fem zabaras dynamics

8/9/2019 FEM Zabaras Dynamics

1/50

1

MAE 4700 FE Analysis for Mechanical & Aerospace Design

N. Zabaras (11/10/2009)

MAE4700/5700Finite Element Analysis for

Mechanical and Aerospace DesignCornell University, Fall 2009

Nicholas ZabarasMaterials Process Design and Control Laboratory

Sibley School of Mechanical and Aerospace Engineering101 Rhodes Hall

Cornell UniversityIthaca, NY 14853-3801
http://mpdc.mae.cornell.edu/Courses/MAE4700/MAE4700.htmlhttp://mpdc.mae.cornell.edu/Courses/MAE4700/MAE4700.html


2/50

2


N. Zabaras (11/10/2009)

Elastic solids subject to dynamic loads

The stress equilibrium equations considered in Lecture 15

considered a body in static equilibrium.

If a body is in motion, then at any instant of time Newtons

law of motion implies that the sum of all forces must be

equal to the inertial force.

t

Therefore, the governing equations lead to a dynamics or

transient problem.


3/50

3


N. Zabaras (11/10/2009)

Strong form of dynamic linear elasticity

3

Denote the mass density of the material by and the

accelerations in the coordinate directions by and .Then the equation of motion in 2D can be written as follows:

Find the displacements on such that

2

2xuu

t=

y

u

u

,x x x

y y y

b u

b u on

+ =

+ =

i

i

Swhere D u=

, x x y y t

u

with

n t n t on

u u on

= = =

i i

Plane stress:

Plane strain:

2

1 0

1 01

(1 )

0 0 2

vE

D vv

v

=

1 0

1 0(1 )(1 2 )

(1 2 )0 0

2

v vE

D v vv v

v

= +

0

0

T

S

x y

y x

=


4/50

4


N. Zabaras (11/10/2009)

Weak form of dynamic linear elasticity

Comparing with the static problem, we only have the extra

inertial terms. Thus the weak form corresponding to thesedifferential equations can be obtained following the same

steps as before.

We pre-multiply the equilibrium equations inxand ydirections and the two natural boundary conditions by the

corresponding weight functions and integrate over the

corresponding domains as follows:

,x y x x y yb u b u on + = + =

i i ( )

( )

0

0

,

.

x x x x x x

y y y y y y

b w d u w d w U

b w d u w d w U

+ =

+ =

i

i

, x x y y tn t n t on= =

i i

( )

( )

0

0

0 ,

0 .

t

t

x xx x

y yy y

w n t d w U

w n t d w U

=

=

i

i


5/50

5


N. Zabaras (11/10/2009)

Weak form of dynamic linear elasticity

Applying Greens theorem (integration by parts) to the first

term in each of these equations and follow the exact

procedures as in the static case, we have the matrix form of

the weak form as follows:

:u U on such that

( )2

02t

TS S

uw d w D ud w td w bd w U

t

+ = +

i i i

{ }

{ }

1

10

: : ,

: , 0

u

u

where U u u H u u on and

U w w H w on

= =

= =

find


6/506


N. Zabaras (11/10/2009)

Finite Element Approximation

Same as in static problem, we apply the finite element

approximation to the derived weak form.

Here ue is the finite element approximation of thedisplacement field in element e.

The element shape function matrix Ne in the matrix form is

where nen is the number of element nodes.

( ){ }

int ( , )

( , ) ( , )e e e

x and y Nodaldisplacements displacements

at po x y

u x y N x y d t =

1 2

1 2

0 0 ... 0

0 0 ... 0

e e enene

e e enen

N N N

N N N N

=


7/507


N. Zabaras (11/10/2009)

Strain displacement matrix

We need to compute the strains in terms of the element

shape functions and the nodal displacements. Applying the

symmetric gradient operator to Ne gives

where the strain-displacement matrixe

is defined as:

{ } { }[ ]

{ }e

ex

e e e e e y S S

e Bxy

u N d

= = =

[ ] [ ]

1 2

1 2

1 1 2 2

0 0 ... 0

0 0 ... 0

...

ee e

nen

ee e

e e

nen

S

e ee e e e

nen nen

NN N

x x xNN N

B N y y y

N N N N N N

y x y x y x

= =


8/508


N. Zabaras (11/10/2009)

Semidiscrete Finite Element equations

Following the same procedure as before, the finite element

equations become

T

e

e e e ed

= K B D B

T T

e e

t

ee

t

e e e

ffbodyboundaryforce vector

force vector

d d

= + f N t N b

( ) ( ) ( )t t t+ =Mu Ku f

where Me is known as an element mass matrix

e

e Td

= M N N

and all other matrices are exactly the same as those for the

static case:


9/509


N. Zabaras (11/10/2009)

Equation of Motion For Beams For Bernoulli-Euler beam theory, the equation of motion is

of the form

( )

2 22

2 2 2 ,y y

u u

EI q x t t x x

+ =

where denotes the mass density per unit length,A is the

area of cross section, Eis the modulus, and I is the second

moment of inertia.


