1
Elasticity
2
3
Chapter 6 The Basic Solution
of Temperature Stress Problems
§ 6-4 Solve plane problem of temperature stresses by
displacement
§ 6-3 The boundary conditions of temperature filed
§ 6-2 The differential equation of heat conduction
§ 6-1 The basic concept of temperature field and
heat conduction
§ 6-5 The introducing of potential function of
displacement
§ 6-6 The plane problems of thermal stresses in
axisymmetric temperature field
4
第六章 温度应力问题的基本解法
§ 6-4 按位移求解温度应力的平面问题
§ 6-3 温度场的边界条件
§ 6-2 热传导微分方程
§ 6-1 温度场和热传导的基本概念
§ 6-5 位移势函数的引用
§ 6-6 轴对称温度场平面热应力问题
5
When the temperature of a elastic body changes,its volume will expand or
contract,If the expansion or contraction can’t happen freely due to the external
restrictions or internal deformation compatibility demands,additional stresses
will be produced in the structure,These stresses produced by temperature
change are called thermal stresses,or temperature stresses,
Neglecting the effects of the temperature change on the material
performance,to solve the temperature stresses,we need two aspects of
calculation,(1) Solve the temperature field of the elastic body by the initial
conditions and boundary conditions,according to heat conduction equations,
And the difference between the former temperature field and the later
temperature field is the temperature change of the elastic body,(2) Solve the
temperature stresses of the elastic body according to the basic equations of the
elastic mechanics,This chapter will present these two aspects of calculation
simply,
6
当弹性体的温度变化时，其体积将趋于膨胀和收缩，若
外部的约束或内部的变形协调要求而使膨胀或收缩不能自由
发生时，结构中就会出现附加的应力。这种因温度变化而引
起的应力称为热应力，或温度应力。
忽略变温对材料性能的影响，为了求得温度应力，需要
进行两方面的计算：（ 1）由问题的初始条件、边界条件，
按热传导方程求解弹性体的温度场，而前后两个温度场之差
就是弹性体的变温。（ 2）按热弹性力学的基本方程求解弹
性体的温度应力。本章将对这两方面的计算进行简单的介绍。
7
§ 6-1 The Basic Concept of Temperature
Field And Heat Conduction
1.The temperature field,The total of the temperature at all the points in a
elastic body at a certain moment,denoted by T,
Unstable temperature filed or nonsteady temperature field,The temperature
in the temperature field changes with time,
i.e,T=T（ x,y,z,t）
Stable temperature filed or steady temperature field,The temperature in the
temperature field is only the function of positional coordinates,
i.e,T=T（ x,y,z）
Plane temperature field,The temperature in temperature field only changes
with two positional coordinates,
i.e,T=T（ x,y,t）
8
§ 6-1 温度场和热传导的基本概念
1.温度场：在任一瞬时，弹性体内所有各点的温度值的总体。用
T表示。
不稳定温度场或非定常温度场：温度场的温度随时间而变化。
即 T=T（ x,y,z,t）
稳定温度场或定常温度场：温度场的温度只是位置坐标的函数。
即 T=T（ x,y,z）
平面温度场：温度场的温度只随平面内的两个位置坐标而变。
即 T=T（ x,y,t）
9
2.Isothermal surface,The surface that
connects all the points with the same
temperature in the temperature field at a
certain moment,Apparently,the
temperature doesn’t changes along the
isothermal surface; The changing rate is
the largest along the normal direction of
the isothermal surface,
T+2△ T
T+△ T T
T-△ T x o
y
3.Temperature gradient:The vector that points to the direction in which
temperature increase along the normal direction of the isothermal surface,
It is denoted by △ T,and its value is denoted by,where n is the
normal direction of the isothermal surface,The components of
temperature gradient at each coordinate are
nT??
10
2.等温面：在任一瞬时，连接温度
场内温度相同各点的曲面。显然，
沿着等温面，温度不变；沿着等温
面的法线方向，温度的变化率最大。
T+2△ T
T+△ T T
T-△ T x o
y
nT??
3.温度梯度：沿等温面的法线方向，指向温度增大方向的矢
量。用 △ T表示，其大小用 表示。其中 n为等温面的法线方
向。温度梯度在各坐标轴的分量为
11
0n
Define to be the unit vector in normal direction of the isothermal
surface,pointing to the temperature increasing direction,
n
Tn
?
??
0
△ T （ 1）
4.Thermal flux speed,The quantity of heat flowing through the area S
on the isothermal surface in unit time,denoted by,
dt
dQ
)c o s (
)c o s (
)(c o s
zn
n
T
z
T
yn
n
T
y
T
xn
n
T
x
T
，
，
，
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
12
0n
取 为等温面法线方向且指向增温方向的单位矢量，则有
n
Tn
?
??
0
△ T （ 1）
4.热流速度：在单位时间内通过等温面面积 S 的热量。用 表示。 dtdQ
)c o s (
)c o s (
)(c o s
zn
n
T
z
T
yn
n
T
y
T
xn
n
T
x
T
，
，
，
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
13
Its value is,
SdtdQq /?
SdtdQnq /0?? (2)
Thermal flux density,The thermal flux speed flowing through unit area
on the isothermal surface,denoted by, Then we have q
5.The basic theorem of heat transfer,The thermal flux density is in
direct proportion to the temperature gradient and in the reverse
direction of it,i.e,
???q (3)
SnTdtdQ ??? /?
is called the coefficient of the heat transfer,Equations (1),(2) and (3)
lead to
?
14
热流密度：通过等温面单位面积的热流速度。用 表示，
则有 q
SdtdQq /?
其大小为
SdtdQnq /0??
(2)
SnTdtdQ ??? /?
? 称为导热系数。由（ 1）、（ 2）、（ 3）式得
5.热传导基本定理：热流密度与温度梯度成正比而方向相反。
即
???q (3) △ T
15
We can see that the coefficient of the heat transfer means,the thermal flux
speed through unit area of the isothermal surface per unit temperature
gradient”,
n
Tq
?
?? ?
From equations (1) and (3),we can see that the value of the thermal flux
density
The projections of the thermal flux density on axes,
x
Tq x
?
??? ?
y
Tq y
?
??? ?
zTq z ???? ?
It is obvious that the component of thermal flux density in any direction is
equal to the coefficient of heat transfer multiplied by the descending rate of
the temperature in this direction,
16
n
Tq
?
?? ?
由（ 1）和（ 3）可见，热流密度的大小
可见，导热系数表示, 在单位温度梯度下通过等温面单位面积
的热流速度, 。
热流密度在坐标轴上的投影
可见：热流密度在任一方向的分量，等于导热系数乘以
温度在该方向的递减率。
z
T
q
y
T
q
x
T
q
z
y
x
?
?
??
?
?
??
?
?
??
?
?
?
17
The principle of heat quantity equilibrium,Within any period of time,
the heat quantity accumulated in any minute part of the object equals the heat
quantity conducted into this minute part plus the heat quantity supplied by
internal heat source,
§ 6-2 The Differential Equation of Heat Conduction
dxxqq xx ???xq
x
y
z
Take a minute hexahedron dxdydz as shown in the above figure,Suppose
that the temperature of this hexahedron rises from T to, The heat
quantity accumulated by temperature is,where is the
density of the object,C is the heat quantity needed when the temperature of the
object with a unit mass rise one degree——specific thermal capability,
dttTT ???
dttTd x d y d zC ??? ?
18
热量平衡原理：在任意一段时间内，物体的任一微小部
分所积蓄的热量，等于传入该微小部分的热量加上内部热源
所供给的热量。
§ 6-2 热传导微分方程
dxxqq xx ???xq
x
y
z
取图示微小六面体 dxdydz。假定该六面体的温度在 dt时间内
由 T 升高到 。由温度所积蓄的热量是,
其中 是物体的密度,C 是单位质量的物体升高一度时所需的
热量 ——比热容。
dttTT ??? dttTd x d y d zC ???
?
19
Within the same period of time dt,the heat quantity qxdydzdt is conducted
into the hexahedron from left,and the heat quantity is
conducted out the hexahedron through right,Hence,the net heat quantity
conducted into is
d y d z d tdxxqq xx )( ???
d x d y d z d txq x???
x
Tq
x ?
??? ? Introduce into it, We can see that
d x d y d z d txT22???
d y d z d x d tyT22???
d z d x d y d tzT22???
The net heat quantity conducted into it from left and right is,
The net heat quantity conducted into it from top and bottom is,
The net heat quantity conducted into it from front and back is,
d x d y d z d tz Ty Tx T )( 222222 ?????????
Hence,the total net heat quantity conducted into the hexahedron is,
T d xd yd z d t2??
which can be abbreviated as,
20
在同一段时间 dt内，由六面体左面传入热量 qxdydzdt，
由右面传出热量 。因此，传入的净热量为
d y d z d tdxxqq xx )( ???
d x d y d z d txq x???
x
Tq
x ?
??? ?将 代入可见,
d x d y d z d txT2
2
?
??
d y d z d x d tyT22???
由左右两面传入的净热量为
由上下两面传入的净热量为
由前后两面传入的净热量为,
因此，传入六面体的总净热量为,
简记为,
d z d x d y d tzT22???
d x d y d z d tz Ty Tx T )( 222222 ?????????
T d xd yd z d t2??
21
Suppose that there is a positive heat resource to supply heat inside the
object,which supply heat quantity W per unit volume in unit time.,Then the
heat quantity that supplied by this heat resource during time dt is Wdxdydzdt,
According to the principle of heat quantity equilibrium,
W d x d y d z d tT d x d y d z d td x d y d z d txTc ????? 2??
??
?
c
WT
ct
T ???
?
? 2
?
?
ca ?
?c
WTa
t
T ???
?
? 2
which can be simplified as,
Let
This is the differential equations of heat transfer,
Thus
22
假定物体内部有正热源供热，在单位时间、单位体积供
热为 W，则该热源在时间 dt内所供热量为 Wdxdydzdt。
根据热量平衡原理得,
W d x d y d z d tT d x d y d z d td x d y d z d txTc ????? 2??
??
?
c
WT
ct
T ???
?
? 2
?
?
ca ?
?c
WTa
t
T ???
?
? 2
化简后得,
记
则
这就是热传导微分方程。
23
§ 6-3 The Boundary Conditions of Temperature Filed
To solve the differential equation,and sequentially solve the temperature
filed,the temperature of the object at initial moment must be known,i.e,the
so-called initial condition,At the same time,the rule of heat exchange
between the object surface and the surrounding medium after the initial
moment must be also known,i.e,the so-called boundary conditions,The
initial condition and the boundary conditions are called by a joint name of the
initial value conditions,
Initial condition,
Boundary conditions are divided into four kinds of forms,
The first kind of boundary condition,The temperature at any point on the
object surface is known at all moments,i.e,
where Ts is the surface temperature of the object,
),,()( 0 zyxfT t ??
)(tfT s ?
24
§ 6-3 温度场的边值条件
初始条件,
边界条件分四种形式,
第一类边界条件 已知物体表面上任意一点在所有瞬时
的温度，即
其中 Ts 是物体表面温度。
),,()( 0 zyxfT t ??
)(tfT s ?
