mechanical springs
DESCRIPTION
okTRANSCRIPT
-
BITS PilaniPilani Campus
VINAYAK KALLURI
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BITS PilaniPilani Campus
-
BITS Pilani, Pilani Campus
Springs
A mechanical spring is an elastic member (generally
metal) whose primary function is to deflect under
load and then to recover its original shape and
position when the load is released.
Used for efficient storage and release of energy
Strength and flexibility are two essential
requirements of spring design.
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1. Helical springs (Tension / Compression)
2. Torsion spring
3. Leaf springs
4. Spiral spring
5. Belleville Springs
Spring Types
Helical springsTorsion spring
Spiral spring
Leaf spring
Belleville Spring
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Stresses in Helical Springs
( )2323max
4842/16
2/
d
F
d
FD
d
F
d
FD
A
F
J
Tr
FDT
+=+=+=
=
inside inside
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Spring Index and shear stress correction factor
d
DC =If we define the spring index to be as :
Then the expression for maximum shear stress can be expressed as:
+=
+=+=Cd
FD
D
d
d
FD
d
F
d
FD
2
11
8
21
8483323max
Where Ks is called as the shear stress correction factor and serves to
correct the shear stress estimated from the torsion alone for the direct
shear. Here Ks is
C
CK s
2
12 +=
For the standard springs, C ranges between 6 and 12.
3max
8
d
FDK s
=
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Curvature effect in fatigue loading
Only in fatigue loading, the curvature of the wire
introduces more shear stress than estimated above
Hence that expression for maximum shear stress needs
correction.
Many factors have been suggested for correction.
Prominent are Wahl factor (Kw) and Bergstrasser factor
(KB).
These must replace Ks when incorporated.
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Curvature effect in fatigue loading:
Wahl factor and Bergstrasser factor
34
24615.0
44
14
+
=+
=C
CKor
CC
CK Bw
Static Loading, only effect of direct shear:
C
CK s
2
12 +=
Fatigue Loading, effect of both direct shear and curvature:
Wahl factor (Kw) or Bergstrasser factor (KB) is used
Fatigue Loading, effect of only curvature:
( )( )( )1234
242
++
==CC
CC
K
KK
s
Bc
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Deflection and Stiffness:
AG
lF
GJ
lTU
22
22
+=
The total strain energy for a helical spring is composed
of a torsional component and a shear component.
The strain energy
4
;32
;
;2
,
2
4
dA
andd
J
DNl
FDT
where
a
=
=
=
=
2
2
4
32 24
Gd
DNF
Gd
NDFU +=
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Deflection and Stiffness:
a
aaa
aa
ND
Gd
y
Fk
Gd
NFC
Gd
NFD
CGd
NFDy
Gd
FDN
Gd
NFD
F
Uy
3
4
3
4
3
24
3
24
3
8
88
2
11
8
D/d,Cindex spring gIntroducin
48
: theoremsecond so'Castiglian
==
=
+=
=
+=
=
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Compression Springs
Manufacturing processing at the ends and effect on total coils
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Compression Springs
Formulas for the Dimensional Characteristics of Compression-Springs
Table 101
If interested:For a thorough discussion and development of these relations, refer
Cyril Samonov, Computer-Aided Design of Helical Compression Springs, ASME
paper No. 80-DET-69, 1980.
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SPRING MATERIALS
Music wire, Oil-tempered wire, Hard drawn wire,
Chrome-vanadium wire and Chrome-silicon wire
Use Table 10-4 for A and m.