10/5010


N. Zabaras (11/10/2009)

Semidiscrete Finite Element equations

Following the same procedure as before, the finite element

equations become( ) ( ) ( )y yt t t+ =Mu Ku f

where Me is known as an element mass matrix

e

e TL

AN Ndx= and all other matrices are exactly the same as those for the

static case: 2 23

2 2

1

1

12 6 12 6

6 4 6 22 12 6 12 6

6 2 6 4

T

e e

e e e ee e e

e e e e e

e ee

e e e e

L L

L L L L L E I K B E I B d L LL

L L L L

= =

1

11

21

2

1

6( )[ ] ( )

12 2

6

e

eu

e ee e T

u

e

NLN L qL

f q x N dx q x d N

N L

= = =


11/5011


N. Zabaras (11/10/2009)

Mass Matrix

e

e T N Nd= Mass matrix

The mass matrix term is the only new term in the element

equations for dynamic problems.

During assembly, the element mass matrices are assembled

to form a global mass matrix in exactly the same manner as

the stiffness matrix.

Explicit mass matrices for commonly used structural

elements are derived in the following slides.


12/5012


N. Zabaras (11/10/2009)

Mass Matrix for Axial deformations

The element extends from to and has a length .

For simplicity the element is assumed to have a uniform

area of cross section

1x 2x 2 1 L x x=

The element is based on the following interpolation functions:

2 2 1 11 2

1 2 2 1

; x x x x x x x x

N N x x L x x L

= = = =

Using these interpolation

functions, the mass matrix can bewritten as follows:

2

1

2

2 1

1

2 1

1 26e

e exe Tx

x x x x x x A LL

M N Nd A dx x x L L

L

= = =


13/5013


N. Zabaras (11/10/2009)

Mass Matrix For Plane Truss Element

For the mass matrix, we must consider displacements both

in the x and y directions, since motion along both axesgenerates the inertia forces.

Therefore, the mass matrix for a truss element is derived by

writing the linear interpolation functions for x and ydisplacements

Assuming a coordinate s along the axis of the element with

s=0at node 1 and s = L, at node 2 the interpolation functions

are as follows:1 2;

s L sN N

L L

= =

1

1 2 1

1 2 2

2

0 0

0 0

u

N N vu

N N uv

v

= =

Nd


14/5014


N. Zabaras (11/10/2009)

Mass Matrix For Plane Truss Element

Using these interpolation functions, the mass matrix can bewritten as follows:

0

0

0 0 0

0 0 0

0

e

Le T

s L

L

s L s L s L L L

d A dss s L s

L L L

sL

= =

N N

Carrying out matrix multiplication and integrating each term,

we obtain 2 0 1 0

0 2 0 1

1 0 2 06

0 1 0 2

e ee A LM

=


15/50

15


N. Zabaras (11/10/2009)

Mass Matrix For Space Truss Element

The mass matrix for a three dimensional space truss element,

can be written using exactly the same arguments as those forthe plane truss.

The interpolation functions for x, y, and z displacements in

terms of a coordinate s along the axis of the element withs=0at node 1 and s = L, at node 2 are as follows:

1 2;s L s

N NL L

= =

1

1

1 2

1

1 2

2

1 2

2

2

0 0 0 0

0 0 0 0

0 0 0 0

u

vN Nu

wN Nvu

N Nwv

w

= =

Nd


16/50

16


N. Zabaras (11/10/2009)

Mass Matrix For Space Truss Element

Using these interpolation functions, the mass matrix can be

written as follows:

2 0 0 1 0 0

0 2 0 0 1 00 0 2 0 0 1

1 0 0 2 0 06

0 1 0 0 2 00 0 1 0 0 2

e

e ee T A LM d

= = N N


17/50

17


N. Zabaras (11/10/2009)

Mass Matrix for Beam Element

2

1

2

1

2

2

2

2

1 (1 ) (2 )4

(1 ) (1 )

