Lecture 2 Fundamentals of Discrete Element Modeling Anthony D. Rosato Granular Science Laboratory ME Department New Jersey Institute of Technology Newark,

Lecture 2 Fundamentals of Discrete Element

Modeling

Anthony D. RosatoGranular Science Laboratory

ME DepartmentNew Jersey Institute of Technology

Newark, NJ, USA

Lecture 2: Fundamentals of Discrete Element Modeling

2

Presentation OutlinePresentation Outline

Part 1: Background & Motivation

Part 2: Particle Collision Models

Part 3: Wave Propagation Ideas

Part 4: Elastic Impact of Spheres


Granular Science Lab - NJIT

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Discrete Element Modeling is an outgrowth of molecular dynamics simulations used in computational statistical physics.

However, the discrete element method was independently developed by P. Cundall in the 1970’s.

B. J. Alder and T. E. Wainwright, “Statistical mechanical theory of transport properties,” Proceedings of the International Union of Pure and Applied Physics, Brussels, 1956.

W. T. Ashurst and W. G. Hoover, “Argon shear viscosity via a Lennard-Jones potential with equilibrium and nonequilibrium molecular dynamics,” Phys. Rev. Lett. 31, 206 (1973).

P. A. Cundall, Symposium of the International Society of Rock Mechanics,” Nancy, France (1971).P. A. Cundall, “Computer model for rock-mass behavior using interactive graphics for input and output of geometrical data,” U.S. Army Corps. Of Engineers, Technical Report No. MRD, 1974, p. 2074.



5

Approximate the collisional interactions between particles using idealized force models that dissipate energy.

Integrate system equations of motion Determine individual particle positions and velocities.

Compute relevant transport quantities, bulk properties and analyze evolving microstructure.

Basic Idea of DEM Modeling



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Initial positions, orientations and velocities

Update particle link-list(find new or broken contacts)

Calculate the force and torque on each particle

Integrate the equation of motion to calculate the newpositions, velocities and orientation

Accumulate statistics to calculate transport properties

Time increment: t = t + dt

LOOP

“F = ma”

“T = I”

Basic Flow Chart

- for spheres -



7

Equations of Motion for the Particle - Integration

Integration time step t approximated from normal force model by dividing twice the time spent in unloading period during a particle collision into n steps

12

et m K

n

tvtF

txPosition, velocity and force at time t

tω

tθ

tT

Angular position, velocity and Torque at time t

0 ,4

0 ,

0 ,2

oo21

211

2/121

tm

tt

ttm

ttt

ttt

Fvv

vxx

Fvv

0 ,4

0 ,

0 ,2

oo21

211

2/121

tm

tt

ttm

ttt

ttt

Tωω

ωθθ

Tωω

- for spheres -



See Appendix L2-C for derivation

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M o d e lM o d e l

Seek qualitative / quantitative agreement with physical experiments.

Critical Goal

Once validation has been established, the simulations provides a means of carrying out an ‘experiment’ on the computer.



9

In contrast to the energy conservation of molecular systems-energy dissipation is a critical characteristic of granular systems, and consequently, it is necessary to employ realistic approximations to model energy loss in colliding particles.

For this purpose, there are essentially two basic approaches.

By considering particles to be infinitely stiff, “hard particle” models assume instantaneous, binary collisions governed by a collision operator, which is a function of particle properties (i.e., friction, normal and tangential restitution coefficients) and the pre- and postcollisional velocities and spins. Such an approach is appropriate in collision-dominated systems, where continuous and/or multiple contacts are not characteristic.

‘Hard’ Particle Models

‘Soft’ Particle Models

The interaction is a function of an allowed overlap between colliding particles that is intended to model the plastic deformation* at the contact.

*0. R. Walton, K. A. Hagen, and J. M. Cooper, “Modeling of inelastic frictional, contact forces in flowing granuIar assemblies,” 16th Znternational Congress of Theoretical and Applied Mechanics, Lyngby, Denmark, 1984.



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11

Elementary Hard Particle Model

When particles collide, some of the kinetic energy is lost – that is, a portion of the initial kinetic energy goes into deforming the objects.

Thus, a ball that is dropped and hits the ground will not rebound to the same height from which is was released.

As a first approach, this energy loss is modeled through a

“coefficient of restitution”, denoted by e.

e = 1 No energy loss (perfectly elastic)

e = 0 Complete energy loss (plastic)

0 < e < 1 A portion of the incident energy is loss



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The coefficient of restitution is related to the velocities of the two particles by the relation

Consider a particle A of mass mA that has a velocity vA ,and collides with a stationary particle B of mass mB.

