CLOSE COILED HELICAL SPRING -AXIAL TORQUE      Close Coiled under axial force      Close Coiled Helical Spring-Axial torque         Open coiled helical spring-axial load          Salient features of springs-2          Laminated springs           Salient features of springs-1           MCQ Springs-2             MCQ Springs-1           Q. ANS. Springs          Flat spiral springs


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 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

Φ=length/radius of curvature=L/R

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


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


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 of spring  = k =  M/Ф Ed4/64Dn


Stress σ = 32M/πd3

Rotation of free end = Ф =64WDn/Ed4

Resilience = u= σ2/8E

Axial deflection = δ =0