Engineering Mechanics: Statics & Dynamics (14th Edition)

Published by Pearson
ISBN 10: 0133915425
ISBN 13: 978-0-13391-542-6

Chapter 10 - Moments of Inertia - Section 10.7 - Mohr's Circle for M oments of Inertia - Problems - Page 558: 58


$ I_{xy}=\frac{a^2b^2}{6}$

Work Step by Step

We can determine the required product of inertia for the shaded area with respect to x and y axes as follows: We know that $dI_{xy}=0+ydx\cdot x\frac{y}{2}=\frac{xy^2}{2}dx$ Now $I_{xy}=\int_0^a \frac{x^{(\frac{bx^{1/2}}{a^{1/2}})^2}}{2}dx$ This simplifies to: $\implies I_{xy}=\frac{a^2b^2}{6}$
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