What is the kinetic energy lost KE during the impact?

It can be shown that …

)1( )(2

22 emm

mmKE A

BA

BA

v

BA

ABevv

vv

A BVA

VB

Before collision

A BBV

AV

After collision



See Appendix L2-A

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A ball is dropped from a height h = 36 inches onto a smooth floor. Knowing that the height of the first bounce is h1 = 32”, determine the coefficient of restitution e and the expected height h2 after the second bounce and after n bounces.

e..

e 913913

vv

v

v

Smooth floor

ho

d1 d2

h1 h2

vft/sec 9.132

2

1 2 oo ghmmgh vv

Conservation of Energy

oo h

hem ghegh(

mmghm 12

12

12 )2

2

2

1v

9428.036

322 ee

It can be shown that "44.2842 ehh o and in general, after k bounces,

ok

k heh 2



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Three Parameter Hard Sphere Collision Model (Walton)

L. Labous, et al. (1997), “Measurement of collisional properties of spheres using high-speed video analysis,” PRE 56(6), 5717-5727.

S.F. Foerster, et al.(1994), “Measurements of the collision properties of small spheres,” Phys. Fluids 6 (3), 1108-1115.

O. R. Walton (1992), in Particulate Two Phase Flow (ed. M. C. Roco), pp. 884-907.

The model mimics the phenomenon explained by R. Mindlin* in his solution for the collision of two elastic spheres – where there is a partial retrieval of elastic energy stored as tangential deformation of the spheres in the contact region.

In Walton’s model, the collision is ‘instantaneous’, and the post-collisional velocities are determined by their pre-collisional values and 3 phenomenological material parameters, i.e.,

e Coefficient of normal restitution Coefficient of tangential restitution Coefficient of sliding friction

*R. D. Mindlin (1949), J. Appl. Mech 16, 249; R. D. Mindlin & H. Deresiewicz (1953), J. Appl. Mech. 21, 237

Comment: This model was motivated by the experiments of Maw Barber & Fawcett and by Savage and Lun’s ‘beta’ parameter in their kinetic theory for particles with rotational degrees of freedom and friction



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sv cv

1m

2m

nvP

Consider 2 homogeneous spheres of masses 21 , mm

moments of inertia about their centers 21 , II and

diameters 21 , dd

21 , rr Locations of the particle centers at the instant of

collision with respect to an inertial reference frame

21

21 , ,vv Translational and rotational velocities of

the particles immediately before collision P

Impulse that sphere 2 exerts on sphere 1

21

21rr

rrn

Unit vector in the direction of the line of centers - from sphere 1 to sphere 2

sn vvcot Angle of incidence



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Apply conservation of linear and angular momentum to determine the velocities after collision (denoted by a ‘prime’)

222111 vvvv

mmP

222

211

1

1 22

d

I

d

IPn

(1)

(2)

Relative velocity of spheres at contact point (sliding velocity) immediately before collision

ndd

2

21

121 22

vvvc (3)

nncn

vv Normal component of relative sliding velocity immediately before collision

nnccncs vvvvv Tangential component of relative sliding

velocity immediately before collision(4)



ss vv t Tangent direction

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nen 2121 vvvv Definition of the normal restitution coefficient e

ss vv Definition of the tangential restitution coefficient

models the collisional energy loss due to the tangential compliance of the particle surface. It reduces the magnitude of the relative surface velocity after collision.

*Maw et al. showed that is a function of the angle of incidence , such that [-1, 1].

* N. Maw, J. Barber, and J. N. Fawcett, Wear 38, 101 (1976).

By using definition (5b) and the conservation of linear momentum (1), it can be shown that

the normal component of P

is given by nv

rn meP 1 where

2121 mmmmmr is the reduced mass.

It now remains to find the tangential component of the momentum change.

(5a)



(5b)

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Walton’s assumption: Collision is either sliding or rolling throughout the contact

Sliding Contact tPP nt

o

o is a measure of the friction at the particle surface

(6)

It can be shown that for collisions that are almost head-on, i.e., is close to , (6) yields -values larger than 1. This is physically impossible since it means that there is an increase in the energy.

Therefore, an apriori known value of is assumed to be known, 1o

such that whenever the collision yields o the collision is in rolling contact.

