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.

Question

Accepted Answer

Clearly, from Figures 14.37(a) and (b), we obtain

[latex] \bar{I}_{z z}=2\left[\frac{1}{12}\left\lgroup \frac{t}{\sin 30^{\circ}} ight group (35)^3+\left\lgroup \frac{t}{\sin 30^{\circ}} ight group (35)(17.5)^2+ ight] \frac{1}{12} t(70)^3 mm ^4 [/latex]

[latex] =857504 t mm ^4 [/latex]

Now, shear stress in leg AB is

[latex] \left( au_{x s} ight)_{ AB }=\frac{V_y Q_z}{t \bar{I}_{z z}} [/latex]

where as shown in Figure 14.38, we obtain from the shaded area

[latex] Q_z=(s t) \frac{s}{2} \sin 30^{\circ}=\frac{s^2 t}{4} [/latex]

Therefore,

[latex] \left( au_{x s} ight)_{ AB }=\left\lgroup \frac{V_y}{4 \bar{I}_{z z}} ight group s^2 [/latex]; 0 ≤ s ≤ 70 mm

Force carried by leg AB is

[latex] F_1=\int_0^{70}\left( au_{x s} ight)_{ AB } t d s=\left\lgroup \frac{V_y t}{4 \bar{I}_{z z}} ight group\int_0^{70} s^2 d s=\frac{V_y t}{12 \bar{I}_{z z}}(70)^3 [/latex]

Noting the symmetry of the section, we show the shear force distribution in Figure 14.39:

Applying Varignon’s theorem, we get

[latex] V_y e=\left(2 F_1 ight)(35) an 60^{\circ} \cdot \sin 30^{\circ} [/latex]

[latex] =70 F_1 an 60^{\circ} \cdot \sin 30^{\circ} [/latex]

[latex] =\left(70 an 60^{\circ} ight) \frac{V_y t}{12 \bar{I}_{z z}}(70)^3 \cdot \sin 30^{\circ} [/latex]

Therefore,

[latex] e=(70)^4 an 60^{\circ} \frac{t}{85750 t} imes \frac{1}{12} imes \sin 30^{\circ} [/latex]

or e = 20.207 mm

Question 14.8: By determining shear stress distribution in leg AB, find the...

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