Explanation Given information: The centroid along y axis y ¯ = 120 mm . Explanation: Show the free-body diagram of the semi-circle as in Figure 1. Differential element: Refer to Figure (1). The value of x can be written as: x = 200 ( y ) 1 / 2 Write the equation to find the area of the differential element parallel to the x -axis as follows: d A = 2 x d y (I) Substitute 200 ( y ) 1 / 2 for x in Equation (I). d A = 2 200 ( y ) 1 / 2 d y Determine the area moment of inertia of the area about the x -axis using the relation: I x = ∫ A y 2 d A (II) Substitute 2 200 ( y ) 1 / 2 d y for d A in Equation (II). I x = ∫ 0 200 mm y 2 × 2 200 ( y ) 1 / 2 d y Integrate the equation and apply the limits. I x = 2 200 ∫ 0 200 mm ( y ) 5 / 2 d y = 2 200 ( 2 7 y 7 / 2 ) | 0 200 mm = 2 200 ( 2 7 ( 200 ) 7 / 2 − 0 ) = 914 , 285 , 714.3 mm 4 Thus, the area moment of inertia of the area about the x -axis is I x = 914 , 285 , 714

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Author: HIBBELER

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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 10.8, Problem 4RP

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The area moment of inertia of the area about the x axis and then by using the parallel-axis theorem, find the moment of inertia about the x′ axis that passes through the centroid C of the area.

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The area moment of inertia of the area about the x axis and then by using the parallel-axis theorem, find the moment of inertia about the x ′ axis that passes through the centroid C of the area.

Solution Summary: The author shows the free-body diagram of the semi-circle as in Figure 1. The area moment of inertia is determined using the parallel-axis theorem.

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The area moment of inertia of the area about the x axis and then by using the parallel-axis theorem, find the moment of inertia about the x′ axis that passes through the centroid C of the area.

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