The above defines a critical value of the incidence angle o above which (6) is no longer valid. In this situation, the is set to its limiting value o



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In summary, the three model parameters which characterize the collisional energy losses, and which must be determined experimentally, are

e Normal coefficient of restitution

o Tangential restitution of coefficient

Coefficient of friction

After some work, the equations (1-6) can be combined to yield the following

(rolling) for ,

(sliding) for ,111

oo

o1

os

neKvv

dmd

IK diameter of spheres homogenousfor

5

242

(7)



21

1

1

oRolling

Energy limit

Sliding (slope = 1)

s

neKvv

11 1

Graphical representation of the Walton 3-Parameter Model (7)

From a single collision experiment between two spheres, one can determine e and either or o.

The factor determining which of the latter two can be found is the ratio sn vv

Perhaps the first experiments on real spheres to determine the collision parameters were done by Foester et al. and reported inS.F. Foerster, et al.(1994), “Measurements of the collision properties of small spheres,” Phys. Fluids 6 (3), 1108-1115.



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S. F. Foerester, M. Y. Louge, H. Chang, K. Allia, Physics of Fluids 6(3), 1108 – 1115, 1994.



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Comprehensive Data can be found on the website of Prof. Michel Louge, Mechanical & Aerospace Engineering, Cornell University, Ithaca, NY, USA.

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The behavior of real colliding particles is a great deal more complex than the model just explained because of the interplay of translational and rotational motion. The behavior of a “superball” demonstrates this ….

The ball reverses spin upon rebounding from the bottom of the table top.

Laboratory experiments to measure particle collision properties



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26

Results from experiments with 2.54 cm nylon spheres



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Soft Particle Models

Linear Spring – Dashpot Model for Two Interacting Spheres

x1

x2 KD

Fig. 1: Sphere interaction modeled as a spring and dashpot.

Consider a sphere of mass m1 and diameter d1 whose collisional

interaction with another sphere (mass m2, diameter d2) is

idealized by a dashpot and spring as depicted in Fig. 1. Define the penetration distance x as:

21212

21

1 21

22ddxx

dx

dxx

Hence the interpenetration velocity and acceleration is given by

21

21

xxx

xxx

(1)

(2)

The material presented in this section was taken from S. Luding, Models and Simulations of Granular

Materials, Ph.D. Dissertation, 1994.



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For each sphere, the equation of motion is written, and the resulting two equations can be combined to obtain the governing differential equation for the penetration x,

0 r rx D / m x K / m x (3)

Dimensions m ~ FT2/LK = ~ F/LK/m ~ 1/T2

D ~ FT/LD/m ~ 1/T________________L = unit of lengthF = unit of forceT = unit of time

Equation (4) is solved for the penetration x in the usual fashion to yield,

(4)

21

21

mm

mmmr

is the reduced mass. where

teAteAtx tt sincos 21

2o 2

r r

K D,

m mwhere22

o and

The natural frequency is

22

22of

Thus, the collision time tc can be approximated as 22

o2

1

ftc

(5b) (5a)



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are applied to (5), we obtaino0v

tx 0

0

txAs an example, if initial conditions and

tetx t

sinvo

ttetx t

cossinvo

(6)

Contact between the spheres is broken when ttx 0

Hence in this case, the contact time is exactly 22

o

ct

Normal Coefficient of Restitution

/

021

21

00v

v-

vv

vve

x

txt cc

t

tt c

/

ov etx cwhere

Therefore, 22

o

e(7)



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Maximum penetration xmax occurs when dx/dt = 0. Differentiation of the first of equation (6) yields

omax arcsin

1arctan

1t from which

arcsinexpv

o

omaxx (8)

Case of low dissipation:

22 1 o r r o o K m D m

Therefore, 22 o

maxctt

Time for maximum penetration is approximately ½ the collision time

From (8), it follows that o

o

oo

omax

v

2exp

v

x (9)

To numerical example



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Numerical example of maximum penetration length for steel spheres

d = 1.5 mm

= 0.28

m = 1.38 x 10-5 kg

E = 2.06 x 1011 N/m2

kg 1069.02

5m

mr

02 20 xxx

redm

K2

0red

n

m

D

2

9.0

sec. 106.4 6

22o

ct

c

e

t

sec 108.6

sec 1029.2

5o

-14

arcsinexpv

o

omaxx For Vo = 1 m/s, %093.0max

d

x



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Nonlinear Spring – Dashpot Model for Two Interacting Spheres

rij

Particle i Particle j

rij

x

½ (di+dj)

ijij xxr

ij

ijij

r

rn

Consider two uniform spheres of diameters d, where the vector from the center of sphere i to j is

and the unit normal is

.

ijijji ddx rn 2

1The penetration distance is jiij vvv and the relative velocity of the spheres is

Note that when x < 0, the spheres are not in contact with each other.

