intensive class

8/3/2019 Intensive Class

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COMSOL MultiphysicsIntensive Training Course

Walter Frei, Ph.D.COMSOL, Inc.

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Syllabus Day 1

Introduction and remarks

Hands on: Microresistor beam

Theoretical Notes: Non-linear single physics problems

Applied Notes: Using COMSOL expressions

LUNCH

Hands on: Large deformation micromirror

Applied Notes: CAD Modeling & Meshing

Day 2 Hands on: Laminar Static Mixer

Theoretical Notes: Solving

Theoretical Notes: Multiphysics problems

LUNCH

Hands on: Fluid-Structure Interaction

Theoretical Notes: Transient problems

Q & A

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Hands on: Joule heating induced thermal

expansion of a microresistorProblem:Voltage applied to this beamcreates resistive heating

Heating the structurecauses thermal expansion

Material properties are

temperature dependent

Want to find the deformationand stresses

0.2V

Ground

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The DC Problem

0)( V

Governing Equation for the

Conductive Media DCApplication Mode

Grounded

V=0.2 V

Conductivity

A steady state voltage difference is applied to the titanium beam in air

All other surfaces

are insulated

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The Heat Transfer Problem

2VQ

Governing Equation for the

Heat Transfer Application Mode

QTk

Natural Convection

h=5 W/m2K

Text=300 K

T=300 K

T=300 K

Thermal

Conductivity

Ohmic heat generation

calculated by Conductive Media DC

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)(T

The Coupled DC+Heat Problem

QTk

0)( V

The material properties of

titanium are temperature

dependent

)(Tkk

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The nonlinear material functions

7

k

T T

(1-T/1000[K])*7.407e5[S/m] exp(-T/1000[K])*7.5[W/m/K]

Consider these one at a time


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The Structural Mechanics Problem

Feet are fixed

to substrate

The material experiences avolumetric thermal expansion

due to the temperature given by:),(x

T,VT

0TT

Assume small strain linear deformation theory

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The finite element method

V+ T

Before you begin:1) Physics2) CAD3) Boundary Conditions4) Material Properties

bAx

This course focuses on:1) Efficient model set-up2) Meshing3) Solving

4) Finding mistakes

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Saving COMSOL Files

When saving a file, the *.mph format (default) is a binary format

The *.m file is a MATLAB script format that logs all commands

If you go to File... Reset... you will delete the mesh, solution, and log This is an efficient way of emailing a file among colleagues

File... Generate Report... is a good way of getting a simple summaryof all data inside the file

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Lecture 1: Introduction to linear and

non-linear single physics steady-state finite element problems

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A very simple finite element problem...

p= 2 Nk2= 4 N/m k3= 4 N/m

u1 u2u3

We can write a force balance on each node:

Node 1

c

k1(u3-u1)

Node 2

k3(u3-u2)-k2(u2-u1)

k1= 6 N/m

k2(u2-u1)

Constraint Force

Node 3

p

-k1(u3-u1)

-k3(u3-u2)

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Write the force balance as three equations:

However, we now have three equations in four unknowns, c, u1, u2, u3

k1(u3-u1) + k2(u2-u1) = c

k2(u2-u1) -k3(u3-u2) = 0

k1(u3-u1) +k3(u3-u2) =p

(k2)u

2+ (k

1)u

3= c

(k2+k3)u2 + (-k3)u3 = 0

(-k3)u2 + (k1+k3)u3 =p

Algebra

Constraint Equations

System Equations

k1(u

3- 0 ) + k

2(u

2- 0 ) = c

k2(u2 - 0) -k3(u3 - u2) = 0

k1(u3 - 0) + k3(u3 - u2) =p

Three equations, three

unknowns: c,u2, u3

But we also have a constraint, u1=0

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Can now solve the problem

(k2)u2 + (k1)u3 = c

(k2+k3)u2 + (-k3)u3 = 0

(-k3

)u2

+ (k1

+k3

)u3

=p

cu

ukk

3

2

12

pu

u

kkk

kkk 0

3

2

313

332

Solve the systemequations first

Ku=b

u=K-1b

k1= 6 N/m

k2=k

3= 4 N/m

p=2 N

250.0

125.0

3

2

u

u

Can also solve the

constraint equations,if desired

2250.0

125.064

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So far, we have only solved a linear problem

A linear problem must have all three of these properties:

1) Applying zero loads results in a zero solution2) Doubling the magnitude of the load doubles the magnitude of the solution3) Changing the sign of the load changes only the sign of the solution

load

solution

load

solution

Linear Non-Linear 15


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What makes a problem non-linear?

Material properties that depend upon the solution, k(u)

Loads that depend upon the solution, b(u)

u

k,b(u)

Linear

u

k,b(u)

Weakly Non-linear

u

k,b(u)

Strongly Non-linear

u

k,b(u)

OutrageouslyNon-linear

There are no clear distinctionsbetween these three cases

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Not all non-linear problems have a (unique) solution

load

solution

No solution

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load

solution

No unique solution


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Two very simple finite element examples

k= 4 N/m

u

p = 2 N

Force balance on node:

f(u) = p - ku = 0

f(u) = 2- 4 u = 0

f(u)

usolution= 0.5

u

k= exp(u) N/m

u

p = 2 N


f(u) = p - k u = 0

f(u) = 2- exp(u) u = 0

u

f(u)

usolution 0.853

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First, solve the linear problem

f(u) = 2- 4 u = 0

f(u)

usolution=0.5

Algorithm:

1) Start at a point, e.g. u0 = 0

2) Find the slope, f (u0)

3) Solve the problem:

0

00

' uf

ufuu

solution

5.0

4

20

solutionu

(1)(2)

(3)

i

iii

uf

ufuu

'1 This is a single Newton-Raphson iteration:

For a linear problem, the starting point does not matter 19


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Solving a FE system matrix is equivalent to takinga single Newton-Raphson iteration

f(u) = 2- 4 u = 0

00

0 ' uf

uf

uusolution

5.04

20

solutionu

pu

u

kkk

kkk 0

3

2

313

332

0

00

3

2

313

332

u

u

kkk

kkk

puf

f(u) = b- Ku, u0 = 0

K

b

K

Kub0

uf

ufuu

0

0

0

0 'solution

bKu1solution

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The non-linear problem is solved the same way

Algorithm:

