英漢雙語彈性力學2

1、1 23Chapter 2 The Basic theory of the Plane Problem2-11 Stress function.Inverse solution method and semi-inverse method2-1 Plane stress problem and plane strain problem2-2 Differential equation of equilibrium2-3 The stress on the incline.Principal stress2-4 Geometrical equation.The displacement of t

2、he rigid body2-5 Physical equation2-6 Boundary conditions2-7 Saint-Venants principle2-8 Solving the plane problem according to the displacement2-9 Solving the plane problem according to the stress.Compatible equation2-10 The simplification under the circumstances of ordinary physical forceExercise L

3、esson4第二章第二章 平面問題的基本理論平面問題的基本理論2-11 2-11 應力函數(shù)逆解法與半逆解法應力函數(shù)逆解法與半逆解法2-1 2-1 平面應力問題與平面應變問題平面應力問題與平面應變問題2-2 2-2 平衡微分方程平衡微分方程2-3 2-3 斜面上的應力主應力斜面上的應力主應力2-4 2-4 幾何方程剛體位移幾何方程剛體位移2-5 2-5 物理方程物理方程2-6 2-6 邊界條件邊界條件2-7 2-7 圣維南原理圣維南原理2-8 2-8 按位移求解平面問題按位移求解平面問題2-9 2-9 按應力求解平面問題。相容方程按應力求解平面問題。相容方程2-10 2-10 常體力情況下的簡化

4、常體力情況下的簡化習題課習題課 51.Plane stress problem2-1 Plane stress problem and plane strain problem In actual problem,it is strictly saying that any elastic body whose external force for suffering is a space system of forces is generally the space object.However,when both the shape and force circumstance of th

5、e elastic body for investigating have their own certain characteristics.As long as the abstraction of the mechanics is handled together with appropriate simplification,it can be concluded as the elasticity plane problem. The plane problem is divided into the plane stress problem and plane strain pro

6、blem. Equal thickness lamella bears the surface force that parallels with plate face and dont change along the thickness.At the same time,so does the volumetric force. z = 0 zx = 0 zy = 0Fig.216一、平面應力問題一、平面應力問題2-1 2-1 平面應力問題與平面應變問題平面應力問題與平面應變問題 在實際問題中，任何一個彈性體嚴格地說都是空間物體，它所受的外力一般都是空間力系。但是，當所考察的彈性體的形狀和

7、受力情況具有一定特點時，只要經(jīng)過適當?shù)暮喕土W的抽象處理，就可以歸結為彈性力學平面問題。 平面問題分為平面應力問題和平面應變問題。 等厚度薄板，板邊承受平行于板面并且不沿厚度變化的面力，同時體力也平行于板面并且不沿厚度變化。z = 0 zx = 0 zy = 0圖217xy Characteristics：1) The dimension of length and breadth is far larger than that of thickness.2) The force along the plate face for suffering is the face force in

8、parallel with plate face,and along the thickness even,the volumetric force is in parallel with plate force and doesnt change along the thickness, and has no external force function on the surface front and back of the flat panel.0zAttention: Plane stress problem z =0,but ,this is contrary to plane s

9、train problem. 8xy 特點：1) 長、寬尺寸遠大于厚度2) 沿板邊受有平行板面的面力，且沿厚度均布，體力平行于板面且不沿厚度變化，在平板的前后表面上無外力作用。問題相反。0z注意：平面應力問題z =0，但，這與平面應變92.Plane strain problem Very long column bears the face force in parallel with plate face and doesnt change along the length on the column face,at the same time,so does the volumetric

10、 force.z = 0 zx = 0 zy = 0 x Fig. 22For example:dam,circular cylinder piping by the internal air pressure and long level laneway etc.0zAttention: Plane strain problemz = 0，but ,this is contrary to plane stress problem.xyP10二、平面應變問題二、平面應變問題 很長的柱體，在柱面上承受平行于橫截面并且不沿長度變化的面力，同時體力也平行于橫截面并且不沿長度變化。z = 0 zx =

