Explanation Given: The total height of the cross-section area is 200 mm . The total width of the cross-section area is 160 mm . To determine the moment of inertia for the cross-sectional area about the x and y axis, subdivide the cross-section area into 3 segments as: Part 1: rectangle Part 2: rectangle Part 3: rectangle Show the free body diagram of the subdivided cross-section area as in Figure 1. From Figure (1), The width and height of rectangles 2 and 3 is same, thus the calculated value for both of the rectangles will also be the similar. Write the height and width of the rectangle in part 1 about the x axis. b 1 = 140 mm h 1 = 10 mm Here, the height of rectangle in part 1 is h 1 and width of rectangle in part 1 is b 1 . Write the height and width of the rectangle in part 1 about the y axis. b 1 = 10 mm h 1 = 140 mm Write the height and width of the rectangle in part 2 about the x axis. b 2 = 10 mm h 2 = 100 mm Here, the height of rectangle in part 2 is h 2 and width of rectangle in part 2 is b 2 . Write the height and width of the rectangle in part 2 about the y axis. b 2 = 100 mm h 2 = 10 mm Calculations: Calculate the moment of inertia for the cross section about the x axis. I x = ∑ ( I ¯ x ′ + A d y 2 ) = 1 12 b 1 ( h 1 ) 3 + 2 [ 1 12 b 2 ( h 2 ) 3 + b 2 h 2 { h 2 2 − b 2 2 } 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 the area is A . Calculate the moment of inertia for the cross section about the y axis. I y = ∑ ( I ¯ y ′ + A d x 2 ) = 1 12 b 1 ( h 1 ) 3 + 2 [ 1 12 b 2 ( h 2 ) 3 + b 2 h 2 { h 1 2 + h 2 2 } 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 the 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 for the cross section about the x and y axis. I x y = ∑ ( I ¯ x ′ y ′ + A d x d y ) = 0 + [ 0 + 10 ( 100 ) ( − 75 ) ( 45 ) ] + [ 0 + 10 ( 100 ) ( 75 ) ( − 45 ) ] = − 3 , 375 ( 10 3 ) + [ − 3 , 375 ( 10 3 ) ] = − 6.75 ( 10 6 ) mm 4 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 (III) Compute the orientation of the principal axes. tan 2 θ p 2 = − 2 I x y I x − I y (IV) Compute the orientation of the principal axes. θ p 1 = θ p 2 − 90 ° (V) Conclusion: Substitute 140 mm for b 1 , 10 mm for h 1 , 10 mm for b 2 , and 100 mm for h 2 in Equation (I). I x = 1 12 ( 140 ) ( 10 ) 3 + 2 [ 1 12 ( 10 ) ( 100 ) 3 + ( 10 ) ( 100 ) { 100 2 − 10 2 } 2 ] = 1 12 ( 140 ) ( 10 ) 3 + 2 [ 1 12 ( 10 ) ( 100 ) 3 + ( 10 ) ( 100 ) ( 45 ) 2 ] = 11 , 666

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 74P

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The orientation (θp1,θp2) of the principal axes having an origin at point C, and the principal moments of inertia (Imax,Imin) of the cross section about these axes.

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Chapter 10.7, Problem 73P

Chapter 10.7, Problem 75P

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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 orientation ( θ p 1 , θ p 2 ) of the principal axes having an origin at point C , and the principal moments of inertia ( I max , I min ) of the cross section about these axes.

Solution Summary: The author explains how to determine the moment of inertia for the cross-sectional area about the x and y axis.

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The orientation (θp1,θp2) of the principal axes having an origin at point C, and the principal moments of inertia (Imax,Imin) of the cross section about these axes.

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