## 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 :

**1) LOAD-STRESS EQUATION**

**2) LOAD-DEFLECTION EQUATION**

### 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:**

**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= *π*D. 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):*

*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(M_{t}) given by; ** M _{t}=P.D/2**

Therefore; Torsional shear stresses (**τ _{1}**) on the bar given by;

**τ _{1}=16 M_{t}/πd^{3}= 16 /πd^{3} {P.D/2}=8PD/πd^{3}** —————-

**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):*

*ANALYSIS (CONSIDERING OTHER STRESS EFFECTS):*

Now, Considering** these stress effects** , we define two assumed factors , namely **K _{s} and K_{c}**.

**K _{s} **= Factor to account for Direct Shear Stress

**K _{c}** = Factor to account for Stress Concentration due to coil curvature.

*COMBINED EFFECT OF THESE TWO FACTORS GIVES * K=K_{s}.K_{c} * Where, K= Factor on account for combined effect.

** Let Direct Shear Stress= **τ _{2}**

**τ _{2}=P/A = P/(πd^{2}/4)=4P/πd^{2} **

**τ _{2}=8PD/πd^{3}(0.5d/D)**

Therefore,

**τ**_{total}= **τ**_{1}+** τ _{2}**

=**8PD/πd ^{3}+ 8PD/πd^{3}(0.5d/D)**

=**8PD/πd ^{3}{1+(0.5d/D)}**

**where, 1+(0.5d/D)= K_{s} =Shear stress concentration factor**

* K_{s}*=

**1+(0.5d/D)**=

**1+(0.5/C)**

Therefore , equation now becomes as:

**τ**= **8PD/πd ^{3}{K_{s}}**

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

**τ**= **8PD/πd ^{3}{K}**

**= Stress factor/Wahl Factor**

**K***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).