1) Start at a point, e.g. u0 = 0

2) Take repeated Newton-Raphson iterations

3) Terminate when:

f(u) = 2- exp(u) u

u

f(u)

toleranceuuff iiii /&/ 11

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Similarly, if we have a non-linear system matrix

f(u) = b(u)- K(u)u

iiiiii

iii ubuuKuSu

uf

ufuu

1

1'

f(u)= S(u)

Iterate until: toleranceff iiii uu /&/ 11

p= 2k2 = exp(u2-u1) k3 = 4

u1 u2u3

k1 = 6 + (u3-u1)2

S(u) is the known as the JACOBIAN

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The possibility of convergence of a non-linearproblem is dependent upon the starting point, u

0

Non-linear problems have a radius of convergence

f(u) = 2- exp(u) u

u

f(u)

Non-convergentif starting on thisside of the line

u0= -1

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Finding a good starting point, u0,canimprove convergence

f(u) = 2- exp(u) u

u

f(u)

slowconvergence

fast

convergence

It is not strictly possibleto define slow or fast

convergence.

Finding a good startingpoint is a matter ofexperience, and luck.

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A damping factor, , is used to improve convergence

u

f(u)

i

iii

uf

ufuu

'1

choose (< 1) & ( |fi| )

recomputefi+1

(ui+1

)

end

|fi|

undamped |fi+1|

intermediate |fi+1|

final |fi+1|

f(u) = 2- exp(u) u

Problem: This requires many more evaluations 25


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A better solution is to ramp up the load

f(u)

f(u) = 2- exp(u) u k= exp(u) N/m

u

p = 2 N

u

Recall that:LoadLoad

f(u) = 1- exp(u) u = 0

Almost all problems have a

zero solution at zero loadRamping up the load is aphysically reasonable approach

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Ramping up the non-linearity can also work

f(u)

f(u) = 2- exp(u) u = 2 - ku

u

Non-linearity

f(u) = 2- 1u

f(u) = 2- {(1-0.5)+(0.5exp(u)}u

f(u) = 2- exp(u) u

Identify non-linearity

- kNL = exp(u)

Linearize around chosen u0

- kLIN = exp(u0=0) = 1

Use an intermediate value

- k= (1-)kLIN+kNL- starts at 0

Use intermediate solution as new u0

Ramp from 0 to 1

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Review

At this point, you should know: What is meant by a finite element problem

The difference between a linear and non-linear problem

How linear and non-linear finite element problems are solved

Why a non-linear problem may not converge

Strategies for getting a non-linear problem to converge

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Hands-on: heat flow through a square

T=0K

q = 100 W/m2

k, Thermal conductivity

1x1m square

Build this 2D model in BASE Heat Transfer,Save this model, we will build upon this

Mesh

Solution

Tinitial=0K

Insulated

Insulated

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Study the following cases:

T

k

T

k= 0.1+exp(-(T/25[K])^2)[W/(m*K)]

= 1[W/(m*K)]

T

k= (1+T/200[K])[W/(m*K)]

T

=(0.1+10*(T>25[K]))[W/(m*K)]k


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Questions:

How many iterations did each case take? Was damping used?

What does the temperature and thermal conductivity solution look like?

Did they all converge? Why not?

Try solving case #2 with an initial temperature of 50[K], how does thisaffect the solution? What about the other cases? Set back to T=0 afterwards

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Ramp up the heat load and monitor peak temperature

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T

k=(1-T/200[K])[W/(m*K)]

Qin

Tmax

?

Why does this happen?


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Non-linear solver settings

Stop conditions forNewton iterations

Toggling off damping is

rarely, if ever, useful

Can choose differentdefaults for the damping

Augmented Lagrangiansolver settings, used instructural contactproblems only

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Advanced solver settings

All other settings arethere for historicalreasons and are relatedto memory management

Wave-type problems usecomplex numbers, andrequire this be on

Stores time-dependentsolution to temp fileduring the solve, andgets loaded into RAMonce the solve is done

Wave-type problems withperiodic boundary

conditions require this tobe checked on

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Using the parametric solver with one parameter

Consider again, the outrageously non-linear problem, k=0.1+10*(T>25)Try instead: k=0.1+10*flc2hs(T-25,STEP)

Try the parametric solver with STEP = 8, 4, 2, 1, 0.5, 0.25

Try with STEP = 8, 0.25

When using the parametric solver with one parameter, COMSOL will take additionalintermediate steps in an attempt to ramp towards the final solution. This is similar to,but more computationally demanding than, damping. The intermediate steps do notget stored, and they can require more time.

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Using the parametric solver with severalparameters

Now try: k = 0.1+DK*flc2hs(T-25,STEP)

Try: DK, STEP = 0 8 5 4 7.5 2 10 0.5 10 0.25

STEP

DK

(0,8)

(5,4)

(7.5,2)

(10,0.5)

(10,0.25)

When using several

parameters, the previoussolution is still used asthe initial condition, butthe software does nottake intermediate steps. T

k

(0,8)

(5,4)

(7.5,2)

(10,0.5)

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Other methods for getting convergence

Observe the solution during the non-linear iterations

Toggle ON Plot while solving on the Solver Parameters form

Toggle ON Highly non-linear problem This also ramps up Dirichlet boundary conditions

Run examplefrom booklet

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Achieving convergence for non-linear problems

Assuming that the problem iswell-posed, try solving it

1) Check the initial condition

2) Use parametric solver to ramp load1

3) Use parametric solver to ramp non-linearity

4) Refine the mesh

Does the problem have a steady-state solution?Are there additional effects? Check all assumptions.

Perform a mesh

refinement study.Try a finer meshand check that thesolution is similar.

DONE

Not converging?

1) Very similar to toggling on Highly non-linear problem

Not converging?

Not converging?

Not converging?

Not converging?

Converging?

Not converging?

Converging?

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Linear problems are much easier to debugThe error is usually some variation of these mistakes:

What are the stresses in the eggshell for these two cases?


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Linear problems are much easier to debugThe error is usually some variation of these mistakes:

If balanced forces are acting upon

an object, where is the object?


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COMSOL Expressions

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Application notes: Problem constants

These constants can be entered into anything underneath the Physics... menu

Units are tracked and converted to the default unit system42


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COMSOL expressions

EXPRESSIONS

Expressions are defined via the Options... Expressions... submenu

These expressions can be entered into anything underneath the Physics... menu

Units are tracked and converted to the default unit system

The same everywhere

The same within a geometry tabThe same within a subdomainetc...

