1_ch 04 Mechanical Design budynas_SM_ch04

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

Info iconThis preview shows page 1. Sign up to view the full content.

View Full Document Right Arrow Icon
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 ² 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 k = k
Background image of page 1
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: 1 + k 2 = K d 4 l 1 + K d 4 l 2 = K d 4 1 x + 1 l x And = T k = T K d 4 1 x + 1 l x Then T = k = K d 4 x + K d 4 l x k 2 k 1 k 3 F k 2 k 1 k 3 y F k 1 k 2 k 3 y...
View Full Document

This note was uploaded on 12/13/2011 for the course EML 3013 taught by Professor Shingley during the Fall '11 term at UNF.

Ask a homework question - tutors are online