1_ch 04 Mechanical Design budynas_SM_ch04

1_ch 04 Mechanical Design budynas_SM_ch04 - 1 k 2 = K d 4 l...

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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 ) + (1 / k 2 ) + (1 / 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 1 4-2 For a torsion bar, k T = T = Fl , and so θ = Fl / 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 = Fl 2 k T + F k C Or k = 1 ( l 2 / k T ) + (1 / 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
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Unformatted text preview: 1 + k 2 = K d 4 l 1 + K d 4 l 2 = K d 4 ± 1 x + 1 l − x ² And θ = T k = T K d 4 ± 1 x + 1 l − x ² Then T = k θ = K d 4 x θ + K d 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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