工程力学的题目求解答

Question

如题所述

Answer 1

A点切应力:
TA =Mk.ρA/I =(2.5KN.m)(0.055m/4)/(0.1x0.055m^4) ≈37566KN/m^2 =37.566MPa
.
最大切应力：
Tmax =Mk.R/I =(2.5KN.m)(0.055m/2)/(0.1x0.055m)^4 ≈75.132MPa
.
单位长度扭转角：
θ =Mk/GI =(2.5KN.m)/[(82000000KPa)(0.1x0.055m^4)] ≈0.0333rad/m
.
A点切应变：
γA = ρA.θ =(0.055m/4)(0.0333rad/m) ≈0.00183