Mechanical Springs Guide

Prepared By:
Suraj Kumar Chand
Assistant Professor
02-12-2017 Departmentof ME, GEC Raipur 1

CONTENT
• Introduction
• Function andApplications
• Types of spring
• Terminology of helical spring
• Styles of end
• Stress and Deflection equation
• Series and Parallel equation
• Spring materials and manufacturing process
• Design of helical spring
• Surge in spring
• Helical torsion springs
• Multileaf spring
• Equalised stress in spring leaves (nipping)

INTRODUCTION
• Springs are flexible machine elements used primarily to deflect under load with the
ability to return to its original shape when unloaded.
• They are designed to store energy, measure a force, or absorb shocks and vibrations.
• The popular types of mechanical springs are helical compression spring, helical
extension spring, helical torsion spring, and multileaf spring.
• There are three criterions in design of springs which is sufficient strength to withstand
external load, desired load deflection characteristic and sufficient buckling strength.

FUNCTIONS AND APPLICATIONS
• Springs are used to absorb shocks and vibrations, e.g., vehicle suspension springs,
railway buffer springs, buffer springs in elevators and vibration mounts for
machinery.
• Springs are used to store energy, e.g., springs used in clocks, toys, movie-cameras,
circuit breakers, and starters.
• Springs are used to measure the force, e.g., springs used in weighing balance and
scales.
• Springs are used to apply force and control motion, e.g., in cam follower
mechanism to maintain the contact between them, in clutch it provides the
required force to engage the clutch.

TYPES OF SPRINGS
1. Helical spring : the helical spring is made from wire, usually of circular cross-section,
which is bent in the form of a helix. There are two types of helical spring-
I. Helical compression spring II. Helical extension spring

• In helical compression spring, the external force tends to shortens the spring while in
helical extension spring, the external force tends to lengthen the spring.
• In both the cases, the external force acts along the axis of the spring and induces torsional
shear stress in the spring wire. The words ‘compression’ and ‘extension’ are related to
total spring and not to stresses in spring wire.
• The helical springs are sometimes classified as closely-coiled helical spring and open-
coiled helical spring.
• A helical spring is said to be closely-coiled helical spring, when the spring wire is coiled so close that
the plane containing each coil is almost at right angle to the axis of helix. (Helix angle ≤ 𝟏𝟎𝒐)
• A helical spring is said to be closely-coiled helical spring, when the spring wire is coiled in such a way,
that there is a large gap between adjacent coil. (Helix angle > 𝟏𝟎𝒐)
ADVANTAGES
• They are easy to manufacture.
• They are cheaper than other types of springs.
• Their reliability is high.
• The deflection of spring is linearly proportional to the force acting on the spring.

2. Helical torsion spring :
• The construction of this spring is similar to that of compression or extension spring, except that the ends
are formed in such a way, that the spring is loaded by a torque, about the axis of the coils.
• It is used to transmit a torque (P x r) to a particular component in a machine or a mechanism.
• It is used in door-hinges, brush holders, automobile starters and door locks.
• The helical torsion spring resists the bending moment (P x r), which tends to wind up the spring. The
bending moment induces bending stresses in the bending wire.

3. Multileaf spring : A multileaf or laminated spring consists of a series of flat plates,
usually of semi-elliptical shape. The flat plates, called leaves, have varying lengths. The
leaves are held together by means of U-bolts and center clip. The longest leaf, called
master leaf, is bent at the two ends to form spring eyes. The leaves of multileaf spring
are subjected to bending stresses. They are widely used in automobile and railroad
suspensions.

TERMINOLOGY OF HELICAL SPRING
The main dimensions of helical spring subjected to compressive force are as follows:
d = wire diameter of spring (mm)
Di = inside diameter of spring coil (mm)
Do = outside diameter of spring coil (mm)
D = mean coil diameter (mm)
Therefore,
𝐷 = 𝐷𝑖+𝐷0
2

• There is an important parameter in spring design called spring index. It is denoted by
letter C. the spring index is defined as the ratio of mean coil diameter to the wire
diameter,
𝐷
𝐶 =
𝑑
• The spring index indicates the relative sharpness of the curvature of the coil-
A low spring index means high sharpness of curvature. When the spring index is low (C < 3),
the actual stresses in the wire are excessive due to curvature effect. Such a spring is difficult to
manufacture and special care in coiling is required to avoid cracking in some wires.
 when the spring index is high (C > 15), it results in large variation in coil diameter. Such a
spring is prone to buckling and also tangles easily during handling.
Therefore spring index from 4 to12 is considered better for manufacturing consideration.

