Budynas ch04

Shigley's Mechanical Engineering Design (McGraw-Hill Series in Mechanical Engineering)

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Chapter 4 4-1 (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 = ± 1 k 1 + 1 k 2 + k 3 ² 1 4-2 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. 4-3 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 ± 1 x + 1 l x ² And θ = T k = T 4 ± 1 x + 1 l x ² 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 4 71 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 ± 1 x + 1 l x ² = π 4 32 ± 1 x + 1 l x ² Ans. Then the maximum torque is found to be T max = π d 3 x τ all 16 ± 1 x + 1 l x ² Ans. 4-4 Both legs have the same twist angle. From Prob. 4-3, for equal shear, d is linear in x . Thus, d 1 = 0 . 2 d 2 Ans. k = π G 32 ³ ( 0 . 2 d 2 ) 4 0 . 2 l + d 4 2 0 . 8 l ´ = π G 32 l µ 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. 4-5 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 · l 0 dx ( r 1 + x tan α ) 2 = F π E ± 1 tan α ( r 1 + x tan α ) ² l 0 = F π E 1 r 1 ( r 1 + l tan α ) l x ± dx F F r 1
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72 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 ± 1 + 2 l d 1 tan α ² Ans. 4-6 ³ 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 infinitesmal stretch of the free body of original length is d δ = Tdx AE = P + w ( l x ) Integrating, δ = ´ l 0 [ P + w ( l x )] δ = Pl + w l 2 2 Ans. 4-7 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 ± dT w dx T
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Chapter 4 73 4-8 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. 4-9 ds = ± 2 + 2 = ² 1 + ³ ´ 2 Expand right-hand term by Binomial theorem µ 1 + ³ ´ 2 1 / 2 = 1 + 1 2 ³ ´ 2 +··· Since / is small compared to 1, use only the first two terms, d λ = = µ 1 + 1 2 ³ ´ 2 = 1 2 ³ ´ 2 ± λ = 1 2 · l 0 ³ ´ 2 Ans. This contraction becomes important in a nonlinear, non-breaking extension spring. 4-10 y = Cx 2 (4 lx x 2 6 l 2 ) where C = w 24 = (12 4 x 2 12 l 2 ) = 4 (3 x 2 3 l 2 ) ³ ´ 2 = 16 C 2 (15 l 2 x 4 6 5 18 x 3 l 3 + x 6 + 9 l 4 x 2 ) λ = 1 2 l · 0 ³ ´ 2 = 8 C 2 l · 0 l 2 x 4 6 5 18 x 3 l 3 + x 6 + 9 l 4 x 2 ) = 8 C 2 ³ 9 14 l 7 ´ = 8 ¸ w 24 ¹ 2 ³ 9 14 l 7 ´ = 1 112 ¸ w ¹ 2 l 7 Ans.
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Budynas ch04 - Chapter 4 4-1(a k1 k2 k3 F y k= so(b k1 k2 k3 F y F y y= F F F k1 k2 k3 Ans k= 1(1/k1(1/k2(1/k3 F = k1 y k2 y k3 y k = F/y = k1 k2

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