最新弹性力学双语版-西安交通大学-8PPT课件.ppt

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1、弹性力学双语版弹性力学双语版-西安交通西安交通大学大学-8-82Similarly,we get Here we prove the relation of the equality of cross shears againfrom List the equations,cancel terms,we getThese are differential equations of equilibrium under rectangular coordinate of spaceTwo.Geometric Equations For spatial problems,deformation co

2、mponents and displacement components should satisfy following geometric equations 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.9同理可得这只是又一次证明了剪应力的互等关系。由 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。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

3、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。10Three.Physical Equations For an isotropic body,the relations between deformation components and stress components are as follows:These are physical equations for spatial problems.If stress components are denoted by strain components,physical equations can be w

4、ritten as:where：11三 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：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：其中：12Four Equations of Compatibility Differentiate the second and the third formula of geometric equations at the left.Adding these two,we get Substitute the fourth formula of geometric equations into the above equa

5、tion,we get（a）Similarly（b）13四 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，得将几何方程第四式代入，得（a）同理（b）14 Differentiate the late three formulas of geometric equations separately for X,Y,Z,we get From the above equations,we get15 将几何方程中的后三式分别对x、y、z求导，得并由此而得16Similarly（d）The equations of(a),(b),(c),(d)are called co

6、mpatibility conditions of deformation,also known as equations of compatibility.Substituting physical 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:Namely（c）17同理（

7、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（c）18We call them Michel compatibility equations.19称其为密切尔相容方程。20 Among spatial problems,if the elasticity bodys geometric shape,restraint condition and any external factors are symmetrical in a certain axis(any plane which

8、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.The forms of elastomers of axial symmetry problem are generally divided into two kinds:cylinder or half space body.Acc

9、ording 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 and displacement components will be only the function of r and z,with the coordinate have nothing to do with.8-3 Axia

10、lly Symmetric Problems for Space21 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。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22One.Differential Equations of Equilibrium Consider a

11、 small element as shown in figure.For axial symmetry,the elements two cylindrical planes exist only normal stresses and axial shear stresses;its two horizontal planes exist only normal stresses and radial shear stresses;its two perpendicular planes exist only round normal stresses,which are shown in

12、 figure.According to the assumption of continuity,stress components of the small element s positive planes have a small increase compared with the negative ones.Attention:the increase of round normal stresses are zero at this moment.For equilibrium at radial direction and axial direction and from ,c

13、anceling terms and ignoring the high order small values,we get23一 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：24 These are the differential equations of equilibrium for axial symmetry probl

14、ems in terms of cylindrical coordinates.Two.Geometric Equations Similar to the analysis of plane problem in term of polar coordinates,we get,the strain components caused by radial displacement are:The strain components caused by axial displacement are:From the principle of superposing,namely we get

15、the geometric equations for spatial axial symmetry problems:25 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：由轴向位移引起的形变分量为：由叠加原理，即得空间轴对称问题的几何方程：26Three.Physical Equations Because the cylindrical coordinates are orthogonal coordinates as the rectangular ones,we can get the physical equ

16、ations directly from Hookes law:If stress components are expressed with strain components,the above equations can be written as:Where:27三 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：应力分量用形变分量表示的物理方程：其中：28Four.Solution of Axial Symmetry Problems Substitute the geometric equations into the physical equ

17、ations which stress components are expressed with strain components,we get the elastic equations:Where:Substitute the above equations into the differential equations of equilibrium,and use the notation:We get These are known as basic differential equations for solving the spatial axial symmetry prob

18、lems in terms of displacement 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:29四 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：其中：再将弹性方程代入平衡微分方程，并记：得到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

19、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：30Five.Displacement Tendency Function For simplicity,ignoring the body force,the basic differential equations in term of displacement components can be simplified as:Supposing now the displacement has tendency,we use displacement tendency function to denote the dis

20、placement components:Thus we get:Substitute with the basic differential equations which ignoring the body force,we get:Namely 31五 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：现在假设位移是有势的，把位移分量用位移势函数 表示为：从而有代入不计体力的基本微分方程，得即32 is a mediation function.The solving representations of stress components from displacemen

21、t tendency function are:If only ,we get .Namely 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.In order to solve axial symmetry pro

22、blems,Lame introduces a displacement function Attention:not all the displacement functions of spatial problems have tendency.But if they have,the volumetric strain .Six Lame Displacement FunctionDefine Where 33取 ，则 。即 为调和函数，由位移势函数求应力分量的表达式为：为求解轴对称问题，拉甫引用一个位移函数 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

23、力分量能够满足边界条件，就得到该问题的正确解答。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 。六 拉甫位移函数令其中34 Substitute the above functions into the basic differential functions which in the absence of body force,we get:Namely is a repeated mediation function,we call it Lame displacement function.The representations of stress compon

