The hollow cylinder shown in Fig. P.14.22 is built-in at one end and carries a vertical load of 2 kN offset a distance e from its vertical plane of symmetry. If the material of the cylinder fails in simple tension at a stress of $150 N / mm ^{2}$ calculate the maximum allowable value of e using the Tresca and von Mises theories of elastic failure.
Ans. 0.8 m (Tresca), 0.9 m (von Mises).

Question

The hollow cylinder shown in Fig. P.14.22 is built-in at one end and carries a vertical load of 2 kN offset a distance e from its vertical plane of symmetry. If the material of the cylinder fails in simple tension at a stress of [latex]150 N / mm ^{2}[/latex] calculate the maximum allowable value of e using the Tresca and von Mises theories of elastic failure.
Ans. 0.8 m (Tresca), 0.9 m (von Mises).

Accepted Answer

The maximum bending moment applied to the cylinder is 2×1 = 2 kNm and the torque, which is constant along the length of the cylinder, is 2e kNm in which e is measured in m. Then, from Eq. (9.9)

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

[latex]\sigma_{\max }=\frac{2 imes 10^{6} imes 30}{\pi\left(60^{4}-40^{4} ight) / 64}=117.6 N / mm ^{2}[/latex]

and, from Eq. (11.4)

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

[latex] au_{\max }=\frac{2 e imes 10^{6} imes 30}{\pi\left(60^{4}-40^{4} ight) / 32}=58.8 e N / mm ^{2}[/latex]

Substituting these values in Eqs (14.8) and (14.9)

[latex]\sigma_{ I }=\frac{\sigma_{x}+\sigma_{y}}{2}+\frac{1}{2} \sqrt{\left(\sigma_{x}-\sigma_{y} ight)^{2}+4 au_{x y}^{2}}[/latex] (14.8)

[latex]\sigma_{ II }=\frac{\sigma_{x}+\sigma_{y}}{2}-\frac{1}{2} \sqrt{\left(\sigma_{x}-\sigma_{y} ight)^{2}+4 au_{x y}^{2}}[/latex] (14.9)

and simplifying

[latex]\sigma_{ I }=58.8\left[1+\sqrt{\left(1+e^{2} ight)} ight][/latex]

[latex]\sigma_{ II }=58.8\left[1-\sqrt{\left(1+e^{2} ight)} ight][/latex]

Clearly [latex]\sigma_{ I }[/latex] is positive while [latex]\sigma_{ II }[/latex] is negative so that, for the Tresca theory, Eq. (14.42) applies.

[latex]\sigma_{ I }-\sigma_{ II }=\sigma_{ Y }[/latex] (14.42)

Then

[latex]\sigma_{ I }-\sigma_{ II }=2 imes 58.8 \sqrt{\left(1+e^{2} ight)}=150[/latex]

from which
e = 0.8 m.

The expressions for [latex]\sigma_{ I }[/latex] and [latex]\sigma_{ II }[/latex] may be written as [latex]\sigma_{ I }=58.8( A + B ), \sigma_{ II }=58.8( A - B )[/latex]. Substituting in Eq. (14.55)

[latex]\sigma_{ I }^{2}+\sigma_{ II }^{2}-\sigma_{ I } \sigma_{ II }=\sigma_{ Y }^{2}[/latex] (14.55)

and simplifying gives

[latex]A ^{2}+3 B ^{2}=150^{2} / 58.8^{2}[/latex]

That is

[latex]1+3\left(1+e^{2} ight)=150^{2} / 58.8^{2}[/latex]

from which
e = 0.9 m .

Question 14.P.22: The hollow cylinder shown in Fig. P.14.22 is built-in at one...

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