英汉双语弹性力学7

nullnullnullnullChapter 7 Difference Solution to the Questions of Plane§7-1 Derivation of Difference Formula§7-2 Difference Solution to Steady Temperature Field§7-3 Difference Solution to Unsteady Temperature Field§7-4 Difference Solution to Stress Function§7-5 Example of Difference Solution to Stress Function§7-6 Difference Solution of Stress Function to the Question of Temperature Stress§7-7 Difference Solution to Displacement§7-8 Example of Difference Solution to Displacement§7-9 Displacement Difference Solution to more Continuous ObjectExercise of《 Difference Solution to Plane Questions 》null第七章 平面问题的差分解§7-1 差分公式的推导§7-2 稳定温度场的差分解§7-3 不稳定温度场的差分解§7-4 应力函数的差分解§7-5 应力函数差分解的实例§7-6 温度应力问题的应力函数差分解§7-7 位移的差分解§7-8 位移差分解的实例§7-9 多连体问题的位移差分解习题课null The typical solutions to the theory of elasticity have a certain limits. When the elastic objects’ boundary conditions and loads are a little complex ，always the rigorous solution to boundary questions of the partial differential equations can’t be found。Thus the numerical solutions have an important practical meaning。Difference solution is one of the numerical solutions。 Difference solution is a method that uses difference equations (algebra equations) instead of basic equations and boundary conditions (sometimes they are differential equations), and translates the solutions to differential equations into algebra equations. null 弹性力学的经典解法存在一定的局限性，当弹性体的边界条件和受载情况复杂一点，往往无法求得偏微分方程的边值问题的解析解。因此，各种数值解法便具有重要的实际意义。差分法就是数值解法的一种。 所谓差分法，是把基本方程和边界条件（一般均为微分方程）近似地改用差分方程（代数方程）来表示，把求解微分方程的问题改换成为求解代数方程的问题。null§7-1 Derivation of Difference FormulationWe make a square grid on the surface of elastic object，by using two group lines which are parallel to the coordinate axes and the distance of two parallel lines is h . Shown in Fig. 7-1.Suppose f=f(x,y) is a continue function in elastic object . This function is in a line which is parallel to x axes. Fig.7-1For example it is in 3-０-１. It only changes with the change of coordination of x axes . function f can be opened up into taylor series in the neighbor of point 0：null§7-1 差分公式的推导 我们在弹性体上，用相隔等间距h而平行于坐标轴的两组平行线织成正方形网格，如图7-1。 设f=f(x,y)为弹性体内的某一个连续函数。该函数在平行于x轴的一根网线上，例如在3-０-１上，它只随x坐标的改变而变化。在邻近结点０处，函数f可展为泰勒级数如下：图7-1nullWe will only think of those points which are very near to point 0. It means that x-x0 is sufficient small. So three or more power of（x-x0）can be eliminated .The above formulation can be simplified as：At point 3，x=x0-h；at point 1, x=x0+h.We can get from (b)：We can get the difference formula from (c) and (d)： null 我们将只考虑离开结点０充分近的那些结点，即（x-x0）充分小。于是可不计（x-x0）的三次及更高次幂的各项，则上式简写为：在结点３，x=x0-h；在结点1， x=x0+h。代入(b) 得：联立（c）、（d），解得差分公式： nullSimilarly,we can get difference formula in the line 4-0-2:The above（１）—（４）are the basic difference formulas，thus we can get other difference formulas from them as follows ：Fig.7-2null同理，在网线4-0-2上可得到差分公式： 以上（１）—（４）是基本差分公式，从而可导出其它的差分公式如下：图7-2nullDifference formulas of (１) and (３) can be called as midpoint derivative formulas. Because they use the function value of two crunodes whose interval is 2h to express the first derivative value of the midpoint.The formula which uses the function value of three border upon crunodes to express the first derivative value of a endpoint can be called endpoint derivative formula.We must point out that midpoint derivative has a higher precision than endpoint. Because the former reflects the change of function of both sides of the crunodes. But the later only reflects one side of the crunodes. So we always try our best to use the former , and only use the later because we can’t use the former.null 差分公式(１)及(３)是以相隔2h的两结点处的函数值来表示中间结点处的一阶导数值，可称为中点导数公式。 以相邻三结点处的函数值来表示一个端点处的一阶导数值，可称为端点导数公式。 应当指出：中点导数公式与端点导数公式相比，精度较高。因为前者反映了结点两边的函数变化，而后者却只反映了结点一边的函数变化。因此，我们总是尽可能应用前者，而只有在无法应用前者时才不得不应用后者。