Course in Nonlinear FEM - homes.civil.aau.dk -...

Computational Mechanics, AAU, EsbjergNonlinear FEM

Course inNonlinear FEM

Dynamics

Dynamics 2Computational Mechanics, AAU, EsbjergNonlinear FEM

Outline

Lecture 1 – IntroductionLecture 2 – Geometric nonlinearityLecture 3 – Material nonlinearityLecture 4 – Material nonlinearity continuedLecture 5 – Geometric nonlinearity revisitedLecture 6 – Issues in nonlinear FEALecture 7 – Contact nonlinearityLecture 8 – Contact nonlinearity continuedLecture 9 – DynamicsLecture 10 – Dynamics continued


Nonlinear FEMLecture 1 – Introduction, Cook [17.1]:

– Types of nonlinear problems– Definitions

Lecture 2 – Geometric nonlinearity, Cook [17.10, 18.1-18.6]:– Linear buckling or eigen buckling– Prestress and stress stiffening– Nonlinear buckling and imperfections– Solution methods

Lecture 3 – Material nonlinearity, Cook [17.3, 17.4]:– Plasticity systems– Yield criteria

Lecture 4 – Material nonlinearity revisited, Cook [17.6, 17.2]:– Flow rules– Hardening rules– Tangent stiffness


Nonlinear FEMLecture 5 – Geometric nonlinearity revisited, Cook [17.9, 17.3-17.4]:

- The incremental equation of equilibrium- The nonlinear strain-displacement matrix- The tangent-stiffness matrix- Strain measures

Lecture 6 – Issues in nonlinear FEA, Cook [17.2, 17.9-17.10]:– Solution methods and strategies– Convergence and stop criteria– Postprocessing/Results– Troubleshooting


Elements for Mass-Spring-Damper Systems


MASS21: Structural Mass

Large deflectionSpecial Features

Key for element coordinate systemKEYOPT(2)

0 - 3-D mass with rotary inertia2 - 3-D mass without rotary inertia3 - 2-D mass with rotary inertia4 - 2-D mass without rotary inertia

KEYOPT(3)

NoneBody Loads

NoneSurface Loads

NoneMaterial Properties

MASSX, MASSY, MASSZ, IXX, IYY, IZZ if KEYOPT(3) = 0MASS if KEYOPT(3) = 2MASS, IZZ if KEYOPT(3) = 3MASS if KEYOPT(3) = 4

Real Constants

UX, UY, UZ, ROTX, ROTY, ROTZ if KEYOPT(3) = 0UX, UY, UZ if KEYOPT(3) = 2UX, UY, ROTZ if KEYOPT(3) = 3UX, UY if KEYOPT(3) = 4

Degrees of Freedom

INodes

MASS21Element Name


COMBIN14: Spring/Damper

Nonlinear (if CV2 is not zero), Stress stiffening, Large deflection, etc.

Special Features

0 - Linear Solution (default)1 - Nonlinear solution (required if CV2 is non-zero)

KEYOPT(1)

0 - 3-D longitudinal spring-damper1 - 3-D torsional spring-damper2 - 2-D longitudinal spring-damper (2-D elements must lie in an X-Y plane)

KEYOPT(3)

NoneBody Loads

NoneSurface Loads

NoneMaterial Properties

K, CV1, CV2Real Constants

UX, UY, UZ if KEYOPT(3) = 0ROTX, ROTY, ROTZ if KEYOPT(3) = 1UX, UY if KEYOPT(3) = 2etc.

Degrees of Freedom

I, JNodes

COMBIN14Element Name


Equivalent Mass/Spring/Damper

Dissipated Energy

Potential Energy

Kinetic Energy

Rotational SystemsTranslational Systems

2

21 yMKE eq=

2

21 yKPE eq=

2yDdt

dDEeq=

2

21 θeqJKE =

2

21 θeqKPE =

2

21 θeqD

dtdDE

=


Design of Spring/Damper in a Recoil Landing System


Problem Description

Vehicle mass M

Piston

Cylinder

Damper D

Spring K

Rod

Foot


Modeling Considerations

y

D

K

M

0V(0)y ,0)0(0

===++

yKyyDyM


Equivalent Spring/Damper

y

ΔL

K

θ

( )2cos213 θyKPE ×=

( )2cos3 θKKeq =

( )2cos3 θyDdt

dDE×=

( )2cos3 θDDeq =


ANSYS Procedure (1/2)FINISH/CLEAR/TITLE, Recoil Landing System (SI)/PREP7

K = 72000M = 2000D = 24000 ! Critically dampedV0 = 5 ! 18 km/Hr

ET, 1, COMBIN14,,, 2ET, 2, MASS21,,, 4R, 1, K, DR, 2, M

N, 1, 0 ! GroundN, 2, 1 ! VehicleTYPE, 1 $ REAL, 1 $ E, 1, 2TYPE, 2 $ REAL, 2 $ E, 2FINISH

