Answer to Question #230160 in Molecular Physics | Thermodynamics for Anuj

Question

Two steel rings of radial thickness 30 mm, common radius 70 mm and length 40 mm are 

shrunk together to form a compound ring. It is found that the axial force required to separate 

the rings, i.e. to push the inside ring out, is 150 kN. Determine the shrinkage pressure at the 

mating surfaces and the shrinkage allowance. E = 208 GN/mZ. The coefficient of friction 

between the junction surfaces of the two rings is 0.15. 

Answer 1

Let the pressure set up between the rings be $p\;MN/m^2$ .

Then, normal force between rings

$= p \times 2 \pi rL = N \\ = p \times 10^6 \times 2 \pi \times 70 \times 10^{-3} \times 40 \times 10^{-3} \\ = 5600 \pi p \;newtons.$

Friction force between rings $= \mu N = 0.15 \times 5600 \pi p$

$0.15 \times 5600 \pi p = 150 \times 10^3 \\ p = \frac{150 \times 10^3}{0.15 \times 5600 \pi} \\ = 57 MN/m^2$

Now, for liner tube:

$\sigma_r = -57$ at r = 0.07

and

$\sigma_r=0$ at r = 0.04

$-57 = A -\frac{B}{0.07^2} = A -204B \\ 0 = A -\frac{B}{0.04^2} = A -625B \\ 57 = -421B \\ B = -0.135 \\ A=625B \\ A=-84.5$

Therefore at the common radius the hoop stress in the inner tube is given by

$\sigma_{Hi}= A + \frac{B}{0.07^2} = A + 204B = -112.1 \;Mn/m^2$

For the outer tube:

$\sigma_r = -57$ at r =0.07

and

$\sigma_r = 0$ at r= 0.1

$-57 = A- 204B \\ 0= A-100B \\ 57 = 104B \\ B= 0.548 \\ A= 100B \\ A= 54.8$

Therefore at the common radius the hoop stress in the outer tube is given by

$\sigma_{H0}= A + \frac{B}{0.07^2}= A + 204B = 166.8 \;MN/m^2$

Shrinkage allowance $= \frac{r}{E}(\sigma_{H0} -\sigma_{Hi})$

$= \frac{70 \times 10^{-3}}{208 \times 10^9}[166.8 -(-112.1)]10^6 \\ = \frac{70 \times 278.9}{208} \times 10^{-6} \;m \\ = 93.8 \times 10^{-6} = 0.094 \;mm$