A hollow cylindrical shaft has an internal diameter of 100 mm, an external diameter of 130 mm and is required to support a bending moment and a torque each having a value of 20 kNm. If the equivalent stress in the shaft must not exceed 230 N/mm^2 verify that the dimensions of the shaft are

Question

A hollow cylindrical shaft has an internal diameter of 100 mm, an external diameter of 130 mm and is required to support a bending moment and a torque each having a value of 20 kNm. If the equivalent stress in the shaft must not exceed [latex]230 N / mm ^{2}[/latex] verify that the dimensions of the shaft are satisfactory and also determine the minimum thickness the shaft must have for the same external diameter and loading.
Ans. [latex]\sigma_{ e }=188.8 N / mm ^{2}[/latex] therefore dimensions are satisfactory.
Minimum thickness = 11.3 mm.

Accepted Answer

The direct stress due to the bending moment is given by Eq. (9.9)

[latex]\sigma_{x}=-\frac{M y}{I}[/latex] (9.9)

and is

[latex]\sigma_{x}=\frac{-20 imes 10^{6} imes 65}{\pi\left(130^{4}-100^{4} ight) / 64}=-142.7 N / mm ^{2}[/latex]

The shear stress due to the applied torque is given by Eq. (11.4),

[latex]\frac{T}{J}=\frac{ au}{r}=G \frac{ heta}{L}[/latex] (11.4)

that is

[latex] au_{x y}=\frac{20 imes 10^{6} imes 65}{\pi\left(130^{4}-100^{4} ight) / 32}=71.4 N / mm ^{2}[/latex]

The design criterion is specified in Eq. (14.56)

[latex]\sigma_{ Y }=\sqrt{\sigma_{x}^{2}+3 au_{x y}^{2}}[/latex] (14.56)

where the equivalent stress, [latex]\sigma_{ e }[/latex], is

[latex]\sigma_{ e }=\sqrt{\left[(-142.7)^{2}+3 imes 71.4^{2} ight]}=188.8 N / mm ^{2}[/latex]

This is less than the allowable value of [latex]230 N / mm ^{2}[/latex] so that the dimensions of the shaft are satisfactory.
Suppose that the maximum allowable internal diameter of the shaft is d. Then the direct stress due to bending is, from Eq. (9.9)

[latex]\sigma_{x}=-\frac{20 imes 10^{6} imes 65}{\pi\left(130^{4}-d^{4} ight) / 64}=-\frac{2.65 imes 10^{10}}{\left(130^{4}-d^{4} ight)}[/latex]

and the shear stress due to the applied torque is, from Eq. (11.4)

[latex] au_{x y}=\frac{20 imes 10^{6} imes 65}{\pi\left(130^{4}-d^{4} ight) / 32}=\frac{1.33 imes 10^{10}}{\left(130^{4}-d^{4} ight)}[/latex]

Substituting these expressions in Eq. (14.56) in which [latex]\sigma_{Y}=230 N / mm ^{2}[/latex] and evaluating gives d = 107.5 mm. The minimum allowable thickness is then

[latex]t_{\min }=\frac{130-107.5}{2}=11.3 mm[/latex]

Question 14.P.24: A hollow cylindrical shaft has an internal diameter of 100 m...

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