A cold-formed, thin-walled beam section of constant thickness has the profile shown in Fig. P.9.24. Calculate the position of the neutral axis and the maximum direct stress for a bending moment of 3.5 kNm applied about the horizontal axis Gz.

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Referring to Fig. S.9.24

[latex]\begin{aligned}&I_{z}=2 \int_{0}^{\pi} 6.4(50-50 \cos heta)^{2} imes 50 d heta=7.4 imes 10^{6} mm ^{4} \&I_{y}=2 \int_{0}^{\pi} 6.4(50 \sin heta)^{2} imes 50 d heta=2.51 imes 10^{6} mm ^{4} \&I_{z y}=2 \int_{0}^{\pi} 6.4(50 \sin heta)(50-50 \cos heta) imes 50 d heta=3.2 imes 10^{6} mm ^{4}\end{aligned}[/latex]

Since [latex]M_{y}=0[/latex], Eq. (9.33) reduces to

[latex] an \alpha=\frac{\left(M_zI_{zy}-M_yI_{z} ight) }{\left(M_yI_{zy}-M_zI_{y} ight) }[/latex] (9.33)

[latex] an \alpha=-\frac{I_{z y}}{I_{y}}=-\frac{3.2}{2.51}=-1.275[/latex]

i.e.

[latex]\alpha=-51.9^{\circ}[/latex]

which means that the neutral axis slopes downwards from left to right and makes an angle of 51.9° with the z axis.
The denominator in Eq. (9.31) is equal to [latex] \left(7.54 imes 2.51-3.2^{2} ight) imes 10^{12}=8.69 imes 10^{12} [/latex] and since the bending moment is [latex] M_{z}=3.5 kNm [/latex] the direct stress distribution is given by

[latex]\sigma _x=-\left(\frac{M_yI_z-M_zI_{zy}}{I_zI_y-I_{zy}^2} ight)z-\left(\frac{M_zI_y-M_yI_{zy}}{I_zI_y-I_{zy}^2} ight) y [/latex] (9.31)

[latex]\sigma_{x}=1.29 z-1.01 y[/latex] (i)

Inspection of Eq. (i) shows that the direct stress will be a maximum when z=0 and when y=±100 mm. Then

[latex] \sigma_{x}(\max )=\pm 101.0 N / mm ^{2} [/latex]

Question 9.24: A cold-formed, thin-walled beam section of constant thicknes...

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