17 springs - Springs ME250 Design and Manufacturing I 2...

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Unformatted text preview: Springs ME250 Design and Manufacturing I 2 What? Why? Types of springs Nomenclature Types of ends Spring stiffness Stress in springs Strength of springs Natural frequency & resonance Summary Outline 3 Flexible element used to exert a force or a torque and, at the same time, to store energy What? 4 What? Force can be linear or radial 5 Transform kinetic energy to potential energy Can be used to absorb undesired motion/impact Why? 6 Transform stored potential energy to kinetic energy Stored potential energy can be released as force/torque at a later time Why? 7 Types of springs Compression helical(coil) spring 8 Extension helical(coil) springs Types of springs 9 Flat (cantilever) spring Types of springs 10 Volute spring Types of springs 11 Flat spiral spring Types of springs 12 Torsion coil spring Types of springs 13 Leaf spring Types of springs 14 Belleville spring Types of springs 15 Torsion bar Types of springs 16 D m = mean coil diameter d = wire diameter D o = D m +d = Outside diameter D i = D m -d = Inside diameter N t = Number of coils L = Free length L s = Solid length p = Pitch Nomenclature p D m 17 Compression springs Type of ends affect number of active coils, N a (< N t ) Types of ends Plain end, right hand N a = N t Squared and ground end, left hand N a = N t-2 Squared or closed end, right hand N a = N t-2 Plain end, ground, left hand N a = N t-1 18 Compression springs Formula for compression spring dimensions (Table 10-1 in course pack p. 236) Types of ends 19 Extension springs Types of ends Machine half loop-open Raised hook Full twisted loop Short twisted loop 20 A measure of the resistance offered by an elastic body Anything is spring, but normally very stiff Winding reduces stiffness Spring stiffness L L + δ L-δ F F F F F k 21 Straight wire Stress : Strain : Stiffness ζ = Eε ( Hooke’s law) where E = Young’s modulus Spring stiffness d l F F 2 / 4 F F A d l 2 2 4 / 4 straight F F E d E k l l d 22 Wound wire (helical spring) λ = Pitch angle Number of active coils Assume λ is small ( tan λ << 1 ) Stiffness C = spring index = D m / d , G = shear modulus = E /2( 1+ν ) See Table 10-5 in course pack p. 241 for E and G for various spring materials Spring stiffness m a D l N d l C d E N C Gd F k a helix 2 2 3 ) 1 ( 16 8 23 Straight vs. Helix “Wound” wire has much less k than straight wire Why? Straight line carries load through normal stress Wound line carries load through shear stress Adjust the spring index C to get a desired stiffness ( is a material property – limited choices) Spring stiffness 1 ) 1 ( 4 1 2 C k k straight helix since ν ≈ 0.3 and C = D m / d >> 1 Small C Large k Large C Small k 24...
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This note was uploaded on 03/21/2010 for the course MECHENG 250 taught by Professor Stevis during the Fall '09 term at University of Michigan.

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17 springs - Springs ME250 Design and Manufacturing I 2...

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