为了能够求解热传导微分方程，从而求得温度场，必须
已知物体在初瞬时的温度，即所谓初始条件；同时还必须已
知初瞬时以后物体表面与周围介质之间热交换的规律，即所
谓边界条件。初始条件和边界条件合称为初值条件。
25
The second kind of boundary condition,The normal thermal flux
density at any point on the object surface is known,i.e,
where the subscript s means,surface”,and n means
“normal”,
)()( tfq sn ?
The third kind of boundary condition,The heat release situation of
convection at any point on the boundary of the object is known at all
moments,According to the convection theorem of heat quantity,the thermal
flux density transmitting from the object surface to the surrounding medium
per unit time is in direct proportion to the temperature difference between
them,i.e,
)()( essn TTq ?? ?
Where Te is the temperature of the surrounding medium; is called the
coefficient of the heat release of convection,or heat coefficient for short,?
The forth kind of boundary condition,It is known that the two objects
contact completely,and exchange heat through the form of heat conduction,i.e,
es TT ?
26
第三类边界条件 已知物体边界上任意一点在所有瞬时
的运流（对流）放热情况。按照热量的运流定理，在单位时
间内从物体表面传向周围介质的热流密度，是和两者的温差
成正比的，即
)()( essn TTq ?? ?
?
es TT ?
其中 Te是周围介质的温度； 称为运流放热系数，或简称热
系数。
第四类边界条件 已知两物体完全接触，并以热传导方
式进行热交换。即
第二类边界条件 已知物体表面上任意一点的法向热流密度,
即 其中角码 s 表示, 表面,,角码 n 表示法向 。 )()( tfq sn ?
27
§ 6-4 Solve Plane Problem of
Temperature Stress by Displacement
Suppose the temperature change of every point in the elastic body is T,
For an isotropic body,if there is no constricts,then the minute length at every
point of the elastic body will generate normal strain,(where is the
coefficient of expansion of the elastic body),Thus,the components of strain at
every point of the elastic body are
T? ?
0,?????? xyzxyzzyx T ???????
However,because the elastic body is restricted by the external restrictions
and mutual restrictions among each section in the object,the above-mentioned
deformations can not happen freely,Then the stress is produced,i.e,the so-
called temperature stress.This temperature stress will result in additional strain
due to the elasticity of the object,as expressed by Hooke’s law,Therefore,the
components of the total strain of the elastic body are
28
§ 6-4 按位移求解温度应力的平面问题
设弹性体内各点的温变为 T。 对于各向同性体，若不受约束，
则弹性体内各点的微小长度，都将产生正应变 （ 是弹性体
的膨胀系数），这样，弹性体内各点的形变分量为
T? ?
0,?????? xyzxyzzyx T ???????
但是，由于弹性体所受的外在约束以及体内各部分之间的
相互约束，上述形变并不能自由发生，于是就产生了应力，即
所谓温度应力。这个温度应力又将由于物体的弹性而引起附加
的形变，如虎克定理所示。因此，弹性体总的形变分量是,
29
T
E
T
E
T
E
yxzz
xzyy
zyxx
??????
??????
??????
????
????
????
)]([
1
)]([
1
)]([
1
xyxy
zxzx
yzyz
E
E
E
?
?
?
?
?
?
?
?
?
)1(2
)1(2
)1(2
?
?
?
?
?
?
For the temperature change problems of plane stress,the above equations are
simplified as
xyxy
xyy
yxx
E
T
E
T
E
?
?
?
?????
?????
)1(2
][
1
][
1
?
?
???
???
They are the physical equations of thermal elastic mechanics of the
problems of plane stress,
30
对于平面应力的变温问题，上式简化为
xyxy
zxzx
yzyz
E
E
E
?
?
?
?
?
?
?
?
?
)1(2
)1(2
)1(2
?
?
?
?
?
?
xyxy
xyy
yxx
E
T
E
T
E
?
?
?
?????
?????
)1(2
][
1
][
1
?
?
???
???
这就是平面应力问题热弹性力学的物理方程。
T
E
T
E
T
E
yxzz
xzyy
zyxx
??????
??????
??????
????
????
????
)]([
1
)]([
1
)]([
1
31
Express the components of stress by the components of strain and the
temperature change T,then the physical equations become
xyxy
xyy
yxx
E
TEE
TEE
?
?
?
?
?
???
?
?
?
?
???
?
?
)1(2
1
)(
1
1
)(
1
2
2
?
?
?
??
?
?
?
??
?
?
The geometric equations still are
y
u
x
v
y
v
x
u
xyyx ?
??
?
??
?
??
?
?? ???,,
Introducing the geometric equations into the physical equations yields
the components of stress which are expressed by the components of
displacement and temperature change T
32
将应力分量用形变分量和变温 T表示的物理方程为,
xyxy
xyy
yxx
E
TEE
TEE
?
?
?
?
?
???
?
?
?
?
???
?
?
)1(2
1
)(
1
1
)(
1
2
2
?
?
?
??
?
?
?
??
?
?
几何方程仍然为,
y
u
x
v
y
v
x
u
xyyx ?
??
?
??
?
??
?
?? ???,,
将几何方程代入物理方程，得用位移分量和变温 T 表示的应
力分量
33
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
）（
）（
）（
）（
y
u
x
vE
TE
x
u
y
vE
TE
y
v
x
uE
xy
y
x
?
?
?
?
?
?
?
?
?
?
?
?
12
11
11
2
2
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
0
0
xy
yx
xvy
yxx
??
??
Introducing the above equations into the differential equations of
equilibrium ignoring body forces
34
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
）（
）（
）（
）（
y
u
x
vE
TE
x
u
y
vE
TE
y
v
x
uE
xy
y
x
?
?
?
?
?
?
?
?
?
?
?
?
12
11
11
2
2
将上式代入不计体力的平衡微分方程
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
0
0
xy
yx
xvy
yxx
??
??
35
and simplifying yelds
01
2
1
2
1
01
2
1
2
1
2
2
2
2
2
2
2
2
2
2
?
?
?
??
??
??
?
?
??
?
?
?
?
?
?
??
??
??
?
?
??
?
?
?
y
T
yx
u
x
v
y
v
x
T
yx
v
y
u
x
u
??
??
??
??
）（
）（
（ 1）
??
???
??
??
0
0
sxysy
syxsx
lm
ml
）（）（
）（）（
??
??
These are the differential equations solving the problems of plane
stress of temperature stress by displacement,
In the same way,introducing the components of the stresses into
stress boundary conditions without surface force
36
简化得,
这就是按位移求解温度应力平面应力问题的微分方程。
同理，将应力分量代入无面力的应力边界条件
??
???
??
??
0
0
sxysy
syxsx
lm
ml
）（）（
）（）（
??
??
01
2
1
2
1
01
2
1
2
1
2
2
2
2
2
2
2
2
2
2
?
?
?
??
??
??
?
?
??
?
?
?
?
?
?
??
??
??
?
?
??
?
?
?
y
T
yx
u
x
v
y
v
x
T
yx
v
y
u
x
u
??
??
??
??
）（
）（
（ 1）
37
?
?
?
??
?
?
??
?
?
?
?
??
?
?
?
?
?
?
??
?
?
?
?
??
?
?
?
?
?
?
Tm
y
u
x
v
l
x
u
y
v
m
Tl
x
v
y
u
m
y
v
x
u
l
ss
ss
??
?
?
??
?
?
）（）（）（
）（）（）（
1
2
1
1
2
1
（ 2）
and simplifying yields
vvuu ss ??,
These are the stress boundary conditions to solve plane stress
problems of temperature stress by displacement,
The boundary conditions of displacement still are
Compare equations (1),(2) with the equations (1),(2) in § 2-8,chapter
2,We can see that the components X and Y of the body forces are
displaced by
38
简化后得,
这是按位移求解温度应力平面应力问题的应力边界条件。
位移边界条件仍然为,
vvuu ss ??,
将式 (1),(2)与第二章 § 2-8中式 (1),(2)对比，可见
?
?
?
?
?
?
?
??
?
?
?
?
??
?
?
?
?
?
?
??
?
?
?
?
??
?
?
?
?
?
?
Tm
y
u
x
v
l
x
u
y
v
m
Tl
x
v
y
u
m
y
v
x
u
l
ss
ss
??
?
?
??
?
?
）（）（）（
）（）（）（
1
2
1
1
2
1
（ 2）
39
y
TE
x
TE
?
?
???
?
?? ?
?
?
?
1 a n d 1
?
?
?
?
?? 1 a n d 1
TEmTEl
X YWhile the components and of the surface forces are displaced by
For plane strain problems of temperature stress,it is only needed that in
the plane stress problems of temperature stress
） （
is displaced by
? ? ?
?
? ?
?
?
?
?
1
1
1 2
E E
is displaced by
is displaced by
Then the corresponding equations under the conditions of plane strain are
obtained,
40
代替了体力分量 X 及 Y,而,
则得到在平面应变条件下的相应方程。
y
TE
x
TE
?
?
???
?
?? ?
?
?
?
11 及
?
?
?
?
?? 11
TEmTEl 及
代替了面力分量 及 。 X Y
对于温度应力的平面应变问题，只须将温度应力平面应
力问题的
）（换成
换成
换成
???
?
?
?
?
?
?
?
1
1
1
2
E
E
41
§ 6-5 The introduction of displacement potential function
From last section we know that when solving the problems of temperature
stress by displacement under the situation of plane stress,we must let the
components of displacement u and v satisfy the differential equations
?
?
?
?
?
?
?
?
?
?
??
??
??
?
?
??
?
?
?
?
?
?
??
??
??
?
?
??
?
?
?
01
2
1
2
1
01
2
1
2
1
2
2
2
2
2
2
2
2
2
2
y
T
yx
u
x
v
y
v
x
T
yx
v
y
u
x
u
??
??
??
??
）（
）（
And the boundary conditions of displacement and stress must be satisfied
also on boundaries,We should do it by two steps when solving the problems,
(1) Figure out an arbitrary group of particular solution of the above differential
equations.It need only satisfy the differential equations,but not always satisfy
the boundary conditions,(2) Figure out a group of supplementary solution of
the differential equations ignoring temperature change T,which can satisfy the
boundary conditions after being superposed with the particular solution,
42
§ 6-5 位移势函数的引用
由上一节知：在平面应力的情况下按位移求解温度应力问
题时，须使位移分量 u 和 v 满足微分方程,
?
?
?
?
?
?
?
?
?
?
??
??
??
?
?
??
?
?
?
?
?
?
??
??
??
?
?
??
?
?
?
01
2
1
2
1
01
2
1
2
1
2
2
2
2
2
2
2
2
2
2
y
T
yx
u
x
v
y
v
x
T
yx
v
y
u
x
u
??
??
??
??
）（
）（
并在边界上满足位移边界条件和应力边界条件。实际求解时，
宜分两步进行,（ 1）求出上述微分的任意一组特解，它只需
满足微分方程，而不一定要满足边界条件。（ 2）不计变温 T，
求出微分方程的一组补充解，使它和特解叠加以后，能满足
边界条件。
43
Introduce into a function,and take the particular
solution of displacement as
），（ yx?
yvxu ?
??
?
?? ?? '',
y
T
y
x
T
x
?
????
?
?
?
????
?
?
???
???
）（
）（
1
1
2
2
?
u? v?
The function is called the potential function of displacement,
Introducing and into the differential equations instead of u and v
respectively and simplifying yields,
Because and are both constants,so when let ? ?
