# p5_sol - Problem Set 5 Problem 5.1 Product of Inertia of a...

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Problem Set 5 Problem 5.1 Product of Inertia of a Cross Section Determine the product of inertia for the cross sectional area with respect to the x - and y -axes with origin located at the centroid C . Figure 5.1: Problem 5.1

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Problem 5.2 Product of Inertia Transformations Determine the moments of inertia I u and I v as well as the product of inertia I u v for the channel’s cross section. Use θ = 45 . Figure P5.2: Problem 5.2
Problem 5.3 The polar moment of inertia of the area shown in Fig. P5.3, is I Czz about the z -axis passing through the centroid C . If the moment of inertia about the y 0 axis is I y 0 y 0 and the moment of inertia about the x -axis is I xx . Determine the area A . Numerical application: I Czz = 548 × 10 6 mm 4 , I y 0 y 0 = 383 × 10 6 mm 4 , I xx = 856 × 10 6 mm 4 , and h = 250 mm. x y 0 C x 0 h Figure P5.3: Problem 5.3 Solution ICzz = 548*10^6; % mm^4 Iyyp = 383*10^6; % mm^4 Ixx = 856*10^6; % mm^4 h = 250; % mm Ixxp = Ixx - A*h^2; % ICzz = Ixxp + Iyyp % ICzz = Ixx - A*h^2 + Iyyp % => A = (Ixx + Iyyp - ICzz)/h^2; A = (Ixx + Iyyp - ICzz)/h^2; % A = 1.11e+04 (mm^2)

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Problem 5.4
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## This note was uploaded on 08/29/2011 for the course MECH 2110 taught by Professor Clark,b during the Spring '08 term at Auburn University.

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p5_sol - Problem Set 5 Problem 5.1 Product of Inertia of a...

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