21_ch 04 Mechanical Design budynas_SM_ch04

21_ch 04 Mechanical Design budynas_SM_ch04 - Fl 3 16 E I 1...

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90 Solutions Manual • Instructor’s Solution Manual to Accompany Mechanical Engineering Design Plots for M and M / I are shown below M / I can be expressed by singularity functions as M I = F 2 I 1 x Fl 2 I 1 Fl 4 I 1 ± x l 2 ² 0 + F 2 I 1 ± x l 2 ² 1 where the step down and increase in slope at x = l / 2 are given by the last two terms. Since E d 2 y / dx 2 = M / I , two integrations yield E dy dx = F 4 I 1 x 2 Fl 2 I 1 x Fl 4 I 1 ± x l 2 ² 1 + F 4 I 1 ± x l 2 ² 2 + C 1 Ey = F 12 I 1 x 3 Fl 4 I 1 x 2 Fl 8 I 1 ± x l 2 ² 2 + F 12 I 1 ± x l 2 ² 3 + C 1 x + C 2 At x = 0, y = dy / dx = 0 . This gives C 1 = C 2 = 0, and y = F 24 EI 1 ³ 2 x 3 6 lx 2 3 l ± x l 2 ² 2 + 2 ± x l 2 ² 3 ´ At x = l / 2 and l, y | x = l / 2 = F 24 EI 1 µ 2 l 2 · 3 6 l l 2 · 2 3 l (0) + 2(0) ¸ =− 5 Fl 3 96 EI 1 Ans. y | x = l = F 24 EI 1 µ 2( l ) 3 6 l ( l ) 2 3 l l l 2 · 2 + 2 l l 2 · 3 ¸ =− 3
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Unformatted text preview: Fl 3 16 E I 1 Ans. The answers are identical to Ex. 4-11. 4-43 Dene i j as the deection in the direction of the load at station i due to a unit load at station j . If U is the potential energy of strain for a body obeying Hookes law, apply P 1 rst. Then U = 1 2 P 1 ( P 1 11 ) O F F Fl Fl Fl 2 Fl 4 I 1 Fl 2 I 1 Fl 2 I 1 A O l 2 l 2 2 I 1 x x y x B I 1 C M M I...
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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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