Table 104
mut d
AS = strength, tensileMinimum
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SPRING MATERIALS
Table 105
Mechanical Properties of Some Spring Wires
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Set removal or presetting
Is a process used in the manufacture of compression springs
to induce useful residual stresses
The spring is made to a longer free length than required and
then is compressed beyond the elastic limit by 30% of the
length
When the spring tries spring back, the plastic strain induced
opposes the same resulting in residual stress being set up that
are opposite in direction to the working stresses
Hence the springs behave stronger in service
Set removal must NOT be used for springs used in fatigue
loading
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SPRING MATERIALS
Table 106
Maximum Allowable Torsional Stresses for Helical Compression Springs in Static
Applications
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Check for stability:
A compression spring is stable if it does not buckle under the
load
=
2/1
2
'
2'
10 11eff
cr
CCLy
D
Leff
0 =( )GE
EC
=
2
'
1
( )EG
GEC
+
=2
2 2'2
( ) 2/102
'
2
2
21
+
EG
GEDL
C
eff
For absolute stability and buckling not to occur,
Slenderness ratio Elastic constants
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End -condition constants () for helical compression springs
Table 102, page 522
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A helical compression spring subjected to a force of 500
N, which causes the deflection of 20 mm. The spring
index is 6. The spring is made of music wire A228 (Sut=
2000 MPa, G = 81GPa, & E= 196 GPa) and has square
and ground ends. Calculate (i)wire diameter (d should be
round-up to next standard diameter), (ii) mean coil
diameter (D), (iii) active number of coils (Na, should be
round-up to the next integer), (iv) total number of coils
(Nt) (v) Solid length and (vi) pitch of the coils (p). (v)
check for stability of the spring.
Problem:
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).......(..........56.6,
56.65.381000
21500820
8
)...(....................21
)...(....................5.3032.3
65008083.1900
8
083.112
13
2
12
900200045.045.0
4
3
4
3
23max
iiiNcoilsofnumberActive
NN
Gd
NFDy
iimmCdD
immmmd
dd
FDK
C
CK
MPaSS
a
aaa
s
s
utsy
=
=
=
==
=
==
==+
=
===
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BITS Pilani, Pilani Campus
( )
( )76.10596.49
196812
811962
5.0
2196.49
2
2
).......(..........55.656.6
796.492,
96.49)(,
).......(..........96.29,
).......(..........56.8256.6,
110
2/1
2/1
0
0
0
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A load of 2kN is dropped axially from a height of 250 mm
on a plain and ground helical compression spring of wire
diameter 20 mm. The active number of coils in spring is 20
and spring index is10. Determine the stiffness, the pitch
and the shear stress induced in the spring. Assume G =
84x103 MPa for spring material.
Problem:
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A safety valve operated by a helical tension
spring through the lever mechanism is
schematically shown in Figure. The
diameter of the valve is 25mm. In normal
operating conditions, the maximum pressure
inside the chamber is 2MPa and the
corresponding lift of the valve is 24 mm.
The spring is made of cold drawn steel of
ultimate strength 1.5GPa and modulus of
rigidity of 81GPa. The spring index is 8.
Find the standard spring wire diameter,
stiffness and the active number of coils in
the spring.
Problem:
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A safety valve operated by a helical tension spring
through the lever mechanism is schematically
shown in figure 4. The diameter of the valve is 50
mm. In normal operating conditions, the valve is
closed and the pressure inside the chamber is 0.5
MPa. The valve opens when the pressure inside the
chamber increases to 0.6 MPa. The maximum lift
of the valve is 5 mm. The spring is made of cold
drawn steel of ultimate strength 1200 MPa and
modulus of rigidity of 81370 N/mm2. Allowable
shear stress for the spring wire is 30% of the
ultimate tensile strength. The spring index is 8.
Find the spring wire diameter, stiffness and the
active number of turns in the spring.
Problem:
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Springs are almost always subject to fatigue loading.
Automotive engine valves are supported by compression
springs that are subjected to millions of cycles of
operation without failure.
Shot peening is used to improve the fatigue strength of
dynamically loaded springs. Shot peening can increase
the torsional fatigue strength by 20 percent or more.
Springs are designed for infinite life based on
Zimmerlis data.
Design for Fatigue Load:
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Zimmerlis Data: Shot Peening
A cold working process used to produce a compressive
residual stress layer and modify mechanical properties of
metals
Entails impacting a surface with shot (round metallic,
glass or ceramic particles of 1/64 inch diameter) with
force sufficient to create plastic deformation
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Zimmerlis Data:
The best data on the torsional endurance limits of spring steels are
those reported by Zimmerli and discovered the surprising fact that
size, material, and tensile strength have no effect on the endurance
limits (infinite life only) of spring steels in sizes under 10 mm.
