Explanation Given: The total height of the cross-section area is 0 .45 m The total width of the cross-section area is 1 .1 m . Explanation: To determine the distance y ¯ and the moment of inertia I ¯ x ′ in the x ′ axis to subdivide the cross-sectional area into 3 segments such as: Part 1: rectangle Part 2: triangle Part 3: rectangle Show the subdivided cross-sectional area along with their respective centroids to determine the distance y ¯ as in Figure (1). Show the subdivided cross-sectional area along with their respective centroids to determine the moment of inertia in the x ′ axis as in Figure (2). From Figure (1), Write the height and width of the rectangle in part 1. b 1 = 0.3 m h 1 = 0.4 m Here, the height of the rectangle in part 1 is h 1 and the width of the rectangle in part 1 is b 1 . Write the height and width of the triangle in part 2. b 2 = 2 × 0.2 m = 0.4 m h 2 = 0.4 m Here, the height of the triangle in part 2 is h 2 and the width of the triangle in part 2 is b 2 . Write the height and the width of the rectangle in part 3. b 3 = 1.1 m h 3 = 0.05 m Here, the height of the rectangle in part 3 is h 3 and the width of the rectangle in part 3 is b 3 . Conclusion: Calculate the distance of the centroid in y axis for the rectangle in part 1. y ¯ 1 = 0.05 + h 1 2 (I) Substitute 0 .4 m for h 1 in Equation (I). y ¯ 1 = 0.05 + 0.4 2 = 0.25 m Calculate the distance of the centroid in y axis for the rectangle in part 2. y ¯ 2 = 0.05 + h 2 3 (II) Substitute 0 .4 m for h 2 in Equation (II). y ¯ 2 = 0.05 + 0.4 3 = 0.1833 m Calculate the distance of the centroid in y axis for the rectangle in part 3. y ¯ 3 = h 3 2 (III) Substitute 0 .05 m for h 3 in Equation (III). y ¯ 3 = 0.05 2 = 0.025 m Calculate the distance y ¯ to the centroid C of the beam's cross-sectional area. y ¯ = A 1 y ¯ 1 + A 2 y ¯ 2 + A 3 y ¯ 3 A 1 + A 2 + A 3 = ( b 1 h 1 ) y ¯ 1 + ( b 2 h 2 ) 2 y ¯ 2 + ( b 3 h 3 ) y ¯ 3 ( b 1 h 1 ) + ( b 2 h 2 ) 2 + ( b 3 h 3 ) (IV) Substitute 0 .3 m for b 1 , 0 .4 m for h 1 , 0 .4 m for b 2 , 0 .4 m for h 2 , 1 .1 m for b 3 , 0 .05 m for h 3 , 0 .25 m for y ¯ 1 , 0 .1833 m for y ¯ 2 , and 0 .025 m for y ¯ 3 in Equation (IV). y ¯ = ( 0.3 ) ( 0.4 ) ( 0.25 ) + ( 0.4 ) ( 0.4 ) 2 ( 0.1833 ) + ( 1.1 ) ( 0.05 ) ( 0.025 ) ( 0.3 ) ( 0.4 ) + ( 0.4 ) ( 0.4 ) 2 + ( 1.1 ) ( 0.05 ) = 0.03 + 0.014664 + 0.001375 0.12 + 0.08 + 0.055 = 0.046039 0.255 = 0.1806 m ≈ 0.181 m Hence, the centroid y ¯ of the cross-section is 0 .181 m _ . Calculate the distance of the centroid in x ′ and y axis of the rectangle in part 1

INTERNATIONAL EDITION---Engineering Mechanics: Statics, 14th edition (SI unit)

14th Edition

ISBN: 9780133918922

Author: Russell C. Hibbeler

Publisher: PEARSON

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Chapter 10.4, Problem 50P

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The centroid y¯ and the moment of inertia I¯x′ in the x′ axis.

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INTERNATIONAL EDITION---Engineering Mechanics: Statics, 14th edition (SI unit)

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The centroid y ¯ and the moment of inertia I ¯ x ′ in the x ′ axis.

Solution Summary: The author explains how to determine the distance and moment of inertia of the cross-section.

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The centroid y¯ and the moment of inertia I¯x′ in the x′ axis.

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