Springs [Compatibility Mode]

I/C: KALLURI VINAYAK

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• Used for efficient storage and release of energy

• Strength and flexibility are two essential

requirements of spring design.

Spring Types

1. Helical springs (Tension / Compression)

2. Torsion spring

3. Leaf springs

4. Spiral spring

5. Belleville Springs

Leaf spring

5. Belleville Springs

Helical springsTorsion spring

Spiral spring

Belleville Spring

Stresses in Helical Springs

( )2323max

4842/16

2/

d

F

d

FD

d

F

d

FD

A

F

J

Tr

FDT

ππππτ +=+=+=

=

inside inside

Spring Index and shear stress correction factor

d

DC =

If we define the spring index to be as follows:

Then the foregone expression for maximum shear stress

can be expressed as:

323max

848

d

FDK

d

F

d

FDs πππ

τ =+=323 ddd πππ

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.

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 (K ) and BergstrasserProminent are Wahl factor (Kw) and Bergstrasser

factor (KB).

• These must replace Ks when incorporated.

Curvature effect in fatigue loading: Wahl factor and

Bergstrasser factor

24615.014 +− CC

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

34

24615.0

44

14

−+

=+−−

=C

CKor

CC

CK Bw

Fatigue Loading, effect of only curvature:

( )( )( )1234

242

+−+

==CC

CC

K

KK

s

Bc

Deflection and Stiffness:

aa

a

Gd

FDN

Gd

NFD

F

Uy

dAdJDNlFDT

Gd

DNF

Gd

NDF

AG

lF

GJ

lTU

24

3

24

2

2

4

3222

48

: theoremsecond so'Castiglian

.4/;32/;;2/

24

22

+=∂∂

=

====

+=+=

πππQ

a

aa

ND

Gd

y

Fk

Gd

NFD

CGd

NFDy

GdGdFy

3

4

4

3

24

3

24

8

8

2

11

8

D/d,Cindex spring gIntroducin

==

≅

+=

=

+=∂

=

Manufacturing processing at the ends and effect on total

coils

Compression Springs

Compression Springs

Formulas for the Dimensional Characteristics of Compression-Springs

Table 10–1

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.

SPRING MATERIALS

• Music wire, Oil-tempered wire, Hard drawn wire, Chrome-

vanadium wire and Chrome-silicon wire

mutd

AS = strength, tensileMinimum

Use Table 10-4 for “A” and “m”.

Table 10–4

SPRING MATERIALS

Table 10–5

Mechanical Properties of Some Spring Wires

SPRING MATERIALS

Unless otherwise specified, use MSS criterion for static design for springs

because the primary loading nature is shear.

Table 10–6

Maximum Allowable Torsional Stresses for Helical Compression Springs

in Static Applications

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

Critical Frequency of Helical Springs: Surging

• Spring surge or surging of springs is the problemand it is similar to the wave propagating in water

• If one end of springs is held stationary and otherend is disturbed, the springs vibrates violently

• Failure resulting from the resonance inherent insurging is found to be purely due to torsionalsurging is found to be purely due to torsionalshear and occurs at 45o to the wire axis.

• The governing equation for spring surging is thewave equation:

2

2

22

2

t

u

kgl

W

y

u

∂∂

=

∂∂

Here, u is the displacement, k is the spring stiffness, l is the length of the

spring, g is the acceleration due to gravity, W is the weight of the active part

of the spring

Contd.

The solution to this differential equation give the natural frequency of

vibration:

W

kgmπω =

W

kgmf

2=

m = harmonic number

222 γππ DNdd ( )( )

weightSpecific

44

222

=

===

γ

γπγπ

πγ a

a

DNdDN

dALW

The fundamental frequency, for m=1, should be from 15 to 20 times the

forcing frequency to avoid the resonance and hence in turn the surging of

the spring

Redesign to effect this normally involves increasing k or decreasing the W

STABILITY:• A compression spring is stable if it does not buckle under the

load

−−=

2/1

2

'

2'

10 11eff

cr

CCLy

λ

D

Leff

0αλ =

( )GE

EC

−=

2

'

1

( )EG

GEC

+−

=2

2 2'

2

π

( ) 2/1

02

'

2

2

21

+−

<⇒>EG

GEDL

C

eff απ

λ

For absolute stability and buckling not to occur,

Slenderness ratio Elastic constants

End -condition constants (α) for helical compression springs

Table 10–2, page 522

Helical Compression Spring Design for Static Service

• A helical coil spring force-deflection characteristic

• is ideally linear.

• For very small deflections, and near closure,

nonlinear behavior begins as the number of active

turns diminishes as coils begin to touch.

• The spring’s operating point to the central 75• The spring’s operating point to the central 75

percent of the curve between no load, F = 0, and

closure, F = Fs .

Fs = (1 + ξ )Fmax

Helical Compression Spring Design for Static Service

• In addition to the relationships and material

properties for springs, the recommended design

conditions are:

Spring index range : 6 ≤ C ≤ 12

No of active turns range : 3 ≤ Na ≤ 15No of active turns range : 3 ≤ Na ≤ 15

Robust linearity :ξ ≥ 0.15

Factor of safety at closure : ns ≥ 1.2

4 cost) material (relative- fom merit, of figure The

22 DNd tγπ=

Helical coil compression spring design for static loading.

