Burger’s Model

Burger’s model simulates creep mechanisms by using a Kelvin model and a Maxwell model connected in series in both the normal and shear directions.

Introduction

The Burger’s model provides a Kelvin model acting in series with a Maxwell model, in both the normal and shear directions. The Kelvin model is the combination of linear spring and dashpot components that act in parallel. The Maxwell model, on the other hand, is the combination of linear spring and dashpot components that act in series. Burger’s model acts over a vanishingly small area, and thus transmits only a force.

Behavior Summary

The rheological components of the Burger’s model are shown in Figure 1 for both the normal and shear directions. In both directions, the model combines Kelvin and Maxwell models acting in series. In the normal direction, the Kelvin model provides a linear spring with stiffness Kkn and a dashpot with viscosity Ckn, and the Maxwell model provides a linear spring with stiffness Kmn and a dashpot with viscosity Cmn. The Burger’s model can sustain tensile forces (Mt=0) or not (Mt=1). In the shear direction, , the Kelvin model provides a linear spring with stiffness Kks and a dashpot with viscosity Cks, and the Maxwell model provides a linear spring with stiffness Kms and a dashpot with viscosity Cms. A slider with friction coefficient fs limits the value of the shear force according to a Coulomb law.

../../../../../_images/cmburger_fig1.png

Figure 1: Rheological components of the Burger’s model.

In each direction, the total displacement of the Burger’s model, u, is the sum of the displacement of the Kelvin section (uk) and that of the Maxwell section ( umK , umC ) of the model:

(1)u=uk+umK+umC

The first and second derivatives of the equation above are given by :

(2)˙u=˙uk+˙umK+˙umC¨u=¨uk+¨umK+¨umC

The contact forces, f, using the Kelvin section and the first derivative, are given by (3). Note that symbols ± and correspond to the cases of normal and shear direction, respectively (For example, ± means + for normal direction and for shear direction.).

(3)f=±Kkuk±Ck˙uk˙f=±Kk˙uk±Ck¨uk

Also, using stiffness Km and viscosity of the Maxwell section:

(4)f=±KmumK˙f=±Km˙umK¨f=±Km¨umKf=±Cm˙umC˙f=±Cm¨umC

Using Eqs (2) through (4), the second-order differential equation for contact force f is given by:

(5)f+[CkKk+Cm(1Kk+1Km)]˙f+CkCmKkKm¨f=±Cm˙u±CkCmKk¨u

The force at a given step can be updated based on its value at the previous step and on values of the displacements at the current step and at the previous steps, using a finite dfifference scheme described below.

Activity-Deletion Criteria

A contact with the Burger’s model is active if and only if the contact gap (gc) is less than or equal to zero. The force-displacement law is skipped for inactive contacts.

Force-Displacement Law

The force-displacement law for the Burger’s model updates the contact force and moment:

(6)Fc=F,Mc0

where F combines the contributions from the Kelvin and Maxwell models acting in series. F is resolved into normal and shear forces:

(7)F=Fnˆnc+Fs

where Fn>0 is tension. The shear force lies on the contact plane, and is expressed in the contact plane coordinate system:

(8)Fs=Fssˆsc+Fstˆtc(2Dmodel:Fss0).

From the second equation in (3) of the Kelvin section:

(9)˙uk=Kkuk±fCk

By using a central difference approximation of the finite difference scheme for the time derivative and taking average values for uk and f:

(10)ut+1kutkΔt=1Ck[Kk(ut+1k+utk)2±ft+1ft]

Therefore,

(11)ut+1k=1A[Butk±Δt2Ck(ft+1+ft)]

where:

(12)A=1+KkΔt2CkB=1KkΔt2Ck

For the Maxwell section, the displacement and the first derivative are given by:

(13)um=umK+umC˙um=˙umK+˙umC

Substituting the second and fourth equations of (4) into the second equation above:

(14)˙um=±˙fKm±fCm

By using a central difference approximation of the finite difference scheme and taking the average value for f,

