?
?
?
?
?
?
?
?
?
?
trc
r
rrr
??
57
Its general solution is,
Integrate r once,we get,
? ? ? ?tFtcrrr ?????? 222
1
2
2 11 ?
?
Because we let F(t)=0,so it doesn’t affect displacement and
above formula can be simplified as,
ru
? ? ? ??? rrcrt 222122 ?????
? ? ? ?tcrftcrfr 1211 ?????
It is obvious that transmitting speed of spherical wave is
(spherical wave is without rotating wave),expresses
spherical wave transmitting form inside to outside,and
expresses spherical wave transmitting from outside to inside,
1c
1f
2f
58
它的通解是,
对 r积分一次，得,
? ? ? ?tFtcrrr ?????? 222
1
2
2 11 ?
?
由于令 F(t)=0，并不会影响位移,因此上式可简写成为,
ru
? ? ? ??? rrcrt 222122 ?????
? ? ? ?tcrftcrfr 1211 ?????
显然，球面波的传播速度等于 (球面波是无旋波）。
表示由内向外传播的球面波,表示由外向内传播的球面
波。
1c 1f
2f
59
Exercise 11.1 What is elastic wave? What is
the meaning to study elastic wave?
Solution,(omitted)
Exercise11.2 Given,elasticity module of steel E=210GPa，
density ?=7950kg/m3,elasticity module of concrete E=30GPa，
density ?=2400kg/m3,calculate the transmitting speed of
vertical wave in the two material,
Solution, transmitting speed formula of vertical wave in one
dimension strait pole is,
?Ev ?
so s m v s m v / 3500,/ 5130 ? ? concrete steel
60
练习 11.1 什么是弹性波？研究弹性波有何意义？
答：（略）
练习 11.2 已知钢的弹性模量 E=210GPa，密度 ?=7950kg/m3,
混凝土的弹性模量 E=30GPa,密度 ?=2400kg/m3,问在此两
种材料杆中纵波的传播速度。
解,由纵波在一维直杆中的传播速度公式
smvsmv /3 5 0 0,/5 1 3 0 ?? 混凝土钢
?Ev ?
得
61
62
