Mechanical Springs

MECHANICAL SPRINGS
“Spring is an elastic m/c element, which deflects
under the action of the load and returns to its
original shape when the load is removed.”
Spring Functions:
1) To absorb shocks: vehicle suspension springs,
railway buffer springs, buffer springs in
elevators & vibration mounts for machinery.
2) To store energy: Springs used in clocks, toys,
movie cameras, circuit breakers and starters.
© Dr. V.R Deulgaonkar

3) To measure force: weighing balances and
scales.
4) To apply force and control the motion: e.g.
spring in cam and follower, in rocker arm, in
clutches.

Types of springs
Springs are classified according to their shape,
which can be helical coil of a wire or a piece of
stamping or flat wound-up strip.
Commonly used coil is helical coil which is
further classified as
a) Compression Spring: Spring is compressed
b) Extension Spring: Spring is elongated
c) Close coiled : Helix angle is <10 deg.

Compression and extension Springs

d) Open Coil : Helix angle is > 100
Advantages of Helical Springs:
1) Easy to manufacture.
2) Cheaper than other types of coils.
3) High reliability.
4) Deflection is linearly proportional to the
force acting on the spring.

Spring terminology

Spring Index (C)
It is defined as the ratio of mean coil diameter
to wire diameter.
C = D/d
C indicates the relative sharpness of the
curvature of the coil. Low spring index means
high curvature.
In practical applications spring index varies from
4 to 12.

1) Solid Length: It is defined as the axial length
of the spring which is so compressed, that
the adjacent coils touch each other. In this
case the spring is completely compressed
and no further compression is possible
Solid Length = Nt d
Nt is total number of coils

2) Compressed length: It is defined as the axial
length of the spring, that is subjected to
maximum compressive force. In this case
spring is subjected to maximum deflection δ.
When the spring is subjected to maximum force
there should be some gap to prevent clashing
of coils. (15% of total deflection) or
Total gap = (Nt -1) x gap between adjacent coils

3) Free length : It is defined as the axial length
of an unloaded helical compression spring. In
this case no external force acts on the spring.
Free length = solid length + total axial gap + δ
Pitch p = (Free length) / (Nt -1)

Active and inactive coils
1) Active coils (N): are those coils in the spring,
which contribute to the spring action, support
and deflect under the action of external force.
2) Inactive coils: is the portion of end coils
which is in contact with the seat and does not
contribute to the spring action.
Inactive coils = Nt - N

Helical Torsion Spring
It is similar to helical
compression or extension
spring, except the ends
are formed in such a way
that the spring is loaded
by a torque about the axis
of coils.
Uses:
It is used to transmit torque
to a m/c component .

It is used in door hinges ,brush holders,
automobile starters and door locks.
It is noted that this spring is subjected to
bending stress and not torsional stress.
For a wire of circular c/s y = d/2 and I = πd4 /64
The bending stress σb = k (Mb y /I); k is stress
concentration factor due to curvature.

Stress concentration factors for inner and outer
fibers of the coil by AM Whal are
Ki = (4C2 – C-1)/4C(C-1)
Ko = (4C2 + C-1)/4C(C+1)
Strain energy stored in spring is
U = ʃ (Mb)2 dx /2EI, which is integrated over the
whole length of the wire i.e. 0 to πDN

• Stiffness of helical torsion spring is defined as
the bending moment required to produce unit
angular displacement.
k = Mb / θ = Pr/θ = Ed4/64DN

Spring Materials
Selection of spring material depends on factors
1) Load acting on the spring.
2) Range of stress through which the spring
operates.
3) Environmental Conditions: temperature and
corrosive atmosphere.
4) Severity of deformation while making the
spring.

