# FLAT SPIRAL SPRING CLASS NOTES FOR MECHANICAL ENGINEERING

# FLAT SPIRAL SPRING

# CLASS NOTES FOR

# MECHANICAL ENGINEERING

**This spring is an important part in almost**

** every machinery. These springs perform **

**different functions in different**

** applications. These absorb shocks in**

** vehicles running on uneven roads. These**

**store energy in toys and circuit breakers.**

** These give cushion effect in sofas and cars.**

** Springs are of different types namely close**

**& open coiled ****helical springs, laminated**

** and flat spiral springs.**** **

### Introduction

#### This spring stores energy. Used in watches, toys, wall clocks, galvanometers. Further, used in steering wheels and balancing springs. Recently pencil cells have replaced these in watches and toys. These are compact. Spiral springs are like a Jalebi. Hence also called Jalebi springs*.* It releases energy as it unwinds.

**Refer Fig.**

#### it consists of a flat rectangular thin metallic strip (width b and thickness t as shown). The spring is wound in a spiral shape in one plane. Inner end named ‘I’ is fixed at the center. The outer end ‘o’ is hinged. The horizontal distance between the fixed end and pinned end is ‘R’ radius of spring. A torque acts on the spring to wind it to storing energy. There will be two reactions at the outer end as R_{y} and R_{x}. Now consider a small length ’dl’ of spring for analysis at distances x and y as shown. Bending moment on the small length element will be

#### dM= R_{y} x—R_{x} y

#### This bending moment tries to close the spring. It tends to increase the number of turns. Thus, bending moment is positive.

#### Rotation caused at point (x, y) will be

#### dФ = (dM/EI)dl

#### dФ = ((R_{y} x—R_{x} y )/EI)dl

#### Integrating over the entire length

#### Ф = =

#### Ф = = (R_{y}/EI) — (R_{x}/EI)

#### The centroid of the spring is at the fixed end.

#### Because x_{1} dl_{1} + x_{2} dl_{2} + x_{3} dl_{3} ——– = LR

#### Therefore

#### M = R_{y} R

#### R_{y} = M/R

#### Ф = (R_{y}/EI) LR=(M/REI) LR = ML/EI

#### Ф = ML/EI

#### Therefore increase in number of turns

#### n _{increase = Ф/2π}

#### The maximum bending moment will occur at point ‘B’ as shown. Point ‘B’ is at 2R from the pinned end.

#### M _{max} = R_{y} x 2R= 2M

** STRESS IN THE SPIRAL SPRING**

#### Maximum σ_{max} = (M _{max} t/2)/I= 12 Mt/bt^{3} = 12M/bt^{2}

#### σ_{max} =12M/bt^{2}

** ENERGY STORED IN THE SPIRAL SPRING**

#### Energy U = (1/2) M Ф = (1/2) M( ML/EI)= 6M^{2}L/Ebt^{3}

#### = 6M^{2}L/Ebt^{3} = 6 (σ_{max} bt^{2}/12)L/Ebt^{3}

#### U = σ^{2}_{max} b t L/24 E= (σ^{2}_{max}/24 E) Volume of the spring

**Resilience is the energy stored per unit volume**

#### u = σ^{2}_{max}/24 E

https://www.mesubjects.net/wp-admin/post.php?post=7602&action=edit Salient features of springs