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1 1.021, 3.021, 10.333, 22.00 Introduction to Modeling and Simulation Part I – Continuum and particle methods Markus J. Buehler Laboratory for Atomistic and Molecular Mechanics Department of Civil and Environmental Engineering Massachusetts Institute of Technology Review session—preparation quiz I Lecture 12

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Page 1: Part I Lecture 12 Review session—preparation quiz I · PDF fileContinuum modeling approaches and solution approaches 3. ... Visualization and data analysis 6. Mechanical properties

1

1.021, 3.021, 10.333, 22.00 Introduction to Modeling and Simulation

Part I – Continuum and particle methods

Markus J. BuehlerLaboratory for Atomistic and Molecular MechanicsDepartment of Civil and Environmental EngineeringMassachusetts Institute of Technology

Review session—preparation quiz I Lecture 12

Page 2: Part I Lecture 12 Review session—preparation quiz I · PDF fileContinuum modeling approaches and solution approaches 3. ... Visualization and data analysis 6. Mechanical properties

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Content overview

I. Particle and continuum methods1. Atoms, molecules, chemistry2. Continuum modeling approaches and solution approaches 3. Statistical mechanics4. Molecular dynamics, Monte Carlo5. Visualization and data analysis 6. Mechanical properties – application: how things fail (and

how to prevent it)7. Multi-scale modeling paradigm8. Biological systems (simulation in biophysics) – how

proteins work and how to model them

II. Quantum mechanical methods1. It’s A Quantum World: The Theory of Quantum Mechanics2. Quantum Mechanics: Practice Makes Perfect3. The Many-Body Problem: From Many-Body to Single-

Particle4. Quantum modeling of materials5. From Atoms to Solids6. Basic properties of materials7. Advanced properties of materials8. What else can we do?

Lectures 2-13

Lectures 14-26

Page 3: Part I Lecture 12 Review session—preparation quiz I · PDF fileContinuum modeling approaches and solution approaches 3. ... Visualization and data analysis 6. Mechanical properties

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Overview: Material covered so far…Lecture 1: Broad introduction to IM/S

Lecture 2: Introduction to atomistic and continuum modeling (multi-scale modeling paradigm, difference between continuum and atomistic approach, case study: diffusion)

Lecture 3: Basic statistical mechanics – property calculation I (property calculation: microscopic states vs. macroscopic properties, ensembles, probability density and partition function)

Lecture 4: Property calculation II (Monte Carlo, advanced property calculation, introduction to chemical interactions)

Lecture 5: How to model chemical interactions I (example: movie of copper deformation/dislocations, etc.)

Lecture 6: How to model chemical interactions II (EAM, a bit of ReaxFF—chemical reactions)

Lecture 7: Application to modeling brittle materials I

Lecture 8: Application to modeling brittle materials II

Lecture 9: Application – Applications to materials failure

Lecture 10: Applications to biophysics and bionanomechanics

Lecture 11: Applications to biophysics and bionanomechanics (cont’d)

Lecture 12: Review session – part I

Page 4: Part I Lecture 12 Review session—preparation quiz I · PDF fileContinuum modeling approaches and solution approaches 3. ... Visualization and data analysis 6. Mechanical properties

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Check: goals of part IYou will be able to …

Carry out atomistic simulations of various processes (diffusion, deformation/stretching, materials failure)

Carbon nanotubes, nanowires, bulk metals, proteins, silicon crystals, …

Analyze atomistic simulations

Visualize atomistic/molecular data

Understand how to link atomistic simulation results with continuum models within a multi-scale scheme

Page 5: Part I Lecture 12 Review session—preparation quiz I · PDF fileContinuum modeling approaches and solution approaches 3. ... Visualization and data analysis 6. Mechanical properties

Topics covered – part I

5

Atomistic & molecular simulation algorithms

Property calculation

Potential/force field models

Applications

Page 6: Part I Lecture 12 Review session—preparation quiz I · PDF fileContinuum modeling approaches and solution approaches 3. ... Visualization and data analysis 6. Mechanical properties

Lecture 12: Review session – part I

1. Review of material covered in part I1.1 Atomistic and molecular simulation algorithms1.2 Property calculation1.3 Potential/force field models1.4 Applications

2. Important terminology and concepts

Goal of today’s lecture: Review main concepts of atomistic and molecular dynamics, continuum modelsPrepare you for the quiz on Thursday

6

Page 7: Part I Lecture 12 Review session—preparation quiz I · PDF fileContinuum modeling approaches and solution approaches 3. ... Visualization and data analysis 6. Mechanical properties

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1.1 Atomistic and molecular simulation algorithms

1.2 Property calculation1.3 Potential/force field models

1.4 Applications

Goals: Basic MD algorithm (integration scheme), initial/boundary conditions, numerical issues (supercomputing)

Page 8: Part I Lecture 12 Review session—preparation quiz I · PDF fileContinuum modeling approaches and solution approaches 3. ... Visualization and data analysis 6. Mechanical properties

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Differential equations solved in MD

Page 9: Part I Lecture 12 Review session—preparation quiz I · PDF fileContinuum modeling approaches and solution approaches 3. ... Visualization and data analysis 6. Mechanical properties

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Basic concept: atomistic methods

)(),(),( tatvtr iii

Results of MD simulation:

PDE solved in MD:i

ii rd

rdUdt

rdm )(2

2

−= }{ jrr =

Ni ..1=

Ni ..1=)( Fma =

Page 10: Part I Lecture 12 Review session—preparation quiz I · PDF fileContinuum modeling approaches and solution approaches 3. ... Visualization and data analysis 6. Mechanical properties

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Complete MD updating scheme

( ) ...)()(2)()( 20000 +Δ+Δ+Δ−−=Δ+ ttattrttrttr iiii

ijjij amf ,, =

Positions at t0

Accelerationsat t0

Positions at t0-Δt

jijij mfa /,, =

rx

Ff ijij

,, =

(4) Obtain force vectors from potential (sum over contributions from all neighbor of atom j)

Potential

(2) Updating method (integration scheme - Verlet)

(3) Obtain accelerations from forces

⎟⎟⎠

⎞⎜⎜⎝

⎛⎥⎦⎤

⎢⎣⎡−⎥⎦

⎤⎢⎣⎡=

612

4)(rr

r σσεφ

(1) Initial conditions: Positions & velocities at t0 (random velocities so that initial temperature is represented)

rrF

d)(dφ

−=

Nj ..1=∀

Page 11: Part I Lecture 12 Review session—preparation quiz I · PDF fileContinuum modeling approaches and solution approaches 3. ... Visualization and data analysis 6. Mechanical properties

11

Algorithm of force calculation

6 5

j=1

2 3

4

cutr

i

for i=1..N

for j=1..N (i ≠ j)

[add force contributions]

Page 12: Part I Lecture 12 Review session—preparation quiz I · PDF fileContinuum modeling approaches and solution approaches 3. ... Visualization and data analysis 6. Mechanical properties

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Summary: Atomistic simulation – numerical approach “molecular dynamics – MD”

