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 10.4 - Moments of Inertia for Composite Areas - Problems - Page 544: 25



Work Step by Step

We can find the required moment of inertia as follows: The area of the first rectangle is given as $A_1=\frac{1}{2}hb$ $\implies A_1=\frac{1}{2}(3)(3)=4.5in^2$ The area of the second rectangle is $A_2=\frac{1}{2}hb$ $A_2=\frac{1}{2}(6)(6)=18in^2$ The area of the third rectangle is $A_3=hb$ $\implies A_3=9in^2$ Now $I=\Sigma (I+Ady^2)$ We plug in the known values to obtian: $I=\frac{3\cdot 3^3}{36}+4.5(4)^2+(\frac{6\times 6^3}{36}+18\times (2)^2)+(\frac{3\times(3)^3}{12}+9(15)^2)$ This simplifies to: $I=209in^4$
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