Explanation Calculation: Divide the circular section into number of identical circular sectors. The sections are approximated by isosceles triangles. Large number of sectors approximate dimension of one of the isosceles triangle is shown below. Sketch the circular sector as shown in Figure 1. Refer to problem 9.28. The polar moment of inertia ( J O ) is b h 48 ( 12 h 2 + b 2 ) . Sketch the triangle section as shown in Figure 2. Refer to Figure 2. b = r Δ θ And h = r Find the polar moment of inertia sector as shown below: ( Δ J O ) sector ≃ 1 48 ( b ) h [ J O ] (1) Here, b is the width of section, J O is polar moment of inertia, and h is height of section. Substitute r Δ θ for b , r for h , and b h 48 ( 12 h 2 + b 2 ) for J O in Equation (1). ( Δ J O ) sector ≃ 1 48 ( r Δ θ ) r [ b h 48 ( 12 h 2 + b 2 ) ] = 1 48 r 4 Δ θ [ ( 12 + Δ θ 2 ) ] (2) Differentiate Equation (2) with respect to θ . d J O d θ = lim Δ θ → 0 ( Δ J O Δ θ ) (3) Substitute 1 48 r 4 Δ θ [ ( 12 + Δ θ 2 ) ] for ( Δ J O ) sector in Equation (3)

Vector Mechanics for Engineers: Statics and Dynamics

11th Edition

ISBN: 9780073398242

Author: Ferdinand P. Beer, E. Russell Johnston Jr., David Mazurek, Phillip J. Cornwell, Brian Self

Publisher: McGraw-Hill Education

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Chapter 9.1, Problem 9.29P

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Prove the centroidal polar moment of inertia of a circular area of radius r is πr42.

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Chapter 9.1, Problem 9.28P

Chapter 9.1, Problem 9.30P

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Prove the centroidal polar moment of inertia of a circular area of radius r is π r 4 2 .

Solution Summary: The author calculates the centroidal polar moment of inertia of a circular area of radius r. The sections are approximated by isosceles triangles.

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Prove the centroidal polar moment of inertia of a circular area of radius r is πr42.

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