Explanation Calculation: Sketch the cross section 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 The moment of inertia is 0, when 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 139.3 mm for b 1 and 12.7 mm for h 1 in Equation (1). A 1 = 139.3 × 12.7 = 1 , 769.11 mm 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 12.7 mm for b 2 and 102 mm for h 2 in Equation (2). A 2 = 12.7 × 102 = 1 , 295.4 mm 2 Find the centroid ( x 1 ) as shown below: ( x 1 ) = ( 152 2 ) − ( 12.7 ) = 63.3 2 = 32.0 mm . Find the centroid ( x 2 ) as shown below: ( x 2 ) = − ( 37

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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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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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