1

Granular Flows

2

Lecture 2: Contact forces

Contact forces between individual grains follow the

solid constitutive relations.

– Are they applicable for large-scale flows?

Contact forces:

Normal elastic stresses due to weight

Electrostatic attraction

Hertzian normal forces (+ impulsive case + liquid bridges)

Coulomb friction (+ impulsive case)

3

Newton’s cradle (1)

History:

Christiaan Huygens (1703): discusses Newton’s first law &

collisions of suspended bodies

Demonstrates conservation of momentum and energy with

swinging spheres

Assumptions:

Conservation of kinetic energy before and after collision

Conservation of total momentum before and after collision

Let’s take a look!

4

Newton’s cradle (2)

Let’s scale it up!

Mythbusters episode “Newton’s crane cradle” (Oct. 2011)

http://dsc.discovery.com/tv-

shows/mythbusters/videos/newtons-cradle-

high-speed-1.htm

http://www.wired.com/wiredscience/2011/1

0/what-went-wrong-with-the-mythbusters-

newton-cradle/

5

Forces from elastic stresses

Normal elastic stress on a cube:

In z-direction: zz = g h = E = E w/h

Example:

Density = 2.5.103 kg/m3

Gravity g = 9.8 m/s2 zz = 24.5 N/m2

Size h = 1 mm

Young’s modulus for glass E = 7.1010 N/m2

displacement w = h = h zz /E = 3.5.10-13 m

Elastic stresses are small (picometers), compared to

stresses due to electrostatic forces!

6

Electrostatic forces

Very dry conditions creates solid-solid rubbing:

Friction creates triboelectric charging

Long-range interactions manifest as Coulomb repulsion

Charge per grain scales with surface area

smaller particles have more static electricity

Causes: agglomeration,

adhesion and segregation

In experiments:

Provide grounding

Use metal surface, not plastic

Keep an eye on the humidity

From: http://www.physik.uni-wuerzburg.de/~hinrichsen

7

Forces from Hertzian contacts

From Wikipedia: http://en.wikipedia.org/wiki/Contact_mechanics

8

Focus: Hertzian contact force

Assumptions:

Frictionless contacts

R1, R2 >> l

Non-conforming surface

Hertzian theory:

Geometrical relation:

Hertzian pressure distribution:

Radius of contact area:

Depth of indentation:

9

Focus: Hertzian contact force

Assumptions:

Frictionless contacts

R1, R2 >> l

Non-conforming surface

Hertzian theory:

Geometrical relation:

Hertzian pressure distribution:

Radius of contact area:

Depth of indentation:

10

Speed of sound

Sound speed:

Sphere-on-sphere geometry:

high confining pressures

Conical geometry:

low confining pressures

What happens on

the free surface?

From: Jia et al., Phys. Review Letters, 1999

11

Impulsive Hertzian forces (1)

Drop a ball with a velocity v1 against a stationary ball:

Conservation of kinetic energy: ½ m1 v12 = ½ m2 v2

2

Momentum conservation: m1 v1 = m1 vr + m2 v2 = Fn t

Coefficient of restitution: = (-vr + v2)/v1

assume vr = 0, v1 = v2 & = 1 (perfectly elastic, no losses)

12

Impulsive Hertzian forces (2)

Coefficient of restitution :

= 1: perfectly elastic, no energy loss

= 0: plastic, complete energy loss

0 < < 1: a proportion of the energy is lossed

Bouncing ball:

= (h1/h0)½

hk = 2k h0

13

Processes during contact:

As soon as collision occurs, compressive forces act &

velocities change

Time of collision >> period of lowest mode of vibration

Calculate collision time during contact:

Normal force:

Acceleration (Fn t = m1 v1):

Collision time:

Details in: Theory of elasticity, Timoshenko & Goodier

Impulsive Hertzian forces (3)

14

A practical example of the collision time:

R = 1 mm

v = 1 m/s

E = 50 GPa, = 0.3

= 2.5.103 kg/m3

… and the maximum penetration depth:

R = 1 mm

v = 1 m/s

E = 50 GPa, = 0.3

= 2.5.103 kg/m3

Applied strain: = /l = (/R)½ = 4.8.10-2

Impulsive Hertzian forces (4)

t = 6.8.10-6 s

= 2.3.10-6 m c3 = 2.3.10-3 (m/s)-4/5

c2 = 6.8.10-3 (m/s)-4/5

15

What happens in the idealized Newton’s cradle?

Just after contact, all balls are in contact with each other

Net impulses of balls must be zero, for balls that end up with

the same velocity (also for stationary balls)

Net impulses of ball 1 & N are not zero, as forces are not

balanced during contact!

The kinetic energy of ball N was stored as compressed energy

at the interfaces

Two released particles?

Ball 1: initially at V, impulse slows it to rest

Ball 2: zero net impulse, equal and opposite compression force

Ball 3: initially at rest, impulse speed it up to V

Newton’s cradle (1)

16

What can go wrong (Mythbuster experiment?):

Balls are not perfectly elastic: vr ≠ 0

Flansburg and Hudnut [1979]: three steel balls

Final velocity ideal case: 0, 0, and 1

Final velocity real case: -0.06, +0.09 and +0.97

Air gaps between balls, alignment incorrect

Mass or dimensions of balls are not perfectly the same

As always, scale is an issue!

Newton’s cradle (2)

17

Liquid bridges between spheres:

Same principles apply!

As spheres approach they:

Experience resistance due to lubrication

Attain a minimum separation

Experience resistance upon rebound (not enough fluid left)

Fluid resistance cannot be neglected

Outcome depends on Stokes number (= inertial-viscous forces)

& liquid thickness between ball 2 & 3

Stokes’s cradle

From: Donahue et al., PRL, 2010

18

Granular medium: microscopic and macroscopic friction

Microscopic: in contacts between grains

Macroscopic: friction angle (later lectures)

Use Coulomb (solid-solid) friction as a model:

Stick: Ff < s Fn

Slip: Ff = d Fn

with the static s and dynamic d

coefficient of friction,

typically: 0 < d < s < 1

Force is opposite of motion

Frictional forces (1)

19

Stick-slip motion:

Alternating behavior: move – stop – move - stop

Spring elongates, mass moves, spring jumps back

Frictional forces (2)

From: Les Milieux granulaires, O. Pouliquen

20

Indeterminacy:

4 unknown forces (Ff, Fn for each wall)

2 force balances

1 momentum balance

Indeterminacy, solve by knowing history!

For a pile of sand: how was it build?

Frictional forces (3)

21

Ball on a surface:

slips and decelerates by friction force

decrease in linear momentum increase angular momentum

starts rolling with rolling friction (empirical)

when ball rolls without slip rolling velocity: Vr = 5/7 V0

Rolling or slipping?

22

Bowling ball is thrown with backspin and velocity V0:

assume mass m, radius R, moment of inertia Icm = 2/5 m R2

furthermore, assume V0 > R 0

calculate speed Vf when it rolls without slipping

Example: Bowling ball (1)

23

Equate angular momentum:

Linitial = m V0 R – Icm 0

Lfinal = m Vf R + Icm f

Substitute: vf = f R vf = 5/7 (v0 – 2/5 R 0)

Example: Bowling ball (2)

Linitial = Lfinal