fem zabaras dynamics
TRANSCRIPT
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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 -
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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.
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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
=
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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
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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
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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
=
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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
= =
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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:
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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.
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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
= = =
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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.
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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
= = =
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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
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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
=
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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
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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
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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:
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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
=
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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
N x y a b x c y x y x y y y x x x yA A
N x y a b x c y x y x y y y x x x yA A
= + + = + +
= + + = + +
= + + = + +
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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
=
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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.
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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.
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Free Dynamic Vibration Analysis
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
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Free Dynamic Vibration Analysis
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
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Free Dynamic Vibration Analysis
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
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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
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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
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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
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Newmark Time Integration Methods
( )[ ]( )[ ]
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+
N k Ti I t ti M th d
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Newmark Time Integration Methods
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
N k Ti I t ti M th d
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Newmark Time Integration Methods
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
N k Ti I t ti M th d
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Newmark Time Integration Methods
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
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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
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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.
Stability and Accuracy
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Stability and Accuracy
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.
Stability and Accuracy
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Stability and Accuracy
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
Stability and Accuracy
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Stability and Accuracy
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= =
Stability and Accuracy
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Stability and Accuracy
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
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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
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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
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MAE 4700 FE Analysis for Mechanical & Aerospace Design
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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
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MAE 4700 FE Analysis for Mechanical & Aerospace Design
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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
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MAE 4700 FE Analysis for Mechanical & Aerospace Design
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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
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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
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MAE 4700 FE Analysis for Mechanical & Aerospace Design
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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
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MAE 4700 FE Analysis for Mechanical & Aerospace Design
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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
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MAE 4700 FE Analysis for Mechanical & Aerospace Design
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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
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MAE 4700 FE Analysis for Mechanical & Aerospace Design
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Numerical Time Integration Scheme
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
Numerical Time Integration Scheme
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MAE 4700 FE Analysis for Mechanical & Aerospace Design
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Numerical Time Integration Scheme
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
Numerical Time Integration Scheme
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MAE 4700 FE Analysis for Mechanical & Aerospace Design
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