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

Published by Pearson

# Chapter 10 - Moments of Inertia - Section 10.3 - Radius of Gyration of an Area - Problems - Page 539: 18

#### Answer

$I_y=\frac{b^3h}{6}$

#### Work Step by Step

We can find the required moment of inertia as follows: $dA=ydx$ Given that $y=\frac{h}{b^3}x^3$ $\implies dA=\frac{h}{b^3}x^3 dx$ Now $I_y=\int x^2 dA$ $\implies I_y=\int_0^b x^2 \frac{h}{b^3}x^3 dx$ $\implies I_y=\int_0^b \frac{h}{b^3}x^5dx$ $\implies I_y=\frac{h}{b^3}\frac{x^6}{6}|_0^b$ $\implies I_y=\frac{b^3h}{6}$

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