In the derivation that follows, we consider only a normal force

model. ije kx nF 1

Elastic force: Damping force: ijijijd xD nnvF

the linear model is recovered When 0



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The equations of motion are:

221

111

xmxDxkx

xmxDxkx

(10)

Multiplication the first of these by m2 and the second by m1 followed by a subtraction of the resulting equations from each other yields desired model for the evolution of the penetration x, i.e.,

01 xxm

Dx

m

kx

rr (11)

21

21

mm

mmmr

is the reduced mass. where

By using definitions (5a), the equation can be written as

02 12 xxxx o (12)

The solution of equation (12) is the penetration x between two spheres, whose behavior is governed by an elastic and dissipative normal contact force models.



2o 2

r r

K D,

m m22

o and

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Equation (12) can be re-expressed in terms of the Hertzian* contact force between two elastic spheres of radii r1 and r2 with elastic moduli E1, E2 and Poisson’s ratios 1 and 2.

3/ 21 2 1 22 21 1 2 2 1 2

4

3 (1 ) (1 )

H

E E r rF x

E E r r

*S. P. Timoshenko and J. N. Goodier, Theory of Elasticity, 3rd edition, pg. 412, McGraw Hill, 1970.

When the spheres have the same elastic properties (E, ) and diameter d=2r, the above relation reduces to 3 2

2

3(1 )

H

EF d x

Upon equating FH to the assumed form for the elastic interaction force

1F ne ijkx one finds

2, where

3(1 )

E

k E d E

Following S. Luding**, assume the forms

**S. Luding, “Models and Simulations of Granular Materials, Ph.D. Dissertation, 1994.

1k Ed 1D d and

Observe that when = ½, Hertzian contact is recovered.



2/3 xkF HH or

Details

36

A substitution of the latter two forms into equation (11) yields the re-expressed model for the penetration x, i.e.,

0

r r

d x Ed xx x x

m d m d

(13)

Maximum Penetration Depth, Contact Time and Normal Restitution Coefficient

To determine the maximum overlap or depth of penetration of the two spheres xmax, assume that all of the incident kinetic energy is stored as elastic energy.

Hence, if vo is the incident relative velocity, then, 21 v2 r o elasm E

Eelas is approximated for the case of weak or no dissipation for which 0

x

Multiply (13) by x and integrate the resulting equation.

(14)



37

222

1

2

10 x

dt

d

d

xdExm

dt

d

d

xxdxExxm rr

2 2

2max1

011

00

2

max

maxmax xdEdxxdEdtxx

d

xdExmdE

x

xx

relas

A substitution of (15) into (14) yields the maximum penetration,

(15)

)2(12)2(1

1

)2(1

max )(v2

1

o

r

dE

mx (16)

The contact time tc can be estimated from energy conservation (i.e., little or no dissipation)

xdd

xdEeFdxFxmd e

r

where,02

2

2/

0

)2/()2/()2/(1

1

)2/(1cos2v

2

2/12 d

dE

mt o

rc



38

This latter expression is simplified using the beta function* to produce the result…

G

dE

mt o

rc

)2/()2/(1

1

)2/(1

v2

1 (17a)

244

2

1

21

G

(17b)

Reduction to Hertzian model when = ½: 5/1v~ oct(18)

*Abramowitz & Stegun, Handbook of Mathematical Functions, pg. 258, Dover, 1972.

Normal Coefficient of Restitution

The coefficient of restitution can be determined from the energy dissipated in a collision max xFE ddis

ood xdDxF vv 1 where

)22()24()2()1(

11 v~

o

rdis

dE

mdE (19)



39

energy ofn dissipatio smallfor 2

111

oo

E

EEE dis

dis

After substituting Edis into the above expression, one obtains )2()2(v2

11~ o

For Hertzian contact, = ½ and = 0 corresponding to a constant damping force

From this, one can find the approximate energy loss in a collision

)2()2(ov ~ Loss

In this case, the restitution coefficient o

5/1o vas 0 v

2

11~

This is physically unrealistic behavior as it is known from experiments* and finite element calculations** that the restitution coefficient decreases as a function of the incident velocity.

