Springs - DESIGN OF MACHINE ELEMENTS-II

UNIT-1 SPRINGS
Dr. L K Bhagi
Associate Professor
School of Mechanical Engineering
Lovely Professional University

SPRINGS
➢ A spring is basically defined as an ELASTIC BODY
whose function is to deformed when loaded and to
recover its original shape when the load is removed
➢ A spring is a machine member capable of providing
large elastic deformation.
2
Design of Machine Elements II MEC306
(Dr. L K Bhagi)

Objectives of Spring
Following are the objectives of a spring when used as a machine member:
1. Cushioning , Absorbing , or Controlling of energy due to shock
and vibration.
✓ Car springs or
✓ Railway buffers
✓ To control energy, springs-supports and vibration dampers.
2. Control of Motion
✓ Maintaining contact between two elements
(cam and its follower)
✓ Creation of the necessary pressure in a
friction device (a brake or a clutch)
3
(Dr. L K Bhagi)

Objectives of Spring
3. Measuring Forces
Spring balance
4. Storing of Energy
In clocks or Toys
The clock has spiral type of spring which is wound to coil and then
the stored energy helps gradual recoil of the spring when in
operation.
4
(Dr. L K Bhagi)

Before considering the design aspects of springs let’s
have a quick look at the SPRING MATERIALS and
MANUFACTURING METHODS.

Selection of Material for Spring Depends Upon the
Following Factors
✓ The Load acting on the spring
✓ The range of stress through which spring operates.
✓ The limitations of mass and volume of spring.
✓ The environmental conditions in which spring will operate.
6
(Dr. L K Bhagi)

Material of the spring should have
• High fatigue strength
• High ductility
• High resilience and
• It should be creep resistant.
Mainly depends upon the service for which they are used i.e.
• Severe service (automotive engine valve springs)
• Average service or (automobile suspension springs)
• Light service (safety valve springs)
7
(Dr. L K Bhagi)

Four basic varieties of steel wire used in
majority of applications
1. Patented and cold-drawn steel wire(unalloyed)
(0.85-0.95% C)
2. Oil-hardened and tempered spring steel wires
(0.55-0.75% C) (moderate fluctuating stresses) and valve
spring wires (0.60 – 0.75% C) (high magnitude of fluctuating
stresses)
8
(Dr. L K Bhagi)

Four basic varieties of steel wire used in
majority of applications
3. Oil-hardened and tempered steel wires(alloyed)
Chromium-Vanadium steel (0.48-0.53% C 0.8-1.1% Cr 0.15% V)
Chromium-silicon steel (0.51-0.59% C 0.6-0.8% Cr 1.2-1.6% Si)
Used for high stress conditions and at high temperature up to
220 -250oC
4. Stainless steel spring wires (Cr 17-19 % and Ni 8 -10 %)
Good tensile strength, High corrosion resistance and Good heat
resistance
It maintains its strength at temperatures up to 300oC
9
(Dr. L K Bhagi)

Spring Manufacturing Processes
1. Cold Winding Process
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.
2. Hot Winding
For very large springs having also large coil diameter and wire
diameter one has to go for manufacture
10
(Dr. L K Bhagi)

3. Heat Treatment
Whether the steel has been coiled hot or cold, the process has
created stress within the material. To relieve stress and allow the
steel to maintain its characteristic resilience, the spring must be
tempered by heat treating it.
4. Grinding
If the design requires for flat ends on the spring, the ends are
ground at this stage of the manufacturing process.
11
(Dr. L K Bhagi)

3. Shoot Peening
This process strengthens the steel to resist metal fatigue and
cracking during its lifetime of repeated loadings.
4. Coating
To prevent corrosion, the entire surface of the spring is protected
by coating it with another metal such as zinc or chromium.
5. Packaging
12
(Dr. L K Bhagi)

Types of Springs
1) Helical springs Load Type/stresses induced
a) Tension helical spring (Extension/Torsional shear stress)
b)Compression helical spring (Compression/Torsional shear stress)
c) Torsion spring (Torsion/Bending stress)
d)Flat Spiral spring (Torsion/Bending stress)
2) Leaf springs (Bending/Bending stress))
13
(Dr. L K Bhagi)

Types of Springs Helical Springs
✓ Closely coiled helical spring: Helix angle is usually less than 10
degrees
✓ Open coiled helical spring: Helix angle is usually more than 10
degrees.
Open coiled springs have a few applications as compared with
Closed coiled.
14
(Dr. L K Bhagi)

Terminology of Helical Springs
15
(Dr. L K Bhagi)
Solid Length = Nt × d
Nt = Total number of turns (coils)
d = Wire diameter of spring

Terminology of Helical Springs
16
(Dr. L K Bhagi)
Solid Length = Nt × d
Nt = Total number of turns (coils)
d = Wire diameter of spring

17
(Dr. L K Bhagi)
Total Axial Gap = (Nt -1)× gap b/w adjacent coils
or
= 15% of δ = 0.15× δ
(Clashing allowance)
Free Length = Compressed Length + δ
= Solid Length + Total Axial Gap + δ
Pitch (p) =
1N
LengthFree
t −
Rate of spring or Spring Constant or
Stiffness (k) =
( )

PforceAxial

End Connections for Compression Helical
Springs
18
(Dr. L K Bhagi)

End Connections for Compression Helical
Springs
19
(Dr. L K Bhagi)

Number of Active Turns in Helical Springs
20
(Dr. L K Bhagi)

Number of Active Turns in Helical Springs
21
(Dr. L K Bhagi)

22
(Dr. L K Bhagi)
Spring Index (C) is defined as the ratio of mean coil diameter to wire
diameter:
Spring index indicates the relative sharpness of the curvature of the coil.
C will determine the
Strength of the spring,
The stress induced on the spring, and
The manufacturability of the spring.
d
D
C =
Helical Springs>Spring Index

23
(Dr. L K Bhagi)
d
D
C =

24
(Dr. L K Bhagi)

25
(Dr. L K Bhagi)

26
(Dr. L K Bhagi)
For low spring index (C < 3) the sharpness of the curvature of the coil
increases predominantly. Thereby increases the stresses in the spring &
also difficult to manufacture.
When spring index is very high (C >15) then spring will prone to buckle.
Best range of C for manufacturing is 4-12 and 6-9 for those subjected to
cyclic loading.

27
(Dr. L K Bhagi)
Compression coil springs will BUCKLE when the free length of the
spring is larger and the end conditions are not proper to evenly distribute
the load all along the circumference of the coil.
The coil compression springs will have a tendency to BUCKLE when the
deflection (for a given free (length) becomes too large.

Stresses in Helical Spring of Circular Wire
28
(Dr. L K Bhagi)
torsional shear stress
twisting moment

Resultant Shear Stress Induced in the Wire
29
(Dr. L K Bhagi)

30
(Dr. L K Bhagi)

31
(Dr. L K Bhagi)
33
8
2
1
1
8
d
PD
K
Cd
PD
S

=





+=
Ks= Factor to account for direct shear stress
So, it can be observed that effect of direct shear stress
is appreciable for springs of small spring index C.







