Determine the centroid (x,y) for the triangle shown in Fig. 3.23.

Question

Determine the centroid [latex](\overline{x},\overline{y} )[/latex] for the triangle shown in Fig. 3.23.

Accepted Answer

Although there are multiple ways to perform the requisite integration, let us first do so with a differential area dA=dxdy noting that y=(h/b)x (i.e., the slope, or rise over run, is h/b and the intercept is zero). Hence, FIGURE 3.23 Determine a general formula for the first moment of area for this triangular cross section.

[latex] A=\int_{0}^{b}{\left(\int_{0}^{hx/b}{dy} \right)dx }=\int_{0}^{b}{\frac{hx}{b} }dx=\frac{h}{b}\left(\frac{x^{2}}{2}\mid^{b}_{0} \right)=\frac{1}{2}bh,[/latex]

as expected. Similarly,

[latex] \iint{xdA}=\int_{0}^{b}{x\left(\int_{0}^{hx/b}{dy} \right)dx }=\int_{0}^{b}{\frac{hx^{2}}{b} }dx=\frac{h}{b}\left(\frac{b^{3}}{3} \right)=\frac{1}{3}b^{2}h[/latex]

and

[latex] \iint{ydA}=\int_{0}^{b}{\left(\int_{0}^{hx/b}{ydy} \right)dx }=\int_{0}^{b}{\frac{1}{2}\left(\frac{h^{2}x^{2}}{b^{2}} \right)dx }=\frac{1}{2}\frac{h^{2}}{b^{2}}\left(\frac{b^{3}}{3} \right)=\frac{1}{6}bh^{2}.[/latex]

Consequently,

[latex] \overline{x}=\frac{\iint{xdA} }{\iint{dA} }=\frac{\frac{1}{3}b^{2}h }{\frac{1}{2}bh }=\frac{2}{3}b, \overline{y}=\frac{\iint{ydA} }{\iint{dA} }=\frac{\frac{1}{6}bh^{2} }{\frac{1}{2}bh }=\frac{1}{3}h,[/latex]

as expected. Show that the same result is obtained by considering a differential area dA=ydx=(h/b)x dx.

Question A3.1: Determine the centroid (x,y) for the triangle shown in Fig...

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