Top Banner

of 7

Analysis of Traction Forces in a... Kannel 1986

Feb 25, 2018

Download

Documents

Paulo Negrão
Welcome message from author
This document is posted to help you gain knowledge. Please leave a comment to let me know what you think about it! Share it to your friends and learn new things together.
Transcript
  • 7/25/2019 Analysis of Traction Forces in a... Kannel 1986

    1/7

    J. W . Kanne l

    Battelle Columbus Laboratory,

    Columbus, Ohio 43201

    T. A. Dow

    Precision Engineering Laboratory,

    North Carolina State University,

    Raleigh, N.C. 27695-79 10

    Analysis of Traction Forces in a

    Precision Traction Drive

    A theory for the shear stress between a rough elastic cylinder and a cylinder w ith a

    soft laye r has been developed. The theory is based on a Fourier tran sform approach

    for the elasticity equations coupled with surface deflection equations for transient

    contacts. For thick layers (h > .001 in.) the shear stress on the surface approaches

    the shear of the layer alone. The elastic shear deflection (-100 \xin.) as a result of

    the tangential load is significant and increases if a surface layer such as a thin

    coating is added to one or both cylinders. The predicted interfacial shear stresses are

    considerably altered by surface roughn ess on uncoated surfaces and these effects

    are ameliorated by the addition of a thin soft surface coating.

    Intr oduc t ion

    Two crit ical aspects of a precision machine are l inear

    location of one part relative to another and smooth motion

    between lim its. One m eth od of achieving this result [1, 2] is to

    use a traction drive on the slideway as i l lustrated in Fig. 1.

    Positioning accuracy below the microinch level is typically

    required. W hen accu racies of this level are involved, virtually

    all factors which affect motio n must be considered in order to

    minimize errors in the system. One such factor is the shear

    deformation of the drive system, especially the elasticity of

    the traction interface. The elasticity is affected by many

    factors, including the Young's modulus of the traction

    components, surface layers on the rollers (such as solid fi lm

    layers) and the roughn ess of the rollers and slideway.

    The most extensive work reported on the analysis of the

    traction interface is by Kalker [ 3 , 4 ] . Kalker traces the traction

    interface between two extremes: the Cattaneo [5] problem and

    the Carter [6] problem. The Cattaneo problem occurs when a

    cylinder is rotated slightly, while in contact with a stationary

    surface. The Carter problem occurs when both the cylinder

    and the mating surface are moving but at sl ightly different

    speeds. Kalker 's study traces the traction forces through the

    transients between the two extremes.

    Bentall and Johnson [7] analyzed the slip between two

    dissimilar cylinders in rolling contact. This research allowed

    for tangential deflections due to microslip. Barber [8] con

    ducted research similar to Kalker 's, only he analyzed three-

    dimensional contacts of rollers under misalignment. Poritsky

    [9] derived basic equations for cylinders in contact and

    discussed the problem of rough surfaces. Krause and Senuma

    [10] did experimental studies with rollers which developed

    surface corrugations. The surface corrugations notably af

    fected the tractio n beh avior of the cylinders.

    The work presented here is an extension of the Kalker and

    Poritsky work with allowances for surface layers. The surface

    layer algorithm is developed from the work of Sneddon [11],

    and Gupta and Walowit [12].

    Contributed by the Tribology Division of THE

    AM ERICAN SOCIETY

    OF

    MECHANICAL ENGINEERS and presented at the A SM E/A SL E Joint Tr ibology

    Conference, A tlanta, Ga ., October 8-10, 1985. M anuscript received by the

    Tribology Division, A pril 19, 1985. Paper N o. 85-Trib-45.

    Leaf spring

    Hydrostatic bearings

    Drive motor

    Traction bar

    Idler roller

    Capstan

    Hydrostatic

    bearings

    The resolution of the slide drive will be

    0.2pin. (0.005yumj. This is the equivalent to

    0.05 arc.

    Fig. 1 Illustration of traction control system for precision engineering

    A p p r o a c h

    The same general approach used for the normal stress

    analysis [13] can be used for the shear stress computa tions. A s

    will be shown subsequently, the matrix equation is almost

    identical to that developed by Gupta-Walowit for the normal

    stress computations. That is the deformation equations can be

    put in the matrix form:

    CF=e

    (1)

    where C is the matrix equation with elements c,y that relate the

    tangential de flection, e, to the tangential forces. A s with the

    normal stress equations, a relationship between point-loads

    and po int stresses can be developed.