T??? ）（ ??? 12
），（ yx? satisfy the differential equations,So and can be a group
of particular solution of the differential equations,'u 'v
Introducing,and into the expression of
the components of stress expressed by the components of displacement and
the temperature change T
??? 21 1 ???Tyvxu ?????? ?? ','
44
引用一个函数,将位移特解取为,
yvxu ?
??
?
?? ?? '',
），（ yx?
函数 称为位移势函数。以 和 分别作为 u和 v代入微分
方程，简化后得,
? u? v?
y
T
y
x
T
x
?
?
???
?
?
?
?
???
?
?
???
???
）（
）（
1
1
2
2
由于 和 都是常量，所以取,? ?
T??? ）（ ??? 12
时,满足微分方程。因此, 可以作为微分方程
的一组特解。将
），（ yx? 'u 'v
以及 ?
?? 21
1 ?
??Tyvxu ?
??
?
?? ?? ','
代入位移分量和变温 T表示的应力分量表达式
45
)(
)1(2
1
)(
1
1
)(
1
2
2
y
u
x
vE
TE
x
u
y
vE
TE
y
v
x
uE
xy
y
x
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
yields the components of stress of corresponding particular solutions of
displacement
?
?
?
?
?
??
?
?
?
?
??
?
?
?
?
?
?
??
?
?
?
??
yx
E
x
E
y
E
xy
y
x
?
?
?
?
?
?
?
?
?
2
2
2
2
2
1
1
'
1
'
46
)(
)1(2
1
)(
1
1
)(
1
2
2
y
u
x
vE
TE
x
u
y
vE
TE
y
v
x
uE
xy
y
x
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
可得相应位移特解的应力分量是,
?
?
?
?
?
??
?
?
?
?
??
?
?
?
?
?
?
??
?
?
?
??
yx
E
x
E
y
E
xy
y
x
?
?
?
?
?
?
?
?
?
2
2
2
2
2
1
1
'
1
'
47
Suppose and are the supplementary solution of displacement,
Then, and must satisfy the homogeneous differential equations "u "v"u "v
?
?
?
?
?
?
?
?
??
??
?
?
??
?
?
?
?
??
??
?
?
??
?
?
?
0
"
2
1"
2
1"
0
"
2
1"
2
1"
2
2
2
2
2
2
2
2
2
2
yx
u
x
v
y
v
yx
v
y
u
x
u
??
??
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
)
""
(
)1(2
"
)
""
(
1
"
)
""
(
1
"
2
2
y
u
x
vE
x
u
y
vE
y
v
x
uE
xy
y
x
?
?
?
?
?
?
?
?The components of stress corresponding to the supplementary solution of
displacement are (Notice that the temperature change is ignored,i.e,T=0.)
48
设, 为位移的补充解，则, 需满足齐次微
分方程,
"u "u"v "v
?
?
?
?
?
?
?
?
??
??
?
?
??
?
?
?
?
??
??
?
?
??
?
?
?
0
"
2
1"
2
1"
0
"
2
1"
2
1"
2
2
2
2
2
2
2
2
2
2
yx
u
x
v
y
v
yx
v
y
u
x
u
??
??
相应于位移补充解的应力分量为（注意不计变温，即 T=0）,
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
)
""
(
)1(2
"
)
""
(
1
"
)
""
(
1
"
2
2
y
u
x
vE
x
u
y
vE
y
v
x
uE
xy
y
x
?
?
?
?
?
?
?
?
49
Thus the components of the total displacement are,"',"' vvvuuu ????
They should satisfy the boundary conditions of displacement,The
components of the total stress are,,
They should satisfy the boundary conditions of the stress,In the problems of
the stress boundary (no boundary condition of displacement),the
components of stress corresponding to the supplementary solution of
displacement can be expressed by stress function directly,i.e,
"',"',"' zzzyyyxxx ????????? ??????
yxxy xyyx ??
???
?
??
?
?? ?????? 2
2
2
2
2 ",","
in which the stress function can be chosen according to the request of the
boundary conditions of stress,
?
In the case of plane strain,for the above-mentioned equations,
） （
is displaced by
? ? ?
?
? ?
?
?
?
?
1
1
1 2
E E
is displaced by
is displaced by
50
总的应力分量是,"',"',"'
zzzyyyxxx ????????? ??????
需满足应力边界条件。在应力边界问题中（没有位移边界条
件），可以把相应于位移补充解的应力分量直接用应力函数
来表示，即
其中的应力函数 可以按照应力边界条件的要求来选取。
yxxy xyyx ??
???
?
??
?
?? ?????? 2
2
2
2
2
",","
?
在平面应变条件下，将上述各方程中的
）（换成
换成
换成
???
?
?
?
?
?
?
?
1
1
1
2
E
E
这样总的位移分量是,"',"' vvvuuu ???? 需满足位移边界条件
51
Solution,The differential equation that the potential function of
displacement need satisfy is ?
）（）（ 2202 11 byT ???? ???
Let 22 ByAy ???
Introducing it into the above equation yields,
）（）（ 220 11122 2 byTByA ???? ??
Example 1,The temperature change of the rectangular thin plate shown in
the figure is,
）（ 220 1 byTT ??
where T0 is a constant,If a>>b,evaluate the
temperature stress,y
o
a a
b
b x
212
1
2
1 00
b
TBTA ???? ）（，）（ ?????
Comparing the coefficient of the two sides yields,
52
例 1：图示矩形薄板中发生如下的变温,
其中的 T0 是常量。若,试求其温
度应力。
）（ 2
2
0 1 b
yTT ??
ba》
x
y
o
a a
b
b
解：位移势函数 所应满足的微分方程为 ?
）（）（ 2
2
0
2 11
b
yT ???? ???
22 ByAy ???
）（）（ 2
2
0 11122 2 b
yTByA ???? ??
212
1
2
1 00
b
TBTA ???? ）（，）（ ?????比较两边系数，得
代入上式，得
取
53
Substituting A and B back yields the potential function of displacement,
）（）（ 2420 1221 byyT ??? ???
So the components of stress corresponding to the particular solution of
displacement are,
0',0'),1(' 220 ????? xyyx byTE ????
To obtain the supplementary solutions,let and we can arrive
at the needed components of stress corresponding to the supplementary
solutions of displacement
2cy??
0",0",2" 222 ??????????? yxcy xyyx ?????
0"'
0"'
)1(2"' 2
2
0
???
???
?????
xyxyxy
yyy
xxx b
y
TEc
???
???
????Therefore,the total components of stress are,
0)(,0)(
0)(,0)(
??
??
????
????
byxybyy
axxyaxx
??
??
The boundary conditions require,
54
将 A,B回代，得位移势函数
于是相应于位移特解的应力分量为
为求补充解，取 可得所需要的相应于位移补充解
的应力分量,
）（）（ 2420 1221 byyT ??? ???
2cy??
0',0'),1(' 2
2
0 ????? xyyx b
yTE ????
0",0",2"
2
2
2
??????????? yxcy xyyx ?????
0"'
0"'
)1(2"'
2
2
0
???
???
?????
xyxyxy
yyy
xxx
b
y
TEc
???
???
????
因此，总的应力分量为
0)(,0)(
0)(,0)(
??
??
????
????
byxybyy
axxyaxx
??
??
边界条件要求
55
It is obvious that the last three conditions are satisfied,while the first
condition can’t be satisfied,But due to a>>b,the first condition can be
transformed to equivalent static condition by utilizing Saint-Venant
principle,i.e,the principal vector and principal moment of equal to
zero at the boundaries of, ax ??
x?
0)(,0)( ?? ?????? ?? y d ydy axb b xaxb b x ??
)1(2 220 byTEcx ??? ??
03
22 TEc ??
Introducing
into the above equation yields
0
0
)
3
1
(
2
2
0
?
?
??
xy
y
x
b
y
TE
?
?
??
So the temperature stresses of the rectangular plate are,
56
显然，后三个条件是满足的；而第一个条件不能满足，但由
于,可应用圣维南原理，把第一个条件变换为静力等效
条件，即，在 的边界上,的主矢量及主矩等于零,
将
ba》
ax ?? x?
0)(,0)( ?? ?????? ?? y d ydy axb b xaxb b x ??
)1(2 2
2
0 b
yTEc
x ??? ??
代入上式，求得
于是矩形板的温度应力为,
03
22 TEc ??
0
0
)
3
1
(
2
2
0
?
?
??
xy
y
x
b
y
TE
?
?
??
57
§ 6-6 The Plane thermal stress Problems
of Axisymmetric Temperature Field
For the elastic body of axisymmetric structure such as circle,annulus and
cylinder etc.,if the temperature change of them is also axisymmetric T=T(r),
then they can be simplified as the plane problems of thermal stress of
axisymmetric temperature field,which are suitable to be solved with polar
coordinate,
0
21
0
1
??
?
?
?
?
?
?
?
?
?
?
?
?
?
rrr
rrr
rr
rrr
???
??
??
?
?
??
?
??
are simplified in axisymmetric problems.The second equation is satisfied of
course.While the first one becomes
0????? rr rr ????
The equilibrium equations of plane stress problems ignoring body forces
58
§ 6-6 轴对称温度场平面热应力问题
对于圆形、圆环及圆筒等这类轴对称结构弹性体，若其
变温也是轴对称的 T=T（ r），则可简化为轴对称温度场平面
热应力问题。轴对称温度场平面热应力问题，宜采用极坐标
求解。
不考虑体积力平面应力问题平衡方程
0
21
0
1
??
?
?
?
?
?
?
?
?
?
?
?
?
?
rrr
rrr
rr
rrr
???
??
??
?
?
??
?
??
在轴对称问题中得到简化，其第二式自然满足；而第
一式成为
0????? rr rr ????
59
The geometric equations are simplified as
r
u
dr
du rr
r ?? ???,
T
E
T
E
r
rr
?????
?????
??
?
???
???
)(
1
)(
1
The physical equations are simplified as
?
?
???
?
?
?
?
???
?
?
??
?
?
??
?
?
?
??
?
?
1
)(
1
1
)(
1
2
2
TEE
TEE
r
rr
Expressing the stress with strains
60
几何方程简化为
r
u
dr
du rr
r ?? ???,
物理方程简化为
T
E
T
E
r
rr
?????
?????
??
?
???
???
)(
1
)(
1
将应力用应变表示
?
?
???
?
?
?
?
???
?
?
??
?
?
??
?
?
?
??
?
?
1
)(
1
1
)(
1
2
2
TEE
TEE
r
rr
61
Introducing the geometric equations into the above equations,then
introducing it into the equilibrium equations,yields the basic equation to
solve the axisymmetric thermal stress by displacement
dr
dT
r
u
dr
du
rdr
ud rrr ?? )1(1
22
2 ????
drdTrudrdrdrd r ?? )1()](1[ ??
Or it can be written as
Integrating it twice can yield the components of the displacement of the
axisymmetric problems
rBArT r d rru rar ???? ??? )1(
in which A and B are arbitrary constants,and the lower limit of the
integration is a,
0
])1()1[(
1
])1()1[(
1
222
222
?
????
?
??
???
?
???
?
?
?
?
?
???
?
?
?
??
?
?
