theoryofelasticity弹性力学内容摘要:
of the other two leaving two constants to be determined by the boundary conditions.(由上式 就可求出所有系数 ) 4 2 2 44 0 2 2 4 0f A x A x y A y 4 2 2 440 22 40 0A x A x y A y 02 4422444 yyxx 满足双调各方程 : Cartesian Coordinate Solutions Using Polynomials (直角坐标下的多项式解答) 00mnf m nmnx y A x y ( , )Chapter Page 8 10 the general relation that must be satisfied to ensure that the polynomial grouping is biharmonic(对于任意阶多项式要满足双调和方程 ) 02 4422444 yyxx 满足双调各方程 : 22222 1 1 2 1 12 1 1 0m n m mmnm m m m A m m n n An n n n A ,( ) ( ) ( ) ( ) ( )( ) ( ) ( ) Uniaxial Tension of a Beam (单轴拉伸梁 ) Chapter Page plane stress case Saint Venant approximation to the more general case with nonuniformly distributed tensile forces at the ends x =177。 l. (由圣维南原理可知对于在 x =177。 l处拉力分布不均但静力等效的情况也适用 ) Solution( inverse method) (逆解法 ): The boundary conditions (边界条件 ): 0)(,)(0)(,0)(lxxylxxcyxycyyT 8 11 Problem: Chapter Page constant stresses on each of the beam’s boundaries: 求应力函数 Φ 0)(,)( lxxylxx T Therefore, this problem is given by: 8 12 202Ay 2 2 222x y x yy x x y ,022 , 0x y x yA 02 2AT0x y x yT ,Boundary condition polynomial is biharmonic Uniaxial Tension of a Beam (单轴拉伸梁 ) Chapter Page 求 u ,ε //xyTETEyETvxETu11x x yy y xuTvx E EvTvvy E E ()()Tu x f yETv v y g xE ()()2 0 ( ) ( ) 0xyxyuv f y g xyx ( ) ( ) c ons t a n tg x f y ()()oooof y y ug x x v 0fg0x y x yT ,Integral(积分 ) Uniaxial Tension of a Beam (单轴拉伸梁 ) 022 , 0x y x yA Chapter Page inverse method(逆解法 ) 0)(,)( lxxylxx T Physical Equations e u Geometrical Equations 02 4422444 yyxx 8 14 202Ay Uniaxial Tension of a Beam (单轴拉伸梁 ) Chapter Page Bending of a Beam by Uniform Transverse Loading(受均匀横向载荷的梁弯曲问题 ) Airy stress function(艾里应力函数 ) x y l l ql ql 1 y z h/2 h/2 q stress field(应力场 ) Boundary Conditions(边界条件 ) pare with elementary strength of materials(和材力结果相比 ) 8 15 Chapter Page 8 16 plane stress conditions (semiinverse method 半逆解法 ) x y l l ql ql 1 y z h/2 h/2 q 1, Airy stress function yxxy。theoryofelasticity弹性力学
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