    Matrix Coeff ic ients

    For a solid body, Po ritsky derived a similar relationship for

    load-d eflec tion coefficients as for nor ma l stress coefficients,

    that is

    4(1

    v

    2

    )

    7rE

    In I/-1

    (2)

    Journal of Tr ibology JULY 1986, Vol . 108/4 03Copyright 1986 by ASME

    wnloaded From: http://tribology.asmedigitalcollection.asme.org/ on 01/16/2016 Terms of Use: http://www.asme.org/about-asme/terms-of-use

  • 7/25/2019 Analysis of Traction Forces in a... Kannel 1986

    2/7

    1

    -s

    2

    1

    - z

    Z( v

    x

    ' -

    _

    (1 + " '

    where

    z-

    ? =

    1)

    )

    3 - z -

    (1 +

    1-

    =lsl

    =

    exp(z)

    0

    I

    \-z

    \'(\-z)

    )-2z/s

    2

    1

    s

    1

    7

    2

    zt

    1

    ( l - v , ' ) z ?

    2

    (1 + I V ) ?

    2

    v'=(2-v)/(\-v)

    v

    =v/(\-

    V

    )

    Table 1

    0

    s

    -,2

    ( 1 + z ) ?

    2

    [2+ ( l + z ) ( l - V ) J '

    2

    [(l + v

    x

    ) + 2z/s

    2

    ]l

    2

    0

    0

    - ?

    z?

    ( 1 - K

    2

    ' ) Z ? ( y

    2

    '

    - ( 1 - K

    2

    " ) 7 ? . 2z7? /s'

    3)7?

    s

    2

    c,

    Di

    A

    2

    B

    J

    --

    -i

    =

    =

    1

    0

    0

    0

    ^oj

    /3= ( 1 - ^

    2

    ) E , / ( 1 - . ,

    2

    ) E

    3

    7 = ( l -

    J

    ' 2

    2

    ) E i / ( l - " i

    2

    ) E

    2

    wherer is the distance between a point tangential load and the

    corresp onding tange ntial deflections. A s with the norm al

    stress equations, a relationship between point loads and

    stresses over a small region is:

    F:

    rdx

    (3)

    where Tis the surface shear stress acting over a distance dx (or

    A x in finite term inology). Then

    -4(1

    -v

    2

    )

    7rE

    j -

    \n\x:x\dx (4)

    Inlet

    F ig .2 S l ippage in the contac t zone

    which allows a solution to be found whenx

    ;

    = x

    h

    A n expression has been developed for the influence-

    coefficient matrix for shear stresses in a layered solid (see

    A ppendix A ) . The so lu t ion (assuming no normal pressure on

    the surface) can be written as:

    7rE

    [ J o

    2 5

    =i

    where e,(0) is the assumed initial deflection relative to a fixed

    poin t (x

    0

    = - 106). For subsequent t imes equation (7) is used

    with AV At being a constant that is added to each time step to

    produce a given traction.

    In the computations the coefficients c,y were set and the

    shear stress computed by a matrix solution of equation (12).

    A t some poin ts

    Tj >f'Pj

    wh ere / is a coefficient of friction an dP j is the local pressure

    computed using the technique given in reference [12 and 13].

    For these points the matrix was adjusted as follows:

    n

    Y, l--8jj

    )(c

    u

    -c

    0J

    )TjAx=e

    i

    / =

    n

    -H h

    f

    fPj^iJ-

    c

    oj)-r

    Jf

    =fPj

    f

    (13)

    where

    I

    1

    J=Jj

    j

    are the po in ts where r

    y

    >

    fpj.

    The essential size of the

    matrix will be reduced by one row for each value ofjf. The

    computations involved a simple i teration starting with a full

    matrix and subsequently reducing the matirx for each value of

    Tj > fPj until further i terations produced no changes.

