Let the pressure set up between the rings be pMN/m2 .
Then, normal force between rings
=p×2πrL=N=p×106×2π×70×10−3×40×10−3=5600πpnewtons.
Friction force between rings =μN=0.15×5600πp
0.15×5600πp=150×103p=0.15×5600π150×103=57MN/m2
Now, for liner tube:
σr=−57 at r = 0.07
and
σr=0 at r = 0.04
−57=A−0.072B=A−204B0=A−0.042B=A−625B57=−421BB=−0.135A=625BA=−84.5
Therefore at the common radius the hoop stress in the inner tube is given by
σHi=A+0.072B=A+204B=−112.1Mn/m2
For the outer tube:
σr=−57 at r =0.07
and
σr=0 at r= 0.1
−57=A−204B0=A−100B57=104BB=0.548A=100BA=54.8
Therefore at the common radius the hoop stress in the outer tube is given by
σH0=A+0.072B=A+204B=166.8MN/m2
Shrinkage allowance =Er(σH0−σHi)
=208×10970×10−3[166.8−(−112.1)]106=20870×278.9×10−6m=93.8×10−6=0.094mm
Comments
Leave a comment