Explanation Calculation: Consider that two case of moments and product of inertia. It has different orientation and coordinate axes. The origin point ( O ) of the two cases is same. Case 1: I x ′ = I x I y ′ = I y I x ′ y ′ = I x y Case 2: I x ′ = I max I y ′ = I min I x ′ y ′ = 0 Apply the invariance property. I x I y − I x y 2 = I max I min I x y = ± I x I y − I max I min (1) Refer to Figure 9.13A and 9.13B, “Properties of rolled steel shapes” in the textbook. The section properties of L3 × 2 × 1 4 as follows: The moment of inertia I ¯ x about x axis is 1.09 in 4 . The moment of inertia I ¯ y about y axis is 0.390 in 4 . Refer to Equation 9.21 in section, “9.3B Principal Axes and principal Moments of inertia” in the textbook. I ¯ x + I ¯ y = I ¯ max + I ¯ min (2) Here, I ¯ x is moment of inertia about x axis I ¯ y is moment of inertia about y axis, I ¯ max is maximum moment of inertia, and I ¯ min is minimum moment of inertia. Substitute 1.09 in 4 for I ¯ x , 0.390 in 4 for I ¯ y , and 1

Vector Mechanics for Engineers: Statics and Dynamics

12th Edition

ISBN: 9781259638091

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

Publisher: McGraw-Hill Education

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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 9.4, Problem 9.110P

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Find the product of inertia (Ixy) of the L3×2×14 in. angle cross section.

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Chapter 9.4, Problem 9.109P

Chapter 9.5, Problem 9.111P

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Vector Mechanics for Engineers: Statics and Dynamics

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Find the product of inertia ( I x y ) of the L 3 × 2 × 1 4 in . angle cross section.

Solution Summary: The author calculates the invariance property of two cases of moments and product of inertia, which has different orientation and coordinate axes.

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Find the product of inertia (Ixy) of the L3×2×14 in. angle cross section.

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