Explanation Show the free body diagram of the beam’s cross-sectional area, as in Figure (1). From Figure (1), Calculate the moment of inertia about the x axis. I x = 2 [ 1 12 a t 3 + a t { b − t 2 } 2 ] + 1 12 t ( 2 b − 2 t ) 3 (I) Calculate the moment of inertia about the y axis. I y = 2 [ 1 12 t ( a − t ) 3 + t ( a − t ) { a − t 2 + t 2 } 2 ] + 1 12 2 b t 3 (II) Express the product of inertia about the x and y axis. I x y = − 2 [ a − t 2 + t 2 ] [ b − t 2 ] t ( a − t ) (III) Compute the principal moments of inertia. I max / min = I x + I y 2 ± [ ( I x − I y ) / 2 ] 2 + ( I x y ) 2 (IV) Conclusion: Substitute 4 in . for a , 4 in . for b and 3 8 in . for t in Equation (I). I x = 2 [ 1 12 ( 4 ) [ 3 8 ] 3 + ( 4 ) [ 3 8 ] { 4 − 3 8 × 2 } 2 ] + 1 12 [ 3 8 ] ( 2 × 4 − 2 × 3 8 ) 3 = 2 [ 1 3 ( 0.375 ) 3 + ( 4 ) ( 0.375 ) ( 3.8125 ) 2 ] + 1 12 ( 0.375 ) ( 7.25 ) 3 = 55.55 in . 4 Substitute 4 in . for a , 4 in . for b and 3 8 in . for t in Equation (II)

13th Edition

ISBN: 9780132915540

Author: Russell C. Hibbeler

Publisher: Prentice Hall

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Chapter 10.7, Problem 74P

To determine

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

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Chapter 10.7, Problem 73P

Chapter 10.7, Problem 75P

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Chapter 10 Solutions

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

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

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Chapter 10 Solutions