Cd
PD
2
18
3


Stresses in Helical Spring
With Curvature Effect
32
(Dr. L K Bhagi)
a b
Direction of
view
d
a b
d c
c
After Twisting
b
c
So, Shearing Strain at ab surface is more than that of cd surface and that is
what curvature effect is.

33
(Dr. L K Bhagi)
Same Wire with having two different diameters then the curvature effect will be
more in the wire having small diameter. So, more and more having spring
index (D/d) the lesser will be the curvature effect, e.g. railway carriage springs
are made with large wire diameter relatively small coil diameter then the
curvature effect will be predominately very high.

34
(Dr. L K Bhagi)
3
8
d
PD
K

=
KS= Factor to account for direct shear stress
KC= Factor to account for stress concentration due to curvature effect
When we taken into the account of the effect of curvature along
with direct shear, we get
In order to consider the effects of both direct shear as well as the
curvature of the wire, a Wahl’s stress factor (K) is introduced for
calculating the Maximum Shearing Stress at the inner fiber:

Deflection of Helical Spring of Circular Wire
35
(Dr. L K Bhagi)
θ = Angular deflection/Angle
of twist of the wire when acted
upon by the torque T.
θ δ
P
Equation is called the load-deflection equation.

P
kStiffnessSpring =)(
When a helical spring is cut into two
parts, the parameters G, d and D remain
same and stiffness (k) will be double
when N becomes (N/2).

(Dr. L K Bhagi)
36

Helical Spring >Numerical Problem (N1)
A compression coil spring made of an alloy steel is having the following
specifications :
Mean diameter of coil = 50 mm ; Wire diameter = 5 mm ; Number of
active coils = 20.
If this spring is subjected to an axial load of 500 N ; calculate the
maximum shear stress (neglect the curvature effect) to which the
spring material is subjected.
37
(Dr. L K Bhagi)

10
5
50
d
D
C ===
( ) 051
102
1
1
C2
1
1 .KfactorstressShear S =

+=+=
38
(Dr. L K Bhagi)
MPa7.534ormmN7534
5
505008
051
d
PD8
K
C2
1
1
d
PD8
2
3
3S3
/.. =


=

=





+

=
,curvature)wireofeffecttheg(neglectinstressshearMaximum

A helical spring is made from a wire of 6 mm diameter and has outside
diameter of 75 mm. If the permissible shear stress is 350 MPa and
modulus of rigidity 84 kN/mm2 .
Find the axial load which the spring can carry and the deflection per
active turn.
39
(Dr. L K Bhagi)

511
6
69
d
D
C .===
( ) 0431
5112
1
1
C2
1
1 .
.
KfactorstressShear S =

+=+=
40
(Dr. L K Bhagi)
Mean coil diameter of the spring,
D = Do – d = 75 – 6 = 69 mm
1. Neglecting the effect of curvature
wire,theininducedstressshearMaximum
3S3
d
PD8
K
C2
1
1
d
PD8

=





+

=

41
(Dr. L K Bhagi)
N7412
8480
350
P
P8480
6
69P8
0431350
d
PD8
K
3
3S
.
.
..
==
=


=

=
( ) 4
3
Gd
NPD8
=springtheofDeflection
mm.
.
turnactiveperDeflection 969
61084
6974128
Gd
PD8
N 43
3
4
3
=


==

=

42
(Dr. L K Bhagi)
wire,theininducedstressshearMaximum
2
d
PC8
K

=
2. Considering the effect of curvature
( )
( )
1231
511
6150
45114
15114
C
6150
4C4
1C4
K .
.
.
.
..
=+
−
−
=+
−
−
=Wahl’s stress factor (K)
N4383
9130
350
P
P9130
6
511P8
1231350 2
.
.
.
.
.
==
=


=

43
(Dr. L K Bhagi)
( ) 4
3
Gd
NPD8
=springtheofDeflection
( )
( ) ( )
mm.
.
turnactiveperDeflection 269
61084
6943838
Gd
PD8
N 43
3
4
3
=


==

=

A wagon weighing 50 kN moving at a speed of 8 kmph has to be
brought to rest. Springs made of wire of diameter 25 mm with a
mean diameter of 250 mm and with 24 turns are available. Find the
number of springs required in the buffer to stop the wagon at a
compression of 180 mm.
G = 84 Gpa.
44
(Dr. L K Bhagi)

V = 8 kmph
K.E. =
m/s22.2m/s
3600
8000
==
( ) Nmmmv 52
3
2
106.1222.2
81.92
1050
2
1
=


=
43
3
4
3
251084
24250P8
180
Gd
NPD8


== NP 75.1968=
NmmP 5
1018.018075.1968
2
1
2
1
doneWork === 
70
1018.0
106.12
SpringsofNo. 5
5
=


=
45
(Dr. L K Bhagi)

A closed coiled helical spring is made out of wire of 6 mm diameter with
a mean diameter of 80 mm.
what axial pull will produce a shear stress of 140 Mpa?
If the spring has 20 coils, how much the spring will extend under pull?
G = 80 Gpa. What work will be done in producing this extension?
46
(Dr. L K Bhagi)

d = 6 mm; D = 80 mm; C = D/d = 80/6 = 13.33
0375.1
66.26
1
1
2
1
1 =





+=





+=
C
K
33
6
80P8
03751140
d
PD8
K


=→

= . NP 143=
mm113
61084
20801438
Gd
NPD8
43
3
4
3
=


==
Nmm5.8079113143
2
1
2
1
doneWork === P
47
(Dr. L K Bhagi)

A railway wagon moving at a velocity of 1.5 m/s is brought to rest by a
bumper consisting of two helical springs arranged in parallel. The mass
of wagon is 1500 kg. the springs are compressed by 150 mm in bringing
the wagon to rest. The spring index can be taken as 6. the springs are
made of oil hardened and tempered steel wire with ultimate tensile
strength of 1250 N/mm2 and modulus of rigidity of 81370 N/mm2. The
permissible shear stress for the spring wire can be taken as 50% of the
ultimate tensile strength. Design the spring and calculate:
(i) Wire diameter
(ii) Mean coil diameter
(iii) Number of active coils
(iv) Total number of coils
(v) Solid length
(vi) Free length
(vii) Pitch
(viii) required spring rate
(ix) Actual spring rate 48
(Dr. L K Bhagi)

Given
m = 1500 Kg v = 1.5 m/s δ = 150 mm C = 6
Sut = 1250 N/mm2 G = 81370 N/mm2 τ = 0.5 Sut
The K.E. of wagon =
The Strain energy absorbed by two springs
The strain energy absorbed by the two springs is equal to the K.E of the
wagon;
( ) Nmm101687.5Nm5.16875.11500
2
1
2
1 322
===mv
( ) ( )Nmm150150
2
1
2
2
1
2 PPPE =