Unpeened springs were tested from a minimum torsional stress of
138 MPa to a maximum of 620 MPa and peened springs in the range
138 MPa to 930 MPa . The corresponding endurance strength
components for infinite life were found to be
utsyututsu
smsa
smsa
SSSorSS
MPaSMPaS
Peened
MPaSMPaS
Unpeened
557.035.067.0
534398
:
379241
:
=
==
==
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BITS Pilani, Pilani Campus
Design for Fatigue Loading Based on Zimmerlis Data
criterion. failure fatigue aapply then and
,,,,,,,, minmax susyemama SorSSFFfindgivenFF
33
minmaxminmax
88
22
d
DFKand
d
DFK
FFFand
FFF
mWm
aWa
ma
==
+=
=
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Zimmerlis Data (Gerber criteria)
a
saf
Sn
=Safety, ofFactor
m
a
m
a
ut
se
se
susa
su
sm
sase
su
sm
se
sa
F
Fr
rS
S
S
SrS
lineloadgivenforcordinatetionInter
S
S-
SS
S
S
S
S
==
++=
==
+
222
2
2
211
2
sec
1
1
limit endurance thefindThen
Refer Table 6-7 ; page 307
Ssa and Ssm are
from Zimmerlis
data.
Ssu= 0.67Sut
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Zimmerlis Data (Goodman criteria)
a
saf
Sn
= Safety, ofFactor
m
a
m
a
sesu
susesa
su
sm
sase
su
sm
se
sa
F
Fr
SrS
SrSS
lineloadgivenforcordinatetionInter
S
S-
SS
S
S
S
S
==
+=
==
+
sec
1
1
limit endurance thefindThen
Refer Table 6-6 ; page 307
Ssa and Ssm are from
Zimmerlis data.
Ssu= 0.67Sut
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Zimmerlis Data (ASME- Elliptic criteria)
a
saf
Sn
= Safety, ofFactor
m
a
m
a
syse
syse
sa
sy
sm
sase
sy
sm
se
sa
F
Fr
SrS
SSrS
lineloadgivenforcordinatetionInter
S
S-
SS
S
S
S
S
==
+=
==
+
222
222
2
22
sec
1
1
limit endurance thefindThen
Refer Table 6-8 ; page 308
Ssa and Ssm are from
Zimmerlis data.
utsyut SSS 557.035.0
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BITS Pilani, Pilani Campus
In an automotive plate clutch, six identical helical compression springs
(squared and ground end) arranged in between two parallel plates to provide
the axial thrust on the plate, which is varying from 600 N to 3000 N. The
minimum deflection of the springs corresponding to the maximum thrust
force is 20mm. The spring index is 6. The spring is made of cold drawn steel
(G= 81370 MPa) with ultimate strength of 1250 MPa and allowable shear
stress of 450 MPa. There should be a gap of 1 mm between adjacent turns
when the spring is subjected to maximum thrust force.
(i) Assuming the springs are subjected to static maximum thrust force, find
the standard spring wire diameter and free length.
(ii) By using the above spring wire diameter, determine the factor of safety
for fluctuating force based on Gerber criteria (unpeened) using Zimmerlis
data.
Problem
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Tension/ Extension springs: end preparation
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Combined axial tension and bending stress at A
Only torsion at B
Improved design
Side view
Tension /Extension springs:
Side view
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Analysis of stresses in tension springs
( ) dr
CCC
CCK
K
dd
DKF
A
A
AA
11
11
1
2
1
23
2,
14
14
bygiven curvature,for factor correction strss Bending
416
moment bending tension axial combined todueA at stress tensilemaximum The
=
=
=
+=
d
rC
C
CKB
K
d
FDK
w
B
22
2
2
3
2,
44
14
bygiven curvature,for factor correction strss Torsional
8
bygiven is Bat stressshear torsionalmaximum The
=
=
=
=
B
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Extension spring
Stress is to be computed at three locations
F
A
B
C
( ) dr
CCC
CCK
dd
DKF
A
AA
11
11
1
2
1
23
2,
14
14
416
=
=
+=
d
rC
C
CK
d
FDK
B
BB
22
2
2
3
2,
44
14
8
=
=
=
C
CK
d
FDK
s
sC
2
12
83
+=
=
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Initial tension in close-wound tension springs
When extension springs are made with coils in contact with one
another, they are said to be close-wound.
