This is a power point presentation on the design of Helical springs subjected to Static and Fluctuating load. It is part of Design of Machine elements subject.

1. DESIGN OF HELICAL SPRING

2. SPRING SUBJECTED TO STATIC LOAD
OBJECTIVES FOR DESIGN
•It should possess sufficient strength to
withstand the external load .
•It should have the required load
deflection characteristic’s.
•It should not buckle under the external
load.

3. DESIGN PROCEDURE
• For given application, estimate the maximum spring force (P) and the
corresponding required deflection (δ) of the spring. In some case
max. spring force and stiffness k, which is (P/δ) are specified.
• Select fs i.e. factor of safety as 1.5 or less if not specified.
• Select a suitable spring material and find out ultimate tensile
strength Sut from the data. Calculate the Permissible shear stress in
following manner :
Permissible shear stress (τ)
τ =
𝑆𝑠𝑦
1.5
, Assume, Syt = 0.75Sut and Ssy = 0.577Syt
τ =
(0.577)(0.75) 𝑆 𝑢𝑡
1.5
τ ≈ 0.3Sut
In general,
τ ≅ 0.3Sut - 0.5Sut

4. • Assume the suitable value of spring index (C). For industrial
applications, the spring index varies from 8 to 10. A spring index of 8
is considered good value. C can be taken as 5 in valves and clutches.
C should never be less than 3.
• Calculate the Wahl correction factor by following relation :
K =
4𝐶−1
4𝐶−4
+
0.615
𝐶
• Determine wire diameter by following relation :
τ = K
8𝑃𝐶
𝜋𝑑3
• Determine mean coil diameter by following relation :
D = Cd
• Determine the no. of active coils (N) by following relation :
δ=
8𝑃𝑑3 𝑁
𝐺𝑑4 , G = 81370 N/mm2

5. • Determine the style end and find out no. of active coils. Adding total
no. of active and inactive coils find total no. of coils (Nt).
• Determine the solid length by following relation :
Ls = Ntd
• Determine the actual deflection of the spring by following relation:
δ=
8𝑃𝑑3 𝑁
𝐺𝑑4
• Assume gap of 0.5 to 2 mm between adjacent coils, when the spring
is under the action of maximum load, the total axial gap between the
coils is given by :
Total gap = (Nt – 1) * gap between two adjacent coils
• Determine free length of the spring by following relation :
Free length = solid length = total gap = δ
• Determine rate of spring by following relation :
k =
𝐺𝑑4
8𝐷3 𝑁

6. • Prepare list of spring specifications.

7. SPRING SUBJECTED TO FLUCTUATING LOAD
• In many applications the spring is subjected to fluctuating load.
• In such cases spring is designed on the basis of two criteria -
design for infinite life.
- design for finite life.
• Let us consider a spring subjected to an external fluctuating
force, which changes its magnitude Pmax. to Pmin in the load
cycle.
• The mean force Pm and the force amplitude Pa are given by,
Pm =
1
2
(Pmax – Pmin)
Pa =
1
2
(Pmax + Pmin)

8. • The mean stress is calculated from mean force (Pm) by using shear
stress correction factor (Ks). It is given by,
τ= Ks (
8𝑃𝐶
𝜋𝑑2), where Ks = 1 +
0.5
𝐶
Ks is the correction factor for direct shear stress only, it is
applicable only to mean stress only.
• For torsional shear amplitude (τa), it is necessary to also consider
the effect of stress concentration but to curvature in addition to
direct shear stress. Therefore,
τa = KsKc
8𝑃 𝑎 𝐷
𝜋𝑑3
τa = K
8𝑃 𝑎 𝐷
𝜋𝑑3
Where K is the Wahl correction, which takes into account the effect
of direct shear stress as well as of stress correction due to
curvature.

9. • In general the spring wires are subjected to pulsating shear
stresses, which vary from 0 to S’se (endurance limit).
For Patented and cold-drawn steel wires :
S’se = 0.21Sut
Ssy = 0.42Sut
For oil-hardened and tempered steel wires:
S’se = 0.22Sut
Ssy = 0.45Sut
• The general equation used for spring design is :
𝜏 𝑎
𝑆 𝑠𝑦
𝑓 𝑠
− 𝜏 𝑚
=
1
2
𝑆′ 𝑠𝑒
𝑆 𝑠𝑦−
1
2
𝑆′ 𝑠𝑒