Explanation Given: The moment of inertia about the x axes, I ¯ x = 450 in . 4 The moment of inertia about the y axes, I ¯ y = 1730 in . 4 . The moment of inertia about the x and y axes, I ¯ x y = 138 in . 4 . Explanation: Show the free body diagram of the cross-section of an airplane as in Figure (1). From the Figure 1, Compute the orientation of the principal axes. tan 2 θ p = − 2 I ¯ x y I ¯ x − I ¯ y (I) Compute the principal moments of inertia. I max / min = I ¯ x + I ¯ y 2 ± [ ( I ¯ x − I ¯ y ) / 2 ] 2 + ( I ¯ x y ) 2 (II) Conclusion: Substitute 450 in . 4 for I ¯ x , 1 , 730 in . 4 for I ¯ y , and 138 in . 4 for I ¯ x y in Equation (I). tan 2 θ p = − 2 ( 138 ) 450 − 1730 tan 2 θ p = 0

INTERNATIONAL EDITION---Engineering Mechanics: Statics, 14th edition (SI unit)

14th Edition

ISBN: 9780133918922

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 78P

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The orientation θp of the principal axes and the principal moments of inertia (Imax,Imin) .

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Chapter 10.7, Problem 77P

Chapter 10.7, Problem 79P

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INTERNATIONAL EDITION---Engineering Mechanics: Statics, 14th edition (SI unit)

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The orientation θ p of the principal axes and the principal moments of inertia ( I max , I min ) .

Solution Summary: The author shows the free body diagram of the cross-section of an airplane as in Figure (1).

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The orientation θp of the principal axes and the principal moments of inertia (Imax,Imin) .

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