11、 0 zy = 0 x 圖 22如：水壩、受內壓的圓柱管道和長水平巷道等。0z注意平面應變問題z = 0，但問題相反。，這恰與平面應力xyP112-2 Differential Equation of Equilibrium Whether plane stress problem or plane strain problem, is the research problem in plane xy,all the physics quantity has nothing to do with z. Discuss below the correlation between any poin

12、t stress and volumetric force when the object is placed in the state of equilibrium,and lead an equilibrium differential equation from here.From the lamella shown in Fig.2-1,we take out a small and positive parallelepiped PABC,and take for an unit length in the directional dimension in z.yoxydyyyyxd

13、xxxxxydxxxyxyyxdyyyxyxPABCXYDFig.23),(yxxxdx Establishing the function of the positive stress force in an unit on the left side is ,the coordinate on the right side x gets the increment ,the positive stress on the face is ,spreading the formula above will be Taylors series: ),(ydxxxnnxnxxxxdxxyxndxx

14、yxdxxyxyxydxx)(),(!1)(),(! 21),(),(),(222122-2 2-2 平衡微分方程平衡微分方程 無論平面應力問題還是平面應變問題，都是在xy平面內研究問題，所有物理量均與z無關。 下面討論物體處于平衡狀態(tài)時，各點應力及體力的相互關系，并由此導出平衡微分方程。從圖21所示的薄板取出一個微小的正平行六面體PABC(圖23），它在z方向的尺寸取為一個單位長度。yoxydyyyyxdxxxxxydxxxyxyyxdyyyxyxPABCXYD圖23),(yxxxxdx 設作用在單元體左側面上的正應力是 ，右側面上坐標 得到增量 ，該面上的正應力為 ，將上式展開為泰勒級數(shù)：

15、),(ydxxxnnxnxxxxdxxyxndxxyxdxxyxyxydxx)(),(!1)(),(! 21),(),(),(22213After omitting small quantity of the two rank and above the two rank,can get ,at the same time, , , are get the state of stress from the drawing show.dxxyxyxxx),(),(yxyyx While considering the volumetric force to the plane stress st

16、ate,still prove mutual and equal theory of shearing strength.Regard the center D and straight line in parallel with the shaft of z as the moment shaft, list the equilibrium equation of the moment shaft : 0DM02121)(2121)(dydxdydxdyydxdydxdydxxyxyxyxxyxyxyThe both sides of the formula above divide get

17、:dxdydyydxxyxyxxyxy2121Cause0, 0dydx,Omitting small quantity isnt accounted,can get:yxxy14略去二階及二階以上的微量后便得 同樣 、 、 都一樣處理，得到圖示應力狀態(tài)。dxxyxyxxx),(),(yxyyx 對平面應力狀態(tài)考慮體力時，仍可證明剪應力互等定理。以通過中心D并平行于z軸的直線為矩軸，列出力矩的平衡方程 ： 0DM02121)(2121)(dydxdydxdyydxdydxdydxxyxyxyxxyxyxy將上式的兩邊除以 得到：dxdydyydxxyxyxxyxy2121令0, 0dydx，即

18、略去微量不計，得：yxxy15 Deduce the equilibrium differential equation of the plane stress problem below,list the equilibrium equation to the unit:0111)(11)(:0dydxXdxdxdyydydydxxFyxyxyxxxxx0111)(11)(:0dydxYdydydxxdxdxdyyFxyxyxyyyyy16 下面推導平面應力問題的平衡微分方程，對單元體列平衡方程：0111)(11)(:0dydxXdxdxdyydydydxxFyxyxyxxxxx0111)(

19、11)(:0dydxYdydydxxdxdxdyyFxyxyxyyyyy17 Sorting them gets:00YxyXyxxyyyxx These two differential equation include three unknown functions .Therefore,deciding the problem of the stress weight is exceedingly and statically determinate;And still must consider the deformation and displacement,then the pro

20、blem can be solved. For the plane strain problem,the faces front and back still have But they do not affect completely the establishes of the equation above.So the equation above applies two kinds of plane problem alike. zyxxyyx,18 整理得:00YxyXyxxyyyxx 這兩個微分方程中包含著三個未知函數(shù) 。因此決定應力分量的問題是超靜定的；還必須考慮形變和位移，才能

