mechanical springs

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BITS Pilani Pilani Campus VINAYAK KALLURI

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  • BITS PilaniPilani Campus

    VINAYAK KALLURI

  • 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.

  • BITS Pilani, Pilani Campus

    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

  • BITS Pilani, Pilani Campus

    Stresses in Helical Springs

    ( )2323max

    4842/16

    2/

    d

    F

    d

    FD

    d

    F

    d

    FD

    A

    F

    J

    Tr

    FDT

    +=+=+=

    =

    inside inside

  • BITS Pilani, Pilani Campus

    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

    =

  • BITS Pilani, Pilani Campus

    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.

  • BITS Pilani, Pilani Campus

    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

  • BITS Pilani, Pilani Campus

    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 +=

  • BITS Pilani, Pilani Campus

    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

    ==

    =

    +=

    =

    +=

    =

  • BITS Pilani, Pilani Campus

    Compression Springs

    Manufacturing processing at the ends and effect on total coils

  • BITS Pilani, Pilani Campus

    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.

  • BITS Pilani, Pilani Campus

    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

  • BITS Pilani, Pilani Campus

    SPRING MATERIALS

    Table 105

    Mechanical Properties of Some Spring Wires

  • BITS Pilani, Pilani Campus

    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

  • BITS Pilani, Pilani Campus

    SPRING MATERIALS

    Table 106

    Maximum Allowable Torsional Stresses for Helical Compression Springs in Static

    Applications

  • BITS Pilani, Pilani Campus

    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

  • BITS Pilani, Pilani Campus

    End -condition constants () for helical compression springs

    Table 102, page 522

  • BITS Pilani, Pilani Campus

    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:

  • BITS Pilani, Pilani Campus

    ).......(..........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

    =

    =

    =

    ==

    =

    ==

    ==+

    =

    ===

  • 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

  • BITS Pilani, Pilani Campus

    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:

  • BITS Pilani, Pilani Campus

  • BITS Pilani, Pilani Campus

  • BITS Pilani, Pilani Campus

    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:

  • BITS Pilani, Pilani Campus

  • BITS Pilani, Pilani Campus

    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:

  • BITS Pilani, Pilani Campus

  • BITS Pilani, Pilani Campus

  • BITS Pilani, Pilani Campus

    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:

  • BITS Pilani, Pilani Campus

    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

  • BITS Pilani, Pilani Campus

    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

    :

    =

    ==

    ==

  • 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

    ==

    +=

    =

  • BITS Pilani, Pilani Campus

    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

  • BITS Pilani, Pilani Campus

    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

  • BITS Pilani, Pilani Campus

    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

  • 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

  • BITS Pilani, Pilani Campus

  • BITS Pilani, Pilani Campus

  • BITS Pilani, Pilani Campus

    Tension/ Extension springs: end preparation

  • BITS Pilani, Pilani Campus

    Combined axial tension and bending stress at A

    Only torsion at B

    Improved design

    Side view

    Tension /Extension springs:

    Side view

  • BITS Pilani, Pilani Campus

    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

  • BITS Pilani, Pilani Campus

    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

    +=

    =

  • BITS Pilani, Pilani Campus

    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

  • BITS Pilani, Pilani Campus

    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.

  • BITS Pilani, Pilani Campus

    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.

  • BITS Pilani, Pilani Campus

    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

    =+=+=

    =

    =

    =

    =++=++=

    =

    ==

    =

    +=+=

    ===

    ===

  • BITS Pilani, Pilani Campus

    ( ) ( )( )( )

    ( ) [ ]

    ( ) ..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)

  • 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)

  • 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:

  • BITS Pilani, Pilani Campus

    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

  • 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

    =

    ===

    ++

    =

    =

    =

  • 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,

  • BITS Pilani, Pilani Campus

    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.

  • 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

  • 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

  • BITS Pilani, Pilani Campus

    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 =

  • BITS Pilani, Pilani Campus

    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

  • 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

    /

  • 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

    =

    ==

    ==

    ==

    =

    ==