1

ov

* W. Goldsmith, Impact, E. Arnold Publishers (1960).** O. R. Walton, “Numerical Simulation of Inelastic, Frictional Particle-Particle Interactions,” in Particulate Two-Phase Flows (Ed. M.C. Roco), Buterworth-Heinemann, Boston (1993).

xDxxDFd



40



J.C. Mccrae & W. A. Gray, “Significance of the properties of materials in the packing of realspherical particles,” Brit. J. Appl. Phys. 12 [4], 164-171 (1964).

41

A further problem with using linear damping and Hertzian contact is a discontinuous force-displacement curve.

dampingelas FFF

F

x



42

Selection of a ‘Proper’ Value of

t

K

m

DAt

m

K

m

DAex

rmrrr

tmD

r2

1

2

2

22

1

2

2

1

)2

(

4exp

4exp

Critical Damping: KmDm

K

m

Drcrit

rr

2 4 2

2

(from O.R. Walton, “Force Models for Particle-Dynamics Simulations of Granular Materials”, in Mobile Particulate Systems, ed. E. Guazzell & L. Oger, NATO AS1 Series E: Applied Sciences, Vol. 287, Kluwer Academic Publishers, (1995), Chpt. 20, Pg. 367)

We want Deff to scale with the slope of the Hertzian spring force in order to maintain a fixed fraction of critical damping for each incremental step along the force-displacement curve. Recall the solution to the linear spring-dashpot model 0 KxxDxmr

Recall the general non-linear model 01 xxm

Dx

m

Kx

rr

Define an ‘effective damping coefficient as .effDDx

whose solution is given by



43

For Hertzian contact, = ½ and = 1/4, corresponding to 4/1xxDxxDFd 2/3kxFelas and

.02/34/1 kxxDxxmr the model becomes

kmxD

kxdx

dmdxdFmmKD

r/

crit

elascrit

6

222

41

2/12/32/1

1/4 criteff DDxD

In this case, 1v2

11~ )2()2(

o

Thus, the model produces a constant restitution coefficient (in contrast to the physically unrealistic behavior that occurs with linear damping and Hertzian contact), which is acceptable for incident (relative) collision velocities that are less than a critical value (which depends on the particle material).



44

Two-Dimensional Soft Particle ModelsInclude the effects of friction and particle rotation

Idealized contact between an elastic disk and a planar surface:

1. No sliding ( = )

2. Stress in the ‘normal’ direction is independent of the tangential force. (This is the usual assumption in most theoretical & numerical treatments of contacts for the local stress field.

m Mass of disk

Io = mR2/2 Moment of inertiay

x



45

y

x KxKy

The model consists of two springs

Kx Normal spring stiffness

Ky Tangential spring stiffness

Assume that the spring contact forces are establish as soon as the disk encounters the planar surface.

x Tangential displacement of the disk’s center of gravity measured from its initial contact point

y Normal displacement of the disk’s center of gravity from its initial contact point

Orientation of the disk relative to when the contact is first established



46

Equations of Motion

RxRKRFTRxKFyKF xxxxyy

(3)

(2) 1

(1) 1

oo

θ

RxKI

R

I

N

yKmm

Fy

RxKmm

Fx

x

yy

xx

Upon adding equations (1) and (3), we obtain

Rx

I

mR

m

KRx x

o

21 (4)



47

The objective is to design the system so that the spring in the normal direction will return to its equilibrium position at the same time as the tangential spring. In order to do this, the period of oscillation for the simple harmonic motion described by equations (2) and (4) must be the same. Hence,

o

21

I

mRKK xy (5)

Tuning the Springs

xy

xy

KKmRI

KKmRI

72 52 :Spheres

3 2 :Disks

2o

2o

Mindlin’s theory for real elastic materials (R.D. Mindlin, J. Appl. Mech. 16, 259 (1949)) indicates that

13

2

y

x

K

K



48

Implications

• For most real contacts, the frequency of rotational motion is higher than that of the normal motion

• If the contacting disk or spheres is initially in rotation only slightly, it could experience a reversal in the direction of rotation during the time it is in contact with the surface.

• (cf. Walton’s 3-parameter hard sphere model)

(3)

(2) 1

(1) 1

oo

θ

RxKI

R

I

N

yKmm

Fy

RxKmm

Fx

x

yy

xx

The solution of the system will be discussed in the next slides.

