弹性力学双语版西安交通大学8

1、1 23chapter8 space problem8-4 the spherical symmetric problem of space8-3 the axially symmetric problem of space8-2 the basic equation unde rectangular coordinate8-1 introduction4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 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5in this chapter we f

2、irst 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 peculiar boundary conditions, we discuss the axial symmetric problems and the ba

3、ll symmetric problems of space emphatically.8-1 introductionball symmetric problemxzyaxial symmetric problemxzyp6 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xzy轴对称问题xzyp78-2 basic equations under rectangular coordinateone. differential equation

4、s 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 centers of two faces of the hexahedron, then from we get 0abm22dydxdzdydxdzdyyyzyzyz022dzdxdydzdxdydzzz

5、yzyzy canceling terms and neglecting higher order small variables,we get88-2 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 0abm22dydxdzdydxdzdyyyzyzyz022dzdxdydzdxdydzzzyzyzy化简并略去高阶微量，得9yxxyxzzxzyyzsimilarly,we get here we prove the relation of

6、the equality of cross shears againfrom 0, 0, 0zyxlist the equations,cancel terms,we getthese are differential equations of equilibrium under rectangular coordinate of space000zyxzyxzyxzyxyzxzzxyzyyzxyxxtwo. geometric equations for spatial problems, deformation components and displacement components

7、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 out with the same method.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。000zyxzyxzyxzyxyzxzzxyz

8、yyzxyxx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。11three. physical equations for an isotropic body, the relations between deformation components and stress components are as follows:yxzzxzyyzyxxeee111xyxyzxzxyzyzggg111 these are physical eq

9、uations for spatial problems. if stress components are denoted by strain components, physical equations can be written as:zzyyxxgegege222xyxyzxzxyzyzgggwhere：zyxe211e12三 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：zzyyxxgegege222xyxyzxzxyzyz

10、ggg其中：zyxe211e13four equations of compatibility differentiate the second and the third formula of geometric equations at the left.adding these two,we getywzvzyyzwzyvyzzy22323222 substitute the fourth formula of geometric equations into the above equation, we getzyyzyzzy2222（a）similarlyyxxyxzzxxyyxzx

11、xz2222222222（b）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yxxyxzzxxyyxzxxz2222222222（b）15 differentiate the late three formulas of geometric equations separately for x,y,z,we getzyuzxvzyxwyzuyxzvxywxxyzxyz222222 from the above equations,we getzyxu

12、zyzyuxzyxxxxyzxyz22222216 将几何方程中的后三式分别对x、y、z求导，得zyuzxvzyxwyzuyxzvxywxxyzxyz222222并由此而得zyxuzyzyuxzyxxxxyzxyz22222217yxyxzzxzxzyyzzxyzxyyyzxyzx2222similarly（d） the equations of (a),(b),(c),(d)are called compatibility conditions of deformation, also known as equations of compatibility. substituting phy

13、sical equations into the above equations,and canceling terms according to differentiate equations of equilibrium,we get the compatibility equations which are expressed with stress components:namelyzyzyxxxxyzxyz22（c）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，

14、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）19we call them michel compatibility equations.yyxxzzzxxzzyyyzzyyxxxzyx211121112111222222222yxxyyxxzzxxzzyyzzyxyzxyz11111122222220称其为密切尔相容方程。yyxxzzzxxzzyyyzzyyxxxzyx211121112111222222222yxxyyxxzzxxzzyyzzyxyzxyz11111122222221 among spatial problems, if the

15、 elasticity 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 probl

16、em of space. the 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,s

17、train components 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 space22 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

18、的应力分量、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 two horizo

19、ntal 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 negati

20、ve 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,22sinddd24一 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；在两个水平面上只有正应力和径向剪应力；在两个垂直面上

21、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,22sinddd250zrrzrzrzz0rrzrrkrzr these are the differential equations of equilibrium for axial symmetry problems in terms of cylindrical coordinates.two . geometric equations similar to the analysis of plane pr

22、oblem 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 symmetry problems:zwruruzrrrrwz

23、urzr260zrrzrzrzz0rrzrrkrzr 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：zwruruzrrrrwzurzr27three. physical equations because the cylindrical coordinates are orthogonal coordinates as the rectangular ones, we can get the physica

24、l equations directly from hookes law:rzzrzzrreee111zrzrzreg121 if stress components are expressed with strain components, the above equations can be written as:zzrreeeeee211211211zrzre12where :zre28三 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定律得物理方程：rzzrzzrreee111zrzrzreg121 应力分量用形变分量表示的物理方程：zzrreeeeee2112

25、11211zrzre12其中：zre29four. 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:zweerueerueezrrrr111111rwzuerzr12where :zwruruerr substitute the above equations int

26、o 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 displacement components. obviously, the displacement components in above equati

27、ons are functions coordinates r and z,they cant be solved directly. so we introduce the following method: 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：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

