T=Fr
F=2T/D=800/20=400
剪應力=2T/DWL
400/0.02x0.1=2Mpa
軸承受400N-m之扭矩
T=400N-m=400000N-mm
Ss=2*T/DWL
=>2*400000/20*200*100=2Mpa
265.有一鍵2×2×10cm裝於直徑20cm之軸上,該軸承受400N-m之扭矩..-阿摩線上測驗