?
r
r
a
r
ar
TE
r
B
A
E
Tr d r
r
E
r
B
A
E
Tr d r
r
E The components of stress can be obtained through the above equation,
62
将几何方程代入上式，然后将其代入平衡方程，得按位
移求解轴对称热应力的基本方程,
或写成,dr
dT
r
u
dr
du
rdr
ud rrr ?? )1(1
22
2
????
dr
dTru
dr
d
rdr
d
r ?? )1()](
1[ ??
积分两次可得到轴对称问题位移分量,
r
BArT r d r
ru
r
ar ??
?? ??? )1(
式中 A,B为任意常数，积分下限取为 a。由上式可得应力分
量,
0
])1()1[(
1
])1()1[(
1
222
222
?
????
?
??
???
?
???
?
?
?
?
?
???
?
?
?
??
?
?
?
r
r
a
r
a
r
TE
r
B
A
E
Tr d r
r
E
r
B
A
E
Tr d r
r
E
63
where the constants A and B are decided by the boundary conditions,
In the case of plane strain,it need only that
is displaced by
? ? ? ? 1
? ? 1 2 E E
is displaced by
? ） （ ? ? ? 1 is displaced by
a
b Ta T
b
Example 2 Suppose that there is a cylinder with
thick wall,It’s internal radius is a,external radius is
b, Heat up it from a homogeneous temperature,The
temperature rise of its internal surface is Ta,and that
of its external surface is Tb,as shown in the figure,
Evaluate the thermal stress after the thermal flux is
steady without heat resource in the cylinder,
?? c
W
z
T
y
T
x
T
t
T ?
?
??
?
??
?
??
?
? )(
2
2
2
2
2
2
02 ?? T
Solution,Evaluate temperature field at first,From the differential equation
of heat conduction
we arrive at the differential equation of heat conduction after the thermal
flux is steady without heat resource
64
其中常数 A,B由边界条件确定。
在平面应变的情况下，只需在以上各式中将
）（换成
换成
换成
???
?
?
?
?
?
?
?
1
1
1 2
E
E
例 2,设有一厚壁圆筒，内半径为 a，外半
径为 b。从一均匀温度加热，内表面增温 Ta,
外表面增温 Tb，如图所示。试求筒内无热源，
热流稳定后的热应力。
得无热源，热流稳定后的热传导微分方程为
解：首先求温度场。由热传导微分方程
?? c
W
z
T
y
T
x
T
t
T ?
?
??
?
??
?
??
?
? )(
2
2
2
2
2
2
02 ?? T
a
b Ta T
b
65
bbr
aar
TT
TT
?
?
?
?
）（
）（
b
a
r
a
T
a
b
r
b
TT ba
ln
ln
ln
ln
??
From the boundary conditions
we figure out A and B and substitute them back,then we obtain the
temperature field
Integrating two times yields
0)(
1
0
1
2
2
?
??
dr
dT
r
dr
d
r
T
dr
d
rdr
d
）（
or
For axisymmetric temperature field
BrAT ?? ln
66
对于轴对称温度场有
积分两次得,
0)(
1
0
1
2
2
?
??
dr
dT
r
dr
d
r
T
dr
d
rdr
d
）（
或
由边界条件,
求出 A,B后回代，得温度场,
BrAT ?? ln
bbr
aar
TT
TT
?
?
?
?
）（
）（
b
a
r
a
T
a
b
r
b
TT ba
ln
ln
ln
ln
??
67
Integrating yields
TE
a
b
a
b
r
b
TTE
a
b
r
b
a
b
r
b
TTE
a
b
r
b
a
b
r
b
TTE
ba
z
ba
ba
r
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
??
?
?
?
?
?
?
??
?
?
?
?
?
??
]
1
2
ln
1ln2
[
12
]
1
1
ln
1ln
[
12
]
1
1
ln
ln
[
12
2
2
2
2
2
2
2
2
2
2
）（
）（
）（
）（
）（
）（
]
2
[
1
][
1
][
1
22
2
22
22
2
22
22
2
TT r d r
ab
E
TrT r d rT r d r
ab
ar
r
E
T r d rT r d r
ab
ar
r
E
b
a
z
r
a
b
a
r
a
b
a
r
?
??
?
??
?
?
?
?
?
?
?
?
?
?
??
??
）（
）（
）（
?
?
?
?
?
?
?
?
?
?
Substitute T into the stress expressions of the problems of plane strain
68
积分后得
TE
a
b
a
b
r
b
TTE
a
b
r
b
a
b
r
b
TTE
a
b
r
b
a
b
r
b
TTE
ba
z
ba
ba
r
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
??
?
?
?
?
?
?
??
?
?
?
?
?
??
]
1
2
ln
1ln2
[
12
]
1
1
ln
1ln
[
12
]
1
1
ln
ln
[
12
2
2
2
2
2
2
2
2
2
2
）（
）（
）（
）（
）（
）（
]
2
[
1
][
1
][
1
22
2
22
22
2
22
22
2
TT r d r
ab
E
TrT r d rT r d r
ab
ar
r
E
T r d rT r d r
ab
ar
r
E
b
a
z
r
a
b
a
r
a
b
a
r
?
??
?
??
?
?
?
?
?
?
?
?
?
?
??
??
）（
）（
）（
?
?
?
?
?
?
?
?
?
?
将 T代入平面应变问题应力表达式
69
Solution ? ? ? ?
b
yTT
2c o s11 0
2 ?????? ?????
Let
We can arrived at
Therefore
b
yA
2c o s
?? ?
? ?
2
0
2 14
?
?? TbA ???
? ? byTb 2c o s14 02 2 ????? ???
x
y
o
a a b
b
b
yTT
2c o s0
??
Exercise 6.1 Show that the rectangular thin plate
undergoes temperature change
Please evaluate the temperature stress (assume a>>b),
70
练习 6.1 图示矩形薄板中发生变温
试求温度应力（假定 a远大于 b）
b
yTT
2c o s0
??
解,? ? ? ?
b
yTT
2c o s11 0
2 ?????? ?????
取
可解得
所以
b
yA
2c o s
?? ?
? ?
2
0
2 14
?
?? TbA ???
? ? byTb 2c o s14 02 2 ????? ???
x
y
o
a a b
b
71
Hence,we can arrived at
0
1
0
1
2
c o s
1
2
2
2
02
2
'
?
??
?
?
?
?
?
?
?
??
??
?
?
?
??
yx
E
x
E
b
y
ET
y
E
xy
y
x
?
?
?
?
?
?
?
?
?
?
?
’
’
2cy??
0
0
2
2
''
2
2
''
2
2
''
?
??
?
??
?
?
?
?
?
?
?
?
yx
x
c
y
xy
y
x
?
?
?
?
?
?Let
Then
72
由此得
0
1
0
1
2
c o s
1
2
2
2
02
2
'
?
??
?
?
?
?
?
?
?
??
??
?
?
?
??
yx
E
x
E
b
y
ET
y
E
xy
y
x
?
?
?
?
?
?
?
?
?
?
?
’
’
取 2cy??
则
0
0
2
2
''
2
2
''
2
2
''
?
??
?
??
?
?
?
?
?
?
?
?
yx
x
c
y
xy
y
x
?
?
?
?
?
?
73
0
0
2
2
c o s
'''
'''
0
'''
???
???
?????
xyxyxy
yyy
xxx c
b
y
ET
???
???
?
????Therefore
The boundary conditions
? ? ? ? ? ? 0,0,0 ??? ?????? axxybyxybyy ???
are satisfied apparently,
From ? ? 0??
? ?? dy
b
b axx?
0422c o s 00 ??
?
??
?
? ????
?
??
?
? ???
?
cETbdycbyETb
b ?
???i.e,
74
所以
0
0
2
2
c o s
'''
'''
0
'''
???
???
?????
xyxyxy
yyy
xxx c
b
y
ET
???
???
?
????
边界条件
? ? ? ? ? ? 0,0,0 ??? ?????? axxybyxybyy ???
显然满足
由 ? ? 0??
? ?? dy
b
b axx?
即
0422c o s 00 ??
?
??
?
? ????
?
??
?
? ???
?
cETbdycbyETb
b ?
???
75
So
0
0
2
c o s
2
0
?
?
?
?
?
?
?
?
??
xy
y
x
b
y
ET
?
?
?
?
??
?
? ETc 0?We can arrived at
While the boundary condition
? ? 0?
????
y d y
ax
b
b x?
is satisfied permanently,
76
得
?
? ETc 0?
而边界条件
? ? 0?
????
y d y
ax
b
b x?
恒成立。
故
0
0
2
c o s
2
0
?
?
?
?
?
?
?
?
??
xy
y
x
b
y
ET
?
?
?
?
??
77
? ? ??
?
?
???
? ???
2
2
01 1 b
rTTT
Exercise 6.2 It is known that a homogeneous disk with a radius b is placed
in a isothermal rigid hoop,The disk and the hoop are made of same material,
Suppose the disk is heated with the following rule,
The temperature of the hoop is kept normal temperature T0,And the strain
generated due to this temperature can be ignored,Evaluate the value of the
compression stress with a distance r from the center of the disk,
? ? ? ?a1 20 1 rcrcT r d rru r ???? ???
Solution,The expression of displacement of the plane stress problems of the
axisymmetric plate is
Because u is a finite value at the center,i.e,r=0,so c2=0,
78
练习 6.2 已知半径为 b的均质圆盘，置于等温刚性套箍内，
圆盘和套箍由相同的材料制成，设圆盘按如下规律加热
? ? ??
?
?
???
? ???
2
2
01 1 b
rTTT
套箍温度则保持为常温 T0，而由此温度所引起的应变可以
忽略，试求距圆盘中心为 r处的压应力值。
解,轴对称平板的平面应力问题的位移的表达式为
? ? ? ?a1 2
0 1 r
crcT r d r
ru
r ???? ???
由于在中心处，即 r=0处,u为有限值，因此 c2=0。
79
When r=b,u=0,Substitute it into（ a）,i.e,
? ? ? ? ? ?b0
42
1 12
42
01 ???
?
?
??
? ??? bc
b
bb
b
TT??
? ? ? ?
011 4
1 TTc ???? ??
Substituting the value of c1
? ? ? ?
? ? ? ?
?
?
?
?
?
?
???
?
???
?
?
?
?
?
?
???
?
???
?
?
22122
22122
1
11
1
1
1
11
1
1
r
cc
E
ETTr d r
r
E
r
cc
E
Tr d r
r
E
r
a
r
a
r
??
?
???
??
?
??
?
It can be obtained from expression (b) that
80
当 r=b时,u=0，代入（ a）即
? ? ? ? ? ?b0
42
1 12
42
01 ???
?
?
??
? ??? bc
b
bb
b
TT??
由（ b）式得
? ? ? ?
011 4
1 TTc ???? ??
将 c1的值代入
? ? ? ?
? ? ? ?
?
?
?
?
?
?
???
?
???
?
?
?
?
?
?
???
?
???
?
?
22122
22122
1
11
1
1
1
11
1
1
r
cc
E
ETTr d r
r
E
r
cc
E
Tr d r
r
E
r
a
r
a
r
??
?
???
??
?
??
?
81
yields
? ?
? ?
?
?
?
?
?
?
?
?
??
??
?
?
?
?
?
?
?
?
??
??
2
2
01
2
2
01
3
1
3
4
1
3
4
b
rTTE
b
rTTE
r
?
