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Design of helical spring against static loading

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akashpatel281996

This document discusses the design of helical springs against static loading. It defines what a helical spring is and its functions of storing and releasing energy and absorbing shock. The key design considerations for helical springs are described such as required space, forces, tolerances, costs and environment. Formulas are provided for calculating stresses in the spring from torsional and direct shear forces. Common spring materials and effects of end treatment are also summarized. Buckling is discussed and the formula provided. Parameters calculated by the design module are outlined such as spring dimensions, load rating and stresses. Spring testing machines are also briefly mentioned.

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DESIGN OF HELICAL SPRING AGAINST STATIC LOADING BIRLA VISHVAKARMA MAHAVIDHYALYA MECHANICAL ENGINEERING V.V.NAGAR Batch D-6 Created by:- 130070119052-Parikh Manir 130070119053-Parmar Dinesh 130070119054 -Patel Akash 130070119057 -Patel Harsh Guided By:- Prof. R.G. Jivani
Helical spring is a spiral wound wire with a constant coil diameter and uniform pitch. WHAT IS HELICAL SPRING? FUNCTION OF HELICAL SPRING  Used to store energy and subsequently release it  To absorb shock  To maintain a force between contacting surfaces
DESIGN CONSIDERATION OF HELICAL SPRING • The design of new helical involves the following consideration. – Space into which the spring must fit and operate – Values of working forces and deflections – Accuracy and reliability needed. – Tolerances and permissible variations in specifications. – Environmental conditions such as temperature, pressure of corrosive atmosphere. – Cost and qualities needed
C = Spring Index D/d d = wire diameter (m) D = Spring diameter (m) Di = Spring inside diameter (m) E = Young's Modulus (N/m2) F = Axial Force (N) G = Modulus of Rigidity (N/m2) L 0 = Free Length (m) L s = Solid Length (m) n t = Total number of coils n = Number of active coils p = pitch (m) τ = shear stress (N/m2) τ max = Max shear stress (N/m2) θ = Deflection (radians) K W = Wahl Factor = [(4C-1)/(4C+5)}]+ (0,615/C) NOMENCLATURE OF HELICAL SPRING
SPRING MATERIAL 1. High carbon steel • Music Wire (ASTM A228) • Hard Drawn (ASTM A227) • High Tensile Hard Drawn (ASTM A679) • Oil Tempered (ASTM A229) • Carbon Valve (ASTM A230) 2. Alloy spring steel • Chrome Vanadium (ASTM A231) • Chrome Silicon (ASTM A401) 3. Stainless steel • AISI 302/304 - ASTM A313 • AISI 316 - ASTM A313 • 17-7 PH - ASTM A313(631) 4. Copper based alloy • Phosphor Bronze (Grade A) - ASTM B159 • Beryllium Copper - ASTM B197 • Monel 400 (AMS 7233) • Monel K500 (QQ-N-286 5. Nickel based alloy • A286 Alloy • Inconel 600 (QQ-W-390) • Inconel 718 • Inconel X-750 (AMS 5698, 5699)
STRESSES IN HELICAL SPRING • A round wire helical compression is subjected to axial force. • The effect of axial force is equivalent to: 1. Torsional T=FD/2, acting on the wire cross section 2. Direct shear force F acting on wire cross section • So stresses induced in spring wire are, 1. Torsional shear stress 2. Direct shear stress
1. Torsional Shear Stress: - The torsional shear stress induced in a spring wire,due to torsional moment T=FD/2 is given by, 2. Direct Shear Stress: - The direct shear stress induced in a springwire,due to direct shear force F is given by, 33 816 d FD d T t      22 4 4 d F d F d   
3. Resultant Shear Stress: - Hence, the maximum resultant shear stress in the spring wire is given by, dt   23 48 d F d FD        D d d FD 2 1 8 3       Cd FD 5.0 1 8 3         2 8 d FC Ks  
EFFECT OF END TREATMENT
1. Outer Diameter (Do) 2. Spring Index (C) 3. Slenderness ratio 4. Spring Deflection 5. Maximum load 7. Maximum shear stress 6. Corrected maximum stress 8. Wahl correction factor THE MODULE CALCULATES THE FOLLOWING DESIGN PARAMETERS: dDD 0 dDDi  DL / Gd nFC3 8  maxmax kP  max ' max  wK 2 8 d FcKs   
 For steel springs BUCKLING If free length of helical compression spring is too large as compared to the mean coil dia., the spring act as flexible column and it may buckle under the action of axial force. In order to avoid buckling, the spring must satisfy following condition   2 1 2 2          EG GE D L    63.2  D L
Design of helical spring against static loading
Design of helical spring against static loading
SPRING TESTING MACHINE
REFERENCES • www.wikipedia.com • A Textbook of Machine Design by R.S.KHURMI AND J.K.GUPTA, S. Chand publication • https://centroidworks.files.wordpress.com/2008/11/design -of-helical-spring • http://s3.amazonaws.com/pptdownload/springppt- 140129092311-phpapp02.ppt?response-content- disposition=attachment&Signature=7aeV8Uo1rykGO3UTq9 ZK2XdKv38%3D&Expires=1444133422&AWSAccessKeyId=A KIAIA7QTBOH2LDUZRTQ
Design of helical spring against static loading