=





= 
( ) N11250105.1687150 3
== PP
( ) 2
N/mm62512505.0 ==
49
(Dr. L K Bhagi)

Given
Wire diameter
Mean coil diameter
Number of active coils
( )
( )
2525.1
6
615.0
464
164615.0
44
14
=+
−
−
=+
−
−
=
CC
C
K
mmord
dd
PC
K 2056.18
6112508
2525.1625
8
22
=→






=→=


120206 ===CdD
coilsor
Gd
NPD
1356.12
2081370
2012011508
150
64
4
3
4
3
=


=→=
50
(Dr. L K Bhagi)

Total Number of coils
Assume square and ground ends
Solid Length of Spring
Assume a gap 2 mm between adjacent coils when the spring is subjected
to the maximum force of 11250 N.
Total axial gap
Free Length = solid length + total axial+δ
= 300+28+155.29 = 483.29 or 485 mm
coils512132 =+=+= NNt
mm29.155
2081370
1312011250864
4
3
4
3
=


==
Gd
NPD

mm282)115( =−=
51
(Dr. L K Bhagi)

Pitch of coils
Required Stiffness
Actual Stiffness
mm64.34
115
485
1
LengthFree
=
−
=
−
=
tN
N/mm20
15
300
===

P
k
N/mm87.19
5408
581370
8 3
4
3
4
=


==
ND
Gd
k
52
(Dr. L K Bhagi)

Series and Parallel Connection
of Helical Spring
Spring in series
Total deflection of the springs
21  +=
21
21
111
kkk
k
P
k
P
k
P
+=
+=
53
(Dr. L K Bhagi)

Series and Parallel Connection
of Helical Spring
Spring in Parallel
Total force acting on the spring
21 PPP +=
21
21
kkk
kkk
+=
+= 
54
(Dr. L K Bhagi)

Design of Helical Springs
Objectives for design
✓ It should have sufficient strength to withstand the external load.
✓ It should have the required load-deflection characteristic.
✓ It should not buckle under the external load.
55
(Dr. L K Bhagi)

Number of Active TurnsMean Coil DiameterWire Diameter
Load-Stress Equation
or
3max
8
d
PD
K

 =
2max
8
d
PC
K

 =
Load-Deflection Equation
TurnsActive
ofNo.theisNWhere
8
4
3
Gd
NPD
=
56
(Dr. L K Bhagi)

Factor of Safety (fos)
While designing a component, it is necessary to provide
sufficient reserve strength in case of failure.
The factor of safety in the design of springs is usually 1.5 or less.
That means
Where, Ssy is torsional yield strength
Assuming
&
So,
loadWorkingstressAllowableStressePermissibl
loadfailurestressUltimatestressfailure
fos
//
//
=
5.1
syS
=
57
(Dr. L K Bhagi)
ytsy SS = 577.0 utyt SS = 75.0
( )( )
ut
ut
S
S
3.0
5.1
75.0577.0
=

Factor of Safety (fos)
But the Indian Standard 4454–1981 has recommended a much
higher value for the permissible shear stress.
Therefore, in design of helical springs, the permissible shear
stress (τ) is taken from 30% to 50% of the ultimate tensile strength
(Sut).
58
(Dr. L K Bhagi)
utS5.0=

Design Procedure of Helical Springs
Step 1
Step 2
Select suitable material and find Sut from data table & calculate
permissible shear stress,
τ = 0.3 Sut or τ = 0.5 Sut
Step 3
Assume appropriate spring index, 3 < C < 10 (for Valves and
Clutches, C = 5, is perfect)
59
(Dr. L K Bhagi)
( )
)(
)(
Deflection
PforcespringMaximum
kStiffness =

Step 4
Calculate Wahl factor
Step 5
Calculate spring wire diameter (d) using,
Step 6
Find mean coil diameter using, D = Cd
Step 7
Find number of active coils (N) using
60
(Dr. L K Bhagi)
2
8
d
PC
K

 =
4
3
8
Gd
NPD
=

Step 8
Decide the end style condition and then calculate total number
of coils (Nt).
Step 9
Then find, Solid length of the helical spring = Nt×d
Step 10
Find actual deflection of the spring using again
Step 11
Under maximum load, assume axial gap of 0.5 to 2 mm b/w
adjacent coils & then find total axial gap,
Total gap = (Nt-1)×gap between adjacent coil
61
(Dr. L K Bhagi)
4
3
8
Gd
NPD
=

Step 12
and then calculate, free length= Solid length + total gap+δ
Step 13
Find the pitch (p) of the coil using
Step 14
And Finally calculate rate of spring using
Check to avoid buckling
[Guides not required]
[Guides required]
62
(Dr. L K Bhagi)
ND
Gd
k 3
4
8
=
( )1−
=
tN
lengthFree
p
6.2
diametercoilMean
lengthFree
6.2
diametercoilMean
lengthFree

It is required to design a helical compression spring subjected to a max. force
of 1250 N. The deflection of the spring corresponding to the maximum force
should be approx. 30mm. The spring index can be taken as 6. The spring is
made of patented and cold drawn steel wire. The ultimate tensile strength &
modulus of rigidity of the spring material are 1090 and 81370 N/mm2. The
permissible shear stress for the spring wire should be taken as 50% of the
ultimate tensile strength. Design the spring and calculate:
1. Wire diameter (ii) Mean coil diameter
(iii) No. of active coils (iv) Total number of coils
(v) Free length of the spring (vi) Pitch of the coil
63
(Dr. L K Bhagi)
Design of Helical Spring >Numerical Problem
(N6)

64
(Dr. L K Bhagi)
(N6)

65
(Dr. L K Bhagi)
(N6)

66
(Dr. L K Bhagi)
(N6)

67
(Dr. L K Bhagi)
(N6)

68
(Dr. L K Bhagi)
It is required to design a helical compression spring subjected to a
maximum force of 7.5 kN. The mean coil diameter should be 150 mm
from space consideration. The spring rate is 75 N/mm. The spring is
made of oil-hardened and tempered steel wire with ultimate tensile
strength of 1250 N/mm2. The permissible shear stress for the spring
wire is 30% of the ultimate tensile strength (G = 81 370 N/mm2).
Calculate
(i) wire diameter; and
(ii) number of active coils.
(N7)

69
(Dr. L K Bhagi)
P = 7.5 kN ; D = 150 mm ; k = 75 N/mm
The spring is made of oil-hardened and tempered steel wire with
ultimate tensile strength Sut = 1250 N/mm2.
The permissible shear stress (τ) for the spring wire is 30% of the
ultimate tensile strength i.e. τ = 0.3 Sut
G = 81 370 N/mm2.
Calculate
(i) wire diameter; and
(ii) number of active coils.
(N7)

