1、1 23Chapter 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 The introdu

2、cing of potential function of displacement6-6 The plane problems of thermal stresses in axisymmetric temperature field4第六章第六章 溫度應力問題的基本解法溫度應力問題的基本解法6-4 6-4 按位移求解溫度應力的平面問題按位移求解溫度應力的平面問題6-3 6-3 溫度場的邊界條件溫度場的邊界條件6-2 6-2 熱傳導微分方程熱傳導微分方程6-1 6-1 溫度場和熱傳導的基本概念溫度場和熱傳導的基本概念6-5 6-5 位移勢函數的引用位移勢函數的引用6-6 6-6 軸對稱溫度場

3、平面熱應力問題軸對稱溫度場平面熱應力問題5 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. These st

4、resses 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

5、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 equ

6、ations of the elastic mechanics. This chapter will present these two aspects of calculation simply.6 當彈性體的溫度變化時，其體積將趨于膨脹和收縮，若外部的約束或內部的變形協調要求而使膨脹或收縮不能自由發生時，結構中就會出現附加的應力。這種因溫度變化而引起的應力稱為熱應力，或溫度應力。 忽略變溫對材料性能的影響，為了求得溫度應力，需要進行兩方面的計算：（1）由問題的初始條件、邊界條件，按熱傳導方程求解彈性體的溫度場，而前后兩個溫度場之差就是彈性體的變溫。（2）按熱彈性力學的基本方程求解彈性體的溫

7、度應力。本章將對這兩方面的計算進行簡單的介紹。76-1 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 the tempera

8、ture 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 positiona

9、l coordinates. i.e. T=T（x，y，t）86-1 6-1 溫度場和熱傳導的基本概念溫度場和熱傳導的基本概念1.溫度場：在任一瞬時，彈性體內所有各點的溫度值的總體。用T表示。 不穩定溫度場或非定常溫度場：溫度場的溫度隨時間而變化。 即 T=T（x，y，z，t） 穩定溫度場或定常溫度場：溫度場的溫度只是位置坐標的函數。 即 T=T（x，y，z） 平面溫度場：溫度場的溫度只隨平面內的兩個位置坐標而變。 即 T=T（x，y，t）92.Isothermal surface: The surface that connects all the points with the same

10、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 direction in wh

11、ich temperature increase along the normal direction of the isothermal surface. It is denoted by T 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 arenT102.等溫面：在任一瞬時，連接溫度場內溫度相同各點的曲面。顯然，沿著等溫面，溫度不變；沿著等

12、溫面的法線方向，溫度的變化率最大。T+2TT+TTT-TxoynT3.溫度梯度：沿等溫面的法線方向，指向溫度增大方向的矢量。用T T表示，其大小用 表示。其中n n為等溫面的法線方向。溫度梯度在各坐標軸的分量為110nDefine to be the unit vector in normal direction of the isothermal surface, pointing to the temperature increasing direction.nTn0T T（1）4.Thermal flux speed: The quantity of heat flowing throu

13、gh the area S on the isothermal surface in unit time, denoted by .dtdQ)cos()cos()(cosznnTzTynnTyTxnnTxT，120n取 為等溫面法線方向且指向增溫方向的單位矢量，則有nTn0T T（1）4.熱流速度：在單位時間內通過等溫面面積S 的熱量。用 表示。dtdQ)cos()cos()(cosznnTzTynnTyTxnnTxT，13Its value is: SdtdQq/SdtdQnq/0(2) Thermal flux density: The thermal flux speed flowing

14、 through unit area on the isothermal surface, denoted by . Then we haveq5.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

15、), (2) and (3) lead to14熱流密度：通過等溫面單位面積的熱流速度。用 表示，則有 qSdtdQq/其大小為SdtdQnq/0(2)SnTdtdQ/稱為導熱系數。由（1）、（2）、（3）式得5.熱傳導基本定理：熱流密度與溫度梯度成正比而方向相反。即 q(3)T15We 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”.

