819 Inverted T-section | Moment of Inertia

$\bar{I}_1 = \dfrac{8(2^3)}{12} = \dfrac{16}{3} \, \text{ in.}^4$

$y_1 = 1 \, \text{ in.}$
 

$A_2 = 8(2) = 16 \, \text{ in.}^2$

$\bar{I}_2 = \dfrac{2(8^3)}{12} = \dfrac{256}{3} \, \text{ in.}^4$

$y_2 = 2 + 4 = 6 \, \text{ in.}$
 

$A = A_1 + A_2 = 16 + 16$

$A = 32 \, \text{ in.}^2$
 

$A\bar{y} = A_1y_1 + A_2y_2$

$32\bar{y} = 16(1) + 16(6)$

$\bar{y} = 3.5 \, \text{ in.}$
 

$\bar{I} = [ \, \bar{I}_1 + A_1(\bar{y} - y_1)^2 \, ] + [ \, \bar{I}_2 + A_2(y_2 - \bar{y})^2 \, ]$

$\bar{I} = [ \, \frac{16}{3} + 16(3.5 - 1)^2 \, ] + [ \, \frac{256}{3} + 16(6 - 3.5)^2 \, ]$

$\bar{I} = 290.67 \, \text{ in.}^4$           answer
 

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