英汉双语弹性力学12.pptVIP

1 Exercise 12.3 Given a circular plate, its radius is a and it is simply supported at r=b, the load is as the figure shown. Evaluate the maximal deflection. q q r z b a Solution: The deflection function of plate can be divided into two parts 练习12.3 有一半径为a 的圆板，在 r=b 处为简支，荷载如图所示。求其最大挠度。 q q r z b a 解： 板的挠度函数可分两部分表达 Every-order derivative of and are as follows: 和 的各阶导数如下： The boundary condition or continuous condition at r=a and r=b gives Substituting deflection and its derivative into the above six equations, we can get six constants as follows 在 r=a 及 r=b 处的边界条件或连续条件为 将挠度及其导数代入上述六式，可解出六个常数如下： Substituting the solved constants into the expression of , and compare and ,the larger one is the maximal deflection of circular plate. 把求出的常数代入 的表达式，并将 与 进行比较，较大者即为圆板的最大挠度。 挠度的表达式： 若a=b，则 可见，在级数中仅取两项，就可以达到较高的精度。 § 12－6 Axisymmetric Bending of Circular Sheet When solving the problems of bending of circular sheet, it will be more convenience to choose polar coordinates. If the transverse load of circular sheet is about z axis symmetry (axis z is vertical to sheet and towards underside), thus the displacement of elastic sheet is also about z axis symmetry ,i.e. is only the function of r and don’t change according to . 1. Differential Equation of Elastic Curved Face Referring to the differential equation of elastic curved face in rectangular coordinate .In polar coordinate, when it is the axisymmetric bend of circular sheet, differential equation of elastic curved face can be written as: or § 12－6 圆形薄板的轴对称弯曲 求解圆板弯曲问题时，采用极坐标较方便。如果圆形薄板所受的横向载荷是绕z 轴对称的（z 轴垂直板面向下），则该弹性薄板的位移也将是绕z 轴对称的，即 只是r 的函数，不随 而变。 一、弹性曲面微分方程 参照直角坐标下的弹性曲面微分方程。极坐标下，圆形薄板轴对称弯曲时，曲面微分方程可写成： 或 2.Internal Force expanding yields: so the general solution of differential equation is where is a arbitrary particular solution. From the sheet, take out a differential cell, as the figure shown. In the