70
(Dr. L K Bhagi)
Step 1 Wire Diameter
The permissible shear stress τ = 0.3 Sut
τ = 0.3 (1250) = 375 N/mm2
C = D/d ; d = 150/C
Equation is to be solved by the trial and error method.
2max
8
d
PC
K

 =
( )
( ) ( ) ( )
( )75008
375150
P8
150
KC
150
PC8
K
22
3
2
3

=

=

=
79441KC3
.=
(N7)

71
(Dr. L K Bhagi)
C K KC3
5 1.311 163.88
6 1.253 270.65
7 1.213 416.06
8 1.184 606.21
7.5 1.197 505.98
7.1 1.210 433.07
7.2 1.206 450.14
7.3 1.203 467.99
It is observed from the above table that the spring index should
be between 7.1 and 7.2 to satisfy Eq.
(N7)

72
(Dr. L K Bhagi)
C = 7.2
d = 15/C = 150/7.2 = 20.83 or 21 mm
For Number of active coils
ND8
Gd
k
Gd
NPD8
3
4
4
3
==
( )
( )
coils8or.817
N1508
2181370
ND8
Gd
75 3
4
3
4
=


==
(N7)

Design Against Fluctuating Load
73
(Dr. L K Bhagi)
( )
2
minmax
m


+
=StressMean ( )
2
minmax
a


−
=amplitudeStress

Repeated stress and Reversed stress are special cases of fluctuating
stress with (σmin=0) and (σm=0)
74
(Dr. L K Bhagi)
max =a

Design against fatigue failure
Fluctuating stress
Infinite life
(Endurance Limit)
75
(Dr. L K Bhagi)
Finite life
(S-N Curve)
Gerber parabola
Soderberg line
Goodman line
Completely reversed stress 0=m
FoS
Se
a =
FoS
Sse
a =

76
(Dr. L K Bhagi)

77
(Dr. L K Bhagi)
The endurance limit (Sse) of a component subjected to fluctuating
torsional shear stress is obtained from the endurance limit in
reversed bending (Se) using theories of failures.
According to max shear stress theory Sse=0.5 Se
According to distortion energy theory Sse=0.577 Se

78
(Dr. L K Bhagi)
When the component is designed for finite life, the S-N curve
can be used. S-N curve for steel is

(Dr. L K Bhagi)
79

80
(Dr. L K Bhagi)
When the component is subjected to fluctuating stresses

Helical Spring Design Against
Fluctuating Load
81
(Dr. L K Bhagi)
When the spring is subjected to repeated shear stress
max
a
m a
0min =
2
max
 == am

Fluctuating Load
82
(Dr. L K Bhagi)
The mean stress (τm) is calculated using shear stress correction factor
(Ks).
For torsional stress amplitude (τa) stress concentration due to curvature
(Kc) along with factor Ks also considered.
3
8
d
DP
K m
Sm

 = 





+=
C
KS
2
1
1






=





= 33
88
d
DP
K
d
DP
KK aa
cSa



Fluctuating Load (Soderberg Failure Criterion)
83
(Dr. L K Bhagi)
valuemaximumsometozero
fromvariationstressforshearinlimitendurance=seS
AlternatingStress(τa)
Mean Stress
(τm)
A
B
C
DE
τa
τm
F
X





 
2
,
2
sese SS
Line of Failure
Safe stress line
syS
FoSSsy

Fluctuating Load (Soderberg Failure Criterion)
Considering similar ∆s XFD and AEB,
The above eqn is used in the design of springs subjected to fluctuating
stresses. In the section “Design steps of helical spring” it was explained
that the FoS in spring design is usually 1.5 or less.
EB
AE
FD
XF
= or;
OE-OB
AE
OF-OD
XF
=
se
se
sy
S
S
FoS
S
−

=
−
2
1
S
2
1
sym
a


84
(Dr. L K Bhagi)

Helical Spring Design Against Fluctuating Load (Soderberg
Failure Criterion) >Numerical Problem(N8)
A helical compression spring made of oil tempered carbon steel, is
subjected to a load which varies from 400 N to 1000 N. The spring
index is 6 and the design factor of safety is 1.25. If the yield stress in
shear is 770 MPa and endurance stress in shear is 350 MPa, find : 1.
Size of the spring wire, 2. Diameters of the spring, 3. Number of
turns of the spring, and 4. Free length of the spring.
The compression of the spring at the maximum load is 30 mm. The
modulus of rigidity for the spring material may be taken as 80
kN/mm2.
85
(Dr. L K Bhagi)

Given :
Pmin = 400 N ; Pmax = 1000 N ; C = 6 ; FoS = 1.25 ; Ssy = 770 Mpa; =
770 N/mm2 ; Se = 350 MPa = 350 N/mm2 ; δ = 30 mm ; G = 80
kN/mm2 = 80 × 103 N/mm2
86
(Dr. L K Bhagi)
se
se
sy
S
S
FoS
S
−

=
−

2
1
S
2
1
sym
a
se
se
se
se
sy S
S
S
S
FoS
S −
=
−
=
−

sy
sym
a
2S
2
1
S
2
1
( )( ) ( )( ) ( )se
sesy
se S
FoS
SS
S maasy2S −







=− ( )( ) ( )( ) ( )sese
sesy
SS
FoS
SS
maasy2S +−=







( )( ) ( )( ) ( )
se
se
se
se
se S
S
S
S
SFoS
1
sy
m
sy
a
sy
asy
SSS
2S 
+

−

=
( )( ) ( ) ( )
sy
m
sy
aa
SS
2 
+

−

=
seSFoS
1

87
(Dr. L K Bhagi)
3
m
Sm
d
DP8
K

=







= 3
a
a
d
DP8
K
( )( ) ( ) ( )
sy
m
sy
aa
SS
2 
+

−

=
seSFoS
1
( ) N700
2
4001000
2
PP
PLoadMean =
+
=
+
= minmax
m
( ) N300
2
4001000
2
PP
PLoad =
−
=
−
= minmax
aAmplitudeorVariable
0831
62
1
1KS .=






+= 25251
6
6150
464
164
C
6150
4C4
1C4
K .
..
=





+
−
−
=





+
−
−
=

88
(Dr. L K Bhagi)
( )( ) ( ) ( )
sy
m
sy
aa
SS
2 
+

−

=
seSFoS
1
( )
2
222
d
440d
11582
d
5740
350
d
5740
251
1 .
770770
2
.
=






+






−






=
( ) 2
23m mmN
d
11582
d
d67008
0831 /. =

=
2
23a mmN
d
5740
d
d63008
25251 /. =







=
mm7.1dor... === 550440251d2

89
(Dr. L K Bhagi)
Diameters of the spring
D = C.d = 6 × 7.1 = 42.6 mm
Outer diameter of the spring,
Do = D + d = 42.6 + 7.1 = 49.7 mm
and inner diameter of the spring,
Di = D – d = 42.6 – 7.1 = 35.5 mm
Number of turns of the spring
( )
( )43
3
4
3
171080
N64210008
30
Gd
NPD8
.
.


