Explanation Calculation: Draw the subject area with vertical differential strip element as in Figure (1). Refer to Figure (1), Show the line equation follows: y = m x + C Substitute ( b a ) for m . y = b a x Calculate the area of the given section using the relation: d A = ( y + 2 b ) d x Substitute b a x for y . d A = ( b a x + 2 b ) d x = ( b x + 2 a b a ) d x = b a ( x + 2 a ) d x Calculate the moment of inertia ( I x ) about x axis using the formula: I x = ∫ d I x = ∫ − a a ( 1 3 y t o p 3 − 1 3 y b o t t o m 3 ) d x From Figure (1), I x = ∫ − a a ( 1 3 ( b a x ) 3 − 1 3 ( − 2 b ) 3 ) d x = ∫ − a a 1 3 [ ( b a x ) 3 − ( − 2 b ) 3 ] d x = ∫ − a a 1 3 [ ( b 3 x 3 a 3 + 8 b 3 ) ] d x = ∫ − a a 1 3 [ ( b 3 x 3 + 8 b 3 a 3 a 3 ) ] d x = ∫ − a a 1 3 b 3 a 3 ( x 3 + 8 a 3 ) d x = 1 3 b 3 a 3 [ x 4 4 + 8 a 3 x ] − a a = 1 3 b 3 a 3 [ ( a 4 4 + 8 a 3 a ) − ( ( − a ) 4 4 + 8 a 3 ( − a ) ) ] Simplify the Equation, I x = 1 3 b 3 a 3 [ ( a 4 4 + 8 a 4 ) − ( a 4 4 − 8 a 4 ) ] = 1 3 b 3 a 3 [ a 4 4 + 8 a 4 − a 4 4 + 8 a 4 ] = 1 3 b 3 a 3 ( 16 a 4 ) = 16 3 a b 3 Calculate the moment of inertia ( I y ) about y axis using the formula: I y = ∫ d I y = ∫ − a a x 2 d A Substitute b a ( x + 2 a ) d x for dA

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 9.22P

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The polar moment of inertia (JP) and polar radius of gyration (kP) of the shaded area with respect to point P.

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Chapter 9.1, Problem 9.21P

Chapter 9.1, Problem 9.23P

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The polar moment of inertia ( J P ) and polar radius of gyration ( k P ) of the shaded area with respect to point P .

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The polar moment of inertia (JP) and polar radius of gyration (kP) of the shaded area with respect to point P.

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Chapter 9 Solutions