弹性力学双语版西安交通大学课件

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1、弹性力学双语版西安交通大学课件1 弹性力学双语版西安交通大学课件2弹性力学双语版西安交通大学课件3Chapter8 Space Problem8-4 The Spherical Symmetric Problem of Space8-3 The Axially Symmetric Problem of Space8-2 The Basic Equation unde Rectangular Coordinate8-1 Introduction弹性力学双语版西安交通大学课件4第八章第八章 空间问题空间问题8-4 8-4 空间球对称问题空间球对称问题8-3 8-3 空间轴对称问题空间轴对称问题8-

2、2 8-2 直角坐标下的基本方程直角坐标下的基本方程8-1 8-1 概概 述述弹性力学双语版西安交通大学课件5In this chapter we first give out the equations of equilibrium,the geometric equations and the physical equations under rectangular coordinate for spatial problems.For the analytic solutions of spatial problems can only be obtained under peculia

3、r boundary conditions,we discuss the axial symmetric problems and the ball symmetric problems of space emphatically.8-1 IntroductionBall Symmetric ProblemxzyAxial Symmetric ProblemxzyP弹性力学双语版西安交通大学课件6 本章首先给出空间问题直角坐标下的平衡方程、几何方程和物理方程。针对空间问题的解析解一般只能在特殊边界条件下才可以得到，我们着重讨论空间轴对称问题和空间球对称问题。8-1 8-1 概概 述述球对称问题

4、xzy轴对称问题xzyP弹性力学双语版西安交通大学课件78-2 Basic Equations under Rectangular CoordinateOne.Differential Equations of Equilibrium Consider an arbitrary point inside the body and fetch a small parallel hexahedron,which stress components on each side are shown as figure.If ab denotes the line which joins the cent

5、ers of two faces of the hexahedron,then from we get 0abm22dydxdzdydxdzdyyyzyzyz022dzdxdydzdxdydzzzyzyzy Canceling terms and neglecting higher order small variables,we get弹性力学双语版西安交通大学课件88-2 8-2 直角坐标下的基本方程直角坐标下的基本方程一 平衡微分方程 在物体内任意一点 P，取图示微小平行六面体。微小平行六面体各面上的应力分量如图所示。若以连接六面体前后两面中心的直线为ab，则由 得 0abm22dydx

6、dzdydxdzdyyyzyzyz022dzdxdydzdxdydzzzyzyzy化简并略去高阶微量，得弹性力学双语版西安交通大学课件9yxxyxzzxzyyzSimilarly,we get Here we prove the relation of the equality of cross shears againfrom 0,0,0ZYXList the equations,cancel terms,we getThese are differential equations of equilibrium under rectangular coordinate of space000

7、ZyxzYxzyXzyxyzxzzxyzyyzxyxxTwo.Geometric Equations For spatial problems,deformation components and displacement components should satisfy following geometric equationszwyvxuzyxyuxvxwzuzvywxyzxyz Of which the first two and the last have been obtained among plane problems,the other three can be led ou

8、t with the same method.弹性力学双语版西安交通大学课件10yxxyxzzxzyyz同理可得这只是又一次证明了剪应力的互等关系。由 0,0,0ZYX立出方程，经约简后得这就是空间直角坐标下的平衡微分方程。000ZyxzYxzyXzyxyzxzzxyzyyzxyxx二 几何方程 在空间问题中，形变分量与位移分量应当满足下列 6 个几何方程zwyvxuzyxyuxvxwzuzvywxyzxyz其中的第一式、第二式和第六式已在平面问题中导出，其余三式可用相同的方法导出。弹性力学双语版西安交通大学课件11Three.Physical Equations For an isotrop

9、ic body,the relations between deformation components and stress components are as follows:yxzzxzyyzyxxEEE111xyxyzxzxyzyzGGG111 These are physical equations for spatial problems.If stress components are denoted by strain components,physical equations can be written as:zzyyxxGeGeGe222xyxyzxzxyzyzGGGwh