== 10or.
.
879
043
30
N ==
12210 =+=tN

90
(Dr. L K Bhagi)
Free length of the spring
LF = free length = Solid length + total gap + δ
= Solid length +0.15× δ+ δ
= 12 × 7.1 + 30 + 0.15 × 30 mm
= 119.7 say 120 mm

Concentric Springs
Concentric spring (nested spring) consist of two helical compression
springs, one inside the other. Two springs have opposite hand of
helices. If outer spring has a right hand helix then the inner spring
always has a left hand helix. Just to prevent the locking of coils, in the
event of axial misalignment or buckling of springs.
92
(Dr. L K Bhagi)

Concentric Springs
Advantages of Concentric springs
1. The load carrying capacity is increased
or in other words to get greater
spring force within a given space
2. The operation of the mechanism continues
even if one of the springs break.
3. The spring vibrations are eliminated.
93
(Dr. L K Bhagi)

Concentric Springs
In the concentric springs
a) Maximum shear stress developed in the concentric springs is same
for same material and free length
b) Deflection is same for both the springs
c) Solid length of both the springs is same
94
(Dr. L K Bhagi)

Concentric Springs
The following notations are used in the analysis:
d1 = wire diameter of outer spring
d2 = wire diameter of inner spring
D1 = mean coil diameter of outer spring
D2 = mean coil diameter of inner spring
P1 = axial force transmitted by outer spring
P2 = axial force transmitted by inner spring
P = total axial force
δ1 = deflection of outer spring
δ2 = deflection of inner spring
N1 = number of active coils in outer spring
N2 = number of active coils in inner spring
c = radial clearance between the springs.
95
(Dr. L K Bhagi)

Concentric Springs
The design analysis of concentric spring is
based on the following assumptions:
(i) The springs are made of the same material.
(ii) The maximum torsional shear stresses
induced in outer and inner springs are
equal.
(iii) They have the same free length.
(iv) Both springs are deflected by the same
amount and therefore, have same solid
length.
96
(Dr. L K Bhagi)

Concentric Springs
Since the maximum torsional shear stresses induced in both springs are
equal
τ1= τ2
If the same material is used, the concentric springs are designed for the
same stress. In order to get the same stress factor (K), it is desirable to
have the same spring index (C).
K1 = K2
97
(Dr. L K Bhagi)






=





3
2
22
23
1
11
1
88
d
DP
K
d
DP
K







=





3
2
22
3
1
11
d
DP
d
DP

Concentric Springs
Since the deflection of two springs are equal
δ1= δ2
When both springs compressed completely
Solid length of outer spring = Solid length of inner spring
98
(Dr. L K Bhagi)
4
2
2
3
22
4
1
1
3
11 88
Gd
NDP
Gd
NDP
=
4
2
2
3
22
4
1
1
3
11
d
NDP
d
NDP
=
2211 NdNd =

Concentric Springs
It is assumed that there are no inactive coils. The total number of coils
is equal to the number of active coils.
99
(Dr. L K Bhagi)
5
2
22
3
22
5
1
11
3
11 )(8)(8
Gd
dNDP
Gd
dNDP
=
5
2
3
22
5
1
3
11
d
DP
d
DP
=
2211 NdNd =






=





3
2
22
3
1
11n
eqFrom
d
DP
d
DP
5
2
3
22
5
1
3
11
bydivide
d
DP
d
DP
=
indexSpringor
2
2
1
1
2
2
2
2
2
1
2
1
==== C
d
D
d
D
d
D
d
D

Concentric Springs
Substituting Eqn
in
Then we get
That means the total load shared by each spring is proportional to the
cross-sectional area of the wire.
100
(Dr. L K Bhagi)
indexSpring
2
2
1
1
=== C
d
D
d
D






=





3
2
22
3
1
11
d
DP
d
DP
2
1
2
2
2
1
2
1
2
2
2
1
2
1
2
2
2
2
1
1
a
a
d
d
P
P
d
d
P
P
d
P
d
P
===





=








Concentric Springs
Radial clearance (c) is an important parameter in the design of
concentric springs.
In the design of concentric springs, the diametral clearance (2c) is taken
101
(Dr. L K Bhagi)






+++





++=
22
2
22
1122
21
dd
c
dd
DD
( ) ( )21212 ddDDc +−−=
( ) ( )
22
2121 ddDD
c
+
−
−
=
( ) ( )
2
2 21
21
dd
cddc
−
=−=

Concentric Springs
So,
Substituting
We get
Equations and are used to find out the force
transmitted by each spring.
102
(Dr. L K Bhagi)
( ) ( )
222
212121 ddDDdd +
−
−
=
−
( ) ( ) 1212121 2dddddDD =++−=−
2211 and CdDCdD ==
( ) ( ) 21121 2or2 CddCdddC =−=−
22
1
−
=
C
C
d
d
2
2
2
1
2
1
d
d
P
P
=
22
1
−
=
C
C
d
d

Optimum Design of Helical Spring
In certain applications springs are designed for specific objective,
✓Minimum Weight
✓Minimum volume or
✓Maximum energy storage capacity
103
(Dr. L K Bhagi)

Design based upon certain objectives
like in valve spring.
1. Minimum force (Pmin) required at
most extended position to keep
the valve closed.
2. Max shear stress (τmax) when
spring is fully compressed.
104
(Dr. L K Bhagi)

For the purpose of analysis, the effect of inactive coils is neglected and
the spring is designed on the basis of minimum weight.
Cross sectional area of wire
Length of one coil
Length of all active coils
Volume of spring wire
If (ρ) as density then weight of the spring is
105
(Dr. L K Bhagi)
2
4
d

=
D=
DN=
( )DNd 
 2
4
=
( )DNdW 2
2
4 





=



Substituting
Primary design equation
106
(Dr. L K Bhagi)
( ) ND
C
DN
C
D
W 3
2
2
2
22
44

 





=





=






=
C
D
d
( ) 









=





= 3
max
3
max
max
88
C
D
DP
K
d
DP
K


3
max
2
max
8KC
D
P

=
GD
NPC
GDd
NPD
Gd
NPD 4
4
4
4
3
888
===

The deflection of the spring (∆) for movement of the valve is
Substitute Pmax in above
107
(Dr. L K Bhagi)
( )minmax
44
min
4
max
minmax
888
PP
GD
NC
GD
NCP
GD
NCP
−=−=
−= 






−= min3
max
24
8
8
P
KC
D
GD
NC 
( )min
3
max
2
8 PKCDC
GDK
N
−

=


Substitute N in
For minimum weight right hand factor should be minimum
108
(Dr. L K Bhagi)
ND
C
W 3
2
2
4







=
( )min
3
max
2
4
3
2
84 PKCD
D
C
GK
W
−





 






=


 
  0)4(8)2(
08
5
min
33
max
4
min
32
max
=−−−
=−
−−
−−
DPKCD
DPKCD
dD
d



109
(Dr. L K Bhagi)
)8(2 3
max
2
min
KC
D
P

=
2
max
min
P
P =
So, if the spring is to have minimum weight, it must be designed
in such a way that the Pmin should be 50% of the Pmax.

