Question 14.11: Locate the shear centre of the section shown below in Figure...

The section is symmetric. Therefore, by putting α = π/2 in Eq. (1) of Example 14.10 we obtain

\bar{I}_{z z}=\frac{\pi}{2} R^3 t+2(b t) R^2=\left\lgroup \frac{\pi}{2} R^3+2 b R^2 \right\rgroup t

or          \bar{I}_{z z}=\frac{R^2 t}{2}(\pi R+4 b)              (1)

Let us now find stress distribution in leg AB

Considering the shaded area in Figure 14.47, we obtain

\left(\tau_{x z}\right)_{ AB }=\frac{V_y Q_z}{t \bar{I}_{z z}}

\left(\tau_{x z}\right)_{ AB }=\left\lgroup \frac{V_y R}{\bar{I}_{z z}} \right\rgroup s ; \quad 0 \leq s \leq b             (2)

Now, force carried by leg AB is

F_1=\int_0^b\left(\tau_{x z}\right)_{ AR } t  d s=\left\lgroup \frac{V_y R T}{\bar{I}_{z z}} \right\rgroup \int_0^b s  d s

Thus,

F_1=\left\lgroup\frac{b^2 R t}{2 \bar{I}_{z z}} \right\rgroup V_y              (3)

From Example 14.10, we can find the shear stress \tau_{x \phi} distribution in the semicircular leg by putting α = π/2 in Eq. (2) and adding Q_z of A given by

Q_z=R^2 t\left[\cos \left\lgroup \frac{\pi}{2}-\phi \right\rgroup -\cos \frac{\pi}{2}\right]+b t R

and so

\tau_{x \phi}=R\left\lgroup \frac{V_y}{\bar{I}_{z z}} \right\rgroup (R \sin \phi+b) ; \quad 0 \leq \phi \leq \frac{\pi}{2}               (4)

We show the force distribution in Figure 14.48:

By computing moment of all forces about O, we obtain

V_y e=2 \int_0^{\pi / 2} \tau_{x \phi} t R^2 d \phi+2 R F_1

From Eqs. (2)–(4), we get

V_y e=2 \frac{V_y R^3 t^{\pi / 2}}{\bar{I}_{z z}} \int_0^2(R \sin \phi+b) d \phi+\frac{b^2 R^2 t}{\bar{I}_{z z}} V_y

Therefore,

e=\frac{2 R^4 t}{\bar{I}_{z z}}+\frac{b^2 R^2 t}{\bar{I}_{z z}}+\frac{\pi R^3 t b}{\bar{I}_{z z}}

or            e=\frac{R^2 t}{\bar{I}_{z z}}\left(2 R^2+b^2+\pi R b\right)

From Eq. (1), putting \bar{I}_{z z} in the above equation, we obtain

e=\frac{2\left(2 R^2+b^2+\pi R b\right)}{(\pi R+4 b)}