10、ere：zyxe211E弹性力学双语版西安交通大学课件12三 物理方程对于各向同性体，形变分量与应力分量之间的关系如下：yxzzxzyyzyxxEEE111xyxyzxzxyzyzGGG111 这就是空间问题的物理方程。将应力分量用应变分量表示，物理方程又可表示为：zzyyxxGeGeGe222xyxyzxzxyzyzGGG其中：zyxe211E弹性力学双语版西安交通大学课件13Four Equations of Compatibility Differentiate the second and the third formula of geometric equations at the

11、left.Adding these two,we getywzvzyyzwzyvyzzy22323222 Substitute the fourth formula of geometric equations into the above equation,we getzyyzyzzy2222（a）Similarlyyxxyxzzxxyyxzxxz2222222222（b）弹性力学双语版西安交通大学课件14四 相容方程 将几何方程第二式左边对z的二阶导数与第三式左边对y的二阶导数相加，得ywzvzyyzwzyvyzzy22323222将几何方程第四式代入，得zyyzyzzy2222（a）同理

12、yxxyxzzxxyyxzxxz2222222222（b）弹性力学双语版西安交通大学课件15 Differentiate the late three formulas of geometric equations separately for X,Y,Z,we getzyuzxvzyxwyzuyxzvxywxxyzxyz222222 From the above equations,we getzyxuzyzyuxzyxxxxyzxyz222222弹性力学双语版西安交通大学课件16 将几何方程中的后三式分别对x、y、z求导，得zyuzxvzyxwyzuyxzvxywxxyzxyz222222

13、并由此而得zyxuzyzyuxzyxxxxyzxyz222222弹性力学双语版西安交通大学课件17yxyxzzxzxzyyzzxyzxyyyzxyzx2222Similarly（d）The equations of(a),(b),(c),(d)are called compatibility conditions of deformation,also known as equations of compatibility.Substituting physical equations into the above equations,and canceling terms according

14、 to differentiate equations of equilibrium,we get the compatibility equations which are expressed with stress components:Namelyzyzyxxxxyzxyz22（c）弹性力学双语版西安交通大学课件18yxyxzzxzxzyyzzxyzxyyyzxyzx2222同理（d）方程(a)、(b)、(c)、(d)称为变形协调条件，也称相容方程。将物理方程代入上述相容方程，并利用平衡微分方程简化后，得用应力分量表示的相容方程：即zyzyxxxxyzxyz22（c）弹性力学双语版西安交

15、通大学课件19We call them Michel compatibility equations.yYxXzZzxXzZyYyzZyYxXxzyx211121112111222222222yXxYyxxZzXxzzYyZzyxyzxyz111111222222弹性力学双语版西安交通大学课件20称其为密切尔相容方程。yYxXzZzxXzZyYyzZyYxXxzyx211121112111222222222yXxYyxxZzXxzzYyZzyxyzxyz111111222222弹性力学双语版西安交通大学课件21 Among spatial problems,if the elasticity

16、bodys geometric shape,restraint condition and any external factors are symmetrical in a certain axis(any plane which passes this axis is all symmetrical one),then all stresses,deformations and displacements are symmetrical in this axis.This kind of problem is called axial symmetry problem of space.T

17、he forms of elastomers of axial symmetry problem are generally divided into two kinds:cylinder or half space body.According to the characteristic of axial symmetry,we should adopt the cylindrical coordinates .if we take z axis as the axis of symmetry,then all the stress components,strain components

18、and displacement components will be only the function of r and z,with the coordinate have nothing to do with.zr,8-3 Axially Symmetric Problems for Space弹性力学双语版西安交通大学课件22 在空间问题中，若弹性体的几何形状、约束情况以及所受的外来因素，都对称于某一轴（通过这个轴的任一平面都是对称面），则所有的应力、形变和位移也对称于这一轴。这种问题称为空间轴对称问题。根据轴对称的特点，应采用圆柱坐标 表示。若取对称轴为 z 轴，则轴对称问题的应力

