Explanation Given: The total height of the beam's cross-sectional area is 200 mm . The total width of the beam's cross-sectional area is 200 mm . To determine the distance y ¯ and the moment of inertia I ¯ x ′ about the x ′ axis, subdivide the cross-sectional area into 3 segments as follows: Part 1: Rectangle Part 2: Rectangle Part 3: Rectangle Show the subdivided beam's cross-sectional area along with their respective centroids as in Figure 1: From Figure (1), Write the height and the width of the rectangle in part 1: b 1 = 17 0 mm h 1 = 3 0 mm Here, 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 2. b 2 = 3 0 mm h 2 = 17 0 mm Here, the height of the rectangle in part 2 is h 2 and the width of the rectangle in part 2 is b 2 . Write the height and width of the rectangle in part 3: b 3 = 10 0 mm h 3 = 3 0 mm Here, the height of rectangle in part 3 is h 3 and the width of rectangle in part 3 is b 3 : Centroid calculation: Calculate the distance between the centroid and y axis for rectangle in part 1: y ¯ 1 = h 1 2 (I) Substitute 30 mm for h 1 in Equation (I). y ¯ 1 = 30 2 = 15 mm Calculate the distance between centroid and y axis for the rectangle in part 2: y ¯ 2 = h 2 2 (II) Substitute 170 mm for h 2 in Equation (II). y ¯ 2 = 170 2 = 85 mm Calculate the distance between centroid and y axis for the rectangle in part 3: y ¯ 3 = h 2 + h 3 2 (III) Substitute 170 mm for h 2 , and 30 mm for h 3 in Equation (III). y ¯ 3 = 170 + 30 2 = 185 mm Conclusion: 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 ) y ¯ 2 + ( b 3 h 3 ) y ¯ 3 ( b 1 h 1 ) + ( b 2 h 2 ) + ( b 3 h 3 ) (IV) Substitute 170 mm for b 1 , 30 mm for h 1 , 30 mm for b 2 , 170 mm for h 2 , 100 mm for b 3 , 30 mm for h 3 , 15 mm for y ¯ 1 , 85 mm for y ¯ 2 , and 185 mm for y ¯ 3 in Equation (IV). y ¯ = ( 170 ) ( 30 ) ( 15 ) + ( 170 ) ( 30 ) ( 85 ) + ( 100 ) ( 30 ) ( 185 ) ( 170 ) ( 30 ) + ( 170 ) ( 30 ) + ( 100 ) ( 30 ) = 76 , 500 + 433 , 500 + 555 , 000 5 , 100 + 5 , 100 + 3 , 000 = 106 , 5000 13 , 200 = 80.68 mm ≈ 80.7 mm Hence, the distance y ¯ to the centroid C of the beam's cross-sectional area is 80 .7 mm _ . Moment of inertia calculation: Calculate the distance between centroid and x ′ axis about y axis for rectangle in part 1. d y 1 = y ¯ − h 1 / 2 (V) Substitute 80

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Chapter 10, Problem 44P

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The distance y¯ and the moment of inertia I¯x′ about the x′ axis.

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Chapter 10, Problem 43P

Chapter 10, Problem 45P

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The distance y ¯ and the moment of inertia I ¯ x ′ about the x ′ axis.

Solution Summary: The author explains how to determine the distance and moment of inertia of the beam's cross-sectional area.

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The distance y¯ and the moment of inertia I¯x′ about the x′ axis.

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