hw2_soln - d D C 3 1 C C 4 1 C C 4 K 2 d M 32 K 1 in 75 D...

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Problem 1

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Problem 2
Problem 2

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Problem 3 E d MDN 64 ) 5 ( M DN 64 E d k ) 4 ( d / D C ) 3 ( ) 1 C ( C 4 1 C C 4 K ) 2 ( d M 32 K ) 1 ( 4 a a 4 2 i 3 i = θ θ = = = = π = σ σ – Max Bending Stress (psi) K i – Stress Augmentation Factor M – Max Torque (in-lb) D – Coil Diameter (in) d – Wire Diameter (in) C – Spring Index E – Wire Modulus (psi) N a – Number of Active Coils k – Spring Stiffness (in-lb/rad) θ – Spring Rotation (rad) 12in 1.4 lb 180 Known D – .75in ΔY – 12 in F – 1.4 lb θ – 180 = π rad Music Wire σ – 230 kpsi FOS 1.5 E – 28.5 Mpsi r Need torsion spring on shaft to drop ball 12 inches and return to original position .75in Spring
Problem 3

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Unformatted text preview: d / D C ) 3 ( ) 1 C ( C 4 1 C C 4 K ) 2 ( d M 32 K ) 1 ( in 75 . D lb in 3 . 5 8 . 3 4 . 1 Fr M in 8 . 3 r r in 12 2 i 3 i = − − − = π = σ = = ⋅ = = = π = UnKnown M – Max Torque (in-lb) d – Wire Diameter (in) N a – Number of Active Coils k – Spring Stiffness (in-lb/rad) a 4 DN 64 E d k ) 4 ( rad lb in 69 . 1 3 . 5 M k = = π = θ = Solve for Na = 5.54 – Round to 6 Solve for d = (0.0325 0.053i),(0.064),(0.750) Guess 0.064 in, reasonable wire diameter Problem 3...
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This note was uploaded on 07/22/2010 for the course ME Mecheng 35 taught by Professor 350 during the Winter '08 term at University of Michigan-Dearborn.

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hw2_soln - d D C 3 1 C C 4 1 C C 4 K 2 d M 32 K 1 in 75 D...

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