From Table A-28; 1051

• A music wire helical compression spring is needed to

support an 89 N load after being compressed 50.8 mm.

Because of assembly considerations the solid height

cannot exceed 25.4 mm and the free length cannot be

more than 101.6 mm. Design the spring.

• 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

Design for Fatigue Load:

• 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

Zimmerli’s data.

Zimmerli’s 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) withglass or ceramic particles of 1/64 inch diameter) with

force sufficient to create plastic deformation

Zimmerli’s 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 becomponents for infinite life were found to be

utsyututsu

smsa

smsa

SSSorSS

MPaSMPaS

Peened

MPaSMPaS

Unpeened

557.035.067.0

534398

:

379241

:

≤≤=

==

==

Design for Fatigue Loading Based on Zimmerli’s

Data

criterion. failure fatigue aapply then and

,,,,,,,, minmax susyemama SorSSFFfindgivenFF ττ

minmaxminmax

22

FFFand

FFF ma

+=

−=

33

88

22

d

DFKand

d

DFK

FandF

mWm

aWa

ma

πτ

πτ ==

==

Zimmerli’s Data (Gerber criteria)

su

sm

sase

su

sm

se

sa

lineloadgivenforcordinatetionInter

S

S-

SS

S

S

S

S

=⇒=

+

2

2

sec

1

1

limit endurance thefindThen Ssa and Ssm are from

Zimmerli’s data.

Ssu= 0.67Sut

Factor of Safety,

a

saf

Sn

τ=

m

a

sm

sa

ut

se

se

susa

F

F

S

Sr

rS

S

S

SrS

==

++−=

222 2

112

Refer Table 6-7 ; page 307

Zimmerli’s Data (Goodman criteria)

su

sm

sase

su

sm

se

sa


S

S-

SS

S

S

S

S

=⇒=

+

sec

1

1


Zimmerli’s data.

Ssu= 0.67Sut

Factor of Safety,

a

saf

Sn

τ=

m

a

sm

sa

sesu

susesa

F

F

S

Sr

SrS

SrSS


==

+=

sec


Zimmerli’s Data (ASME- Elliptic criteria)

sy

sm

sase

sy

sm

se

sa


S

S-

SS

S

S

S

S

=⇒=

+

2

22

sec

1

1


Zimmerli’s data.

Ssu= is to calculated

from Table 10-5

page 526

Factor of Safety,

a

saf

Sn

τ=

m

a

sm

sa

syse

syse

sa

F

F

S

Sr

SrS

SSrS


==

+=

222

222

sec


Tension/ Extension springs: end preparation

Combined axial tension and

bending stress at A

Only

torsion at

B

Side

views

Tension /Extension springs:

Improved design

views

Analysis of stresses in tension springs

( )

( )

( )( ) d

rC

CC

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

=−−−

=

=

+=ππ

σ

( )

( )

( ) ( )( ) d

rC

C

CK

K

d

FDK

B

B

BB

22

2

2

3

2,

44

14

bygiven curvature,for factor correction strss Torsional

8

bygiven is Bat stressshear torsionalmaximum The

=−−

=

=

=π

τ

Extension springF

A

B

( )

( )( ) d

rC

CC

CCK

dd

DKF

A

AA

11

11

1

2

1

23

2,

14

14

416

=−−−

=

+=ππ

σ

( )d

FDK BB 3

8=

πτ

Stress is to be computed at three locations

C

( ) ( )( ) d

rC

C

CK

d

B2

2

2

2 2,

44

14=

−−

=

π

34

24

83

−+

=

=

C

CK

d

FDK

B

BC πτ

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

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.

• When the spring is completed and removed from the

mandrel, the initial tension is locked in because the spring

cannot get any shorter.

( )MPa

C

e Ci

−−±=

5.6

349.6

231

is, stress torsionalduncorrecte of range Preferred

105.0τ

PROBLEM

• A hard-drawn steel wire 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.

The number of body turns is 12.17. From the

given information: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.

Solution:

( ) ( )

( )( )mmN

Gdk

turnsE

GNN

CCK

dDC

mmdODD

ba

B

/885.39.01079

57.1210198

107917.12

254.134/24

67.59.0/1.5/

1.59.06

434

3

3

=×

==

=××

+=+=

=−+=

===

=−=−=

( )( )( ) ( )

( ) ( ) ( ) ( )( )

mmyLL

mmk

FFy

mmdNdDL

mmNND

Gdk

o

i

bo

a

14.2589.425.20

89.4885.3

524

25.209.0117.129.01.5212

/885.357.121.58

9.01079

8

11

11

33

=+=+=

=−

=−

=

=++−=++−=

=×

==

( ) ( )( )( )

( ) [ ]

( ) range. preferred in thenot isIt .1.1027.248.126

5.1517.248.1265.6

367.549.6

231

07.899.0

1.5588

is stress initial duncorrecte The

min

67.5*105.0max

33

MPa

MPae

MPad

DF

i

i

i

uncorri

=−=

=+=

−−+=

===

τ

τ

ππτ

( )( )( )

15.5369.0

1.5248254.1

8

33

11 ===

ππτ B MPa

d

DFK

( )

( )

( )

526.115.536

55.818S

55.818181945.0S45.0S

18199.0

1783

d

AS

9.0

1

sy

utsy

190.0mut

331

===

===

===

τ

ππ

s

B

n

MPa

MPa

d

Contd.

