694_Mechanics SolutionInstructors_Sol.Manual-Mechanics_Materials_7e.book_Gere_light.1

694_Mechanics SolutionInstructors_Sol.Manual-Mechanics_Materials_7e.book_Gere_light.1

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Unformatted text preview: 08Ch08.qxd 9/18/08 688 11:06 AM Page 688 CHAPTER 8 Applications of Plane Stress A PRINCIPAL STRESSES AND MAXIMUM SHEAR STRESS sx + sy s1, 2 ; 2 a sy sx 2 sc 2 b + t2 xy A 4324 psi s2 sx sy tmax a 2 ; 1870 psi MAXIMUM SHEAR STRESS: 1227 psi ; 3097 psi s1 MAXIMUM COMPRESSIVE STRESS: tmax 1870 psi NOTE: Since the principal stresses have opposite signs, the maximum in-plane shear stress is larger than the maximum out-of-plane shear stress. 2 b + t2 xy ; 3100 psi 3097 psi MAXIMUM TENSILE STRESS: st 4320 psi ; Problem 8.5-12 A semicircular bar AB lying in a horizontal plane is O supported at B (see figure). The bar has centerline radius R and weight q per unit of length (total weight of the bar equals pqR). The cross section of the bar is circular with diameter d. Obtain formulas for the maximum tensile stress st, maximum compressive stress sc, and maximum in-plane shear stress tmax at the top of the bar at the support due to the weight of the bar. A B R d Solution 8.5-12 Semicircular bar STRESSES AT THE TOP OF THE BAR AT B sB tB d diameter of bar q weight of bar per unit length W weight of bar R radius of bar pqR Weight of bar acts at the center of gravity From Case 23, Appendix D, with b y 2R p ‹c 2R p Bending moment at B: MB Torque at B: TB p/2, we get WR Wc 2qR2 pqR2 (Shear force at B produces no shear stress at the top of the bar.) M B(d/2) I TB(d/2) IP (2qR2)(d/2) 64qR2 pd 4/64 (pqR2)(d/2) pd 3 16qR2 pd 4/32 d3 ...
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This note was uploaded on 12/22/2011 for the course MEEG 310 taught by Professor Staff during the Fall '11 term at University of Delaware.

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