20_ch 04 Mechanical Design budynas_SM_ch04

20_ch 04 Mechanical Design budynas_SM_ch04 -...

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Unformatted text preview: budynas_SM_ch04.qxd 11/28/2006 20:50 FIRST PAGES Page 89 89 Chapter 4 yB = 1 10(106 )(0.1667) = −0.045 87 in 4-41 Ans . π (1.54 ) = 0.2485 in4 64 I1 = I2 = 200 (12) = 1200 lbf 2 R1 = For 0 ≤ x ≤ 16 in, M = 1200x − M 1200x − 4800 = I I1 1 1 − I1 I2 = 4829x − 13 204 x − 4 E [−26.67(183 ) + 40(18 − 6) 3 + 552.58(18)] 0 π4 (2 ) = 0.7854 in4 64 MI 200 x −4 2 x −4 2 1 1 1 − I1 I2 − 1200 0 − 3301.1 x − 4 dy = 2414.5x 2 − 13 204 x − 4 dx x − 1651 x − 4 1 x −4 − 127.32 x − 4 3 − 100 x −4 I2 2 2 − 42.44 x − 4 1 2 + C1 dy = 0 at x = 10 in dx 0 = 2414.5(102 ) − 13 204(10 − 4) 1 − 1651(10 − 4) 2 − 42.44(10 − 4) 3 + C1 C1 = −9.362(104 ) Boundary Condition: E y = 804.83x 3 − 6602 x − 4 2 − 550.3 x − 4 3 − 10.61 x − 4 4 − 9.362(104 ) x + C2 y=0 at x = 0 ⇒ C2 = 0 For 0 ≤ x ≤ 16 in y= 1 [804.83x 3 − 6602 x − 4 2 − 550.3 x − 4 30(106 ) − 10.61 x − 4 4 − 9.362(104 ) x ] Ans. 3 at x = 10 in y |x =10 = 1 [804.83(103 ) − 6602(10 − 4) 2 − 550.3(10 − 4) 3 30(106 ) − 10.61(10 − 4) 4 − 9.362(104 )(10)] = −0.016 72 in Ans. 4-42 q=F x −1 − Fl x −2 − F x −l −1 Integrations produce V=F x 0 − Fl x −1 M=F x 1 − Fl x 0 − F x −l − F x −l 1 0 = F x − Fl ...
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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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