24、ents from this function are: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 problem.35 将上式代入不计体力位移分量的基本微分方程，可见：即 是重调和函数，称为拉甫位移函数。由拉

25、甫位移函数求应力分量的表达式为：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。36Seven 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

26、 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 order power of length coordinate of which F multiplies R、z、.After preliminary calculation,we set d

27、isplacement function as:According to the relations of displacement components and stress components and displacement function:xzyPRz37七 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，设位移函数为根据位移分量和应力分量与位移函数的关系：xzyPRz38We can obtain the displacem

28、ent components and the stress components39可以求得位移分量和应力分量40The boundary conditions are(a)(b)According to the Saint-Venants Principle,we have(c)The boundary condition(a)is satisfied.From boundary condition(b),we get(d)From condition(c),we get(e)Solving in terms of(d)and(e),we get41边界条件是(a)(b)根据圣维南原理，有(

29、c)边界条件(a)是满足的。由边界条件(b)得(d)由条件(c)得(e)由(d)及(e)二式的联立求解，得42Substitute the obtained A1 and A2 into the forgoing representations,we get 43将得出的A1及A2回代，得44 Among spatial problems,if the elasticity bodys geometric shape,restraint condition and any external factors are symmetrical in a certain point(any plane

30、 which passes this point is all symmetrical one),then 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 tak

31、e elasticity bodys symmetrical point as the coordinates 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.Obviously,spherically symmetric problems can only exist

32、 in hollow or solid round spheroid.8-4 Spherically Symmetric Problem For Space45 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。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

33、题46One.Differential Equations of Equilibrium 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:Fetch a small element.Fet

34、ch a small hexahedron 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.47一 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

35、面体。由于球对称，各面上只有正应力，其应力情况如图所示。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：48Two Geometric Equations 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 directi

36、on.The geometric equations for spherically symmetric problems are:Three Physical Equations The physical equations for spherically symmetric problems can directly be led out from Hookes law If stress components are expressed with strain components,we get49二 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

37、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：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：将应力用应变表示为：50Four.The Basic Differential Equation in Terms of Displacement Substitute the geometric equations into the physical equations,we get the elastic equations:Substitute the above equations into the differential equations of equilibriu

38、m,we get This is known as the basic differential equations for solving the spherically symmetric problems in terms of displacement.51四 位移法求解的基本微分方程 将几何方程代入物理方程，得弹性方程再代入平衡微分方程，得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。52Example:a hollow pellet which is under action of the even distributed pressure consider a hollow

39、 pellet.Its interior radius is a,the exterior is b,the 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:Solution:for ignoring the body force,the differential equation for spherically symmetr

40、ic problems can be simplified as xzyFive 53五 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。其解为得应力分量解：由于体力不计，球对称问题的微分方程简化为xzy54Substitute the boundary conditions into the above formulas,we getAnd then we get the radial displacement of the problem:The stress expressions are:55将边界条件代入上式解得于是

41、得问题的径向位移应力表达式56Exercise 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 components be suitable for any condition.zySolution:the stress components are:The body force components are:

42、57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。试证明应力分量能满足所有一切条件。zy解：已知应力分量为体力分量为58One The Inspection of Differential Equations of EquilibriumObviously they are satisfied.Two.The Inspection of Compatibility Because the body force is a constant,the compatibility equations are59一 检验平衡微分方程显然满足。二.检验相容性因为体力为常量，相

43、容方程为：60Substitute into the stress components,obviously they are satisfied.Three.The Inspection of Boundary ConditionsOn the lower endSubstitute into the boundary conditions:61将应力分量代入，显然均能满足。三.检验边界条件下端面：代入边界条件62They are all satisfied.On the left,right profile:On the front,back profileSubstitute them

44、into formula(a),obviously they are satisfied.In sum,the given stress components satisfy the equations of equilibrium,the compatibility equations and boundary conditions under the action of extern forces63均满足。左、右侧面：前、后侧面：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。64Exercise 8.2 Try using the Love stres

45、s function to solve the column poles stress components.The column poles two ends are under the action of even distributed forces zxyLSolution:first we inspect whether the stress function satisfies the compatibility condition Differentiating the function ,we get 65练习8.2 试用Love应力函数 求解圆柱杆的两端受均匀分布作用的各应力

46、分量。zxyL解：首先检查应力函数是否满足相容条件对函数 进行求导，得66Obviously The stress components67显然应力分量68The constants of stress components are determined by the boundary conditionsSubstitute the stress expressions into the boundary conditions,we get69应力分量中的常数由边界条件决定将应力表达式代入边界条件，得70From formulas(7),(8),we get Substitute c1,c2 into the stress components expressions(1),(2),(3)and(4),we get71由式(7),(8)得将c1,c2代入应力分量表达式(1),(2),(3)和(4)，得727374结束语结束语谢谢大家聆听！谢谢大家聆听！75