1. Solid length : Solid length 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 = 𝑁𝑡d
Where 𝑁𝑡 = total number of coils
2. Compressed length : compressed length is defined as the axial length of the spring, which is
subjected to maximum compressive force. In this case, the spring is subjected to maximum
deflection δ. There should be some gap between the coils at the maximum deflected position to
avoid clashing between them. The clashing allowance or total axial gap is usually taken as 15%
of the maximum deflection. Sometimes an arbitrary decision is taken and it is assumed that there
is a gap of 1 mm or 2 mm between adjacent coils under maximum load condition, in this case-
Total gap = (𝑁𝑡-1) x gap between adjacent coils
3. Free length: Free length is defined as the axial length of an unloaded helical compression spring.
In this case there is no external force acts on the spring.
Free length = compressed length + δ
= solid length + total axial gap + δ

4. Pitch : The pitch of the coil is defined as the axial distance between adjacent coils in
uncompressed state of spring. It is denoted by p.
p = 𝐹𝑟𝑒𝑒 𝑙𝑒𝑛𝑔𝑡ℎ
(𝑁𝑡−1)
5. Stiffness : The stiffness of the spring is defined as the force required to produce unit
deflection. Therefore,
k = 𝑃
δ
6. Active and inactive coils: Active coils are the coils in the spring, which contribute to
spring action, support the external force and deflect under the action of force. A portion
of the coil at each end, which is in contact with the seat, does not contribute to spring
action and called inactive coil. These coils do not support the load and do not deflect
under the action of external force. The number of inactive coils is given by,
Inactive coils = (N𝑡 - N)

STYLES OF END : Helical compression spring
Type of ends Numbers of active turns
Plain ends 𝑁𝑡
Plain ends (ground) (𝑁𝑡 − 1/2)
Square ends (𝑁𝑡 − 2)
Square ends (ground) (𝑁𝑡 − 2)

STYLES OF END : Helical extension spring
• For helical extension springs, all coils are active coils. The number of active coils (N) is the same as
the total number of coils (𝑁𝑡).

STRESS AND DEFLECTION EQUATION
There are two basic equations for the design of helical springs :
1. Load-stress equation
2. Load-deflection equation
Load-stress Equation

The dimensions of equivalent bar are as follows:
• The diameter of the bar is equal to the wire diameter of the spring (d).
• The length of one coil in the spring is (πD). There are N such active coils. Therefore, the length of
equivalent bar is (πDN).
• The bar is fitted with bracket at each end. The length of this bracket is equal to mean coil radius of the
spring (D/2).
Torsional moment, 𝑀𝑡 = 𝑃𝐷
2
Torsional shear stress, τ1 = 16𝑀𝑡
π𝑑3
or
8𝑃𝐷
π𝑑3
…..(a)
When the equivalent bar is bent in the form of helical coil, there is additional stresses on
account of following two factors:
i. There is a direct or transverse shear stress in the spring wire.
ii. When the bar is bent in the form of coil, the length of inner fibre is less than the length of outside fibre.
This results in stress concentration at the inside fibre of the coil.

The resultant stress consists of superimposition of torsional shear stress, direct shear stress
and additional stresses due to the curvature of the coil.