Standards of spring
IS 4454-1981: Specifications of steel wires for
cold formed steels.
IS 7906-1975 : Helical Compression Springs.
IS 7907-1976: Helical Extension springs

Materials
1) Patented and Cold-drawn steel wires
2) Oil hardened and tempered steel wires and
valve spring wires
3) Oil hardened and tempered steel wires
(alloyed)
4) Stainless steel spring wires.
Commonly used material is high-carbon hard-
drawn or patented and cold-drawn steel wire

Patented and cold-drawn steel wires are
prominently used in springs subjected to static
and moderate fluctuating forces. There are four
grades of this wire;
1) Grade 1: is used in springs subjected to static or
low load cycles.
2) Grade 2 : is used in springs for moderate load
cycles.
3) Grade 3 : is used for moderate dynamic load or
highly stressed static springs

Grade 4 : is suitable for springs subjected to
severe stresses
Their G = 81370MPa
Second group of steel wires is unalloyed oil-
hardened and tempered spring steel wires
(0.55-0.75 % carbon) and valve spring wires
(0.60 to 0.75 % carbon). There are two grades
of them viz. SW & VW

Grade SW is suitable for springs subjected to
moderate fluctuating stresses and Grade VW
is used when the spring is subjected to high
magnitude of fluctuating stresses. Their
limiting temperatures are 1000 and 800 C
respectively.

Varieties of alloy steel wires are
1) Chromium-Vanadium steel (0.48-0.53%
Carbon & 0.80-1.10% Chromium and 0.15%
Vanadium).
2) Chromium-Silicon steel (0.51-0.59% carbon ,
0.60-0.80% chromium and 1.2-1.6% silicon).
Used for applications involving higher
stresses , impact or shock loads e.g. in
pneumatic hammers.

Stress and Deflection Equations for
helical springs
For the design of helical springs there are two
basic equations viz. Load-stress and Load-
deflection equation.
From fig. D = mean coil diameter, d = wire
diameter.
N = Number of active coils in this spring.
P = Axial force to which this spring is subjected.

• Fig • Fig

• When the wire of helical spring is uncoiled and
straightened, it takes the shape of the bar.
• The dimensions of this equivalent bar are as
under:
1) Diameter of bar =wire diameter of spring = d
2) Length of one coil in the spring = π D
For N active coils length = π DN
3) Bar is fitted with a bracket at each end. Length of
this bracket = D/2

a) Helical Spring b) Helical Spring : Unbent

• Torsional shear stresses are induced in the
bracket due to force P acting at the end of
bracket.
• Torsional moment Mt is given by Mt = PD/2
• Torsional Shear stresses in the bar is
τ1 = (16Mt )/πd3 = (16PD/2)/ πd3
τ1 = 8PD/ πd3 ----------- (a)

• When the bar is bent in the form of a helical
coil there are additional stresses due to
• 1) direct or transverse shear stress in the
spring wire.
• 2) when the bar is bent in the form of a coil
the length of inside fibre is less than the
length of outside fibre, which results in stress
concentration at the inside fibre.

• Resultant shear stress consists of
superimposition of torsional shear stress,
direct shear stress and additional stresses due
to curvature of the coil. Fig.
• To account for this equation (a) is modified by
assuming two factors as KS = factor to account
for direct shear stress & KC = factor to account
for stress concentration due to curvature.

• The combined effect of these two factors is
given as K = KS KC.
• From fig. the direct shear stress in the bar is
given as τ2 = P/A = 4P/(πd2)
• τ2 = 4P/(πd2) = [8PD/ πd3 ] x[0.5d/D] ----- (b)
• Superimposing the two stresses, we get τ as
τ = τ1 +τ2 = [8PD/ πd3 ] + [8PD/ πd3 ] x[0.5d/D]

τ = [8PD/ πd3 ] {1+ (0.5d/D)} -----(c)
The shear stress concentration factor is Ks
defined as Ks = 1+(0.5d/D) = 1+(0.5/C) --- (d)
Using in (c), it becomes τ = Ks [8PD/ πd3 ] --(e)
AM Whal derived equation for resultant stress
that includes torsional shear stress, direct
shear stress and stress concentration due to
curvature given as

• τ = K[8PD/ πd3 ]---(f) K is Stress or Whal Factor
• K = {(4C-1)/(4C-4)} +[0.615/C] ---- (g)
• C is the spring index.
for normal applications spring is designed by
Whal factor. When the spring is subjected to
fluctuating stresses the two factors Ks & Kc are
separately used.