19、分量、形变分量和位移分量都将只是 r 和 z 的函数，而与 坐标无关。zr,轴对称问题的弹性体的形状一般为圆柱体或半空间体。8-3 8-3 空间轴对称问题空间轴对称问题弹性力学双语版西安交通大学课件23One.Differential Equations of Equilibrium Consider a small element as shown in figure.For axial symmetry,the elements two cylindrical planes exist only normal stresses and axial shear stresses;its tw

20、o horizontal planes exist only normal stresses and radial shear stresses;its two perpendicular planes exist only round normal stresses,which are shown in figure.According to the assumption of continuity,stress components of the small element s positive planes have a small increase compared with the

21、negative ones.Attention:the increase of round normal stresses are zero at this moment.For equilibrium at radial direction and axial direction and from ,canceling terms and ignoring the high order small values,we get12cos,22sinddd弹性力学双语版西安交通大学课件24一 平衡微分方程 取图示微元体。由于轴对称，在微元体的两个圆柱面上，只有正应力和的轴向剪应力；在两个水平面上

22、只有正应力和径向剪应力；在两个垂直面上只有环向正应力，图示。根据连续性假设，微元体的正面相对负面其应力分量都有微小增量。注意：此时环向正应力的增量为零。由径向和轴向平衡，并利用 ，经约简并略去高阶微量，得：12cos,22sinddd弹性力学双语版西安交通大学课件250Zrrzrzrzz0rrzrrKrzr These are the differential equations of equilibrium for axial symmetry problems in terms of cylindrical coordinates.Two.Geometric Equations Simil

23、ar to the analysis of plane problem in term of polar coordinates,we get,the strain components caused by radial displacement are:zurururzrrrr,The strain components caused by axial displacement are:rwzwzrz,From the principle of superposing,namely we get the geometric equations for spatial axial symmet

24、ry problems:zwruruzrrrrwzurzr弹性力学双语版西安交通大学课件260Zrrzrzrzz0rrzrrKrzr 这就是轴对称问题的柱坐标平衡微分方程。二 几何方程 通过与平面问题及极坐标中同样的分析，可见，由径向位移引起的形变分量为：zurururzrrrr,由轴向位移引起的形变分量为：rwzwzrz,由叠加原理，即得空间轴对称问题的几何方程：zwruruzrrrrwzurzr弹性力学双语版西安交通大学课件27Three.Physical Equations Because the cylindrical coordinates are orthogonal coordi

25、nates as the rectangular ones,we can get the physical equations directly from Hookes law:rzzrzzrrEEE111zrzrzrEG121 If stress components are expressed with strain components,the above equations can be written as:zzrreEeEeE211211211zrzrE12Where:zre弹性力学双语版西安交通大学课件28三 物理方程 由于圆柱坐标，是和直角坐标一样的正交坐标，所以可直接根据虎克

26、定律得物理方程：rzzrzzrrEEE111zrzrzrEG121 应力分量用形变分量表示的物理方程：zzrreEeEeE211211211zrzrE12其中：zre弹性力学双语版西安交通大学课件29Four.Solution of Axial Symmetry Problems Substitute the geometric equations into the physical equations which stress components are expressed with strain components,we get the elastic equations:zweEru

27、eErueEzrrrr111111rwzuErzr12Where:zwruruerr Substitute the above equations into the differential equations of equilibrium,and use the notation:222221zrrrWe get02111222rrrKruureE0211122ZwzeE These are known as basic differential equations for solving the spatial axial symmetry problems in terms of dis

28、placement components.Obviously,the displacement components in above equations are functions coordinates r and z,they cant be solved directly.So we introduce the following method:弹性力学双语版西安交通大学课件30四 轴对称问题的求解 将几何方程代入应力分量用应变分量表示的物理方程，得弹性方程：zweErueErueEzrrrr111111rwzuErzr12其中：zwruruerr 再将弹性方程代入平衡微分方程，并记：