There are two factors to account for these effects:
𝐾𝑆= Factor to account for direct shear stress
𝐾𝐶= factor to account for stress concentration due to curvature effect
The combined effect of these two factors is given by,
K = 𝐾𝑆𝐾𝐶
The direct shear stress in the bar is given by
2
τ =
4
=
π 2 3
𝑑2 π𝑑 π𝑑 𝐷
𝑃 4𝑃
=
8𝑃𝐷
(0.5𝑑
) ......(b)
Resultant shear stress, τ = τ1 + τ2
π𝑑3 𝐷
=
8𝑃𝐷
(1 + 0.5𝑑
) …...(c)
𝑆 𝑆 𝐷
𝐶
The shear stress correction factor (𝐾 ) is defined as, 𝐾 = (1 + 0.5𝑑
) or (1 + 0.5
)
Hence
8𝑃𝐷
τ = K (π𝑑3) …..(d)
Where K is called stress factor or Wahl factor and is given by
K =
4𝐶−1
+ 0.615
4𝐶−4 𝐶

Load-deflection equation:
The angle of twist for the equivalent bar is given by,
𝐽
𝐺
θ = 𝑀𝑡 𝑙
= (𝑃𝐷/2)(π𝐷𝑁)
π𝑑4
32
𝐺
or
2
θ = 16 𝑃𝐷 𝑁
𝐺𝑑4
…..(e)
The axial deflection ‘δ’of the spring, for
small value of θ, is given by,
3
δ = θ x (D/2) = 8𝑃𝐷 𝑁
𝐺𝑑4
…..(f)
The rate of spring (k) is given by,
k = 𝑃
=
𝐺𝑑4
δ 8𝐷3𝑁
…..(g)
The energy stored in the spring E =Area under load-deflection line
2
= 1
𝑃δ

SERIES AND PARALLEL CONNECTIONS
Series connection Parallel connection
=
𝑘 𝑘2
1 1 1
+
𝑘2
k = 𝑘1+ 𝑘2

SPRING MATERIALS
There are four basic varieties of steel wire which are used in springs in the majority of
applications-
I. Patented and cold-drawn steel wires (unalloyed)
• There are two terms related to this type of spring wire which is ‘patenting’ and ‘cold drawing’.
Patenting is defined as heating the steel above the critical range and followed by rapid cooling.
This operation produces a tough uniform structure that is suitable for severe cold working.
• They are made of high carbon steel and contain 0.85-0.95% carbon.
• Least expensive.
• Patented and cold-drawn steel wires are mainly used in springs subjected to static forces and
moderate fluctuating forces.

II. Oil-hardened and tempered spring steel wires and valve spring wires
• Oil hardened and tempered spring steel wire contains 0.55-0.75% carbon. The wire is cold drawn and
then hardened and tempered.
• Valve spring wire contains 0.60-0.75% carbon. It is the highest quality of oil hardened and tempered steel
wire.
• There are two grades of unalloyed, oil-hardened and tempered spring steel wire and valve spring wire
which is SW and VW.
• Grade SW is suitable for spring subjected to moderate fluctuating stresses, where as grade VW is
recommended when the spring is subjected to a high magnitude of fluctuating stresses.
III. Oil-hardened and tempered steel wires (alloyed)
• There are two varieties of alloy steel wires, namely chromium-vanadium steel and chromium-silicon
steel.
• Chromium-vanadium steel contains 0.48-0.53% carbon, 0.80-1.10% chromium, and 0.15% vanadium.
These wires are used for applications involving higher stresses and for springs subjected to impact or
shock loads.
• Chromium-silicon spring steel contains 0.51-0.59% carbon, 0.60-0.80% chromium and 1.2-1.6% silicon.
These are also used for shock or impact loading.

IV. Stainless steel spring wires.
• Stainless steel springs, which exhibit an excellent corrosion resistance, are ideal to work in steam or
other corrosive medium.
• Costly.
SPRING MANUFACTURING PROCESSES
• If springs are of very small diameter and the wire diameter is also small then the springs
are normally manufactured by a cold drawn process through a mangle.
• However, for very large springs having also large coil diameter and wire diameter (greater
than 6 mm) one has to go for manufacture by hot processes. First one has to heat the wire
and then use a proper mangle to wind the coils.

