英汉双语弹性力学6.ppt

1、1,Chapter 6 The Basic Solution of,The Temperature Stress Problems,2,第六章 温度应力问题的基本解法,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 con

2、duction,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,The Basic Solution of The Temperature Stress Problems,4,温度应力问题的基本解法,第六章 温度应力问题的基本解法,5,When the

3、 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. These stresses produced by temperature

4、 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 boundar

5、y 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 mechanic

6、s. This chapter will present these two aspects of calculation simply.,Chapter 6 The Basic Solution of The Temperature Stress Problems,The Basic Solution of The Temperature Stress Problems,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

7、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。,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,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

8、 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 co

9、ordinates. 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）,The Basic Solution of The Temperature Stress Problems,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）

10、 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）,温度应力问题的基本解法,9,3.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

11、 n is the normal direction of the isothermal surface. The components of temperature gradient at each coordinate are,The Basic Solution of The Temperature Stress Problems,10,温度应力问题的基本解法,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。,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

12、方向。温度梯度在各坐标轴的分量为,11,Define to be the unit vector in normal direction of the isothermal surface, pointing to the temperature increasing direction.,4.Thermal flux speed: The quantity of heat flowing through the area S on the isothermal surface in unit time, denoted by .,The Basic Solution of The Tempe

13、rature Stress Problems,12,温度应力问题的基本解法,取 为等温面法线方向且指向增温方向的单位矢量，则有,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。,13,Its value is:,(2),Thermal flux density: The thermal flux speed flowing through unit area on the isothermal surface, denoted by . Then we have,5.The basic theorem of heat transfer: The thermal flu

14、x density is in direct proportion to the temperature gradient and in the reverse direction of it. i.e.,(3),The Basic Solution of The Temperature Stress Problems,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,温度应力问题的基本解法,其大小为,(2),称为导热

15、系数。由（1）、（2）、（3）式得,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”.,From equations (1) and (3), we can see that the value of the thermal flux density,The projections of the thermal flux de

16、nsity on axes:,The Basic Solution of The Temperature Stress Problems,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,温度应力问题的基本解法,由（1）和（3）可见，热流密度的大小,可见，导热系

17、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。,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 sup

18、plied by internal heat source.,6-2 The Differential Equation of Heat Conduction,The Basic Solution of The Temperature Stress Problems,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 tempe

19、rature 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 degreespecific thermal capability.,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,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

20、 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。,温度应力问题的基本解法,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,The Basic Solution of The Temp

21、erature Stress Problems,Introduce into it . We can see that,Hence, the total net heat quantity conducted into the hexahedron is:,which can be abbreviated as:,20,温度应力问题的基本解法,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：,21,Suppose that there is a positive heat resource to supply heat i

22、nside 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,The Basic Solution of The Temperature Stress Problems,22,假定物体内部有正热源供热，在单位时间、单位体

23、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：,温度应力问题的基本解法,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 sam

24、e 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

25、: 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.,The Basic Solution of The Temperature Stress Problems,24,6-3 温度场的边值条件

26、,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。,温度应力问题的基本解法,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. wh

27、ere the subscript s means “surface”, and n means “normal”.,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.,The Basic Solution of The Temperature Stress Problems,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

28、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,温度应力问题的基本解法,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。,27,6-4 Solve Plane Problem of Temperature Stress by Displacement,Suppose the temperature change of every point in the el

29、astic 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,However, becau

30、se 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

31、 due to the elasticity of the object, as expressed by Hookes law. Therefore, the components of the total strain of the elastic body are,The Basic Solution of The Temperature Stress Problems,Suppose the temperature change of every point in the elastic body is T. For an isotropic body, if there is no

32、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,However, because the elastic body is restricted by the external restr

33、ictions 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 H

34、ookes 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,温度应力问题的基本解法,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

35、,The Basic Solution of The Temperature Stress Problems,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 .,30,对于平面应力的变温问题，上式简化为,温度应力问题的基本解法,这就是平面应力问题热弹性力学的物理方程。,31,Ex