Deflection equation of spring
• The angle of twist for the bar shown is given
by θ = Mt l /JG;
θ = angle of twist (radians)
Mt = torsional Moment PD/2,
l = length of the bar = πDN,
J = Polar M.I = π d4/32
G= Modulus of Rigidity

• Using these values we get,
θ = {(PD/2)(πDN)}/(π d4/32)(G)
θ = 16PD2N/Gd4 ----- (*)
• The axial deflection ‘δ’ of the spring for small
values of θ is
δ = θ x length of bracket i.e. θx(D/2)
Using in * we get,
δ = 8PD3 N/Gd4 ----(1)
---- is load deflection equation.

• The rate of spring is k = P/δ using this in
equation (1), we get
K = Gd4 / 8D3 N------(2)
• Strain energy stored in the spring is E given as
E = Pδ/2

Styles of End
Types of ends
Number of
active
turns (N)
Plain Ends Nt
Plain Ends (Ground) (Nt - 0.5)
Square Ends ( Nt - 2 )
Square ends
(ground)
( Nt - 2 )

Design of helical & tension springs
• Objectives for the design of helical spring are
(i) It should posses sufficient strength to
withstand external load
(ii) It should possess the required load-
deflection characteristics.
(iii) It should not buckle under external load.
Number of springs can be designed for given
applications by changing the three basic
parameters d, D & N

• As there are practical limitations on these
parameters like space limitations e.g. spring
fits in a hole of certain diameter (Do is
restricted), spring fits over a rod (Di is
restricted), designer should specify the
limitations on these parameters before design.

• Dimensions to be calculated in the spring
design are d, D, & N.
• d & D are calculated by load-stress equation
and N is calculated by load-deflection
equation.

• Using equation τ = K [8PD/ πd3] and D/d = C, we
get
τ = K[8PC/ πd2]-------- (1)
• FACTOR OF SAFETY
The f.o.s used in spring design is 1.5 or less.
Justification:
1) In most of the applications springs operate with
well defined deflections, so the forces and
corresponding stresses on the spring are precisely
calculated.

2) For helical compression springs an overload
will simply close the gaps between the coils
without a dangerous increase in deflection
and stresses.
3) For helical extension springs, generally
overload stops are provided to prevent
excessive deflection and stresses.
4) The spring material is carefully controlled at
all stages of manufacturing.

• So f.o.s based on torsional yield strength (Ssy )
is 1.5 for springs subjected to static force.
τ = (Ssy ) /1.5 ------ (a)
• Assuming Syt = 0.75 Sut & Ssy = 0.577 Syt ;
(a) becomes τ = {(0.577 x 0.75) Sut }/1.5
τ = 0.3 Sut ------ (b)
• IS 4454-1981 suggests τ = 0.5 Sut to be used
in design

Steps in design
1) Estimate the max. force P, and deflection δ of
spring for given application.
2) Select a suitable material and find out the
ultimate tensile strength from data. Calculate
permissible shear stress for spring wire by using
relation τ = 0.3 Sut = 0.5 Sut
3) Assume suitable spring index ‘C’ (Varies form 8
to 10 for industrial applications). C for valves
and clutches is 5. C should never be less than 3.

4) Calculate Whal Factor by equation
K = {(4C-1)/(4C-4)} +[0.615/C]
5) Determine wire diameter by
τ = K [8PC/ πd2]
6) Determine the mean coil diameter D by
D = C d
7) Find number of active coils N by using
δ = 8PD3 N/Gd4 (G= 81370MPa)

8) Decide the styles of end according to
configuration of spring & find the number of
inactive coils. Adding active & inactive coils
find the total number of coils.
9) Find the solid length of the spring by using
equation Solid Length = Nt d
10) Determine the actual deflection of the
spring by δ = 8PD3 N/Gd4

11) Assume a gap of 0.5 to 2 mm between the
adjacent coils, when the spring is under the
action of maximum load. The total axial gap
between the coils :
Total gap = (Nt -1) x (gap between two adjacent
coils)
For few cases total axial gap is taken as 15% of
the maximum deflection

12) Find the free length of the spring by
Free length = solid length + total gap + δ
13) Find the pitch of the coil as
p = (free length) / (Nt -1)
14) Determine the rate of the spring as
k = Gd4 / 8D3 N
15) Make a list of spring specifications