Atomistic model; requires atomistic microstructure and atomic positions at beginning (initial conditions), initial velocities from random distribution according to specified temperatureStep through time by integration scheme (Verlet)Repeat force calculation of atomic forces based on their positionsExplicit notion of chemical bonds – captured in interatomic potential

Loop

Page 13: Part I Lecture 12 Review session—preparation quiz I · PDF fileContinuum modeling approaches and solution approaches 3. ... Visualization and data analysis 6. Mechanical properties

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Scaling behavior of MD code

Page 14: Part I Lecture 12 Review session—preparation quiz I · PDF fileContinuum modeling approaches and solution approaches 3. ... Visualization and data analysis 6. Mechanical properties

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Force calculation for N particles: pseudocode

Requires two nested loops, first over all atoms, and then over all other atoms to determine the distance

for i=1..N:

for j=1..N (i ≠ j):

determine distance between i and j

calculate force and energy (if rij < rcut, cutoff radius)

add to total force vector / energy

time ~ N2: computational disaster

Δt1~NΔt0Δt0

Δt to

tal~

N·N

Δt 0=

N2 Δ

t 0

Page 15: Part I Lecture 12 Review session—preparation quiz I · PDF fileContinuum modeling approaches and solution approaches 3. ... Visualization and data analysis 6. Mechanical properties

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Strategies for more efficient computation

Two approaches

1. Neighbor lists: Store information about atoms in vicinity, calculated in an N2 effort, and keep information for 10..20 steps

Concept: Store information of neighbors of each atom within vicinity of cutoff radius (e.g. list in a vector); update list only every 10..20 steps

2. Domain decomposition into bins: Decompose system into small bins; force calculation only between atoms in local neighboring bins

Concept: Even overall system grows, calculation is done only in a local environment (have two nested loops but # of atoms does notincrease locally)

Page 16: Part I Lecture 12 Review session—preparation quiz I · PDF fileContinuum modeling approaches and solution approaches 3. ... Visualization and data analysis 6. Mechanical properties

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Force calculation with neighbor lists

Requires two nested loops, first over all atoms, and then over all other atoms to determine the distance

for i=1..Nneighbors,i:

for j=1..N (i ≠ j):

determine distance between i and j

calculate force and energy (if rij < rcut, cutoff radius)

add to total force vector / energy

time ~ N: computationally tractable even for large N

Δt1~NΔt0Δt0 Δ

t tota

l~N

Δt 0

Page 17: Part I Lecture 12 Review session—preparation quiz I · PDF fileContinuum modeling approaches and solution approaches 3. ... Visualization and data analysis 6. Mechanical properties

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Periodic boundary conditionsPeriodic boundary conditions allows studying bulk properties (no free surfaces) with small number of particles (here: N=3), all particles are “connected”

Original cell surrounded by 26 image cells; image particles move in exactly the same way as original particles (8 in 2D)

Image by MIT OpenCourseWare. After Buehler.

Particle leaving box enters on other side with same velocity vector.

Page 18: Part I Lecture 12 Review session—preparation quiz I · PDF fileContinuum modeling approaches and solution approaches 3. ... Visualization and data analysis 6. Mechanical properties

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Parallel computing – “supercomputers”

Supercomputers consist of a very large number of individual computing units (e.g. Central Processing Units, CPUs)

Images removed due to copyright restrictions. Please see http://www.sandia.gov/ASC/images/library/ASCI-White.jpg

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Domain decomposition

Each piece worked on by one of the computers in the supercomputerFig. 1 c from Buehler, M., et al. "The Dynamical Complexity of Work-Hardening: A Large-Scale Molecular Dynamics Simulation.”Acta Mech Sinica 21 (2005): 103-11. © Springer-Verlag. All rights reserved. This content is excluded from our Creative Commons license.For more information, see http://ocw.mit.edu/fairuse.

Page 20: Part I Lecture 12 Review session—preparation quiz I · PDF fileContinuum modeling approaches and solution approaches 3. ... Visualization and data analysis 6. Mechanical properties

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Modeling and simulation

Page 21: Part I Lecture 12 Review session—preparation quiz I · PDF fileContinuum modeling approaches and solution approaches 3. ... Visualization and data analysis 6. Mechanical properties

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Modeling and simulation

What is a model? What is a simulation?

• Modeling: developing a mathematical representation of a physical situation

•Simulation: solving the equations that arose from the development of the model.

Modeling vs Simulation

© Google, Inc. All rights reserved. This content is excluded from our Creative Commons license. For more information, see http://ocw.mit.edu/fairuse.

© Massachusetts Bay Transportation Authority. All rights reserved. This content is excluded from our Creative Commons license. For more information, see http://ocw.mit.edu/fairuse.

Page 22: Part I Lecture 12 Review session—preparation quiz I · PDF fileContinuum modeling approaches and solution approaches 3. ... Visualization and data analysis 6. Mechanical properties

22

Modeling and simulation

M

S

M

S

M

Page 23: Part I Lecture 12 Review session—preparation quiz I · PDF fileContinuum modeling approaches and solution approaches 3. ... Visualization and data analysis 6. Mechanical properties

23

Atomistic versus continuum viewpoint

Page 24: Part I Lecture 12 Review session—preparation quiz I · PDF fileContinuum modeling approaches and solution approaches 3. ... Visualization and data analysis 6. Mechanical properties

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Diffusion: Phenomenological description

Ink dot

Tissue2

2

dxcdD

tc

=∂∂ 2nd Fick law (governing

equation)

Physical problem

BC: c (r = ∞) = 0IC: c (r=0, t = 0) = c0

© source unknown. All rights reserved. This content is excluded from our Creative Commons license. For more information, see http://ocw.mit.edu/fairuse

Page 25: Part I Lecture 12 Review session—preparation quiz I · PDF fileContinuum modeling approaches and solution approaches 3. ... Visualization and data analysis 6. Mechanical properties

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Continuum model: Empirical parameters

Continuum model requires parameter that describes microscopic processes inside the material

Need experimental measurements to calibrate

Microscopicmechanisms

???

Continuummodel

Length scale

Time scale“Empirical”or experimentalparameterfeeding

Page 26: Part I Lecture 12 Review session—preparation quiz I · PDF fileContinuum modeling approaches and solution approaches 3. ... Visualization and data analysis 6. Mechanical properties

How to solve continuum problem: Finite difference scheme

26

2

2

dxcdD

tc

=∂∂

Concentration at iat new time

==

ji space

time

Concentration at iat old time

Concentration at i+1at old time

Concentration at iat old time

Concentration at i-1at old time

“explicit” numerical scheme(new concentration directly from concentration at earlier time)

( )jijijijiji cccDxtcc ,1,,12,1, 2 −++ −−+=

δδ

Other methods: Finite element method

Page 27: Part I Lecture 12 Review session—preparation quiz I · PDF fileContinuum modeling approaches and solution approaches 3. ... Visualization and data analysis 6. Mechanical properties

27

Atomistic model of diffusion

Diffusion at macroscale (change of concentrations) is result of microscopic processes, random motion of particle

Atomistic model provides an alternative way to describe diffusion

Enables us to directly calculate the diffusion constant from thetrajectory of atoms

Follow trajectory of atoms and calculate how fast atoms leave their initial position

txpD

ΔΔ

−=2

Concept:follow this quantity overtime

Page 28: Part I Lecture 12 Review session—preparation quiz I · PDF fileContinuum modeling approaches and solution approaches 3. ... Visualization and data analysis 6. Mechanical properties

28

Atomistic model of diffusion

2rΔ

Average squares of displacements of all particles

( )22 )0()(1)( ∑ =−=Δi

ii trtrN

tr

Particles Trajectories

Mean Squared Displacement function

i

ii rd

rdUdt

rdm )(2

2

−= }{ jrr = Ni ..1=

Images courtesy of the Center for Polymer Studies at Boston University. Used with permission.

