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 x ¯ and the moment of inertia I ¯ y ′ about the y ′ 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 width of the rectangle in part 1: b 1 = 3 0 mm h 1 = 17 0 mm Here, the height of rectangle in part 1 is h 1 and the width of rectangle in part 1 is b 1 . Write the height and the width of the rectangle in part 2: b 2 = 17 0 mm h 2 = 3 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 = 3 0 mm h 3 = 10 0 mm Here, the height of rectangle in part 3 is h 3 and the width of the rectangle in part 3 is b 3 . Centroid calculation: Calculate the distance between centroid and x axis for the rectangle in part 1: x ¯ 1 = h 1 2 + h 2 (I) Substitute 30 mm for h 2 and 170 mm for h 1 in Equation (I). x ¯ 1 = 170 2 + 30 = 85 + 30 = 115 mm Calculate the distance between centroid and x axis for the rectangle in part 2: x ¯ 2 = h 2 2 (II) Substitute 30 mm for h 2 in Equation (II). x ¯ 2 = 30 2 = 15 mm Calculate the distance between centroid and x axis for rectangle in part 3: x ¯ 3 = h 3 2 (III) Substitute 100 mm for h 3 in Equation (III). x ¯ 3 = 100 2 = 50 mm Conclusion: Calculate the distance x ¯ to the centroid C of the beam's cross-sectional area: x ¯ = A 1 x ¯ 1 + A 2 x ¯ 2 + A 3 x ¯ 3 A 1 + A 2 + A 3 = ( b 1 h 1 ) x ¯ 1 + ( b 2 h 2 ) x ¯ 2 + ( b 3 h 3 ) x ¯ 3 ( b 1 h 1 ) + ( b 2 h 2 ) + ( b 3 h 3 ) (IV) Substitute 30 mm for b 1 , 170 mm for h 1 , 170 mm for b 2 , 30 mm for h 2 , 30 mm for b 3 , 100 mm for h 3 , 115 mm for x ¯ 1 , 15 mm for x ¯ 2 , and 50 mm for x ¯ 3 in Equation (IV). x ¯ = ( 30 ) ( 170 ) ( 115 ) + ( 170 ) ( 30 ) ( 15 ) + ( 100 ) ( 30 ) ( 50 ) ( 30 ) ( 170 ) + ( 170 ) ( 30 ) + ( 100 ) ( 30 ) = 586 , 500 + 76 , 500 + 150 , 000 5 , 100 + 5 , 100 + 3 , 000 = 813 , 000 13 , 200 = 61.6 mm Hence, the distance x ¯ to the centroid C of the beam's cross-sectional area is 61 .6 mm _ . Moment of inertia calculation: Calculate the distance between centroid and y ′ axis about the x axis for rectangle in part 1: d x 1 = h 1 2 + h 2 − x ¯ (V) Substitute 30 mm for h 2 , 170 mm for h 1 , and 61

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Chapter 10, Problem 45P

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

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Chapter 10, Problem 44P

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

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