05-04ChapGere.0003 - 2 5 61 41 r 1 5 d 2 2 t 5 d 2 2 d 10 5...

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Solution 5.9-3 Wind load on a sign 340 CHAPTER 5 Stresses in Beams (Basic Topics) b 5 width of sign b 5 10 ft P 5 75 lb/ft 2 s allow 5 7500 psi t allow 5 2000 psi d 5 diameter W 5 wind force on one pole (a) R EQUIRED DIAMETER BASED UPON BENDING STRESS ( d 5 inches) 5 1164.6 in. 3 d 5 10.52 in. d 3 5 17.253 M max s allow 5 (17.253)(506,250 lb-in.) 7500 psi s 5 Mc I 5 M ( d / 2) 369 p d 4 / 40,000 5 17.253 M d 3 c 5 d 2 I 5 p 64 B d 4 2 ¢ 4 d 5 4 R 5 p d 4 64 ¢ 369 625 5 369 p d 4 40,000 (in. 4 ) I 5 p 64 ( d 2 4 2 d 2 4 ) Ê d 2 5 d Ê d 1 5 d 2 2 t 5 4 5 d M max 5 W ¢ h 1 1 h 2 2 5 506,250 lb-in. t 5 d 10 Ê W 5 ph 2 ¢ b 2 5 1875 lb (b) R EQUIRED DIAMETER BASED UPON SHEAR STRESS V max 5 W 5 1875 lb d 5 2.56 in. (Bending stress governs.) d 2 5 7.0160 V max t allow 5 (7.0160)(1875 lb) 2000 psi 5 6.5775 in. 2 t 5 4 V 3 ¢ 61 41 ¢ 100 9 p d 2 5 7.0160 V d 2 A 5 p 4 ( d 2 2 2 d 1 2 ) 5 p 4 B d 2 2 ¢ 4 d 5 2 R 5 9 p d 2 100 r 2 2 1 r 2 r 1 1 r 1 2 r 2 2 1 r 1 2 5 ¢ d 2 2 1 ¢ d 2 ¢ 2 d 5 1 ¢ 2 d 5 2 ¢ d 2 2 1 ¢ 2 d 5
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Unformatted text preview: 2 5 61 41 r 1 5 d 2 2 t 5 d 2 2 d 10 5 2 d 5 t 5 4 V 3 A r 2 2 1 r 2 r 1 1 r 1 2 r 2 2 1 r 1 2 r 2 5 d 2 d h 2 = 5ft h 1 = 20 ft W Problem 5.9-4 Solve the preceding problem for a sign and poles having the following dimensions: h 1 5 6.0 m, h 2 5 1.5 m, b 5 3.0 m, and t 5 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. A-PDF Split DEMO : Purchase from www.A-PDF.com to remove the watermark...
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This note was uploaded on 09/20/2009 for the course COE 3001 taught by Professor Armanios during the Spring '08 term at Georgia Institute of Technology.

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