??
?
?
??
?
?
82
得
? ?
? ?
?
?
?
?
?
?
?
?
??
??
?
?
?
?
?
?
?
?
??
??
2
2
01
2
2
01
3
1
3
4
1
3
4
b
rTTE
b
rTTE
r
?
??
?
?
??
?
?
83
84
Elasticity
2
3
Chapter 6 The Basic Solution
of Temperature Stress Problems
§ 6-4 Solve plane problem of temperature stresses by
displacement
§ 6-3 The boundary conditions of temperature filed
§ 6-2 The differential equation of heat conduction
§ 6-1 The basic concept of temperature field and
heat conduction
§ 6-5 The introducing of potential function of
displacement
§ 6-6 The plane problems of thermal stresses in
axisymmetric temperature field
4
第六章 温度应力问题的基本解法
§ 6-4 按位移求解温度应力的平面问题
§ 6-3 温度场的边界条件
§ 6-2 热传导微分方程
§ 6-1 温度场和热传导的基本概念
§ 6-5 位移势函数的引用
§ 6-6 轴对称温度场平面热应力问题
5
When the temperature of a elastic body changes,its volume will expand or
contract,If the expansion or contraction can’t happen freely due to the external
restrictions or internal deformation compatibility demands,additional stresses
will be produced in the structure,These stresses produced by temperature
change are called thermal stresses,or temperature stresses,
Neglecting the effects of the temperature change on the material
performance,to solve the temperature stresses,we need two aspects of
calculation,(1) Solve the temperature field of the elastic body by the initial
conditions and boundary conditions,according to heat conduction equations,
And the difference between the former temperature field and the later
temperature field is the temperature change of the elastic body,(2) Solve the
temperature stresses of the elastic body according to the basic equations of the
elastic mechanics,This chapter will present these two aspects of calculation
simply,
6
当弹性体的温度变化时，其体积将趋于膨胀和收缩，若
外部的约束或内部的变形协调要求而使膨胀或收缩不能自由
发生时，结构中就会出现附加的应力。这种因温度变化而引
起的应力称为热应力，或温度应力。
忽略变温对材料性能的影响，为了求得温度应力，需要
进行两方面的计算：（ 1）由问题的初始条件、边界条件，
按热传导方程求解弹性体的温度场，而前后两个温度场之差
就是弹性体的变温。（ 2）按热弹性力学的基本方程求解弹
性体的温度应力。本章将对这两方面的计算进行简单的介绍。
7
§ 6-1 The Basic Concept of Temperature
Field And Heat Conduction
1.The temperature field,The total of the temperature at all the points in a
elastic body at a certain moment,denoted by T,
Unstable temperature filed or nonsteady temperature field,The temperature
in the temperature field changes with time,
i.e,T=T（ x,y,z,t）
Stable temperature filed or steady temperature field,The temperature in the
temperature field is only the function of positional coordinates,
i.e,T=T（ x,y,z）
Plane temperature field,The temperature in temperature field only changes
with two positional coordinates,
i.e,T=T（ x,y,t）
8
§ 6-1 温度场和热传导的基本概念
1.温度场：在任一瞬时，弹性体内所有各点的温度值的总体。用
T表示。
不稳定温度场或非定常温度场：温度场的温度随时间而变化。
即 T=T（ x,y,z,t）
稳定温度场或定常温度场：温度场的温度只是位置坐标的函数。
即 T=T（ x,y,z）
平面温度场：温度场的温度只随平面内的两个位置坐标而变。
即 T=T（ x,y,t）
9
2.Isothermal surface,The surface that
connects all the points with the same
temperature in the temperature field at a
certain moment,Apparently,the
temperature doesn’t changes along the
isothermal surface; The changing rate is
the largest along the normal direction of
the isothermal surface,
T+2△ T
T+△ T T
T-△ T x o
y
3.Temperature gradient:The vector that points to the direction in which
temperature increase along the normal direction of the isothermal surface,
It is denoted by △ T,and its value is denoted by,where n is the
normal direction of the isothermal surface,The components of
temperature gradient at each coordinate are
nT??
10
2.等温面：在任一瞬时，连接温度
场内温度相同各点的曲面。显然，
沿着等温面，温度不变；沿着等温
面的法线方向，温度的变化率最大。
T+2△ T
T+△ T T
T-△ T x o
y
nT??
3.温度梯度：沿等温面的法线方向，指向温度增大方向的矢
量。用 △ T表示，其大小用 表示。其中 n为等温面的法线方
向。温度梯度在各坐标轴的分量为
11
0n
Define to be the unit vector in normal direction of the isothermal
surface,pointing to the temperature increasing direction,
n
Tn
?
??
0
△ T （ 1）
4.Thermal flux speed,The quantity of heat flowing through the area S
on the isothermal surface in unit time,denoted by,
dt
dQ
)c o s (
)c o s (
)(c o s
zn
n
T
z
T
yn
n
T
y
T
xn
n
T
x
T
，
，
，
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
12
0n
取 为等温面法线方向且指向增温方向的单位矢量，则有
n
Tn
?
??
0
△ T （ 1）
4.热流速度：在单位时间内通过等温面面积 S 的热量。用 表示。 dtdQ
)c o s (
)c o s (
)(c o s
zn
n
T
z
T
yn
n
T
y
T
xn
n
T
x
T
，
，
，
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
13
Its value is,
SdtdQq /?
SdtdQnq /0?? (2)
Thermal flux density,The thermal flux speed flowing through unit area
on the isothermal surface,denoted by, Then we have q
5.The basic theorem of heat transfer,The thermal flux density is in
direct proportion to the temperature gradient and in the reverse
direction of it,i.e,
???q (3)
SnTdtdQ ??? /?
is called the coefficient of the heat transfer,Equations (1),(2) and (3)
lead to
?
14
热流密度：通过等温面单位面积的热流速度。用 表示，
则有 q
SdtdQq /?
其大小为
SdtdQnq /0??
(2)
SnTdtdQ ??? /?
? 称为导热系数。由（ 1）、（ 2）、（ 3）式得
5.热传导基本定理：热流密度与温度梯度成正比而方向相反。
即
???q (3) △ T
15
We can see that the coefficient of the heat transfer means,the thermal flux
speed through unit area of the isothermal surface per unit temperature
gradient”,
n
Tq
?
?? ?
From equations (1) and (3),we can see that the value of the thermal flux
density
The projections of the thermal flux density on axes,
x
Tq x
?
??? ?
y
Tq y
?
??? ?
zTq z ???? ?
It is obvious that the component of thermal flux density in any direction is
equal to the coefficient of heat transfer multiplied by the descending rate of
the temperature in this direction,
16
n
Tq
?
?? ?
由（ 1）和（ 3）可见，热流密度的大小
可见，导热系数表示, 在单位温度梯度下通过等温面单位面积
的热流速度, 。
热流密度在坐标轴上的投影
可见：热流密度在任一方向的分量，等于导热系数乘以
温度在该方向的递减率。
z
T
q
y
T
q
x
T
q
z
y
x
?
?
??
?
?
??
?
?
??
?
?
?
17
The principle of heat quantity equilibrium,Within any period of time,
the heat quantity accumulated in any minute part of the object equals the heat
quantity conducted into this minute part plus the heat quantity supplied by
internal heat source,
§ 6-2 The Differential Equation of Heat Conduction
dxxqq xx ???xq
x
y
z
Take a minute hexahedron dxdydz as shown in the above figure,Suppose
that the temperature of this hexahedron rises from T to, The heat
quantity accumulated by temperature is,where is the
density of the object,C is the heat quantity needed when the temperature of the
object with a unit mass rise one degree——specific thermal capability,
dttTT ???
dttTd x d y d zC ??? ?
18
热量平衡原理：在任意一段时间内，物体的任一微小部
分所积蓄的热量，等于传入该微小部分的热量加上内部热源
所供给的热量。
§ 6-2 热传导微分方程
dxxqq xx ???xq
x
y
z
取图示微小六面体 dxdydz。假定该六面体的温度在 dt时间内
由 T 升高到 。由温度所积蓄的热量是,
其中 是物体的密度,C 是单位质量的物体升高一度时所需的
热量 ——比热容。
dttTT ??? dttTd x d y d zC ???
?
19
Within the same period of time dt,the heat quantity qxdydzdt is conducted
into the hexahedron from left,and the heat quantity is
conducted out the hexahedron through right,Hence,the net heat quantity
conducted into is
d y d z d tdxxqq xx )( ???
d x d y d z d txq x???
x
Tq
x ?
??? ? Introduce into it, We can see that
d x d y d z d txT22???
d y d z d x d tyT22???
d z d x d y d tzT22???
The net heat quantity conducted into it from left and right is,
The net heat quantity conducted into it from top and bottom is,
The net heat quantity conducted into it from front and back is,
d x d y d z d tz Ty Tx T )( 222222 ?????????
Hence,the total net heat quantity conducted into the hexahedron is,
T d xd yd z d t2??
which can be abbreviated as,
20
在同一段时间 dt内，由六面体左面传入热量 qxdydzdt，
由右面传出热量 。因此，传入的净热量为
d y d z d tdxxqq xx )( ???
d x d y d z d txq x???
x
Tq
x ?
??? ?将 代入可见,
d x d y d z d txT2
2
?
??
d y d z d x d tyT22???
由左右两面传入的净热量为
由上下两面传入的净热量为
由前后两面传入的净热量为,
因此，传入六面体的总净热量为,
简记为,
d z d x d y d tzT22???
d x d y d z d tz Ty Tx T )( 222222 ?????????
T d xd yd z d t2??
21
Suppose that there is a positive heat resource to supply heat inside the
object,which supply heat quantity W per unit volume in unit time.,Then the
heat quantity that supplied by this heat resource during time dt is Wdxdydzdt,
According to the principle of heat quantity equilibrium,
W d x d y d z d tT d x d y d z d td x d y d z d txTc ????? 2??
??
?
c
WT
ct
T ???
?
? 2
?
?
ca ?
?c
WTa
t
T ???
?
? 2
which can be simplified as,
Let
This is the differential equations of heat transfer,
Thus
22
假定物体内部有正热源供热，在单位时间、单位体积供
热为 W，则该热源在时间 dt内所供热量为 Wdxdydzdt。
根据热量平衡原理得,
W d x d y d z d tT d x d y d z d td x d y d z d txTc ????? 2??
??
?
c
WT
ct
T ???
?
? 2
?
?
ca ?
?c
WTa
t
T ???
?
? 2
化简后得,
记
则
这就是热传导微分方程。
23
§ 6-3 The Boundary Conditions of Temperature Filed
To solve the differential equation,and sequentially solve the temperature
filed,the temperature of the object at initial moment must be known,i.e,the
so-called initial condition,At the same time,the rule of heat exchange
between the object surface and the surrounding medium after the initial
moment must be also known,i.e,the so-called boundary conditions,The
initial condition and the boundary conditions are called by a joint name of the
initial value conditions,
Initial condition,
Boundary conditions are divided into four kinds of forms,
The first kind of boundary condition,The temperature at any point on the
object surface is known at all moments,i.e,
where Ts is the surface temperature of the object,
),,()( 0 zyxfT t ??
)(tfT s ?
