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Implicit and Explicit finite element method
Submitted by vijay310105 on Mon, 2009-05-04 13:24. education
i am not clear about implicit and explicit FEM
kindly let me know the difference of two and where to use which
one
vijay310105's blog Login or register to post comments 89249
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difference between explicit and implicit FEMSubmitted by yawlou
on Mon, 2009-05-04 14:34.
Hello,
The following is how I understand it. I have done some of both
in
graduate school. The following is a brief answer, since this
would
take a lot to answer thoroughly.
1. Preliminary comments regarding the incremental nature of
Explicit and
Implicit Analysis
A geometric and/or material nonlinear analysis requires
incremental load (or
displacement) steps. At the end of each increment the structure
geometry
changes and possibly the material is nonlinear or the material
has yielded.
Each of these things, geometry change or material change, may
then need to
be considered as you update your stiffness matrix for the next
increment in the
analysis.
2. Explicit
An Explicit FEM analysis does the incremental procedure and at
the end of each
increment updates the stiffness matrix based on geometry changes
(if
applicable) and material changes (if applicable). Then a new
stiffness matrix is
constructed and the next increment of load (or displacement) is
applied to the
system. In this type of analysis the hope is that if the
increments are small
enough the results will be accurate. One problem with this
method is that you
do need many small increments for good accuracy and it is time
consuming. If
the number of increments are not sufficient the solution tends
to drift from the
correct solution. Futhermore this type of analysis cannot solve
some
problems. Unless it is quite sophisticated it will not
successfully do cyclic
loading and will not handle problems of snap through or snap
back. Perhaps
most importantly, this method does not enforce equilibrium of
the internal
structure forces with the externally applied loads.
3. Implicit
An Implicit FEM analysis is the same as Explicit with the
addition that after each
increment the analysis does Newton-Raphson iterations to enforce
equilibrium
of the internal structure forces with the externally applied
loads. The
equilibirium is usually enforced to some user specified
tolerance. So this is the
primary difference between the two types of anlysis, Implicit
uses Newton-
Raphson iterations to enforce equilibrium. This type of analysis
tends to be
more accurate and can take somewhat bigger increment steps.
Also, this type
of analysis can handle problems better such as cyclic loading,
snap through,
and snap back so long as sophisticated control methods such as
arc length
control or generalized displacement control are used. One draw
back of the
method is that during the Newton-Raphson iterations one must
update and
reconstruct the stiffness matrix for each iteration. This can be
computationally
costly. (As a result there are other techniques that try to
avoid this cost by
using Modified Newton-Raphson methods.) If done correctly the
Newton-
Raphson iterations will have a quadratic rate of convergence
which is very
desireable.
A suggestion. If you'd like to learn further about these two
techniques it
would be instructive for you to use both techniques and compare
on the same
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problem. Explicit can be done by simply turning off
Newton-Raphson iterations
in an Implicit routine, or by setting the equilibrium tolerance
to a large number
in an implicit routine.
As to the question of which method to use, the answer is that it
depends. The
type of analysis that is sufficient for your needs will depend
on the type of
problem that you are trying to solve. Often times since dynamic
analyses are
computationally intensive they are done with the explicit
method. However,
for static problems now days it is becoming more common to do
the full Implicit
type of analysis.
Nonlinear analysis takes lots of experience and a careful
understanding of
what you want to accomplish and also a careful understanding of
the anlaysis
capabilities of the software you are trying to use. As I
mentioned I have
worked with the above methods of analysis in graduate school and
know a
little about it, however, I would be happy for others here at
iMechanica who
have more experience than me to give their thoughts on this as
well.
It is indeed a very big topic that is difficult to cover in just
a brief blog. You
should consider looking at Crisfield's book volume 1 for
additional information.
Also, look at the following location for nonlinear fem
information
http://www.colorado.edu/engineering/CAS/courses.d/NFEM.d/Home.html
I hope this helps,
Louie
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goodSubmitted by safaei on Mon, 2009-11-09 08:04.
thanks
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thanks a lot for theSubmitted by sriramk on Fri, 2012-02-24
23:54.
thanks a lot for the information that u have posted.
using abaqus/explicit is that possible to do machining
simulation?