    D isc uss ion

    Figure 3 i l lustrates the shear stress at the interface for th e

    case of a stationary lower cylinder (Vt/b = 0) and for a series

    of relative slip values

    (Vt/b

    > 0). The slip values are ex

    pressed in terms of the half width, b. I f Vt/b = 0.8, a point

    on the upper cylinder has moved a distance equivalent to 80

    percent of the half-width of con tact between the cylinders.

    For the case of a stationary lower disk (analogous to the

    Cattaneo problem), the shear stresses are the lowest in the

    center of contact and rise toward the edges. This rise is due to

    the contribution of each element to support the shear

    deformation outside the contact region. The rise in shear

    stress is limited at

    x/b =

    0.6 by slip between the cylinder

    surfaces; tha t is, the shear stress becomes equal to the friction

    coefficient t imes the normal stress. Thus, the shear stress

    curve has the same shape as the normal stress distribution

    where slip is present. For this example the horizontal load was

    105 N /m m (600 lb / in . ) , and to produ ce th is force, the upper

    cylinder was rotated 2 X

    10

    ~

    4

    rad (the upper surface moved 2

    /mi).

    Cont inued ro tat ion of the upper cy l inder would produce

    rotation of the lower disk because the loading was assumed to

    remain constant (analogous to Carter problem). Different

    values of the upper cylinder motion are also i l lustrated in Fig.

    3 . For the largest rotation of the upper cylinder shown in Fig.

    3 (Vt/b = 1.2, correspon ding to 3.2 x 10

    2

    rad (about 2 deg)

    of rotation), a point on the upper cylinder has moved a total

    of 2.5 ^m from a point on the lower cylinder which was

    adjacent at no load. This means that the driving cylinder will

    rota te 2.7 x 1 0

    4

    rad more than the driven cylinder for an

    average rotation of 3.2 X 10 ~

    2

    ra d against a load of 6001b.

    One purpose of the analytical traction model was to

    Journal of Tr ibology JULY 1986, Vol . 1 08/40 5

    wnloaded From: http://tribology.asmedigitalcollection.asme.org/ on 01/16/2016 Terms of Use: http://www.asme.org/about-asme/terms-of-use

  • 7/25/2019 Analysis of Traction Forces in a... Kannel 1986

    4/7

    -0. 8 -0.6 -0.4 -0. 2 0 0.2 0.4 0.6 0.8

    Distance From Contact Center, x/ b

    Fig .

    4 Con tact zone shear stress distribution (0.01 in. layer)

    b = .89 mm (.035 in.) Load = 543 N/mm (3099 lb/in.)

    F, = 14.5 N/mm (83 lb/in.) R=9.44 m m (.37 in.)

    E T =2 GPa

    290 ksi)

    p

    h

    =.44 GPa

    64

    ksi)

    e

    = 2.5 pm (100 ^in.)

    evalaute the role of surface layers on traction drive per

    forma nce. Figures 4 to 6 show the shear stress distribution for

    the condition of a uniform tangential displacement of 2.5 xim

    but with a soft surface layer (E = 2 GP a) on one of the

    cylinders. Surface layer thicknesses of 250 /xm, 25 ttm and 2.5

    /xm are illustrated.

    A ll curves show peak shear stresses at the edges of contac t

    and reasonably uniform shear stress in the center. For the

    thicker layers this center shear stress should approach the

    shearing of a soft layer of known thickness a known amount.

    That is:

    T

    G

    m

    y G

    m

    e/h (14)

    where

    7 is the shear strain

    G, is the shear modulus of the layer (G, = E /2 ( l + v) =

    7 80 M Pa)

    For the 250 /xm layer shown in Fig. 4, the predicted shear

    stress using equation 14 is 7.8 M Pa (1.125 ksi). This predicted

    stress is consistent with the stress near the center of contact of

    Fig. 4. A t the edges of the contact (x = b) the stress rises

    considerably above this level to compensate for the forces

    required to tange ntially deflect the layer outside of the contac t

    region. Very near the edges, the shear stress is l imited by the

    coefficient of friction times the normal pressure.