UTS-EXPONENTIAL RELATIONSHIP
• It is observed that the ultimate tensile strength of spring material varies significantly with
wire diameter. Therefore, ultimate tensile strength cannot be specified unless the wire
diameter is known.
• Anumber of experiment is carried out to find
out the relationship between the ultimate tensile
strength and the wire diameter and it is
concluded that the graph of ultimate tensile
strength versus wire diameter is almost a straight
line when plotted on a log-log paper. The equation
of this straight line is given by,
𝑢
𝑡
𝑆 =
𝐴
𝑑𝑚
where,A= intercept of straight line
m = slope of straight line

DESIGN OF HELICAL SPRINGS
There are three objectives for design of the helical spring-
• It should possess sufficient strength to withstand the external load.
• It should have the required load-deflection characteristics; and
• It should not buckle under the external load.
The main dimensions to be calculated in the spring design are- wire diameter, mean coil
diameter and the number of active coils. The first two are calculated by load-stress equation,
while the third is calculated by load-deflection equation.
The factor of safety in design of spring is usually 1.5 or less.
τ =
𝑆𝑠𝑦
1.5
Assuming, 𝑆𝑦𝑡 = 0.75 𝑆𝑢𝑡 and 𝑆𝑠𝑦 = 0.577 𝑆𝑦𝑡
τ ≈ 0.3 𝑆𝑠𝑦
Generally in design of helical springs, the permissible shear stress (τ) is taken from 30% to
50% of ultimate tensile strength (𝑆𝑢𝑡)
τ = 0.3 𝑆𝑢𝑡 or 0.5 𝑆𝑢𝑡

Procedure…
1. For the given application, estimate the maximum spring force (P) and the corresponding required
deflection (δ) of the spring. In some cases, maximum spring force (P) and stiffness k, which is
(P/δ), are specified.
2. Select suitable spring material and find out ultimate tensile strength (𝑆𝑢𝑡 ) from the data.
Calculate the permissible shear stress for the spring wire by following relationship:
τ = 0.3 𝑆𝑢𝑡 or 0.5 𝑆𝑢𝑡
3. Assume a suitable value for spring index (C). For industrial applications, the spring index varies
from 8 to 10. A spring index of 8 is considered as a good value. The spring index for springs in
valves and clutches is 5. the spring index should never be less than 3.
4. Calculate the Wahl factor by following equation:
K = 4𝐶−1
+ 0.615
4𝐶−4 𝐶
π𝑑3
8𝑃𝐷
τ = K ( )
5. Determine wire diameter (d) by,
6. Determine mean coil diameter (D) by following relationship
D = Cd

7. Determine the number of active coils (N) by
3
δ = 8𝑃𝐷 𝑁
𝐺𝑑4
the modulus of rigidity (G) for steel wires is 81370 N/𝑚𝑚2
8. Decide the style of ends for the spring depending upon configuration of the application.
Determine the number of inactive coils. Adding active coils and inactive coils, find out the total
number of coils (𝑁𝑡).
9. Determine the solid length of the spring by following relationship,
Solid length = 𝑁𝑡d
10. Determine the actual deflection of the spring by,
3
δ = 8𝑃𝐷 𝑁
𝐺𝑑4
11. Assume a gap of 0.5 to 2 mm between adjacent coils, when the spring is under the action of
maximum load. The total axial gap between the coil is given by,
Total gap = (𝑁𝑡-1) x gap between two adjacent coils
In some cases, the total axial gap is taken as 15% of the maximum deflection.
12. Determine the free length of the spring by following relationship
Free length = solid length + total gap + δ

13. Determine the pitch of the coil by following relationship
p = 𝑓𝑟𝑒𝑒 𝑙𝑒𝑛𝑔𝑡ℎ
(𝑁𝑡−1)
14. Determine the rate of spring by
k = 𝑃
=
𝐺𝑑4
δ 8𝐷3𝑁
15. Prepare a list of spring specifications.
A helical compression spring, that is too long compared to the mean coil diameter, acts as a
flexible column and may buckle as a comparatively low axial force. The spring should be preferably
designed buckle-proof. The thumb rules for provision of guide are as follows:
𝑓𝑟𝑒𝑒 𝑙𝑒𝑛𝑔𝑡ℎ
𝑚𝑒𝑎𝑛 𝑐𝑜𝑖𝑙 𝑑𝑖𝑎𝑚𝑒𝑡𝑒𝑟
𝑓𝑟𝑒𝑒 𝑙𝑒𝑛𝑔𝑡ℎ
𝑚𝑒𝑎𝑛 𝑐𝑜𝑖𝑙 𝑑𝑖𝑎𝑚𝑒𝑡𝑒𝑟
≤ 2.6 [Guide not necessary]
> 2.6 [Guide required]