81

(1 ) (2 )4

(1 ) ( 1)8

u

e

u

e

N

LN

N

L

N

= +

= +

= +

= +

12( ) 1, 1 1e

e

x x

L

=

1(1 ) , 1 1

2

eeL

x x = + +

1

1

1 1 2 2

2

2

( ) [ ] Nd

y

e

y u u

y

u

u x N N N N

u

= =

The two-node beam element shown in the figure was

developed in Lecture 4. The interpolation functions are as

follows:


18/50

18


N. Zabaras (11/10/2009)

Mass Matrix for Beam Element

The mass matrix can be written as follows:

2

2

12 2 2 2

12

2

1(1 ) (2 )

4

(1 ) (1 )1 18

(1 ) (2 ) (1 ) (1 ) (1 ) (2 ) (1 ) ( 1)1 4 8 4 8(1 ) (2 )

4

(1 ) ( 1)8

e

e

e ee

e

e T

L

L L

d

L

M d A

+

+

+ + + + + +

= = N N

Carrying out matrix multiplication and integrating each term,we get

2 2

2 2

156 22 54 13

22 4 13 3420 54 13 156 22

13 3 22 4

e e

e e e ee e

e e

e e e e

e

L L

L L L LA LL L

L L L L

M

=


19/50

19


N. Zabaras (11/10/2009)

Triangular Element for Plane Stress/Strain

The three-node triangular element shown in figure was

developed for plane stress and plane strain problems in

Lecture 16. The interpolation functions are as follows:

1 2 3

1 2 3

0 0 0

0 0 0

e e e ex e

e e e ey

u N N Nd

u N N N

=

where:[ ] [ ]

[ ] [ ]

[ ] [ ]

1 1 1 1 2 3 3 2 2 3 3 2

2 2 2 2 3 1 1 3 3 1 1 3

3 3 3 3 1 2 2 1 1 2 2 1

1 1( , ) ( ) ( ) ( )

2 2

1 1

( , ) ( ) ( ) ( )2 2

1 1( , ) ( ) ( ) ( )

2 2

e

e e

e

e e

e

e e

N x y a b x c y x y x y y y x x x yA A



= + + = + +

= + + = + +

= + + = + +


20/50

20


N. Zabaras (11/10/2009)

Triangular Element for Plane Stress/Strain

With constant thickness tand using these interpolation

functions, the mass matrix can be written as follows:

e

A

e TdA M d t

= = T

N NN N

Carrying out matrix multiplication and integrating over the

triangular as explained before, we get

2 0 1 0 1 00 2 0 1 0 1

1 0 2 0 1 0

0 1 0 2 0 1121 0 1 0 2 0

0 1 0 1 0 2

e ee A tM

=


21/50

CCOORRNNEELLLLU N I V E R S I T Y 21


N. Zabaras (11/10/2009)

Mass Matrix for Isoparametric Element

The mass matrix for an isoparametric element may be

computed by numerical integrations as described before. For

example, for two-dimensional elements the mass matrix is

given by

( ) ( )1

, , | |i

e

N

i i i i i

i

e T J wM d =

= = T

N NN N

Now since it is the product of shape functions which areintegrated, the order of quandrature used for standard

integration will suffice to accurately compute the mass matrix.


22/50



N. Zabaras (11/10/2009)

Free Dynamic Vibration Analysis

Most structural dynamics and vibration problems typically

start with the analysis of the free vibration motion of the

system. There is no externally applied load in this situation,

and thus the global system of equations for the system is of

the following form:

0+ =Mu Ku

where the global mass matrix M and the global stiffness matrixK are assembled from the corresponding element matricesusing the usual assembly procedure.

This equation expresses the condition of natural vibration(simple harmonic motion), where at any instant the

restoration influences in the system balance the inertia

influences.


23/50



N. Zabaras (11/10/2009)


This system of equation is satisfied by a harmonic solution of

the following form

cos t=u where and are parameters that will be later identified as the

mode shape (eigenvectors) and the natural frequency.

Substituting this into the differential equation, we have

2 cos cos 0t t + =M K 2 0 + =M K


24/50



N. Zabaras (11/10/2009)


Rearranging terms, we have what is known as the

generalized eigenvalue problem

= MK where is introduced for convenience.2=

From the solution of this system, we get n natural frequencies

and corresponding free-vibration mode shapes when

the size of the matrix and is .n ni i

=

K M

i

The eigenvectors are called normal modes of the system if the

eigenvectors are normalized such that

1, 1,2, ,Ti i i n= =M


25/50

25


N. Zabaras (11/10/2009)


Note that we must apply the essential boundary conditions

on the mass and stiffness matrix before solving the

eigenvalue problem.

