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 10.3 - Radius of Gyration of an Area - Problems - Page 538: 12



Work Step by Step

We can find the required moment of inertia as follows: $dA=(8-y)dx$ $\implies dA=(8-\frac{1}{8}x^3)dx$ Now $I_y=\int_0^4 x^2 dA$ $\implies I_y=\int_0^4 x^2 (8-\frac{1}{8}x^3)dx$ $\implies I_y=8\int_0^4x^2 dx-\frac{1}{8}\int_0^4 x^5 dx$ $\implies I_y=8\frac{x^3}{3}|_0^4-\frac{1}{8}\frac{x^6}{6}|_0^4$ This simplifies to: $I_y=85.3m^4$
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