24
§ 6-3 温度场的边值条件
初始条件,
边界条件分四种形式,
第一类边界条件 已知物体表面上任意一点在所有瞬时
的温度，即
其中 Ts 是物体表面温度。
),,()( 0 zyxfT t ??
)(tfT s ?
为了能够求解热传导微分方程，从而求得温度场，必须
已知物体在初瞬时的温度，即所谓初始条件；同时还必须已
知初瞬时以后物体表面与周围介质之间热交换的规律，即所
谓边界条件。初始条件和边界条件合称为初值条件。
25
The second kind of boundary condition,The normal thermal flux
density at any point on the object surface is known,i.e,
where the subscript s means,surface”,and n means
“normal”,
)()( tfq sn ?
The third kind of boundary condition,The heat release situation of
convection at any point on the boundary of the object is known at all
moments,According to the convection theorem of heat quantity,the thermal
flux density transmitting from the object surface to the surrounding medium
per unit time is in direct proportion to the temperature difference between
them,i.e,
)()( essn TTq ?? ?
Where Te is the temperature of the surrounding medium; is called the
coefficient of the heat release of convection,or heat coefficient for short,?
The forth kind of boundary condition,It is known that the two objects
contact completely,and exchange heat through the form of heat conduction,i.e,
es TT ?
26
第三类边界条件 已知物体边界上任意一点在所有瞬时
的运流（对流）放热情况。按照热量的运流定理，在单位时
间内从物体表面传向周围介质的热流密度，是和两者的温差
成正比的，即
)()( essn TTq ?? ?
?
es TT ?
其中 Te是周围介质的温度； 称为运流放热系数，或简称热
系数。
第四类边界条件 已知两物体完全接触，并以热传导方
式进行热交换。即
第二类边界条件 已知物体表面上任意一点的法向热流密度,
即 其中角码 s 表示, 表面,,角码 n 表示法向 。 )()( tfq sn ?
27
§ 6-4 Solve Plane Problem of
Temperature Stress by Displacement
Suppose the temperature change of every point in the elastic body is T,
For an isotropic body,if there is no constricts,then the minute length at every
point of the elastic body will generate normal strain,(where is the
coefficient of expansion of the elastic body),Thus,the components of strain at
every point of the elastic body are
T? ?
0,?????? xyzxyzzyx T ???????
However,because the elastic body is restricted by the external restrictions
and mutual restrictions among each section in the object,the above-mentioned
deformations can not happen freely,Then the stress is produced,i.e,the so-
called temperature stress.This temperature stress will result in additional strain
due to the elasticity of the object,as expressed by Hooke’s law,Therefore,the
components of the total strain of the elastic body are
28
§ 6-4 按位移求解温度应力的平面问题
设弹性体内各点的温变为 T。 对于各向同性体，若不受约束，
则弹性体内各点的微小长度，都将产生正应变 （ 是弹性体
的膨胀系数），这样，弹性体内各点的形变分量为
T? ?
0,?????? xyzxyzzyx T ???????
但是，由于弹性体所受的外在约束以及体内各部分之间的
相互约束，上述形变并不能自由发生，于是就产生了应力，即
所谓温度应力。这个温度应力又将由于物体的弹性而引起附加
的形变，如虎克定理所示。因此，弹性体总的形变分量是,
29
T
E
T
E
T
E
yxzz
xzyy
zyxx
??????
??????
??????
????
????
????
)]([
1
)]([
1
)]([
1
xyxy
zxzx
yzyz
E
E
E
?
?
?
?
?
?
?
?
?
)1(2
)1(2
)1(2
?
?
?
?
?
?
For the temperature change problems of plane stress,the above equations are
simplified as
xyxy
xyy
yxx
E
T
E
T
E
?
?
?
?????
?????
)1(2
][
1
][
1
?
?
???
???
They are the physical equations of thermal elastic mechanics of the
problems of plane stress,
30
对于平面应力的变温问题，上式简化为
xyxy
zxzx
yzyz
E
E
E
?
?
?
?
?
?
?
?
?
)1(2
)1(2
)1(2
?
?
?
?
?
?
xyxy
xyy
yxx
E
T
E
T
E
?
?
?
?????
?????
)1(2
][
1
][
1
?
?
???
???
这就是平面应力问题热弹性力学的物理方程。
T
E
T
E
T
E
yxzz
xzyy
zyxx
??????
??????
??????
????
????
????
)]([
1
)]([
1
)]([
1
31
Express the components of stress by the components of strain and the
temperature change T,then the physical equations become
xyxy
xyy
yxx
E
TEE
TEE
?
?
?
?
?
???
?
?
?
?
???
?
?
)1(2
1
)(
1
1
)(
1
2
2
?
?
?
??
?
?
?
??
?
?
The geometric equations still are
y
u
x
v
y
v
x
u
xyyx ?
??
?
??
?
??
?
?? ???,,
Introducing the geometric equations into the physical equations yields
the components of stress which are expressed by the components of
displacement and temperature change T
32
将应力分量用形变分量和变温 T表示的物理方程为,
xyxy
xyy
yxx
E
TEE
TEE
?
?
?
?
?
???
?
?
?
?
???
?
?
)1(2
1
)(
1
1
)(
1
2
2
?
?
?
??
?
?
?
??
?
?
几何方程仍然为,
y
u
x
v
y
v
x
u
xyyx ?
??
?
??
?
??
?
?? ???,,
将几何方程代入物理方程，得用位移分量和变温 T 表示的应
力分量
33
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
）（
）（
）（
）（
y
u
x
vE
TE
x
u
y
vE
TE
y
v
x
uE
xy
y
x
?
?
?
?
?
?
?
?
?
?
?
?
12
11
11
2
2
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
0
0
xy
yx
xvy
yxx
??
??
Introducing the above equations into the differential equations of
equilibrium ignoring body forces
34
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
）（
）（
）（
）（
y
u
x
vE
TE
x
u
y
vE
TE
y
v
x
uE
xy
y
x
?
?
?
?
?
?
?
?
?
?
?
?
12
11
11
2
2
将上式代入不计体力的平衡微分方程
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
0
0
xy
yx
xvy
yxx
??
??
35
and simplifying yelds
01
2
1
2
1
01
2
1
2
1
2
2
2
2
2
2
2
2
2
2
?
?
?
??
??
??
?
?
??
?
?
?
?
?
?
??
??
??
?
?
??
?
?
?
y
T
yx
u
x
v
y
v
x
T
yx
v
y
u
x
u
??
??
??
??
）（
）（
（ 1）
??
???
??
??
0
0
sxysy
syxsx
lm
ml
）（）（
）（）（
??
??
These are the differential equations solving the problems of plane
stress of temperature stress by displacement,
In the same way,introducing the components of the stresses into
stress boundary conditions without surface force
36
简化得,
这就是按位移求解温度应力平面应力问题的微分方程。
同理，将应力分量代入无面力的应力边界条件
??
???
??
??
0
0
sxysy
syxsx
lm
ml
）（）（
）（）（
??
??
01
2
1
2
1
01
2
1
2
1
2
2
2
2
2
2
2
2
2
2
?
?
?
??
??
??
?
?
??
?
?
?
?
?
?
??
??
??
?
?
??
?
?
?
y
T
yx
u
x
v
y
v
x
T
yx
v
y
u
x
u
??
??
??
??
）（
）（
（ 1）
37
?
?
?
??
?
?
??
?
?
?
?
??
?
?
?
?
?
?
??
?
?
?
?
??
?
?
?
?
?
?
Tm
y
u
x
v
l
x
u
y
v
m
Tl
x
v
y
u
m
y
v
x
u
l
ss
ss
??
?
?
??
?
?
）（）（）（
）（）（）（
1
2
1
1
2
1
（ 2）
and simplifying yields
vvuu ss ??,
These are the stress boundary conditions to solve plane stress
problems of temperature stress by displacement,
The boundary conditions of displacement still are
Compare equations (1),(2) with the equations (1),(2) in § 2-8,chapter
2,We can see that the components X and Y of the body forces are
displaced by
38
简化后得,
这是按位移求解温度应力平面应力问题的应力边界条件。
位移边界条件仍然为,
vvuu ss ??,
将式 (1),(2)与第二章 § 2-8中式 (1),(2)对比，可见
?
?
?
?
?
?
?
??
?
?
?
?
??
?
?
?
?
?
?
??
?
?
?
?
??
?
?
?
?
?
?
Tm
y
u
x
v
l
x
u
y
v
m
Tl
x
v
y
u
m
y
v
x
u
l
ss
ss
??
?
?
??
?
?
）（）（）（
）（）（）（
1
2
1
1
2
1
（ 2）
39
y
TE
x
TE
?
?
???
?
?? ?
?
?
?
1 a n d 1
?
?
?
?
?? 1 a n d 1
TEmTEl
X YWhile the components and of the surface forces are displaced by
For plane strain problems of temperature stress,it is only needed that in
the plane stress problems of temperature stress
） （
is displaced by
? ? ?
?
? ?
?
?
?
?
1
1
1 2
E E
is displaced by
is displaced by
Then the corresponding equations under the conditions of plane strain are
obtained,
40
代替了体力分量 X 及 Y,而,
则得到在平面应变条件下的相应方程。
y
TE
x
TE
?
?
???
?
?? ?
?
?
?
11 及
?
?
?
?
?? 11
TEmTEl 及
代替了面力分量 及 。 X Y
对于温度应力的平面应变问题，只须将温度应力平面应
力问题的
）（换成
换成
换成
???
?
?
?
?
?
?
?
1
1
1
2
E
E
41
§ 6-5 The introduction of displacement potential function
From last section we know that when solving the problems of temperature
stress by displacement under the situation of plane stress,we must let the
components of displacement u and v satisfy the differential equations
?
?
?
?
?
?
?
?
?
?
??
??
??
?
?
??
?
?
?
?
?
?
??
??
??
?
?
??
?
?
?
01
2
1
2
1
01
2
1
2
1
2
2
2
2
2
2
2
2
2
2
y
T
yx
u
x
v
y
v
x
T
yx
v
y
u
x
u
??
??
??
??
）（
）（
And the boundary conditions of displacement and stress must be satisfied
also on boundaries,We should do it by two steps when solving the problems,
(1) Figure out an arbitrary group of particular solution of the above differential
equations.It need only satisfy the differential equations,but not always satisfy
the boundary conditions,(2) Figure out a group of supplementary solution of
the differential equations ignoring temperature change T,which can satisfy the
boundary conditions after being superposed with the particular solution,
42
§ 6-5 位移势函数的引用
由上一节知：在平面应力的情况下按位移求解温度应力问
题时，须使位移分量 u 和 v 满足微分方程,
?
?
?
?
?
?
?
?
?
?
??
??
??
?
?
??
?
?
?
?
?
?
??
??
??
?
?
??
?
?
?
01
2
1
2
1
01
2
1
2
1
2
2
2
2
2
2
2
2
2
2
y
T
yx
u
x
v
y
v
x
T
yx
v
y
u
x
u
??
??
??
??
）（
）（
并在边界上满足位移边界条件和应力边界条件。实际求解时，
宜分两步进行,（ 1）求出上述微分的任意一组特解，它只需
满足微分方程，而不一定要满足边界条件。（ 2）不计变温 T，
求出微分方程的一组补充解，使它和特解叠加以后，能满足
边界条件。
43
Introduce into a function,and take the particular
solution of displacement as
），（ yx?
yvxu ?
