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/m² whilst the assembly rotates in service. If the maximum circumferential stress on

Question

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 [latex]30 MN/m^{2}[/latex] whilst the assembly rotates in service. If the maximum circumferential stress on the inside surface of the ring is limited to [latex] 240 MN/m^{2}[/latex], 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, [latex]ρ = 7470 kg/m^{3}[/latex], ν = 0.3, [latex]E = 208 GN/m^{2}[/latex]

Accepted Answer

From eqn. (4.7)

[latex]\sigma _{r} =A-\frac{B}{r^{2} }-\frac{\left(3+ u ight) }{8} ho \omega ^{2} r^{2} [/latex] (1) (4.7)
Now when [latex] r=0.15 [/latex] [latex]\sigma _{r}=0[/latex]
[latex]0=A-\frac{B}{0.15^{2} }-\frac{3.3 }{8} ho \omega ^{2} (0.15)^{2} [/latex] (2)

Also, when [latex] r=0.1 [/latex] [latex]\sigma _{r}=-30 MN/m^{2}[/latex]
[latex]-30 imes 10^{6} =A-\frac{B}{0.1^{2} }-\frac{3.3 }{8} ho \omega ^{2} (0.1)^{2} [/latex] (3)

(2)- (3 ) [latex] 30 imes 10^{6} =B(100-44.4)-\frac{3.3 }{8} ho \omega ^{2} (0.0225-0.01)[/latex]

[latex]B=\frac{30 imes 10^{6} }{55.6} +3.3 imes \frac{0.0125 imes 7470}{8 imes 55.6} \omega ^{2}[/latex]
[latex]B=0.54 imes 10^{6}+0.693 \omega ^{2}[/latex]

and from (3),

[latex]A=100(0.54 imes 10^{6}+0.693 \omega ^{2} )+\frac{3.3 imes 7470 imes 0.01\omega ^{2} }{8} -30 imes 10^{6}[/latex]
[latex]=54 imes 10^{6} +69.3\omega ^{2}+30.8\omega ^{2}-30 imes 10^{6}[/latex]
[latex]=24 imes 10^{6}+100.1\omega ^{2}[/latex]

But since the maximum hoop stress at the inside radius is limited to [latex]240 MN/m^{2}[/latex], from eqn. (4.8)

[latex]\sigma _{H} =A+\frac{B}{r^{2} } -\frac{(1+3 u ) }{8} ho \omega ^{2}r^{2}[/latex] (4.8)

i.e
[latex]240 imes 10^{6}=(24 imes 10^{6}+100.1\omega ^{2} ) +\frac{(0.54 imes 10^{6} +0.693\omega ^{2} )}{0.1^{2} }-\frac{1.9}{8} imes 7470 imes 0.01\omega ^{2}[/latex]
[latex]240 imes 10^{6}=78 imes 10^{6} +169.3\omega ^{2} -17.7\omega ^{2}[/latex]
[latex]151.7\omega ^{2}=162 imes 10^{6}[/latex]
[latex]\omega ^{2}=\frac{162 imes 10^{6} }{151.7} =1.067 imes 10^{6}[/latex]
[latex]\omega =1.33 rad/s=9860 rev/min[/latex]

Question 4.1: A steel ring of outer diameter 300 mm and internal diameter ...

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