0102030405060708091011121314151617181920


ANSYS Procedure (2/2)/SOLU

ANTYPE, TRANSTRNOPT, FULLD, ALL, UY, 0D, 1, UX, 0IC, 2, UX, 0, -V0DELTIM, 0.01TIME, 2OUTRES, NSOL, ALLSOLVEFINISH

/POST26

NSOL, 2, 2, U, X, DISP/GRID, 1/AXLAB, Y, DISPLACEMENTPLVAR, 2

22232425262728293031323334353637383940


Design of Damper in a Spring Scale


Problem DescriptionItem placed on platform mass m

Plateform mass Mp

Spring K mass Mk

Rack mass Mr Damper Dmass Md

Calibrated dial Jc

Gear ratio n

Geared magnifier Jg

R



Itemm

ScaleM

SpringK

DamperD

y

( )0(0)y ,0)0( ===+++

ymgKyyDyMm

2

2

23 RnJ

RJMMMMM cgk

drp +++++=


Lumped Masses (1/4)Lumped Mass for a Spring

L

Vo

V

xdm

Equivalent lumped mass Meq

LxVV o=

621

21 2

0

22 okL ko

L

VMLdxM

LxVdmVKE =⎟⎠⎞

⎜⎝⎛== ∫∫

3k

eqMM =


Lumped Masses (2/4)Lumped Mass for a Gear-and-Rack Set

R

Mr

JG

Vo

Rolls without slipping

θ

22

2222

21

21

21

21

21

oG

ro

GorGor VRJM

RVJVMJVMKE ⎟

⎠⎞

⎜⎝⎛ +=⎟

⎠⎞

⎜⎝⎛+=+= θ

2RJMM G

req +=


Lumped Masses (3/4)Lumped Mass for a Geared Shafts Set

J1

J2

N1

N2Jeq

Gears

Shafts

222

2221 2

121

21

21

21

gc

gcg

c

gggggccgg N

NJJ

NN

JJJJKE θθ

θθθ⎥⎥⎦

⎤

⎢⎢⎣

⎡⎟⎟⎠

⎞⎜⎜⎝

⎛+=⎟⎟

⎠

⎞⎜⎜⎝

⎛+=+=

2

⎟⎟⎠

⎞⎜⎜⎝

⎛+=

c

gcgeq N

NJJJ


Lumped Masses (4/4)Lumped Mass for the Spring Scale System

2

2

RnJJ

MM cgreq

++=

2

2

23 RnJ

RJMMMMM cgk

drp +++++=

Gear set


ANSYS Procedure (1/2)FINISH/CLEAR/TITLE, Spring Scale (cgs)/PREP7

m = 1500 ! GMeq = 500 ! GK = 3.2E6 ! dyn/cmD = 6.4E4 ! dyn-s/cm

ET, 1, COMBIN14,,, 2ET, 2, MASS21,,, 4R, 1, K, DR, 2, m+Meq

N, 1, 0N, 2, 1TYPE, 1 $ REAL, 1 $ E, 1, 2TYPE, 2 $ REAL, 2 $ E, 2FINISH

0102030405060708091011121314151617181920



ANTYPE, TRANSTRNOPT, FULLDELTIM, 0.02TIME, 2KBC, 1D, ALL, UY, 0D, 1, UX, 0F, 2, FX, -m*981OUTRES, NSOL, ALLSOLVEFINISH

/POST26

NSOL, 2, 2, U, X, DISP/GRID, 1PLVAR, 2

22232425262728293031323334353637383940


Quenching of a Shaft


Problem Description

T(0)

Tb(0)T(t)

Tb(t)

