Explanation Given: The width of the triangular area is 9 in . The height of the triangular area is 6 in . Show the free body diagram of the triangular area as in Figure 1. From the Figure 1, Compute the moment of inertia of the triangular area about the x axes using the parallel-axis theorem. I x = I ¯ x ′ + A d y 2 = 1 36 b a 3 + [ 1 2 b a ] [ 1 3 a ] 2 (I) Here, the moment of inertia for the beam’s cross-sectional area about the x axis is I x , the first integral of the moment of inertia of the area about the centroidal axis in x direction is I ¯ x ′ , half of the distance from the centroid is d y , and an area is A . Compute the moment of inertia of the triangular area about the y axes using the parallel-axis theorem. I y = I ¯ y ′ + A d x 2 = 1 36 a b 3 + [ 1 2 a b ] [ 2 3 b ] 2 (II) Here, the moment of inertia for the beam’s cross-sectional area about the y axis is I y , half of the distance from the centroid is d x , and first integral of the moment of inertia of the area about the centroidal axis in y direction is I ¯ y ′ . Calculate the moment of inertia of the triangular area about the x and y axes. I x y = b 2 a 2 8 (III) Calculate the maximum and minimum principal of moments. I max / min = I x + I y 2 ± [ ( I x − I y ) / 2 ] 2 + ( I x y ) 2 (IV) Compute the orientation of the principal axes. tan 2 θ p 2 = − 2 I x y I x − I y (V) Compute the orientation of the principal axes. θ p 1 = θ p 2 − 90 ° (VI) Conclusion: Substitute 6 in . for a and 9 in . for b in Equation (I). I x = 1 36 ( 9 ) ( 6 ) 3 + [ 1 2 ( 9 ) ( 6 ) ] [ 1 3 ( 6 ) ] 2 = 54 + 108 = 162 in . 4 Substitute 6 in . for a and 9 in . for b in Equation (II). I y = 1 36 ( 6 ) ( 9 ) 3 + [ 1 2 ( 6 ) ( 9 ) ] [ 2 3 ( 9 ) ] 2 = 121

Engineering Mechanics: Statics Plus Mastering Engineering with Pearson eText -- Access Card Package (14th Edition) (Hibbeler, The Engineering Mechanics: Statics & Dynamics Series, 14th Edition)

14th Edition

ISBN: 9780134160689

Author: Russell C. Hibbeler

Publisher: PEARSON

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Chapter 10.7, Problem 72P

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The direction (θp1,θp2) of the principal axes having an origin at point O, and the principal moments of inertia (Imax,Imin) for the triangular area about the axes.

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Chapter 10.7, Problem 71P

Chapter 10.7, Problem 73P

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Engineering Mechanics: Statics Plus Mastering Engineering with Pearson eText -- Access Card Package (14th Edition) (Hibbeler, The Engineering Mechanics: Statics & Dynamics Series, 14th Edition)

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The direction ( θ p 1 , θ p 2 ) of the principal axes having an origin at point O , and the principal moments of inertia ( I max , I min ) for the triangular area about the axes.

Solution Summary: The author explains the moment of inertia of the triangular area about the x axes using the parallel-axis theorem.

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The direction (θp1,θp2) of the principal axes having an origin at point O, and the principal moments of inertia (Imax,Imin) for the triangular area about the axes.

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