A steel shaft, 20 cm external diameter and 7.5 cm internal, is subjected to a twisting moment of 30 kNm, and a thrust of 50 kN. Find the shearing stress due to the torque alone and the percentage increase when the thrust is taken into account. (RNC)

Question

Accepted Answer

For this case, we have

[latex]r_1 = 0.100 m, r_2 = 0.0375 m \\ A = \pi(r_1^2 - r_2^2) = 0.0270 m^2[/latex]

The compressive stress is

[latex]\sigma = -\frac{P}{A} = -\frac{50 imes 10^3}{0.0270} = -1.85 MN/m^2[/latex]

Now

[latex]J = \frac{\pi}{2}(r_1^4 - r_2^4) = 0.00247 m^4[/latex]

The shearing stress due to torque alone is

[latex] au = \frac{Tr_1}{J} = \frac{(30 imes 10^3)(0.100)}{0.00247} = 1.22 MN/m^2[/latex]

The maximum shearing stress due to the combined loading is

[latex] au_{\max} = \frac{1}{2}[\sigma^2 + 4 au^2]^{\frac{1}{2}} = 1.53 MN/m^2[/latex]

Question 16.P.5: A steel shaft, 20 cm external diameter and 7.5 cm internal, ......

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