Spring manufacturers prefer some initial tension in close-wound
springs in order to hold the free length more accurately.
( ) ( ) ( )
E
GNNcoilsofnumberActive
dNCdNdDL
kyFF
ba
bbo
i
+=
+=++=
+=
,
1212 :length Free
No of body coils
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INITIAL TENSION IN CLOSE-WOUND TENSION SPRINGS:
The initial tension in an extension spring is created in the
winding process by twisting the wire as it is wound onto
the mandrel.
( )MPa
C
e Ci
=5.6
349.6
231
is, stress torsionalduncorrecte of range Preferred
105.0
When the spring is completed and
removed from the mandrel, the
initial tension is locked in because
the spring cannot get any shorter.
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PROBLEM
A hard-drawn steel wire (G=79 GPa, E= 198GPa)
extension spring has a wire diameter of 0.9 mm, an
outside coil diameter of 6 mm, hook radii of r1=2.55
mm and r2= 2.1 mm, and an initial tension of 5 N and
subjected to a static load of 24 N. The number of body
turns is 12.17. From the given information:
(a) Determine the physical parameters of the spring
(b) Check the initial preload stress conditions
(c) Find the factors of safety under a static 24 N load.
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Solution:
( )( )( ) ( )
( ) ( ) ( ) ( )( )
mmyLL
mmk
FFy
mmdNdDL
mmNND
Gdk
turnsE
GNN
dDC
mmdODD
o
i
bo
a
ba
14.2589.425.20
89.4885.3
524
25.209.0117.129.01.5212
/885.357.121.58
9.01079
8
57.1210198
107917.12
67.59.0/1.5/
1.59.06
11
11
3
43
3
4
3
3
=+=+=
=
=
=
=++=++=
=
==
=
+=+=
===
===
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( ) ( )( )( )
( ) [ ]
( ) ..1.1027.248.126
5.1517.248.1265.6
367.549.6
231
07.899.0
1.5588
is stress initial duncorrecte The
range preferred in thenot isIt min
67.5*105.0max
33
MPa
MPae
MPad
DF
i
i
i
uncorri
==
=+=
+=
===
( )( )( )
( )( )
737.13.465
55.818S
55.818181945.0S45.0S
18199.0
1783
d
AS
3.4659.0
1.5248088.1
8
1
sy
utsy
190.0mut
33
11
===
===
===
===
s
s
n
MPa
MPa
MPad
DFK 088.1
12=
+=
C
CK si)
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BITS Pilani, Pilani Campus
( )( )
( )( )
( ) ( ) ( )( ) ( )( )
( )( )
( ) ( )( )
( ) ( ) ( )( )
( ) 595.1513
55.818,513
9.0
1.52420.18
2.1466.44
166.44
44
14
66.49.0
9.01.5
d
2r
is Bat hook -endin situation The
336.11021
25.1364
25.1364181975.0S75.0S
,10219.0
4
9.0
1.51615.124
15.1167.567.5*4
167.567.54
14
14
67.5d
D
d
2rC
isA at bendinghook end in thesituation The
31
2
2
22
1
uty
231
2
11
1
2
1
11
====
=
=
=
=
=
==
===
===
=
+=
=
=
=
====
ByB
B
A
y
Ay
A
A
nMPa
C
CK
d
dDC
Sn
MPa
MPa
CC
CCK
C
ii)
iii)
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BITS Pilani, Pilani Campus
Close wound like helical coil
extension spring
Negligible initial tension
The ends connect a force at a
distance from coil axis to apply
a torque
Wound with a pitch that just
separates the body coils to avoid
intercoil friction.
