Question 14.3: Locate the shear centre of the section shown in Figure 14.23...

First we determine \bar{I}_{z z} of the section as follows:

\bar{I}_{z z}=2\left(\bar{I}_{ AB }+\bar{I}_{ BD }\right)_{z z}+\left(\bar{I}_{ DE }\right)_{z z}

since the section is symmetric with respect to z-axis. So,

\bar{I}_{z z}=2\left[\frac{1}{12} b_1^3 t+\left(b_1 t\right)\left\lgroup \frac{b_1+b_3}{2} \right\rgroup^2+\left(b_2 t\right)\left\lgroup \frac{b_3}{2} \right\rgroup^2\right]+\frac{1}{12} b_3^3 t

or          \bar{I}_{z z}=\frac{t}{12}\left[b_3^2\left(b_3+6 b_2+6 b_1\right)+4 b_1^2\left(2 b_1+3 b_3\right)\right]            (1)

Next, we determine the shear stress distribution as follows:

(a) Leg AB (refer Figure 14.24)

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

where

Q_z=(s t)\left\lgroup \frac{b_3}{2}+b_1-\frac{s}{2} \right\rgroup

Therefore,

\left|\left(\tau_{x y}\right)_{ AB }\right|=\left|\frac{V_y s}{\bar{I}_{z z}}\left\lgroup (b_1+\frac{b_3}{2}-\frac{s}{2} \right\rgroup\right| ; \quad 0 \leq s \leq b_1               (2)

(b) Leg BD (refer Figure 14.25)

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

where

Q_z=(s t)\left\lgroup \frac{b_3}{2} \right\rgroup+\left(b_1 t\right) \frac{1}{2}\left(b_1+b_3\right)

Therefore,

\left|\left(\tau_{x z}\right)_{ BD }\right|=\left|\frac{V_y}{2 \bar{I}_{z z}}\left[b_1\left(b_1+b_3\right)+s b_3\right]\right| ; \quad 0 \leq s \leq b_2            (3)

(c) Leg DE (refer Figure 14.26)

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

where

Q_z=\left(b_1 t\right) \frac{1}{2}\left(b_1+b_3\right)+\left(b_2 t\right) \frac{b_2}{2}+(s t) \frac{1}{2}\left(b_3-s\right)

Therefore,

\left|\left(\tau_{x y}\right)_{ DE }\right|=\left|\left\lgroup \frac{V_y}{2 \bar{I}_{z z}} \right\rgroup\left[b_1\left(b_1+b_3\right)+b_2 b_3+s b_3-s^2\right]\right|                 (4)

(d) Leg EF

\left|\left(\tau_{x z}\right)_{ EF }\right|=\left|\left(\tau_{x z}\right)_{ BD }\right|

(e) Leg FG

\left|\left(\tau_{x y}\right)_{ FG }\right|=\left|\left(\tau_{x y}\right)_{ AB }\right|

The above shear stress distributions can be shown in Figure 14.27(a):

Now, we can find shear forces in different legs of the section shown in Figure 14.27(b) as follows:
From Eqs. (2)–(4), on integration, we get

F_1=\int_0^{b_1}\left(\tau_{x y}\right)_{ AB } t  d s=\left\lgroup \frac{V_y t}{\bar{I}_{z z}} \right\rgroup \int_0^{b_1}\left[\left\lgroup b_1+\frac{b_3}{2} \right\rgroup s-\frac{s^2}{2}\right] d s

=\left\lgroup \frac{V_y t}{\bar{I}_{z z}} \right\rgroup\left[\frac{b_1^2}{2}\left(b_1+\frac{b_3}{2}\right)-\frac{b_1^3}{6}\right]

=\frac{V_y t b_1^2}{2 \bar{I}_{z z}}\left[b_1+\frac{b_3}{2}-\frac{b_1}{2}\right]=\frac{V_y t b_1^2}{2 \bar{I}_{z z}}\left\lgroup \frac{2}{3} b_1+\frac{b_3}{2} \right\rgroup

or          F_1=\frac{V_y b_1^2 t}{12 \bar{I}_{z z}}\left(4 b_1+3 b_3\right)        (5)

Similarly,

F_2=\frac{V_y t}{2 \bar{I}_{z z}} \int_0^{b_2}\left[s b_3+b_1\left(b_1+b_3\right)\right] d s

or          F_2=\frac{V_y t}{2 \bar{I}_{z z}}\left[\frac{b_2^2 b_3}{2}+b_1 b_2\left(b_1+b_3\right)\right]          (6)

and

F_3=\left\lgroup \frac{V_y t}{2 \bar{I}_{z z}} \right\rgroup\left\{\left[b_1\left(b_1+b_3\right)+b_2 b_3\right] b_3+\frac{b_3^3}{2}-\frac{b_3^3}{3}\right\}

or          F_3=\left\lgroup \frac{V_y t}{2 \bar{I}_{z z}} \right\rgroup\left\{\left[b_1\left(b_1+b_3\right)+b_2 b_3\right] b_3+\frac{b_3^3}{6}\right\}         (7)

We note from Eqs. (5) and (7),

Thus, 2 F_1+F_3=V_y from Eq. (1) above, and this is expected from equilibrium. Thus, we show in Figure 14.27(c) above the resultant force diagram, where resultant force V_y passes through S. Clearly, from Varignon’s theorem of moments

V_y e=b_3 F_2-2 b_2 F_1

From Eqs. (5) and (6),

V_y e=\frac{V_y t}{2 \bar{I}_{z z}}\left[\frac{b_2^2 b_3^2}{2}+b_1 b_2 b_3\left(b_1+b_3\right)\right]-\frac{V_y b_1^2 b_2 t}{6 \bar{I}_{z z}}\left(4 b_1+3 b_3\right)

Therefore,

Note: Compare the above result with that of Problem 14.1.