The number of eigenvalues obtained in the finite elementmethod is always equal to the number of unknown nodal

values. As the mesh is refined, not only do we increase the

number of eigenvalues but we also improve the accuracy of

the preceding eigenvalues.

S l i f Ei l bl


26/50

26


N. Zabaras (11/10/2009)

Solution of Eigenvalue problem

In Matlab the eigenvalues and corresponding eigenvectors

are obtained by using the eigcommand as follows:

[V,Lam]=eig(K,M)

In the result, Lam is a diagonal matrix of eigenvalues and V isa matrix whose columns are the eigenvectors (mode shapes)

which are scaled so that the norm is 1.

If K and M are sparse matrices and you want to return the

eigenvectors, you may use eigs instead. Or you can first

convert them to full matrix, and then use eig.

T i R


27/50

27


N. Zabaras (11/10/2009)

Transient Response

For structures subjected to dynamic loads the global system

of equations for the system is of the following form:

( ) ( ) ( )t t t+ =Mu Ku f

where the global mass matrix M, the global stiffness matrix Kand the global load vectorf are assembled from thecorresponding element matrices using the usual assembly

procedure.

For a unique solution initial displacements and velocities at

time 0 at all degrees of freedom must also be specified.

( ) 0 00 , (0)u v= =u u

N k Ti I i M h d


28/50

28


N. Zabaras (11/10/2009)

Newmark Time Integration Methods

Suppose that we are able to get estimates for the

acceleration and both at the start and end of a

general time step .

( )[ ]( )[ ]

2

1 2 2 1

1 1 1 1

12

1

n n n n n

n n n n

t

t

t

+ +

+ +

= + + += + +

u u u u u

u u u u

( )tu ( )t t+ u

We could then use a Taylor series expansion to obtain

estimates of displacement and velocity at time t t+

t

where means the value of the function at the kth step in

time.nu u

N k Ti I t ti M th d


29/50

29


N. Zabaras (11/10/2009)


( )[ ]( )[ ]

2

1 2 2 1

1 1 1 1

12

1

n n n n n

n n n n

t

t

t

+ +

+ +

= + + +

= + +

u u u u u

u u u u

Here and are two adjustable parameters that determinethe nature of the time integration scheme.

1 2

If we set the acceleration is estimated based on its

value at time . This is known as an explicittime integration

scheme.

1 2 0= =t

Alternatively, If we set the acceleration is estimated

from its value at time . This is known as an implicittimeintegration scheme.

1 2 1= =t t+



30/50

30


N. Zabaras (11/10/2009)


The acceleration at the start and end of the time step is

computed using the finite element equation of motion.

At time , we havet

( )1

n n n

= +u M Ku f At time , we havet t+

( )[ ]

( )

1 1 1

2

1 2 2 1 1

222 1 2 1

0

1 02

11

2 2

n n n

n n n n n n

n n n n n

tt

tt t

+ + +

+ + +

+ +

+ =

+ + + + =

+ = + + +

Mu Ku f

Mu K u u u u f

M K u K u u u f



31/50

31


N. Zabaras (11/10/2009)


We can write the recurrence formula in terms of an effective

stiffness and load vector:

1 1

n n+ +=Ku f

where the effective stiffness matrix is2

2

12

t= + K M K

and the effective load vector is

( )2

1 2 1 1

2n n n n n

tt+ +

= + + +

f K u u u f

After this step, the values of and can be found from

( )[ ]

( )[ ]

2

1 2 2 1

1 1 1 1

12

1

n n n n n

n n n n

tt

t

+ +

+ +

= + + +

= + +

u u u u u

u u u u

1n+u 1n+u



32/50

32


N. Zabaras (11/10/2009)


Note that the calculation of and requires the knowledge of

the initial conditions and . In practice, we do notknow . As an approximation, it can be calculated from

K f

0 00,u u 0u

0u

( )10 0 0= +u M Ku f

The algorithm isInitial Calculations

1. Form the global matrices M and K;

2. Assemble the effective stiffness matrix . Modify it foressential boundary conditions. Factor .

3. Initialize and . Select .

K

K

At Each Time Step1. Assemble the effective load vector . Modify it for

essential boundary conditions.