29、222221zrrr得到02111222rrrKruureE0211122ZwzeE这就是按位移求解空间轴对称问题所需要的基本微分方程。显然，上述基本微分方程中的位移分量是坐标r、z 的函数，不可能直接求解，为此介绍下列方法：弹性力学双语版西安交通大学课件31Five.Displacement Tendency Function For simplicity,ignoring the body force,the basic differential equations in term of displacement components can be simplified as:021122

30、ruurerr02112wze Supposing now the displacement has tendency,we use displacement tendency function to denote the displacement components:zr,zGwrGur21,21Thus we get:221Gzwruruerr222121zGzerGre22221rGruurr2221zGw0,022zrC2 Substitute with the basic differential equations which ignoring the body force,we

31、 get:Namely 弹性力学双语版西安交通大学课件32五 位移势函数 为简单起见，不计体力。位移分量的基本微分方程简化为：021122ruurerr02112wze 现在假设位移是有势的，把位移分量用位移势函数 表示为：zr,zGwrGur21,21从而有221Gzwruruerr222121zGzerGre22221rGruurr2221zGw0,022zrC2代入不计体力的基本微分方程，得即弹性力学双语版西安交通大学课件33 is a mediation function.The solving representations of stress components from dis

32、placement tendency function are:If only ,we get .Namely020Crrrr1,22zrzrzz222,So for an axial symmetry problem,if we find a suitable mediation function ,from which the displacement components and stress components satisfy the boundary conditions,then we get the correct solution of the problem.zr,In o

33、rder to solve axial symmetry problems,Lame introduces a displacement function zr,Attention:not all the displacement functions of spatial problems have tendency.But if they have,the volumetric strain .Ce2Six Lame Displacement FunctionDefine zrGur2212221221zGwWhere 222221zrrr弹性力学双语版西安交通大学课件3402取 ，则 。即

34、 0C 为调和函数，由位移势函数求应力分量的表达式为：rrrr1,22zrzrzz222,为求解轴对称问题，拉甫引用一个位移函数zr,这样，对于一个轴对称问题，如果找到适当的调和函数 ，使得由此给出的位移分量和应力分量能够满足边界条件，就得到该问题的正确解答。zr,注：并不是所有问题中的位移函数都是有势的。若位移势函数有势，则体积应变 。Ce2六 拉甫位移函数令zrGur2212221221zGw其中222221zrrr弹性力学双语版西安交通大学课件35 Substitute the above functions into the basic differential functions w

35、hich in the absence of body force,we get:04rrzrzr12222 Namely is a repeated mediation function,we call it Lame displacement function.The representations of stress components from this function are:22222212zrzzzrz So for an axial symmetry problem,if we find a suitable repeated mediation function ,fro

36、m which the displacement components and stress components satisfy the boundary conditions,then we get the correct solution of the problem.zr,弹性力学双语版西安交通大学课件36 将上式代入不计体力位移分量的基本微分方程，可见：04rrzrzr12222即 是重调和函数，称为拉甫位移函数。由拉甫位移函数求应力分量的表达式为：22222212zrzzzrz 可见，对于一个轴对称问题，只须找到恰当的重调和的拉甫位移函数 ，使得该位移函数给出的位移分量和应力分量能

37、够满足边界条件，就得到该问题的正确解答。zr,弹性力学双语版西安交通大学课件37Seven Example:half space body which is under the action of outward drawn concentrated forces in the boundaryConsider a half space body,which body forces are ignored.It receives outward drawn concentrated forces in the boundary,as shown in figure.Please solve i

38、ts stresses and displacements.Solution:choose the coordinate system as fig.Through the dimensional analysis,Lames displacement function is positive one order power of length coordinate of which F multiplies R、z、.After preliminary calculation,we set displacement function as:zzrzAzrAAzRzRARA222222121l

39、nln According to the relations of displacement components and stress components and displacement function:2222)1(221,21zGzrGurxzyPRz弹性力学双语版西安交通大学课件38七 举例：半空间体在边界上受法向集中力 设有半空间体，体力不计，在其边界上受有法向集中力，如图所示。试求其应力与位移。解：取坐标系如图。通过量纲分析，拉甫位移函数应是F乘以R、z、等长度坐标的正一次幂，试算后，设位移函数为zzrzAzrAAzRzRARA222222121lnln根据位移分量和应力分量