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Expressions are evaluated both before and

during the solution

Square brackets are used to specify units, they arenot strictly necessary, but are encouraged

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How to enter more complicated expressions

1) Options... Functions... Table...

Pick appropriate extrapolationand interpolation method

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How to enter more complicated expressions

2) Options... Functions... Piecewise Analytic Functions...

Use General Function type

Pick appropriate extrapolationmethod and smoothing function

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Entering x-y data from a file

(1)

(2)

(3)

(4)

% Grid-2.00 0.00 2.00-1.00 2.00% Data0.25 0.00 -0.25-0.50 0.00 0.50

The data can be on a

square grid, or read infrom another analysispackage on anunstructured mesh.

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Functions with a discontinuity

In COMSOL such functions can bewritten as Boolean expressions:

(t>0)(t>=0)(t>0)&&(t1)||(t


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Material Properties

Can also use:

www.matweb.com

www.efunda.com

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Reserved COMSOL variable names

t

x, y, z, r, X, Y, Z, R

s, s1, s2

n, nx, ny, nz, nr

tx, ty, tz, tr

t1x, t1y, t1z, t2x, t2y, t2z

un, unx, uny, unz

dn, dnx, dny, dnz

eps, i, j, pi

lambda

h, dom, meshtype, meshelement, dvol,qual, reldetjac, reldetjacmin

Time:

Position:

Edge/Surface Parameter:

Edge/Surface Normal:

Edge Tangent:

Surface Tangents:

Edge/Surface Up Normal:

Edge/Surface Down Normal:

Numerical Constants:

Eigenvalues:

Mesh Information:

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COMSOL derivative variables

Given solution variables: T, V, etc...The spatial derivatives are given by: Tx, Ty, Tz, Vx, Vy, Vz, etc...

Tx

Tx T

y

Ty T

z

Tz

The time derivatives (if solving atransient problem) are: Tt, Vt, etc... Tt

Tt

These can be mixed:

yx

T

t

2

2

2

Txytt

The edge and surface tangentderivatives are: TTx, TTy, TTz, etc...

xT TnnITTx

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Application Mode Variables

Each application model has its own automatically defined variablesthat have an underscore and a suffixThey are all listed underneath Phyics... Equation System...

For example:

fluxx_ht =-kxx_ht*Tx-kxy_ht*Ty-kxz_ht*Tz

z

Tk

y

Tk

x

Tkq

z

Tk

y

Tk

x

Tkq

z

Tk

y

Tk

x

Tkq

zzzyzxz

yzyyyxy

xzxyxxx

T kq52


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Hands on: pre-stressed micromirror

Aspect ratio is >10, so sweptmeshing is reasonable

Large deformation analysis

Intermediate solutions are required tohelp the model converge to solution

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CAD and Meshing

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Geometry modeling choices

COMSOL geometry modeling tools

Suitable for simple geometries

Foreign CAD tools Import STEP, ACIS, or PARASOLID format, IGES and STL are poor choices

Specialized CAD packages have greater flexibility in geometry definition New ECAD Import, Imports 2D layout files, builds 3D geometry

CAD import is one-way

Solidworks and Inventor live-link (after the geometry is modifiedduring modeling, with this, you can connect the two) Bi-directional interface,

build CAD, build FEM, update CAD, automatic FEM update

Can drive dimensions from within COMSOL

Limited to CAD capabilities in those packages

Scan data (MRI, geophysical) Simpleware(native), MIMICS(native), Avizo, Amira... 55


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COMSOL Draw Mode

Draw 2 rectanglesGet 3 domains

Continuity is automatically assumed at interior boundaries

Overlapping geometry is OK, once you switch tosubdomain mode you will get what you see on the screen(assign subdomain into defined groups, group

definition will remain even the objectives are removed)56


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Deleting interior boundaries and splitting objects

Draw a square & circle Union into one object Delete interior boundaries

Split into three objects

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Using COMSOL Groups

APPROACH #1

1) Select each object2) Apply properties

3) Switch to Groups tab4) Verify groups visually

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Using COMSOL Groups

APPROACH #2

1) Switch to Groups tab2) Create a Group name3) Give properties for

group

4) Switch to subdomain tab5) Select each object6) Assign it to a group

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MESHING: The free mesh parameters

Usually want to adjust each one of these fields,both globally and locally

The Subdomain,Boundary, Edge, andPoint tabs locallyoverwrite the Global

settings

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Maximum element size

This sets the maximum length of the edge of any element If you do not pick a value, COMSOL uses L/10, where L is the

maximum dimension of the model

Do not use maximum element size scaling factor

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Element growth rate

This controls the maximum element size between adjacent elements

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Mesh curvature factor

radius=0.5

Local element size will be:

(radius of curvature)x(mesh curvature factor)

Maximum element size is 0.05

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Mesh curvature cutoff radius=0.25

If: redge < (rcutoff) x (Lmax)

Then: r = (rcutoff) x (Lmax)

As a result: The element size of these edges is:

(Mesh Curvature Cutoff) x (Mesh Curvature Factor)

Maximum element size is 0.05

Lmax

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Resolution of narrow regions

Resolution=1 Resolution=5

Puts approximately the specified numberof elements into the narrow regions 65


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2D Triangular, Quadrilateral, Mapped, andBoundary Layer Meshes, and when to use them

Triangle Quad Mapped Boundary Layer

66Transient thermal heating, high gradients

in the solution normal to boundary

Which will be the best mesh to resolve the gradient?

Gradient need to be more resolved close to the inner

curve and less/no at outer curve, also gradient is

only perpendicular to curve not along the curve.So triangle and quad is too much considering the direction

along the curve. So Map is the optimum.

Boundary is specially good for modeling skin problem

because it give special meshes on the boundary


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3D Tetrahedral, Extruded, Revolved, Swept, andBoundary Layer Meshes

Tetrahedral Extruded Revolved

Swept Boundary Layer67

Tetrahedral is most useful. Swept: you work out the condition for surface, or

some subdomains, than swept for the rest (that is not so important,

or not of that interest)


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The wrong way to work with 3D CAD files:

CADDesigners

AnalysisStaff

CAD File

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3D CAD pitfalls

Working with overly detailed geometry, extraneous features

Sliver faces (drafts), short edges

Fillets, fillets, fillets

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Downloaded from thomasnet.com


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Making the reverse of the part is usually

easier to do in the original CAD program

COMSOL can cap rectangular andcircular planar faces on 3D objects, othercapping operations are more difficult


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Include a fillet to remove a singularity

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Consider DC current flow. Cuta notch out of a square block,

apply insulating boundaryconditions, because it is

insulating, so current can onlygo in the direction shown but

not perpendicular to thesurface.

