圆轴的抗扭矩
Ip=πd^4/32,
Wp=πd^3/16
最大
剪切应力:τ=T/Wp
扭转角φ=Tl/GIp
本题:
τ=T/Wp=200Nm/(π40mm^3/16)=200×1000Nmm/(π40mm^3/16)
=15.92N/mm²=15.92MPa<[τ]=40MPa
φ=Tl/GIp=200Nm×1m/(80GPa×π40mm^4/32)
=200×10^6Nmm²/(80000MPa×π40mm^4/32)
=0.00994718394rad
=0.00994718394rad/πrad×180°
=0.57°<1°/m
满足要求。