• A helical compression spring , too long as
compared with mean coil diameter acts as a
flexible column and may buckle at
comparatively low axial load. So the spring
should be buckle proof. The compression
springs which cannot be designed buckle
proof must be guided in sleeve or over an
arbor. Thumb rule to provide guides is as

If
(Free length) /(Mean coil diameter) <= 2.6
------------- Guide is not necessary
(Free length) /(Mean coil diameter) > 2.6
----------------- Guide is necessary

Springs in series and Parallel
Connections
• Objectives of series and parallel connections
are a) to save the space
b) to change the rate of the spring at certain
deflection
c) to provide a fail-safe design

• Fig. shows two springs
connected in series with
spring rates k1 and k2 ,for this
i) Force acting on each spring is
same and equal to external
force & ii) total deflection is
the sum of individual
deflections of each spring

δ = δ1 + δ2 --- (a)
δ1 , δ2 being the deflections of two springs.
We know that
δ = P/k ; hence δ1 = P/ k1 δ2 = P/ k2 --(b)
Using (b) in (a), P/k = P/k1 + P/ k2
1/k = 1/ k1 + 1/ k2 --- (c)
k is combined stiffness of the springs

fig. shows springs connected in
parallel.
For this 1) force acting on the spring
combination is the sum of forces
of individual springs 2)
Deflection of individual springs is
same and equal to the deflection
of the combination.
P = P1 + P2 ---- (d)
But P = kδ hence kδ = k1δ + k2δ so
k = k1 + k2 ---(e)

Concentric helical (Nested) spring
• It consists of two helical compression springs
one inside the other having same axis.
• In general there are two springs, but in certain
applications it consists of three coaxial springs
namely inner, middle and outer springs. If the
outer spring has RH.helix, the inner spring
always has LH. Helix &vice versa.
• Adjacent springs have opposite hand to
prevent locking of coils

Advantages
• As there are two springs the load carrying
capacity is increased and heavy load can be
transmitted in a restricted space.
• Operation of the mechanism continues if one
of the springs breaks. This results in fail safe
design.
• The spring vibrations called surge is
eliminated.

• Such springs are used as valve springs in heavy
duty diesel engines, aircraft engines and rail
road suspensions, governors of variable speed
engines to account for fluctuating centrifugal
force
• Radial clearance between the two springs is
given as c = (d1-d2)/2
• From fig. c = [(D1-d1)/2] –[(D2+d2)/2]

i.e. c = [(D1-D2)/2] – [(d1+d2)/2]
But c = (d1-d2)/2
(d1-d2)/2 = [(D1-D2)/2] – [(d1+d2)/2]
d1 = [(D1-D2)/2] -----(*)
We know that C = D/d
D = C x d and D1 = C x d1 ; D2 = C x d2

Using in (*), we get
d1 = (Cd1 - Cd2 )/2
2d1 = C d1 - Cd2
( d1 /d2 ) = (C/C-2)------(a) is used in design of
concentric springs

Multi-Leaf springs

Construction
• It consists of a series of flat plates of semi-
elliptical shape as shown in fig. above.
• Flat plates are known as leaves of the spring
and they have graduated lengths.
• The length gradually decreases from top leaf
to the bottom leaf.
• Longest leaf at the top is called master leaf,
which is bent at both the ends to form spring
eye.

• Two bolts are inserted through these two eyes
to fix the leaf spring to the automobile body.
• Leaves are held together by means of two U-
bolts and a centre clip.
• Rebound clips are provided to keep the leaves
in alignment and prevent lateral shifting of
leaves during operation.
• Leaf spring is supported on axle at the centre.

• Multi-leaf springs are provided with one or
two full length leaves in addition to master
leaf.
• Extra full length leaves are provided to
support the transverse shear force.
• For analysis leaves are divided into two groups
viz. 1) Master leaf along with graduated-
length leaves 2) Extra full length leaves.

