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: 82


$I_{max}=309in^4$ $I_{min}=42.1in^4$ $\theta_1=-31.4^{\circ}$ $\theta_2=58.6^{\circ}$

Work Step by Step

We can determine the directions of the principal axes and the principal moments of inertia as follows: $I_x=\Sigma (I+Ad^2_y)$ $\implies I_x=\frac{4(6)^3}{12}+4(6)(3)^2-\frac{\pi (1)^4}{4}-(1)^2\pi(4)^2$ $\implies I_x=236.95in^4$ The moment of inertia about the y-axis is given as $I_y=\Sigma (I+Ad^2_x)$ $\implies I_y=\frac{(6)(4)^3}{12}+(4)(6)(2)^2-\frac{\pi(1)^4}{4}-(1)^2\pi (2)^2=114.65in^4$ The product of inertia about the x and y axes is given as $I_{xy}=\Sigma (I+Ad_x d_y)$ $\implies I_{xy}=0+4(6)(3)(2)-0-(1)^2\pi (4)(2)=118.87in^4$ Now the minimum and maximum moment of inertia can be calculated as $I_{max}=\frac{I_x+I_y}{2}+\sqrt{(\frac{I_x-I_y}{2})^2+I^2_{xy}}$ We plug in the known values to obtain: $I_{max}=\frac{236.95+114.65}{2}-\sqrt{(\frac{236.95-114.95}{2})^2+(118.87)^2}$ $\implies I_{max}=309in^4$ and the minimum moment of inertia is given as $I_{min}=\frac{I_x+I_y}{2}-\sqrt{(\frac{I_x-I_y}{2})^2+I^2_{xy}}$ We plug in the known values to obtain: $I_{min}=\frac{236.95+114.65}{2}-\sqrt{(\frac{236.95-114.95}{2})^2+(118.87)^2}$ $\implies I_{max}=42.1in^4$
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