Here, the insulating boundary conditionmeans that current can only flowtangentially to the boundary, as shown

by the arrows.

This is enforced no matter how muchyou zoom in on the notch.

This geometry implies that the current

takes a sharp bend over zero radius.

This is independent of the mesh!


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Although the fillet/ sharp angle introducesa singularity, it may not matter for the

quantity of interest

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With mesh refinement, a derived quantity, such as resistive heating,

may diverge at a sharp corner, but the actual solution variable, in thiscase voltage, will be the same, regardless of the mesh

V=0.5 V=0.5 V=0.5


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An example from structural mechanics

73

Some fillets should stay


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If a corner introduces gradients into the solution,then consider adding a realistic fillet

74

FLUID FLOW EXAMPLEAdding a small, physicallyrealistic, fillet at interior cornerscan be helpful for convergence


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Boundary conditions can introduce singularities

75

P

A cantilever beam fixed at one end

A fixed boundary

condition meansthe material isinfinitely stiff

Where will there be a singularity? How is it manifested?

Is this model convergent? If so, with respect to what?

Point loadStiffness:left infinite stiffness, right, finite


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Boundary conditions can introduce singularities

76

T = 50K

T

=

0K

T

=

100Kx

TkAq

q?


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CAD Pitfalls: Glancing contact

77

Introduce a slightintersection to avoid

the sharp interiorangle of the element


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CAD Pitfalls: Narrow regions

78

- Approximate this via a boundary condition- Consider using a reduced dimension model


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CAD Pitfalls: High aspect ratios

79

2D aspect ratios of up to 1000:1 are possible (but difficult)

In 3D, due to memory are accuracy constraints, try to stay below 100:1

Lmax

Lmin


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Demo: CAD meshing and de-featuring

Import: repair_demo_1.x_b fromC:\COMSOL\caddata directory

93,429 elements

using defaultNormal mesh

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Fillets and small surfaces are present

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Try repairing the part with a tolerance of 1e-3

45,517 elements

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Draw CAD repairrepair change from 1e-5 to 1e-3


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Try changing the resolution of narrow regions

By changing from 0.5 to 0.1, resulting in

50,797 elements, almost 50% less!

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Tricks for diagnosing CADIf you are unable to put a 3D mesh on the part, 99% of thetime, it is one of the problems we discussed in the previouscases, (small edges and small intersections, etc), but it

is hard to find. In that case, instead of doing a 3D mesh,just select the surface of the geometry and do a 2D meshto help find the problem.

Try interactivesurface meshing

This is a mesh onthe surface only,but it can visuallyguide you to theproblem areas

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Good CAD data hasjust enough detail

to capture the

physics, but notone detail more.85


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Meshing notes: The more elements, the better,

but this has some practical limits

P Displacement

0.990

0.995

1.000

0 1 2 3 4 5

Refinement Iteration

NormalizedM

ax.

Disp.

Make sure to study thesolution variable, andnot a derived variable

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Numerical Error

k= exp(u) N/m

u

p = 2 N


f(u) = p - k u = 0f(u) = 2- exp(u) u = 0

u

f(u)

usolution 0.853

This problem has an exactsolution, but we are solving itnumerically via Newton-Raphson iterations.

The error can be minimized bytaking many iterations, but a

computer usually cannot findan exact answer for tworeasons:

1) Numerical approximations,e.g. 1 2-1+2-2+2-3+2-4+...

2) Operations such as exp(u)are approximate

This is NUMERICAL error

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Discretization Error

A finite element basis function can onlyapproximately represent a true solution.

The error can be minimized by usingsmaller elements, or increasing elementorder.

This is MESH or DISCRETIZATIONerror

10 1st order elements

16 1st order elements

4 2nd order elements

uexact-umodel

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1.E-10

1.E-08

1.E-06

1.E-04

1.E-02

1.E+00

1 10 100 1000 10000 100000 1000000

Number of Elements per Wavelength

TotalError

Discretization error and numerical error

Mesh errordecreases withmore elements

Numerical errorincreases withmore elements

Total error will be about 10-6

1

st

order elements

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Square root of instrument precision


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Element shape functions discretize curved edges

Domain

Single 1st

ordertriangular element Single 2nd

ordertriangular element

Isoparametric elements use polynomialexpressions to approximate the domain shape

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Geometric discretization error

Four linear elementsdiscritizing a unit circle

Error: (-2)/

1.E-06

1.E-04

1.E-02

1.E+00

1 10

Elements per Chord

Error

1st order

2nd order3rd order

This mesh adequatelydescribes the geometry, butnot necessarily the solution!

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Total Error vs. memory requirements

n: # of elementsper side (5 in this

example)

n=1

n=1 n=2

n=20

Insulatedquartercylinderwith heatgeneration 92


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Memory requirements depend on mesh type

18 elements

49 nodes

~3 adjoining elements

9 elements

40 nodes

~4 adjoining elements

[Kelement]6x6[Kelement]8x8

Error is a related to how far apart the nodes are,and is usually not a function of element type. 93


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Tetrahedral vs. prismatic elements

2,500 elements

83K d.o.f.

17s solve time

150 MB for K matrix

23,000 elements

186k d.o.f.

22s solve time

150 MB for K matrix

Solvers are tuned to take advantage of tetrahedral elements

currentcarrying

wire

94


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When to use quad/hex/prism elements:

95

ALE moving mesh formulations High aspect ratio geometry Known small gradients in one direction Large deformation structural analysis

If also solving a contact problem, use linear elements

For almost all other cases, use triangular ortetrahedral elements

Use boundary layer meshing when thereare high gradients normal to a boundary


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When to use linear elements?

Almost never

If you already have a very fine mesh due to geometric complexity

Very large models where you can only get low accuracy

Transient problems

Higher than 2nd order is usually only of academic interest

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The stationary adaptive mesher

Triangular and tetrahedral elementsONLY are subdivided based on theL2 norm error estimate.

