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——此处研究一块园板,该板支承在一个同心园的园周上,承受均布荷载而弯曲,对于这一具有实际意义的问题,本文应用一种特殊的连续近似法求此非线性问题的解。以比率γ=b/a为参量,得出相应的数值解,并且发现此解与实验结果极为吻合。符号:a=板的外半径, b=板的支承半径,h=板的厚度, q=作用于板上荷载,r=板的任意半径, w=板的挠度,w_0=板中心挠度, x=(a~2-r~2)/a~2,D=板的弯曲刚度Eh~3/12(1-v~2)。E=弹性模量;N_r=单位宽度的径向薄膜张力;N_t=单位宽度的切向薄膜张力;Q=[3(1-v~2)]~(3/2)qa~4/4Eh~4;S=3(1-v~2)a~2 N_r/Eh~3;T=3(1-v~2)a~2N_t/Eh~3;W=■w/h;β=1-b~2/a~2;γ=b/a;v=波桑比。
- Here a study of a circular plate, which is supported on a circumference of a concentric garden, is subject to uniform load and bending. For this practical problem, this paper applies a special continuous approximation method to find the nonlinearity. Solution to the problem. With the ratio γ=b/a as a parameter, the corresponding numerical solution is obtained, and it is found that this solution is in good agreement with the experimental results. Symbols: a = outer radius of the board, b = bearing radius of the board, h = thickness of the board, q = load acting on the board, r = arbitrary radius of the board, w = deflection of the board, w_0 = board center deflection, x =(a~2-r~2)/a~2, D= bending stiffness of the panel Eh~3/12(1-v~2). E=elastic modulus; N_r=radial film tension per unit width; N_t=tangential film tension per unit width; Q=[3(1-v~2)]~(3/2)qa~4/4Eh~ 4;S=3(1-v~2)a~2 N_r/Eh~3;T=3(1-v~2)a~2N_t/Eh~3;W=■w/h;β=1- b~2/a~2; γ=b/a; v=Possambi.