Question 14.6: Locate the shear centre S of the following thin-walled secti...

Noting the symmetry of the given thin-walled section, we can write

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

Putting values, we get

\bar{I}_{z z}=2\left[\frac{1}{12}\left\lgroup \frac{t}{0.6} \right\rgroup (75)^3+\left\lgroup \frac{t}{0.6} \right\rgroup (75)(100)^2\right]+\frac{1}{12} t(125)^3  mm ^4

or      \bar{I}_{z z}=2779947.917 t  mm ^4              (1)

Let us now find the shear stress in leg BD and the shear load F_2 carried by it from Figure 14.34:

Clearly,

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

where Q_z is the first moment of shaded area about z-axis as shown in Figure 14.34(a). So,

Q_z=\left\lgroup\frac{t}{0.6} \right\rgroup (75)(100)+(s t)(62.5-0.5 s)

= (12500 + 62.5s − 0.5s²)t

Therefore,

\left|\left(\tau_{x y}\right)_{ BD }\right|=\frac{V_y}{\bar{I}_{z z}}\left|\left(12500+62.5 s-0.5 s^2\right)\right| ; \quad 0 \leq s \leq 125  m

By noting symmetry of the section, we conclude that force distribution in the section will be as shown in Figure 14.34(b).

Now, we calculate the distance of point of intersection of lines of action of F_1 in legs AB and DE on z-axis from the leg BD as (137.5 cot θ − 100) mm = 83.33 mm.
Taking moment with respect to 0 and applying Varignon’s theorem, we get

(83.33-e) V_y=83.33 F_2

From Eq. (1),

83.33-e=(83.33)(1725260.417) \frac{t}{2779947.917 t}

or             e = 31.615 mm