16、nTq From equations (1) and (3), we can see that the value of the thermal flux densityThe projections of the thermal flux density on axes: xTqxyTqyzTqzIt is obvious that the component of thermal flux density in any direction is equal to the coefficient of heat transfer multiplied by the descending ra

17、te of the temperature in this direction.16nTq由（1）和（3）可見，熱流密度的大小可見，導熱系數表示“在單位溫度梯度下通過等溫面單位面積的熱流速度”。熱流密度在坐標軸上的投影可見：熱流密度在任一方向的分量，等于導熱系數乘以溫度在該方向的遞減率。zTqyTqxTqzyx17 The principle of heat quantity equilibrium: Within any period of time, the heat quantity accumulated in any minute part of the object equals

18、the heat quantity conducted into this minute part plus the heat quantity supplied by internal heat source.6-2 The Differential Equation of Heat Conductiondxxqqxxxqxyz Take a minute hexahedron dxdydz as shown in the above figure. Suppose that the temperature of this hexahedron rises from T to . The h

19、eat 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 degreespecific thermal capability.dttTTdttTdxdydzC18 熱量平衡原理：在任意一段時間內，物體的任一微小部分所積蓄的熱量，等于傳入該微小部分的熱量加上內部熱源所供給的熱量。6-2 6-2 熱傳導微分方程熱傳導

20、微分方程dxxqqxxxqxyz 取圖示微小六面體dxdydz。假定該六面體的溫度在dt時間內由T 升高到 。由溫度所積蓄的熱量是 ，其中 是物體的密度，C 是單位質量的物體升高一度時所需的熱量比熱容。dttTTdttTdxdydzC19 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

21、, the net heat quantity conducted into isdydzdtdxxqqxx)(dxdydzdtxqxxTqx Introduce into it . We can see thatdxdydzdtxT22dydzdxdtyT22dzdxdydtzT22The 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

22、 into it from front and back is:dxdydzdtzTyTxT)(222222 Hence, the total net heat quantity conducted into the hexahedron is:Tdxdydzdt2which can be abbreviated as:20 在同一段時間dt內，由六面體左面傳入熱量qxdydzdt，由右面傳出熱量 。因此，傳入的凈熱量為dydzdtdxxqqxx)(dxdydzdtxqxxTqx將 代入可見：dxdydzdtxT22dydzdxdtyT22由左右兩面傳入的凈熱量為由上下兩面傳入的凈熱量為由前后

23、兩面傳入的凈熱量為：因此，傳入六面體的總凈熱量為：簡記為：dzdxdydtzT22dxdydzdtzTyTxT)(222222Tdxdydzdt221 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 Wdxdydz

24、dt.According to the principle of heat quantity equilibrium,WdxdydzdtTdxdydzdtdxdydzdtxTc2cWTctT2ca cWTatT2which can be simplified as: LetThis is the differential equations of heat transfer.Thus22 假定物體內部有正熱源供熱，在單位時間、單位體積供熱為W，則該熱源在時間dt內所供熱量為Wdxdydzdt。 根據熱量平衡原理得：WdxdydzdtTdxdydzdtdxdydzdtxTc2cWTctT2ca

25、cWTatT2化簡后得：記則這就是熱傳導微分方程。236-3 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 exchan

26、ge 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 divid

27、ed 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.),()(0zyxfTt)(tfTs246-3 6-3 溫度場的邊值條件溫度場的邊值條件 初始條件： 邊界條件分四種形式： 第一類邊界條件 已知物體表面上任意一點在所有瞬時的溫度，即 其中Ts 是物體表

28、面溫度。 ),()(0zyxfTt)(tfTs 為了能夠求解熱傳導微分方程，從而求得溫度場，必須已知物體在初瞬時的溫度，即所謂初始條件；同時還必須已知初瞬時以后物體表面與周圍介質之間熱交換的規律，即所謂邊界條件。初始條件和邊界條件合稱為初值條件。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 “n

29、ormal”.)()(tfqsn 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 m

30、edium per unit time is in direct proportion to the temperature difference between them, i.e.)()(essnTTq 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

31、 that the two objects contact completely, and exchange heat through the form of heat conduction, i.e.esTT26 第三類邊界條件 已知物體邊界上任意一點在所有瞬時的運流（對流）放熱情況。按照熱量的運流定理，在單位時間內從物體表面傳向周圍介質的熱流密度，是和兩者的溫差成正比的，即 )()(essnTTqesTT其中Te是周圍介質的溫度； 稱為運流放熱系數，或簡稱熱系數。 第四類邊界條件 已知兩物體完全接觸，并以熱傳導方式進行熱交換。即 第二類邊界條件 已知物體表面上任意一點的法向熱流密度，即 其

