# EQUATIONS FOR HELICAL SPRING DESIGN:STUDY ## EQUATIONS INVOLVING HELICAL SPRINGS(COMPR./EXT.) DESIGN :

Equations came in derivation from many corners of the world ; so that Helical Springs’ Design can be done.

From the following, two equations came in selection, namely :

### ANALYSIS ,STUDY AND OBSERVATIONS

Consider a helical spring that is made from circular cross section wire as shown.

Here, D and d are Mean coil diameter and wire diameter respectively.

No. of active coils in the spring=N ;

and the spring subjected to Axial force =P

Now, let’s uncoil this spring and straighten it in linear direction. Therefore, it takes the shape of a bar as shown below.

In this analogy, stress equation derived for the bar to be taken equivalent to the actual helical spring.

#### DIMENSIONS OF EQUIVALENT BAR:

1. The diameter of the bar always taken equal to the Wire diameter of the spring (d).

2. We know; length of one coil in the spring= πD. And ,we have N active coils. Therefore, length of equivalent bar = πDN

3.The bar equipped with brackets on either ends. The length of this bracket equal to the Mean coil radius of spring (D/2).

#### ANALYSIS (BASIC):

##### A) RELATION BETWEEN STRESS AND LOAD :

**Force P acts at the bracket end , due to which Torsional shear stress induces in the bar. Torsional Moment(Mt) given by; Mt=P.D/2

Therefore; Torsional shear stresses (τ1) on the bar given by;

τ1=16 Mt/πd3= 16 /πd3 {P.D/2}=8PD/πd3 —————- Eq.1

During uncoiling of spring and converting it to a bar , additional stresses are induced in the system due to following two factors:

1) Presence of direct or transverse shear stress in the spring wire.

2)During the bending of bar in the form of a coil, the length of inner fibres are less than the length of outside fibres ;which leads to stress concentration.

In practical study, the Resultant stress developed in the spring consists of Superimposition of Torsional shear stress, direct shear stress , and additional stresses occurring due to curvature of the coil.

The Stress Distribution under various conditions have been shown below:

The Eq.1 calculated above was derived on the assumption that there is no effect of Direct Shear Stress and Stress Concentration due to coil curvature.

###### ANALYSIS (CONSIDERING OTHER STRESS EFFECTS):

Now, Considering these stress effects , we define two assumed factors , namely Ks and Kc.

Ks = Factor to account for Direct Shear Stress

Kc = Factor to account for Stress Concentration due to coil curvature.

*COMBINED EFFECT OF THESE TWO FACTORS GIVES K=Ks.Kc Where, K= Factor on account for combined effect.

** Let Direct Shear Stress= τ2

τ2=P/A = P/(πd2/4)=4P/πd2

τ2=8PD/πd3(0.5d/D)

Therefore,

τtotal= τ1+ τ2

=8PD/πd3+ 8PD/πd3(0.5d/D)

=8PD/πd3{1+(0.5d/D)}

where, 1+(0.5d/D)= Ks =Shear stress concentration factor

Ks= 1+(0.5d/D)=1+(0.5/C)

Therefore , equation now becomes as:

τ= 8PD/πd3{Ks}

This Equation was later experimentally verified by A.M.WAHL. Therefore equation now becomes:

τ= 8PD/πd3{K} K= Stress factor/Wahl Factor

This Equation Came to be known as the LOAD STRESS EQUATION.

K=(4C-1)/(4C-4) + 0.615 /C where , C= D/d= spring index

#When Spring subjected to fluctuating stresses ; Ks and Kc are separately used in the equation.

##### B)RELATION BETWEEN STRESS AND DEFLECTION :

This Eq.3 is called LOAD DEFLECTION EQUATION.

#### VARIATION OF SPRING PARAMETERS:

#When a spring is cut into 2 parts ; following observed changes in the equations ;

1.Parameters G, d and D remains same.

2.N becomes N/2.

3.Stiffness (k) becomes 2k.

#### ANALYSIS OF STRAIN ENERGY IN SPRINGS:

As observes from Eq.3 , we can draw a plot between Load(P) and Deflection(δ).