Question 3.10: Shrink/interference fit A pair of mild steel cylinders (E = ...

Conditions
• After assembly, the radial interference pressure, p, will be the same on both cylinders, i.e. cylinder 1 will have an external pressure, p, and cylinder 2 will have an internal pressure, p, as indicated in Figure 3.99.
• The decrease in the outside radius of cylinder 1, $i_{1}$, plus the increase in the inside radius of cylinder 2, $i_{2}$, will be equal to the radial interference, i.e. $i = i_{1} +  i_{2}$.
• Axial stresses are assumed to be zero (or negligible).

For cylinder 1:

and

at r = 20 mm, $\sigma _{r} =0$,

∴$B_{1} = 400A_{1}$

at r = 40 mm (no significant difference with 40.03 mm), $\sigma _{r} = –  p$

∴$–  p = A_{1}  –  \frac{20^{2}}{40^{2}} A_{1} = A_{1}  –  \frac{400}{1600} A_{1}$

∴$A_{1} = – \frac{4}{3} p$

and

Thus

and

At the outside of cylinder 1, r = 40 mm,

i.e.$\frac{-  i_{1}}{40} = \frac{1}{200   000} \left(-  \frac{4p}{3} \right)\left(1 + \frac{400}{1600}  –  \nu \left(1 – \frac{400}{1600} \right) \right)$

∴$i_{1} = \frac{2p}{30  000} \left(5  –  3 \nu \right)$

For cylinder 2

and

at r = 60 mm, $\sigma _{r} =0$

∴ $B_{2} = 3600 A_{2}$

at r = 40 mm, $\sigma _{r} = –  p$

∴$–  p = A_{2}  –  \frac{60^{2}}{40^{2}}A_{2} = A_{2}  –  \frac{3600}{1600}A_{2}$
i.e.$A_{2} = \frac{4}{5} p$

and

Thus,

and

At the inside of cylinder 2, r = 40 mm,

i.e. $i_{2} = \frac{8p}{50  000} \left(\frac{13}{4} + \frac{5\nu }{4} \right)$

∴$i_{2} = \frac{2p}{50  000} \left(13 + 5 \nu \right)$

But $i_{1} + i_{2} = i = 0.03  mm$

∴$\frac{2p}{30  000}(5  –  3\nu ) + \frac{2p}{50  000}(13 + 5\nu ) = 0.03$

i.e. $p = \frac{4500}{128} N/mm^{2}$

For cylinder 1,

and for cylinder 2,