(新)弹性力学 平面问题的极坐标解答3——精品.pptVIP

将两部分解答叠加，即得q1、q2作用下的应力分量 例4、带圆孔的矩形薄板，受均布拉力，求应力分量 x y a q q x y a q/2 q/2 q/2 q/2 x y a q/2 q/2 q/2 q/2 + 其应力分量为： 讨论： 1、孔边应力：r=a，?r=0，?r?=0 x y a q q 3q 2q q 0 -q ?? 90 ? 60? 45? 30? 0? ? * * 第四章 平面问题的极坐标解答 4—1 Differential Equations of Equilibrium in Polar Coordinates 4—2 Geometrical Equations and Physical Equations in Polar Coordinates 4—3 Stress Function and Compatibility Equations in Polar Coordinates 4—9 Effect of circular holes on stress distribution 4. Solution of Plane Problems in Polar Coordinates 4—4 Coordinates Transformation of Stress Components 4—5 Axisymmetrical Stresses and Cooresponding Displacements 4—6 Hollow Cylinder Subjected to Uniform Pressures 4—11 Concentrated Normal Load on a Straight Boundary 4—6 圆环或圆筒受均布压力(P64) qa qb a b Hollow Cylinder Subjected to Uniform Pressures 例1. 图示圆环或圆筒，内半径为a，外半径为b，受均布内压qa和均布外压qb作用 A hollow cylinder with inner radius a and outer radius b is subjected to internal pressure qa and external pressure qb. 由于结构轴对称，荷载轴对称，所以应力分量也是轴对称，直接利用公式 Evidently the stress distribution must be axisymmetrical. 利用边界条件求待定常数A、B、C 恒满足 边界条件： qa qb 考察环向位移(for circumferential displacement)： 是多值项，对于同一r，当?不同，该项有不同的值，而在此轴对称问题中，这是不可能的(P65) 由上述两个边界条件不能完全确定A、B、C，因此要考虑位移单值条件 Now the two equations are not sufficient to determine the three constants A,B,C and we have to consider the condition of single-valued displacement. In the hollow cylinder, this is physically impossible. 根据位移单值条件，可知：B=0 上述边界条件变为： 最后求得： 讨论只有内压或只有外压单独作用的情况 a、只有内压力的作用：qb=0 If qb=0 (only internal pressure qa=0 acts), 结论：?r总为压应力，??为拉应力 应力分布如图所示： ?r ?? qa 当圆环或圆筒的外半径b ? 时，它就成为具有圆孔的无限大薄板，或具有孔道的无限大弹性体，应力分量为 ?r is always compression and ?? is always tension. When the outer radius b is much large than the inner radius a, the cylinder becomes a large body with a small circular hole of radius a . 应力与a2/r2成正比，当r??a时，应力分量很小，这也证实了圣维南原理 At the edge of the hole(r=a), ?r =- qa and ?? =- qa. At a large distance from the hole, the stresses are negligibly small