最佳答案
计算支反力：ΣMA =0， FB.2m -(2KN/m).2m.1m =0 FB =2KN（向上）ΣFy =0, FAy +2KN -(2KN/m).2m =0 FAy =2KN（向上）ΣFx =0, FAx =0.计算最大弯矩：设A为坐标原点，X轴正向向右，横坐标为x梁截面弯矩方程:M(x) =(FAy)x -(qx)(x/2) =2x -x^2M(x)对x的一阶导数 M'(x) =2-2x当x =1, 导数为0，弯矩有极值：Mmax =2(1) -(1)^2 =1 KN.m.计算梁抗弯截面模量：Wz =(1/6)bh^2 =(1/6)(0.03m)(0.05m)^2 =0.0000125m^3 .校核梁抗弯强度：σmax =Mmax/Wz =(1KN.m)/(0.0000125m^3) =80000KPa =80MPa σmax。