28、坐标r、z 的函数，不可能直接求解，为此介绍下列方法：31five. displacement tendency function for simplicity, ignoring the body force, the basic differential equations in term of displacement components can be simplified as: 021122ruurerr02112wze supposing now the displacement has tendency, we use displacement tendency functio

29、n to denote the displacement components: zr,zgwrgur21,21thus we get:221gzwruruerr222121zgzergre22221rgruurr2221zgw0, 022zrc2 substitute with the basic differential equations which ignoring the body force,we get:namely 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

30、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33 is a mediation function. the solving representations of stress components from displacement tendency function are: if only , we get .namely020crrrr1,22zrzrzz222, so for an axial symmetry problem, if we f

31、ind 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 order to solve axial symmetry problems, lame introduces a displacement function zr, attention: not all the displacem

32、ent functions of spatial problems have tendency. but if they have, the volumetric strain .ce2six lame displacement functiondefine zrgur2212221221zgwwhere 222221zrrr3402取 ，则 。即 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

33、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222221zrrr35 substitute the above functions into the basic differential functions which in the absence of body force, we get:04rrzrzr12222 namely is a repeated mediation function, we call it lame displacement function. t

34、he representations of stress components from this function are:22222212zrzzzrz so for an axial symmetry problem, if we find a suitable repeated mediation function ,from which the displacement components and stress components satisfy the boundary conditions, then we get the correct solution of the pr

35、oblem.zr,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够满足边界条件，就得到该问题的正确解答。zr,37seven example: half space body which is under the action of outward drawn concentrated forces in the boundaryconsider a ha

36、lf space body, which body forces are ignored. it receives outward drawn concentrated forces in the boundary, as shown in figure. please solve its stresses and displacements. solution:choose the coordinate system as fig. through the dimensional analysis, lames displacement function is positive one or

37、der power of length coordinate of which f multiplies r、z、. after preliminary calculation, we set displacement function as:zzrzazraazrzrara222222121lnln according to the relations of displacement components and stress components and displacement function:2222)1 (221,21zgzrgurxzyprz38七 举例：半空间体在边界上受法向集

38、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与位移函数的关系：2222)1 (221,21zgzrgurxzyprz392222222222)1 (,)2(1,zrzzrrzrzzrzrwe can obtain the displacement components and the stress components3252313253312313253312321231

39、3)21 (3)21 (,)()21 (,)(13)21 (,2432,)(22rrarrzrrarzarzrzazrrarzazrrrzarzrrzagrarzrgazrgrragrrzauzrzrr402222222222)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,)(22rrarrzrrarzarzrzazrrarzazrrrzarzrrzagrarzrgazrgrragrrzauzrzrr41the boundary conditi

40、ons are0)(0)(0,00,0rzrzrzz(a)(b)according to the saint-venants principle,we have00)d2(prrz(c)the boundary 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 (,22142边界条件是0)(0)(0,00,0rzrzrzz(a)

41、(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papa2)21 (,22143substitute the obtained a1 and a2 into the forgoing representations, we get 525323222pr3,232)21 (3)21 (2rzrpzzrrrzrprzrzrrrprzzrzr222)1 ( 22)1 ()21 (2)1 (rzerpzrrrrzerpur44将得出的a1及a2回代，得5253

42、23222pr3,232)21 (3)21 (2rzrpzzrrrzrprzrzrrrprzzrzr222)1 ( 22)1 ()21 (2)1 (rzerpzrrrrzerpur45 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

43、 all stresses,strains and displacements are symmetrical in this point. 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 coordi

44、nates origin , then all the stress components,strain components and displacement 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 sphe

45、rically symmetric problem for space46 在空间问题中，如果弹性体的几何形状、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47one. differential equations of equi

46、librium for symmetry, the small element only has 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

47、 from the elastomer. it is formed by two pellet faces, which 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.drd02rrrkrdrd48一 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

48、。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：rk22sindd02rrrkrdrd49two 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 stra

49、in along the coordinate direction. 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,

50、 we gettrre21211rtte21150二 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：trre21211rtte21151four. the basic differential equation in terms of displacement substitute the geometric equations into the physical eq

51、uations, 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.

52、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。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 inner pressure is

53、 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 2rbuarr33212

54、1121rteebareebarxzyfive 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2rbarur331211221rbeaerbeaetrxzy55substitute the boundary conditionsrarbrarbqq into the above formulas, we get3 333333312,12ababab qqaqbqabe bae baand then we get t

55、he radial displacement of the problem: the stress expressions are:3333333312121112211rabbarrruqqebaab3333333333333333111122,1111rabtabbabarrrrqqqqbabaabab 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barqbaraqabrberu3333333311212112121batbarqbaraqabrbqbaraqabrb3333333333333

56、333121112,111157exercise 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 suitable for any condition.zysolution : the stress components are:0，0，

57、0，0，0 xyzxyyzyzzthe body force components are:0,xyz 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, 059one the inspection of differential equations of equilibriumobviously they are satisfied.000yxxzxxyyzyyzxzzxxyzyxyzzxyztwo. the inspection of compatibility because the body force is a constant, the compatibility equations are60一 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：61222222222222222