Page 29: Part I Lecture 12 Review session—preparation quiz I · PDF fileContinuum modeling approaches and solution approaches 3. ... Visualization and data analysis 6. Mechanical properties

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2rΔ

Calculation of diffusion coefficient

1D=1, 2D=2, 3D=3

slope = D

( )22 )0()(1)( ∑ =−=Δi

ii trtrN

tr

Position of atom i at time t

Position of atom i at time t=0

( ))(lim21 2 tr

dtd

dD

tΔ=

∞→

Einstein equation

(nm)

(ps)

Page 30: Part I Lecture 12 Review session—preparation quiz I · PDF fileContinuum modeling approaches and solution approaches 3. ... Visualization and data analysis 6. Mechanical properties

30

???

SummaryMolecular dynamics provides a powerful approach to relate the diffusion constant that appears in continuum models to atomistictrajectories

Outlines multi-scale approach: Feed parameters from atomistic simulations to continuum models

MD

Continuummodel

Length scale

Time scale“Empirical”or experimentalparameterfeeding

Page 31: Part I Lecture 12 Review session—preparation quiz I · PDF fileContinuum modeling approaches and solution approaches 3. ... Visualization and data analysis 6. Mechanical properties

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1.1 Atomistic and molecular simulation algorithms

1.2 Property calculation1.3 Potential/force field models

1.4 Applications

Goals: How to calculate “useful” properties from MD runs (temperature, pressure, RDF, VAF,..); significance of averaging; Monte Carlo schemes

Page 32: Part I Lecture 12 Review session—preparation quiz I · PDF fileContinuum modeling approaches and solution approaches 3. ... Visualization and data analysis 6. Mechanical properties

32

Property calculation: IntroductionHave:

“microscopic information”

Want: Thermodynamical properties (temperature, pressure, ..)Transport properties (diffusivities, shear viscosity, ..)Material’s state (gas, liquid, solid)

)(),(),( tatvtr iii Ni ..1=

Page 33: Part I Lecture 12 Review session—preparation quiz I · PDF fileContinuum modeling approaches and solution approaches 3. ... Visualization and data analysis 6. Mechanical properties

33

Micro-macro relation

)(131)(

1

2 tvmNk

tTN

iii

B∑

=

=

)(tT

t

Which to pick?

Specific (individual) microscopic states are insufficient to relate to macroscopic properties

Images courtesy of the Center for Polymer Studies at Boston University. Used with permission.

Page 34: Part I Lecture 12 Review session—preparation quiz I · PDF fileContinuum modeling approaches and solution approaches 3. ... Visualization and data analysis 6. Mechanical properties

34

Ergodic hypothesis: significance of averaging

Ergodic hypothesis:

Ensemble (statistical) average = time average

All microstates are sampled with appropriate probability density over long time scales

∑∑==

=>=<>=<tA Nit

TimeEnsNiA

iAN

AAiAN ..1..1

)(1)(1

Monte CarloAverage over Monte Carlo steps

NO DYNAMICAL INFORMATION

MDAverage over time steps

DYNAMICAL INFORMATION

drdprprpAAp r∫ ∫>=< ),(),( ρ property must

be properlyaveraged

Page 35: Part I Lecture 12 Review session—preparation quiz I · PDF fileContinuum modeling approaches and solution approaches 3. ... Visualization and data analysis 6. Mechanical properties

35

Monte Carlo schemeConcept: Find convenient way to solve the integral

Use idea of “random walk” to step through relevant microscopic states and thereby create proper weighting (visit states with higher probability density more often)

Monte Carlo schemes: Many applications (beyond what is discussed here; general method to solve complex integrations)

drdprprpAAp r∫ ∫>=< ),(),( ρ

Page 36: Part I Lecture 12 Review session—preparation quiz I · PDF fileContinuum modeling approaches and solution approaches 3. ... Visualization and data analysis 6. Mechanical properties

36

Monte Carlo scheme: area calculation

Ω

∫Ω

Ω= dxfA )(

Method to carry out integration (illustrate general concept)

Want:

E.g.: Area of circle (value of π)

4

2dACπ

=4π

=CA 1=d

Page 37: Part I Lecture 12 Review session—preparation quiz I · PDF fileContinuum modeling approaches and solution approaches 3. ... Visualization and data analysis 6. Mechanical properties

37

Monte Carlo schemeStep 1: Pick random point in Step 2: Accept/reject point based on criterion (e.g. if inside or outside of circle and if in area not yet counted)Step 3: If accepted, add to the total sum otherwise

Ω

1)( =ixf

ix

∫Ω

Ω= dxfAC )(

∑=i

iA

C xfN

A )(1

See also: http://math.fullerton.edu/mathews/n2003/MonteCarloPiMod.html

AN : Attemptsmade

0)( =ixf

2/1=d

Courtesy of John H. Mathews. Used with permission.

Page 38: Part I Lecture 12 Review session—preparation quiz I · PDF fileContinuum modeling approaches and solution approaches 3. ... Visualization and data analysis 6. Mechanical properties

38

Area of Middlesex County (MSC)

∑=i

iA

SQMSC xfN

AA )(1

100 km

100

km 2km000,10=SQA

Fraction of points that lie within MS County

Expression provides area of MS County

x

y

Image courtesy of Wikimedia Commons.

Page 39: Part I Lecture 12 Review session—preparation quiz I · PDF fileContinuum modeling approaches and solution approaches 3. ... Visualization and data analysis 6. Mechanical properties

39

Detailed view - schematic

NA=55 points (attempts)12 points within

22 km8.2181km22.010000)(1=⋅== ∑

ii

ASQMSC xf

NAA

12

Page 40: Part I Lecture 12 Review session—preparation quiz I · PDF fileContinuum modeling approaches and solution approaches 3. ... Visualization and data analysis 6. Mechanical properties

40

Results U.S. Census Bureau

Taken from: http://quickfacts.census.gov/qfd/states/25/25017.html

2km8.2181=MSCAMonte Carlo result:

823.46 square miles = 2137 km2 (1 square mile=2.58998811 km2 )

Page 41: Part I Lecture 12 Review session—preparation quiz I · PDF fileContinuum modeling approaches and solution approaches 3. ... Visualization and data analysis 6. Mechanical properties

41

Analysis of satellite images

200 m

Spy Pond (Cambridge/Arlington)

Image removed due to copyright restrictions.