32、中角碼 s 表示“表面”，角碼n 表示法向。)()(tfqsn276-4 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 stra

33、in , (where is the coefficient of expansion of the elastic body). Thus, the components of strain at every point of the elastic body areT0,xyzxyzzyxT However, because the elastic body is restricted by the external restrictions and mutual restrictions among each section in the object, the above-mentio

34、ned 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 Hookes law. Therefore, the components of the total strain of the elastic body are S

35、uppose 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

36、 at every point of the elastic body areT0,xyzxyzzyxT 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 temperat

37、ure 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 are286-4 6-4 按位移求解溫度應力的平面問按位移求解溫度應力的平面問題題 設彈性體內各點的溫變為T。對于各向同性體，若不受約束，則彈性體內各點的微小長度，都將產生正應變 （ 是彈性體的膨脹系數

38、），這樣，彈性體內各點的形變分量為T0,xyzxyzzyxT 但是，由于彈性體所受的外在約束以及體內各部分之間的相互約束，上述形變并不能自由發生，于是就產生了應力，即所謂溫度應力。這個溫度應力又將由于物體的彈性而引起附加的形變，如虎克定理所示。因此，彈性體總的形變分量是：29TETETEyxzzxzyyzyxx)(1)(1)(1xyxyzxzxyzyzEEE)1(2)1(2)1(2 For the temperature change problems of plane stress, the above equations are simplified asxyxyxyyyxxETETE)1

39、(211 They are the physical equations of thermal elastic mechanics of the problems of plane stress . 30對于平面應力的變溫問題，上式簡化為xyxyzxzxyzyzEEE)1(2)1(2)1(2xyxyxyyyxxETETE)1 (211這就是平面應力問題熱彈性力學的物理方程。TETETEyxzzxzyyzyxx)(1)(1)(131 Express the components of stress by the components of strain and the temperature c

40、hange T, then the physical equations becomexyxyxyyyxxETEETEE)1 ( 21)(11)(122 The geometric equations still areyuxvyvxuxyyx, 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 3

41、2將應力分量用形變分量和變溫T表示的物理方程為：xyxyxyyyxxETEETEE)1 ( 21)(11)(122幾何方程仍然為：yuxvyvxuxyyx,將幾何方程代入物理方程，得用位移分量和變溫T 表示的應力分量33）（）（）（）（yuxvETExuyvETEyvxuExyyx1211112200 xyyxxvyyxx Introducing the above equations into the differential equations of equilibrium ignoring body forces 34）（）（）（）（yuxvETExuyvETEyvxuExyyx1211

42、1122將上式代入不計體力的平衡微分方程00 xyyxxvyyxx35and simplifying yelds0121210121212222222222yTyxuxvyvxTyxvyuxu）（）（1）00sxysysyxsxlmml）（）（）（）（ 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

43、into stress boundary conditions without surface force36簡化得：這就是按位移求解溫度應力平面應力問題的微分方程。 同理，將應力分量代入無面力的應力邊界條件00sxysysyxsxlmml）（）（）（）（0121210121212222222222yTyxuxvyvxTyxvyuxu）（）（1）37TmyuxvlxuyvmTlxvyumyvxulssss）（）（）（）（）（）（121121（2）and simplifying yieldsvvuuss ， These are the stress boundary conditions to

44、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 by38簡化后得：這是按位移求解溫度應力平面應力問題的應力邊界條件。 位移邊界

45、條件仍然為：vvuuss ， 將式(1)、(2)與第二章2-8中式(1)、(2)對比，可見TmyuxvlxuyvmTlxvyumyvxulssss）（）（）（）（）（）（121121（2）39yTExTE1 and 11 and 1TEmTElXYWhile 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 temperat

46、ure stress）（is displaced by1112EEis displaced byis displaced byThen the corresponding equations under the conditions of plane strain are obtained. 40代替了體力分量 X 及 Y ，而：則得到在平面應變條件下的相應方程。yTExTE11及11TEmTEl及代替了面力分量 及 。XY 對于溫度應力的平面應變問題，只須將溫度應力平面應力問題的）（換成換成換成1112EE416-5 The introduction of displacement pote

47、ntial 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 equations0121210121212222222222yTyxuxvyvxTyxvyuxu）（）（And the boundary conditi