For x < 0:



49

0

0

0

0

00

0

o2

o

y

x

IKRIRK

mK

mRKmK

y

x

xx

y

xx

:0When y

:0When y

0

0

o

I

gym

xm

Superball Motion

How to set the parameters so that the disk jumps back and forth without deviation ?



50

Superball Motion Continued

Find the initial conditions so that the direction of the surface velocity is reversed after the collision

oo 1

RvRvVV

V

Vxxss

s

s

Conservation of momentum oxxx vvmdtF

xFjj xx FFRR

and

Conservation of angular momentum

ooo kk IIdtRFx o

o R

IdtFx

By equating the above two relations, one obtains oo

o

R

Ivvm xx

Spin reversal also requires that o , ,oo

yyxx vvvv

A substitution into * gives the desired condition

*

mR

IVx o

o

o



51


Implementation of Superball Model

Superball.Exe


Copyrighted application by Y. Chung

52

Walton and Braun Soft Particle ModelsWalton, O. R., Numerical simulation of inelastic, frictional particle-particle interactions, in M. C. Roco, ed., Particulate Two-Phase Flow, Butterworths, Boston, 1992, pp. 884-911.

fc

d

b

a

K1

K2

Fn

Normal Interaction

Model produces a constant restitution coefficient

1 2e K K

Linear Loading – Unloading ModelF1 = K1F2 = K2

Initial Loading: a b with slope K1

Unloading initiated at b, follows b c with slope K2

Reloading from c follows path c b d

Subsequent unloading from d follows path d f c a

K1 estimate



See Appendix L2-B

54

Tangential Interaction

O. R. Walton, R. L. Braun, J. Rheol. 30(5), 949-980, 1986.

KT = Tangential stiffness of a contact, which decreases with tangential displacement. Full sliding occurs when KT = 0

* 0 initially when relative tangential slip reverses direction

TT

T = total tangential forceKo = initial tangential stiffness ~ K1

= coefficient of sliding friction

decreasingfor 1

increasingfor 1

3/1

*

*

3/1

*

*

TTN

TTK

TTN

TTK

K

o

o

T

At each new time step, compute TnewsKTT Toldnew

s = Relative surface displacement between contacting particles

Full sliding of the contact occurs at the friction limit when KT becomes zero.

Based on. R. D. Mindlin, H. Deresiewicz, “Elastic spheres in contacts under varying oblique forces,” J. Appl. Mech. 20, 327 (1953).



55

Notes taken from O. R. Walton, “Force Models for Particle-Dynamics Simulations of Granular Materials,” in Mobile Particulate Systems (eds. E. Guazzelli & L. Oger), NATO ASI Series (Series E: Applied Sciences), Vol. 287, pp. 370 (1994).



56

Consider a 1-D “Chain of Spheres”

Spheres of mass m and diameter d are held in contact by externally applied boundary load. Assume that there is a small disturbance so that we can model contacts as linear springs of stiffness K.

1iu iu 1iu

Let x be the equilibrium spacing between the point masses, s.t. x = dLet ui-1, ui, ui+1 be displacements from equilibrium of 3 adjacent masses.

Fi Force acting on ith particle

ii uuK 1 1 ii uuK

11 iiiii uuKuuKF

Newton’s 2nd Law:

2112

112

2x

uuuxKuuuKum iii

iiii

Wave speed mKdmKxc



57

Wave Speed for Rigid Particles Not in Contact

Systems which are sufficiently energetic so that time between collisions >> tc Then, if this time scale separation is large enough, system can be modeled as rigid spheres which experience instantaneous, rigid-body collisions.

“Information” is propagated through the motion of particles.

s

doV

Equally spaced chain of spheres

Particle strikes one end of chain at velocity Vo.

velocity V0.



58

Suppose that collisions are perfectly elastic, i.e., e = 1. Then kinetic energy is preserved, and hence, at any instant, there is only one particle moving at velocity Vo.

collision after 2 particle ofVelocity 0 2 particle of velocity Initial

collisionafter 1 particle ofVelocity 1 particle of velocity Initial

2o2

1o1

VV

VV

VVVVVmmVmV 1o121o2o1

12o1

o1

21 1 VVVV

VVe

o121 ,0 VVV

travels that waveDistance sd

s)(d V

sΔ distance the travel tofor wave Time

o

fraction void theis Δ

Δ where,

Δvoid

void

oo s

sd f

f

VV

s

sd c

csf 0), that meaning( ,0 If void



59

Elastic Particle – Finite Contact Time

Account for finite contact time of elastic particles in estimating wave speed.Consider a 1-D chain of elastic spheres equally spaced.