40、与位移函数的关系：2222)1(221,21zGzrGurxzyPRz弹性力学双语版西安交通大学课件392222222222)1(,)2(1,zrzzrrzrzzrzrWe can obtain the displacement components and the stress components32523132533123132533123212313)21(3)21(,)()21(,)(13)21(,2432,)(22RrARrzRrARzARzRzAzRRARzAzRRRzARzrRzAGRARzRGAzRGRrAGRrzAuzrzrr弹性力学双语版西安交通大学课件402222222

41、222)1(,)2(1,zrzzrrzrzzrzr可以求得位移分量和应力分量32523132533123132533123212313)21(3)21(,)()21(,)(13)21(,2432,)(22RrARrzRrARzARzRzAzRRARzAzRRRzARzrRzAGRARzRGAzRGRrAGRrzAuzrzrr弹性力学双语版西安交通大学课件41The boundary conditions are0)(0)(0,00,0rzrzrzz(a)(b)According to the Saint-Venants Principle,we have00)d2(Prrz(c)The bou

42、ndary condition(a)is satisfied.From boundary condition(b),we get0)21(21AAr(d)From condition(c),we getPAA21214(e)Solving in terms of(d)and(e),we getPAPA2)21(,221弹性力学双语版西安交通大学课件42边界条件是0)(0)(0,00,0rzrzrzz(a)(b)根据圣维南原理，有00)d2(Prrz(c)边界条件(a)是满足的。由边界条件(b)得0)21(21AAr(d)由条件(c)得PAA21214(e)由(d)及(e)二式的联立求解，得PA

43、PA2)21(,221弹性力学双语版西安交通大学课件43Substitute the obtained A1 and A2 into the forgoing representations,we get 525323222Pr3,232)21(3)21(2RzRPzzRRRzRPRzrzRRRPrzzrzr222)1(22)1()21(2)1(RzERPzRrRrzERPur弹性力学双语版西安交通大学课件44将得出的A1及A2回代，得525323222Pr3,232)21(3)21(2RzRPzzRRRzRPRzrzRRRPrzzrzr222)1(22)1()21(2)1(RzERPzRrR

44、rzERPur弹性力学双语版西安交通大学课件45 Among spatial problems,if the elasticity bodys geometric shape,restraint condition and any external factors are symmetrical in a certain point(any plane which passes this point is all symmetrical one),then all stresses,strains and displacements are symmetrical in this point.

45、This kind of problem is called spherically symmetry problem of space.According to the characteristic of spherically symmetry,we should adopt the spherical coordinates .if we take elasticity bodys symmetrical point as the coordinates origin ,then all the stress components,strain components and displa

46、cement components will be only the function of radial coordinate r,with the other two coordinates have nothing to do with.,rO Obviously,spherically symmetric problems can only exist in hollow or solid round spheroid.8-4 Spherically Symmetric Problem For Space弹性力学双语版西安交通大学课件46 在空间问题中，如果弹性体的几何形状、约束情况以

47、及所受的外来因素，都对称于某一点（通过这一点的任意平面都是对称面），则所有的应力、形变和位移也对称于这一点。这种问题称为空间球对称问题。根据球对称的特点，应采用球坐标 表示。若以弹性体的对称点为坐标原点 ，则球对称问题的应力分量、形变分量和位移分量都将只是径向坐标 r 的函数，而与其余两个坐标无关。,rO 显然，球对称问题只可能发生于空心或实心的圆球体中。8-4 8-4 空间球对称问题空间球对称问题弹性力学双语版西安交通大学课件47One.Differential Equations of Equilibrium For symmetry,the small element only has

48、radial volume force .From radial equilibrium,and considering ,Neglecting the higher order small variables,we get the differential equations of equilibrium for spherically symmetric problems:rK22sindd Fetch a small element.Fetch a small hexahedron from the elastomer.It is formed by two pellet faces,w