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Machining simulation is indeed possible
withABAQUS/Explicit.Submitted by nas on Wed, 2012-09-12 07:24.
Hi, It is ofcourse possible to simulate machining process. I
have
performed simulation of drilling in ABAQUS/Explicit
Login or register to post comments
Thanks a lot for yourSubmitted by nikohj on Mon, 2009-05-04
20:58.
Thanks a lot for your explanation.
It is also quite useful for me.
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5/21/2014 Implicit and Explicit finite element method |
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Implicite vs Explicit solutionSubmitted by Peyman Khosravi on
Fri, 2009-05-08 18:56.
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Hi Vijay;
Generally there are two methods to solve a dynamic
equilibrium equation at every time step. One method is to
predict the solution at
time t+dt by using the solution at time t. This is called
explicit. In this
method, one does not need to inverse the stiffness matrix (see
the book Finite
Element Procedures, 1996,
Bathe, page 770 for explanation). This may seem at first a good
method, however
one should note that it is not stable (i.e. it diverges from the
correct
answer) unless the time step is very small. This is why it is
called
conditionally stable. So it is used only when the time duration
of the problem
is short (like crash problems).
On the other hand, one can solve the equation at time t+dt
based on itself, and also using the solution which has been
found for time t. This is
called
implicit, and the most famous one is Newmark method. In this
case you need to
inverse the stiffness matrix (because of
the nature of the equations), however since it can be
unconditionally stable you
may be able to choose a larger time step. This is a great
advantage, which
enables us to finish the problem faster.
For more information refer to the above book.
Cheers
Peyman
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the words explicit and implicit used in dynamicsand
staticsSubmitted by yawlou on Thu, 2009-05-07 20:29.
Thanks Peyman
I'm glad that Peyman has added his comments. My experience
using the words explicit and implicit has been in the context of
static
problems. However, Peyman has wisely added comments for the case
when
the words explicit and implicit are used in a dynamic context.
This added
information provided by Peyman makes the discussion more
complete.
regards,
Louie
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Question for LouieSubmitted by kajalschopra on Fri, 2009-05-08
16:57.
Dear Sir,
While you were talking about "Explicit" analysis above, you
said:
"Perhaps most importantly, this method does not enforce
equilibrium of the
internal structure forces with the externally applied loads.
"
I have a simple question (pleae excuse me if fundamentally
wrong)
While we upgrade the stiffness matrix, following geometry or/and
material
change (if applicable), we solve F = ku, which indeed is the
equilibrium
equation, then, how ican we say equilibrium is not enforced?
Thanks,
kajal
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answer for kajalSubmitted by yawlou on Fri, 2009-05-08
19:04.
Hi Kajal,
In an incremental analysis the equation F=kU is really F=
(kcurrent)u, so you apply an incremental force and get an
incremental corresponding displacement using the current
stiffness
matrix. The current stiffness matrix is based on the current
material properties
and possibly the current geometry if significant geometry
changes have taken
place.
If you sum up all the incremental externally applied forces and
compare them
to the current internal forces they will not be in
equilibrium.
You are asking a very good question that is one of the very
fundamental
issues of a nonlinear static analysis. The fact that the forces
are not in
equilibrium, between internal and external forces, is the very
reason that we
must use Newton-Raphson iterations to correct each u until
external and
internal forces are in equilibrium for the load step we are
currently working on.
As a first step to understanding this it would be helpful for
you to go over the
Newton-Raphson method for a single nonlinear equation f(x). You
may find
this in standard numerical analysis text books. Then usually the
books also
discuss how to extend the method to a set of simultaneous
nonlinear
equations. Essentially, we are solving F=F(u), where if we
linearize this
equation the stiffness matrix k(u) is a nonlinear function of u.
We keep
updating it and it does not stay constant. Hence, the sum of our
incremental
internal forces are not equal to the sum of the incrementally
applied external
forces.
