弹性力学平面问题地直角坐标解答.pptVIP

* * 第三章 平面问题的直角坐标解答 3—1 逆解法与半逆解法 3—2 矩形梁的纯弯曲 3—3 位移分量的求出 3—4 简支梁受均布荷载 3—5 楔形体受重力和液体压力 Solution of Plane Problems in Rectangular Coordinates Inverse Method and Semi-Inverse Method Pure Bending of a Rectangular Beam Determination of Displacments Bending of a Simple Beam Under Uniform Loads Triangular Gravity Wall 3—4 简支梁受均布荷载作用 例2. 图示矩形截面简支梁，梁高为h，长为2 ，受均布荷载作用，取板厚度为1，两端的支反力为q ，体力不记，求应力函数?及应力分量（P41）。 x h 1 y h/2 h/2 o q q q Simple beam under uniform load Consider a simple beam, with length 2l and depth h, subjected to a uniformly distributed load of intensity q. for convenience, only a unit width of the beam is considered, so the reaction at each end will be ql. 1、求应力分量（用半逆解法） 由材料力学已知：弯曲应力?x主要由弯矩引起，剪应力?xy由剪力引起，挤压应力?y由荷载q引起。由于q不随x而变化，所以?y不随x而变，可以假设： ?y=f（y） 而 Just as the bending stress ?x and the shearing stress ?xy are mainly produced by the moment and the shearing force respectively, the crushing stress ?y is mainly produced by the direct load on the beam. Since the direct load q does not vary with x, we may assume that ?y does not vary with x either and consequently it is only a function of y： f1(y)、f2（y）是待定函数 where f1(y) and f2(y) are arbitrary functions. 考察?是否满足相容方程？ To determine the function f(y), f1(y) and f2(y), we substitute the expression for ? into compatibility equation, obtaining 代入相容方程得 （1） （2） 这是关于x的二次方程，要使此方程在x为任何值时恒满足，则此方程中x前的系数和自由项必须为零，即 This is a quadratic equation of x, but it must be satisfied for all values of x between –l and l, as the condition of compatibility requires. This is possible only when the coefficients of x2 and x, as well as the term independent of x, are zero: （3） 由（2）得： 略去常数项 由（1）得： Integration of (1) and (2) yields： 由（3）得： 略去一次式和常数项 Substituting f into (3) and integrating, we have： The constant term and the term linear in y are neglect, because they will not affect the stress. The stress components will be: These expressions satisfy the differential equations of equilibrium and the compatibility equation. Hence, if the arbit