49、hich distance is ,and two pairs of radial planes,which angle is respectively.For spherical symmetry,each plane only has normal stress.Its stress situations are shown in fig.drd02rrrKrdrd弹性力学双语版西安交通大学课件48一 平衡微分方程 取微元体。用相距 的两个圆球面和两两互成 角的两对径向平面，从弹性体割取一个微小六面体。由于球对称，各面上只有正应力，其应力情况如图所示。drd 由于对称性，微元体只有径向体积

50、力 。由径向平衡，并考虑到 ，再略去高阶微量，即得球对称问题的平衡微分方程：rK22sindd02rrrKrdrd弹性力学双语版西安交通大学课件49Two Geometric Equationsdrdurr For symmetry,it can only exist radial displacement ;for the same reason,it can only exist radial normal strain and tangent normal strain ,it cant exist shear strain along the coordinate direction.

51、The geometric equations for spherically symmetric problems are:rurtrurtThree Physical Equations The physical equations for spherically symmetric problems can directly be led out from Hookes law trrE21rttE11 If stress components are expressed with strain components,we gettrrE21211rttE211弹性力学双语版西安交通大学

52、课件50二 几何方程 由于对称，只可能发生径向位移 ；又由于对称，只可能发生径向正应变 及切向正应变 ，不可能发生坐标方向的剪应变。球对称问题的几何方程为：rurtdrdurrrurt三 物理方程 球对称问题的物理方程可直接根据虎克定律得来：trrE21rttE11将应力用应变表示为：trrE21211rttE211弹性力学双语版西安交通大学课件51Four.The Basic Differential Equation in Terms of Displacement Substitute the geometric equations into the physical equations

53、,we get the elastic equations:rudrduErudrduErrtrrr21121211Substitute the above equations into the differential equations of equilibrium,we get0222111222rrrrKurdrdurdrudE This is known as the basic differential equations for solving the spherically symmetric problems in terms of displacement.弹性力学双语版西

54、安交通大学课件52四 位移法求解的基本微分方程 将几何方程代入物理方程，得弹性方程rudrduErudrduErrtrrr21121211再代入平衡微分方程，得0222111222rrrrKurdrdurdrudE这就是按位移求解球对称问题时所需要用的基本微分方程。弹性力学双语版西安交通大学课件53Example:a hollow pellet which is under action of the even distributed pressure consider a hollow pellet.Its interior radius is a,the exterior is b,the

55、 inner pressure is qa,outer pressure is qb.At the absence of body force,please find its stresses and displacements.Its solution isAnd the stress components are:222220rrrd uduur drdrrSolution:for ignoring the body force,the differential equation for spherically symmetric problems can be simplified as

56、 2rBuArr332121121rtEEBArEEBArxzyFive 弹性力学双语版西安交通大学课件54五 举例：空心圆球受均布压力 设有空心圆球，内半径为a，外半径为b，内压为qa，外压为qb，体力不计，试求其应力及位移。其解为得应力分量022222rrrurdrdurdrud解：由于体力不计，球对称问题的微分方程简化为2rBArur331211221rBEAErBEAEtrxzy弹性力学双语版西安交通大学课件55Substitute the boundary conditionsrarbrarbqq into the above formulas,we get3 333333312,1

57、2abababqqaqbqABE baE baAnd then we get the radial displacement of the problem:The stress expressions are:3333333312121112211rabbarrruqqEbaab3333333333333333111122,1111rabtabbabarrrrqqqqbabaabab 弹性力学双语版西安交通大学课件56将边界条件bbrraarrqq代入上式解得12,2133333333abEqqbaBabEqbqaAbaba于是得问题的径向位移应力表达式barqbaraqabrbEru3333

58、333311212112121batbarqbaraqabrbqbaraqabrb3333333333333333121112,1111弹性力学双语版西安交通大学课件57Exercise 8.1 suppose there is a equal section pole with arbitrary shape.its density is ,with its upper end hung and lower end free,which is shown as fig.Try to prove the stress components0，0，0，0，0 xyzxyyzyzz be suit

