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1、 Elasticity12Chapter 6 The Basic Solution of Temperature Stress Problems6-4 Solve plane problem of temperature stresses by displacement6-3 The boundary conditions of temperature filed6-2 The differential equation of heat conduction 6-1 The basic concept of temperature field and heat conduction6-5 Th
2、e introducing of potential function of displacement6-6 The plane problems of thermal stresses in axisymmetric temperature field3第六章第六章 温度应力问题的基本解法温度应力问题的基本解法6-4 6-4 按位移求解温度应力的平面问题按位移求解温度应力的平面问题6-3 6-3 温度场的边界条件温度场的边界条件6-2 6-2 热传导微分方程热传导微分方程6-1 6-1 温度场和热传导的基本概念温度场和热传导的基本概念6-5 6-5 位移势函数的引用位移势函数的引用6-6 6
3、-6 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4 When the temperature of a elastic body changes, its volume will expand or contract. If the expansion or contraction cant happen freely due to the external restrictions or internal deformation compatibility demands, additional stresses will be produced in the structure.
4、 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 bod
5、y 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
6、basic equations of the elastic mechanics. This chapter will present these two aspects of calculation simply.5 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
7、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。66-1 The Basic Concept of Temperature Field And Heat Conduction1.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 th
8、e 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
9、positional coordinates. i.e. T=T（x，y，t）76-1 6-1 温度场和热传导的基本概念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）82.Isothermal surface: The surface that connects all the points with
10、the same temperature in the temperature field at a certain moment. Apparently, the temperature doesnt changes along the isothermal surface; The changing rate is the largest along the normal direction of the isothermal surface. T+2TT+TTT-Txoy3.Temperature gradient:The vector that points to the direct
11、ion in which temperature increase along the normal direction of the isothermal surface. It is denoted by TT, and its value is denoted by , where n is the normal direction of the isothermal surface. The components of temperature gradient at each coordinate are92.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
12、变；沿着等温面的法线方向，温度的变化率最大。T+2TT+TTT-Txoy3.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TT表示，其大小用 表示。其中n n为等温面的法线方向。温度梯度在各坐标轴的分量为10Define to be the unit vector in normal direction of the isothermal surface, pointing to the temperature increasing direction.TT（1）4.Thermal flux speed: The quantity of heat flowing through th
13、e area S on the isothermal surface in unit time, denoted by .11取 为等温面法线方向且指向增温方向的单位矢量，则有TT（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。12Its value is: (2) Thermal flux density: The thermal flux speed flowing through unit area on the isothermal surface, denoted by . Then we have5.The basic theorem of heat trans
14、fer: The thermal flux density is in direct proportion to the temperature gradient and in the reverse direction of it. i.e.(3) is called the coefficient of the heat transfer. Equations (1), (2) and (3) lead to13热流密度：通过等温面单位面积的热流速度。用 表示，则有 其大小为(2)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 (3)T
15、14We 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”. From equations (1) and (3), we can see that the value of the thermal flux densityThe projections of the thermal flux density on axes: It is
16、 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.15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。16 The princip
17、le 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 Conductionxyz Take a minut
18、e 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 degrees
19、pecific thermal capability.17 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 6-2 热传导微分方程热传导微分方程xyz 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。18 Within the same period of time dt, the heat quantity qxdydzdt is conducted into the hexahedron from
20、left, and the heat quantity is conducted out the hexahedron through right. Hence, the net heat quantity conducted into is Introduce into it . We can see thatThe 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 quan
21、tity conducted into it from front and back is: Hence, the total net heat quantity conducted into the hexahedron is:which can be abbreviated as:19 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，传入的净热量为将 代入可见：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：20 Suppose that there is a positive heat re
22、source 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,which can be simplified as: LetThis is the differential e
23、quations of heat transfer.Thus21 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：化简后得：记则这就是热传导微分方程。226-3 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
24、 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 jo
25、int 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.236-3 6-3 温
26、度场的边值条件温度场的边值条件 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。 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。24 The second kind of boundary condition: The normal thermal flux density at any point on the object surface is known, i.e.
