Question 3.10: Shrink/interference fit A pair of mild steel cylinders (E = ...
Shrink/interference fit
A pair of mild steel cylinders (E = 200 GPa) of equal length have the following dimensions:
(1) 40 mm bore and 80.06 mm outside diameter
(2) 80 mm bore and 120 mm outside diameter
i.e. there is a diametral interference of 0.06 mm. The larger cylinder is heated, placed around, and allowed to shrink onto, the smaller cylinder. Calculate the stresses after assembly.
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:
\sigma _{r} = A_{1} – \frac{B_{1}}{r^{2}}and
\sigma _{\theta } = A_{1} + \frac{B_{1}}{r^{2}}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
B_{1} = – \frac{1600}{3}pThus
(\sigma _{r})_{1} = – \frac{4p}{3} \left(1 – \frac{400}{r^{2}} \right)and
(\sigma _{\theta })_{1} = – \frac{4p}{3} \left(1 + \frac{400}{r^{2}} \right)\varepsilon _{\theta } = \frac{u}{r} = \frac{1}{E}(\sigma _{\theta } – \nu (\sigma _{r } + \sigma _{z })) = \frac{1}{E}(\sigma _{\theta } – \nu \sigma _{r })
At the outside of cylinder 1, r = 40 mm,
\frac{- i_{1}}{40} = \frac{1}{200 000}(\sigma _{\theta } – \nu \sigma _{r})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{8p}{30 000} \left(\frac{5}{4} – \frac{3\nu }{4} \right)
∴ i_{1} = \frac{2p}{30 000} \left(5 – 3 \nu \right)
For cylinder 2
\sigma _{r} = A_{2} – \frac{B_{2}}{r^{2}}and
\sigma _{\theta } = A_{2} + \frac{B_{2}}{r^{2}}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
B_{2} = 3600 \times \frac{4}{5}pThus,
(\sigma_{r})_{2} = \frac{4p}{5} \left(1 – \frac{3600 }{r^{2}} \right)and
(\sigma_{\theta })_{2} = \frac{4p}{5} \left(1 + \frac{3600 }{r^{2}} \right)\\[0.5 cm]\varepsilon _{\theta } = \frac{u}{r} = \frac{1}{E}(\sigma _{\theta } – \nu (\sigma _{r } + \sigma _{z })) = \frac{1}{E}(\sigma _{\theta }-\nu \sigma _{r })At the inside of cylinder 2, r = 40 mm,
\frac{+ i_{2}}{40} = \frac{1}{200 000} \left( \frac{4p}{5} \right)\left(1 + \frac{3600}{1600} – \nu \left(1 – \frac{3600}{1600} \right) \right)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
\frac{10p}{30 000} – \frac{2\nu p}{10 000}+ \frac{26p}{50 000} + \frac{2\nu p}{10 000} = 0.03
\frac{50p + 78p}{150 000} = 0.03
i.e. p = \frac{4500}{128} N/mm^{2}
For cylinder 1,
(\sigma _{r})_{1} = – 46.9\left( 1 – \frac{400}{r^{2}} \right)(\sigma _{\theta })_{1} = – 46.9\left( 1 + \frac{400}{r^{2}} \right)
and for cylinder 2,
(\sigma _{r})_{2} = 28.2\left( 1 – \frac{3600}{r^{2}} \right)(\sigma _{\theta })_{2} = – 46.9\left( 1 + \frac{3600}{r^{2}} \right)