    In Fig. 5 the predicted shear stress using equation (14)

    would be 78 M Pa (11.25 ksi). This stress is higher than the

    stress at the center of contact given in Fig. 5. That is, for thin

    films, the shear stresses tend to be high enough to deflect the

    substrate as well as the surface. When the substrate is

    deflected this deflection must be subtracted from

    e

    in com

    puting the surface shear. For very thin fi lms (h = 2.5 /xm) as

    shown in Fig. 6 the center shear stress is considerably lower

    than predicted by shearing of the layer alone. For this case the

    equatio n (14) shear stress would be 780 M Pa (112.5 ksi) or

    about four t imes that given in Fig. 6. Clearly then the

    evaluation of this surface films requires the use of com

    prehensive theories and cannot be achieved by simple

    analyses.

    Figures 7 and 8 i l lustrate the compariso n between a layered

    and a nonlayered body. For the conditions given here a

    surface "wind-up" of 2.5 / t in produces a tangential load of

    105 N /m m w hen the soft layer (E = 2 GP a) is in place. If

    there were no layer, 105 N /m m could be obtained with a

    - 0 .8 - 0 .6 - 0 .4 - 0 .2 0 0 .2 0 .4 0 .6 0 .8

    Distance From Contact Center, x/ b

    Fig . 5 Con tact zone shear stress distribution (1000 in. layer)

    b

    = .44 mm (.0173 in.) Load = 490 N/mm (2800 lb/in.)

    F

    (

    = 45 N/mm (258 lb/in.)

    R =

    9.4(.37 in.)

    E = 2 GPa 290 ksi) p

    h

    =.8 GPa 117ksi)

    e= 2 . 5 / i m ( 1 0 0 x 1 0 % i n . )

    45

    40

    35

    p

    s

    " o

    30

    S

    t

    e

    S 20

    CO

    15

    10

    5

    -

    ~ / ^\

    I I I

    -

    y \ -

    \ -

    i i i

    - I .0 -0 .8 -0.6 -0. 4 -0. 2 0 0.2 0.4 0.6 0.8 I .O

    Distance From Contact Center, x/ b

    Fig . 6 Con tact zone shear stress distribution (100 p in. layer)

    b = .29 mm (.0115 in.) Load = 544 N/mm (3110 lb/in.)

    F

    (

    = 103 N/mm

    591

    lb/in.) R= 9.4 mm (.37 in.)

    E=2G Pa 290ksi) p

    h

    = 1.28 GPa 187ksi)

    c

    = 2 .5 / im(100/i in.)

    "wind-up" of 2/xm, as indicated in Fig. 8. The presence of the

    soft layer also increases the amount of "wind-up" required to

    move the driven cylinder against the load. For example, for

    the bare cylinder (Fig. 8) a " w in d- up " of 2.5 /xm occurs in

    traversing through 1.2 half widths 0.3 mm). For the layered

    cylinder (Fig. 3) a "wind-up" of 3.5 tun occurs in traversing

    the same distance.

    For the above examples the surface of the drive cylinder

    would move 2.5 t tm farther than the driven cylinder over 0.3

    mm traverse. If one of the cylinders contained a soft layer the

    differential tra verse w ould be increased by ab out 1 /xm. In a

    precision control system, all errors must be minimized to

    reduce the level of error compensation required of the control

    system. Based on error minimization alone then, i t would

    seem that bare cylinders would be superior to coated cylin

    ders. However, when surface roughness factors are included

    40 6/V ol . 108, JULY 1986 Transact ions of the ASME

    wnloaded From: http://tribology.asmedigitalcollection.asme.org/ on 01/16/2016 Terms of Use: http://www.asme.org/about-asme/terms-of-use

  • 7/25/2019 Analysis of Traction Forces in a... Kannel 1986

    5/7

    -1.0 -0 8 -0, 6 -0.4 -0. 2 0 0.2 0.4 0.6 0.8 10

    Distance From Contact Center, x/b

    Fig.

    7 Conta ct zone shear stress distribution (100p in. layer)

    b = .29 mm (.0115 in.) Load = 544 N/mm (3110 lb/in.)

    F

    t

    =105 N/mm (600 lb/in.) R=9.4 mm (.37 in.)

    D

    =

    2

    G Pa

    290

    ksi)

    p

    h

    =

    1.28 GPa

    187

    ksi)

    -1.0 -0.B -0.6 -0.4 -0. 2 0 0.2 0.4 0.6 0.8 I.0

    Distance From Contact Center, x/ b

    Fig.