??
?
?? ?? '',
y
T
y
x
T
x
?
????
?
?
?
????
?
?
???
???
）（
）（
1
1
2
2
?
u? v?
The function is called the potential function of displacement,
Introducing and into the differential equations instead of u and v
respectively and simplifying yields,
Because and are both constants,so when let ? ?
T??? ）（ ??? 12
），（ yx? satisfy the differential equations,So and can be a group
of particular solution of the differential equations,'u 'v
Introducing,and into the expression of
the components of stress expressed by the components of displacement and
the temperature change T
??? 21 1 ???Tyvxu ?????? ?? ','
44
引用一个函数,将位移特解取为,
yvxu ?
??
?
?? ?? '',
），（ yx?
函数 称为位移势函数。以 和 分别作为 u和 v代入微分
方程，简化后得,
? u? v?
y
T
y
x
T
x
?
?
???
?
?
?
?
???
?
?
???
???
）（
）（
1
1
2
2
由于 和 都是常量，所以取,? ?
T??? ）（ ??? 12
时,满足微分方程。因此, 可以作为微分方程
的一组特解。将
），（ yx? 'u 'v
以及 ?
?? 21
1 ?
??Tyvxu ?
??
?
?? ?? ','
代入位移分量和变温 T表示的应力分量表达式
45
)(
)1(2
1
)(
1
1
)(
1
2
2
y
u
x
vE
TE
x
u
y
vE
TE
y
v
x
uE
xy
y
x
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
yields the components of stress of corresponding particular solutions of
displacement
?
?
?
?
?
??
?
?
?
?
??
?
?
?
?
?
?
??
?
?
?
??
yx
E
x
E
y
E
xy
y
x
?
?
?
?
?
?
?
?
?
2
2
2
2
2
1
1
'
1
'
46
)(
)1(2
1
)(
1
1
)(
1
2
2
y
u
x
vE
TE
x
u
y
vE
TE
y
v
x
uE
xy
y
x
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
可得相应位移特解的应力分量是,
?
?
?
?
?
??
?
?
?
?
??
?
?
?
?
?
?
??
?
?
?
??
yx
E
x
E
y
E
xy
y
x
?
?
?
?
?
?
?
?
?
2
2
2
2
2
1
1
'
1
'
47
Suppose and are the supplementary solution of displacement,
Then, and must satisfy the homogeneous differential equations "u "v"u "v
?
?
?
?
?
?
?
?
??
??
?
?
??
?
?
?
?
??
??
?
?
??
?
?
?
0
"
2
1"
2
1"
0
"
2
1"
2
1"
2
2
2
2
2
2
2
2
2
2
yx
u
x
v
y
v
yx
v
y
u
x
u
??
??
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
)
""
(
)1(2
"
)
""
(
1
"
)
""
(
1
"
2
2
y
u
x
vE
x
u
y
vE
y
v
x
uE
xy
y
x
?
?
?
?
?
?
?
?The components of stress corresponding to the supplementary solution of
displacement are (Notice that the temperature change is ignored,i.e,T=0.)
48
设, 为位移的补充解，则, 需满足齐次微
分方程,
"u "u"v "v
?
?
?
?
?
?
?
?
??
??
?
?
??
?
?
?
?
??
??
?
?
??
?
?
?
0
"
2
1"
2
1"
0
"
2
1"
2
1"
2
2
2
2
2
2
2
2
2
2
yx
u
x
v
y
v
yx
v
y
u
x
u
??
??
相应于位移补充解的应力分量为（注意不计变温，即 T=0）,
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
)
""
(
)1(2
"
)
""
(
1
"
)
""
(
1
"
2
2
y
u
x
vE
x
u
y
vE
y
v
x
uE
xy
y
x
?
?
?
?
?
?
?
?
49
Thus the components of the total displacement are,"',"' vvvuuu ????
They should satisfy the boundary conditions of displacement,The
components of the total stress are,,
They should satisfy the boundary conditions of the stress,In the problems of
the stress boundary (no boundary condition of displacement),the
components of stress corresponding to the supplementary solution of
displacement can be expressed by stress function directly,i.e,
"',"',"' zzzyyyxxx ????????? ??????
yxxy xyyx ??
???
?
??
?
?? ?????? 2
2
2
2
2 ",","
in which the stress function can be chosen according to the request of the
boundary conditions of stress,
?
In the case of plane strain,for the above-mentioned equations,
） （
is displaced by
? ? ?
?
? ?
?
?
?
?
1
1
1 2
E E
is displaced by
is displaced by
50
总的应力分量是,"',"',"'
zzzyyyxxx ????????? ??????
需满足应力边界条件。在应力边界问题中（没有位移边界条
件），可以把相应于位移补充解的应力分量直接用应力函数
来表示，即
其中的应力函数 可以按照应力边界条件的要求来选取。
yxxy xyyx ??
???
?
??
?
?? ?????? 2
2
2
2
2
",","
?
在平面应变条件下，将上述各方程中的
）（换成
换成
换成
???
?
?
?
?
?
?
?
1
1
1
2
E
E
这样总的位移分量是,"',"' vvvuuu ???? 需满足位移边界条件
51
Solution,The differential equation that the potential function of
displacement need satisfy is ?
）（）（ 2202 11 byT ???? ???
Let 22 ByAy ???
Introducing it into the above equation yields,
）（）（ 220 11122 2 byTByA ???? ??
Example 1,The temperature change of the rectangular thin plate shown in
the figure is,
）（ 220 1 byTT ??
where T0 is a constant,If a>>b,evaluate the
temperature stress,y
o
a a
b
b x
212
1
2
1 00
b
TBTA ???? ）（，）（ ?????
Comparing the coefficient of the two sides yields,
52
例 1：图示矩形薄板中发生如下的变温,
其中的 T0 是常量。若,试求其温
度应力。
）（ 2
2
0 1 b
yTT ??
ba》
x
y
o
a a
b
b
解：位移势函数 所应满足的微分方程为 ?
）（）（ 2
2
0
2 11
b
yT ???? ???
22 ByAy ???
）（）（ 2
2
0 11122 2 b
yTByA ???? ??
212
1
2
1 00
b
TBTA ???? ）（，）（ ?????比较两边系数，得
代入上式，得
取
53
Substituting A and B back yields the potential function of displacement,
）（）（ 2420 1221 byyT ??? ???
So the components of stress corresponding to the particular solution of
displacement are,
0',0'),1(' 220 ????? xyyx byTE ????
To obtain the supplementary solutions,let and we can arrive
at the needed components of stress corresponding to the supplementary
solutions of displacement
2cy??
0",0",2" 222 ??????????? yxcy xyyx ?????
0"'
0"'
)1(2"' 2
2
0
???
???
?????
xyxyxy
yyy
xxx b
y
TEc
???
???
????Therefore,the total components of stress are,
0)(,0)(
0)(,0)(
??
??
????
????
byxybyy
axxyaxx
??
??
The boundary conditions require,
54
将 A,B回代，得位移势函数
于是相应于位移特解的应力分量为
为求补充解，取 可得所需要的相应于位移补充解
的应力分量,
）（）（ 2420 1221 byyT ??? ???
2cy??
0',0'),1(' 2
2
0 ????? xyyx b
yTE ????
0",0",2"
2
2
2
??????????? yxcy xyyx ?????
0"'
0"'
)1(2"'
2
2
0
???
???
?????
xyxyxy
yyy
xxx
b
y
TEc
???
???
????
因此，总的应力分量为
0)(,0)(
0)(,0)(
??
??
????
????
byxybyy
axxyaxx
??
??
边界条件要求
55
It is obvious that the last three conditions are satisfied,while the first
condition can’t be satisfied,But due to a>>b,the first condition can be
transformed to equivalent static condition by utilizing Saint-Venant
principle,i.e,the principal vector and principal moment of equal to
zero at the boundaries of, ax ??
x?
0)(,0)( ?? ?????? ?? y d ydy axb b xaxb b x ??
)1(2 220 byTEcx ??? ??
03
22 TEc ??
Introducing
into the above equation yields
0
0
)
3
1
(
2
2
0
?
?
??
xy
y
x
b
y
TE
?
?
??
So the temperature stresses of the rectangular plate are,
56
显然，后三个条件是满足的；而第一个条件不能满足，但由
于,可应用圣维南原理，把第一个条件变换为静力等效
条件，即，在 的边界上,的主矢量及主矩等于零,
将
ba》
ax ?? x?
0)(,0)( ?? ?????? ?? y d ydy axb b xaxb b x ??
)1(2 2
2
0 b
yTEc
x ??? ??
代入上式，求得
于是矩形板的温度应力为,
03
22 TEc ??
0
0
)
3
1
(
2
2
0
?
?
??
xy
y
x
b
y
TE
?
?
??
57
§ 6-6 The Plane thermal stress Problems
of Axisymmetric Temperature Field
For the elastic body of axisymmetric structure such as circle,annulus and
cylinder etc.,if the temperature change of them is also axisymmetric T=T(r),
then they can be simplified as the plane problems of thermal stress of
axisymmetric temperature field,which are suitable to be solved with polar
coordinate,
0
21
0
1
??
?
?
?
?
?
?
?
?
?
?
?
?
?
rrr
rrr
rr
rrr
???
??
??
?
?
??
?
??
are simplified in axisymmetric problems.The second equation is satisfied of
course.While the first one becomes
0????? rr rr ????
The equilibrium equations of plane stress problems ignoring body forces
58
§ 6-6 轴对称温度场平面热应力问题
对于圆形、圆环及圆筒等这类轴对称结构弹性体，若其
变温也是轴对称的 T=T（ r），则可简化为轴对称温度场平面
热应力问题。轴对称温度场平面热应力问题，宜采用极坐标
求解。
不考虑体积力平面应力问题平衡方程
0
21
0
1
??
?
?
?
?
?
?
?
?
?
?
?
?
?
rrr
rrr
rr
rrr
???
??
??
?
?
??
?
??
在轴对称问题中得到简化，其第二式自然满足；而第
一式成为
0????? rr rr ????
59
The geometric equations are simplified as
r
u
dr
du rr
r ?? ???,
T
E
T
E
r
rr
?????
?????
??
?
???
???
)(
1
)(
1
The physical equations are simplified as
?
?
???
?
?
?
?
???
?
?
??
?
?
??
?
?
?
??
?
?
1
)(
1
1
)(
1
2
2
TEE
TEE
r
rr
Expressing the stress with strains
60
几何方程简化为
r
u
dr
du rr
r ?? ???,
物理方程简化为
T
E
T
E
r
rr
?????
?????
??
?
???
???
)(
1
)(
1
将应力用应变表示
?
?
???
?
?
?
?
???
?
?
??
?
?
??
?
?
?
??
?
?
1
)(
1
1
)(
1
2
2
TEE
TEE
r
rr
61
Introducing the geometric equations into the above equations,then
introducing it into the equilibrium equations,yields the basic equation to
solve the axisymmetric thermal stress by displacement
dr
dT
r
u
dr
du
rdr
ud rrr ?? )1(1
22
2 ????
drdTrudrdrdrd r ?? )1()](1[ ??
Or it can be written as
Integrating it twice can yield the components of the displacement of the
axisymmetric problems
rBArT r d rru rar ???? ??? )1(
in which A and B are arbitrary constants,and the lower limit of the
integration is a,
0
])1()1[(
1
])1()1[(
1
222
222
?