Toggle this on, solve, then toggle off.

Do not change the settings.

Discuss

Initial mesh

Final mesh

Go back to the square model, case 3

k = 0.1+exp(-(T/25[K])^2)[W/(m*K)]

Turn on adaptive mesh refinement

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Guidelines for meshing refinement

1. Start with a mesh that you believe will resolve the gradients on thesolution that you expect, use the previously mentioned techniquesto get an answer

2. If it is easy to manually put more elements in regions where you

observe steep gradients in the solution, then do so. Monitor theconvergence of the solution as you proceed

3. Use adaptive mesh refinement to automatically refine the mesh,but be prepared to devote time and RAM to the solution

4. If the solution itself does not change, to a tolerance that you

consider acceptable, then the solution is converged

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End of Day 1

Questions?

99


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Hands on: Laminar Static Mixer

1) Solve fluid flow problem

2) Re-mesh with a finer mesh

3) Map solution onto new mesh

4) Solve convection-diffusion problem

100


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Lecture 2: Different ways of

solving the linear system matrix

Ku=b101


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Lets take another look at the system equations

0Kubuf

0

00

3

2

313

332

u

u

kkk

kkk

p

Define a quadratic function: r(u) = bu-Kuu

u2

u3r(u)

SolutionThe solution, f(usolution) = 0, isthe point where r(u), is at aminimum

102


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Finding the minimum of a quadratic function

r(u) = 2u2 - 3u + 1

r(u)

u

Newtons method: 0

00

u

uuu

r

rsol

For a quadratic function, thisconverges in one iteration, for any u0

r(u) = 4u- 3

r(u) = 4

75.04

3040

0

00

ur

uruusol

u0 = 0

Via our choice of r(u), this reduces to thesame equation from the previous section:

r(u) = b - Ku

r(u) = -K

r(u) = bu-Kuu

K

Kubu

u

uuu

00

0

00

r

rsol

0u 0

bK

K

b0u

1sol

usol

103

Finding the minim m of the q adratic f nction


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Finding the minimum of the quadratic function,r(u), by the direct method means solving u=K-1b

This is known as Gaussian Elimination, or LU factorization

The numerical algorithms are beyond the scope of thiscourse, but they have the following important properties:

For 3D, requires O(n2

)-O(n3

) numeric operations, where n is the length of u Requires O(n2)-O(n3) memory

Robustness of the algorithm is only very weakly dependent upon K

The direct solvers in COMSOL are: PARDISO: fast, multi-core capable, BEST FIRST CHOICE

SPOOLES: slow, uses the least memory

UMFPACK: fast, robust, uses the most memory

PARDISO out-of-core, uses disk memory, VERY SLOW

104

Introduction to iterative methods for finding the


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Introduction to iterative methods for finding theminimum of a quadratic function, a naive approach

1) Start here

2) Search along coordinate axis

3) Find the minimum along that axis

4) Repeat until converged

This iterative method requires onlythat we can repeatedly evaluate r(u)as opposed to the direct methods,

which get to the minimum by in onestep by evaluating [r(u)]-1

105

Th i l l i K ff t th l ith


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The numerical values in K can affect the algorithm

u2

u3r(u)

Quadraticsurface

r(u) = bu-Kuu

Conditionnumber = 1

Conditionnumber = 2

Conditionnumber = 10

The higher the condition number the morenumerical error creeps into the solution 106


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Ill-conditioned matrices require more iterations

Numerical error becomes significant107

A better iterative method for finding the minimum:


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A better iterative method for finding the minimum:The Conjugate Gradient (CG) method

3) Find the minimum along that vector

The CG method requires that we canevaluate r(u),r(u) andr(u)

CG converges in at mostniterations

CG does NOT compute K-1

2) Initially, find the gradient vector

1) Start here

4) Find the conjugate gradient vector*5) Repeat 3-4 until converged

*)

See e.g. Wikipedia, or Scientific Computing by Michael Heath

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Condition number matters for iterative methods

Numerical error becomes significant

for ill-conditioned matrices

A matrix with a condition number of 1

will converge in one iteration, (rare!)

All iterative methods in COMSOL are some


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All iterative methods in COMSOL are somevariation upon the CG method

Conjugate Gradient,F/GMRES, BiCGStab, Geometric MultigridThe details of these algorithms are beyond the scope of this course

These methods all make use of PRECONDITIONERS

The system equation, Ku=f, is multiplied by a preconditioner matrix, M,to improve the condition number

Exercise: Show that

the best possiblepreconditioner is thematrix M=K-1

Ku = b MKu = Mb 110

Wh t d t k b t it ti l


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What you need to know about iterative solvers

They converge in at most n iterations (good) Solution time is O(n1-n2)(good)

Solution time does depend on condition number

Memory requirements are O(n1-n2) (very good)

They are less robust that the direct solvers (neutral) Convergence depends upon condition number

An ill-conditioned problem is often set up incorrectly

Different physics require different iterative methods (bad) This is an ongoing research topic

Often cannot solve two physics with the same solver

Improvements in methods are ongoing

We have tried to find the best combination of iterative solvers andpreconditioners for many physics, to find these settings, read the manualsor open a new model file, select a space dimension of 3D and the physicsyou want to solve

111


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Review

At this point, you should know: What a direct solver does

Advantages and disadvantages of a direct solver

What an iterative solver does What a preconditioner does

Advantages and disadvantages of an iterative solver

112


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The linear system solver GUI

113

Monitor the memory requirements of the direct


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y qsolvers and convergence of iterative solvers for

the unit square as you refine the mesh

In class exercise:

Build a 3D heat transfer model

Try out different direct and iterativesolvers and compare to the default

114

This curve shows the iterativesolver approaching thebottom of the residual

surface

Th S l M


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The Solve Menu

Solve Problem Uses the settings in the Solver Parameter and Solver Manager

Restart Try to avoid this, it is better to use the Solver Manager

Update Model Maps old solution onto a new mesh Evaluates any additional expressions that you have added

Get Initial Value Displays, as the solution, the specified initial condition

These are specified either via the Init tab on subdomain propertiesor via the Solver Manager form

Solver Parameters Allows you to select the solver type and linear system solver

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Using the parametric solvers

Solve > Solver Parameters > Parametric solver This scans over a set of expressions (frequency, load) but not time

All results are saved in the same model file (and can be compared)

