A thin-walled cylinder of diameter 40 mm is subjected to an internal pressure of $5 N / mm ^{2}$ and a torque of 100 Nm. If the material of the cylinder has a yield stress in simple tension of $150 N / mm ^{2}$ determine the required wall thickness using the Tresca and von Mises theories of elastic failure.
Ans. 0.81 mm (Tresca), 0.74 mm (von Mises).

Question

A thin-walled cylinder of diameter 40 mm is subjected to an internal pressure of [latex]5 N / mm ^{2}[/latex] and a torque of 100 Nm. If the material of the cylinder has a yield stress in simple tension of [latex]150 N / mm ^{2}[/latex] determine the required wall thickness using the Tresca and von Mises theories of elastic failure.
Ans. 0.81 mm (Tresca), 0.74 mm (von Mises).

Accepted Answer

From Eq. (7.62)

[latex]\sigma_{ L }=\frac{p d}{4 t}[/latex] (7.62)

the longitudinal stress due to the internal pressure is given by

[latex]\sigma_{ L }=\frac{5 imes 40}{4 t}=\frac{50}{t}=\sigma_{x}[/latex]

and the circumferential stress is, from Eq. (7.63)

[latex]\sigma_{ C }=\frac{p d}{2 t}[/latex] (7.63)

[latex]\sigma_{ C }=\frac{5 imes 40}{2 t}=\frac{100}{t}=\sigma_{y}[/latex]

The shear stress produced by the applied torque is given by Eq. (11.22)

[latex] au_{\max }=\frac{T_{\max }}{2 A t_{\min }}[/latex] (11.22)

and is

[latex] au_{x y}=\frac{100 imes 10^{3}}{2\left(\pi imes 40^{2} / 4 ight) t}=\frac{39.8}{t}[/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)

gives

[latex]\sigma_{ I }=\frac{122}{t}, \quad \sigma_{ II }=\frac{28}{t}[/latex]

Both these values are positive and [latex]\sigma_{ I }>\sigma_{ II }[/latex] so that, for the Tresca theory, Eq. (14.41) applies.

[latex]\sigma_{ I }=\sigma_{ Y }[/latex] (14.41)

Then

[latex]\frac{122}{t}=150[/latex]

which gives
t = 0.81 mm .
Substituting in Eq. (14.55)

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

[latex](122 / t)^{2}+(28 / t)^{2}-\left(122 imes 28 / t^{2} ight)=150^{2}[/latex]

from which
t = 0.74 mm .

Question 14.P.23: A thin-walled cylinder of diameter 40 mm is subjected to an ...

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