A beam of circular cross section is 3000 mm long and is attached at each end to supports which allow rotation of the ends of the beam in the longitudinal vertical plane of symmetry but prevent rotation of the ends in vertical planes perpendicular to the axis of the beam (Fig. P.14.17). The beam

Question

A beam of circular cross section is 3000 mm long and is attached at each end to supports which allow rotation of the ends of the beam in the longitudinal vertical plane of symmetry but prevent rotation of the ends in vertical planes perpendicular to the axis of the beam (Fig. P.14.17). The beam supports an offset load of 40 000 N at mid-span.

If the material of the beam suffers elastic breakdown in simple tension at a stress of [latex]145 N / mm ^{2}[/latex], calculate the minimum diameter of the beam on the basis of the Tresca and von Mises theories of elastic failure.
Ans. 136 mm (Tresca), 135 mm (von Mises).

Accepted Answer

The loading on the beam is equivalent to a central concentrated load of 40 kN together with a torque of 40 kNm applied at mid-span. The maximum bending moment on the beam is [latex]\left(40 imes 10^{3} imes 3 imes 10^{3} ight) / 4=30 imes 10^{6} Nmm[/latex] . The direct stress due to bending is given by Eq. (9.9),

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

i.e.

[latex]\sigma_{x}=\frac{30 imes 10^{6}\left(\frac{D}{2} ight)}{\left(\frac{\pi D^{4}}{64} ight)}=\frac{306 imes 10^{6}}{D^{3}}[/latex]

From symmetry the reaction torques are equal and of magnitude [latex]20 imes 10^{6} Nmm[/latex] is therefore the maximum torque on the beam. Then, from Eq. (11.4)

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

[latex] au_{x y}=\frac{20 imes 10^{6}\left(\frac{D}{2} ight)}{\left(\frac{\pi D^{4}}{32} ight)}=\frac{102 imes 10^{6}}{D^{3}}[/latex]

Now 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)

for [latex]\sigma_{x}[/latex] and [latex] au_{x y}[/latex] gives

[latex]\sigma_{ I }=\frac{336.9 imes 10^{6}}{D^{3}}, \quad \sigma_{ II }=\frac{-30.9}{D^{3}}[/latex]

Since [latex]\sigma_{ II }[/latex] is negative Eq. (14.42)

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

for the Tresca theory applies. Then

[latex](336.9+30.9) imes \frac{10^{6}}{D^{3}}=145[/latex]

which gives
D = 136 mm
Substituting now in the two-dimensional case 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(336.9^{2}+30.9^{2}+336.9 imes 30.9 ight)\left(\frac{10^{6}}{D^{3}} ight)^{2}=145^{2}[/latex]

from which
D = 135 mm.

Question 14.P.17: A beam of circular cross section is 3000 mm long and is atta...

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