21、解決問題。 對于平面應變問題,雖然前后面上還有 ,但它們完全不影響上述方程的建立。所以上述方程對于兩種平面問題都同樣適用。zyxxyyx,192-3 The stress on the Inclined Plane.Principal stress1.The stress on the inclined plane Having known the stress weight of any point P inside the elastic body,we try to get the stress which pass the point P on the arbitrarily incl

22、ined cross section.From neighborhood of point P taking a plane AB,which is in parallel with the inclined plane above,and draws a small set square or three column PAB on two planes which pass point P and have perpendicularity in the shaft of x and y.When the plane AB approaches point P infinitely,the

23、 mean stress on the plane AB will become the stress on the inclined plane above. yxxyyx, Establish the length of the face AB in the plane xy is dS,N is the exterior normal direction,and its direction cosine is:myNlxN),cos(,),cos(PABxyxyNyxNNXNYSNyxFig.24o202-3 2-3 斜面上的應力、主應力斜面上的應力、主應力一、斜面上的應力一、斜面上的應

24、力 已知彈性體內任一點P處的應力分量 ，求經(jīng)過該點任意斜截面上的應力。為此在P點附近取一個平面AB，它平行于上述斜面，并與經(jīng)過P點而垂直于x軸和y軸的兩個平面劃出一個微小的三角板或三棱柱PAB。當平面AB與P點無限接近時，平面AB上的應力就成為上述斜面上的應力。 yxxyyx, 設AB面在xy平面內的長度為dS，厚度為一個單位長度，N 為該面的外法線方向，其方向余弦為：myNlxN),cos(,),cos(PABxyxyNyxNNXNYSNyx圖24o21The projection of the whole stress on the inclined plane AB is XN and

25、YN respectively along with the shaft of x and y.From the PAB equilibrium term can get: 0 xFmdSldSdSXyxxNDivide and get:dSyxxNmlXSame from and get: 0yFxyyNlmYThe positive stress on the inclined plane AB,from the projection can get:NxyyxNNNlmmlmYlX222The shearing strength on the inclined plane AB,from

26、 the projection can get:NxyxyNNNmllmmXlY)()(2222 斜面AB上全應力沿x軸及y軸的投影分別為XN和YN。由PAB的平衡條件 可得： 0 xFmdSldSdSXyxxN除以 即得：dSyxxNmlX同樣由 得出： 0yFxyyNlmY斜面AB上的正應力 ，由投影可得：NxyyxNNNlmmlmYlX222斜面AB上的剪應力 ，由投影可得：NxyxyNNNmllmmXlY)()(22233.Principal stress If the shearing stress of some inclined plane through point P is

27、equal to zero,then the positive stress of that inclined plane calls a principal stress of point P,but that inclined plane calls the main plane of the stress at point P,and the normal direction of that inclined plane calls the main direction of the stress at point P.1.The size of the principal stress

28、2221)2(2xyyxyx2.The direction of the principal stress is in the perpendicularity with for each other. 1224二、主應力二、主應力 如果經(jīng)過P點的某一斜面上的切應力等于零，則該斜面上的正應力稱為P點的一個主應力，而該斜面稱為P點的一個應力主面，該斜面的法線方向稱為P點的一個應力主向。1.主應力的大小2221)2(2xyyxyx2.主應力的方向 與 互相垂直。12252-4 Geometrical Equation.The Displacement of the Rigid Body In pl

29、ane problem,every point inside the elastic body can produce the arbitrarily directional displacement.Take an unit PAB through any point P inside the elastic body,such as Fig.2-5 show.After the elastic body suffers force,the point P,A,B move to the point P、A、Brespectively.PoxyABPABuvdxxuudyyvvdyyuudx

30、xvvFig.25一、一、The positive strain at pointThe positive strain at point P Pxudxudxxuux)( Here because of small deformation, PA for causing stretch and shrink from the y direction displacement v is the small quantity of a high rank and this small quantity may be omitted.262-4 2-4 幾何方程、剛體位移幾何方程、剛體位移 在平面

