Explanation Given: The height of the shaded area is 5 in . The width of the shaded area is 5 in . Explanation: Show the shaded area as in Figure (1). From Figure 1, As the shaded area is symmetrical about the x- axis, write the product of inertia for the shaded area with respect to the x and y axes as 0 . I x y = 0 Compute the product of inertia for the shaded area with respect to the x -axis. I x = 1 12 2 a c 3 + 1 12 b ( 2 a ) 3 (I) Compute the product of inertia for the shaded area with respect to the y -axis. I y = 1 12 2 a b 3 + 2 a b ( a + b 2 ) 2 + 1 12 c ( 2 a ) 3 (II) Compute the moment of inertia of the shaded area with respect to the u -axis. I u = [ I x + I y 2 ] + [ I x − I y 2 ] cos 2 θ − I x y sin 2 θ (III) Compute the moment of inertia of the shaded area with respect to the v -axis. I v = [ I x + I y 2 ] − [ I x − I y 2 ] cos 2 θ + I x y sin 2 θ (IV) Conclusion: Substitute 0 .5 in . for a , 5 in . for c , and 4 in . for b in Equation (I). I x = 1 12 ( 2 ) ( 0.5 ) ( 5 ) 3 + 1 12 ( 4 ) [ 2 ( 0.5 ) ] 3 = 10.416 + 0.334 = 10.75 in . 4 Substitute 0 .5 in . for a , 5 in . for c , and 4 in . for b in Equation (II). I y = 1 12 2 ( 0.5 ) ( 4 ) 3 + 2 ( 0.5 ) ( 4 ) ( 0

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Author: Russell C. Hibbeler

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Inertia can be termed as a resistance offered by a body to change its state. A rigid body initially at rest will apply resistance when acted by an external force that tends to change its state of rest or velocity, or a body initially at motion will provi…

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Chapter 10.7, Problem 64P

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Engineering Mechanics: Statics

Ch. 10.3 - Determine the moment of inertia of the shaded area...Ch. 10.3 - Determine the moment of inertia of the shaded area...Ch. 10.3 - Determine the moment of inertia of the shaded area...Ch. 10.3 - Determine the moment of inertia of the shaded area...Ch. 10.3 - Prob. 1P Ch. 10.3 - Prob. 2P Ch. 10.3 - Prob. 3P Ch. 10.3 - Prob. 4P Ch. 10.3 - Prob. 5P Ch. 10.3 - Prob. 6P

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The moments of inertia of the shaded area with respect to the u and v axes.

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