SPRING DESIGN – TRIALAND ERROR METHOD
From mechanical properties of spring wire it is observed that the tensile strength decreases as wire diameter
increases. Therefore, tensile strength is inversely proportional to wire diameter.
𝑢
𝑡
𝑑
𝑆 𝖺 1
……….(a)
τ = 0.3 𝑆𝑢𝑡 or 0.5 𝑆𝑢𝑡…………(b)
but
hence 𝑑
τ 𝖺 1
……………….(c)
we know that τ = K (8𝑃𝐷
)……………(d)
therefore τ 𝖺
1
𝑑2
π𝑑3
…………(e)
From expressions (c) and (e), it is observed that permissible stress is proportional to (1/d), while induced stress is
proportional to (1/ 𝑑2). The trial and error procedure consists of following steps:
i. Assume some wire diameter (d).
ii. Find out the corresponding tensile strength from data book and using this value find out permissible stress by eq.
(b).
iii. Find out induced stress by eq. (d).
iv. Check up whether permissible stress is more than induced stress. If not, increase the wire diameter and repeat the
above steps.
v. The above mentioned procedure is repeated till the value of induced stress comes out to be less than the
02-12-p
20
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1
r7
missible stress. Department of ME, GEC Raipur 31

DESIGN AGAINST FLUCTUATING LOAD
• The springs subjected to fluctuating stresses are designed on the basis of two criteria- design for infinite life
and design for finite life.
let us consider a spring subjected to an external fluctuating force, which changes its magnitude from
𝑃𝑚𝑎𝑥 𝑡𝑜 𝑃𝑚𝑖𝑛 in the load cycle.
𝑃𝑚= ½ (𝑃𝑚𝑎𝑥 + 𝑃𝑚𝑖𝑛)
𝑃𝑎= ½ (𝑃𝑚𝑎𝑥 - 𝑃𝑚𝑖𝑛)
Mean stress is given by,
τ𝑚 = 𝐾𝑠( 𝑚
8𝑃 𝐷
𝜋𝑑3
)
𝑠 𝐶
where, 𝐾 = (1 + 0.5
)
𝐾𝑠 is the correction factor for direct shear stress and it is applicable to mean stress only. For torsional
stress amplitude (τ𝑎), it is necessary to also consider the effect of stress concentration due to curvature in
addition to direct shear stress. Therefore,
τ𝑎 = 𝐾𝑠𝐾𝑐 𝑎
; τ = 𝐾
8𝑃𝑎𝐷 8𝑃𝑎𝐷
𝜋𝑑3 𝜋𝑑3
where K is the Wahl factor, which takes into consideration the effect of direct shear stress as well as
of stress concentration due to curvature.

• A helical compression spring is subjected to purely compressive force, on the other hand, a helical
extension spring is subjected to purely tensile force. In general, the spring wires are subjected to
𝑠
𝑒
pulsating shear stresses, which vary from zero to 𝑆′ .
• 𝑆′ is the endurance limit in shear for the stress variation from zero to some maximum value.
𝑠𝑒
• The data regarding the experimental values of endurance strength of spring wire is not readily
available. In absence of such values, the following relationships suggested by H.J. Elmendorf can
be used,
For patented and cold-drawn steel wires (Grade -1 to 4),
𝑆′ 𝑆′
𝑠𝑒 𝑢𝑡 𝑠𝑦 𝑢
𝑡
= 0.21 𝑆 ; = 0.42 𝑆 ……………………………..(a)
for oil-hardened and tempered steel wires (SW and VW grade),
𝑠𝑒 𝑢
𝑡
𝑆′ 𝑆′
𝑠𝑦 𝑢
𝑡
= 0.22 𝑆 ; = 0.45 𝑆 ……………………………..(b)