Optimum Design of Helical Spring > Numerical
Problem (N9)
A helical compression spring of the exhaust valve mechanism is
initially compressed with a pre-load of 375 N. When the spring is
further compressed and the valve is fully opened, the torsional shear
stress in the spring wire should not exceed 750 N/mm2. Due to space
limitations, the outer diameter of the spring should not exceed 42 mm.
The spring is to be designed for minimum weight.
Calculate the wire diameter and the mean coil diameter of the spring.
(Dr. L K Bhagi)
110

Problem (N9)
(Dr. L K Bhagi)
111

Problem (N9)
The problem is solved by trial and error method. In practice, the
spring index varies from 6 to 10. Considering values of C in this
range, the results are tabulated in the following manner.
Comparing Eq. (a) and
the values in above table,
C = 8
… (a)
(Dr. L K Bhagi)
112

Problem (N9)
(Dr. L K Bhagi)
113

Problem (N9)
(Dr. L K Bhagi)
114

Helical Torsion Springs
115
(Dr. L K Bhagi)
A helical torsion spring is a device used to transmit the torque to a
particular component of a machine or mechanism.

116
(Dr. L K Bhagi)
A helical torsion spring is a device used to transmit the torque to a
particular component of a machine or mechanism.

117
(Dr. L K Bhagi)

118
(Dr. L K Bhagi)

119
(Dr. L K Bhagi)
Ends of torsion spring are formed in such a way that the spring is
loaded by a torque about the axis of the coils. The helical torsion
spring resists the bending moment (P× r), which tends to wind up the
spring.
Using the curved beam theory, the bending stresses are given by
where, Mb=P×r





=
I
yM
K b
b

120
(Dr. L K Bhagi)
K Stress concentration factor due to curvature (Wahl’s stress factor).
For circular wire, and
Therefore,













=
4
64
2
d
d
rP
Kb

 




 
= 3
32
d
rP
Kb


2
d
y =
64
4
d
I

=
)1(4
14
K&
)1(4
14 2
O
2
+
−+
=
−
−−
=
CC
CC
CC
CC
Ki

121
(Dr. L K Bhagi)
The strain energy stored in the spring
The deflection in the direction of force P is
Approximately r×θ
So from strain energy formula, we get the angular deflection or total
angle of twist
( )
EI
DNrP
EI
dxrP
U
DN
22
22
0
22

== 
( )
4
22
)(64
22
1
Ed
DNrP
EI
DNrP
rPU

=

== 



122
(Dr. L K Bhagi)
The stiffness of helical torsion spring is the bending moment required
to produce unit angular deflection.
Since the diameter of torsion spring reduces as the coil wind up under
applied load, therefore a small clearance between the adjacent coils
must be provided in order to prevent sliding friction.
DN
EdrP
k
64
4
=

=


Helical Torsion Springs>Numerical Problem (N10)
123
(Dr. L K Bhagi)
It is required to design a helical torsion spring for a window shade.
The spring is made of patented and cold-drawn steel wire of Grade-4.
The yield strength of the material is 60% of the ultimate tensile
strength and the factor of safety is 2. From space considerations, the
mean coil diameter is kept as 18 mm. The maximum bending moment
acting on the spring is 250 N-mm. The modulus of elasticity of the
spring material is 207 000 N/mm2. The stiffness of the spring should
be 3 N-mm/rad.
Determine the wire diameter and the number of active coils.

124
(Dr. L K Bhagi)

125
(Dr. L K Bhagi)

126
(Dr. L K Bhagi)

127
(Dr. L K Bhagi)

Spiral Spring (Flat Spiral Spring)
128
(Dr. L K Bhagi)
A flat spiral spring is long thin strip of rectangular cross-section
having elastic material wound like a spiral.

129
(Dr. L K Bhagi)
The spring is wound by rotating the axle (arbor) and during the
winding process, the energy is stored in the spring.

130
(Dr. L K Bhagi)
The following notations are used in the analysis of spiral spring:
P = Force induced at the outer end A due to winding of the arbor (N)
r = Distance of center of gravity of spiral from outer end (mm)
t = Thickness of strip (mm)
b = Width of strip perpendicular to plane of paper (mm)
l = Length of strip from outer end to inner end (mm).

131
(Dr. L K Bhagi)
The outer end A is pulled by force P and point B is at a maximum
distance from the point of application of P.
So, maximum bending moment at point B is
When both ends are clamped, the angular deflection (in radians) of
spring is
23
12
12
2
2
)2(
bt
M
bt
t
M
I
yrP
I
yMb
b =

=

==
3
12
Ebt
Ml
IE
lM
=


=

132
(Dr. L K Bhagi)
And the deflection
The strain energy stored in the spring
Or in terms of bending stress
r
Ebt
Ml
r 





== 3
12

3
2
3
612
2
1
2
1
2
1
Ebt
lM
Ebt
Ml
MrPPU =





=== 
( ) springtheofvolume
2424
144
24
246
2
42
2
3
2
=


==
EEtb
btlM
bt
bt
Ebt
lM
U b

Flat Spiral Spring >Numerical Problem (N11)
133
(Dr. L K Bhagi)
A flat spiral spring is required to provide a maximum torque of
1200 N-mm. The spring is made of a steel strip and the
maximum bending stress should not exceed 800 N/mm2. When
the stress in the spring decreases from 800 to 0 N/mm2, the arbor
turns through three complete revolutions with respect to the
retaining drum. The thickness of the steel strip is 1.25 mm and
the modulus of elasticity is 207000 N/mm2. Calculate the width
and length of the steel strip.

134
(Dr. L K Bhagi)

135
(Dr. L K Bhagi)

Leaf Springs
137
(Dr. L K Bhagi)
In order to have an idea of working principle of a leaf spring,
let us think of the diving board in a swimming pool. The diving board
is a cantilever with a load, the diver, at its free end. The diver initiates
a to and fro swing of the board at the free end and utilizes the spring
action of the board for jumping.
The diving board basically is a leaf spring.

Leaf Springs
138
(Dr. L K Bhagi)
The leaf springs are widely used in suspension system of railway
carriages and automobiles. But the form in which it is normally seen is
laminated leaf spring or Multi-leaf Spring.