Google Satellite image of Spy Pond, in Cambridge, MA.

Page 42: Part I Lecture 12 Review session—preparation quiz I · PDF fileContinuum modeling approaches and solution approaches 3. ... Visualization and data analysis 6. Mechanical properties

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Metropolis Hastings algorithm

drdprprpAAp r∫ ∫>=< ),(),( ρ

Images removed due to copyright restrictions.

Please see: Fig. 2.7 in Buehler, Markus J. Atomistic Modeling of Materials Failure. Springer, 2008.

Images removed due to copyright restrictions.

Please see: Fig. 2.8 in Buehler, Markus J. Atomistic Modeling of Materials Failure. Springer, 2008.

Page 43: Part I Lecture 12 Review session—preparation quiz I · PDF fileContinuum modeling approaches and solution approaches 3. ... Visualization and data analysis 6. Mechanical properties

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Molecular Dynamics vs. Monte Carlo

MD provides actual dynamical data for nonequilibrium processes (fracture, deformation, instabilities)

Can study onset of failure, instabilities

MC provides information about equilibrium properties(diffusivities, temperature, pressure)

Not suitable for processes like fracture

∑∑==

=>=<>=<tA Nit

TimeEnsNiA

iAN

AAiAN ..1..1

)(1)(1

Page 44: Part I Lecture 12 Review session—preparation quiz I · PDF fileContinuum modeling approaches and solution approaches 3. ... Visualization and data analysis 6. Mechanical properties

44

Property calculation: Temperature and pressure

Temperature

><= ∑=

N

iii

B

vmNk

T1

2131

iii vvv ⋅=2

2~ vT

Tv ~

Page 45: Part I Lecture 12 Review session—preparation quiz I · PDF fileContinuum modeling approaches and solution approaches 3. ... Visualization and data analysis 6. Mechanical properties

Temperature control in MD (NVT ensemble)Berendsen thermostat

45

...)()()(2)( 20000 +Δ+Δ−−=Δ+ ttattrtrttr iiii

Calculate temperature ><= ∑=

N

iii

B

vmNk

T1

2131

Rescale velocities

Tv ~

Calculate ratio of current vs. desired temperature

Page 46: Part I Lecture 12 Review session—preparation quiz I · PDF fileContinuum modeling approaches and solution approaches 3. ... Visualization and data analysis 6. Mechanical properties

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Interpretation of RDF

Page 47: Part I Lecture 12 Review session—preparation quiz I · PDF fileContinuum modeling approaches and solution approaches 3. ... Visualization and data analysis 6. Mechanical properties

47

Formal approach: Radial distribution function (RDF)

ρρ /)()( rrg =

The radial distribution function is defined as

Provides information about the density of atoms at a given radius r; ρ(r) is the local density of atoms

ρ1

)()()(

2

2r

r

rrNrg

Δ

Δ

±Ω>±<

=

=drrrg 22)( π Number of particles that lie in a spherical shellof radius r and thickness dr

Local density

Density of atoms (volume)

Volume of this shell (dr)

Number of atoms in the interval 2rr Δ±

Page 48: Part I Lecture 12 Review session—preparation quiz I · PDF fileContinuum modeling approaches and solution approaches 3. ... Visualization and data analysis 6. Mechanical properties

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Radial distribution function: Solid versus liquid versus gas

Note: The first peak corresponds to the nearest neighbor shell, the second peak to the second nearest neighbor shell, etc.

In FCC: 12, 6, 24, and 12 in first four shells

Image by MIT OpenCourseWare.

5

4

3

2

1

00 0.2 0.4 0.6 0.8

g(r)

distance/nm

5

4

3

2

1

00 0.2 0.4 0.6 0.8

g(r)

distance/nm

Solid Argon

Liquid Argon

Liquid Ar(90 K)

Gaseous Ar(90 K)

Gaseous Ar(300 K)

Page 49: Part I Lecture 12 Review session—preparation quiz I · PDF fileContinuum modeling approaches and solution approaches 3. ... Visualization and data analysis 6. Mechanical properties

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RDF and crystal structure

Peaks in RDF characterize NN distance, can infer from RDF about crystal structure

http://physchem.ox.ac.uk/~rkt/lectures/liqsolns/liquids.html

Image by MIT OpenCourseWare.

5

4

3

2

1

00 0.2 0.4 0.6 0.8

g(r)

distance/nm

Solid Argon

Liquid Argon

1st nearest neighbor (NN)

2nd NN3rd NN

Page 50: Part I Lecture 12 Review session—preparation quiz I · PDF fileContinuum modeling approaches and solution approaches 3. ... Visualization and data analysis 6. Mechanical properties

50Images courtesy of Mark Tuckerman. Used with permission.

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Interpretation of RDF

X

1st nearest neighbor (NN)

2nd NN3rd NN

copper: NN distance > 2 Å

not a liquid (sharp peaks)

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52

Graphene/carbon nanotubes

Graphene/carbon nanotubes (rolled up graphene)NN: 1.42 Å, second NN 2.46 Å …

RDF

Images of graphene/carbon nanutubes:

http://weblogs3.nrc.nl/techno/wp-content/uploads/080424_Grafeen/Graphene_xyz.jpg

http://depts.washington.edu/polylab/images/cn1.jpg

Page 53: Part I Lecture 12 Review session—preparation quiz I · PDF fileContinuum modeling approaches and solution approaches 3. ... Visualization and data analysis 6. Mechanical properties

53

><= ∑=

N

iii

B

vmNk

T1

2131

iii vvv ⋅=2

>±Ω±

=<Δ

Δ

ρ)()()(

2

2r

r

rrNrg

( )22 )0()(1)( ∑ =−>=Δ<i

ii trtrN

tr

Temperature

MSD

RDF

Direct

Diffusivity

Atomic structure (signature)

Property Definition Application

Summary – property calculation

Page 54: Part I Lecture 12 Review session—preparation quiz I · PDF fileContinuum modeling approaches and solution approaches 3. ... Visualization and data analysis 6. Mechanical properties

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Material properties: Classification

Structural – crystal structure, RDF

Thermodynamic -- equation of state, heat capacities, thermal expansion, free energies, use RDF, temperature, pressure/stress

Mechanical -- elastic constants, cohesive and shear strength, elastic and plastic deformation, fracture toughness, use stress

Transport -- diffusion, viscous flow, thermal conduction, use MSD, temperature

Page 55: Part I Lecture 12 Review session—preparation quiz I · PDF fileContinuum modeling approaches and solution approaches 3. ... Visualization and data analysis 6. Mechanical properties

Bell model analysis (protein rupture)

55

Page 56: Part I Lecture 12 Review session—preparation quiz I · PDF fileContinuum modeling approaches and solution approaches 3. ... Visualization and data analysis 6. Mechanical properties

