Centroid of Composite Area: Rectangle and Two Triangles

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Centroid of Composite Area: Rectangle and Two Triangles

I want the solution to this question img_20170307_084352.jpgplease

Romel's picture
$A_1 = 126(54) = 6,804 ~

$A_1 = 126(54) = 6,804 ~ \text{mm}^2$

$x_1 = \frac{1}{2}(126) - 54 = 9 ~ \text{mm}$

$y_1 = \frac{1}{2}(54) = 27 ~ \text{mm}$
 

$A_2 = \frac{1}{2}(126)(30) = 1,890 ~ \text{mm}^2$

$x_2 = \frac{2}{3}(126) - 54 = 30 ~ \text{mm}$

$y_2 = 54 + \frac{1}{3}(30) = 64 ~ \text{mm}$
 

$A_3 = \frac{1}{2}(72)(48) = 1,728~ \text{mm}^2$

$x_3 = \frac{2}{3}(72) = 48 ~ \text{mm}$

$y_3 = -\frac{1}{3}(48) = -16 ~ \text{mm}$
 

mech_009-centroid-composite-areas.gif

 

$A = \Sigma A_n = 6,804 + 1,890 + 1,728$

$A = 10,422 ~ \text{mm}^2$
 

$\bar{x} = \dfrac{\Sigma A_n x_n}{A} = \dfrac{6804(9) + 1890(30) + 1728(48)}{10\,422}$

$\bar{y} = \dfrac{\Sigma A_n y_n}{A} = \dfrac{6804(27) + 1890(64) - 1728(16)}{10\,422}$
 

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