A thin-walled, cantilever beam of unsymmetrical cross section supports shear loads at its free end as shown in Fig. P.9.21. Calculate the value of direct stress at the extremity of the lower flange (point A) at a section half-way along the beam if the position of the shear loads is such that no

Question

A thin-walled, cantilever beam of unsymmetrical cross section supports shear loads at its free end as shown in Fig. P.9.21. Calculate the value of direct stress at the extremity of the lower flange (point A) at a section half-way along the beam if the position of the shear loads is such that no twisting of the beam occurs.

Accepted Answer

The bending moments halfway along the beam are

[latex]M_{z}=800 imes 1000=800000 Nmm , \quad M_{y}=400 imes 1000=400000 Nmm[/latex]

By inspection the centroid of area is midway between the flanges. Its distance [latex]\bar{z}[/latex] from the vertical web is given by

[latex](40 imes 2+100 imes 2+80 imes 1) \bar{z}=40 imes 2 imes 20+80 imes 1 imes 40[/latex]

i.e.

[latex]\bar{z}=13.33 mm[/latex]

The second moments of area of the cross section are calculated using the approximations for thinwalled sections described in Section 9.6. Then

[latex]\begin{gathered}I_{z}=40 imes 2 imes 50^{2}+80 imes 1 imes 50^{2}+\frac{2 imes 100^{3}}{12}=5.67 imes 10^{5} mm ^{4} \I_{y}=100 imes 2 imes 13.33^{2}+\frac{2 imes 40^{3}}{12}+2 imes 40 imes 6.67^{2}+\frac{1 imes 80^{3}}{12}+1 imes 80 imes 26.67^{2} \=1.49 imes 10^{5} mm ^{4}\end{gathered}[/latex]

[latex] I_{z y}=40 imes 2(-6.67)(50)+80 imes 1(-26.67)(-50)=0.8 imes 10^{5} mm ^{4} [/latex]

The denominator in Eq. (9.31) is then [latex] \left(5.67 imes 1.49-0.8^{2} ight) imes 10^{10}=7.81 imes 10^{10} [/latex]. From Eq. (9.31)

[latex]\sigma_{x}=-\left(\frac{M_{y} I_{z}-M_{z} I_{z y}}{I_{z} I_{y}-I_{z y}^{2}} ight) z-\left(\frac{M_{z} I_{y}-M_{y} I_{z y}}{I_{z} I_{y}-I_{z y}^{2}} ight) y[/latex] (9.31)

[latex]\begin{aligned}\sigma=&-\left(\frac{400000 imes 5.67 imes 10^{5}-800000 imes 0.8 imes 10^{5}}{7.81 imes 10^{10}} ight) z \&-\left(\frac{800000 imes 1.49 imes 10^{5}-400000 imes 0.8 imes 10^{5}}{7.81 imes 10^{10}} ight) y\end{aligned}[/latex]

i.e.

[latex]\sigma=-2.08 z-1.12 y[/latex]

and at the point A where z=-66.67 mm, y=-50 mm

[latex]\sigma( A )=194.7 N / mm ^{2}[/latex]

Question 9.21: A thin-walled, cantilever beam of unsymmetrical cross sectio...

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