E$4\mathrm{.}9$ Example $4\mathrm{.}9$ Locate the centroids of the profiles as shown in Fig. $4\mathrm{.}16.$ The measurements are given in $mm.$

Solution a$)$ The $y-$axis coincides with the axis of symmetry. Therefore, $x_c=0.$ In order to calculate $y_c,$ we consider the area to be

composed of three rectangles with known centroids. Then we obtain

$y_c{?}_{{?}_{?}}=\frac{{\displaystyle \underset{?}{\overset{?}{}}}{y}_i{A}_i}{{\displaystyle \underset{?}{\overset{?}{}}}{A}_i}=\frac{2(4*45)+14(5*20)+28(8*15)}{4*45+5*20+8*15}=\frac{5120}{400}$

$=12\mathrm{.}8(mm{)}_{{?}_{?}}.$

b$)$ We again consider the area to be composed of three rectangles:

$x_c{?}_{{?}_{?}}{?}_{?}=\frac{22\mathrm{.}5(4*45)+2\mathrm{.}5(5*20)+7\mathrm{.}5(8*15)}{4*45+5*20+8*15}=\frac{5200}{400}=m{m}_{{?}_{?}}{?}_{,}$

$y_c{?}_{{?}_{?}}={?}_{?}=\frac{2(4*45)+14(5*20)+28(8*15)}{400}=12\mathrm{.}8(mm{)}_{.}{?}_{?}$

Note that a displacement of the areas in the $x-$direction does not change the $y-$coordinate of the centroid.