461_Mechanics SolutionInstructors_Sol.Manual-Mechanics_Materials_7e.book_Gere_light.1

# 461_Mechanics SolutionInstructors_Sol.Manual-Mechanics_Materials_7e.book_Gere_light.1

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Unformatted text preview: 05Ch05.qxd 9/25/08 2:29 PM Page 455 SECTION 5.9 Shear Stresses in Circular Beams Solution 5.9-3 b b p Wind load on a sign (b) REQUIRED DIAMETER BASED UPON SHEAR STRESS width of sign 10 ft 75 lb/ft2 7500 psi 2000 psi sallow tallow d diameter t d 10 W t wind force on one pole b ph2 a b 2 W 1875 lb r1 W a h1 + h2 b 2 p (d 4 64 2 d24) d2 d d1 I p4 cd 64 4d 4 a bd 5 pd 4 369 a b 64 625 d 2t d 2 s Mc I d3 (d M(d/2) d 2 d 10 r2 + r1 t p2 cd 4 A p2 (d2 4 100 4V 61 a ba b 3 41 9pd2 3 10.52 in. ; Problem 5.9-4 Solve the preceding problem for a sign and poles having the following dimensions: h1 6.0 m, h2 1.5 m, b 3.0 m, and t d/10. The design wind pressure is 3.6 kPa, and the allowable stresses in the aluminum are 50 MPa in bending and 14 MPa in shear. d2) 1 r2 d 2 2d 5 d a 4d 2 bd 5 7.0160 V d2 7.0160 Vmax tallow (7.0160)(1875 lb) 2000 psi 17.253 M 369pd /40,000 d (17.253)(506,250 lb-in.) 17.253 Mmax sallow 7500 psi d b d2 2d 2 a b +a b 2 5 d2 4 1164.6 in. 2 t inches) 3 2 4 d 5 369pd 4 (in.4) 40,000 c d 2 + r2r1 + r12 r2 2 + r1 2 2d 2 d 2d d2 a b + a ba b + a b 2 2 5 5 506,250 lb-in. I 4V a 3A 1875 lb r22 r2 + r2r1 + r2 2 1 (a) REQUIRED DIAMETER BASED UPON BENDING STRESS Mmax W Vmax 2.56 in. ; (Bending stress governs.) 6.5775 in.2 61 41 9pd2 100 455 ...
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