31_ch 04 Mechanical Design budynas_SM_ch04

31_ch 04 Mechanical Design budynas_SM_ch04 -...

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100 Solutions Manual • Instructor’s Solution Manual to Accompany Mechanical Engineering Design Solve simultaneously or use software R A =− 3885 N, F BE = 15 830 N, F DF = 8058 N, C 1 62 . 045 N · m 2 σ = 15 830 ( π/ 4)(12 2 ) = 140 MPa Ans. , σ = 8058 ( 4)(12 2 ) = 71 . 2MPa Ans. EI = 209(10 9 )(8)(10 7 ) = 167 . 2(10 3 )N · m 2 y = 1 167 . 2(10 3 ) ± 3885 6 x 3 + 15 830 6 ± x 0 . 5 ² 3 10 3 (10 3 ) ± x 1 ² 3 62 . 045 x ² B : x = 0 . 5m, y B 6 . 70(10 4 )m 0 . 670 mm Ans. C : x = 1m, y C = 1 167 . 2(10 3 ) ± 3885 6 (1 3 ) + 15 830 6 0 . 5) 3 62 . 045(1) ² 2 . 27(10 3 2 . 27 mm Ans. D : x = 1 . 5, y D = 1 167 . 2(10 3 ) ± 3885 6 . 5 3 ) + 15 830 6 . 5 0 . 5) 3 10 3 3 )(1 . 5 1) 3 62 . 045(1 . 5) ² 3 . 39(10 4 0 . 339 mm Ans. 4-62 = 30(10 6 )(0 . 050) = 1 . 5(10 6 ) lbf · in 2 (1) R C + F F FD = 500 ( a ) 3 R C + 6 F = 9(500) = 4500 ( b ) (2) M 500 x + F ± x 3 ² 1 + R C ± x 6 ² 1 dy dx 250 x 2 + F 2 ± x 3 ² 2 + R C 2 ± x 6 ² 2 + C 1 EIy 250 3 x 3 + F 6 ± x 3 ² 3 + R C 6 ± x
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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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