• The fatigue diagram for the spring is shown in figure.
The line GH is constructed in such a way that its slope θ is given by, tan θ = 𝑐𝑎
𝑐𝑚
Consider similar triangles XFD andAEB,
𝑋𝐹
= 𝐴𝐸
; or 𝑐
𝐹𝐷 𝐸𝐵
𝑎
𝑆𝑠𝑦
(𝑓𝑠)
− 𝑐𝑚
= 2 𝑠
𝑒
1
𝑆𝘍
1
𝑆𝑠𝑦− 2𝑆𝑠𝑒
𝘍

SURGE IN SPRING
• When the natural frequency of vibrations of spring coincides with the frequency of external
periodic force, which acts on it, resonance occurs. In this state, the spring is subjected to a wave of
successive compressions of coils that travels from one end to the other and back. This type of
vibratory motion is called ‘surge’ of spring. Surge is found in valve springs, which are subjected to
periodic force.
Methods to avoid surge in springs:
• The spring is designed in such a way that the natural frequency of the spring is 15 to 20 times the frequency
of excitation of the external force. This prevents the resonant condition to occur.
• The spring is provided with friction dampers on central coils. This prevents propagation of surge valve.
• A spring made of stranded wire reduces the surge. In this case, the wire of spring is made of three strands. The
direction of winding of strands is opposite to the direction of winding of the coils while forming the spring. In
case of compression of coils, the spring tends to wind the individual wires closure together, which introduces
friction. This frictional damping reduces the possibility of surge.

HELICAL TORSION SPRINGS
• A helical torsion spring is a device used to transmit the torque to a particular component of a
machine or mechanism. It is widely used in door hinges, brush holders, automobile starters and
door locks.
Each individual section of the torsion spring is, in effect, a portion of a curved beam. Using the
curved beam theory, the bending stresses are given by,
σ𝑏 = 𝐾(
𝑀𝑏𝑦
𝐼
) ……….(a)
y = (d/2) and I = (π𝑑4/64)
Hence, 𝑏 π𝑑3
32𝑀𝑏
σ = 𝐾( )
K is stress concentration factor and is given by,
𝐾𝑖 =
4𝐶2 − 𝐶 − 1
𝐾𝑜 =
4𝐶(𝐶 − 1)
4𝐶2 + 𝐶 − 1
4𝐶(𝐶 + 1)
Where 𝐾𝑖 and 𝐾𝑜 are stress concentration factors at the inner and outer fibres of the coil respectively.

𝑀𝑏 = 𝑃𝑟
2𝑑𝑥
U = ∫ 𝑀𝑏
2𝐸𝐼
or
2 2
U = ∫𝑃 𝑟 (𝜋𝐷𝑁)
2𝐸𝐼
We have, θ = 64 𝑃𝑟𝐷𝑁
𝐸 𝑑4
And stiffness of the helical torsion spring
𝑘 = 𝑃𝑟
=
𝐸 𝑑4
𝜃 64 𝐷𝑁

MULTILEAF SPRING
• Multileaf springs are widely used for the suspension of cars, trucks, and railway wagons.
• It consists of flat plates, usually of semi-elliptical shape. The flat plates are called leaves of spring.
• The leaves have graduated lengths. The leaf at the top has maximum length. The length gradually
decreases from the top leaf to the bottom leaf. The longest leaf at the top is called master leaf. It is
bent at both ends to form the spring eyes. Two bolts are inserted through these eyes to fix the leaf
spring to the automobile body. The leaves are held together by means of two U-bolts and a center
clip.
• Rebound clips are provided to keep the leaves in alignment and prevent lateral shifting of the
leaves during operation.

• At the center, the leaf spring is supported on the axle. Multileaf springs are provided with one or
two extra full length leaves in addition to master leaf. They are provided to support the transverse
shear force.
For the purpose of analysis, the leaves are divided into two groups namely, master leaf along with
graduated length leaves forming one group and extra full length leaves forming the other.
𝑛𝑓 = number of extra full-length leaves
𝑛𝑔 = number 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 the 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 the extra full-length leaves (N)
𝑃𝑔= portion of P taken by the graduated-length leaves (N)