27、where the subscript s means “surface”, and n means “normal”. 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
28、from the object surface to the surrounding medium per unit time is in direct proportion to the temperature difference between them, i.e. 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 o
29、f boundary condition: It is known that the two objects contact completely, and exchange heat through the form of heat conduction, i.e.25 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 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
30、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。266-4 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
31、 strain , (where is the coefficient of expansion of the elastic body). Thus, the components of strain at every point of the elastic body are However, because the elastic body is restricted by the external restrictions and mutual restrictions among each section in the object, the above-mentioned defo
32、rmations 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 Hookes law. Therefore, the components of the total strain of the elastic body are Suppose t
33、he 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 ever
34、y point of the elastic body are 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 tempe
35、rature stress will result in additional strain due to the elasticity of the object, as expressed by Hookes law. Therefore, the components of the total strain of the elastic body are276-4 6-4 按位移求解温度应力的平面问题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 但是，
36、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：28 For the temperature change problems of plane stress, the above equations are simplified as They are the physical equations of thermal elastic mechanics of the problems of plane stress . 29对于平面应力的
37、变温问题，上式简化为这就是平面应力问题热弹性力学的物理方程。30 Express the components of stress by the components of strain and the temperature change T, then the physical equations become The geometric equations still are Introducing the geometric equations into the physical equations yields the components of stress which are e
38、xpressed by the components of displacement and temperature change T 31将应力分量用形变分量和变温T表示的物理方程为：几何方程仍然为：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32 Introducing the above equations into the differential equations of equilibrium ignoring body forces 33将上式代入不计体力的平衡微分方程34and simplifying yelds（1） These are the differe
39、ntial 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 force35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（1）36（2）and simplifying yields These are the
40、 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 by37简化后得
41、：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： 将式(1)、(2)与第二章2-8中式(1)、(2)对比，可见（2）38While 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 bymaammmm+-1112EEis displa
42、ced byis displaced byThen the corresponding equations under the conditions of plane strain are obtained. 39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。代替了面力分量 及 。 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406-5 The introduction of displacement potential function From last section we know that when solving the problems of t
43、emperature stress by displacement under the situation of plane stress, we must let the components of displacement u and v satisfy the differential equationsAnd the boundary conditions of displacement and stress must be satisfied also on boundaries. We should do it by two steps when solving the probl
44、ems: (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 chang
45、e T ,which can satisfy the boundary conditions after being superposed with the particular solution.416-5 6-5 位移势函数的引用位移势函数的引用 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：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。42 Introduce in
46、to a function , and take the particular solution of displacement as 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 satisfy the different
47、ial equations. So and can be a group of particular solution of the differential equations.Introducing .and into the expression of the components of stress expressed by the components of displacement and the temperature change T43 引用一个函数 ，将位移特解取为：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：由于 和 都是常量，所以取：时， 满足微
48、分方程。因此 ， 可以作为微分方程的一组特解。将以及代入位移分量和变温T表示的应力分量表达式44yields the components of stress of corresponding particular solutions of displacement 45可得相应位移特解的应力分量是：46 Suppose and are the supplementary solution of displacement. Then . and must satisfy the homogeneous differential equationsThe components of stress
49、 corresponding to the supplementary solution of displacement are (Notice that the temperature change is ignored, i.e. T=0.)47 设 ， 为位移的补充解，则 ， 需满足齐次微分方程：相应于位移补充解的应力分量为（注意不计变温，即T=0）：48 Thus the components of the total displacement are: They should satisfy the boundary conditions of displacement. The c
50、omponents 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 dir