This is indeed a difficult thing to explain in words. You will
need to consult
some nonlinear finite element textbooks and ponder this for a
while. Perhaps
do a single bar in tension with a geometrically nonlinear
stiffness equal to
k(u)=1+u^2,(therefore since df/du=k(u), by integration it
follows that
f(u)=u+u^3/3). Apply three equal load increments of 1 unit each.
The results
are as follows by repeatedly using the relation f=k(u)u:
Step1, uo=0, k(uo)=1, f=1, u1=f/k(uo)=1
Step2, u1=uo+u1=1, k(u1)=2, f=1, u2=f/k(u1)=1/2
Step3, u2=u1+u2=1.5, k(u2)=3.25, f=1, u3=f/k(u2)=1/3.25
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u3=u2+u31.81
The sum of external applied forces is 1+1+1=3=Fext
The current internal forces are f(u3)=Fint=(u3+(u3)^3/3)3.79
Hence it is clear that FextFint, not in equilibrium.
To be clear, my example is an illustration of an explicit
analysis with just 3 load
steps. If instead we applied 6 load steps of 0.5 units of force
Fint and Fext
would be closer to each other in value. For an Implicit
analysis, using Newton-
Raphson iterations, Fint would be equal to Fext to the precision
that we
specify/ or at best to the precision of the computer.
I hope this helps and hopefully I didn't make any computational
mistakes
above.
regards,
Louie
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A correction to my previous commentSubmitted by Peyman Khosravi
on Fri, 2009-05-08 19:11.
Thanks Louie;
I would like to correct my previous comment (I edited that):
Regarding the implicit and explicit dynamic analysis methods,
no
Newton-Raphson iteration is required unless we deal with a
problem which is nonlinear in its nature e.g. geometrically
nonlinear problems.
So the main difference as I said is in the stability of the
method and required
time step and inversing stiffeness matrix. Overall, explicite
methods are not
usually recommended.
Another good reference is chapter 20 of the book written by
"Edward L.
Wilson" which is available on his website for free, however
instead of t and
t+dt he uses t-dt and t. If you want to use this reference,
remember that there
used to be some typos in the coefficients of table 20.1. So
check it before
using it.
Thanks
Login or register to post comments
questions on louie's exampleSubmitted by kajalschopra on Sat,
2009-05-09 13:02.
Dear Sir,
Thank you very much for the reply.
You said:
"----Perhaps do a single bar in tension with a geometrically
nonlinear stiffness
equal to k(u)=1+u^2,(therefore since df/du=k(u), by integration
it follows that
f(u)=u+u^3/3).----"
Sorry, if what i am asking is too stupid:
I understand that by integrating we get f(u) = u + u^3/3--here
f(u) is the
internal force.
1) Can you show me the internal force in the bar at every step
of incremental
displacement- without using the integrated formula.In other
words I do not
want that f(u1),f(u2),f(u3) to come from using the integrated
formula-I want to
physically use summation to get final internal force.Please
excuse me for
asking a entry level question.
2)Basically, with increase in displacement, the stiffness of the
bar should
reduce-where in your example is this degradation of stiffness
reflected?Can
you show me the reduction in stiffness at Step 1, step 2, step
3?
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Respects,
kajal
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just an example function to illustrate explicitanalysisSubmitted
by yawlou on Sun, 2009-05-10 14:27.
Hi Kajal,
Answers to your questions:
1) For the example as given, one cannot calculate the internal
force without
using the equation. The problem would need to be more elaborate
and have
strains calculated and stress strain relations prescribed as is
done in a normal
solid mechanics finite element problem. Then the internal
stresses could be
calculated from those relations. From the internal stresses the
internal forces
may be obtained. This is often done by using the common
expression "integral
of B transpose times sigma".
2)There is no law or requirement that says the stiffness must
degrade. In fact,
for the example I have given, the stiffness for this problem
increases with
displacement. Stiffening with increasing displacement is
actually possible in
real problems, so the problem is not too farfetched.