Leaf Springs
139
(Dr. L K Bhagi)
In the cantilever beam type leaf spring, for the same leaf thickness t,
leaf of uniform width b (case 1)
b

Leaf Springs
140
(Dr. L K Bhagi)
and, leaf of width, which is uniformly reducing from b (case 2) is
considered for having same thickness t and length of beam L.
b

Leaf Springs
141
(Dr. L K Bhagi)
From the basic equations of bending stress and deflection, the
maximum bending stress, σb in the leaf spring at the support and tip
deflection, δ at the load point can be derived.
For case 1(uniform width)
23
6
12
2
)(
bt
PL
bt
t
LP
I
yMb
b =

==
3
3
4
Ebt
PL
= =
L
EI
dxM
U
0
2
2

Leaf Springs
142
(Dr. L K Bhagi)
For case 2(Non-uniform width)
2
6
bt
PL
b =
3
3
6
Ebt
PL
=

Leaf Springs
143
(Dr. L K Bhagi)
Leaf spring of simply supported beam type, for which the stress and
deflection equation are
P
P/2
P/2

Leaf Springs
144
(Dr. L K Bhagi)
For Case 1 (uniform width)
For Case 2 (non uniform width)
2
6
bt
PL
b =
3
3
2
Ebt
PL
=
2
6
bt
PL
b =
3
3
3
Ebt
PL
=
=
L
EI
dxM
U
0
2
2
2
I
yMb
b =

(Dr. L K Bhagi)
145
2b
bt
PL6
= 3
3
4
Ebt
PL
=
P
P/2 P/2
2
6
bt
PL
b = 3
3
2
Ebt
PL
=
b
2
6
bt
PL
b = 3
3
6
Ebt
PL
=
2
6
bt
PL
b = 3
3
3
Ebt
PL
=
Uniform Width
Non-Uniform Width

Multi-Leaf Springs
146
(Dr. L K Bhagi)
A multi-leaf spring consists of a series of flat plates, usually of semi-
elliptical shape.
The flat plates are called leaves of the 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 from the spring eyes. Two bolts are inserted through these eyes
to fix the leaf spring to the automobile body.

Multi-Leaf Springs
147
(Dr. L K Bhagi)

Multi-Leaf Springs
148
(Dr. L K Bhagi)

Multi-Leaf Springs
149
(Dr. L K Bhagi)
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 centre, the leaf spring is supported on the axle.
Multi-leaf springs are provided with extra full-length leaves, to
support the transverse shear force.

Multi-Leaf Springs
150
(Dr. L K Bhagi)
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.
The following notations are used in the analysis:
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)

Multi-Leaf Springs
151
(Dr. L K Bhagi)
The group of graduated-length leaves along with the master leaf placed
at the center can be treated as a triangular plate. The maximum width at
the support as ngb.
The bending stress in the plate at the
support and the deflection at the load
point are,
( )
( )
2
3
6
)(
12
1
2)(
btn
LP
tbn
tLP
I
yM
g
g
g
gb
gb =

==
( ) 3
3
3
33
6
)(
12
1
2
2
and
tbnE
LP
tbnE
LP
EI
LP
g
g
g
gg
g =




==

Multi-Leaf Springs
152
(Dr. L K Bhagi)
The bending stress in the plate at the support and the deflection at the
load point are,
( )
( )
2
3
6
)(
12
1
2)(
btn
LP
tbn
tLP
I
yM
f
f
f
fb
fb =

==
( ) 3
3
3
33
4
)(
12
1
3
3
and
tbnE
LP
tbnE
LP
EI
LP
f
f
f
ff
f =




==

Multi-Leaf Springs
153
(Dr. L K Bhagi)
Since the deflection of full-length leaves is equal to the deflection of
graduated-length leaves.
gf  =
( ) ( ) 3
3
3
3
64
tbnE
LP
tbnE
LP
g
g
f
f
=
g
f
g
f
n
n
P
P
2
3
=
gf PPP +=LoadTotal

Multi-Leaf Springs
154
(Dr. L K Bhagi)
( )gf
f
f
nn
Pn
P
23
3
+
=
( )gf
g
g
nn
Pn
P
23
2
+
=
( ) 2
6
btn
LP
f
f
fb = ( ) 2
6
and
btn
LP
g
g
gb =
( )
( ) 2
23
18
So,
btnn
PL
gf
fb
+
= ( )
( ) 2
23
12
and
btnn
PL
gf
gb
+
=
( ) ( )gbfb 
2
3
=
The bending stresses in full-length
leaves are 50% more than those in
graduated-length leaves.

Multi-Leaf Springs
155
(Dr. L K Bhagi)
Et
L
Ebt
PL
3
24
and
2
3
3

 =2
6
bt
PL
b =
( ) ( )fgfg
f
nnEbt
PL
nnbt
PL
Et
L
Et
L
32
12
32
18
3
2
3
2
deflectionTotal 3
3
2
22
+
=








+
==



Multi-Leaf Springs
Nipping
156
(Dr. L K Bhagi)
Stresses due to support hinges
• Stresses due to longitudinal forces
• Possibility of stresses arising due to twist
Disadvantage of this is that full length leaves get more and more
stressed than the graduated leaves.
Methods to reduce additional stress
• Full length leaf is made of stronger material than the other leaves
• Full length leaf is thinner than the other leaves
• Radius of curvature of the full length leaf is larger than the
graduated leaf

Multi-Leaf Springs
Nipping
157
(Dr. L K Bhagi)
Radius of curvature of the full length leave is larger than the
graduated leaf.

Multi-Leaf Springs
Nipping
158
(Dr. L K Bhagi)
Tightening of leaves by a center bolt:
(a) condition before tightening and (b) deformation caused by
tightening.

Multi-Leaf Springs
Nipping
159
(Dr. L K Bhagi)
Assuming, under maximum load condition the stress in all leaves will
be same.
22
66
btn
LP
btn
LP
g
g
f
f
=
( ) ( )gbfb  =
f
g
f
g
f
f
g
g
n
n
P
P
n
P
n
P
==
PPP fg =+knowweAs
n
Pn
P
n
n
P
P
n
n
P
P f
f
fff
g
f
g
==+=+ 11
n
Pn
P
g
g =so,and

Multi-Leaf Springs
Nipping
(Dr. L K Bhagi)
Under maximum force, P condition the stress in all leaves will be
same.
fg  
( ) 3
3
4
tbnE
LP
f
f
f =
( ) 3
3
6
tbnE
LP
g
g
g =
( ) ( ) 







−=−=
f
f
g
g
f
f
g
g
n
P
n
P
Ebt
L
tbnE
LP
tbnE
LP
C
23246
3
3
3
3
3
3
fgC  −=
( ) ( ) 





−=−=
n
P
n
P
Ebt
L
tbnE
LP
tbnE
LP
C
f
f
g
g 23246
3
3
3
3
3
3
3
3
2
Enbt
PL
C =
160

Multi-Leaf Springs
Nipping
(Dr. L K Bhagi)
The load on the centre clip bolts i.e. initial preload required to close
the gap C is determined by considering initial deflection of leaves.
Under the action of Preload Pi
( ) ( )ifigC +=
( )
( ) 3
3
2
6
tbnE
L
P
g
i
ig = ( )
( ) 3
3
2
4
tbnE
L
P
f
i
if =
3
3
3
3
3
3
22
4
2
6
Enbt
PL
btEn
L
P
btEn
L
P
C
f
i
g
i
=