Bell model analysis (protein rupture)

56

a =kBTxb

b = −kbTxb

ln ω 0xb exp −Eb

kbT⎛

⎝ ⎜

⎠ ⎟

⎝ ⎜

⎠ ⎟

sec/1101 130 ×=ω

Get a and b from fitting to the graphs

Solve (1) for a, use (2) to solve for Eb

a=(800-400)/(0-(-4.6))) pN=86.86 pN xb=0.47 Å

(1)

(1)

bvaExvf bb +⋅= ln),;(

b=800 pN a=86.86 pN

kB=1.3806505E-23 J/K

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Determining the energy barrier

57

( )

( )

( )

( ) TkTk

bxxE

Tkbxx

TkE

TkEx

Tkbx

TkEx

xTk

TkEx

xTkb

bb

bbb

b

bb

b

b

b

bb

b

b

b

bb

b

b

b

bb

b

b

⎥⎦

⎤⎢⎣

⎡+=

+=

−=−

⎥⎦

⎤⎢⎣

⎡−−=⎟

⎟⎠

⎞⎜⎜⎝

⎛⎟⎟⎠

⎞⎜⎜⎝

⎛−−=

0

0

0

00

ln

ln

ln

lnexpln

ω

ω

ω

ωω

b=800 pN Eb≈9 kcal/mol

Possible mechanism: Rupture of 3 H-bonds (H-bond energy 3 kcal/mol)

1 kcal/mol=6.9477E-21 J

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58

Scaling with Eb : shifts curve

bvaExvf bb +⋅= ln),;(

↑bE

0ln vx

Tkbx

Tkab

B

b

B ⋅⋅

−=⋅

= ⎟⎟⎠

⎞⎜⎜⎝

⎛⋅

−⋅⋅=Tk

Exvb

bb exp00 ω

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59

bvaExvf bb +⋅= ln),;(

↓bx

0ln vx

Tkbx

Tkab

B

b

B ⋅⋅

−=⋅

= ⎟⎟⎠

⎞⎜⎜⎝

⎛⋅

−⋅⋅=Tk

Exvb

bb exp00 ω

Scaling with xb: changes slope

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60

1.1 Atomistic and molecular simulation algorithms

1.2 Property calculation1.3 Potential/force field models

1.4 Applications

Goals: How to model chemical interactions between particles (interatomic potential, force fields); applications to metals, proteins

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61

Concept: Repulsion and attraction

“point particle” representation

Attraction: Formation of chemical bond by sharing of electronsRepulsion: Pauli exclusion (too many electrons in small volume)

r

Electrons

Core

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e

Energy Ur

Repulsion

Attraction

1/r12 (or Exponential)

1/r6

Radius r (Distancebetween atoms)

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62

Generic shape of interatomic potential

Many chemical bonds show this generic behavior

Attraction: Formation of chemical bond by sharing of electronsRepulsion: Pauli exclusion (too many electrons in small volume)

Image by MIT OpenCourseWare.Image by MIT OpenCourseWare.

e

Energy Ur

Repulsion

Attraction

1/r12 (or Exponential)

1/r6

Radius r (Distancebetween atoms)

Harmonic oscillatorr0

φ

r~ k(r - r0)2

Effective interatomic potential

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63

Atomic interactions – different types of chemical bonds

Primary bonds (“strong”)Ionic (ceramics, quartz, feldspar - rocks) Covalent (silicon) Metallic (copper, nickel, gold, silver)(high melting point, 1000-5,000K)

Secondary bonds (“weak”)Van der Waals (wax, low melting point) Hydrogen bonds (proteins, spider silk)(melting point 100-500K)

Ionic: Non-directional (point charges interacting)Covalent: Directional (bond angles, torsions matter)Metallic: Non-directional (electron gas concept)

Wea

ker b

ondi

ng

Know models for all types of chemical bonds!

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64

How to choose a potential

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65

How to calculate forces from the “potential”or “force field”

Define interatomic potentials, that describe the energy of a set of atoms as a function of their coordinates

“Potential” = “force field”

)(rUU totaltotal =

NirUF totalri i..1)( =−∇=

⎟⎟⎠

⎞⎜⎜⎝

∂∂

∂∂

∂∂

=∇iii

r rrri,3,2,1

,,

Depends on position ofall other atoms

Change of potential energydue to change of position ofparticle i (“gradient”)

{ } Njrr j ..1==

Position vector of atom i

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66

Force calculation

Pair potentialNote cutoff!

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67

Pair potentials: energy calculation

Pair wise summation of bond energies

∑ ∑≠= =

=N

jii

N

jijrU

,1 121

tot )(φ

)( ijrφ

∑=

=N

jiji rU

1

)( φEnergy of atom i

Pair wiseinteraction potential

Simple approximation: Total energy is sum over the energy of all pairs of atoms in the system

=ijr distance betweenparticles i and j

12r

25ravoid double counting

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68

Total energy of pair potential

Assumption: Total energy of system is expressed as sum of the energy due to pairs of atoms

( )32312321131221 φφφφφφ +++++= totalU

∑ ∑≠= =

=N

jii

N

jijtotal rU

,1 121 )(φ

)( ijij rφφ =with

beyond cutoff=0

=0

12r 23r

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69

Force calculation – pair potential

ij

ij

rr

Fd

)(dφ−=

Forces can be calculated by taking derivatives from the potential function

Force magnitude: Negative derivative of potential energy with respect to atomic distance

To obtain force vector Fi, take projections into the three axial directions

x1

x2fij

ii r

xFf =

x1

x2

ijr

j

i

Component i of vector ijr

ijij rr =1f

2f

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70

Force calculation – pair potential

Forces on atom 3: only interactionwith atom 2 (cutoff)

23

23

d)(d

rrF φ

−=23rxFf i

i = 2323 rr =ijr

1. Determine distance between atoms 3 and 2, r232. Obtain magnitude of force vector based on derivative of potential3. To obtain force vector Fi, take projections into the

three axial directionsComponent i of vector

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71

Interatomic pair potentials: examples

Morse potential

Lennard-Jones 12:6 potential (excellent model for noble gases, Ar, Ne, Xe..)

Buckingham potential

Harmonic approximation

( ) ( ))(exp2)(2exp)( 00 rrDrrDr ijijij −−−−−= ααφ

6

exp)( ⎟⎟⎠

⎞⎜⎜⎝

⎛−⎟⎟

⎞⎜⎜⎝

⎛−=

ij

ijij r

Cr

Ar σσ

φ

⎥⎥

⎢⎢

⎟⎟⎠

⎞⎜⎜⎝

⎛−⎟

⎟⎠

⎞⎜⎜⎝

⎛=

612

4)(ijij

ij rrr σσεφ

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72

Lennard-Jones potential – example for copper

LJ potential – parameters for copper (e.g.: Cleri et al., 1997)

0.2

0.1

0

2 3 4 5

-0.1

-0.2r0

φ (e

V)

r (A)o

ε

σ 6 2 = r0

Image by MIT OpenCourseWare.