What is the response of the chain to a single particle hitting chain at one end?

Factors which contribute to wave velocity in non-contacting granular assemblies:

Velocity of the particles in the system.Duration of contacts in collisions.

In order to approximate effect of finite contact time, assume that during contact period, each particle has an average velocity of ½ of the incident colliding particles velocity.

2 mecontact ti during traveledDistance oo Vc

considered are collisions isolated with casesonly that so that Assume cs

ooo contactsbetween flight of Time

VV

s

V

s ccg

oy at velocit contactsbetween moves particle a Distance Vs c



60

ccds in time travelswave'' Distance

2s

2-s Speedn Propagatio

o

o

c

oo

cVcgc

VsdVsdVsdc

2s o

o

cV

Vsdc

Compare the speed to the model for which the contact time was not included, i.e.,

oΔ

Vs

sd c

Observe that when the contact time is included in the rough model, the wave speed is slower because of the “Voc/2” term in the denominator.



61

Material taken from K. Johnson, Contact Mechanics, Chapter 11, Cambridge University Press (1985).



62

1G

2G

1zV

2zV

11 yx RV

22 yx RV

1y

2y

The classical Hertz theory for the impact of frictionless, elastic spheres assumes (1) that the deformation is restricted to the vicinity of the contact area and is given by a static theory and (2) elastic wave motion is ignored.

Collinear Impact of Spheres spheres of Masses , 21 mm

O

Collision occurs at point O

collision of at time centers of line along spheres of Velocities , 21 zz VV

Assume colinear impact so that 02121 xxyy VV

zDisplacement amount by which the sphere centers approach each other due to elastic deformation.

12 zzz VV

dt

d

tP Force acting between the spheres during their contact.

dtzdV

mdt

zdVmtP 2

21

1

(a)

(b)



63

Multiply the 1st by m2 and the second by m1 and then subtract the resulting equations.

1221

21

21

2121

1122

zz VVdt

dmmP

mm

mm-

dtzdV

mmtPm

dtzdV

mmtPm

dt

d z

Therefore, the equation relating P to the evolution of z is given by z

dt

dP

mm

mm

2

2

11

21

Next, assume that P and are related through Hertz’s static, elastic contact law.

21*

2

22

1

21

*

3/1

2**

2

111 and

111 where

16

9

RRR

EEEER

P

(c)



64

Solve the latter equation for P in terms of

2/32/3**2/3221

212

21

21

21 )()(3

4

)1()1(

3

4zz KER

EE

EE

RR

RRP

and substitute the result into equation (c) to obtain 2/32

2)( z

zr K

dt

dm

where

21

21

mm

mmmr

To solve this equation, multiply by sides by z and integrate. This yields,

elocityapproach v relative theis where5

2

2

1012

2/52

2

tzzzz

r

zz VVV

m

K

dt

dV

At the maximum compression, * and 0 zzz dtd and hence,

(d)

(e)5/2

**

2*

16

15

ER

Vm zrz

5/22*

4

5

K

Vm zrz or



65

Time versus Compression Curve

5

2

2

1 2/32

2z

r

zz m

K

dt

dV

Multiply this equation by *z and simplify to yield:

)()(1 *

2/12/5*** zzzzz

z

z

z dV

dt

d

)()(1 *

2/12/5**

zzzzz

z dV

t

The above was numerically integrated by Deresiewicz+ to produce a t vs z curve, which, by using P(t) = Kz

3/2, yields a force versus time curve.

+ H. Deresiewicz, “A note on Hertz impact”, Acta Mechanica 6, 110 ( 1968)

(f)



66

Contact time Tc in terms of the relative incident velocity Vz

ncompressio maximum of at time Load

occursn compressio maximumat which Time *

*

P

t

Equation (d):

2/3

**2/3

2/3**

)(

)(

z

z

z

z

P

P

KP

KP*tt

1 2

*

The total contact time is 2t* because the spheres are perfectly elastic and frictionless. Hence there is no loss of energy (i.e., reversible).

94.2)()(122*1

0

*2/12/5*

**

z

zzzzz

z

zc V

dV

tT

(g)



67

By substituting equation (e), rewritten as

51

2**

51*

)(256

225

z

rzz VER

mV

into (g), we obtain the result that

51

51

2** )(87.2

z

rc V

ER

mT

The question arises under what conditions is the use of Hertz’s theory in the above analysis valid? This is the topic of discussion in Chapter 11, Sections 1-3 of Johnson’s book.