Solve > Parametric Sweep Also scans over a set of expressions

Can wrap this around a transient simulation

Results are saved in different model files

Can take advantage of parallel computing

File > Solidworks/Inventor Connection > Geometric Parameter Sweep Scans over a set of geometric dimensions

Must define COMSOL expressions that drive CAD equations

Results are saved in different model files

This is the way to vary the geometry

Work the electricimpedancesensor model

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Computer architecture issues, 32- vs. 64-bit

A 32-bit architecture can store at most 232 numbers (~4GB RAM) In practice, due to inefficient memory managers, most OSs can only allocate

about 1.5GB per process, and the available memory becomes fragmented

COMSOL allocates memory in chunks, and will fail if it tries to exceed ~1.5GB

A 64-bit system can address virtually unlimited memory

In practice, most desktop OSs address up to 128 GB RAM Still need to have the RAM available for good performance

To solve a large model, a 64-bit computer and 64-bit OS is needed No significant performance difference between OSs (Win, Linux, OSX)

There is no reliable way to predict memory requirements and speed

Too many variables involved, and some can have big effects

COMSOL recommends: 64-bit computer, 4-16GB RAM to start, leave room to expand, good graphics card

117

S


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Solver selection for 1D and 2D problems

Begin with the default solver, UMFPACK or SPOOLES

Try out PARDISO, which may use less memory and solve faster

If running out of memory: Use PARDISO out-of-core

Upgrade your computer, 64-bit & more memory

Make sure that your problem is not overly complicated, typical 2D modelsshould solve easily on a 32-bit computer with 2GB of RAM

If you are seeing differences between the solvers, check the

problem very carefully, this usually indicates a mistake in the model If using PARDISO for fluid-flow type problems, turn off Check Tolerances in

the solver settings

118

S l l i f D bl


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Solver selection for 3D problems

Set up a linear problem first

Use the default solver settings for the physics you are solving

If you run out of memory, upgrade to 64-bit and more RAM

Monitor the memory requirements as you grow the problem size

Setup a non-linear problem only after you have successfully solveda linear problem

119


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Lecture 3: Multiphysics FE Problems

120

Consider two simple physics, on the same geometry


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p p y g y

q=100 W/m2

T=0K

All other surfacesinsulated

0

0100

0

2

T

KTTk

Tk

m

Wn

in the domain

right boundary

left boundary

other boundaries

fT=

KTu

T-

bT

Heat Transfer Conductive Media DC


V=1VV=0V

0

01

0

V

VVVV

V in the domain

right boundary

left boundary

other boundaries

fV= K

Vu

V- b

V

k= 1 W/mK = 1S/m

121

How to solve this linear problem


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This part of the matrix is zero because: fV/uT= 0

This part of the matrix is zero because: fT/uV= 0

How to solve this linear problem

fT(uT) = KTuT- bT

fV(uV) = KVuV- bV

V

T

V

T

V

T

b

b

u

u

K0

0Kuf

Combine the two linear system equations:

bSub

b

u

u

uf0

0ufuf

V

T

V

T

VV

TT

Solve this linear problem with asingle Newton-Raphson iteration: bKuSuu

0

1

0sol

122

Now, consider a coupled multiphysics problem


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p p y p

= (1-0.001T) S/m

The temperature solution now affects thematerial properties of the DC problem

fT= K

Tu

T- b

T f

V

= KV

(uT

)uV

- bV

q=100 W/m2

T=0K




V=1VV=0Vk= 1 W/mK

123

How to solve this coupled multiphysics problem


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fT(uT) = KTuT- bT

fV(uV) = KV(uT)uV- bVTake the two system equations:

buuKbb

u

u

ufuf

0uf

uf

V

T

V

T

VVTV

TT

This coupled problem is solved viaNewton-Raphson iterations:

iiii

iiiii

ufuufuu

buuKuSuu

1

1

1

1

Combine these into a single system equation:

This coupled multiphysics problem is solved in the exactsame way that a non-linear single physics problem is solved

Can no longer use the iterative solvers, since they are tuned

for single physics, and often can not handle the coupled terms

124

Consider more couplings between the physics


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q=100 W/m2

T=0K



V=1VV=0VQ(V)

k= 1 W/mK (T)

fT= KTuT- bT(uv) fV= KV(uT)uV- bV

Volumetric resistiveheating due to current

ubuuKb

ub

u

u

ufuf

ufuf

uf

V

VT

V

T

VVTV

VTTT

iiii

iiiiii

ufuufuu

ubuuKuSuu

1

1

1

1

Same equation

as before

125

Consider all possible couplings between the physics


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(T,V)

q(T,V)

V=V(T,V)

fT= KT(uV,uT)uT- bT(uV,uT)

k(T,V)

),( VTQvol

ubuuK

uub

uub

u

u

ufuf

ufufuf

TVV

TVT

V

T

VVTV

VTTT

,

,

iiii

iiiiii

ufuufuu

ubuuKuSuu

1

1

1

1

fV= KV(uV,uT)uV- bV(uV,uT)

V=V(T,V)T=T(T,V)

General Newton-Raphson multi-physics iteration:

126

Different two-way couplings have different


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computational requirements

ubuuKuub

uub

u

u

ufuf

ufufuf

TVV

TVT

V

T

VVTV

VTTT

,

,

If one or both of these are zero, then the memory requirements are less

If the couplings are weak then the problem converges faster

(T,V)

q(T,V)

V=V(T,V)

k(T,V)

),( VTQvolV=V(T,V)

T=T(T,V)

127

Definitions of various types of couplings


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Definitions of various types of couplings

One-way coupled Information passes from one physics to the next, in one direction

Two-way coupled Information gets passed back and forth between physics

Load coupled The results from one physics affect only the loading on the other physics

Material coupled The results from one physics affect the materials properties of other physics

Non-linear coupled The results of one physics affects both that, and other, physics

Fully coupled

All of the above Weakly coupled

The physics do not strongly affect the loads/properties in other physics

Strongly coupled The opposite of weakly coupled

128

It is possible to solve multiphysics problems in a


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p p y psegregated sense, solving each physics separately

