# A7Sol_new.pdf - Solid Mechanics HW#6 Solution Assignment 7...

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Solid Mechanics HW # 6 Solution Problem 2. Section Properties: The neutral axis passes through centroid C of the cross section as shown in Fig. a. The location of C is ¯ y = ˜ yA A = 0 . 015(0 . 03)(0 . 3) + 0 . 18(0 . 3)(0 . 03) 0 . 03(0 . 3) + 0 . 3(0 . 03) = 0 . 0975 m Thus, the moment of inertia of the cross section about the neutral axis is I = 1 12 (0 . 3)(0 . 03 3 )+0 . 3(0 . 03)(0 . 0975 - 0 . 015) 2 + 1 12 (0 . 03)(0 . 3 3 )+0 . 03(0 . 3)(0 . 18 - 0 . 0975) 2 = 0 . 1907(10 - 3 ) m 4 Allowable Bending Stress: The maximum compressive and tensile stress occurs at the top and bottom-most fibers of the cross section. For the top-most fiber, ( σ allow ) c = Mc I , 150(10 6 ) = M (0 . 33 - 0 . 0975) 0 . 1907(10 - 3 ) , M = 123024 . 19 N · m = 123 kN · m For the bottom-most fiber, ( σ allow ) t = My I , 125(10 6 ) = M (0 . 0975) 0 . 1907(10 - 3 ) , M = 244471 . 15 N · m = 244 kN · m So M = 123024 . 19 N · m = 123 kN · m controls
Problem 4.
Problem 6. For the rod BC: F x = 0 ) P BC + 90 = 0 ) P BC = - 90 KN For the rod AB: F x = 0 ) P AB - 2 30 4 5 + 90 = 0 KN ) P AB = - 42 KN σ max = σ BC = P BC A = - 90 10 3 (30 10 - 3 ) 2 / 4 = - 127 MPa For A992 steel, σ y = 345

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