null§7-2 Difference Solution to Steady Temperature FieldThis section we discuss the no heat source, plane and , steady temperature field and explain the application of difference method.In order to use difference method，we make grids in the temperature field. Just as Fig.7-1。At any node，for example at node 0，we can get the follows from difference formula：（c）（b）null§7-2 稳定温度场的差分解 本节以无热源的、平面的、稳定的温度场为例，说明差分法的应用。（a）为了用差分法求解，在温度场的域内织成网格，如图7-1所示。在任意一个结点，如在结点0，由差分公式有：（c）（b）nullSubstitute it into ，we can get difference equation：（1）null（1）nulloutside （a）（b) is a known heat current density which is parallel to x axes of node 0 . Apply difference formula to ,we get ：Get ，then substitute it into formula（1），we cam get the amendatory difference equation ：（2）null图7-3（a）（b)（2）null（3）Toward boundary node 0 which has third boundary conditions as in Fig 7-3b，we need to make difference equation about unknown 。In order to eliminate the temperature of the void node in the equation,we can use the boundary conditions：（4）Toward the boundary node which has fourth boundary conditions ,if the objects are completely tangent,their temperature field is continue.（3）When the boundary plumbs y axes ,we can get similar amendatory difference equation like abovenull（4）对于具有第四类边界条件的边界结点，在完全接触的情况下，由于两个接触体的温度场是连续的，nullSo when they have the same heat nature constant,the boundary node is the same as inside node.If they are not completely tangent ,or they have different heat nature constant ,the question is very complex ,and we will not discuss it here.Example : Suppose a square sheet(Fig.7-4) is 8 meters in length ,6 meters in width .Its right boundary is adiabatic boundary. The temperature of other three boundaries is signed at every node.（unit is ℃），try to find the temperature from to of nodes in the sheet .Establish difference equations of nodes from a to f according to formula (1):Solution：use a 4x3 grid ,h=2 mFig.7-4null因此只要两个接触体具有相同的热性常数，这个边界结点就和内结点完全一样。如果接触不完全，或者两个接触体具有不同的热性常数，则问题比较复杂，这里不进行讨论。按照（1）式立出结点a至f处的差分方程：图7-4nullEstablish difference equations of nodes g and i according to formula (3):Solve the 8 equations above together,we get(unit is ℃）：When the temperature has curve or declining boundary,irregular inside nodes will happen near the boundary. Like node 0 in the Fig.7-5a. null按照（3）式立出结点g及i处的差分方程：联立求解上列8个方程，得到（单位为℃）： 当温度具有曲线边界或斜边界时，在靠近边界处将出现不规则的内结点，如图7-5a中的结点0。nullnullnullFirst we assume that boundary AB is first boundary. Open up temperature T into taylor series along x axes near node 0,and eliminate three or more power of :nullnullThen the difference equation is :Similarly,we can get difference equation of irregular node 0 in Fig.7-5b:Assume that boundary AB in Fig.7-5a is second boundary. Open up temperature T into taylor series along x axes near node A,and eliminate three or more power of :（4） （5）null于是得到差分方程：同理可以导出图7-5b中不规则结点0的差分方程为：（4） （5）nullEliminate ,and we getSubstitute it into formula （4），simplified as（d）Similarly,we can get difference equation of irregular node 0 in Fig.7-5b:null代入（4）式，简化后得差分方程：同理可以导出图7-5b中不规则结点0的差分方程为：（d）nullAssume boundary AB in Fig 7-5a is third boundary,apply boundary conditions we can get:Substitute it into formula （d），we get：If boundary AB in Fig 7-5a is third boundary ，we can get similar equation besides equation (e):（e）Substitute （e）和（f）into formula （5），we can get difference equation.（f）null代入式（d），得：（e）将式（e）和（f）代入式（5），即可得出差分方程。（f）null§7-3 Difference Solution to Unsteady Temperature Field We will introduce difference solution to unsteady temperature field in this section .The main purpose is to calculate the unsteady temperature in concrete caused by the heat because the concrete coagulates .It can be used by the calculation of temperature stress and the control of temperature.Make a grid in the temperature field (Fig.7-6).Apply the differential equation of plane ,unsteady temperature field into any instantaneous inside node 0, we can get：（a)（b）null§7-3 不稳定温度场的差分解 本节简单介绍平面不稳定温度场的差分解法，主要为了说明，如何计算混凝土体中由于混凝土凝结发热而出现的不稳定温度场，供温度应力的计算及温度控制用。