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 5662: 78


$\theta=6.08^{\circ}$ $I_{max}=1.74\times 10^3in^4$ and $I_{min}=435in^4$

Work Step by Step

We can find the orientation of the principal axes and the required principle moments of inertia as follows: $tan2\theta=\frac{-2I}{I_x-I_y}$ $\implies tan2\theta=\frac{-2(188)}{450-1730}$ $\implies \theta=6.08^{\circ}$ Now $I_{max/min}=\frac{I_x+I_y}{2}\pm \sqrt{(\frac{I_x-I_y}{2})^2+I_{xy}^2}$ We plug in the known values to obtain: $I_{max/min}=\frac{450+1730}{2}\pm \sqrt{(\frac{450-1730}{2})^2++(138)^2}$ This simplifies to: $I_{max}=1.74\times 10^3in^4$ and $I_{min}=435in^4$
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