Question 14.10: Locate the shear centre of the thin-walled section shown bel...

From Figure 14.43(b) shown above, as the section is symmetric about the z-axis, we can say that

\bar{I}_{z z}=2 \int_0^\alpha\left(R^2 \sin ^2 \theta\right)(R t) d \theta

=\left(R^3 t\right) \int_0^\alpha(1-\cos 2 \theta) d \theta

or            \bar{I}_{z z}=R^3 t\left\lgroup \alpha-\frac{\sin 2 \alpha}{2} \right\rgroup             (1)

We then determine the shear stress distribution throughout the section by referring to Figure 14.44.

Clearly, \tau_{x \phi}=V_y Q_z / t \bar{I}_{z z} \text { where } Q_z is the first moment about z-axis of the shaded area as shown in Figure 14.44. Thus,

Q_z=\int_{\alpha-\phi}^\alpha(R \sin \theta)(R t) d \theta

or            Q_z=\left(R^2 t\right)(-\theta)_{\alpha-\phi}^\alpha=R^2 t[\cos (\alpha-\phi)-\cos \alpha]             (2)

and so

\tau_{x \phi}=\left\lgroup \frac{V_y}{\bar{I}_{z z}} \right\rgroup R^2[\cos (\alpha-\phi)-\cos \alpha] ; \quad 0 \leq \phi \leq \alpha            (3)

Taking moment with respect to O of the shear force developed on the beam section, we obtain from Eq. (3)

M_{ O }=\iint_A d M_{ O }=\int_0^{2 \alpha} R \tau_{x \phi}(R d \phi) t=2 \int_0^\alpha\left(R^2 t\right) \tau_{x \phi} d \phi

=2 \int_0^\alpha\left(R^2 t\right)\left\lgroup \frac{R^2 V_y}{\bar{I}_{z z}} \right\rgroup [\cos (\alpha-\phi)-\cos \alpha] d \phi

=2\left\lgroup \frac{R^4 V_y t}{\bar{I}_{z z}} \right\rgroup \int_0^\alpha[\cos (\alpha-\phi)-\cos \alpha] d \phi

=2\left\lgroup \frac{R^4 t V_y}{\bar{I}_{z z}} \right\rgroup [-\sin (\alpha-\phi)-\phi \cos \alpha]_0^\alpha

=2\left\lgroup \frac{R^4 t V_y}{\bar{I}_{z z}} \right\rgroup [\sin \alpha-\alpha \cos \alpha]

From Varignon’s theorem of moments, we get

From Eq. (1)

e=\frac{2 R^4 t}{R^3 t} \frac{(\sin \alpha-\alpha \cos \alpha)}{(2 \alpha-\sin 2 \alpha) / 2}

=4 R \frac{\sin \alpha-\alpha \cos \alpha}{2 \alpha-\sin \alpha}

or            e=2 R \frac{\sin \alpha-\alpha \cos \alpha}{\alpha-\sin \alpha \cos \alpha}