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Design of helical spring against static loading

  • 1. DESIGN OF HELICAL SPRING AGAINST STATIC LOADING BIRLA VISHVAKARMA MAHAVIDHYALYA MECHANICAL ENGINEERING V.V.NAGAR Batch D-6 Created by:- 130070119052-Parikh Manir 130070119053-Parmar Dinesh 130070119054 -Patel Akash 130070119057 -Patel Harsh Guided By:- Prof. R.G. Jivani
  • 2. Helical spring is a spiral wound wire with a constant coil diameter and uniform pitch. WHAT IS HELICAL SPRING? FUNCTION OF HELICAL SPRING  Used to store energy and subsequently release it  To absorb shock  To maintain a force between contacting surfaces
  • 3. DESIGN CONSIDERATION OF HELICAL SPRING • The design of new helical involves the following consideration. – Space into which the spring must fit and operate – Values of working forces and deflections – Accuracy and reliability needed. – Tolerances and permissible variations in specifications. – Environmental conditions such as temperature, pressure of corrosive atmosphere. – Cost and qualities needed
  • 4. C = Spring Index D/d d = wire diameter (m) D = Spring diameter (m) Di = Spring inside diameter (m) E = Young's Modulus (N/m2) F = Axial Force (N) G = Modulus of Rigidity (N/m2) L 0 = Free Length (m) L s = Solid Length (m) n t = Total number of coils n = Number of active coils p = pitch (m) τ = shear stress (N/m2) τ max = Max shear stress (N/m2) θ = Deflection (radians) K W = Wahl Factor = [(4C-1)/(4C+5)}]+ (0,615/C) NOMENCLATURE OF HELICAL SPRING
  • 5. SPRING MATERIAL 1. High carbon steel • Music Wire (ASTM A228) • Hard Drawn (ASTM A227) • High Tensile Hard Drawn (ASTM A679) • Oil Tempered (ASTM A229) • Carbon Valve (ASTM A230) 2. Alloy spring steel • Chrome Vanadium (ASTM A231) • Chrome Silicon (ASTM A401) 3. Stainless steel • AISI 302/304 - ASTM A313 • AISI 316 - ASTM A313 • 17-7 PH - ASTM A313(631) 4. Copper based alloy • Phosphor Bronze (Grade A) - ASTM B159 • Beryllium Copper - ASTM B197 • Monel 400 (AMS 7233) • Monel K500 (QQ-N-286 5. Nickel based alloy • A286 Alloy • Inconel 600 (QQ-W-390) • Inconel 718 • Inconel X-750 (AMS 5698, 5699)
  • 6. STRESSES IN HELICAL SPRING • A round wire helical compression is subjected to axial force. • The effect of axial force is equivalent to: 1. Torsional T=FD/2, acting on the wire cross section 2. Direct shear force F acting on wire cross section • So stresses induced in spring wire are, 1. Torsional shear stress 2. Direct shear stress
  • 7. 1. Torsional Shear Stress: - The torsional shear stress induced in a spring wire,due to torsional moment T=FD/2 is given by, 2. Direct Shear Stress: - The direct shear stress induced in a springwire,due to direct shear force F is given by, 33 816 d FD d T t      22 4 4 d F d F d   
  • 8. 3. Resultant Shear Stress: - Hence, the maximum resultant shear stress in the spring wire is given by, dt   23 48 d F d FD        D d d FD 2 1 8 3       Cd FD 5.0 1 8 3         2 8 d FC Ks  
  • 9. EFFECT OF END TREATMENT
  • 10. 1. Outer Diameter (Do) 2. Spring Index (C) 3. Slenderness ratio 4. Spring Deflection 5. Maximum load 7. Maximum shear stress 6. Corrected maximum stress 8. Wahl correction factor THE MODULE CALCULATES THE FOLLOWING DESIGN PARAMETERS: dDD 0 dDDi  DL / Gd nFC3 8  maxmax kP  max ' max  wK 2 8 d FcKs   
  • 11.  For steel springs BUCKLING If free length of helical compression spring is too large as compared to the mean coil dia., the spring act as flexible column and it may buckle under the action of axial force. In order to avoid buckling, the spring must satisfy following condition   2 1 2 2          EG GE D L    63.2  D L
  • 15. REFERENCES • www.wikipedia.com • A Textbook of Machine Design by R.S.KHURMI AND J.K.GUPTA, S. Chand publication • https://centroidworks.files.wordpress.com/2008/11/design -of-helical-spring • http://s3.amazonaws.com/pptdownload/springppt- 140129092311-phpapp02.ppt?response-content- disposition=attachment&Signature=7aeV8Uo1rykGO3UTq9 ZK2XdKv38%3D&Expires=1444133422&AWSAccessKeyId=A KIAIA7QTBOH2LDUZRTQ