????
?
??
???
?
???
?
?
?
?
?
???
?
?
?
??
?
?
?
r
r
a
r
ar
TE
r
B
A
E
Tr d r
r
E
r
B
A
E
Tr d r
r
E The components of stress can be obtained through the above equation,
62
将几何方程代入上式，然后将其代入平衡方程，得按位
移求解轴对称热应力的基本方程,
或写成,dr
dT
r
u
dr
du
rdr
ud rrr ?? )1(1
22
2
????
dr
dTru
dr
d
rdr
d
r ?? )1()](
1[ ??
积分两次可得到轴对称问题位移分量,
r
BArT r d r
ru
r
ar ??
?? ??? )1(
式中 A,B为任意常数，积分下限取为 a。由上式可得应力分
量,
0
])1()1[(
1
])1()1[(
1
222
222
?
????
?
??
???
?
???
?
?
?
?
?
???
?
?
?
??
?
?
?
r
r
a
r
a
r
TE
r
B
A
E
Tr d r
r
E
r
B
A
E
Tr d r
r
E
63
where the constants A and B are decided by the boundary conditions,
In the case of plane strain,it need only that
is displaced by
? ? ? ? 1
? ? 1 2 E E
is displaced by
? ） （ ? ? ? 1 is displaced by
a
b Ta T
b
Example 2 Suppose that there is a cylinder with
thick wall,It’s internal radius is a,external radius is
b, Heat up it from a homogeneous temperature,The
temperature rise of its internal surface is Ta,and that
of its external surface is Tb,as shown in the figure,
Evaluate the thermal stress after the thermal flux is
steady without heat resource in the cylinder,
?? c
W
z
T
y
T
x
T
t
T ?
?
??
?
??
?
??
?
? )(
2
2
2
2
2
2
02 ?? T
Solution,Evaluate temperature field at first,From the differential equation
of heat conduction
we arrive at the differential equation of heat conduction after the thermal
flux is steady without heat resource
64
其中常数 A,B由边界条件确定。
在平面应变的情况下，只需在以上各式中将
）（换成
换成
换成
???
?
?
?
?
?
?
?
1
1
1 2
E
E
例 2,设有一厚壁圆筒，内半径为 a，外半
径为 b。从一均匀温度加热，内表面增温 Ta,
外表面增温 Tb，如图所示。试求筒内无热源，
热流稳定后的热应力。
得无热源，热流稳定后的热传导微分方程为
解：首先求温度场。由热传导微分方程
?? c
W
z
T
y
T
x
T
t
T ?
?
??
?
??
?
??
?
? )(
2
2
2
2
2
2
02 ?? T
a
b Ta T
b
65
bbr
aar
TT
TT
?
?
?
?
）（
）（
b
a
r
a
T
a
b
r
b
TT ba
ln
ln
ln
ln
??
From the boundary conditions
we figure out A and B and substitute them back,then we obtain the
temperature field
Integrating two times yields
0)(
1
0
1
2
2
?
??
dr
dT
r
dr
d
r
T
dr
d
rdr
d
）（
or
For axisymmetric temperature field
BrAT ?? ln
66
对于轴对称温度场有
积分两次得,
0)(
1
0
1
2
2
?
??
dr
dT
r
dr
d
r
T
dr
d
rdr
d
）（
或
由边界条件,
求出 A,B后回代，得温度场,
BrAT ?? ln
bbr
aar
TT
TT
?
?
?
?
）（
）（
b
a
r
a
T
a
b
r
b
TT ba
ln
ln
ln
ln
??
67
Integrating yields
TE
a
b
a
b
r
b
TTE
a
b
r
b
a
b
r
b
TTE
a
b
r
b
a
b
r
b
TTE
ba
z
ba
ba
r
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
??
?
?
?
?
?
?
??
?
?
?
?
?
??
]
1
2
ln
1ln2
[
12
]
1
1
ln
1ln
[
12
]
1
1
ln
ln
[
12
2
2
2
2
2
2
2
2
2
2
）（
）（
）（
）（
）（
）（
]
2
[
1
][
1
][
1
22
2
22
22
2
22
22
2
TT r d r
ab
E
TrT r d rT r d r
ab
ar
r
E
T r d rT r d r
ab
ar
r
E
b
a
z
r
a
b
a
r
a
b
a
r
?
??
?
??
?
?
?
?
?
?
?
?
?
?
??
??
）（
）（
）（
?
?
?
?
?
?
?
?
?
?
Substitute T into the stress expressions of the problems of plane strain
68
积分后得
TE
a
b
a
b
r
b
TTE
a
b
r
b
a
b
r
b
TTE
a
b
r
b
a
b
r
b
TTE
ba
z
ba
ba
r
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
??
?
?
?
?
?
?
??
?
?
?
?
?
??
]
1
2
ln
1ln2
[
12
]
1
1
ln
1ln
[
12
]
1
1
ln
ln
[
12
2
2
2
2
2
2
2
2
2
2
）（
）（
）（
）（
）（
）（
]
2
[
1
][
1
][
1
22
2
22
22
2
22
22
2
TT r d r
ab
E
TrT r d rT r d r
ab
ar
r
E
T r d rT r d r
ab
ar
r
E
b
a
z
r
a
b
a
r
a
b
a
r
?
??
?
??
?
?
?
?
?
?
?
?
?
?
??
??
）（
）（
）（
?
?
?
?
?
?
?
?
?
?
将 T代入平面应变问题应力表达式
69
Solution ? ? ? ?
b
yTT
2c o s11 0
2 ?????? ?????
Let
We can arrived at
Therefore
b
yA
2c o s
?? ?
? ?
2
0
2 14
?
?? TbA ???
? ? byTb 2c o s14 02 2 ????? ???
x
y
o
a a b
b
b
yTT
2c o s0
??
Exercise 6.1 Show that the rectangular thin plate
undergoes temperature change
Please evaluate the temperature stress (assume a>>b),
70
练习 6.1 图示矩形薄板中发生变温
试求温度应力（假定 a远大于 b）
b
yTT
2c o s0
??
解,? ? ? ?
b
yTT
2c o s11 0
2 ?????? ?????
取
可解得
所以
b
yA
2c o s
?? ?
? ?
2
0
2 14
?
?? TbA ???
? ? byTb 2c o s14 02 2 ????? ???
x
y
o
a a b
b
71
Hence,we can arrived at
0
1
0
1
2
c o s
1
2
2
2
02
2
'
?
??
?
?
?
?
?
?
?
??
??
?
?
?
??
yx
E
x
E
b
y
ET
y
E
xy
y
x
?
?
?
?
?
?
?
?
?
?
?
’
’
2cy??
0
0
2
2
''
2
2
''
2
2
''
?
??
?
??
?
?
?
?
?
?
?
?
yx
x
c
y
xy
y
x
?
?
?
?
?
?Let
Then
72
由此得
0
1
0
1
2
c o s
1
2
2
2
02
2
'
?
??
?
?
?
?
?
?
?
??
??
?
?
?
??
yx
E
x
E
b
y
ET
y
E
xy
y
x
?
?
?
?
?
?
?
?
?
?
?
’
’
取 2cy??
则
0
0
2
2
''
2
2
''
2
2
''
?
??
?
??
?
?
?
?
?
?
?
?
yx
x
c
y
xy
y
x
?
?
?
?
?
?
73
0
0
2
2
c o s
'''
'''
0
'''
???
???
?????
xyxyxy
yyy
xxx c
b
y
ET
???
???
?
????Therefore
The boundary conditions
? ? ? ? ? ? 0,0,0 ??? ?????? axxybyxybyy ???
are satisfied apparently,
From ? ? 0??
? ?? dy
b
b axx?
0422c o s 00 ??
?
??
?
? ????
?
??
?
? ???
?
cETbdycbyETb
b ?
???i.e,
74
所以
0
0
2
2
c o s
'''
'''
0
'''
???
???
?????
xyxyxy
yyy
xxx c
b
y
ET
???
???
?
????
边界条件
? ? ? ? ? ? 0,0,0 ??? ?????? axxybyxybyy ???
显然满足
由 ? ? 0??
? ?? dy
b
b axx?
即
0422c o s 00 ??
?
??
?
? ????
?
??
?
? ???
?
cETbdycbyETb
b ?
???
75
So
0
0
2
c o s
2
0
?
?
?
?
?
?
?
?
??
xy
y
x
b
y
ET
?
?
?
?
??
?
? ETc 0?We can arrived at
While the boundary condition
? ? 0?
????
y d y
ax
b
b x?
is satisfied permanently,
76
得
?
? ETc 0?
而边界条件
? ? 0?
????
y d y
ax
b
b x?
恒成立。
故
0
0
2
c o s
2
0
?
?
?
?
?
?
?
?
??
xy
y
x
b
y
ET
?
?
?
?
??
77
? ? ??
?
?
???
? ???
2
2
01 1 b
rTTT
Exercise 6.2 It is known that a homogeneous disk with a radius b is placed
in a isothermal rigid hoop,The disk and the hoop are made of same material,
Suppose the disk is heated with the following rule,
The temperature of the hoop is kept normal temperature T0,And the strain
generated due to this temperature can be ignored,Evaluate the value of the
compression stress with a distance r from the center of the disk,
? ? ? ?a1 20 1 rcrcT r d rru r ???? ???
Solution,The expression of displacement of the plane stress problems of the
axisymmetric plate is
Because u is a finite value at the center,i.e,r=0,so c2=0,
78
练习 6.2 已知半径为 b的均质圆盘，置于等温刚性套箍内，
圆盘和套箍由相同的材料制成，设圆盘按如下规律加热
? ? ??
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?
???
? ???
2
2
01 1 b
rTTT
套箍温度则保持为常温 T0，而由此温度所引起的应变可以
忽略，试求距圆盘中心为 r处的压应力值。
解,轴对称平板的平面应力问题的位移的表达式为
? ? ? ?a1 2
0 1 r
crcT r d r
ru
r ???? ???
由于在中心处，即 r=0处,u为有限值，因此 c2=0。
79
When r=b,u=0,Substitute it into（ a）,i.e,
? ? ? ? ? ?b0
42
1 12
42
01 ???
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??
? ??? bc
b
bb
b
TT??
? ? ? ?
011 4
1 TTc ???? ??
Substituting the value of c1
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22122
22122
1
11
1
1
1
11
1
1
r
cc
E
ETTr d r
r
E
r
cc
E
Tr d r
r
E
r
a
r
a
r
??
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???
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?
It can be obtained from expression (b) that
80
当 r=b时,u=0，代入（ a）即
? ? ? ? ? ?b0
42
1 12
42
01 ???
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??
? ??? bc
b
bb
b
TT??
由（ b）式得
? ? ? ?
011 4
1 TTc ???? ??
将 c1的值代入
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22122
22122
1
11
1
1
1
11
1
1
r
cc
E
ETTr d r
r
E
r
cc
E
Tr d r
r
E
r
a
r
a
r
??
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???
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81
yields
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2
2
01
2
2
01
3
1
3
4
1
3
4
b
rTTE
b
rTTE
r
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82
得
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2
2
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3
1
3
4
1
3
4
b
rTTE
b
rTTE
r
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83
84