31、問題中，彈性體中各點都可能產(chǎn)生任意方向的位移。通過彈性體內的任一點P，取一單元體PAB，如圖2-5所示。彈性體受力以后P、A、B三點分別移動到P、A、B。PoxyABPABuvdxxuudyyvvdyyuudxxvv圖25一、一、P點的正應變點的正應變 在這里由于小變形，由y方向位移v所引起的PA的伸縮是高一階的微量，略去不計。xudxudxxuux)(27The same can get:yvy2.Shearing strain at point PyuxvxyThe corner of the line segment PA:xvdxvdxxvv)(The same can get the

32、 corner of the line segment PB:yuThus28同理可求得：yvy二、二、P P點的切應變點的切應變yuxvxy線段PA的轉角：xvdxvdxxvv)(同理可得線段PB的轉角：yu所以29Therefore get the geometrical equation of the plane problemyuxvyvxuxyyx From the geometrical equation above,when the displacement weight of the object is completely certain,the deformation we

33、ight is completely certain,unique weight can not be made sure thoroughly.30因此得到平面問題的幾何方程：yuxvyvxuxyyx 由幾何方程可見，當物體的位移分量完全確定時，形變分量即可完全確定。反之，當形變分量完全確定時，位移分量卻不能完全確定。312-5 The Physical Equation In the isotropy of the complete elasticity,the relation between the deformation weight and the stress weight is

34、 established according to the Hookes law as follows:xyxyzxzxyzyzyxzzxzyyzyxxGGGEEE111)(1)(1)(1322-5 2-5 物理方程物理方程 在完全彈性的各向同性體內，形變分量與應力分量之間的關系根據(jù)虎克定律建立如下：xyxyzxzxyzyzyxzzxzyyzyxxGGGEEE111)(1)(1)(133 Inside the formula,the E is a modulus of elasticity;the G is a stiffness modulus;the u is a poisson rati

35、o.The relation of three ones above:)1 (2EG1.The physics equation of the plane stress problemxyxyxyyyxxEEE)1 (2)(1)(1)(yxzEAnd have:34 式中，E為彈性模量；G為剛度模量； 為泊松比。三者的關系：)1 (2EG一、平面應力問題的物理方程一、平面應力問題的物理方程xyxyxyyyxxEEE)1 (2)(1)(1)(yxzE且有：352.The physics equation of the plane strain problemxyxyxyyyxxEEE)1 (2)

36、1(1)1(1223.The transformation relation of the relation type between the stress strain and the plane strain.The relation type of the plane stress:xyxyxyyyxxEEE21)(1)(136二、平面應變問題的物理方程二、平面應變問題的物理方程xyxyxyyyxxEEE)1 (2)1(1)1(122三、平面應力的應力應變關系式與平面應變的關系式之間的三、平面應力的應力應變關系式與平面應變的關系式之間的 變換關系變換關系將平面應力中的關系式：xyxyxy

37、yyxxEEE21)(1)(137For change112EECan get the relation type in the plane strain:xyxyxyyyxxEEE)1 (2111122 Because of the similarity of this kind,while solving plane strain problem,the corresponding equation of the plane problem and the elastic constant in the answer can be exchanged as above,can get th

38、e solution of the homologous plane strain problem.38作代換112EE就可得到平面應變中的關系式：xyxyxyyyxxEEE)1 (2111122 由于這種相似性，在解平面應變問題時，可把對應的平面應力問題的方程和解答中的彈性常數(shù)進行上述代換，就可得到相應的平面應變問題的解。392-6 Boundary Conditions When the object is placed in the state of equilibrium,its internal state of stress at all point should satisfy

39、the equilibrium differential equation and also satisfy the boundary term on the boundary. According to the difference of the boundary condition,the elasticity problem is divided into the displacement boundary problem,stress boundary problem and mixed boundary problem.1.Displacement Boundary Term Whe

40、n the displacement has been known on the boundary,the displacement of the point on the object boundary and the equal term of the fixed displacement should be established.For example,if making the boundary of the fixed displacement is ,and have(on the ):uSuSuusvvsAmong them, and means the displacemen

