Contact Analysis for Drum Brakes and Disk Brakes Using Adina

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  • Contact analysis for drum brakes and disk brakes usingADINA

    C. Hohmann*, K. Schiner, K. Oerter, H. Reese

    Institute of Engineering Mechanics and Control Engineering, University of Siegen, Paul-Bonatz Str. 911, Fachbereich 11,

    Mashinentechnik, 57076, Siegen, Germany

    Abstract

    Brakes in cars and trucks are safety parts. Requirements not only in performance but also in comfort,serviceability and working lifetime are high and rising. Optimal design of todays brake systems is found usingadditional calculations based on finite element methods. For both types of brake systems, drum brakes and disk

    brakes, the dierent parts of brakes, i.e. the brake pad with the friction material, the counter body and calliper, canbe modelled. Two examples are given in this paper: a drum brake of a trailer and a typical disk brake used inpassenger cars. The main problem to be solved is the calculation of the distribution of contact forces between brakepad and counter body (drum or disk). The contact problem includes friction and is solved using the ADINA 7.1

    sparse solver. After the brake pressure is applied, the turning moment on the axle rises constantly until the drum ordisk respectively changes from sticking to sliding condition. It is shown that the sparse solver is highly ecient forthis sophisticated nonlinear problem. Results include deformation, stress distribution, contact pressure and showing

    which regions of the contact area are in sticking or sliding condition. # 1999 Elsevier Science Ltd. All rightsreserved.

    1. Introduction

    1.1. Brake construction

    Brakes in cars are expected to work properly with a

    minimum amount of service. The purpose of brakes isto reduce the velocity or to maintain it when the ve-hicle is driving downhill. Without brakes it would not

    be possible to control the speed of the vehicle.Nevertheless the design of brakes is generally underes-timated. In brakes high amounts of energy are trans-formed during short periods. This is underlined by the

    fact, that often the braking power is several times

    higher than the power of the engine.

    In cars and trucks dierent types of braking systems

    are used. In this paper only wheel brakes based on

    friction are considered. Generally two design forms are

    used: disk and drum brakes. The demands made by

    the vehicle industry are strict. The requirements will

    also rise with future development of lighter and more

    economical cars. This supports the need for ecient

    methods of calculating brake systems. The finite el-

    ement method is an ideal tool for this purpose. It is

    suited for analysis of both stress and temperature. In

    this paper the contact problem is described for disk

    brakes and drum brakes. For example a drum brake

    used in trucks and a disk brake used in small passen-

    ger cars are presented.

    Computers and Structures 72 (1999) 185198

    0045-7949/99/$ - see front matter # 1999 Elsevier Science Ltd. All rights reserved.PII: S0045-7949(99 )00007-3

    * Corresponding author. Fax: +49-271-7402436.

  • In order to transfer the kinetic energy of the vehicleinto heat, friction brakes are commonly used at each

    wheel of the vehicle. The area of contact is the originof the heat sources. Cooling surfaces emit heat into theenvironment. Nevertheless the temperatures can get as

    high as 9008C (16508F) at the contact area.For proper operation the following criteria have to

    be considered

    . high and stable coecient of friction;

    . good thermal capacity;

    . good wear resistance of the tribo system (brake lin-ings and disk respectively drum);

    . mechanical resistance of material;

    Fig. 1. Frequently used types of friction brakes: (a) disk brake and (b) drum brake.

    Fig. 2. Transmission of forces in a brake system [1].

    C. Hohmann et al. / Computers and Structures 72 (1999) 185198186

  • . weight optimized construction;

    . and use of environmental suited materials.

    Despite dierent geometric design (see Fig. 1), both

    types of brakes use the same principle to create thebraking force: fixed brake shoes are pressed against arotating counter body. Due to friction the brake forceis acting contrary to the motion of the counter body

    thereby reducing its velocity. The resulting frictionforce is proportional to the normal force N and thecoecient of friction m. The brake parameter C isdefined as the relation between friction forces and theapplied force and is used in the comparison of dierentbrake designs.

    A typical brake used in cars consists of the operatingdevice (pedal or hand lever), the transfer unit and thewheel brake (Fig. 2). For disk and drum brakes thenormal force N is applied, using mechanical, pneu-

    matic or hydraulic transfer units. A servo unit raisesthe force FBet applied by the driver:

    N iAFBet 1

    The factor iA is the total transmission ratio of the

    brake. The friction force R=mN acts on the rubbingsurface. The distance between the friction force andthe centre of revolution is the eective radius re. The

    origin of the friction force for disk brakes is approxi-mately the middle of the friction surface, depending onthe shape of the friction area. In the case of drum

    brakes the eective radius is the inner radius of thedrum.Ecient software and hardware is needed for the

    simulation of complete brake systems. ADINA

    Version 7.1 simplifies the construction of the complexthree-dimensional structures together with providingreliable algorithms for the solution of the contact pro-

    blem. Using the sparse solver the calculation time isreduced significantly.

