Explanation Given: The total height of the beam's cross-sectional area is 4 in . The total width of the beam's cross-sectional area is 8 in . To calculate the distance y ¯ to the centroid of the beam's cross-sectional area, 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: To calculate the moment of inertia about the centroid 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 2: From Figure (1), Write the height and the width of the rectangle in part 1: b 1 = 1 in . h 1 = 6 in . 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 = 4 in . h 2 = 1 in . Here, the height of 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 = 4 in . h 3 = 1 in . Here, the height of the rectangle in part 3 is h 3 and the width of the rectangle in part 3 is b 3 . Centroid calculations: Calculate the distance between centroid and y axis about the x axis for rectangle in part 1: y ¯ 1 = b 1 2 (I) Substitute 1 in . for b 1 in Equation (I). y ¯ 1 = 1 2 = 0.5 in . Calculate the distance between centroid and y axis about the axis for rectangle in part 2: y ¯ 2 = b 2 2 (II) Substitute 4 in . for b 2 in Equation (II). y ¯ 2 = 4 2 = 2 in . Calculate the distance between centroid and y axis about the x axis for rectangle in part 3: y ¯ 3 = b 3 2 (III) Substitute 4 in . for b 3 in Equation (III). y ¯ 3 = 4 2 = 2 in . Conclusion: Calculate the location y ¯ . y ¯ = ∑ y ¯ A ∑ A = y ¯ 1 A 1 + y ¯ 2 A 2 + y ¯ 3 A 3 A 1 + A 2 + A 3 = y ¯ 1 [ ( b 1 ) ( h 1 ) ] + y ¯ 2 [ ( b 2 ) ( h 2 ) ] + y ¯ 3 [ ( b 3 ) ( h 3 ) ] ( b 1 ) ( h 1 ) + ( b 2 ) ( h 2 ) + ( b 3 ) ( h 3 ) (IV) Here, the area is A , the area of rectangle 1 is A 1 , the area of rectangle 2 is A 2 , and the area of rectangle 3 is A 3 . Substitute 0 .5 in . for y ¯ 1 , 2 in . for y ¯ 2 and y ¯ 3 respectively, 1 in . for b 1 , 6 in . for h 1 , 4 in . for b 2 , 4 in . for b 3 , 1 in . for h 2 , and 1 in . for h 3 in Equation (4). y ¯ = ( 0.5 ) [ ( 1 ) ( 6 ) ] + ( 2 ) [ ( 4 ) ( 1 ) ] + ( 2 ) [ ( 4 ) ( 1 ) ] ( 1 ) ( 6 ) + ( 4 ) ( 1 ) + ( 4 ) ( 1 ) = 3 + 8 + 8 6 + 4 + 4 = 19 in . 3 14 in . 2 = 1.357 in . ≈ 1.36 in . Hence, the distance y ¯ to the centroid of the beam's cross-sectional area is 1 .36 in . _ . From Figure (2), Write the height and width of the rectangle in part 1 about the x ′ axis

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ISBN: 9780135681879

Author: HIBBELER

Publisher: PEARSON

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Moment of Inertia

Inertia can be termed as a resistance offered by a body to change its state. A rigid body initially at rest will apply resistance when acted by an external force that tends to change its state of rest or velocity, or a body initially at motion will provi…

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Chapter 10.4, Problem 40P

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

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Chapter 10.4, Problem 39P

Chapter 10.4, Problem 41P

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

Solution Summary: The author explains how to calculate 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 about the centroid x′ axis.

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