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: 55



Work Step by Step

We can find the product of the inertia of the shaded area with respect to the x and y axes as follows: $I_{xy}=\int_0^3 xy dA$ $\implies I_{xy}=\int_0^3 (x\times \frac{y}{2})(y dx)$ But $y=\frac{1}{9}x^3$ Thus, the above equation becomes $I_{xy}=\frac{1}{2}\int_0^3 x(\frac{1}{9}x^3)^2 dx$ $\implies I_{xy}=\frac{1}{162}\int_0^3 x^7 dx$ $\implies I_{xy}=\frac{1}{162}\frac{x^8}{8}|_0^3$ After applying the limits, we obtain: $I_{xy}=5.06in^4$
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