acoustic emission of sand - damtp
TRANSCRIPT
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Granular Flows
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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)
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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!
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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/
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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!
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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
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Forces from Hertzian contacts
From Wikipedia: http://en.wikipedia.org/wiki/Contact_mechanics
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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:
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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:
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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
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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)
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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
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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)
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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
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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)
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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)
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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
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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)
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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
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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)
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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?
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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)
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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