## Engineering Mechanics: Statics & Dynamics (14th Edition)

Published by Pearson

# Chapter 10 - Moments of Inertia - Section 10.4 - Moments of Inertia for Composite Areas - Problems - Page 546: 39

#### Answer

$2376~in^4$

#### Work Step by Step

We can determine the required moment of inertia as follows: $I_{triangle}=\frac{bh^3}{36}$ and $I_{rectangle}=\frac{bh^3}{12}$ Now the area for each part in the given figure can be calculated as $A_1=\frac{bh}{2}=\frac{3(6)}{2}=9in^2$ $A_2=hb$ $A_2=(6)(6)=36in^2$ and $A_3=\frac{hb}{2}$ $A_3=\frac{(9)(6)}{2}=27in^2$ The required moment of inertia is given as $I_{sum}=\Sigma (I+Ad^2_x)$ We plug in the known values to obtian: $I_{sum}=(\frac{bh^3}{36}+(9)(7))+(\frac{bh^3}{12}+36(3)^2)+(\frac{bh^3}{36}+27(6)^2)$ This simplifies to: $I=2376in^4$

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