CLOSE COILED HELICAL SPRINGS CLASS NOTES FOR MECHANICAL ENGINEERING

CLOSE COILED HELICAL SPRINGS

CLASS NOTES FOR MECHANICAL

ENGINEERING

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

and open coiled helical springs, laminated

and spiral springs. This spring is subjected

to either axial load or axial couple. The

parameters to be studied in springs are

stress, stiffness, deflection and resilience.

Close Coiled Helical Spring

Symbols used

 outer  coil diameter=Do

 inside coil diameter =Di

mean coil diameter = D m

d=diameter of wire

na = number of active turns (circles) in the spring

ni  = Number of inactive turns

C = spring index= D/d  (normally 4 to 12)

k = Wahl’s Stress concentration factor

ԏ = shear stress in spring wire

ϴ = angle of twist in wire

L= length of wire

T = torque= W x R

 Close Coiled Helical Springs – Axial Load

Imagine it as a thread being pulled axially off a spool.

As the helix angle (angle of the turn with the horizontal) is small, axial load only causes twist in the wire of the spring. It is similar to a shaft under pure torsion. It is considered only in pure torsion and the torsion equation is used to analyze it.

Actually stresses present will be

  • Torsion shear stress

  • Bending stress

  • Direct shear stress

Bending stress and direct shear stresses have been found to be negligible. However their small effect has been accounted for by Wahl stress concentration factor.

T = W x R=WxD/2

T/J = ԏ/r = Gϴ/L

J = (π/64)d4

R =radius of wire = d/2

G = Modulus of rigidity

ϴ= Angle of twist in radians

L= length of wire = 2πrn

Using T/J = ԏ /r

(WD/2)/ (π/64)d4 = ԏ /(d/2)

Α = angle of helix = 0 in a closed coil spring

Stress = ԏ= 8WD/πd3

Stress concentration factor accounts for two things

(i) Direct shear stress consideration

(ii) Stress concentration due to change in curvature

Wahl’s stress concentration factor is given by

K = (4C-1)/(4C-4)+ 0.615/C

Where C is spring index=D/d

Now actual shear stress becomes = kԏ= k 8WD/πd(i)

(1/2)W δ =(1/2) Tϴ,

δ = 64 WR3n/Gd4                                                                 (ii)

Stiffness = k= W/δ = Gd4/64R3n                                        (iii)

Resilience = u = ԏ2/2G =(8WD/πd3)2 /2G                          (iv)

Solid length = nd                                                                    (v)

CLOSE COILED HELICAL SPRING -AXIAL TORQUE

This torque will rotate the free end with respect to the fixed end. Thus it will change the coil diameter and will be mainly producing the bending effect. Therefore such springs are designed on the basis of pure bending. Further helix angle ‘α’ is small.

Let φ be the total angle through which the free end of the spring turns relative to the free end when the axial couple is applied.

Use bending equation for its analysis

M/I = σ/y = E/R

σ =(M/I) y

Then M = EI/R

Maximum stress in the spring wire due to axial couple will be

Bending stress, σ

σ= (M/I)y=[M/(π/64d4)] x d/2= 32M/πd3

Shear stress ԏ =0

Principal stress σ1 = σb

Maximum shear stress

ԏmax = σb/2

Angle of twist ϴ =0

Axial deflection δ =0

Resilience

U =M2L/2EI

But M/I =σ/y

M = σI/y = σ I/(d/2)= 2σI/d

M2 =( 2σI/d)2 = 2σ2 I2/d2

Therefore= U =(2σ2 I2/d2 )L/2EI

= 2σ2 (π/64)d4/E d2

= U =(σ2/8E)(π/4)d2 L

=(σ2/8E)(Volume of spring wire)

Resilience = u = U/V =  σ2/8E

Stiffness

Stiffness of spring  = k =  M/Ф Ed4/64Dn

FINAL RESULTS

Stress σ = 32M/πd3

Rotation of free end = Ф =64WDn/Ed4

Resilience = u= σ2/8E

Axial deflection = δ =0

 

https://www.mesubjects.net/wp-admin/post.php?post=7620&action=edit         Open coiled helical spring-axial load & axial couple