36、press the components of stress by the components of strain and the temperature change T, then the physical equations become,The Basic Solution of The Temperature Stress Problems,The geometric equations still are,Introducing the geometric equations into the physical equations yields the components of

37、 stress which are expressed by the components of displacement and temperature change T,32,温度应力问题的基本解法,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,33,Introducing the above equations into the differential equations of equilibrium ignoring body forces,The Basic Solution of The Tempe

38、rature Stress Problems,34,将上式代入不计体力的平衡微分方程,温度应力问题的基本解法,35,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,The Basic Solut

39、ion of The Temperature Stress Problems,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,温度应力问题的基本解法,（1）,37,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) w

40、ith 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,The Basic Solution of The Temperature Stress Problems,38,温度应力问题的基本解法,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：,将式(1)、(2)与第二章2-8中式(1)、(2)对比，可见,（2）,39,While the components and of th

41、e surface forces are displaced by,Then the corresponding equations under the conditions of plane strain are obtained.,The Basic Solution of The Temperature Stress Problems,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。,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,温度应力问题的基本解法,41,6-5 The introduction of displacem

42、ent 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,And the boundary conditions of displacement and stress must

43、 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 gro

44、up of supplementary solution of the differential equations ignoring temperature change T ,which can satisfy the boundary conditions after being superposed with the particular solution.,The Basic Solution of The Temperature Stress Problems,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

45、：,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。,温度应力问题的基本解法,43,Introduce into a function , and take the particular solution of displacement as,The function is called the potential function of displacement. Introducing and into the d

46、ifferential equations instead of u and v respectively and simplifying yields:,The Basic Solution of The Temperature Stress Problems,Introducing .and into the expression of the components of stress expressed by the components of displacement and the temperature change T,44,温度应力问题的基本解法,45,yields the c

47、omponents of stress of corresponding particular solutions of displacement,The Basic Solution of The Temperature Stress Problems,46,温度应力问题的基本解法,可得相应位移特解的应力分量是：,47,Suppose and are the supplementary solution of displacement. Then . and must satisfy the homogeneous differential equations,The Basic Solut

48、ion of The Temperature Stress Problems,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：,相应于位移补充解的应力分量为（注意不计变温，即T=0）：,温度应力问题的基本解法,49,The Basic Solution of The Temperature S

49、tress Problems,in which the stress function can be chosen according to the request of the boundary conditions of stress.,50,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。,温度应力问题的基本解法,在平面应变条件下，将上述各方程中的,51,Solution: The differential equation that the poten

50、tial function of displacement need satisfy is,The Basic Solution of The Temperature Stress Problems,Comparing the coefficient of the two sides yields:,52,温度应力问题的基本解法,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。,解：位移势函数 所应满足的微分方程为,53,Substituting A and B back yields the potential function of displacement

51、:,So the components of 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,The Basic Solution of The Temperature Stress Proble

52、ms,Therefore, the total components of stress are:,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：,温度应力问题的基本解法,因此，总的应力分量为,边界条件要求,55,It is obvious that the last three conditions are satisfied, while the first condition cant be satisfied. But due to ab, th

53、e 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 .,The Basic Solution of The Temperature Stress Problems,Introducing,into the above equation yields,So the temp

54、erature 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,温度应力问题的基本解法,代入上式，求得 于是矩形板的温度应力为：,57,6-6 The Plane thermal stress Problems of Axisymmetric Temperature Field,For the elastic body of axisymmetric structure such as circle,an

55、nulus 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.,The Basic Solution of The Temperature Stress Problems,

56、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,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,温度应力问题的基本解法,59,The geometric equations are simplified as,The physical equations are simplified as,Expressing the stress with strains,The Basic Solution of The Temperature Stress Problems,60,温度应力问题的基本解法,几何方程简化为,物理方程简化为,将应力用应变表示,61,The Basic Solution of The Temperature Stress Problems,The components of stress can be obtained through the above equation.,62,温度应力问题的基本解法,63,where the constants A and B are decided by the b