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: 59



Work Step by Step

We can find the product of the moment of inertia with respect to the x and y axes as follows: $dA=ydx$ $dx=(a^{\frac{1}{2}}-x^{\frac{1}{2}})^2$ As $x^{\prime}=x$ $\implies y^{\prime}=\frac{y}{2}$ $\implies y^{\prime}=\frac{1}{2}(a^{\frac{1}{2}}-x^{\frac{1}{2}})^2$ Now $dI_{XY}=dI_{xy}+dAx{\prime}y^{\prime}$ $\implies dI_{XY}=0+((a^{\frac{1}{2}}-x^{\frac{1}{2}})^2dx) x(\frac{1}{2})(a^{\frac{1}{2}}-x^{\frac{1}{2}})^2$ $\implies dI_{XY}=\frac{1}{2}(x^3+a^2 x+6ax^2-4a^{\frac{3}{2}}x^{\frac{3}{2}}-4a^{\frac{1}{2}}x^{\frac{5}{2}})$dx We know that $I_{xy}=\int dI_{xy}$ $\implies I_{xy}=\frac{1}{2}\int_0^a (x^3+a^2 x+6ax^2-4a^{\frac{3}{2}}x^{\frac{3}{2}}-4a^{\frac{1}{2}}x^{\frac{5}{2}})dx$ This simplifies to: $I_{xy}=\frac{a^4}{280}$
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