Question

还是工程力学的题目

Answer 1

(1)抗弯截面系数

Wz = b(h^2)/6 =(0.02)[(0.06m)^2]/6 =0.000012m^3

(2)无X方向载荷,FAx=0, FA =FAy

ΣMB =0, -FA.L +qL.L/2 =0

FA =qL/2,

弯矩 M(x) =YA.x -qx.x/2 =(qL/2)x -qx^2/2

导数 dM/dx =qL/2 -qLx,

当x=L/2, dM/dx =0, 弯矩有极大值：

Mmax =(qL/2)L/2 - [q(L/2)^2] /2 =(qL^2) / 8 =(24KN/m)(2m^2) / 8 =12KN.m

最大正应力σmax =Mmax / Wz =12KN.m / 0.000012m^3 =1000000KPa =100Mpa

(3) σmax =100MPa < 160Mpa

即 σmax < [σ], 抗弯强度足够.