Explanation Given: The width of the beam’s cross-sectional area is 20 in . The height of the beam’s cross-sectional area is 16 in . The angle is 45 ° . Show the beam’s cross-sectional area as in Figure (1). From Figure 1, As the beam’s cross-sectional area is symmetrical about the x and y axes, write the product of inertia for the beam’s area with respect to the x and y axes as 0 . I x y = 0 Compute the moment of inertia for the beam’s cross sectional area with respect to the x axis. I x = 2 3 ( a + b ) c 3 + 1 12 2 b d 3 + 2 b d ( d / 2 ) 2 (I) Compute the moment of inertia for the beam’s cross sectional area with respect to the y axis. I y = 1 12 [ 2 ( a + b ) ] 3 c + 1 12 ( 2 b ) 3 d (II) Compute the moment of inertia for the channel’s cross-sectional 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 for the channel’s cross-sectional 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) Compute the moment of inertia for the beam’s cross sectional area with respect to the u and v axes. I u v = [ I x − I y 2 ] sin 2 θ + I x y cos 2 θ (V) Conclusion: Substitute 8 in . for a , 2 in . for b , 2 in . for c , and 16 in . for d in Equation (I). I x = 2 3 ( 8 + 2 ) ( 2 ) 3 + 1 12 2 ( 2 ) ( 16 ) 3 + 2 ( 2 ) ( 16 ) ( 16 / 2 ) 2 = 53.334 + 1 , 365.334 + 4 , 096 = 5 , 514.668 in . 4 ≈ 5.515 × 10 3 in . 4 Substitute 8 in . for a , 2 in . for b , 2 in . for c , and 16 in . for d in Equation (II). I y = 1 12 [ 2 ( 8 + 2 ) ] 3 ( 2 ) + 1 12 [ 2 ( 2 ) ] 3 ( 16 ) = 1 , 333.34 + 85.334 = 1 , 418.674 in . 4 ≈ 1.419 × 10 3 in . 4 Substitute 5.515 × 10 3 in . 4 for I x , 1.419 × 10 3 in . 4 for I y , 0 for I x y , and 45 ° for θ in Equation (III). I u = { [ ( 5.515 × 10 3 ) + ( 1.419 × 10 3 ) 2 ] + [ ( 5

Engineering Mechanics: Statics & Dynamics (14th Edition)

14th Edition

ISBN: 9780133915426

Author: Russell C. Hibbeler

Publisher: PEARSON

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Moment of Inertia

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 69P

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The moments of inertia Iu,Iv and the product of inertia Iuv for the beam’s cross-sectional area.

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Chapter 10.7, Problem 68P

Chapter 10.7, Problem 70P

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Engineering Mechanics: Statics & Dynamics (14th Edition)

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The moments of inertia I u , I v and the product of inertia I u v for the beam’s cross-sectional area.

Solution Summary: The author shows the beam's cross-sectional area as in Figure 1. The moment of inertia is computed with respect to the x axis.

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The moments of inertia Iu,Iv and the product of inertia Iuv for the beam’s cross-sectional area.

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Chapter 10 Solutions