41、t weight on the boundary,however, and is the coordinate function we have know the boundary.suusvv402-6 2-6 邊界條件邊界條件 當物體處于平衡狀態(tài)時,其內部各點的應力狀態(tài)應滿足平衡微分方程；在邊界上應滿足邊界條件。 按照邊界條件的不同，彈性力學問題分為位移邊界問題、應力邊界問題和混合邊界問題。一、位移邊界條件一、位移邊界條件 當邊界上已知位移時，應建立物體邊界上點的位移與給定位移相等的條件。如令給定位移的邊界為 ，則有（在 上）：uSuSuusvvs其中 和 表示邊界上的位移分量，而 和 在

42、邊界上是坐標的已知函數(shù)。suusvv412.Stress boundary term When the boundary of the object is given to surface force,then the stress of the object on the boundary should satisfy the equilibrium term of forces with the equilibrium of the surface force.YlmXmlsxysysyxsx)()()()( Among them, and are the surface force we

43、ights and , , , are the stress weights on the boundary.XYsx)(sy)(sxy)(syx)( When the boundary face is in perpendicularity in shaft x,stress boundary term can be changed briefly into:YXsxysx)( ,)( When the boundary face is in perpendicularity in shaft y,stress boundary term can be changed briefly int

44、o:XYsyxsy)( ,)(42二、應力邊界條件二、應力邊界條件 當物體的邊界上給定面力時，則物體邊界上的應力應滿足與面力相平衡的力的平衡條件。YlmXmlsxysysyxsx)()()()(其中 和 為面力分量， 、 、 、 為邊界上的應力分量。XYsx)(sy)(sxy)(syx)( 當邊界面垂直于 軸時，應力邊界條件簡化為：xYXsxysx)( ,)( 當邊界面垂直于 軸時，應力邊界條件簡化為：yXYsyxsy)( ,)(433.Mixed boundary condition1.The displacement has been known on a part of boundari

45、es of the object,the result of which have the displacement boundary term,the boundaries of other parts have the surface force we have know.And then there should be stress boundary term and displacement boundary term respectively on two parts of the boundaries.The left surface of the cantilever conta

46、ins displacement boundary term,such as shown in Fig.2-6.00vvuussTop and bottom surface contains stress boundary term:0)(0)(sysyxYXThe right surface contains stress boundary term:0)()(sxysxYqXlqxyo2h2hFig.2-644三、混合邊界條件三、混合邊界條件1.物體的一部分邊界上具有已知位移，因而具有位移邊界條件，另一部分邊界上則具有已知面力。則兩部分邊界上分別有應力邊界條件和位移邊界條件。如圖2-6，懸

47、臂梁左端面有位移邊界條件：00vvuuss上下面有應力邊界條件：0)(0)(sysyxYX右端面有應力邊界條件：0)()(sxysxYqXlqxyo2h2h圖2-6452.On the same boundary,there are not only stress boundary term but displacement boundary term.Coupler sustains the boundary term,such as shown in Fig.2-7.0)(0sxysYuuThe alveolus boundary term shown in Fig.2-8.0)(0sxs

48、XvvoxyFig.2-7xyoFig.2-8462.在同一邊界上，既有應力邊界條件又有位移邊界條件。如圖2-7連桿支撐邊界條件：0)(0sxysYuu如圖2-8齒槽邊界條件：0)(0sxsXvvoxy圖2-7xyo圖2-8472-7 Saint-Venant Principle1.Saint-Venants Principle If transforming a small part of the surface force on the boundary into the surface force that has equal effect but different distribut

49、ion(The main vector is equal,so is the main quadrature to the same point as well),and then the distribution of the stress force nearby will have prominent changes,but the influence from the distant place can not be accounted.2.Give Examples Establishing the component of the column forms,the centroid

50、 of area in cross sections of both ends suffers the tensible force which is equal in size but contrary in direction,such as shown in Fig.2-9a.If transforming an or both ends of tensile force into the force at the same effect as the static force,such as shown in Fig.2-9b or Fig.2-9c,the distribution