    9 Shear stress distribution with 1 p in. CLA surface roughness (no

    layer)

    b =.25mm (.01 in.)

    F, = 105 N/mm (600 lb/in.)

    E = 200GPa(29Mp s i )

    = 1.9 pm (75 pin.)

    Load:

    R-

    Pma x

    =

    :639 N/mm (3649 lb/in.)

    9.4 mm (.37 in.)

    2 GPa (295 ksi)

    -I.O -0.8 -0.6 -0. 4 -0.2 0 0.2 0.4 0.6

    Distonce From Contact Center, x/b

    Fig .8 Conta ct zone shear stress distribution (no layer)

    b

    = .25 mm (.01 in.) Load = 607 N/mm (3470 lb/in.)

    F, = 105 N/mm (600 lb/in.) R=9.4 mm (.37 in.)

    =

    200

    G Pa

    29M

    psi)

    p

    h

    = 1.5 GPa

    219

    ksi)

    in the shear stress examinations the value of a surface layer

    becomes clear. Figure 9 indicates the shape of the shear stress

    distribution for a stationary lower cylinder without a layer but

    with a 0.025

    pm

    center line average (cla) surface roughness.

    The increased deformations due to the surface roughness

    produces peaks in the shear stress distribution in the slip

    regions

    (x/b >

    0.7). The addition of a soft, thin layer (2.5

    nm )cushions the surface asperities and produces a smoother

    shear stress curve as indicated in Fig. 10. Based on the results

    of Figs. 9 and 10, it would be difficult to compensate for the

    erratic stresses for a nonlayered cylinder. However, it is quite

    reasonable to attempt to predict the stresses where a layer is

    present.

    - 10 - OB - 0 6 - 0 .4 - 0 .2 0 0 .2 0 .4 0 6

    OB

    10

    Distance From Contact Center, x/ b

    Fig . 10 Shear stress distribution with 1 p in. CLA surface roughness

    (100 pi n. layer)

    b = .

    29 mm (.0115 in.)

    F

    (

    = 100 N/mm 601lb/in.)

    =

    2

    G Pa

    29

    ksi)

    e= 2.5pm 100pm)

    Conclusions

    Load = 545 N/mm (3114 lb/in.)

    R = 9.4mm(.37 in.)

    Pmax =

    1

    -

    3 3 G P a

    Traction drive systems represent reasonable devices for

    traversing slideways in precision machining. However, one

    inherent problem with traction is that the traction interface

    must incur sizable elastic "wind-up" before the driven

    cylinder will move against a given load. For one specific case

    analyzed the "wind-up" was on the order of 2 ^m for a 105

    N /mm traction load. The amount of "w ind-u p" increased as

    the driven cylinder moved.

    In order to achieve precision control in a traction drive,

    some type of compensation algorithm must be employed to

    eliminate "wind-up" errors. It would be expected that a

    compensation algorithm of the type presented herein could be

    employed provided good reproducibility of the traction in-

    Journal of Tr ibology JULY1986, Vo l . 108 /407

    wnloaded From: http://tribology.asmedigitalcollection.asme.org/ on 01/16/2016 Terms of Use: http://www.asme.org/about-asme/terms-of-use

  • 7/25/2019 Analysis of Traction Forces in a... Kannel 1986

    6/7

    te r face could b e achieved. O n ep r o b l e m in r e p r oduc i b i l it y is

    surface roughness . Even smal l levels o f roughness (0 .025 t im

    cla)

    c a n

    cause wi ld var ia t ion s

    in

    in ter fac ia l shear s t resses .

    A

    th in coat ing such as a 2 .5fim mo lyb den um disul f ide coat ing

    c a n a bs o r b t h e r oughne s s a n d crea te a muc h s moo t he r ( a n d

    hence more reproducible) shear s t ress d is t r ibut ion . T h e

    pr e s e nc e o f t h e soft layer wou ld cau se a slight (0. 6 i tm)

    increase

    i n

    " w i n d - u p "

    b u t

    p r e s uma b l y

    t h e

    " w i n d - u p " w o u l d

    be predic table .

    .Body 3

    ,v

    . 4 $ 2 b *

    T ( X )

    ) h Body 1

    t

    w

    7

    y

    r

    E..

