Question 4.1: A steel ring of outer diameter 300 mm and internal diameter ...
A steel ring of outer diameter 300 mm and internal diameter 200 mm is shrunk onto a solid steel shaft. The interference is arranged such that the radial pressure between the mating surfaces will not fall below 30 MN/m^{2} whilst the assembly rotates in service. If the maximum circumferential stress on the inside surface of the ring is limited to 240 MN/m^{2}, determine the maximum speed at which the assembly can be rotated. It may be assumed that no relative slip occurs between the shaft and the ring. For steel, ρ = 7470 kg/m^{3}, ν = 0.3, E = 208 GN/m^{2}
From eqn. (4.7)
\sigma _{r} =A-\frac{B}{r^{2} }-\frac{\left(3+\nu \right) }{8} \rho \omega ^{2} r^{2} (1) (4.7)
Now when r=0.15 \sigma _{r}=0
0=A-\frac{B}{0.15^{2} }-\frac{3.3 }{8} \rho \omega ^{2} (0.15)^{2} (2)
Also, when r=0.1 \sigma _{r}=-30 MN/m^{2}
-30\times 10^{6} =A-\frac{B}{0.1^{2} }-\frac{3.3 }{8} \rho \omega ^{2} (0.1)^{2} (3)
(2)- (3 ) 30\times 10^{6} =B(100-44.4)-\frac{3.3 }{8} \rho \omega ^{2} (0.0225-0.01)
B=\frac{30\times 10^{6} }{55.6} +3.3\times \frac{0.0125\times 7470}{8\times 55.6} \omega ^{2}
B=0.54\times 10^{6}+0.693 \omega ^{2}
and from (3),
A=100(0.54\times 10^{6}+0.693 \omega ^{2} )+\frac{3.3\times 7470\times 0.01\omega ^{2} }{8} -30\times 10^{6}
=54\times 10^{6} +69.3\omega ^{2}+30.8\omega ^{2}-30\times 10^{6}
=24\times 10^{6}+100.1\omega ^{2}
But since the maximum hoop stress at the inside radius is limited to 240 MN/m^{2}, from eqn. (4.8)
\sigma _{H} =A+\frac{B}{r^{2} } -\frac{(1+3\nu ) }{8}\rho \omega ^{2}r^{2} (4.8)
i.e
240\times 10^{6}=(24\times 10^{6}+100.1\omega ^{2} ) +\frac{(0.54\times 10^{6} +0.693\omega ^{2} )}{0.1^{2} }-\frac{1.9}{8} \times 7470\times 0.01\omega ^{2}
240\times 10^{6}=78\times 10^{6} +169.3\omega ^{2} -17.7\omega ^{2}
151.7\omega ^{2}=162\times 10^{6}
\omega ^{2}=\frac{162\times 10^{6} }{151.7} =1.067\times 10^{6}
\omega =1.33 rad/s=9860 rev/min