    2. Design of drum brakes

    2.1. Analytical calculation

    The analytical calculation of drum brakes is basedon the assumptions of Koessler [2,3]. He describes thedistribution of the normal pressure between drum and

    brake linings with cosine functions (see Fig. 3a). Theassumed distribution only fits if the curvatures of thebrake lining and of the drum are equal. The radius of

    the brake lining is changed during operation due towear. This results in irregular distributions of the con-tact pressure, which are shown in Fig. 3b and c.

    At each point of the contact area the coecient m ofsliding friction is constant. The distributed frictionforces are caused by the pressure and are proportionalto the coecient of sliding friction:

    dR m dN 2

    The braking moment can be calculated with the dis-tance r of the contact point to the centre of revolution.

    The summated friction forces make up the peripheralforce R1 for the leading shoe and R2 for the trailingshoe. Both act against the rotation of the drum with

    Fig. 3. Normal pressure distribution, assumed by Koessler (a), for a new (b) and an old lining (c).

    C. Hohmann et al. / Computers and Structures 72 (1999) 185198 187

  • the lever arm r:

    R1 a2a1

    dR, R2 a2a1

    dR 3

    The brake shoes are spread by a rotation of the S-cam

    clamping device. Therefore the forces S1 and S2 actupon the rolls (Fig. 4a). They are in balance with the

    normal pressure and peripheral pressure acting on thelining surfaces (Fig. 4b) and the resulting forces F1 and

    F2 acting at the bolts. The static conditions for equili-brium can be formulated for each direction (x, y and

    a ) and both brake shoes in the form:

    Fig. 4. Applied forces and reaction forces for the trailing shoe (a), and normal and tangential pressure for a contact point (b).

    Fig. 5. Parts of an S-cam driven drum brake and FEM discretization.

    C. Hohmann et al. / Computers and Structures 72 (1999) 185198188

  • Fig.6.Boundary

    conditionsandapplied

    displacements.

    C. Hohmann et al. / Computers and Structures 72 (1999) 185198 189

  • S1 cos g1 a2a1

    dN x 1 a2a1

    dR x 1 F1 cos d1 0

    S2 cos g2 a2a1

    dN x 2 a2a1

    dR x 2 F2 cos d2 0

    S1 sin g1 a2a1

    dN y 1 a2a1

    dR y 1 F1 sin d1 0

    S2 sin g2 a2a1

    dN y 2 a2a1

    dR y 2 F2 sin d2 0

    S1 ca2a1

    r dR F1 a 0

    S2 ca2a1

    r dR F2 a 0 4

    These equations have to be solved for both shoes. Thesolution of the integrals results in a relationshipbetween applied forces and the braking moment. With

    a0=a1+a2 the brake parameter C is:

    C R1 R2S1 S2 2

    f a0a0r

    f a0a0r

    2m2

    h

    rm with

    f a0 a0 sina04 sin

    1

    2a0

    5

    Type of drum brake: Simplex Duo-duplex Duo-servo

    Brake parameter: C 1 2.0 C 13.0 C 1 5.0

    The finite element model is used, to calculate a more

    realistic distribution of the contact pressure, consider-ing the elastic deformations of both the linings and thedrum.

    2.2. Finite element calculation

    2.2.1. Description of the model

    A complete three-dimensional structure of the drumbrake has to be modelled. The finite element modelconsists of three independent parts: the drum, the lead-

    ing shoe and the trailing shoe (see Fig. 5). All parts,with the exception of the brake linings, are made ofsteel. Normally the brake linings are riveted to thebrake shoe. In the model, however, uniform connec-

    tion between lining and shoe is considered. Threedimensional elements with eight nodes are used tomodel the solids. The definitions of nodes and elements

    are created with the PATRAN-preprocessor. AFORTRAN program translates the node and elementlists for ADINA-IN. Contact areas are defined for the

    rubbing faces of the drum and the four lining pad sur-faces.The fixed bolts are modelled with truss elements.

    This enables the brake shoes to rotate around the fixedbolts (see Fig. 6). The complete S-cam clamping deviceis neglected i

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