Question 14.5: A thin-walled beam of uniform thickness has the cross-sectio...

Here again we note from Figure 14.31(b) that

\tan \theta=\frac{60}{80}=\frac{3}{4}

Therefore,

DD ^{\prime}=\frac{t}{\sin \theta}=\frac{5 t}{3}

or          \bar{I}_{z z}=528 \times 10^3 t  mm ^4            (1)

Now, we calculate shear load F_1 carried by leg AB as

\left|F_1\right|=\int_0^{40}\left|\left(\tau_{x z}\right)_{ AB }\right| t  d s \quad \text { where }\left|\left(\tau_{x z}\right)_{ AB }\right|=\left|\frac{V_y Q_z}{t \bar{I}_{z z}}\right|

here Q_z=(s t)(60)  mm ^3 \text {. Thus, }\left|F_1\right|=F_1 (denoting the magnitude only) is given by

F_1=\int_0^{40}\left\lgroup \frac{V_y}{t \bar{I}_{z z}} \right\rgroup(60 s t) t  d s=\left\lgroup \frac{V_y t}{\bar{I}_{z z}} \right\rgroup 30 \times 40^2

Thus, the shear centre location can be calculated from Figure 14.32.

In the above figure, from symmetry we note (a) F_1=F_4 \text { and (b) } F_2=F_3 . From Varignon’s principle of moments, we get

V_y e=120 F_1=(120)(30)(40)^2 V_y \frac{t}{\bar{I}_{z z}}

From Eq. (1),

e=(120)(30)(40)^2 \frac{t}{528\left(10^3\right) t}  mm

or         e = 10.91 mm