Shaft

Bath

mb = 1 lb

Before quenching After quenching

ms = 0.069 lb



ms, Cs, Ts mb, Cb, Tbh, A


ANSYS Procedure (1/2)FINISH/CLEAR/TITLE, Unit: lb(mass)-ft-(Btu)-hr-F/PREP7

PI = 4*ATAN(1)ET, 1, MASS71R, 1, PI*(1/4/12)**2/4*(5/12)MP, DENS, 1, 486MP, C, 1, 0.11

ET, 2, LINK34R, 2, 0.028MP, HF, 2, 300

0102030405060708091011121314

ET, 3, MASS71,,, 1R, 3, 1/62MP, DENS, 3, 62MP, C, 3, 1.0

N, 1, 0N, 2, 1

TYPE, 1 $ REAL, 1 $ MAT, 1E, 1TYPE, 2 $ REAL, 2 $ MAT, 2E, 1, 2TYPE, 3 $ REAL, 3 $ MAT, 3E, 2FINISH

161718192021222324252627282930



ANTYPE, TRANSTRNOPT, FULLDELTIM, 0.0001KBC, 1TIME, 0.01IC, 1, TEMP, 1300IC, 2, TEMP, 75

OUTRES, NSOL, ALLSOLVEFINISH

/POST26

NSOL, 2, 1, TEMP,, SHAFTNSOL, 3, 2, TEMP,, WATER/GRID, 1PLVAR, 2, 3

3233343536373839404142434445464748495051


Elements Related to Lumped-Mass Systems


Masses


Springs/Dampers


Thermal Links


Circuit Element


Dynamic Effects

• Inertia force• Damping force• Elastic Force• External force• Dynamic Effects

FKDDCDM =++

FKD =


Transient Dynamic Analysis

FKDDCDM =++


Modal Analysis (1/3)

0KDDCDM =++



0KDDCDM =++

0KDDM =+

21 ξ−= ud ff

πξ2−= eR



• Avoid resonance• Exploit resonance• Assess structural stiffness• Structural modal degrees of freedom• Further dynamic analyses• etc.


Harmonic Response Analysis

( )Φ+=++ tωsinFKDDCDM


Solution Methods


Solution Methods

Solution Methods for Equation of Motion

Direct Integration Mode Superposition

Implicit Explicit

ReduceFull Full Reduce


Solution methods


Direct Integration

• Implicit method (ANSYS)• Explicit method (LS-DYNA)


Implicit vs. Explicit Methods

( )ttttttt Δ+Δ−Δ+ = DDDfD ,,...,

( )ttttttt DDDfD ,,..., 2 Δ−Δ−Δ+ =

Implicit method

Explicit method


Linear vs. nonlinear dynamics• Dynamics or equation of motion:

– The causal relation between the present state and the next state in the future. It is a deterministic rule which tells us what happens in the next time step.

– In the case of a continuous time, the time step is infinitesimally small. Thus, the equation of motion is a differential equation or a system of differential equations:du/dt = F(u), where u is the state and t is the time variable.

– An example is the equation of motion of an undriven and undamped pendulum.

– In the case of a discrete time, the time steps are nonzero and the dynamics is a map:un+1 = F(un), with the discrete time n.

– An example is the baker map. Note, that the corresponding physical time points tn do not necessarily occur equidistantly. Only the order has to be the same. That is, n < m implies tn < tm.

– The dynamics is linear if the causal relation between the present state and the next state is linear. Otherwise it is nonlinear.


Implicit vs. Explicit• What is the difference between implicit and explicit dynamics? (Difference

between regular ANSYS and ANSYS/LS-DYNA?)

• For computers, matrix multiplication is easy. Matrix inversion is the more computationally expensive operation. The equations we solve in nonlinear, dynamic analyses in ANSYS and in LS-DYNA are:[M]{a} + [C]{v} + [K]{x} = {F}

• Hence, in ANSYS, we need to invert the [K] matrix when using direct solvers (frontal, sparse). Iterative solvers use a different technique from direct solvers, however, the inversion of [K] is the CPU-intensive operation for any 'regular' ANSYS solver, direct or iterative.

• We then can solve for displacements {x}. Of course, with nonlinearities, [K(x)] is also a function of {x}, so we need to use Newton-Raphson method to solve for [K] as well (material nonlinearities and contact get thrown into [K(x)])


Implicit vs. explicit• In LS-DYNA, on the other hand, we solve for accelerations {a} first.