( )( )

( )( )

( ) ( ) ( )( ) ( )( )

( ) 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

uty

231

2

11

1

2

1

11

===

=

+=

=−−−

=−−−

=

====

A

A

S

MPa

MPa

CC

CCK

C

ππσ

( )( )

( ) ( )( )

( ) ( ) ( )( )

( ) 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

31

2

2

22

1

====

=−−

=−−

=

=−

=−

==

===

ByB

B

A

y

Ay

nMPa

C

CK

d

dDC

Sn

πτ

σ

�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

TORSION SPRINGS:

separates the body coils to avoid

intercoil friction.

�The wire in the torsion spring is

in bending

Free

End

Free

end

location

Back

angle

Angular

rotation,

proportio

-nal to Fl

TORSION SPRING

Fixed

End

location

angle

For all positions of the moving end θ + α =Σ = constant.

turnspartial;body turns

integer360

integer

==

+=+=

pb

pob

NN

NNβ

Bending Stress :

The bending stress can be expressed as

22 1414

''

CCKand

CCK

factorcorrectionstressisKwhere

I

McKσ

−+=

−−=

=

3

3

32,

32

)1(4

14

)1(4

14

d

FlKisequationbendingthe

d

c

IandFlMngSubstituti

unitythanlessalsoandKthanlessalwaysisK

CC

CCKand

CC

CCK

i

io

oi

πσ =

==

+−+

=−−−

=

Torsional stiffness:

( ) radiansEd

Ml

dE

Fl

EI

Fl

l

ye 44

22

3

64

64/33

:deflection End

ππθ ====

The end deflection is bending of a cantilever beam whereasThe end deflection is bending of a cantilever beam whereas

the coils undergo bending action under M = Fl requiring

application of Castigliano theorem.

Strain energy in bending, ∫= EI

dxMU

2

2

bbc

DNDN

c

Ed

MDN

Ed

FlDN

dI

EI

dxFl

EI

dxlF

FF

Ul

bb

44

4

0

2

0

22

6464

64/

2

==⇒

=

=

∂∂

=∂∂

= ∫∫

θ

π

θππ

The Force ‘F’ will deflect through a distance “lθ”

Torsional stiffness

( ) ( )

atebae

bb

t

eect

NEd

MDNNN

D

llN

Defining

D

llN

Ed

MD

Ed

Ml

Ed

Ml

Ed

MDN

EdEd

4

21

21

44

2

4

1

4

21

64;,

3

3

64

3

64

3

6464

(rad), deflectionangular totalThe

=+=+

=

++=++=

++=

θπ

πππθ

θθθθ

Stiffness expressions in torque/radian units:

bc

cDN

EdMk

64

4

==θ ( )21

4

64

3

ll

EdMk

e

e +==

πθ

Stiffness values in torque/turn values (i.e 2π rad /turn) :

at

sDN

EdMk

64

4

==θ

π24

' ×==EdM

k π24

' ×==EdM

k( )

ππ

23 4

' ×==EdM

k

Torsional stiffness

πθ

264'

' ×==at

sDN

EdMk π

θ2

64'

' ×==bc

cDN

EdMk

( )π

πθ

264

3

21

'

' ×+

==ll

EdMk

e

e

at

sDN

EdMk

8.10

4

'

' ==θ bc

cDN

EdMk

8.10

4

'

' ==θ ( )21

4

'

'

8.10

3

ll

EdMk

e

e +==

πθ

Tests show that the effect of friction between the coils is such that the

constant 10.2 (i.e 64/2π) should be increased to 10.8

Torsion spring supported on round bar or pin:

cb

b

N

DND

+= ndeformatioafter and before balance volumefrom,'

'θ

�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, in which case loss of spring function would

ensue

�The helix diameter of the coil D′ becomes

( ) ( )[ ]b'

cb

cb

NπDAθNπD'A

N

××=+××

+θ

( )pin

pinc

b

pin

cb

bpinpini

DdD

DdN

DdN

DNDdDDD

−−∆−

++∆=

−−+

=−−=−=∆

'

1

' '

θ

θ

The new inside diameter D′i = D′ − d makes the diametral clearance ∆

between the body coil and the pin of diameter Dp

Design of Torsion Springs for Strength:

•Static strength

mutd

AS =

Table 10–6

First column entries in Table 10–6 can be divided by 0.577 (from distortion-

energy theory) to give

Documents

Springs [Compatibility Mode]