Let the centroid of the shaded area be represented by C₁

Distance of C₁ from x-axis (i.e. CC₁) :-

$\begin{aligned} &\bar{y}=\frac{\left(\frac{a}{2}-y\right)}{2}+y \\ &=\frac{a-2 y}{4}+y \\ &=\frac{a-2 y+4 y}{4}=\frac{a+2 y}{4} \\ &\therefore \bar{y}=\frac{1}{2}\left(\frac{a}{2}+y\right) \end{aligned}$

For shaded Area : -

$\begin{aligned} &\rightarrow \mathrm{A}_{\mathrm{s}}=\mathrm{b}\left(\frac{\mathrm{a}}{2}-\mathrm{y}\right) \\ &\rightarrow \mathrm{Q}_{\mathrm{x}}=\overline{\mathrm{y}} \cdot \mathrm{A}_{\mathrm{s}} \\ &=\frac{1}{2}\left(\frac{\mathrm{a}}{2}+\mathrm{y}\right) \cdot \mathrm{b} \cdot\left(\frac{\mathrm{a}}{2}-\mathrm{y}\right) \\ &\mathrm{Q}_{\mathrm{x}}=\frac{1}{2} \cdot \mathrm{b} \cdot\left(\frac{\mathrm{a}^{2}}{4}-\mathrm{y}^{2}\right) \end{aligned}$

Now, for Q_max :

$\begin{aligned} &\rightarrow \frac{\mathrm{dQ}}{\mathrm{dy}}=0 \\ &\rightarrow \frac{\mathrm{d}}{\mathrm{dy}}\left[\frac{1}{2} \cdot \mathrm{b} \cdot\left(\frac{\mathrm{a}^{2}}{4}-\mathrm{y}^{2}\right)\right]=0 \\ &\rightarrow \frac{1}{2} \cdot \mathrm{b} \cdot \frac{\mathrm{d}}{\mathrm{dy}}\left(\frac{\mathrm{a}^{2}}{4}-\mathrm{y}^{2}\right)=0 \\ &\rightarrow \frac{1}{2} \cdot \mathrm{b} \cdot(-2 \mathrm{y})=0 \\ &\rightarrow \mathrm{y}=0 \end{aligned}$

$\begin{aligned} &\therefore\left(Q_{x}\right)_{\max }(\ { at } \ y=0)=\frac{1}{2} \cdot b \cdot\left(\frac{a^{2}}{4}-0^{2}\right) \\ &=\frac{1}{2} \cdot b \cdot \frac{a^{2}}{4} \\ &\rightarrow\left(Q_{x}\right)_{\max }=\frac{1}{8} b \cdot a^{2} \end{aligned}$

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Introduction to Mechanics of Solids

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