48、ons 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 bounda

49、ry 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.426-5 6-5 位移勢函數的引用位移勢函數的引用 由上一節知：在平面應力的情況下按位移求解溫度應力問題時，須使位移分量u 和v 滿足微分方程：01212101

50、21212222222222yTyxuxvyvxTyxvyuxu）（）（并在邊界上滿足位移邊界條件和應力邊界條件。實際求解時，宜分兩步進行：（1）求出上述微分的任意一組特解，它只需滿足微分方程，而不一定要滿足邊界條件。（2）不計變溫T，求出微分方程的一組補充解，使它和特解疊加以后，能滿足邊界條件。43 Introduce into a function , and take the particular solution of displacement as），（yxyvxu,yTyxTx）（）（1122uv The function is called the potential funct

51、ion of displacement. Introducing and into the differential equations instead of u and v respectively and simplifying yields: Because and are both constants, so when letT）（ 12），（yx satisfy the differential equations. So and can be a group of particular solution of the differential equations. u vIntro

52、ducing .and into the expression of the components of stress expressed by the components of displacement and the temperature change T211Tyvxu,44 引用一個函數 ，將位移特解取為：yvxu,），（yx函數 稱為位移勢函數。以 和 分別作為u和v代入微分方程，簡化后得：uvyTyxTx）（）（1122由于 和 都是常量，所以取：T）（ 12時， 滿足微分方程。因此 ， 可以作為微分方程的一組特解。將），（yx u v以及211Tyvxu,代入位移分量和變溫T

53、表示的應力分量表達式45)()1 (21)(11)(122yuxvETExuyvETEyvxuExyyxyields the components of stress of corresponding particular solutions of displacement yxExEyExyyx2222211146)()1 (21)(11)(122yuxvETExuyvETEyvxuExyyx可得相應位移特解的應力分量是：yxExEyExyyx2222211147 Suppose and are the supplementary solution of displacement. Then

54、 . and must satisfy the homogeneous differential equationsuvuv02121021212222222222yxuxvyvyxvyuxu)()1(2)(1)(122yuxvExuyvEyvxuExyyxThe components of stress corresponding to the supplementary solution of displacement are (Notice that the temperature change is ignored, i.e. T=0.)48 設 ， 為位移的補充解，則 ， 需滿足齊次

55、微分方程：uuvv02121021212222222222yxuxvyvyxvyuxu相應于位移補充解的應力分量為（注意不計變溫，即T=0）：)()1(2)(1)(122yuxvExuyvEyvxuExyyx49 Thus the components of the total displacement are: , vvvuuuThey should satisfy the boundary conditions of displacement. The components of the total stress are: . They should satisfy the boundar

56、y 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., , zzzyyyxxxyxxyxyyx22222,in which the stress function c

57、an 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 by1112EEis displaced byis displaced by50總的應力分量是：, , zzzyyyxxx需滿足應力邊界條件。在應力邊界問題中（沒有位移邊界條件），可以把相應于位移補充解的應力分量直接用應力函數來表示，即其中的應力函數 可以按照應力邊界條件的要求來選取。yxx

58、yxyyx22222, 在平面應變條件下，將上述各方程中的）（換成換成換成1112EE這樣總的位移分量是：, vvvuuu需滿足位移邊界條件51Solution: The differential equation that the potential function of displacement need satisfy is）（）（220211byTLet22ByAyIntroducing it into the above equation yields:）（）（220111222byTByAExample 1: The temperature change of the recta

59、ngular thin plate shown in the figure is:）（2201byTTwhere T0 is a constant. If ab, evaluate the temperature stress.yoaabbx21212100bTBTA）（，）（Comparing the coefficient of the two sides yields:52例1：圖示矩形薄板中發生如下的變溫：其中的T0 是常量。若 ，試求其溫度應力。）（2201byTTbaxyoaabb解：位移勢函數 所應滿足的微分方程為）（）（220211byT22ByAy ）（）（220111222

60、byTByA21212100bTBTA）（，）（比較兩邊系數，得代入上式，得取53Substituting A and B back yields the potential function of displacement:）（）（24201221byyTSo the components of stress corresponding to the particular solution of displacement are:0, 0),1 (220 xyyxbyTE To obtain the supplementary solutions, let and we can arrive