shi20396_ch05 - Chapter 5 5-1(a k1 k2 k3 F y k= so(b k1 k2...

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Chapter 5 5-1 (a) k = F y ; y = F k 1 + F k 2 + F k 3 so k = 1 (1 / k 1 ) + / k 2 ) + / k 3 ) Ans. (b) F = k 1 y + k 2 y + k 3 y k = F / y = k 1 + k 2 + k 3 Ans. (c) 1 k = 1 k 1 + 1 k 2 + k 3 k = µ 1 k 1 + 1 k 2 + k 3 1 5-2 For a torsion bar, k T = T = Fl , and so θ = / k T . For a cantilever, k C = F /δ, δ = F / k C . For the assembly, k = F / y , y = F / k = l θ + δ So y = F k = 2 k T + F k C Or k = 1 ( l 2 / k T ) + / k C ) Ans. 5-3 For a torsion bar, k = T = GJ / l where J = π d 4 / 32. So k = π d 4 G / (32 l ) = Kd 4 / l . The springs, 1 and 2, are in parallel so k = k 1 + k 2 = K d 4 l 1 + K d 4 l 2 = 4 µ 1 x + 1 l x And θ = T k = T 4 µ 1 x + 1 l x Then T = k θ = 4 x θ + 4 θ l x k 2 k 1 k 3 F k 2 k 1 k 3 y F k 1 k 2 k 3 y
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Chapter 5 107 Thus T 1 = Kd 4 x θ ; T 2 = 4 θ l x If x = l / 2, then T 1 = T 2 . If x < l / 2, then T 1 > T 2 Using τ = 16 T d 3 and θ = 32 Tl / ( G π d 4 ) gives T = π d 3 τ 16 and so θ all = 32 l G π d 4 · π d 3 τ 16 = 2 l τ all Gd Thus, if x < l / 2, the allowable twist is θ all = 2 x τ all Ans. Since k = 4 µ 1 x + 1 l x = π 4 32 µ 1 x + 1 l x Ans. Then the maximum torque is found to be T max = π d 3 x τ all 16 µ 1 x + 1 l x Ans. 5-4 Both legs have the same twist angle. From Prob. 5-3, for equal shear, d is linear in x . Thus, d 1 = 0 . 2 d 2 Ans. k = π G 32 ( 0 . 2 d 2 ) 4 0 . 2 l + d 4 2 0 . 8 l = π G 32 l ³ 1 . 258 d 4 2 ´ Ans. θ all = 2(0 . 8 l ) τ all 2 Ans. T max = k θ all = 0 . 198 d 3 2 τ all Ans. 5-5 A = π r 2 = π( r 1 + x tan α) 2 d δ = Fdx AE = E π ( r 1 + x tan α ) 2 δ = F π E Z l 0 dx ( r 1 + x tan α ) 2 = F π E µ 1 tan α ( r 1 + x tan α ) l 0 = F π E 1 r 1 ( r 1 + l tan α ) l x a dx F F r 1
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108 Solutions Manual • Instructor’s Solution Manual to Accompany Mechanical Engineering Design Then k = F δ = π Er 1 ( r 1 + l tan α ) l = EA 1 l µ 1 + 2 l d 1 tan α Ans. 5-6 X F = ( T + dT ) + w dx T = 0 =− w Solution is T w x + c T | x = 0 = P + w l = c T w x + P + w l T = P + w ( l x ) The infnitesmal stretch oF the Free body oF original length is d δ = Tdx AE = P + w ( l x ) Integrating, δ = Z l 0 [ P + w ( l x )] δ = Pl + w l 2 2 Ans. 5-7 M = w lx w l 2 2 w x 2 2 EI dy = w 2 2 w l 2 2 x w x 3 6 + C 1 , = 0 at x = 0, [ C 1 = 0 EIy = w 3 6 w l 2 x 2 4 w x 4 24 + C 2 , y = 0 at x = 0, [ C 2 = 0 y = w x 2 24 (4 6 l 2 x 2 ) Ans. l x dx P Enlarged Free body oF length dx w is cable’s weight per Foot T 1 dT w dx T
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Chapter 5 109 5-8 M = M 1 = M B EI dy dx = M B x + C 1 , = 0 at x = 0, [ C 1 = 0 EIy = M B x 2 2 + C 2 , y = 0 at x = 0, [ C 2 = 0 y = M B x 2 2 Ans. 5-9 ds = p 2 + 2 = ± 1 + µ 2 Expand right-hand term by Binomial theorem 1 + µ 2 1 / 2 = 1 + 1 2 µ 2 +··· Since / is small compared to 1, use only the frst two terms, d λ = = 1 + 1 2 µ 2 = 1 2 µ 2 [ λ = 1 2 Z l 0 µ 2 Ans. This contraction becomes important in a nonlinear, non-breaking extension spring. 5-10 y =− 4 ax l 2 ( l x ) µ 4 l 4 a l 2 x 2 µ 4 a l 8 l 2 µ 2 = 16 a 2 l 2 64 a 2 x l 3 + 64 a 2 x 2 l 4 λ = 1 2 Z l 0 µ 2 = 8 3 a 2 l Ans. y ds dy dx l
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110 Solutions Manual • Instructor’s Solution Manual to Accompany Mechanical Engineering Design 5-11 y = a sin π x l dy dx = a π l cos π x l µ 2 = a 2 π 2 l 2 cos 2 π x l λ = 1 2 Z l 0 µ 2 λ = π 2 4 a 2 l = 2 . 467 a 2 l Ans.
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shi20396_ch05 - Chapter 5 5-1(a k1 k2 k3 F y k= so(b k1 k2...

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