Question

直径d等于20mm长L=200mm的钢杆,受拉力F=4KN的作用,已知E=200GPa,泊松比U=0.3求杆的应力、总伸

Answer 1

S=1/4×π×d^2=3.14×10^-4 (m^3)

应力=F/S =12.73 (MN)

△L=F×L/(E×S)=0.0127mm

L=L`+△L=200.0127mm