    '

    Body 2

    References

    1 Bryan, J . B., "De sign and Construction of an Ultraprecision 84 Inch Dia

    mond Turn ing M achine,"

    Precision Engineering,

    Vol. 1, N o. 1, 1979, pp.

    13-17.

    2 Barkman, W. E. , "M achine and Tool Drive Sys tem,"

    Precision Engineer

    ing,

    Vol. 2, N o. 3, 1980, pp. 141-146.

    3 Kalker, J . J . , "Trans ien t Ro l l ing Contact Phenomen a,"

    Trans. ASLE,

    Vol. 14, 1971, pp. 177-184.

    4 K alker, J . J ., "R oll ing With Slip and Spin in the Presence of Dry Fric

    t i o n , "

    Wear,

    9, 1966, pp. 20-38.

    5 Cattan eo, C , "Sul Contatto di du Corpi Elast ici: Destribuzione Locale

    Degli Sfoizi ," fiend. Acad.

    Lincei,

    Series 6, Vol. 27, 1938, pp. 342-348,

    434-436, 474-478.

    6 C ar ter , F . W., "O n the A ct ion of a Locomot ive Driving Whe el ,"

    Proc.

    Royal Soc, a 112, 1926, pp . 151-157.

    7 B entall , R . H., and John son, K. L., "Slip in the Roll ing Contact of Two

    Dissimilar Elast ic R ollers ,"

    J. Mech. Eng. Sci.,

    Vol. 9, 1967, pp. 389-404.

    8 Barber, J . R., " The R oll ing Contact of M isaligned Elast ic Cylin ders, " / .

    Mech.

    Eng. Sci.,

    (I . M ech. E.), Vol. 22, N o. 3, 1980, pp.. 125-128.

    9 Poritsky, H., "Stress and Deflections of Cylindrical Bodies in Contact

    Wi th Appl icat ion to Contact Gears and L ocomot ive Wheels ," A SM E

    Journal

    of Applied Mechanics, 1950, pp . 191-201.

    10 Krause, H., and Senuma, T., " Investigation of the Influence of Dynamic

    Forces on the Tribological Behavior of Bodies in R oll ing/Sliding Contac t W ith

    Par t icu lar Regard to Surface Corrugat ions ," A SM E

    JOURNAL

    OF

    LUBRICATION

    TECHNOLOGY Vol. 103, 1981.

    11 Sneddon, I . N . , Fourier Transforms, M cGraw-Hill , 1951.

    12 Gupta, P . K., and Walowit , J . A ., "C ontac t Stresses Between a Cylinder

    and a L ayered Elast ic Sol id ," A SM E

    JOURNAL

    OF

    LUBRI CATION TECHNOLOGY,

    A pr. 1974, pp. 250-257.

    13 Kannel, J . W., and D ow, T. A ., "E valuatio n of Contact Stresses Between

    a R ough Elastic and a L ayered Cylinder," to be presented at the Leeds-L yon

    Conference, Sept. 1985.

    A P P E N D I X

    Development of Shear-Deflection Equations

    Fourier Transform Equation. Th e objective of this

    analysis is to develop a relationship between surface shear

    stresses and tangential deflections in absence of applied

    normal stresses at the boundary. The analyses are based on

    elasticity theory using the Fourier transform approach given

    by Sneddon and Gupta and Walowit . These equat ions are

    given in the following form (see Fig. A l) .

    d

    2

    ^ I f

    0 0

    = \ c o

    2

    Gexp(- f'cox)da>

    2ir J -oo

    x

    2

    d

    2

    V

    dy

    2

    d

    2

    V

    J

    d

    2

    G

    dy

    2

    exp(- /cox)a ' co

    dxdy 2ir J -

    :

    d G

    t

    w

    (co

    ex p( - /cox) aco

    dy

    ( A l )

    l-v

    2

    (" fd

    3

    G (2-v)\

    2

    dG-\ .

    v = \ n (

    )

    u exp ( - /cox)

    2TTE J-ooLcfy

    3

    \ l-v / dy 1

    doi

    u =

    2TTE

    p [d

    2

    G / v \ , -1. , . du

    whe r e

    V i s the

    A i ry s t ress funct ion which sa t i s f ies

    t h e

    b i ha r mon i c e qua t i on a n d G i s the Fou r i e r t r a n s f o r m of Sf,

    t ha t is

    V

    4

    ^

    = 0

    (A2)

    Fig.

