**Problem 819**

Determine the moment of inertia of the T-section shown in Fig. P-819 with respect to its centroidal X_{o} axis.

**Solution 819**

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

$y_1 = 1 \, \text{ in.}$

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

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

$y_2 = 2 + 4 = 6 \, \text{ in.}$

$A = A_1 + A_2 = 16 + 16$

$A = 32 \, \text{ in.}^2$

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

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

$\bar{y} = 3.5 \, \text{ in.}$

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

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

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

**Another Solution**

$I_x = \frac{1}{3}(8)(10^3) - 2[ \, \frac{1}{3}(3)(10^3) \, ] + 2[ \, \frac{1}{3}(3)(2^3) \, ]$

$I_x = \dfrac{2048}{3} \, \text{ in.}^4$

$I_x = \bar{I} + Ad^2$

$\frac{2048}{3} = \bar{I} + 32(3.5^2)$

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