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73

Derivative of LJ potential ~ force

0.2

0.1

0

-0.1

-0.22 3 4 5r0

-dφ/

dr (

eV/A

)ο

r (A)o

F = _ dφ(r)dr

= -φ'

LJFmax,

rEQ

φ(r)

0

Image by MIT OpenCourseWare.

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74

Cutoff radius

∑=

=neighNj

iji rU..1

)(φ6 5

j=1

2 3

4

cutr

i φ

r

LJ 12:6potential

ε0

cutr

Cutoff radius = considering interactions only to a certain distanceBasis: Force contribution negligible (slope)

∑=

=N

jiji rU

1

)( φ

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75

Derivative of LJ potential ~ force

Beyond cutoff: Changes in energy (and thus forces) small

Image by MIT OpenCourseWare.

0.2

0.1

0

-0.1

-0.22 3 4 5r0

r cut

-dφ/

dr (

eV/A

)ο

r (A)o

Potential shift

F = _ _____dφ(r)dr

force not zero

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76

Crystal structure and potentialThe regular packing (ordering) of atoms into crystals is closely related to the potential detailsMany local minima for crystal structures exist, but materials tend to go to the structure that minimizes the energy; often this can be understood in terms of the energy per atomic bond and the equilibrium distance (at which a bond features the most potential energy)

Square lattice Hexagonal lattice

N=4 bonds N=6 bonds per atomφ

r

LJ 12:6potential

ε

2D example

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77

Example

time

http://polymer.bu.edu/java/java/LJ/index.html

Initial: cubic lattice Transforms into triangular lattice

Images courtesy of the Center for Polymer Studies at Boston University. Used with permission.

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78

All bonds are not the same – metals

Pair potentials: All bonds are equal!Reality: Have environmenteffects; it matter that there is a free surface!

+ differentbond EQ distance

stronger

Surface

Bulk

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79

Embedded-atom method (EAM): multi-body potential

)()(21

..1i

Njiji Fr

neigh

ρφφ += ∑=

Pair potential energy

Embedding energy

as a function of electron density

iρ Electron density at atom ibased on a pair potential:

∑=

=neighNj

iji r..1

)(πρ

new

Models other than EAM (alternatives):Glue model (Ercolessi, Tosatti, Parrinello)Finnis SinclairEquivalent crystal models (Smith and Banerjee)

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80

Effective pair interactions: EAM potential

Can describe differences between bulk and surface

∑∑∑ +=≠ i

ijii j

ij FrU )()(21

,

ρφ

Pair potential energy Embedding energy

as a function of electron density

Embedding term: depends on environment, “multi-body”

EAM=Embedded atom method

Image by MIT OpenCourseWare.

Bulk

Surface

Effe

ctiv

e pa

ir po

tent

ial (

eV )

1

0.5

0

-0.53 42

r (A)o

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81

Effective pair interactions: EAM

Can describe differences between bulk and surface

r

+ + + + + ++ + + + + ++ + + + + +

r

+ + + + + ++ + + + + ++ + + + + +

Daw, Foiles, Baskes, Mat. Science Reports, 1993

Image by MIT OpenCourseWare.

Bulk

Surface

Effe

ctiv

e pa

ir po

tent

ial (

eV )

1

0.5

0

-0.53 42

r (A)o

Change in NN distance

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82

Model for chemical interactions

Similarly: Potentials for chemically complex materials – assume that total energy is the sum of the energy of different types of chemical bonds

bondHvdWMetallicCovalentElec −++++= UUUUUUtotal

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83

Concept: energy landscape for chemically complex materials

Different energy contributions from different kinds of chemical bonds are summed up individually, independently

Implies that bond properties of covalent bonds are not affected by other bonds, e.g. vdW interactions, H-bonds

Force fields for organic substances are constructed based on this concept:

water, polymers, biopolymers, proteins …

bondHvdWMetallicCovalentElec −++++= UUUUUUtotal

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84

Summary: CHARMM-type potential

bondHvdWMetallicCovalentElec −++++= UUUUUUtotal

=0 for proteins

rotbendstretchCovalent UUUU ++=

20stretchstretch )(

21 rrk −=φ

20bendbend )(

21 θθφ −= k

:ElecU Coulomb potentialij

jiij r

qqr

1

)(ε

φ =

:vdWU LJ potential⎥⎥

⎢⎢

⎟⎟⎠

⎞⎜⎜⎝

⎛−⎟

⎟⎠

⎞⎜⎜⎝

⎛=

612

4)(ijij

ij rrr σσεφ

:bondH −U )(cos65)( DHA4

10

bondH

12

bondHbondH θφ

⎥⎥

⎢⎢

⎟⎟⎠

⎞⎜⎜⎝

⎛−⎟

⎟⎠

⎞⎜⎜⎝

⎛= −−

−ijij

ij rR

rRDr

))cos(1(21

rotrot ϑφ −= k

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85

Summary of energy expressions: CHARMM, DREIDING, etc.

….

vdW ONLYbetween atoms of differentmolecules

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Bond stretching

Angle Bending

Bond RotationvdW Interactions

Electrostatic interactions

δ−

δ−

δ+δ+

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sp3

sp2

Energy

s

C-C distance

Transition point

p3sp2

focus on C-C bond

20stretchstretch )(

21 rrk −=φ

r

r

stretchφ

BO=1

BO=2

Identify a strategy to model the change of an interatomic bond under a chemical reaction (based on harmonic potential) – from single to double bond

Harmonic oscillatorr0

φ

r~ k(r - r0)2

Effective interatomic potential

φEnergy

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Solution sketch/concept

20stretchstretch )(

21 rrk −=φ

Make potential parameters kstretch and r0 a function of the chemical state of the molecule (“modulate parameters”)

E.g. use bond order as parameter, which allows to smoothly interpolate between one type of bonding and another one dependent on geometry of molecule See lecture notes (lecture 7): Bond order

potential for silicon

(BO)(BO)

00

stretchstretch

rrkk

==

Fig. 2.21c in Buehler, Markus J. Atomistic Modelingof Materials Failure. Springer, 2008.© Springer. All rights reserved. This content is excluded from our Creative Commons license. For more information, see http://ocw.mit.edu/fairuse.

5

0

-5

-100.5 1 1.5 2 2.5 3 3.5

TripleDoubleSingle

S.C.F.C.C.