• It is assumed that the mass matrix is lumped. This basically forces the use of lower-order elements, i.e. for all explicit dynamics codes (ANSYS/LS-DYNA, MSC.Dytran, ABAQUS/Explicit), lower-order elements are used.

• The benefit of doing lumped mass is, if we solve for {a}, then [M], if lumped, is a diagonal mass matrix. This means that inversion of [M] is trivial (diagonal terms only)

• Another way to view it, is that we now have N set of *uncoupled*equations. Hence, we just have to do matrix multiplication, which is less CPU-intensive. Also, [K] does not need to be inverted, and accounting for material nonlinearties and contact is easier.


Implicit vs. explicit• The terms 'implicit' and 'explicit' refer to time integration

• For example– backward Euler method, that is an example of an implicit time integration scheme– central difference or forward Euler are examples of explicit time integration schemes

• It relates to when you calculate the quantities - either based on current or previous time step.

• In any case, this is a very simplified explanation, and the main point is that implicit time integration is unconditionally stable, whereas explicit time integration is not (there is a critical timestep the timestep delta(t) needs to be smaller than).

• Implicit, e.g. 'regular' ANSYS allows for much larger time steps• Explicit, e.g. LS-DYNA requires much smaller time steps. Also, LS-DYNA requires

very tiny steps, i.e. good for impact/short-duration events, not usually things like maybe creep where the model's time scale may be on the order of hours or more.


Implicit vs. explicit• 'Regular' ANSYS uses implicit time integration. This means that {x}

is solved for, but we need to invert [K], which means that each iteration is computationally expensive. However, because we solve for {x}, it is implicit, and we don't need very tiny timesteps (i.e., each iteration is expensive, but we usually don't need too many iterations total). – The overall timescale doesn't affect us much (although there are

considerations of small enough timesteps for proper momentum transfer, capturing dynamic response).

• ANSYS/LS-DYNA uses explicit time integration. This means that {a} is solved for, and inverting [M] is trivial -- each iteration is very efficient. However, because we solve for {a}, then determine {x}, it is explicit, and we need very small timesteps (many, many iterations) to ensure stability of solution since we get {x} by calculating {a} first. – (i.e., each iteration is cheap, but we usually need many, many iterations

total)


Mode Superposition Method

nnCCCC MMMMD ...332211 +++=


Reduced Method

FKD =

⎭⎬⎫

⎩⎨⎧

=⎭⎬⎫

⎩⎨⎧⎥⎦

⎤⎢⎣

⎡

s

m

s

m

sssm

msmm

FF

DD

KKKK

FDK =m

( )msmssss DKFKD −= −1

smssmsmm KKKKK 1−−=

sssmsm FKKFF 1−−=

where


Methods for Nonlinear Dynamic Analysis

• For nonlinear analysis, the only methods applicable is DIRECT INTEGRATION method.

• Reduced method can not be used for nonlinear analysis.

• Either implicit or explicit methods can be used.


Mass and Damping


Consistent vs. Lumped Mass Matrices

⎥⎥⎥⎥⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢⎢⎢⎢⎢

⎣

⎡

⎥⎥⎥⎥⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢⎢⎢⎢⎢

⎣

⎡

xx

xx

xx

xxxxxxxx

xxxxxxxxxx

xx

ROTZUYUX

ROTZUYUX

j

j

j

i

i

i

000000000000000000000000000000

0000

00000000

0000

Consistent mass matrix

Lumped mass matrix


Damping

• Damping effects is the total of all energy dissipation mechanisms– Hysteresis (solid damping)– Viscous damping– Dry-friction (Coulomb damping)


Idealization of Structural Damping

• Structural dampings are usually small (2%-7%).

• Equivalent viscous damping is assumed in ANSYS, i.e.,

DCF =D


How ANSYS Forms Damping Matrix?

• Alpha damping• Beta damping• Material dependent beta damping• Element damping matrices• Frequency-dependent damping matrix

[ ] [ ] ( )[ ] [ ] [ ] [ ]ξξββββα CCKKMC +⎟⎟⎠

⎞⎜⎜⎝

⎛+⎥

⎦

⎤⎢⎣

⎡⎟⎠⎞

⎜⎝⎛

Ω++++= ∑∑

==

em N

kk

N

jjj

mjc

11

2


Copper Cylinder Impacting on a Rigid Wall


Problem Description

x

y

L D

Initial Velocity Vo


Modeling Consideration

• Material: bilinear plastic model.• VISCO106 (2D viscoplastic solid) is

used.• Use axisymmetric model.