Analysis of Leaf-Spring
Notations used in analysis are;
nf = number of extra full-length leaves
ng = no. of graduated-length leaves including
master leaf.
n = total number of leaves
b = width of each leaf (mm)
t = thickness of each leaf (mm)

L = length of cantilever or half the length of
semi-elliptic spring (mm)
P = force applied at the end of the spring (N)
Pf = portion of P taken by extra full-length
leaves (N)
Pg = portion of P taken by graduated-length
leaves

• Group of graduated leaves along with master
leaves is treated as a triangular plate of
thickness t and width at the support being
given as ng b.
• The bending stress in the plate at the support
is (σb)g = Mby/I
(σb)g = (PgL)(t/2)/(ng bt3 /12)
(σb)g = (6PgL)/ng bt2)---- (a)

The deflection δg at the load point of the
triangular plate is given by δg = Pg L3/2EImax
δg = (6PgL3)/Eng bt3) --- (b)
Similarly ,extra full-length leaves can be treated
as a rectangular plate of thickness t and width
nf b. The corresponding bending stress is
(σb)f = (6PfL)/nf bt2) ---- (c)

The deflection δf at the load point of the
triangular plate is given by δf = Pf L3/3EI
δf = (4PfL3)/(E nf bt3) ---- (d)
we know δf = δg : & P = Pf + Pg -- (e)
(6PgL3)/Eng bt3) = (4PfL3)/(E nf bt3)
Pg /Pf = 2ng/3nf ---- (f)

From (e) & (f)
Pf = 3nf P/(3nf+2ng) ---- (g)
Pg = 2ng P/(3nf+2ng) ---- (h)
Using (g),(h) in (a) & (c), we get
(σb)g = (12PL)/(3nf+2ng)bt2
(σb)f = (18PL)/(3nf+2ng)bt2

• From deflection equation we infer that
(σb)f is 50% more that (σb)g
deflection at the end of spring is
δ = (12PL3)/(3nf+2ng)Ebt3
Multi-leaf springs are designed using load-
stress and load deflection equations
Nominal thickness(mm): 3.2,
4.5,5,6,6.5,7,7.5,8,9,10,11,12,14 & 16

• Nominal width (mm):
32,40,45,50,55,60,65,70,75,80,90,100 &125.
• Materials for leaf springs: These are usually
made of steels 55SiMn90, 50Crl, 50CrlV23.
The plates are hardened and tempered.
• f.o.s based on yield strength is 2 to 2.5 for
automobile suspension

Nipping of leaf spring

• We know that (σb)f is 50% more that (σb)g
• Pre-stressing is done to equalize the stresses
in different leaves of the spring.
• The pre-stressing is achieved by bending the
leaves to different radii of curvature, before
their assembly with the centre clip.
• From fig. we see that, radius of curvature of
full-length leaf > adjacent graduated-length
leaf.

• The radius of curvature decreases with shorter
leaves.
• The initial gap C between the extra full-length
leaf and graduated-length leaf before the
assembly is called a nip.
• The pre-stressing is achieved by a difference in
radii of curvature, known as “Nipping”
• Common in automobile suspension springs

Surge in spring
• When the natural frequency of the vibrations
of spring coincides with the frequency of
external periodic force which acts on it,
resonance occurs. During this the spring is
subjected to a wave of successive
compressions of coils that travels from one
end to other end and back. This type of
vibratory motion is called Surge of spring.

SHOT PEENING
• When springs are subjected to fatigue loading,
poor surface finish reduces the endurance
strength and acts as a source of stress
concentration.
• The fatigue crack begins with some surface
irregularity and propagates due to tensile
stresses.
• To reduce the chances of fatigue failure due to
surface cracks, residual compressive stresses

are induced in the surface of the spring wire.
one of the most commonly used method for
this purpose is shot peening.
In this, small steel balls are impinged on the wire
surface with high velocities either by air-blast
or by centrifugal action.
The balls strike against the surface and induce
residual compressive stresses.

• The depth of layer of residual compressive
stresses depends upon the number of factors
as size of the balls, velocity of striking, original
hardness and ductility of spring wire.
• Shot peening is effective for springs loaded
only in one direction, e.g. helical compression,
helical extension, or torsion bar springs.

Mechanical Springs

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