Pote

ntia

l ene

rgy

(eV)

Distance (A)o

Effective pair-interactions for various C-C (Carbon) bonds

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88

Summary: CHARMM force field

Widely used and accepted model for protein structures

Programs such as NAMD have implemented the CHARMM force field

E.g., problem set 3, nanoHUB stretchmol/NAMD module, study of a protein domain part of human vimentin intermediate filaments

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89

Overview: force fields discussed in IM/S

Pair potentials (LJ, Morse, harmonic potentials, biharmonic potentials), used for single crystal elasticity (pset #1) and fracture model (pset #2)Pair potential

EAM (=Embedded Atom Method), used for nanowire (pset #1), suitable for metalsMulti-body potential

ReaxFF (reactive force field), used for CMDF/fracture model of silicon (pset #2) – very good to model breaking of bonds (chemistry)Multi-body potential (bond order potential)

Tersoff (nonreactive force field), used for CMDF/fracture model of silicon (note: Tersoff only suitable for elasticity of silicon, pset #2)Multi-body potential (bond order potential)

CHARMM force field (organic force field), used for protein simulations (pset #3)Multi-body potential (angles, for instance)

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90

How to choose a potential

ReaxFF

EAM, LJ

CHARMM-type (organic)

ReaxFF

CHARMM

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91

f

x

SMD mimics AFM single molecule experiments

xk

vAtomic force microscope

k xv

pset #3

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92

)( xtvkf −⋅=

Virtual atommoves w/ velocity

Steered molecular dynamics used to apply forces to protein structures

)( xtvkf −⋅=

SMD deformation speed vector

time

Distance between end point of molecule and virtual atom

Steered molecular dynamics (SMD)

kvx

fv

end point of molecule

xtv −⋅SMD spring constant

pset #3

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93

Comparison – CHARMM vs. ReaxFF

M. Buehler, JoMMS, 2007

Covalent bondsnever break in CHARMM

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94

Goals: Select applications of MD to address questions in materials science (ductile versus brittle materials behavior); observe what can be done with MD

1.1 Atomistic and molecular simulation algorithms

1.2 Property calculation1.3 Potential/force field models

1.4 Applications

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95

Application – mechanics of materialstensile test of a wire

IImage by MIT OpenCourseWare. mage by MIT OpenCourseWare.

Brittle Ductile

Strain

Stre

ss

BrittleDuctile

Necking

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96

Brittle versus ductile materials behavior

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97

Ductile versus brittle materials

Difficultto deform,breaks easily

BRITTLE DUCTILE

Glass Polymers Ice...

Shear load

Copper, Gold

Easy to deformhard to break

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98

“Macro, global”

“Micro (nano), local”

)(rσ r

Deformation of materials:Nothing is perfect, and flaws or cracks matter

Griffith, Irwine and others: Failure initiates at defects, such as cracks, or grain boundaries with reduced traction, nano-voids, other imperfections

Failure of materials initiates at cracks

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99

Crack extension: brittle response

Large stresses lead to rupture of chemicalbonds between atoms

Thus, crack extends

)(rσ

r

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100

Basic fracture process: dissipation of elastic energy

δa

Undeformed Stretching=store elastic energy Release elastic energydissipated into breaking chemical bonds

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101

Limiting speeds of cracks: linear elastic continuum theory

• Cracks can not exceed the limiting speed given by the corresponding wave speeds unless material behavior is nonlinear• Cracks that exceed limiting speed would produce energy (physically impossible - linear elastic continuum theory)

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Linear Nonlinear

Mode I

Mode II

Mode III

Limiting speed v

Limiting speed v

Cr Cs

Cs

Cl

Cl

Subsonic Supersonic

SupersonicIntersonicSub-RayleighSuper-

Rayleigh

Mother-daughter mechanism

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102

Summary: mixed Hamiltonian approach

( )TersoffReaxFFReaxFFReaxFFTersoffReaxFF )1()( FwFxwF −+=−

xxwxw ∀=+ 1)()( TersoffReaxFF

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WiW2

R-Rbuf R+Rtrans R+Rtrans +RbufR

W1

x

ReaxFF

ReaxFF ghost atoms

Transitionlayer Tersoff

Tersoff ghost atoms

100%

0%

Transition region

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103

Atomistic fracture mechanism

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104

Lattice shearing: ductile response

Instead of crack extension, induce shearing of atomic latticeDue to large shear stresses at crack tipLecture 9

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τ

τ

τ

τ

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105

Concept of dislocations

Localization of shear rather than homogeneous shear that requires cooperative motion

“size of single dislocation” = b, Burgers vector

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ττ

τ

additional half plane

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106

Final sessile structure – “brittle”

Fig. 1 c from Buehler, M., et al. "The Dynamical Complexity of Work-Hardening: A Large-Scale Molecular Dynamics Simulation." Acta Mech Sinica 21 (2005): 103-11.© Springer-Verlag. All rights reserved. This content is excluded from our Creative Commons license. For more information, see http://ocw.mit.edu/fairuse.

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107

Brittle versus ductile materials behavior

Copper, nickel (ductile) and glass, silicon (brittle)

For a material to be ductile, require dislocations, that is, shearing of lattices

Not possible here (disordered)

© source unknown. All rights reserved. This content is excluded from our Creative Commons license. For more information, see http://ocw.mit.edu/fairuse.

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108

Crack speed analysis

slope=equal tocrack speed

crack has stopped

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109

MD simulation: chemistry as bridge between disciplines

MD

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110

Application – protein folding

ACGT

Four letter code “DNA”

Combinationof 3 DNA letters equals a amino acid

E.g.: Proline –

CCT, CCC, CCA, CCG

Sequence of amino acids“polypeptide” (1D structure)

Transcription/translation

Folding (3D structure)

.. - Proline - Serine –Proline - Alanine - ..

Goal of protein folding simulations:Predict folded 3D structure based on polypeptide sequence

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111

Movie: protein folding with CHARMM

de novo Folding of a Transmembrane fd Coat Proteinhttp://www.charmm-gui.org/?doc=gallery&id=23

Quality of predicted structures quite good

Confirmed by comparison of the MSD deviations of a room temperature ensemble of conformations from the replica-exchange simulations and experimental structures from both solid-state NMR in lipid bilayers and solution-phase NMR on the protein in micelles)

Polypeptide chain

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112

2. Important terminology and concepts

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113

Important terminology and concepts

Force field Potential / interatomic potentialReactive force field / nonreactive force fieldChemical bonding (covalent, metallic, ionic, H-bonds, vdW..)Time stepContinuum vs. atomistic viewpointsMonte Carlo vs. Molecular DynamicsProperty calculation (temperature, pressure, diffusivity [transport properties]-MSD, structural analysis-RDF)Choice of potential for different materials (pair potentials, chemical complexity)MD algorithm: how to calculate forces, energies MC algorithm (Metropolis-Hastings)Applications: brittle vs. ductile, crack speeds, data analysis

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3. Continuum vs. atomistic approaches

114

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Relevant scales in materialsλ(atom) << ξ (crystal) << d (grain) << dx,dy,dz << H ,W , D

Atomistic viewpoint:•Explicitly consider discrete atomistic structure•Solve for atomic trajectories and infer from these about material properties & behavior•Features internal length scales (atomic distance)“Many-particle system with statistical properties” 116

Bottom-up

Courtesy of Elsevier, Inc., http://www.sciencedirect.com. Used with permission.

Fig. 8.7 in: Buehler, M. Atomistic Modeling of Materials Failure. Springer, 2008. ©Springer. All rights reserved. This content is excluded from our Creative Commons license. For more information, see http://ocw.mit.edu/fairuse. Top-down

Image from Wikimedia Commons, http://commons.wikimedia.org.

