Explanation Calculation: Sketch the cross section 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 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 = ( b 1 h 1 ) (1) Here, b 1 is width of the section 1 and h 1 is height of the section 1. Substitute 3 in . for b 1 and 0.25 in . for h 1 in Equation (1). A 1 = 3 × 0.25 = 0.75 in 2 Find the area of rectangle section 2 as shown below: A 2 = ( b 2 h 2 ) (2) Here, b 2 is width of the section 2 and h 2 is height of the section 2. Substitute 0.25 in . for b 2 and 1.75 in . for h 2 in Equation (2). A 2 = 0.25 × 1.75 = 0.4375 in 2 Find the centroid ( x 1 ) as shown below: ( x 1 ) = − ( 1.5 − 0.980 ) = − 0.520 in . Find the centroid ( x 2 ) as shown below: ( x 2 ) = 0.980 − 0.125 = 0.855 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 9.74P

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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 9.73P

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

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