51、of stress force drawn only by broken line has prominent changes,whereas,the influence of the rest parts can not be accounted.If changing both ends of tensile force into that of uniform distribution again,the gathering degree is equal to P/A and among them A is the cross-section area of the component

52、,such as shown in Fig.2-9d,there is still the stress close to both ends under the noticeable influence.482-7 2-7 圣維南原理圣維南原理一、一、圣維南原理圣維南原理 如果把物體的一小部分邊界上的面力，變換為分布不同但靜力等效的面力（主矢量相同，對于同一點的主矩也相同），那么，近處的應力分布將有顯著的改變，但是遠處所受的影響可以不計。二、二、舉例舉例 設有柱形構件，在兩端截面的形心受到大小相等而方向相反的拉力 ，如圖2-9a。如果把一端或兩端的拉力變換為靜力等效的力，如圖2-9b或2-9

53、c，只有虛線劃出的部分的應力分布有顯著的改變，而其余部分所受的影響是可以不計的。如果再將兩端的拉力變換為均勻分布的拉力，集度等于 ，其中 為構件的橫截面面積，如圖2-9d，仍然只有靠近兩端部分的應力受到顯著的影響。PAP/A49PP2/P2/P2/P2/PP2/P2/PAP/AP/PPFig.2-9(a)(b)(c)(d)(e) Under the four kinds of circumstances above,parts of distribution of stress force distant from both ends have no marked difference.Att

54、ention: The application of the Saint-Venants principle is by no means separated from the term of Equal Effect of Static Force. 50PP2/P2/P2/P2/PP2/P2/PAP/AP/PP圖2-9(a)(b)(c)(d)(e) 在上述四種情況下，離開兩端較遠的部分的應力分布，并沒有顯著的差別。注意：注意： 應用圣維南原理，絕不能離開“靜力等效”的條件。512-8 Solving the Plane Problem according to the displaceme

55、nt There are three kinds of basic methods to solve the problem in elasticity:the solution to the problem according to displacement,stress force and admixture. While solving problems using displacement method,we regard displacement weight as the basic function unknown.After getting displacement weigh

56、t from only including the differential equation and boundary term of the displacement weight,then get the deformation weight using geometrical equation, therefore, get the stress weight with the physics equation.1.Plane Stress ProblemxyxyxyyyxxEEE)1 (2)(1)(1In plane stress problem, the physics equat

57、ion is:522-8 2-8 按位移求解平面問題按位移求解平面問題 在彈性力學里求解問題，有三種基本方法：按位移求解、按應力求解和混合求解。 按位移求解時，以位移分量為基本未知函數(shù)，由一些只包含位移分量的微分方程和邊界條件求出位移分量以后，再用幾何方程求出形變分量，從而用物理方程求出應力分量。一、平面應力問題一、平面應力問題xyxyxyyyxxEEE)1 (2)(1)(1在平面應力問題中，物理方程為：53From three formulas above mentioned to solve the stress weight,can get: with the substitution

58、of geometrical equation,we can get the elasticity equation:xyxyxyyyxxEEE)1 (2)(1)(122Again equilibrium differential equation with substitution in formula(a), simplification hereafter, can get:)()1 (2)(1)(122yuxvExuyvEyvxuExyyx（a)This is the equilibrium differential equation to mean with the displace

59、ment, ie, when solving the plane stress problem according to displacement method, we adopt a basic differential equation for needs. 0)2121(10)2121(1222222222222YyxuxvyvEXyxvyuxuE（1）54由上列三式求解應力分量，得：xyxyxyyyxxEEE)1 (2)(1)(122將幾何方程代入，得彈性方程：再將式（a）代入平衡微分方程，簡化以后，即得：)()1 (2)(1)(122yuxvExuyvEyvxuExyyx（a)這是用

60、位移表示的平衡微分方程，也就是按位移求解平面應力問題時所需用的基本微分方程。0)2121(10)2121(1222222222222YyxuxvyvEXyxvyuxuE（1）55The stress boundary term with substitution in formula(a), simplification hereafter, can get:YyuxvlxuyvmEXxvyumyvxulEssss)(21)(1)(21)(122This is the stress force boundary to mean with the displacement, ie, we ado