    A1 Coo rdinate system for shear stress analysis

    oo

    $ exp(, wx)dx (A3)

    Eliminating ^ from the above two_ equations and solving the

    resulting differential equation in G, we get a solution of the

    form

    G= A+By)exp - \u\y) + (C+D y)exp(\o)\y)

    (A4)

    whe r e

    A , B , C ,

    a n d

    D

    a r e

    c ons t a n t s

    o f

    i n t e g r a t i on

    to be

    e va l ua t e d a tt he bo unda r y .

    Boundary Condit ions

    The bounda r y c ond i t i ons fo r the t rac t ion analys isa r e :

    1) bou nda ry s t ress i s t h eapp lied shear stress

    2) stress a n d def lec t ions a r e c on t i nuous a c r o s s t h e layer

    in ter face

    3) stress goes

    t o

    zero a t oo;

    at t h esur face ,

    d

    2

    *

    dxdy

    dG

    p

    = -I 1 /'co exp(iu>x)du

    2 i r J - = dy

    (A6)

    ,dG

    If is an odd function, Sneddon shows that:

    dy

    1 f

    IT

    JO

    d G

    J

    co cos wxaco

    dy

    Based on Fourier Transform theory ;

    dG

    dy

    o

    co s

    OJ Xdx

    (A7)

    (A8)

    a t t he s ur fa c e v = 0 r = - r

    0

    . Let t ing T

    0

    be defined over the

    interval A x and lett ing T

    0

    = 1/Ax then lim we have

    dG

    c o =1 (A9)

    dy

    L et t ing

    s = hu, G

    =

    G/h

    2

    , r\ - y/h,

    the first boun dary

    condition becomes

    dG

    , f n

    57- =

    1

    fo r n= 0

    d{

    (A10)

    ( A l l )

    and

    A lso for the case of no norma l stress on the surface

    G = 0

    Equat ions (A 4) , (A 10), and (A l l ) com bine to y ield

    ^ , + C , = 0

    - ^ 4

    1

    5

    2

    + 5

    1

    s - t - C

    1

    i '

    2

    + i ?

    1

    i ' =l (A12)

    where the co nstant in this equation are defined in body 1 of

    Fig. 1.

    4 0 8

    /V ol . 108, JULY 1986

    Transactions of the ASME

    wnloaded From: http://tribology.asmedigitalcollection.asme.org/ on 01/16/2016 Terms of Use: http://www.asme.org/about-asme/terms-of-use

  • 7/25/2019 Analysis of Traction Forces in a... Kannel 1986

    7/7

    The remaining boundary conditions can be met using the

    same approach as used for the normal stress conditions [12].

    These conditions yield the matrix given in Table 1.

    Green's Function for Shear

    The tangential deflection on the surface from equ ation ( A l)

    can be expressed as:

    ( l - i >

    2

    ) I"

    30

    d

    2

    G cos if

    7rE

    I;

    d

    v

    2

    '-ds

    where

    d

    2

    G

    ~ ~ h f

    -2(5, -D

    {

    s

    (A 13)

    (A 14)

    The Green's function can be written with reference to an

    arbitrary displacement

    V

    x

    at f = 1

    1 _ 2

    u

    u,

    =

    r f H L

    2 ( 5

    '

    -D,)s

    co ssi;- co ss

    efe-2j31nf]

    (A 15)

    It can be shown, using the matrix in Table 1 that for larger

    values of s, (B, - > , ) - \/s. Equ ation (A 15) can be ex

    pressed as two integrals (0 < s < s

    0

    ) an d (s

    0

    < s), as given in

    the text.

    If we let

    then this equation correspo nds to eq uation (5) in the text.

    Because the shear deformation is calculated as a relative

    displacement, a reference point must be selected. The

    displacements c alculated from equation (A 15) becomes sm all

    for large f; therefore a reasonable assumption for the

    reference point is 5 contact widths (x = -1 0 6 ) . Th en t h e

    relative ta nge ntial deflection (e,-) is:

    e