31_ch 03 Mechanical Design budynas_SM_ch03

31_ch 03 Mechanical Design budynas_SM_ch03 -...

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Unformatted text preview: budynas_SM_ch03.qxd 44 11/28/2006 21:22 FIRST PAGES Page 44 Solutions Manual • Instructor’s Solution Manual to Accompany Mechanical Engineering Design R1 = 600 lbf/ft 600(15) 20 + 3000 = 8500 lbf 2 15 R2 = y 3-29 600(15) 5 − 3000 = 3500 lbf 2 15 3000 lbf x 5' 15' R2 R1 a= 5500 3500 = 5.833 ft 600 y= ¯ V (lbf) 1(12) + 5(12) = 3 in 24 a x O 3000 3500 M (lbf • ft) 3500(5.833) O 20420 x 15000 y (a) z 1 Iz = [2(53 ) + 6(33 ) − 4(13 )] = 136 in4 3 y At x = 5 ft, σx = − y = −3 in, y = 5 in, At x = 14.17 ft, σx = − y = −3 in, y = 5 in, Max tension = 6620 psi Ans. Max compression = −9010 psi (b) Vmax = 5500 lbf 5 in −15000(12)5 = 6620 psi 136 σx = − σx = − 20420(12)( −3) = 5405 psi 136 20420(12)5 = −9010 psi 136 Ans. Q n.a. = y A = 2.5(5)(2) = 25 in3 ¯ z τmax = V (c) τmax = −15000(12)( −3) = −3970 psi 136 VQ 5500(25) = = 506 psi Ib 136(2) |σmax | 9010 = = 4510 psi 2 2 Ans. Ans. ...
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This note was uploaded on 12/13/2011 for the course EML 3013 taught by Professor Shingley during the Fall '11 term at UNF.

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