1 Elasticity
弹单性生力学 第七章7一均分解
2
DIRFERENCESOLUTLONTOTHEQUESTONSOFPLAIN Chapter 7 Difference Solution to the Questions of Plane D87-1 Derivation of Difference Formula D8 7-2 Difference Solution to Steady Temperature Field >87-3 Difference Solution to Unsteady Temperature Field 87-4 Difference Solution to Stress Function D 87-6 Difference Solution of Stress Function to the on D87-5 Example of Difference Solution to Stress Function Question of Temperature Stress 87-7 Difference Solution to Displacement 87-8 Example of Difference Solution to Displacement D87-9 Displacement Difference Solution to more Continuous object D Exercise of K Difference Solution to Plane Questions
3 Chapter 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 Object Exercise of《 Difference Solution to Plane Questions 》
平二的盖分 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
4 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
DIEFBRENCE SOLUTION TO THB QUESTIONS OF PLAIN The typical solutions to the theory of elasticity have a certain limits. When the elastic objects boundary conditions and loads are a little complex, al ways the rigorous solution to boundary questions of the partial differential equations cant be found o 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
5 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
平二的盖分 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
6 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
DIRFERENCESOLUTLONTOTHEQUESTONSOFPLAIN 87-1 Derivation of Difference Formulation We 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 ff(x,y) is a continue function in elastic object. This function is in a line which is parallel to x axes Fig-1 For 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: (x-x)+ f=fo+ax o 2!(、ax2 (x-x0)+ (x-x0) 0 7
7 §7-1 Derivation of Difference Formulation We 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.7-1 ( ) ... 3! 1 ( ) 2! 1 ( ) 3 0 0 3 3 2 0 0 2 2 0 0 0 − + − + − + = + x x x f x x x f x x x f f f For 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:
平间二的盏分拿 §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=fxy)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, 函数何展为泰勒级数如下: 图7-1 =+(9)(x-x)+ (x-x0)-+ (x-x0)3+ Ox o 2!(a 3 8
8 §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7-1 ( ) ... 3! 1 ( ) 2! 1 ( ) 3 0 0 3 3 2 0 0 2 2 0 0 0 − + − + − + = + x x x f x x x f x x x f f f
DIRFERENCESOLUTLONTOTHEQUESTONSOFPLAIN We will only think of those points which are very near to point 0. It means that x-xo is sufficient small. So three or more power of (X-xo) can be eliminated. The above formulation can be simplified as f=fo (x-xo+ (x-x0)2 (b) 2!(ax At point 3, xxo-h; at point 1, x-Xo +h We can get from (b) =n/0)h2(03f 十 ax)。2(ax (c) sf=fo+ af hlaf oo We can get the difference formula from(c)and(d) af fi-f 2h 9
9 We 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: ( ) ( ) 2! 1 ( ) 2 0 0 2 2 0 0 0 x x b x f x x x f f f − − + = + At point 3,x=x0 -h;at point 1, x=x0+h.We can get from (b): ( ) 2 0 2 2 2 0 3 0 c x h f x f f f h + = − ( ) 2 0 2 2 2 0 1 0 d x h f x f f f h + = + We can get the difference formula from (c) and (d): (1) 2 1 3 0 h f f x f − =
平二的盖分 我们将只考虑离开结点0充分近的那些结点,即(xx0) 充分小。于是可不计(xx0)的三次及更高次幂的各项,则上 式简写为: f=+(x-x)+ Ox (ar2/(x-x)2 在结点3,xx0-h;在结点1,xx0+h。代入(b)得: f3=f6 0)h2(02f 十 ax)。2(ax (c) =+19)+h2(a2 (d) 联立(c)、(d),解得差分公式: af fi-f 2h 10
10 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: ( ) ( ) 2! 1 ( ) 2 0 0 2 2 0 0 0 x x b x f x x x f f f − − + = + 在结点3,x=x0 -h;在结点1, x=x0+h。代入(b) 得: ( ) 2 0 2 2 2 0 3 0 c x h f x f f f h + = − ( ) 2 0 2 2 2 0 1 0 d x h f x f f f h + = + 联立(c)、(d),解得差分公式: (1) 2 1 3 0 h f f x f − =