ANSYS Procedure (1/4)FINISH/CLEAR/TITLE, UNITS: SI/PREP7

ET, 1, VISCO106,,, 1MP, EX, 1, 117E9MP, NUXY, 1, 0.35MP, DENS, 1, 8930

TB, BISO, 1TBDATA,, 400E6, 100E6TBPLOT, BISO, 1

RECTNG, 0, 0.0032, 0, 0.0324LESIZE, 1,,, 4LESIZE, 2,,, 20MSHAPE, 0, 2DMSHKEY, 1AMESH, ALLFINISH

010203040506070809101112131415161718192021



ANTYPE, TRANSTRNOPT, FULLNLGEOM, ONIC, ALL, UY, 0, -227NSEL, S, LOC, X, 0D, ALL, UX, 0NSEL, S, LOC, Y, 0D, ALL, UY, 0NSEL, ALL/PBC, U,, ONEPLOT

TIME, 80E-6DELTIM, 0.4E-6KBC, 1OUTRES, ALL, 4SOLVEFINISH

2324252627282930313233343536373839404142


ANSYS Procedure (3/4)

/POST26

TOPNODE = NODE(0,0.0324,0)

NSOL, 2, TOPNODE, U, Y, DISPDERIV, 3, 2, 1,, VELO

/GRID, 1/AXLAB, X, TIME s/AXLAB, Y, DISPLACEMENT mPLVAR, 2/AXLAB, Y, VELOCITY m/sPLVAR, 3FINISH

4445464748495051525354555657


ANSYS Procedure (4/4)/POST1

SET, LASTPLDISP, 2PLNSOL, EPTO, EQV

ANTIME, 30

59606162636465


Dynamic Loads


Dynamic Loads: An Example/SOLU...F, ... ! 22.5 at the nodesTIME, 0.5 ! Ending timeDELTIM, ... ! Integration stepKBC, 0 ! Ramped loadingAUTOTS, ON ! OptionOUTRES, ... ! OptionSOLVE ! Load step 1

F, ... ! 10 at the nodesTIME, 1 ! Ending timeSOLVE ! Load step 2

FDELE, ... ! Zero the forceTIME, 1.5 ! Ending timeKBC, 1 ! Stepped loadingSOLVE ! Load step 3

010203040506070809101112131415161718

0 0.5 1.0 1.5Time (s)

Force (N)22.5

10


Initial Conditions


Example: An Stationary Plate Subjected to an Impulse Load

• This is the default initial condition. No input is needed.


Example: Initial Velocity on a Golf Club Head

• This simple initial condition can be specified by using IC command.

NSEL, ALL

IC, ALL, UY, 0, V0


Example: Plucking a Cantilever Beam

/SOLUANTYPE, TRANS...TIMINT, OFF ! Transient effects offTIME, 0.001 ! Small time intervalD, ... ! Apply displacement at desired nodesKBC, 1 ! Stepped loadsNSUBST, 2 ! To avoid non-zero velocitySOLVE

TIMINT, ON ! Transient effects onTIME, ... ! Actual time at end of loadDDELE, ... ! Delete the applied displacementSOLVE

0102030405060708091011121314


Example: Dropping an Object from Rest

/SOLUTIMINT, OFF ! Transient effects offTIME, 0.001 ! Small time intervalNSEL, ... ! Select all nodes on the objectD, ALL, ALL, 0 ! Temporarily fix themNSEL, ALLACEL, ... ! Apply accelerationKBC, 1 ! Stepped loadsNSUBST, 2 ! To avoid non-zero velocitySOLVE ! Load step 1

TIMINT, ON ! Transient effects onTIME, ... ! Actual time at end of loadNSEL, ... ! Select all nodes on the objectDDELE, ALL, ALL ! Release themNSEL, ALLSOLVE ! Load step 2

0102030405060708091011121314151617


Integration time Steps


Response Frequency

Response

Time

Minimumresponse

time

20pt ≤Δ


Abrupt Changes in Loading

0 0.5 1.0 1.5Time (s)

Force (N)

22.5

10


Contact Frequency

30Tt ≤Δ


Wave Propagation

cxt

3Δ

≤Δ


Exercise: Rocket Flight

y

140 in.

Thrust

Time

100 lb

1 sec.1

2

3

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