Question 14.8: By determining shear stress distribution in leg AB, find the...
By determining shear stress distribution in leg AB, find the location of shear centre, S of the section shown in Figure 14.37(a). Assume thickness is small and uniform throughout.
Clearly, from Figures 14.37(a) and (b), we obtain
\bar{I}_{z z}=2\left[\frac{1}{12}\left\lgroup \frac{t}{\sin 30^{\circ}} \right\rgroup (35)^3+\left\lgroup \frac{t}{\sin 30^{\circ}} \right\rgroup (35)(17.5)^2+\right] \frac{1}{12} t(70)^3 mm ^4
=857504 t mm ^4
Now, shear stress in leg AB is
\left(\tau_{x s}\right)_{ AB }=\frac{V_y Q_z}{t \bar{I}_{z z}}
where as shown in Figure 14.38, we obtain from the shaded area
Q_z=(s t) \frac{s}{2} \sin 30^{\circ}=\frac{s^2 t}{4}
Therefore,
\left(\tau_{x s}\right)_{ AB }=\left\lgroup \frac{V_y}{4 \bar{I}_{z z}} \right\rgroup s^2 ; 
0 ≤ s ≤ 70 mm
Force carried by leg AB is
F_1=\int_0^{70}\left(\tau_{x s}\right)_{ AB } t d s=\left\lgroup \frac{V_y t}{4 \bar{I}_{z z}} \right\rgroup\int_0^{70} s^2 d s=\frac{V_y t}{12 \bar{I}_{z z}}(70)^3
Noting the symmetry of the section, we show the shear force distribution in Figure 14.39:
Applying Varignon’s theorem, we get
V_y e=\left(2 F_1\right)(35) \tan 60^{\circ} \cdot \sin 30^{\circ}
=70 F_1 \tan 60^{\circ} \cdot \sin 30^{\circ}
=\left(70 \tan 60^{\circ}\right) \frac{V_y t}{12 \bar{I}_{z z}}(70)^3 \cdot \sin 30^{\circ}
Therefore,
e=(70)^4 \tan 60^{\circ} \frac{t}{85750 t} \times \frac{1}{12} \times \sin 30^{\circ}
or e = 20.207 mm
