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1. ASSIGNMENT 1
CASE ON DESIGN OF SPRING USED IN AUTOMOTIVE APPLICATION

2. PROBLEM ANAYLYSIS
A shock absorber (shock "damper") is a mechanical or
hydraulic device designed to absorb and damp shock
impulses. It does this by converting the kinetic energy of
the shock into heat which is later dissipated.

3. ROLL OF THE PART IN
APPLICATION -
Shock absorber is a mechanical device designed to damp shock
impulses and dissipate kinetic energy. It reduces the effect of
traveling over rough ground which leads to improved ride quality
and increase in comfort because it substantially reduced
amplitude of disturbances

4. Problem definition
Case study on design of spring used in avenger 150cc as a
shock absorber made up of stainless steel material to find
the the deflection of two wheeler and with two person
with a combined weight of 130 kgs sitting on the two
wheeler

5. Solution with justification

6. Modeling of helical
spring
Vechicle – avenger 150cc
Material -stainless stell spring wire
The spring is measure to have a constant
pitch and a variable diameters at its ends

7. Experimental weight
distribution
The wight of the two wheeler will be
distributed among the front and rare wheel
through the shock absorber
Total load on rare suspension – 85.1 + 74.75= 160.05
Total load on front suspension – 117.54
Load on each shock absorber = 80.02 kg (784.8N) (rare
wheel )

8. Helical coil spring
Use all properties for stainless steel material
 Mean diameter of the coil Dm = 48 + 44.5 /2 = 46.25
 Diameter of the spring wire d = 8mm
 Max outer diameter of the spring is Do = 56mm
spring index C = Dm/d = 5.7mm
 Free length of spring LF = 200 mm
 Solid length of the spring LS = 112 mm (d*nt)

9.  Pitch of the coil P = 14.76 mm
 Wals stress factor K = 1.2674
 Max shear stress in the spring Ⴀ =
310.6 Mpa
 Spring stifness = k =
29.7579N/mm
 Deflection of the spring under
load = ẟ = 25.27mm
Helical coil spring

10. ASSIGNMENT 1
CASE ON DESIGN OF SPRING USED IN SPECIAL PURPOSE APPLICATION

11. PROBLEM ANAYLYSIS
when the operations are carried out the spring undergo
setting and does not attains original length as well as failure
occurs with crack generation and distortion of the spring.
Failure of die set spring may result in sudden depreciation in
the resisting power of the die plate against the press force
applied to form the product and consequently may lead to
threat of injury or damage to the component. Hence the
spring must be designed for reliability and to withstand the
compression loading during operation over its life time

12. ROLL OF THE PART IN APPLICATION
The die set system used in the wire straightening and
cutting system is comprises of die plate, bolster plate
and guides with mutually helped by helical compression
spring. This spring plays an important role in smooth
movement of the die set by supplying energy stored in it
as well as keeps sufficient pressure on the plate provided
at the bottom of the rod. It acts as suspension system to
absorb the shocks and stores the mechanical energy.

13. Problem definition
A Company is facing a lot problems due to
premature failure of the die set spring.
Design a spring under the loading of 1000 N and
maximum allowable deflection of 25 mm

14. Solution with justification

15. Design of die set spring
material – oil tempered spring steel wire
Reason – having excellent corrosion resistance
Maximum operating temperature: 250°F (121°C)

16. Sw = 100 ⁰c
Ⴀ = 0.5 sut (ductile)
Assume value of spring index = C = 8mm
Assume spring has square and ground end
Reason - Buckling is reduced when the ends of
the spring are closed and ground.

17. Find dimensions
 wals factor K= 1.84
 wire diameter
 Ⴀ = 0.5 sut = 0.5*1450 = 710N/mm2
 d = 6mm
 Mean coil diameter = D = 48mm
 Number of active coils
 For oil tempered steel wire of SW grade G=
81375 N/mm2
 ….from V.B.Bhandari book table 10.3
 N = 2.97= 4


18. Total no of coil = 6
Solid length = 36mm
Actual deflection of spring = 33.35mm
Gap between adjacent coil = 2mm
Reason – Assume a gap of 0.5 to 2mm between adjacent coil
when spring under max length
Free length of the spring = solid length+ total gap +ẟ
= 36+10+33.55
= 79.55mm
pitch of the coil = free length /total no of coil-1
= 15.91mm
Rating of spring K = Gd^4/8D^3N = 68655.93