TVV

TVT

V

T

VVTV

VTTT

uub

uub

u

u

ufuf

ufufuf

,

,

Assume that these are approximately zero and ignore them

initialize uT,i, uV,i

do {

uT,i+1

= uT,i

+ST(u

T,i, u

V,i)-1b

T(u

T,i, u

V,i)

uV,i+1= uV,i+SV(uT,i+1, uV,i)-1bV(uT,i+1, uV,i)

i=i+1

}

while ( not_converged)

is damping

129

Solving in a segregated sense has some advantages


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Solving in a segregated sense has some advantages

V

T

V

T

VVTV

VTTT

b

b

u

u

ufuf

ufufuf

V

T

V

T

VV

TT

b

b

u

u

uf0

0ufuf

Less memory to store matrices:

Less memory to solve:

The Jacobian is O(2) times smaller in memory

iiiiii ubuuKuSuu

1

1

uT,i+1= uT,i+ST(uT,i, uV,i)-1bT(uT,i, uV,i)

uV,i+1= uV,i+SV(uT,i+1, uV,i)-1bV(uT,i+1, uV,i)

Jacobian is exactly 2 times smaller in degrees of freedom

The optimal iterative solver can be used for each physics

If the problem is strongly coupled, then the segregated approachwill not work, and a direct solver is often necessary since theiterative solvers are not tuned for a general Jacobian matrix.

130

Achieving convergence for multiphysics problems


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Achieving convergence for multiphysics problems

Set up the coupled problem and try solving it with a direct solver

If it is not converging: Check initial conditions

Ramp the loads up

Ramp up the non-linear effects Make sure that the problem is well posed (this can be very difficult!)

If you are running out of memory, or the solution time is very long: Use the segregated solver and select the optimal solver (direct or iterative) for

each physics, or group of physics, in the problem. FOR 3D, START HERE!

Upgrade hardware

Try the PARDISO out-of-core solver

Perform a mesh refinement study

131

Hands on example: heat and current flow


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pthrough a square

J=10 A/m2

VL=0

TL=0

All other surfaces insulated

Set up an Inward Current Flowon the right side

k=1 W/mK =1 S/m

All other surfaces insulated

Identical to previous problem

Conductive Media DC Application ModeHeat Transfer Application Mode

Set up and solve this uncoupled linearproblem

q=100 W/m2

132

Consider the following multi-physics couplings:


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Consider the following multi physics couplings:

VolumetricHeat Load

ThermalConductivity

ElectricConductivity

NormalCurrent

IterationsSolution

Times (sec)

Linear 0 1 1 10 1 6.0

weak non-linear

uncoupled0 1-(T/10000) 1-(V/1000) 10 3 18.0

one-way loadcoupled

Q_dc 1 1 10 3 44.0

weak two-wayload coupled

Q_dc 1 1 10+(T/100) 3 48.0

weak fully

coupled Q_dc 1-(V/1000) 1-(T/10000) 10+(T/100) 3 58.0

133

Thermal problem Electric problem

Hands on exercises: Ramp up the applied


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Hands on exercises: Ramp up the appliedcurrent, and monitor peak temperature

134

exp(-T/600) exp(-T/600)

Jin

Tmax

?

Ramp up inward current flow from 0-20 A/m2 using the parametric solver.Plot the temperature distribution. What happens? Why?

ThermalConductivity


1 1Case 1

Case 2

Hands on exercises: Ramp up the applied


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Hands on exercises: Ramp up the appliedvoltage, and monitor peak temperature

135

exp(-T/600) exp(-T/600)

V

Tmax

?

Ramp up applied voltage, how high can it go? Why?

ThermalConductivity


1 1Case 1

Case 2

Same as before

Using the segregated solver


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Pick the best solverfor each physics

Can choose adifferent dampingfor each physics

136

When in doubt, set each physics to run to


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convergence at each iteration

initialize uT,i, uV,i

do {

do {

non-linear solution for uT,i+1use uT,iand uV,i as initial conditions

}

do {

non-linear solution for uV,i+1use uT,i+1and uV,i as initial conditions

}

i = i+1

}

while ( not_converged)

These settings are equivalent to solving thecomplete non-linear problem at each iteration. 137

Additional physics and couplings


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Additional physics and couplings

Physics... Global Equations... Used to add an equation that must be satisfied during the solution

Adding additional constraints

Adding an Ordinary Differential Equation to the model

Options... Coupling Variables Integration: Integrates quantities over a domain and makes them global

Extrusion: Takes expressions in Ndimensions onto an N+1 dimension

Projection: Takes the integral of a quantity in Ndimensions and projects it ontoan entity that has dimension N-1

Adding your own equations Almost unlimited options, see booklet for example

Coefficient form

General form, simpler syntax than coefficient

Weak form, most flexible, but you need to know the terminology

138

Adding Ordinary Differential Equations (ODEs)


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Adding Ordinary Differential Equations (ODE s)

k

p(t)

u

c

0)( tpkuucumForces

00 0,0 uuuu

m

139

Solving an ODE: Damped harmonic oscillator


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00,10

0

uu

kuuu

dunderdampe,2

overdamped,2

dampedcritically,2

k

k

k

overdamped, =100

underdamped, =1

critically damped, =10

Set up and solve this transient problem.

Solve for t = 0:0.01:10

140

Using a global equation as a constraint


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Using a global equation as a constraint

141

P

F?

Set up this problem:The load, P, on the cantilever beam causes some y-deflection, v, at the end.

What force, F, must be applied to get zero deflection at the end?

v

Adding an integration coupling variable, step 1


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g g p g p

Set up and solve this Incompressible Steady StateNavier-Stokes fluid flow problem, (=1e3, =1e-3):

4 mm1 mm

Wall, no slip

Wall, no slip

Inlet, Pressure,

no viscous stressP = 0.1Pa

Outlet, Pressure,

no viscous stressP = 0 Pa

Solution, exhibiting

entrance effect

In post-processing, do aboundary integration of:

rho_ns*u*1[m]

AvdA

The mass flux integral is taken perunit inch out of the modeling plane

142

Adding an integration coupling variable step 2


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Go to Options... Constants...

Add a constant:MassFluxApplied = 1[g/s]

Go to Options... Integration Coupling Variables... Boundary Variables...

Add a variable to boundary 1 calledMassFlux, as shown below

143

Adding an integration coupling variable step 3


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Modify the appliedpressure at the inlet

Add a global equation

This adds an equation,AppliedP,

that must be satisfied.

COMSOL adjusts the inlet pressureto satisfy the condition on the totalmass flux through the boundary

Solve again, and integrate the mass flux across theboundary. Mass flux, which is a function of fluid velocity,

has be fixed by applying a constraint on pressure.