(h)



68

Section 11.1, Johnson’s book

When is the quasi-static analysis for the contact model valid?

*If the contact time is much larger than the time required for the wave to traverse a rod many times.

A. E. Love, Treatise on the Mathematical Theory of Elasticity, 4 th ed., Cambridge University Press (1952).

Love suggested that the same criterion * for rods is also true for spheres. Let R be the radius of the sphere and let

*o

44

4 diameters two travel to wavesallongitudinfor required time

E

RR/C

R t sphc

(i)

Recall equation (h) for the contact time. 51

51

2** )(87.2

z

rc V

ER

mT



69

514o

c )(

85.5

cV

RT

z

If the two colliding spheres are the same size (R1 = R2 = R), then R* = R/2. Substitute this into

(h), as well as the mass in terms of the density , and 2o

* cE to yield

c-sphc tTLet Upon substituting Tc into this definition, one obtains

51

o

~

c

Vz

1According to Love’s criterion,

in order to the quasi-static analysis to hold for colliding spheres



70



End of Lecture 2

71



Appendix L2-A: Coefficient of Restitution

Am v1( )A Am v Pdt

1( ) ( ) ( )A A Am v Pdt m v 11111111111111

1( )B Bm v Pdt Bm v

1( ) ( ) ( )B B Bm v Pdt m v 11111111111111

Deformation Phase

2( )B Bm vRdtBm v

2( )A Am vAm v Rdt

2( ) ( ) ( )B B Bm v Rdt m v 11111111111111

2( ) ( ) ( )A A Am v Rdt m v 11111111111111

Restitution Phase

2

1

( )

( )A

A

Rdt v v

v vPdt

2

1

( )

( )B

B

Rdt v v

v vPdt

2 2

1 1

( ) ( )

( ) ( )B A

A B

v ve

v v

72



Appendix L2-B: Derivation of Normal Restitution Coefficient for Walton Soft-Sphere Model

Consider the impact of two spheres approaching along the line joining the centers of the spheres. The normal interaction force F1 begins to act and change the velocities of the spheres after their initial contact. Suppose v1 and v2 are the value of these velocities, m1 and m2 sphere masses.

121 21

1 Fdt

dvmF

dt

dvm

Relative velocity of approach 21 vv

overlap nonzero from reloadingor g, Unloadin),(

loading ,

22

11

oKF

KF

o - overlap value where the unloading curve goes to zero along the slope K2.

(2)

(1)

(3)

21

211 mm

mmK

Combine (1) – (3) (4)

21

21Let mm

mm

73




Multiply both sides of (4) by

dK

d 2

1 12 (5)

Let be the velocity of approach (immediately before collision when = 0)

av

Let be the maximum velocity – when . mv 0

Integrate (5) (6)

Now consider the unloading period when the spheres are moving apart. The equations of motion are

2221 21 Fdt

dvmF

dt

dvm (7)

Relative velocity of approach 21 vv

Relative approach velocity )( and o21 22 KFvv

1K

v ma

74



)( )( 2

1oo

2 2

dK

d

Now follow the same procedure used to derive (6)


)( ])[( 2

)( 2

1o

0

α0

2 222

mso

v Kvd

Kd

m

s

A substitution of (6) and (8) into the definition of the restitution coefficient yields,

(8)

1

2 )( o

K

K

v

ve

m

m

a

s

Since

(9)

,2

1

12

22oK

K

KF

KF

m

m

2

1 K

Ke (10)

75



Appendix L2-C: Derivation of the Integration Time Step for Walton’s Soft-Sphere Model

The time step is found from the unloading period.

2o

2o

α0

2 )()( ])[( 2

)( 2

1 222

moK

dt

dαd

Kzd

m

22o

2o

o

2 2

)()(

)( Timen Restitutio

oK

μd

K

μt

m

mR

For a monodisperse system, m1 = m2 = m, so that = m/2. Use this result and (10) to obtain

12 22

22 K

me

K

mtR

We estimate the time step by simply doubling the

restitution period, and then divide this result by ‘n’.

12

K

m

n

et

Documents

Lecture 2 Fundamentals of Discrete Element Modeling Anthony D. Rosato Granular Science Laboratory ME Department New Jersey Institute of Technology Newark,