# CLOSE COILED HELICAL SPRING -AXIAL TORQUE

### 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

Here R is the radius of the coil

L= 2πRn

where n is the number of turns in the spring

Bending equation gives 1/R =M/EI

Thus Φ = ML/EI

Work done =W= (1/2) Mφ = M2L/2EI

I = πd4/64

Substituting the value of ‘I’ in Φ, we get

Φ =M 2πRn/(E πd4/64) =128R n M/Ed4

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

ALTERNATIVELY FROM STRAIN ENERGY

Same expression can also be achieved from the strain energy.

Strain energy in bending is given by

U bending = M2L/2EI

(1/2)M Ф = M2L/2EI

Ф =ML/EI= M 2πRn/(2E (π/64)d4 )

Ф =128MRn/Ed4 = 64WDn/Ed4

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