(15)ut+1mutmΔt=±ft+1ftKmΔt±ft+1+ft2Cm

Therefore,

(16)ut+1m=±ft+1ftKm±Δt(ft+1+ft)2Cm+utm

The total displacement and the first derivative of the Burger’s model are given by:

(17)u=uk+um˙u=˙um+˙um

By using the finite difference scheme for the time derivative,

(18)ut+1ut=ut+1kutk+ut+1mutm

Substituting Eqs (11) and (16) into the equation above, the contact force, ft+1, is given by:

(19)ft+1=±1C[ut+1ut+(1BA)utkDft]

Where:

(20)C=Δt2CkA+1Km+Δt2CmD=Δt2CkA1Km+Δt2Cm

The contact force ft+1 can be calculated from know values for ut+1, ut, utk and ft.


The force-displacement law for the Burger’s model consists of the following steps.

  1. Update the normal force Fn according to Eq. (19).

  2. Update the shear force as follows:

    • Update the shear force Fs according to Eq. (19).

    • Compute the shear strength:

      (21)Fs=fsFn.
    • Update the linear shear force:

      (22)Fs={Fs,FsFsFs(Fs/FsFsFs),otherwise.
    • Update the slip state:

      (23)s={true,Fs=Fsfalse,otherwise.

      If the slip state is true, then the contact is sliding. Whenever the slip states changes, the slip_change callback event occurs.

Energy Partitions

The Burger’s model does not provide any energy partition.

Properties

The properties defined by the Burger’s contact model are listed in the table below as a concise reference; see the Contact Properties section for a description of the information in the table columns.

Table 1: Burger’s Model Properties
Keyword Symbol Description Type Range Default Modifiable Inheritable
burger Model name
bur_knk Kkn Normal stiffness Kelvin section [force/length] FLT [0.0,+) 0.0 YES NO
bur_cnk Ckn Normal viscosity Kelvin section [force×time/length] FLT [0.0,+) 0.0 YES NO
bur_knm Kmn Normal stiffness Maxwell section [force/length] FLT [0.0,+) 0.0 YES NO
bur_cnm Cmn Normal viscosity Maxwell section [force×time/length] FLT [0.0,+) 0.0 YES NO
bur_ksk Kks Shear stiffness Kelvin section [force/length] FLT [0.0,+) 0.0 YES NO
bur_csk Cks Shear viscosity Kelvin section [force×time/length] FLT [0.0,+) 0.0 YES NO
bur_ksm Kms Shear stiffness Maxwell section [force/length] FLT [0.0,+) 0.0 YES NO
bur_csm Cms Shear viscosity Maxwell section [force×time/length] FLT [0.0,+) 0.0 YES NO
bur_fric fs Friction coefficient [-] FLT [0.0,+) 0.0 YES NO
bur_mode Mt Normal-force tensile mode [-] INT {0,1} 0 YES NO
    {0: with tensile force1: without tensile force          
bur_slip s Slip state [-] BOOL {false,true} false NO N/A
    {true: slippingfalse: not slipping          
bur_force F Total force (contact plane coord. system) VEC R3 0 NO NO
    (Fn,Fss,Fst)(2D model: Fss0)          

Surface Property Inheritance

The Burger’s model does not provide property inheritance capabilities.


Methods

The Burger’s model does not provide any method.

Callback Events

Table 2: Burger’s Model Callback Events
Event Array Slot Value Type Range Description
contact_activated Contact has become active
  1 C_PNT N/A Contact pointer
slip_change Slip state has changed
  1 C_PNT N/A Contact pointer
  2 INT {0,1} Slip change mode
        {0: slip has initiated1: slip has ended

Usage and Verification Examples

The Verification Problem “Burger’s Contact Model: Stress Relaxation” compares the time-decay of the normal force to the expected analytical solution.

Model Summary

An alphabetical list of the linear model properties is given here.