The wire in the torsion
spring is in bending
TORSION SPRINGS:
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Fixed End
Free
End
Free end
location angle
Back
angle
Angular rotation,
proportional to Fl
For all positions of the moving end + = = constant.
turnspartial;body turns
integer360
integer
==
+=+=
pb
pob
NN
NN
TORSION SPRING
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BITS Pilani, Pilani Campus
Bending Stress :
The bending stress can be expressed as
3
3
22
32,
32
)1(4
14
)1(4
14
''
d
FlKisequationbendingthe
dz
c
IandFlMngSubstituti
unitythanlessalsoandKthanlessalwaysisK
CC
CCKand
CC
CCK
factorcorrectionstressisKwhere
I
McK
i
io
oi
=
===
++
=
=
=
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BITS Pilani, Pilani Campus
Torsional stiffness:
( )
radiansEd
Ml
radiansEd
Ml
dE
Fl
EI
Fl
l
y
e
e
4
44
22
3
64
3
64
64/33
:deflection End
=
====
The angle subtended by the end deflection of a cantilever,
when viewed from the built-in ends, is y/l rad.
From Table A91,
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Ed
MDN
Ed
FlDN
dI
EI
dxFl
EI
dxlF
FF
Ul
bbc
DNDN
c
bb
44
4
0
2
0
22
6464
64/
2
==
=
=
=
=
The Force F will deflect through a distance l
Torsional stiffness
Strain energy in bending, = EIdxM
U2
2
The end deflection is bending of a cantilever beam whereas
the coils undergo bending action under M = Fl requiring
application of Castigliano theorem.
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BITS Pilani, Pilani Campus
( ) ( )
atebae
bb
t
eect
NEd
MDNNN
D
llN
Defining
D
llN
Ed
MD
Ed
Ml
Ed
Ml
Ed
MDN
4
21
21
44
2
4
1
4
21
64;,
3
3
64
3
64
3
6464
(rad), deflectionangular totalThe
=+=+
=
++=++=
++=
Torsional stiffness
-
BITS Pilani, Pilani Campus
Stiffness expressions in torque/radian units:
bc
cDN
EdMk
64
4
== ( )21
4
64
3
ll
EdMk
e
e +==
Stiffness values in torque/turn values (i.e 2 rad /turn) :
at
sDN
EdMk
64
4
==
264
4
'
' ==at
sDN
EdMk
2
64
4
'
' ==bc
cDN
EdMk
( )
264
3
21
4
'
' +
==ll
EdMk
e
e
at
sDN
EdMk
8.10
4
'
' == bc
cDN
EdMk
8.10
4
'
' == ( )21
4
'
'
8.10
3
ll
EdMk
e
e +==
Tests show that the effect of friction between the coils is such that the constant
10.186 (i.e 64/2) should be increased to 10.8
Torsional stiffness
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BITS Pilani, Pilani Campus
Torsion spring supported on round bar or pin:
( ) ( )[ ]b'cbcb
b
NDAND'A
N
DND
=+
+= ndeformatioafter and before balance volumefrom,'
'
( )pin
pinc
b
pin
cb
bpinpini
DdD
DdN
DdN
DNDdDDD
++=
+
===
'
1
' '
When the load is applied to a torsion spring, the spring winds up, causing
a decrease in the inside diameter of the coil body.
Ensure that the inside diameter of the coil never becomes equal to or less
than the diameter of the pin,
The helix diameter of the coil D becomes
The new inside diameter Di = D d makes the diametral clearance
between the body coil and the pin of diameter Dp
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Table 106
Design of Torsion Springs for Strength:
Static strength
First column entries in Table 106 can be divided by 0.577 (from distortion-energy
theory) to give
mut d
AS =
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Design of Torsion Springs for Strength: Dynamic strength
The dynamic design can not be based on Zimmerlis data (which
was used for compression/tension springs) because the nature of
loading and stresses are due to bending rather than torsion.
We have to use the repeated bending stress data provided by
Associated Spring R=0 fatigue strength (Sr)
Table 1010 (Sr values for fatigue)
Maximum Recommended Bending Stresses (Kw Corrected) for Helical
Torsion Springs in Cyclic Applications as Percent of Sut
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BITS Pilani, Pilani Campus
Design for dynamic (fatigue) strength:
2
2/1
2/
=
ut
r
re
S
S
SS
The value of Sr from the table 10-10.