null（c)Substitute （b）、（c）、（d）into formula（a），we get：nullnulloutsideoutsidenullnull(4) Toward boundary nodes which have fourth boundary condition ，when they are completely tangent ,boundary node is the same as inside node。 Toward inside nodes which have second boundary condition ,we can get convergence condition from difference formula:We can see from the above that is always decided by the boundary nodes which have third boundary condition and is max.null(4)对于具有第四类边界条件的边界结点，在完全接触的情况下,边界结点就和内结点完全一样。nullSolution: suppose the height of the concrete mound is far bigger null解：假定混凝土墩的高度远大于nullSo the difference equations are ：Because all the concrete is of the same period,soThen we can simplify the equations as:null所以差分方程为：nullSuppose the data bellow is from the adiabatic thermal change examination of concrete:We calculate the temperature according to the period:Similarly we can get ：null假定由混凝土绝热温升试验得来的数据如下表所示：nullThe temperature change process of the three nodes are in Fig.7-9：null三结点处温度变化的过程如图7-9所示：null§7-4 Difference Solution to Stress FunctionWhen the physical force is omitted ，the stress component in plane question can be expressed with the second derivative of stress function ：If we make grids in the elasticity object as in Fig.7-10,we can express the stress component of every node by using difference formula.Fig.7-10（b）null§7-4 应力函数的差分解 当不计体力时，平面问题中的应力分量可以用应力函数的二阶导数表示：如果在弹性体上织成如图7-10所示的网格，应用差分公式就可以把任一结点处的应力分量表示成为：（a）图7-10（b)nullAnd we also can get ：（c)nullnullFig.7-11There into，the integral on the right of formula （1） expresses the sum of surface force along x axis between A and B； the integral on the right of formula （2）expresses the negative of sum of surface force along y axis between A and B ； the integral on the right of formula （3）expresses the sum of moment according to node B between A and B.1.The value and derivative value of every node in the boundary are in Fig.7-11.The value of node B are :（d）null图7-11（d）null can be expressed by the derivative value in the boundary of function and the value of every node in a row out of the boundary.3.The process of finite difference method .（1）Choose a node as base point A in the boundary,and we suppose :null（三）有限差分法计算步骤null（5）Calculate all the stress component according to formula (b).（2）We can express the value of void nodes in a row out of the boundary with the value of the corresponding nodes by using formula (e).nullnull§7-5 Example of Difference Solution to Stress FunctionSolution :suppose the reaction is fixate. Make the coordinate axis as in the Fig and the space between of the grids is 1/6 of the border. Because it is symmetric , so we only need to calculate the left of it.（1）Suppose the midpoint A of the girder is base point. And :Suppose there is a square concrete girder,just as in Fig.7-12. There is an even load q which plumbs the above of the girder and is balanced by reaction of the below corner. Try to calculate the stress component by using stress function.Fig.7-12null§7-5 应力函数差分解的实例null (2) Express values of every void node in a row out of the boundary (from to ) with the values of every node inside of the boundary. Because in the above and below boundaries , so :On the left ,every node in the boundary has:so：nullnull（3）Establish difference equations to every node in the boundary.For example ,according to node 1(pay attention to the symmetry):null同样：（3）对边界内的各结点建立差分方程。例如对结点1（注意对称性）：nullWe can get the below from formula (a),(b),(c)（the unite is ） （5）Calculate the stress. For example ,in node M, we can get from formula (b):similarly：Compared with the result of material mechanics,we can get :Therefore, according to the deep girder just as in the example, the stress calculated by the formula of material mechanic can’t reflect the truth.null（5）计算应力。同样可得： 与材料力学的结果比较： 可见，对于象本例题中这样的深梁，用材料力学公式算出的应力远远不能反映实际情况。