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=D_{o}
inside coil diameter =D_{i}
mean coil diameter = D _{m}
d=diameter of wire
n_{a} = number of active turns (circles) in the spring
n_{i } = 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)d^{4}
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)d^{4} = ԏ /(d/2)
Α = angle of helix = 0 in a closed coil spring
Stress = ԏ= 8WD/πd^{3}
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 = (4C1)/(4C4)+ 0.615/C
Where C is spring index=D/d
Now actual shear stress becomes = kԏ= k 8WD/πd^{3 }(i)
(1/2)W δ =(1/2) Tϴ,
δ = 64 WR^{3}n/Gd^{4} (ii)
Stiffness = k= W/δ = Gd^{4}/64R^{3}n (iii)
Resilience = u = ԏ^{2}/2G =(8WD/πd^{3})^{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/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
https://www.mesubjects.net/wpadmin/post.php?post=7620&action=edit Open coiled helical springaxial load & axial couple