Gerber criterion:
1
2
=
+
ut
m
e
a
SS
a
af
ut
e
e
uta
ma
Sn
rS
S
S
SrS
MMr
=
++=
=
222 2
112
/
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BITS Pilani, Pilani Campus
The spring is formed by winding cold
drawn stainless steel wire A302 (E =
196 MPa) around a mandrel to obtain
2.5 body turns with torsion ends as
shown in figure. The wire clip is
arranged with initial torque/moment of
1200 N-mm.
A. Find the maximum static
torque/moment that can be applied,
if the factor of safety is 2.5
B. calculate the total angular deflection
of the spring corresponding to the
applied static torque/moment.
Problem
-
BITS Pilani, Pilani Campus
Solution
( )
MPan
S
CC
CCK
d
DC
mmdDD
MPaSSandMPad
AS
AsteelstainlessdrawnCold
y
i
i
utymut
2.4465.2
5.1115
09.1)1(4
14
9
36432
5.111578.014304
2065
302
max
2
263.0
===
=
=
==
=+=+=
=====
-
BITS Pilani, Pilani Campus
rad
D
llN
Ed
MD
mmNMMmomentappliedMaximum
mmNMd
MK
t
bt
initial
i
2.0363
1151155.2
101964
36137264
3
64
1372
257209.132
42.44632
34
21
4
max
3
max3
maxmax
=
+
+
=
++=
==
=
==
-
BITS Pilani, Pilani Campus
PROBLEM :
Consider the twin torsional springs used in the bicycle
carrier. The spring material is oil quenched and tempered
(OQ & T) carbon steel (A229) having E = 198 GPa with a
spring index of 4. Each spring has lengths of the ends of
l1=100 mm and l2=200 mm. The number of body turns in
each spring is 5.3.
Find the required wire diameter (d) if the spring is to be
safe in bending as well as the maximum total angular
deflection be limited to 70o.
-
BITS Pilani, Pilani Campus
SOLUTION:For OQ&T A229: From Table 10-4: m = 0.187 and A = 1855 MPa.
( ) { } 229.1)1(4/14 2 == CCCCKi
( ) ( )
( ) ( )( )
++
=
++=
=====
=
===
CdCd
dK
d
EdD
llN
Ed
MD
dK
d
K
SdMFl
d
FlKS
n
dd
ASS
i
bt
ii
yt
iyt
d
mutyt
3
2001003.5
185587.0
32
64
180
70
3
64
185587.0
3232
32
1. Assume
;1855
87.087.087.0
187.0
3
4
21
4
187.0
33
3
187.0
Simplifying the above equation, we obtain a non-linear equation.
( )mmd
dd
387.2
327.188351.0222.1187.1
==
+=
-
BITS Pilani, Pilani Campus
PROBLEM :
A stock spring is shown in Fig. It is made from 1.8 -mm-diameter
music wire and has 4 .25 body turns with straight torsion ends. It
works over a pin of 10 mm diameter. The coil outside diameter is 15
mm.
(a) Find the maximum operating torque
and corresponding rotation for static
loading.
(b) Estimate the inside coil diameter and
pin diametral clearance when the spring
is subjected to the torque in part (a).
(c) Estimate the fatigue factor of safety
nf if the applied moment varies between
Mmin = 0.1 to Mmax = 0.5 N-m.
-
BITS Pilani, Pilani Campus
For music wire, from Table 104 we find that A = 2211 MPa
mmm and m = 0.145.
