A thin-walled box beam has the cross section shown in Fig. P.14.21 and is simply supported over a span of 3 m. The beam carries a concentrated torque of 10 kNm at mid-span together with a vertical uniformly distributed load of intensity w N/mm over its complete span. If the yield stress in simple

Question

A thin-walled box beam has the cross section shown in Fig. P.14.21 and is simply supported over a span of 3 m. The beam carries a concentrated torque of 10 kNm at mid-span together with a vertical uniformly distributed load of intensity w N/mm over its complete span. If the yield stress in simple tension of the material of the beam is [latex]160 N / mm ^{2}[/latex] find the maximum allowable value of w using the Tresca and von Mises theories of elastic failure.
Ans. 9.7 N/mm (Tresca), 10.6 N/mm (von Mises).

Accepted Answer

The maximum torque on the beam is 5 kNm so that, from Eq. (11.21),

[latex] au=\frac{q}{t}=\frac{T}{2 A t}[/latex] (11.21)

the maximum shear stress in the beam section is

[latex] au_{\max }=\frac{T_{\max }}{2 A t_{\min }}=\frac{5 imes 10^{6}}{2 imes 80 imes 150 imes 4}=52.1 N / mm ^{2}[/latex]

The maximum bending moment in the beam occurs at mid-span and is [latex]w L^{2} / 8[/latex] (see Ex. 3.9). The direct stress due to bending is given by Eq. (9.9)

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

in which

[latex]I_{z}=2 imes 5 imes 80 imes 75^{2}+2 imes 4 imes 150^{3} / 12=6.75 imes 10^{6} mm ^{4}[/latex]

Then

[latex]\sigma_{x}=\frac{-w imes\left(3 imes 10^{3} ight)^{2} imes 75}{8 imes 6.75 imes 10^{6}}=-12.5 w N / mm ^{2}[/latex] (w in N/mm)

The principal stresses are then, from 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)

[latex]\sigma_{ I }=-\frac{12.5 w}{2}+\frac{1}{2} \sqrt{\left[(-12.5 w)^{2}+4 imes 52.1^{2} ight]}[/latex]

and

[latex]\sigma_{ II }=-\frac{12.5 w}{2}-\frac{1}{2} \sqrt{\left.(-12.5 w)^{2}+4 imes 52.1^{2} ight]}[/latex]

Clearly [latex]\sigma_{ I }[/latex] is positive and [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]\sqrt{\left[(-12.5 w)^{2}+4 imes 52.1^{2} ight]}=160[/latex]

from which
w = 9.7 N/mm
The same approach to determining w using the von Mises theory can be used here. That is

[latex]A ^{2}+3 B ^{2}=\sigma_{ Y }^{2}[/latex]

Then

[latex]\frac{(-12.5 w)^{2}}{4}+\frac{3}{4}\left[(-12.5 w)^{2}+4 imes 52.1^{2} ight]=160^{2}[/latex]

which gives
w = 10.6 N/mm

Question 14.P.21: A thin-walled box beam has the cross section shown in Fig. P...

This Question has been Answered.

Already have an account?

Login

Related Answered Questions