Explanation Show the angle’s cross-sectional area, as in Figure (1). From Figure (1), Calculate the moment of inertia about the x axis. I x = [ 1 12 ( 20 ) ( 100 ) 3 + 100 ( 20 ) ( 50 − 32.22 ) 2 ] + [ 1 12 ( 80 ) ( 20 ) 3 + 80 ( 20 ) ( 32.22 − 10 ) 2 ] = ( 1 , 666 , 666.667 + 633680 ) + ( 53 , 333.33 + 788544 ) = 3 , 142 , 223.99 mm 4 ≈ 3.142 ( 10 6 ) mm 4 Calculate the moment of inertia about the y axis. I y = [ 1 12 ( 100 ) ( 20 ) 3 + 100 ( 20 ) ( 32.22 − 10 ) 2 ] + [ 1 12 ( 20 ) ( 80 ) 3 + 80 ( 20 ) ( 60 − 32.22 ) 2 ] = ( 66 , 666.667 + 987 , 456.8 ) + ( 853 , 333.334 + 1 , 236 , 544 ) = 3 , 142 , 000.801 mm 4 ≈ 3.142 ( 10 6 ) mm 4 Express the product of inertia about the x and y axes. I x y = ∑ x ¯ y ¯ A = − ( 32.2 − 10 ) ( 50 − 32.22 ) ( 100 ) ( 20 ) − ( 60 − 32

Engineering Mechanics: Statics and Study Pack (13th Edition)

13th Edition

ISBN: 9780133027990

Author: Russell C. Hibbeler

Publisher: Prentice Hall

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Question

Chapter 10.7, Problem 78P

To determine

The principal moments of inertia (Imax,Imin) for the angle’s cross-sectional area.

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Chapter 10.7, Problem 77P

Chapter 10.7, Problem 79P

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Chapter 10 Solutions

Engineering Mechanics: Statics and Study Pack (13th Edition)

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 principal moments of inertia ( I max , I min ) for the angle’s cross-sectional area.

The principal moments of inertia (Imax,Imin) for the angle’s cross-sectional area.

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Chapter 10 Solutions