null§7-6 Difference Solution of Stress Function to the Question of Temperature StressAccording to the plane questions of temperature stress, the physics equations in the condition of plane stress are :Where T is a changed temperature(not a temperature in a temperature field),substitute them into deformation compatible equations,we can get :On the other hand ,we suppose body force is 0 in equilibrium equations ,and we can get:（a）（b）null§7-6 温度应力问题的应力函数差分解null（a）and（b）are the basic derivative equations when we solve the question of stress.Derive the first and second formulas in (b) by x and y,then add them together,we can get:Substitute it into formula (a),simplify it ,we can get equilibrium equation:In the questions of temperature stress ,there is no body force. So we can simplify the questions further by citing stress functions. We assume :（c）nullnullIn the questions of temperature stress, because the component of face force ,so all nodes in the boundary have:then to solve the plane questions of temperature stress can be simplified as to solve the derivative equation (e) in formula (f). Calculate the stress component according to formula (d).According to the previous knowledge ,when we solve temperature stress by using difference solution,we have :According to any node 0,we can have the following formula form formula (e):Substitute formulas (g) and (h) into formula (i), we can get the difference equation we need:（1）（f）（g）（h）（i）null（f)nullnull（j)nullBesides,the difference of two former and later no heat- source plane stable temperature fields will not cause any temperature stress to the no boundary restricted single continuous objects (no matter what unstable process the former stable temperature field experiences during its transition to the later stable temperature field.)null 此外，前后两个无热源的平面稳定温度场之差，不会在没有边界约束的单连体中引起任何温度应力（不论前一个稳定温度场经过怎样的不稳定过程而过渡到后一个稳定温度场）。null§7-7 Difference Solution to DisplacementAccording to the plane questions of single continuous objects which only have stress boundary conditions and constant body force,we can simply find the value of stress by using difference solution to stress function. But it is more complex to the more continuous objects. When the elastic objects have stress or mixed boundary conditions,especially when the body force is not constant ,it is more difficult to use difference solution to stress function. On the other hand ,even though we can find the value of stress though difference solution to stress function,it is very complex to find the displacement.We can simply find the value of displacement and stress through difference solution to displacement ,no matter the elastic objects are single continuous objects or more continuous ones ,and no matter what boundary they have or their body force is constant or not. 1. Find derivate value of function f(delegating the displacement component u or v)in the grid .1.1 the derivate of function f along the direction of the grid-line is constant in every node (except the crunodes) of the line.null§7-7 位移的差分解 对于只具有应力边界条件的单连体受有常体力时的平面问题，可以通过应力函数的差分解比较简单地求得应力的数值。但是对于多连体，则求解比较繁。当弹性体具有应力边界条件或混合边界条件时，特别是在体力并非常量的情况下，则更难以利用应力函数的差分解。另一方面，即使通过应力函数差分解求得应力的数值，要进一步求出位移也是很繁的。 利用位移的差分解，则不论弹性体是单连体还是多连体，也不论它具有何种边界，以及它所受的体力是否为常量，总可以比较简便地求得位移的数值，从而求得应力的数值。null1.2 The derivate of function f in the direction of plumbing grid-line changes linearly in every node (except the crunodes )in the grid-line. So :Where ,the derivate of f in crunodes are :nullnullWhere:1.3 According to any node ,which is not in the line ,we assume that:null其中：这样就有：null2. Domain3. Deduce difference solution to inside nodesnull二、领域 某个结点的“领域”，是指环绕该结点的那两段、三段或四段网线的垂直平分线所围成的区域。例如图7-14中，角隅结点1的领域是 的正方形1 ；角隅结点2的领域 是 的矩形 。三、推导内结点处的差分公式nullAccording to difference solution formula, we get:Take examples of plane stress questions ,we get:(1)null由差分公式，有：nullSubstitute them into formula (1),simplify and we can get:In order to simplify calculation, we express the above two difference equations with the difference maps in Fig.7-16 and Fig.7-17.null为了计算方便，将上列两个差分方程分别用图7-16和图7-17的差分图式来表示。 