Explanation Given: The width of the shaded area is 4 in . The height of the shaded area is 6 in . The radius is 1 in . Show the free body diagram of the shaded area as in Figure 1. Calculation: From the Figure 1, Compute the product of inertia for the shaded area with respect to the x axes. I x = 1 3 ( c + d ) ( a + b ) 3 − [ π r 4 4 + π r 2 a 2 ] (I) Here, the radius is r . Compute the product of inertia for the shaded area with respect to the y axes. I y = 1 3 ( a + b ) ( c + d ) 3 − [ π r 4 4 + π r 2 d 2 ] (II) Compute the product of inertia for the shaded area with respect to the x and y axes. I x y = [ a + b 2 ] [ c + d 2 ] ( a + b ) ( c + d ) − d a π r 2 (III) Compute the radius. R = [ I x − ( I x + I y ) / 2 ] 2 + ( I x y ) 2 (IV) Calculate the maximum principle moment of inertia. I max = I x + I y 2 + R (V) Calculate the minimum principle moment of inertia. I min = I x + I y 2 − R (VI) Calculate the direction of principal axes. tan 2 θ p 1 = − I x y ( I x − I y ) / 2 (VII) Calculate the direction of principal axes. θ p 2 = 90 ° + θ P 1 (VIII) Conclusion: Substitute 2 in . for c , 2 in . for d , 4 in . for a , 2 in . for b , and 1 in . for r in Equation (I). I x = 1 3 ( 2 + 2 ) ( 4 + 2 ) 3 − [ π ( 1 ) 4 4 + π ( 1 ) 2 ( 4 ) 2 ] = 288 − 51.050 = 236.95 in . 4 Substitute 2 in . for c , 2 in . for d , 4 in . for a , 2 in . for b , and 1 in . for r in Equation (II). I y = 1 3 ( 4 + 2 ) ( 2 + 2 ) 3 − [ π ( 1 ) 4 4 + π ( 1 ) 2 ( 2 ) 2 ] = 128 − 13.351 = 114.65 in . 4 Substitute 2 in . for c , 2 in . for d , 4 in . for a , 2 in . for b , and 1 in . for r in Equation (III). I x y = [ 4 + 2 2 ] [ 2 + 2 2 ] ( 4 + 2 ) ( 2 + 2 ) − ( 2 ) ( 4 ) π ( 1 ) 2 = ( 3 ) ( 2 ) ( 6 ) ( 4 ) − 25

Engineering Mechanics: Statics & Dynamics (14th Edition)

14th Edition

ISBN: 9780133915426

Author: Russell C. Hibbeler

Publisher: PEARSON

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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 10.7, Problem 82P

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The directions (θp1,θp2) of the principal axes with origin located at point O, and the principal moments of inertia (Imax,Imin) for the area about these axes.

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Chapter 10.7, Problem 81P

Chapter 10.7, Problem 83P

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Engineering Mechanics: Statics & Dynamics (14th Edition)

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The directions ( θ p 1 , θ p 2 ) of the principal axes with origin located at point O , and the principal moments of inertia ( I max , I min ) for the area about these axes.

Solution Summary: The author calculates the product of inertia for the shaded area with respect to the x axes.

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The directions (θp1,θp2) of the principal axes with origin located at point O, and the principal moments of inertia (Imax,Imin) for the area about these axes.

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Chapter 10 Solutions