A beam has a rectangular thin-walled box section 50 mmwide by 100 mmdeep and has walls 2 mm thick. At a particular section the beam carries a bending moment M and a torque T. A rectangular strain gauge rosette positioned on the top horizontal wall of the beam at this section recorded the following

Question

A beam has a rectangular thin-walled box section 50 mmwide by 100 mmdeep and has walls 2 mm thick. At a particular section the beam carries a bending moment M and a torque T. A rectangular strain gauge rosette positioned on the top horizontal wall of the beam at this section recorded the following values of strain: [latex]\varepsilon_{ a }=1000 imes 10^{-6}, \varepsilon_{ b }=-200 imes 10^{-6}, \varepsilon_{ c }=-300 imes 10^{-6}[/latex]. If the strain gauge ‘a’ is aligned with the longitudinal axis of the beamand the strain gauge ‘c’ is perpendicular to the longitudinal axis, calculate the values of M and T. Take [latex]E=200000 N / mm ^{2}[/latex] and v = 0.3.
Ans. M = 3333 Nm, T = 1692 Nm.

Accepted Answer

Substituting the given values of strain in Eqs (14.34) and (14.35)

[latex]\varepsilon_{ I }=\frac{1}{2}\left(\varepsilon_{ a }+\varepsilon_{ c } ight)+\frac{1}{\sqrt{2}} \sqrt{\left(\varepsilon_{ a }-\varepsilon_{ b } ight)^{2}+\left(\varepsilon_{ c }-\varepsilon_{ b } ight)^{2}}[/latex] (14.34)

[latex]\varepsilon_{ II }=\frac{1}{2}\left(\varepsilon_{ a }+\varepsilon_{ c } ight)-\frac{1}{\sqrt{2}} \sqrt{\left(\varepsilon_{ a }-\varepsilon_{ b } ight)^{2}+\left(\varepsilon_{ c }-\varepsilon_{ b } ight)^{2}}[/latex] (14.35)

gives

[latex]\varepsilon_{ I }=1201.5 imes 10^{-6}, \quad \varepsilon_{ II }=-501.5 imes 10^{-6}[/latex]

Substituting these values in Eqs (14.32) and (14.33)

[latex]\sigma_{ I }=\frac{E}{1-v^{2}}\left(\varepsilon_{ I }+v \varepsilon_{ II } ight)[/latex] (14.32)

[latex]\sigma_{ II }=\frac{E}{1-v^{2}}\left(\varepsilon_{ II }+v \varepsilon_{ I } ight)[/latex] (14.33)

gives

[latex]\sigma_{ I }=231.0 N / mm ^{2}, \quad \sigma_{ II }=-31.0 N / mm ^{2}[/latex]

Adding 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 }+\sigma_{ II }=\sigma_{x}=200 N / mm ^{2}[/latex]

This stress is produced by the bending moment M. Therefore, from Eq. (9.9)

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

[latex]200=\frac{M imes 50}{2\left[\left(\frac{2 imes 100^{3}}{12} ight)+2 imes 50 imes 50^{2} ight]}[/latex]

from which
M = 3333 Nm
Subtracting Eqs (14.8) and (14.9)

[latex]\sigma_{ I }-\sigma_{ II }=262.0=\sqrt{\left(200^{2}+4 au_{x y}^{2} ight)}[/latex]

which gives

[latex] au_{x y}=84.6 N / mm ^{2}[/latex]

This shear stress is caused by the applied torque and since the beam is a thin-walled box section the shear stress is given by Eq. (11.22),

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

i.e.

[latex]84.6=\frac{T}{2 imes 50 imes 100 imes 2}[/latex]

from which
T = 1692 Nm .

Question 14.P.12: A beam has a rectangular thin-walled box section 50 mmwide b...

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