59、able for any condition.zySolution:the stress components are:0，0，0，0，0 xyzxyyzyzzThe body force components are:0,XYZ 弹性力学双语版西安交通大学课件58练习8.1 设有任意形状的等截面杆，密度为 ，上端悬挂,下端自由，如图所示。试证明应力分量00000yzyzxyzyxz，能满足所有一切条件。zy解：已知应力分量为00000yzyzxyzyxz，体力分量为ZYX,0弹性力学双语版西安交通大学课件59One The Inspection of Differential Equatio

60、ns of EquilibriumObviously they are satisfied.000yxxzxxyyzyyzxzzXxyzYxyzZxyzTwo.The Inspection of Compatibility Because the body force is a constant,the compatibility equations are弹性力学双语版西安交通大学课件60一 检验平衡微分方程显然满足。000ZzyxYzyxXzyxzyzxzzyyxyzxyxx二.检验相容性因为体力为常量，相容方程为：弹性力学双语版西安交通大学课件6122222222222222210,10

61、10,1010,10 xyzyzxzxyz yxx zyy xz Substitute into the stress components,obviously they are satisfied.Three.The Inspection of Boundary ConditionsOn the lower end0,0,1;0zlmnXYZ Substitute into the boundary conditions:弹性力学双语版西安交通大学课件6201,0101,0101,01222222222222222xyzzxyyzxxyzzxyyzx将应力分量代入，显然均能满足。三.检验边界

62、条件下端面：0;1,0,0ZYXnmlz代入边界条件弹性力学双语版西安交通大学课件63 xxyxzxyyzyxzyzzlmnXlmnYalmnZThey are all satisfied.On the left,right profile:0,1;0lnmXYZ On the front,back profile0,1;0mnlXYZ Substitute them into formula(a),obviously they are satisfied.In sum,the given stress components satisfy the equations of equilibri

63、um,the compatibility equations and boundary conditions under the action of extern forces弹性力学双语版西安交通大学课件64 ZnmlaYnmlXnmlzyzxzzyyxyxzxyx均满足。左、右侧面：0;1,0ZYXmnl前、后侧面：0;1,0ZYXlnm代入(a)式显然满足。综上所述，所给应力分量满足平衡方程、相容方程及外力边界条件。弹性力学双语版西安交通大学课件65Exercise 8.2 Try using the Love stress function to solve the column po

64、les stress components.The column poles two ends are under the action of even distributed forces 2221zcrczzxyLSolution:first we inspect whether the stress function satisfies the compatibility condition 022Differentiating the function ,we get zczzcrrzcr22212216,2,2弹性力学双语版西安交通大学课件66练习8.2 试用Love应力函数 求解圆

65、柱杆的两端受均匀分布作用的各应力分量。2221zcrczzxyL解：首先检查应力函数是否满足相容条件022对函数 进行求导，得zczzcrrzcr22212216,2,2弹性力学双语版西安交通大学课件67zcczcrrzczczrrr2121122222646221Obviously 022The stress components 1621221222ccrzr 262121212ccrrz弹性力学双语版西安交通大学课件68zcczcrrzczczrrr2121122222646221显然022应力分量 1621221222ccrzr 262121212ccrrz弹性力学双语版西安交通大学课

66、件69 31624221222cczzz 401222zrzrThe constants of stress components are determined by the boundary conditions 650qlzzarrSubstitute the stress expressions into the boundary conditions,we get弹性力学双语版西安交通大学课件70 31624221222cczzz 401222zrzr应力分量中的常数由边界条件决定 650qlzzarr将应力表达式代入边界条件，得弹性力学双语版西安交通大学课件71 70621221cc 8162421qccFrom formulas(7),(8),we getqcqc16211221 Substitute c1,c2 into the stress components expressions(1),(2),(3)and(4),we get0,0,0zrzrq弹性力学双语版西安交通大学课件72 70621221cc 8162421qcc由式(7),(8)得qcqc1621122