( )MPanSS
MPaSSandMPad
AS
yyt
utymut
1318/
158278.020298.1
2211145.0
==
=====
( )113.1
)1(4
14 2=
=CC
CCK i
The mean coil diameter is D = 15-1.8 = 13.2 mm. The spring index C = D/d = 7.33
( )
( ) ( )( )
mmNK
SdMFl
d
FlKS
i
yt
iyt
.3.678113.132
13188.1
32
32
33
maxmax
3
maxmax
====
==
Factor of safety is 1
Solution
-
BITS Pilani, Pilani Campus
( )( )( )( ) ( )
00
4
'
4
'4
'
'
72)360(2.02.01960008.1
25.42.133.6788.10
8.10
8.10
====
===
turn
Ed
MDN
DN
EdMk
c
bc
bc
c
( )( )( )( )
( ) 00'
'
44
'
'
21
67.78360218.0218.03104
3.678
.310465.42.138.10
1960008.1
8.10
65.4)2.13(3
252525.4
3
=====
====
=+
+=+
+=
turnk
M
mmNDN
EdMk
springcompleteofrateSpring
turnsD
llNN
s
t
at
s
ba
-
BITS Pilani, Pilani Campus
( )
MPar
MPad
MK
M
Mr
mmNMM
M
mmNMM
M
Fatiguec
mmDdDDD
mmN
DNDb
am
aia
m
a
m
a
pinpini
cb
b
2.583
8.3888.1
)200(32113.1
32
667.0
.3002
100500
2
.2002
100500
2
)(
7.0108.15.12'
5.1224.025.4
)2.13(25.4')(
33
minmax
minmax
'
'
==
===
==
=+
=+
=
=
=
=
====
=+
=+
=
-
BITS Pilani, Pilani Campus
From Table 1010, Sr = 0.50Sut = 0.50(2029)=1014.5MPa
( ) ( )( )
( )
22.18.388
4.474
4.474)2029(667.0
541211
5412
2029667.0
211
2
514
2029
2/5.10141
2/5.1014
2/1
2/
222
222
22
===
=
++=
++=
=
=
=
a
af
ut
e
e
uta
ut
r
re
Sn
MPa
rS
S
S
SrS
MPa
S
S
SS
-
BITS Pilani, Pilani Campus
Leaf Springs:
Consists of a series of flat plates,
usually of semi-elliptical shape.
The leaves are held together by means
of two U-bolts and a centre bolt
Rebound clips are provided to keep
the leaves in alignment and prevent
lateral shifting of the plates during
operation
The master leaf (longest leaf) is bent
to form the spring eye
Leaf springs are capable of exerting
large forces within comparably small
spaces
(Reference: Mechanical Design by Peter R. N. Childs)
-
BITS Pilani, Pilani Campus
Leaf Springs:
L LF F
2F
Without pre-stress
L LF F
2F
Pre-stressed
Extra full length leaves are
provided to protect the ends.
They are not uniform strength
beams and hence 50% more
bending stress exists in them. To
protect them they are sometimes
pre-stressed during the
assembly.
-
BITS Pilani, Pilani Campus
The multi-leaf spring shown and the single triangular section beam both have the same stress
and deflection characteristics with the exceptions that the multi-leaf spring is subject to
additional damping due to friction between the leaves
Source: Mechanical Design by Peter R. N. Childs
-
BITS Pilani, Pilani Campus
The bending stress in the plate which is uniform throughout is givenby
Where ng is no. of graduated leaves including master leaf
The deflection at load point is given by (deflection of a triangularplate)
The Spring rate is given by
( )
( )2
3
6
)(12
1
2/
bhn
FL
hbn
hLF
I
Mc
gg
b ===
( ) ( )3
3
3
33 6
)(12
12
2 hbnE
FL
hbnE
LF
EI
LF
gg
=
==
3
3
6 L
bhEnFk
g===
-
BITS Pilani, Pilani Campus
Problem:
A truck spring has 6 no of leaves. The supports are 1.5 mapart and maximum central load is to be 5.4 kN. Thepermissible stress for spring material is 280 MPa.Determine the thickness and width of the spring, if ratio oftotal width to depth of the spring is 4. Also find thedeflection of the spring.
-
BITS Pilani, Pilani Campus
Solution:
( )( )
( )( )( )
( )mm
hbnE
FL
mmb
t
mmbb
bbb
bhn
FL
g
g
b
156.0)85)(7.56(610210
1075.0107.266
854
6
7.5628010900
10900
)4
6)((6
1075.0107.266
33
333
3
3
2
3
2
3
2
33
2
=
==
==
==
=
==