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 Ry and Rx. 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= Ry x—Rx 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 = ((Ry x—Rx y )/EI)dl

Integrating over the entire length

Ф =  =

Ф =  = (Ry/EI)  —  (Rx/EI)

The centroid of the spring is at the fixed end.

Because x1 dl1 + x2 dl2 + x3 dl3 ——– = LR

Therefore

M = Ry R

Ry = M/R

Ф = (Ry/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 = Ry x 2R= 2M

 STRESS IN THE SPIRAL SPRING

Maximum σmax = (M max t/2)/I= 12 Mt/bt3 = 12M/bt2

σmax =12M/bt2

 ENERGY STORED IN THE SPIRAL SPRING

Energy   U = (1/2) M Ф = (1/2) M( ML/EI)= 6M2L/Ebt3

= 6M2L/Ebt3 = 6 (σmax bt2/12)L/Ebt3

U =  σ2max b t L/24 E= (σ2max/24 E) Volume of the spring

Resilience is the energy stored per unit volume

u  = σ2max/24 E

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