51、ectly, i.e.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 bymaammmm+-1112EEis displaced byis displaced by49总的应力分量是：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
52、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。 在平面应变条件下，将上述各方程中的这样总的位移分量是：需满足位移边界条件50Solution: The differential equation that the potential function of displacement need satisfy isLetIntroducing it into the above equation yields:Example 1: The temperature change of the rectangular thin plate shown in the figure
53、 is:where T0 is a constant. If ab, evaluate the temperature stress.yoaabbxComparing the coefficient of the two sides yields:51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。xyoaabb解：位移势函数 所应满足的微分方程为比较两边系数，得代入上式，得取52Substituting A and B back yields the potential function of displacement:So the components of
54、stress corresponding to the particular solution of displacement are: To obtain the supplementary solutions, let and we can arrive at the needed components of stress corresponding to the supplementary solutions of displacement Therefore, the total components of stress are:The boundary conditions requ
55、ire:53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：因此，总的应力分量为边界条件要求54It is obvious that the last three conditions are satisfied, while the first condition cant be satisfied. But due to ab, the first condition can be transformed to equivalent static condition by utilizing Saint-Venant pri
56、nciple, i.e. the principal vector and principal moment of equal to zero at the boundaries of .Introducing into the above equation yieldsSo the temperature stresses of the rectangular plate are:55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：将代入上式，求得于是矩形板的温度应力为：566-6 The
57、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
58、temperature field, which are suitable to be solved with polar coordinate.are simplified in axisymmetric problems.The second equation is satisfied of course.While the first one becomes The equilibrium equations of plane stress problems ignoring body forces576-6 6-6 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 对于圆形、圆环及
59、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 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58 The geometric equations are simplified as The physical equations are simplified as Expressing the stress with strains59 几何方程简化为 物理方程简化为 将应力用应变表示60 Introducing the geometric equati
60、ons into the above equations, then introducing it into the equilibrium equations, yields the basic equation to solve the axisymmetric thermal stress by displacementOr it can be written asIntegrating it twice can yield the components of the displacement of the axisymmetric problemsin which A and B ar
61、e arbitrary constants, and the lower limit of the integration is a . The components of stress can be obtained through the above equation.61 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： 积分两次可得到轴对称问题位移分量：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：62where the constants A and B are decided by the boundary conditions.
62、In the case of plane strain, it need only that is displaced bymmm-1m-12EEis displaced bya）（ma+1is displaced byabTaTb Example 2 Suppose that there is a cylinder with thick wall. Its internal radius is a ,external radius is b . Heat up it from a homogeneous temperature. The temperature rise of its int
63、ernal 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.Solution: Evaluate temperature field at first. From the differential equation of heat conduction we arrive at the
64、differential equation of heat conduction after the thermal flux is steady without heat resource63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abTaTb 64 From the boundary conditions we figure out A and
65、B and substitute them back, then we obtain the temperature field Integrating two times yieldsorFor axisymmetric temperature field65对于轴对称温度场有 积分两次得：或 由边界条件：求出A，B后回代，得温度场：66Integrating yieldsSubstitute T into the stress expressions of the problems of plane strain67积分后得将T代入平面应变问题应力表达式68SolutionLetWe ca
66、n arrived at ThereforexyoaabbExercise 6.1 Show that the rectangular thin plate undergoes temperature changePlease evaluate the temperature stress (assume ab).69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）解：取可解得所以xyoaabb70Hence,we can arrived atLetThen71由此得取则72ThereforeThe boundary conditions are satisfied appar
67、ently.Fromi.e.73所以边界条件显然满足由即74SoWe can arrived at While the boundary condition is satisfied permanently.75得而边界条件恒成立。故76Exercise 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 wi
68、th 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. Solution: The expression of displacement of the plane stress p
69、roblems of the axisymmetric plate is Because u is a finite value at the center, i.e. r=0, so c2=0.77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由于在中心处，即r=0处，u为有限值，因此c2=0。78When r=b，u=0. Substitute it into（a）, i.e.Substituting the value of c1It can be obtained from expression (b) that79当r=b时，u=0，代入（a）即由（b）式得将c1的值代入80yields81得828384
