# CLOSE COILED HELICAL SPRING -AXIAL TORQUE

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

https://www.mesubjects.net/wp-admin/post.php?post=7611&action=edit Salient features of springs-2

https://www.mesubjects.net/wp-admin/post.php?post=7644&action=edit Laminated springs

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https://www.mesubjects.net/wp-admin/post.php?post=4181&action=edit MCQ Springs-1

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https://www.mesubjects.net/wp-admin/post.php?post=7635&action=edit Flat spiral springs

### SPRINGS

### S**pring 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.**

### 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

Φ=length/radius of curvature=L/R

Bending equation gives 1/R =M/EI

Thus Φ = ML/EI

Work done =W= (1/2) Mφ = M^{2}L/2EI

I = πd^{4}/64

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

**Φ** =M 2πRn/(E πd^{4}/64) =**128R n M/Ed ^{4}**

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

Bending stress, σ

**σ**= (M/I)y=[M/(π/64d^{4})] x d/2=** 32M/πd ^{3}**

**ALTERNATIVELY FROM STRAIN ENERGY**

Same expression can also be achieved from the strain energy.

Strain energy in bending is given by

U _{bending }= M^{2}L/2EI

(1/2)M Ф = M^{2}L/2EI

Ф =ML/EI= M 2πRn/(2E (π/64)d^{4} )

Ф =128MRn/Ed^{4} = 64WDn/Ed^{4}

Bending stress, σ

σ= (M/I)y=[M/(π/64d^{4})] x d/2=** 32M/πd ^{3}**

Shear stress ԏ =0

Principal stress σ_{1} = σ_{b}

Maximum shear stress

ԏ_{max} = σ_{b}/2

Angle of twist ϴ =0

Axial deflection δ =0

Resilience

U =M^{2}L/2EI

But M/I =σ/y

M = σI/y = σ I/(d/2)= 2σI/d

M^{2} =( 2σI/d)^{2} = 2σ^{2} I^{2}/d^{2}

Therefore= U =(2σ^{2} I^{2}/d^{2} )L/2EI = 2σ^{2} (π/64)d^{4}/E d^{2} = U =(σ^{2}/8E)(π/4)d^{2} L =(σ^{2}/8E)(Volume of spring wire)

Resilience = u = U/V = σ^{2}/8E

**Stiffness**

Stiffness of spring = k = M/Ф Ed^{4}/64Dn

**FINAL RESULTS**

Stress σ = 32M/πd^{3}

Rotation of free end = Ф =64WDn/Ed^{4}

Resilience = u= σ^{2}/8E

Axial deflection = δ =0