Caution! This approach will, undersome conditions, introduce a factorof two into the model. If you see this,switch to Non-ideal Constraints

144

Adding your own governing equation


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Adding your own governing equation

145

Coefficient Formof the PDE

General Formof the PDE

Weak Formof the PDE

0othersall,,,

QfTukc

fuauuuc

QFTyk

Txk

xTk

xTk

F

,*

*

Heat TransferPDE

QTk

test(T)*Q_ht+test(Ty)*Ty*k_ht-test(Tx)*Tx*k_ht-

0

dvqdvQukv

Most common PDEsare implemented

Enter thecoefficients for thePDE of interest

More compact wayof entering certainequations

Most flexible,requires that youknow how to get tothe weak form

Review


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Review

At this point, you should know: What a multi-physics problem is

How a multi-physics problem is solved

The definition of a coupled multi-physics problem The difference between a load and a material coupling

The difference between one- and two-way couplings

How the couplings can affect solve time & memory requirements

What the segregated solver does

How to add additional couplings and physics

146

Hands on: Fluid-Structure Interaction


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Hands on: Fluid Structure Interaction

Fluid moves through the domain,solid deforms the domain

The mesh deforms the due to thestructural analysis

Loads are automatically passedbetween the two physics usingthe FSI application mode

147


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Lecture 4: Time-stepping and reducingthe error in the finite element model

148

Introduction to time-dependent solutions


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Introduction to time dependent solutions

bKuufu

)(

t

QTkt

TCp

ttttt

ttbuKuuu

1

ttttt t ubuKu 1

1111

ttt

tt

ttbuKuuu

111 tttt tt buuKI

Governing Equation

Time-dependent finite element formulation

Explicit time-stepping

Implicit time-steppingFaster

More stable

Slower

Simple

149

Backward Difference Formula (BDF) order


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Backward Difference Formula (BDF) order

111 tttt tt buuKI

t+1

k(t)

A polynomial is fit throughn previous points topredict properties att+1

Here, the BDF order is n=3

tt-1t-2150

COMSOL picks its own optimum timesteps


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COMSOL picks its own optimum timesteps

These are the outputtimesteps, the timesyou want to see thesolution at.

They do not affect theinternal timesteps.

The internal timestepsare controlled by theTOLERANCES.

151

Absolute and Relative Tolerance


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Solutions withinRelative tolerance

Solutions withinAbsolute tolerance

t

u

t

u

True solution

t

u

At solutions near

zero, the relativetolerance cannot beused, since it wouldresult in very small

timesteps

t

u

Tighten both the

absolute andrelative tolerance

until you aresatisfied with theresults over time.

152

The absolute tolerance can be different forh l ti i bl


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each solution variable

Temperature:Relative Tolerance: 1%Absolute tolerance: 0.1K

Voltage:Relative Tolerance: 1%Absolute tolerance: 0.001V

Often need to set this

when solving multi-physics problems

153

Other timestepping optionsSelecting the physics at


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pp g p

Specify when to store,

and which time-stepsto solve at.

Give a condition to stopat, when this expressionbecomes negative

Selecting the physics atstartup will usually load

the appropriate settingsinto this form.

For wave type problems,gen-alpha is often better.

BDF order, set max andmin to 2 and 1 for waveproblems.

154

Make sure not to impose any ringing loads,

l th ll th


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unless they are really there

155

Load

time

Response

time

A reasonable starting point is necessary


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The load applied at t = 0 must agree with the initial condition

It is helpful to a have a load with time derivative zero at t = 0 The Init tab of the Subdomain Settings form sets initial conditions

The Solver Manager... Initial Value tab overwrites that information

Most common mistake:

Initial conditions for fluid flow: u = 0

Velocity/Pressure inlet conditionthat does NOT ramp up from zero

Solution:

Use flc1hs function to ramp load

and to keep db/dt(t=0)=0

t

u

t

b

u(tinitial) = [

Ksteady-state]

-1b(tinitial)

156

Convergence of a transient problem


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t

u

Outside envelop is absoluteand relative tolerance

True solution(unknown)

Observe the solution with respect to time andmonitor the magnitude of the changes as the

tolerances are increased.

The problem is converged when you say it is.

157

The mesh must be fine enough with respect tod ti


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space and time

Transient 1-D slab

Tfixed

x

Tinitq

T

increasingtime

Hands on:

1) Go back to heat transfer square model

2) Change the analysis type to Transient

3) Set material properties:=1, Cp=1

4) Solve transiently for 1 second (default)

5) Store all timesteps taken by solver

6) Examine results at initial timesteps

158

There are two ways to minimize this problem


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y p

Modify the mesh

Use boundary layermeshing to better resolvethe step in the solution

Modify the physics

Use a smoothed stepfunction to gradually rampup the heat load:

flc1hs(t-0.01,0.01)

t

qq

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Solving transient problems


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Begin by assuming the problem is well posed, try solving it byramping all loads up from consistent initial conditions Tighten the tolerances and see if you get a converging answer

If it is not converging:

Refine the mesh Smooth out abrupt differences in loads and properties as reasonable

Try linearized properties

Some non-linear properties may require a very fine mesh and small time-step

Check the physics carefully

If the solution takes a long time, or runs out of memory Use the time-dependent segregated solver using all previous guidelines

Upgrade hardware

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What we have learned...


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Theory of non-linear single physics finite element modeling

Meshing techniques and guidelines

Theory of multiphysics modeling, adding couplings to the model

Understanding what the solver is doing

Transient modeling

How to get confidence in your results

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Bibliography & Resources


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The Model Library: www.comsol.com/showroom

COMSOL Papers database:

www.comsol.com/papers

Google Scholar, 5000+ papers:

www.scholar.google.com

Books written with COMSOL: www.comsol.com/academic/books/

COMSOL Support:

http://www.comsol.com/support

COMSOL Exchange:

www.comsol.com/community/exchange/

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COMSOL Documentation&References

Nonlinear Programming, Bertsekas

Scientific Computing, Heath

Fundamentals of Heat and MassTransfer, Incorpera & DeWitt

Nonlinear Finite Elements for Continuaand Structures, Belytchko, Liu, Moran

Whatever FEA book you like


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Closing Remarks and Questions

intensive class

Documents