A cantilever of circular cross section has a diameter of 150 mmand is made from steel, which, when subjected to simple tension suffers elastic breakdown at a stress of 150 N/mm^2. The cantilever supports a bending moment and a torque, the latter having a value numerically equal to twice that of the

Question

A cantilever of circular cross section has a diameter of 150 mmand is made from steel, which, when subjected to simple tension suffers elastic breakdown at a stress of [latex]150 N / mm ^{2}[/latex].
The cantilever supports a bending moment and a torque, the latter having a value numerically equal to twice that of the former. Calculate the maximum allowable values of the bending moment and torque on the basis of the Tresca and von Mises theories of elastic failure.
Ans. M = 22.2 kNm, T = 44.4 kNm (Tresca).
M = 24.9 kNm, T = 49.8 kNm (von Mises).

Accepted Answer

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

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

and is

[latex]\sigma_{x}=\frac{M\left(\frac{D}{2} ight)}{\left(\frac{\pi D^{4}}{64} ight)}=\frac{32 M}{\pi D^{3}}[/latex]

The shear stress due to torsion is, from Eq. (11.4)

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

[latex] au_{x y}=\frac{T\left(\frac{D}{2} ight)}{\left(\frac{\pi D^{4}}{32} ight)}=\frac{32 M}{\pi D^{3}}[/latex]

Substituting 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 }=32 M\left(\frac{1.62}{\pi D^{3}} ight), \quad \sigma_{ II }=32 M\left(\frac{-0.62}{\pi D^{3}} ight)[/latex]

Then, from the Tresca theory, Eq. (14.42)

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

[latex]\left(\frac{32 M}{\pi D^{3}} ight)(1.62+0.62)=150 imes 10^{6}[/latex]

which gives, since D = 150 mm,
M = 22.2 kNm
Therefore
T = 44.4 kNm
Now substituting in the failure criterion for the von Mises theory, Eq. (14.55)

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

[latex]\left(\frac{32 M}{\pi D^{3}} ight)^{2}\left(1.62^{2}+0.62^{2}+1.62 imes 0.62 ight)=\left(150 imes 10^{6} ight)^{2}[/latex]

so that
M = 24.9 kNm
and
T = 49.8 kNm

Question 14.P.18: A cantilever of circular cross section has a diameter of 150...

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