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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 0 + F 2 I 1 x l 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 1 + F 4 I 1 x l 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 + F 12 I 1 x l 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 E I 1 2 x 3 6 lx 2 3 l x l 2 2 + 2 x l 2 3 At x = l / 2 and l, y | x = l / 2 = F 24 E I 1 2 l 2 3 6 l l 2 2 3 l (0) + 2(0) = − 5 Fl 3 96 E I 1 Ans. y | x = l = F 24 E I 1 2( l ) 3 6 l ( l ) 2 3 l l l 2 2 + 2 l l 2 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 Deﬁne δ i j as the deﬂection 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 Hooke’s law, apply P 1 ﬁrst. Then U = 1 2 P 1 ( P 1 δ 11 ) O F F Fl ± Fl ± Fl ² 2 ± Fl ² 4 I 1 ± Fl ² 2 I 1 ± Fl ² 2 I 1 A O l ² 2 l ² 2 2 I 1 x x y x B I 1 C M M ² I...
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