nullnullnull4.Difference equations of boundary nodes.And we get：(2)Fig.7-18null图7-18nullApply difference formula, and we can get:null应用差分公式，有：nullSubstitute them into formula (2),simplify them and we can get:The difference map is in Fig.7-19.Fig.7-19null代入式（2），简化后得：其差分图式如图7-19所示。图7-19nullnullnullboundaryboundaryFig.7-22Fig.7-23null图7-22图7-23nullnullnullFig.7-27nullnullAt the corner of two boundaries, the domain of node 0 is a square. We assume that the outside normals of the two boundaries are along the positive direction, just as in Fig.7-28. In the domain encircled with void lines, we use to express the total outside force along direction. The stress components parallel axis are only and . If is unknown, the difference equations about can be got from the equilibrium condition along direction in the domain. The equilibrium condition is:Apply physic and geometry equations, we can get :According to difference formula, we can get:null 在两个边界的交点(角点),结点0的领域将是 的正方形。假定该二边界的向外法线都沿着坐标轴的正向，如图7-28。在虚线所示的结点领域上，作用于 方向的外力总和仍用 表示，平行于 轴的应力分量只有 和 。如果 是未知的，则相应于 的差分方程可由该领域在 方向的平衡条件得来，该平衡条件为：利用物理方程和几何方程改换为：由差分公式可得：nullnullnullIts map is in Fig.7-29, similarly we can get the difference equation of and its map is in Fig.7-30.nullnullJust as in Fig.7-31 to Fig.7-36.Fig.7-30Fig.7-29null如图7-31至7-36。nullnullnullnullnullnullnull§7-8 Example of Difference Solution to Displacementnull§7-8 位移的差分解的实例nullTowards the nodes in the boundary, use endpoint derivative formulas ,we can get:null对于边界上的结点，利用端点导数公式，得：nullExample 2: suppose the below of the sheet in example 1 is supported by poles (support is lubricity ), as in Fig.7-38. Calculate the displacement and stress caused by its weight. Fig.7-38Simplify and calculate them ,we can get:null例2：设例1中的薄板改在下边受连干支承（光滑支承），如图7-38，试求自重引起的位移及应力。简化后求解，得：图7-38nullThe calculation of is like example 1. Next we will calculate and of some nodes:nullnullExample 3: suppose there is a deep rectangle girder in Fig.7-39. Its left and right boundaries are fixed, and its above boundary bears equal load . Calculate the displacement and stress. We makeSolution: because of its symmetry the unknown values are only 6:Fig.7-39null图7-39nullnullnullSimplify the above 6 equations and solve them together, we can get the component of displacement:The component of stress are :null将上列6个方程简化后联立求解，得位移分量为：应力分量为：nullnullnullIllumination: towards plane questions of single continuous object, which only has stress boundary conditions, we can calculate the stress component with displacement difference solution, but when we use stress function difference solution, we can get more accurate result with the same grids, and the workload of calculation is more less. So if we only need calculate the stress not the displacement we only need use stress function difference solution, not displacement difference solution.null说明：对于只具有应力边界条件的单连体平面问题，虽然也可以用位移差分解求得应力分量，但是，改用应力函数差分解时，同样的网格可以给出较精确的应力数值，而且计算工作量较少。因此，如果不须求出位移而只需求出应力，则毫无疑问的应当用应力函数差分解，而完全不必用位移差分解。null§7-9 Displacement Difference Solution to more Continuous Object1. Displacement difference solution to node of inside tine corner. null§7-9 多连体问题的位移差分解一、内尖角处的结点位移差分方程nullUse derivative of displacement component to express stress component ,and the above formula becomes:According to the equilibrium conditions of direction ,we have :null用位移分量的导数来表示应力分量，则上式成为：nullAccording to difference formula, we get :null按照差分公式，有：nullnullnullSimilarly, we can get difference equations of , its difference formulas are in Fig.7-42.2. Displacement difference solution to more continuous object.Because of its symmetry ,we only need calculate a quarter of it. There are 12 independent unknown displacement component. They are:Fig.7-43null二、多连体的位移差分解图7-43nullStress component can be found as before. But because the grids are too sparse, the stress values calculated are approximate, and there will be great error of stress value of inside node. null应力分量可以和以前一