2. Solve for acceleration . Then solve for and .

1

n+f

1n+u 1n+u 1n+u

0 00,u u 0u t

E l 1D d i bl


33/50

33


N. Zabaras (11/10/2009)

Example: 1D dynamic problem

L

tx

2

2 ( , )

u u

A EA f x t x xt

= 0, 100, 1, 100, 5, 10 f A E L t = = = = = =

As an example, we plot the variation

of the displacement at the end of the

bar (x=L) with time.

St bilit d A


34/50

34


N. Zabaras (11/10/2009)

Stability and Accuracy

Accuracyof a numerical scheme is a measure of the

closeness between the approximation solution and the exactsolution whereas stabilityof a solution is a measure of the

boundedness of the approximation solution with time.

Now we investigate the effects of the two adjustable

parameters in the time integration scheme.

As we might expect, the size of the time step can influence

both accuracy and stability.



35/50

35


N. Zabaras (11/10/2009)


First, we look at the solution with (both velocity and

displacement update are fully explicit). The plots below show

the predicted displacement at the end of the bar for two

values of time step

2 0=

t

The result with smaller time step is good, but with a larger

time step the solution is oscillatory. This is an example of a

numerical instability, which is common problem in explicit

time stepping schemes. Eventually the solution blows up

completely, because the oscillation grows exponentially.



36/50

36


N. Zabaras (11/10/2009)


With larger values of the instability disappears. In fact one

can show that for the equation of motion considered here,setting

2 1 1

1,

2

makes the time stepping scheme unconditionally stable no

oscillations will occur even for very large time steps.

For all schemes in which , it is conditionally stable.

The stability requirement is2 1 1

1,

2

<

( )1/2

2max 1 2

1

2crit t

=

where is the maximum natural frequency of the eigenvalue

problem:max

= MK



37/50

37


N. Zabaras (11/10/2009)


Stability does not necessarily mean accuracy.

Results with a fully implicit integration scheme ( )

and a large time step are shown below:1 2 1= =

0.5t =

This shows a different problem energy is dissipated due to

the numerical time integration scheme.

So a larger value of buys stability by introducing artificial

damping, at the expense of a loss of accuracy. A good

compromise is to set

1 2 1/ 2= =



38/50

38


N. Zabaras (11/10/2009)


The following schemes are special cases:

Constant-average acceleration method (stable)

Linear acceleration method (conditionally stable)

Central difference method (conditionally stable)Galerkin method (stable)

Backward difference method (stable)

1 2

1 1,

2 2 = =

1 2

1 1,

2 3 = =

1 2

1

, 02 = =

1 2

3 8,

2 5 = =

1 2

3, 2

2 = =

Note that the same mesh as that used for the transient

analysis must be used to calculate the critical time step.

Mass Lumping


39/50

39


N. Zabaras (11/10/2009)

Mass Lumping

Recall from the Newmark time integration scheme,

accelerations are computed by solving a set of linearequations:

( )2

22 1 2 1

11

2 2n n n n n

tt t+ +

+ = + + +

M K u K u u u f

The mass matrix derived from the weak formulation of the

governing equation, i.e. is called the consistent

mass matrix, and it is symmetric positive-definite andnondiagonal.

e

e T d

= N N

Solution of this equation is the most time-consuming part of

the procedure. But notice that if we set and somehowfind a way to make the mass matrix diagonal, then the

equation above becomes trivial and thus will lead to saving

computational time.

2 0=

Mass Lumping


40/50

40


N. Zabaras (11/10/2009)

Mass Lumping

There are several ways of constructing diagonal mass

matrices, also known as lumped mass matrices. The row-sum and diagonal scalingare discussed here.

In all of these, the essential requirement of mass

preservation is satisfied, i.e.

e

e

aaa

d = M where is the diagonal component of the lumped mass

matrix .aaM

M

Mass Lumping


41/50

41


N. Zabaras (11/10/2009)

Mass Lumping

Row-Sum Lumping:

The sum of the elements of each row of the consistent mass

matrix is used as the diagonal element:

, 0,aa ab abb a b= = M M M

It can be computed as:

1

e ene e e

aa a b ab

N N d N d = = = M

where the property of the interpolation functions is

used. 11

n

b

b

N

=

=

Mass Lumping


42/50

42


N. Zabaras (11/10/2009)

Mass Lumping

Diagonal scaling procedure:

Here the diagonal elements of the lumped mass matrix are

computed to be proportional to the diagonal elements of the

consistent mass matrix while conserving the total mass of

the element, 0,aa aa abc a b= = M M M

with the constant c to satisfy ee

aaa

d = M It can be computed as:

1

,e e en

e e e eaa a a a a

a

c N N d c d N N d =

= = M For constant , this method gives the same lumped mass

matrix as those obtained in the row-sum technique for the

Lagrange linear and quadratic elements.