Image by MIT OpenCourseWare.

z

x

yσyy

σxy

σyx

σyz

σxz σxx

ny

nx

Ax

Ay

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Relevant scales in materials

Continuum viewpoint:•Treat material as matter with no internal structure •Develop mathematical model (governing equation) based on representative volume element (RVE, contains “enough” material such that internal structure can be neglected•Features no characteristic length scales, provided RVE is large enough 117

“PDE with parameters”

Top-down

RVEλ(atom) << ξ (crystal) d (grain) dx,dy,dz << H ,W , D<<<<

Courtesy of Elsevier, Inc., http://www.sciencedirect.com. Used with permission.

Fig. 8.7 in: Buehler, M. Atomistic Modeling of Materials Failure. Springer, 2008. ©Springer. All rights reserved. This content is excluded from our Creative Commons license. For more information, see http://ocw.mit.edu/fairuse.

Image by MIT OpenCourseWare.

z

x

yσyy

σxy

σyx

σyz

σxz σxx

ny

nx

Ax

Ay

Image from Wikimedia Commons, http://commons.wikimedia.org.

Separation of scales

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Example – diffusion

118

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119Need diffusion coefficient to solve for distribution!

BC: c (x = 0) = c0IC: c (x > 0, t = 0) = 0

Example solution – 2nd Fick’s law

x0c

0c

Concentration

x

time t

),( txc

2

2

dxcdD

tc

=∂∂

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2. Finite difference method

How to solve a PDE

120

2

2

dxcdD

tc

=∂∂

+ BCs, ICs

dxdcDJ −=Note: Flux

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Finite difference method

121

Idea: Instead of solving for continuous field

solve for

2

2

dxcdD

tc

=∂∂

),( xtc

),( ji xtc

==

ji space

time

j

i

t

jiji cxtc ,),( =

xNi ..1=

discrete grid(space and time)

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Finite difference method

122

Idea: Instead of solving for continuous field

solve for

Approach: Express continuous derivatives as discrete differentials

2

2

dxcdD

tc

=∂∂

),( xtc

),( ji xtc

tc

tc

ΔΔ

≈∂∂

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Finite difference method

123

Idea: Instead of solving for continuous field

solve for

Approach: Express continuous derivatives as discrete differentials

2

2

dxcdD

tc

=∂∂

),( xtc

),( ji xtc

tc

tc

ΔΔ

≈∂∂

xc

xc

ΔΔ

≈∂∂

xtc

xc

Δ

⎟⎠⎞

⎜⎝⎛

ΔΔ

Δ≈

∂∂

2

2

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Forward, backward and central difference

124

Forward difference

Backward difference

Central difference

x

c

ji txxc

,∂∂

jic ,1+

jic ,

1+ixix1−ix

jic ,1−

x

c

jic ,1+

jic ,

1+ixix1−ix

jic ,1−

backward

forward

central

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How to calculate derivatives with discrete differentials

125

tc

tc

ΔΔ

≈∂∂

tδ discrete time step

t

c

tδcΔ

tcc

ttcc

tc jiji

jj

jiji

tx jiδ

,1,

1

,1,

,

−=

−−

≈∂∂ +

+

+

ji txtc

,∂∂

1, +jic

jic ,

1+jtjt(1)

2

2

dxcdD

tc

=∂∂

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How to calculate derivatives with discrete differentials

126

xc

xc

ΔΔ

≈∂∂

xcc

xxcc

xc jiji

ii

jiji

tx jiδ

,,1

1

,,1

,21

−=

−−

≈∂∂ +

+

+

+

xcc

xxcc

xc jiji

ii

jiji

tx jiδ

,1,

1

,1,

,21

− −=

−−

≈∂∂

xtc

xc

Δ

⎟⎠⎞

⎜⎝⎛

ΔΔ

Δ≈

∂∂

2

22

2

dxcdD

tc

=∂∂

x

jic ,1+

1+ixix1−ix

jic ,1−

central

c

jic ,

jitxx

c

,21+

∂∂

jitxx

c

,21+

∂∂

21+ix

21−ix

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How to calculate derivatives with discrete differentials

127

xc

xc

ΔΔ

≈∂∂

xxc

xc

Δ

⎟⎠⎞

⎜⎝⎛

∂∂

Δ≈

∂∂

2

2

(2)

( )2

,1,,12

,1,,,1,,

,2

2 2

21

21

21

21

xccc

xcccc

xx

xc

xc

xc jijijijijijiji

ii

txtx

tx

jiji

jiδδ

−+−+

−+

− −−=

−−−=

∂∂−

∂∂

≈∂∂ +

xcc

xxcc

xc jiji

ii

jiji

tx jiδ

,,1

1

,,1

,21

−=

−−

≈∂∂ +

+

+

+

xcc

xxcc

xc jiji

ii

jiji

tx jiδ

,1,

1

,1,

,21

− −=

−−

≈∂∂

2

2

dxcdD

tc

=∂∂

ji txxc

,2

2

∂∂

x

jic ,21' +

21+ixix

21−ix

jic ,21' −

central

ji txxc

,∂∂

Apply central differencemethod again…

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Complete finite difference scheme

128

2

2

dxcdD

tc

=∂∂

2,1,,1

,2

2 2x

cccxc jijiji

tx jiδ

−+ −−≈

∂∂

tcc

tc jiji

tx jiδ

,1,

,

−≈

∂∂ +

⎟⎟⎠

⎞⎜⎜⎝

⎛ −−=

− −++2

,1,,1,1, 2x

cccD

tcc jijijijiji

δδ

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Complete finite difference scheme

129

2

2

dxcdD

tc

=∂∂

2,1,,1

,2

2 2x

cccxc jijiji

tx jiδ

−+ −−≈

∂∂

tcc

tc jiji

tx jiδ

,1,

,

−≈

∂∂ +

⎟⎟⎠

⎞⎜⎜⎝

⎛ −−=

− −++2

,1,,1,1, 2x

cccD

tcc jijijijiji

δδ

( )jijijijiji cccDxtcc ,1,,12,1, 2 −++ −−+=

δδ

Want

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Complete finite difference scheme

130

2

2

dxcdD

tc

=∂∂

Concentration at iat new time

==

ji space

time

Concentration at iat old time

Concentration at i+1at old time

Concentration at iat old time

Concentration at i-1at old time

“explicit” numerical scheme(new concentration directly from concentration at earlier time)

( )jijijijiji cccDxtcc ,1,,12,1, 2 −++ −−+=

δδ

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Summary

Developed finite difference approach to solve for diffusion equation

Use parameter (D), e.g. calculated from molecular dynamics, and solve “complex” problem at continuum scale

Do not consider “atoms” or “particles”

131

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132

Quantummechanics

SummaryMulti-scale approach:Feed parameters from atomistic simulations to continuum models

MD

Continuummodel

Length scale

Time scale“Empirical”or experimentalparameterfeeding

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