+






=
161

Multi-Leaf Springs
Nipping
(Dr. L K Bhagi)
3
3
3
3
3
3
22
4
2
6
Enbt
PL
btEn
L
P
btEn
L
P
C
f
i
g
i
=






+






=
n
P
n
P
n
P
f
i
g
i 223
=+
( )gf
fg
i
nnn
Pnn
P
23
2
+
=
162

Multi-Leaf Springs
Nipping
(Dr. L K Bhagi)
The final stress in the leaf spring will be
( )
222
5.062
66
btn
PPL
btn
L
P
btn
LP
f
if
f
i
f
f
b
−
=−=
( ) ( )







+

−
+
=
gf
fg
gf
f
f
b
nnn
Pnn
nn
Pn
btn
L
23
25.0
23
36
2

( )







+
−
=
gf
g
b
nnn
nn
bt
LP
23
36
2
 2
6
nbt
PL
b =
163

Multi-Leaf Springs >Numerical Problem (N12)
(Dr. L K Bhagi)
Design a graduated leaf spring for the following specifications:
Total Load = 140 kN
Number of springs supporting the load = 4
Maximum number of leaves = 10
Span of the spring = 1000 mm
Permissible deflection = 80 m.
Take Young’s modulus, E = 200 kN/mm2 and allowable stress in spring
material as 600 MPa.
164

(Dr. L K Bhagi)
Load on each spring kN35
4
140
springsofNo.
LoadTotal
2 ===P
N17500kN17.5
2
35
===P
2
6
22
106.655001750066
nbtnbtnbt
PL
b

=

==
32
2
6
105.87
106.65
600 =

= nbt
nbt
( )
( )
( ) 3
6
33
3
3
3
106.65
10200
5001750066
80
nbttnbtnbE
PL 
=


===
6
6
3
1082.0
80
106.65
=

=nbt
165

(Dr. L K Bhagi)
mm10mm37.9
105.87
1082.0
3
6
2
3
=


= t
nbt
nbt
( )
mm5.87
1010
105.87
2
3
=


=b32
105.87 =nbt
63
1082.0 =nbt
( )
mm82
1010
1082.0
3
6
=


=b
Take the larger of the two values so width of leaves, b= 87.5 or 90 mm
166

(Dr. L K Bhagi)
A semi-elliptic leaf spring consists of two extra full-length leaves and
six graduated length leaves, including the master leaf. Each leaf is 7.5
mm thick and 50 mm wide. The centre-to-centre distance between the
two eyes is 1 m. The leaves are pre-stressed in such a way that when
the load is maximum, stresses induced in all the leaves are equal to
350 N/mm2.
Determine the maximum force that the spring can withstand.
167

(Dr. L K Bhagi)
168

(Dr. L K Bhagi)
A semi-elliptical multi-leaf spring is used for suspension of the rear
axle of a truck. It consists of two extra full-length leaves and ten
graduated length leaves including the master leaf. The centre to centre
distance between the spring eyes is 1.2 m. The leaves are made of steel
55Si2Mo90 (Syt=1500 N/mm2 and E= 207000 N/mm2) and the factor
of safety is 2.5. The spring is to be designed for a maximum force of
30 kN. The leaves are pre-stressed so as to equalize the stresses in all
leaves. Determine:
a) The cross-section of leaves; and
b) The deflection at the end of the spring
169

(Dr. L K Bhagi)
Cross-section of the leaves
( ) 22
102
600150006
600
6
btnbt
PL
b
+

==
32
mm7500=bt
2
N/mm600
5.2
1500
===
FoS
Syt
b
mm600m2.12
15000kN302
==
==
LL
NPP
Standard widths are 32, 40, 45, 50, 55, 60, 65, 70, 75, 80, 90, 100 & 125 mm
Assuming standard width of 60 mm
mm1211.18
60
75002
== tt
Cross-section of the leaves = 60×12 mm2
170

(Dr. L K Bhagi)
Deflection at the end of the spring
( ) ( )
mm68.69
231021260207000
6001500012
32
12
Deflection 3
3
3
3
=
+

=
+
=
fg nnEbt
PL

171

(Dr. L K Bhagi)
A semi-elliptical laminated vehicle spring to carry a load of 600 N is to
consists of seven leaves 65 mm wide, two of the leaves extending the full
length of the spring. The spring is to be 1.1 m in length and attached to the
axle by U-bolts 80 mm apart. The bolts hold the central portion of the
springs so rigidly that they may be considered equivalent to a band having
width equal to the distance between the bolts. Inside diameter of eye is 20
mm. Assume a design stress for spring material as 350 MPa. Determine:
1. Thickness of leaves, 2. Deflection of spring, 3. Length of leaves, and 4.
Radius to which leaves should be initially bent.
The standard thickness of leaves are 5, 6, 6.5, 7, 7.5, 8, 9, 10, 11 etc. in
mm
172

(Dr. L K Bhagi)
Thickness of leaves
The effective length(2L) = overall length (span) of the spring (2L1)–
width of band or distance between centres of U-bolts (ineffective length
of spring) (l)
mm102080110022 1 =−=−= lLL mm510
2
1020
==L
527leavesgraduatedofNo. =−=−= fg nnn
( )
( ) ( ) 22
655223
510300018
350
23
18
tbtnn
PL
gf
fb
+

=
+
=
Assuming that the leaves are not initially stressed, the maximum stress
is,
mm97.8 =t
173

(Dr. L K Bhagi)
Deflection of spring (Take E = 210 x 103 N/mm2)
( ) ( )
mm30
235296510210
510300012
32
12
Deflection 33
3
3
3
=
+

=
+
=
fg nnEbt
PL

Length of leaves
Ineffective length of the spring = 80 mm
lengtheineffectiv1
1n
lengtheffective
leaf)(1stleafsmallestoflength +
−
=
mm25080
17
1020
=+
−
=
174

(Dr. L K Bhagi)
mm420802
17
1020
leaf2ndoflength =+
−
=
mm590803
17
1020
leaf3rdoflength =+
−
=
mm760804
17
1020
leaf4thoflength =+
−
=
mm930805
17
1020
−
=
leafmasterasactwill
top)(at theleaf7ththeandleaveslengthfullareleaves7th&6thThe
( )
( ) mm128229201100
2td2leafmasteroflength 1
=++=
++=

L
mm1100806
17
1020
−
=
175

(Dr. L K Bhagi)
bentinitiallybeshouldleaveswhich thetoRadius
( ) ( ) springtheofcamberasknownalso-2R
2
1  L=
( ) ( ) mm5.505655030-2R30
2
== R
176

Springs - DESIGN OF MACHINE ELEMENTS-II

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Springs - DESIGN OF MACHINE ELEMENTS-II