• The group of graduated length leaves along with the master leaf can be treated as a triangular plate.
• In this case, it is assumed that the individual leaves are separated and the master leaf placed at the
center. Then the second leaf is cut longitudinally into two halves, each of width (b/2) and placed on
each side of the master leaf. A similar procedure is repeated for other leaves. The resultant shape is
approximately a triangular plate of thickness t and a maximum width at the support as (𝑛𝑔𝑏).
the bending stress in the plate at the support is given by,
𝑏 𝑔 𝐼
(σ ) = 𝑀𝑏𝑦
=
(𝑃𝑔𝐿)(𝑡/2)
1 3
[12
(𝑛𝑔𝑏)(𝑡 )]
𝑏 𝑔
Or, (σ ) =
6𝑃𝑔𝐿
𝑛𝑔𝑏𝑡2
And deflection, δ
𝑃𝑔𝐿3
𝑔 = 2𝐸𝐼𝑚 𝑎 𝑥
.
δ
6𝑃𝑔𝐿3
𝑔
𝑔 = 𝐸𝑛 𝑏𝑡3

• Similarly, the extra full length leaves can be treated as a rectangular plate of thickness t and a
uniform width (𝑛𝑓𝑏).
The bending stress at the support is given by,
𝑏 𝑔 𝐼
(σ ) = 𝑀𝑏𝑦
=
(𝑃𝑓𝐿)(𝑡/2)
1 3
[12
(𝑛𝑓𝑏)(𝑡 )]
𝑏 𝑓
or, (σ ) =
6𝑃𝑓𝐿
𝑛𝑓𝑏𝑡2
The deflection at the load point is given by,
δ𝑓 =
𝑃𝑓𝐿3
3𝐸𝐼
or, 𝑓
δ =
4𝑃𝑓𝐿3
𝑓
𝐸𝑛 𝑏𝑡 3

The deflection of full length leaves is equal to the deflection of graduated length leaves,
δ𝑔= δ𝑓
From above relationship it can be derived as,
𝑓
𝑃 = 𝑔
and 𝑃 =
3𝑛𝑓𝑃 3𝑛𝑔𝑃
(3𝑛𝑓+2𝑛𝑔) (3𝑛𝑓+2𝑛𝑔)
Using these values the bending stresses can be written as
𝑏 𝑔
(σ ) = 𝑏 𝑓
and (σ ) =
12 𝑃𝐿 18 𝑃𝐿
3𝑛𝑓+2𝑛𝑔 𝑏𝑡2 3𝑛𝑓+2𝑛𝑔 𝑏𝑡2
It is seen from the above equations that bending stresses in full length leaves are 50% more than in
graduated length leaves. The deflection at the end of spring (δ) is given by,
δ =
12 𝑃𝐿3
𝐸𝑏𝑡3(3𝑛𝑓+2𝑛𝑔)

EQUALISED STRESS IN SPRING LEAVES (NIPPING)
• As discussed earlier, the bending stresses in full length leaves are 50% more than in graduated
length leaves. All leaves should be stressed to the same level in order to utilize the material to the
best advantage.
• One of the method of equalizing the stresses in different leaves is to prestress the spring. The
prestressing is achieved by bending the leaves to different radii of curvature, before they are
assembled with the center clip and U bolts.
• The full length leaf is given a greater radius of curvature than the adjacent leaf. This will leave a
gap or clearance space between leaves as shown in figure.

• When the center clip is drawn up by tightening U bolts, the upper leaf will bend back and have an
initial stress opposite to that produced by normal load P. on the contrary, the lower leaf will have
an initial stress of the same nature as that produced by normal load P.
• When the normal load is applied, the stress in the upper leaf is relieved and then increased in
normal way.
• The initial gap between the leaves can be adjusted in such a way that under maximum load
condition, the stress in all leaves will be same. The initial gap C between the extra full-length leaf
and the graduated-length leaf before the assembly is called ‘nip’. Such prestressing, achieved by a
difference in radii of curvature, is known as ‘nipping’.
The initial gap C is given by, C =
2𝑃𝐿3
𝐸𝑛𝑏𝑡3
The initial preload 𝑃𝑖 required to close the gap C is given by, 𝑃𝑖 =
2𝑛𝑔𝑛𝑓𝑃
𝑛(3𝑛𝑓+2𝑛𝑔)
And finally the stresses are equal in all leaves is given by, 𝜎𝑏 = 6 𝑃𝐿
𝑛𝑏𝑡2

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