Explanation Calculation: Sketch the cross section of the given area as shown in Figure 1. Express the product of inertia as shown below: I ¯ x y = ( I x y ) 1 + ( I ¯ x y ) 2 Here, [ ( I x y ) 1 , ( I x y ) 2 ] is product of inertia of section 1 and 2. Applying parallel axis theorem, I x y = I x ′ y ′ + x ¯ y ¯ A When the x and y axis is symmetry. I x ′ y ′ = 0 Refer to Figure 1. Find the area of semicircle section 1 as shown below: A 1 = π 2 ( r 2 ) (1) Here, r is semicircle of section 1. Substitute 6 in . for r in Equation (1). A 1 = π 2 ( 6 2 ) = 18 π Find the area of rectangle section 2 as shown below: A 2 = π 2 ( r 2 ) (2) Here, r is semicircle of section 2. Substitute 6 in . for r in Equation (2). A 2 = π 2 ( 6 2 ) = 18 π Find the centroid ( x 1 ) as shown below: ( x 1 ) = − 6 2 = − 3 in . Find the centroid ( x 2 ) as shown below: ( x 2 ) = 6 2 = 3 in

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Author: BEER

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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 9.3, Problem 9.73P

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Find the product of inertia of the area with respect to x and y axes by using parallel axis theorem.

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Chapter 9.3, Problem 9.72P

Chapter 9.3, Problem 9.74P

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Determine the moments of inertia and the product of inertia of the L3 x 2 x 1/4-14 angle cross section of Prob. 9.74 with respect to new centroidal axes obtained by rotating the x and y axes 30° clockwise.(Reference to Problem 9.75):Using the parallel-axis theorem, determine the product of inertia of the area shown with respect to the centroidal x and y axes.

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Find the product of inertia of the area with respect to x and y axes by using parallel axis theorem.

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Find the product of inertia of the area with respect to x and y axes by using parallel axis theorem.

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