Mass Lumping


43/50

43


N. Zabaras (11/10/2009)

Mass Lumping

The use of a lumped mass matrix in transient analysis can

save computational time in two ways.

First, for explicit scheme, lumped mass matrix results in

explicit algebraic equations not requiring matrix inversions.

Second, the critical time step required for conditionally stable

scheme is larger, and hence less computational time is

required when lumped mass matrix is used. But this may

lead to some accuracy loss.

Explicit time integration is cheap, easy to implement and is

therefore a very popular technique. Its disadvantages arethat it is conditionally stable and can require very small time

steps.

Transient Field Problems


44/50

44


N. Zabaras (11/10/2009)

Transient Field Problems

A variety of transient field problems, such as heat and fluid

flow, are governed by a differential equations of the followingform:

( ) ( , , ), inu

c k u f x y t t

=

with the boundary conditions

( ) or on 0

n nu u q q t = =

The initial conditions (i.e. at t=0) are of the form

( )0( , ,0) , inu x y u x y=

Semidiscrete Finite Element Model


45/50

45


N. Zabaras (11/10/2009)

Semidiscrete Finite Element Model

In selecting the approximation for , once again we assume

that the time dependence can be separated from the spacevariation,

u

( )( , , ) ( , )e e ex y t x y t =u N d

Then the semidiscrete finite element model is

( ) ( ) ( )t t t+ =Mu Ku f where

T

e

e e e ed= K B D B

T T

e e

t

e e enq d fd

= + f N N

e

e Tm d

= M N N

Eigenvalue Analysis


46/50

46


N. Zabaras (11/10/2009)

Eigenvalue Analysis

0+ =Mu Ku

The problem of finding such that the following

equation holds is called an eigenvalue problem

te

=u

We obtain

) 0+ =M K Rearranging terms, we also have the generalized eigenvalueproblem

= MK The order of the matrix equations is , where is thenumber of nodes at which the solution is not known.

N N N

Before solving the eigenvalue problem, you need to modify

the matrices to impose the essential boundary conditions.

Transient Analysis


47/50

47


N. Zabaras (11/10/2009)

Transient Analysis

The most commonly used scheme for transient analysis is

the -family of approximation in which a weighted average ofthe time derivatives at two consecutive time step is

approximated by linear interpolation of the values of the

variable at the two steps:

( ) 111 for 0 1n n

n nt

++

+ =

u uu u

or

( )

1

11

n n n

n n n

+ +

+ +

= +

= +

u u u

u u u

Numerical Time Integration Scheme


48/50

48


N. Zabaras (11/10/2009)


Since the semidiscrete finite element mode is valid for any

t>0, we can write it for times and :nt t= 1nt t +=

1 1 1

n n n

n n n+ + +

+ =

+ =

Mu Ku f

Mu Ku f

Substituting and intonu

1

11 1 1

( )

( )

n n n

n n n

+ + +

= +

= +

u M Ku f

u M Ku f

1n+u

( )1

11n n

n n t

+

+

+ =

u uu u

we arrive at

( )( ) ( )1 1 11 ( )

n n n n n nt

+ + +

+ + + =

Ku f Ku f u u



49/50

49


N. Zabaras (11/10/2009)


Solve for vector , we have1n+u

1 1

n n+ +=Ku f

where

1 n t+ = + K M K

( )[ ] [ ]1 1 1 (1 )n n n nt t+ += + + f f f M K u After assembly and imposing boundary conditions, this

equation is solved at each time step for the nodal values 1n+u



50/50


Numerical Time Integration Scheme For different values of, we obtain the following well-known

time integration schemes:

2

0, the forward difference (or Euler) scheme (conditionally stable);order of accuracy = ( t)

1, the Crank-Nicolson scheme ( unconditonally stable); ( t)

2

2 , the Galerkin method (unconditionally s3

=

2table); ( t)

1, the backward difference scheme (unconditionally stable); ( t)

For all numerical schemes in which < , the -family of

approximations is stable only if the time step satisfies the

following stability condition

( )

2

1 2

crit t =

where is the largest eigenvalue associated with theproblem = MK

fem zabaras dynamics

Documents