Additional Comments/observations
1. The example given was constructed to demonstrate an
incremental explicit
analysis for a simple static problem. The 3 steps provided
illustrate the
incremental approach and the final comparison between the
external applied
loads and the final internal force showing that the explicit
analysis does not
stay in equilibrium. More load steps would be required to
achieve better
equilibrium or an implicit analysis would be needed to enforce
equilibrium by
using Newton-Raphson iterations.
2. I say that the example is for a geometrically nonlinear
analysis, however,
there is really nothing here that establishes it as
geometrically nonlinear. We
could just as easily have said it was materially nonlinear, for
this problem.
What is most important for the purposes of this example is that
F is a
NONLINEAR function of displacement. An explicit analysis is a
way to
approximately analyze the nonlinear problem as illustrated. It
would be
helpful for you to plot load versus displacement comparing the
exact solution
to the explicit data points that are calculated. A simple matlab
problem can be
constructed to do this and it will illustrate how increasing the
number of
increments can improve the results.
3. The problem is just a contrived example where force is GIVEN
as a
nonlinear function of displacement. In general in a nonlinear
finite element
analysis we do not have an exact closed form expression for the
load versus
displacement. Instead we would have to keep track of the strains
in the
structure we are analyzing and from those internal strains we
may calculate
the internal stresses and from the stresses we may calculate the
internal
forces. Theses steps are not shown in the simple problem I have
posed
above. The problem above only illustrates many of the relevant
concepts
associated with force, stiffness, explicit, implicit, the
incremental nature of the
solution procedure, external forces, internal forces, lack of
equilibrium, and if
someone does more steps it illustrates the improvement of
results between
Fext and Fint.
I hope this helps,
Louie
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explicit and implicit at global level in staticcontextSubmitted
by yawlou on Fri, 2009-05-29 20:52.
Several comments.
1. To be complete it may be helpful to recognize that the
explicit
and implicit methods I have described in previous posts above
are for static
problems at the global level. (Peyman has mentioned explicit and
implicit in a
dynamic context) At the constitutive level (at a material point
in the structure)
it is also possible to have either explicit or implicit computer
routines that solve
for the amount of plastic flow that the material point has
undergone (in the
case of plasticity). This process at the constitutive level is
at a local material
point and I have not discussed that in my previous blogs.
Therefore, it is
helpful to realize that explicit and implicit may be used in a
variety of contexts.
2. For the simple problem I have mentioned above I have created
a small
tutorial that describes the explicit method and the implicit
method. There are
also matlab files(which go with the tutorial) I have constructed
which
demonstrate the explicit and implicit methods for the simple
problem. The
tutorial is fairly basic and perhaps I will find time to improve
it over time, but for
now it is short and to the point. The files may be found at the
following
location.
http://people.wallawalla.edu/~louie.yaw/nonlinear/
This is for iMechanica and all individuals who have a sincere
desire to learn.
regards,
Louie
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Solving large plastic deformation quasi-static problemsSubmitted
by Sidhu on Tue, 2010-08-24 07:29.
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Why are implicit
solvers (implicit FEM software) such as MSC.Marc, Deform, Forge,
are preferred
(dominates) to explicit FEM (LS-Dyna) in solving large
deformation quasi-static
problems such as e.g. in case of metal forming with an
expectation sheet metal
forming/IHU?
-
5/21/2014 Implicit and Explicit finite element method |
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Are explicit
solvers (LS-Dyna) are faster in solving large plastic
deformation quasi-static
problems? How about the accuracy of the results when compared
with implicit
solvers?
Are explicit solvers or contact algorithms used in
explicit FEM (LS-Dyna) faster in Elasto-Plastic slide contact
with friction?
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Thank you Prof LouieSubmitted by karthic_newbee on Tue,
2010-08-24 21:21.
Dear Prof Louie
Your handouts are a treat to read. Very nicely written and clear
explanations
are given.
Thank you very much for enhancing our knowledge
Karthic
NTU, singapore
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Few related queriesSubmitted by sagarfea on Tue, 2011-05-24
00:02.
Hello everyone,
Can someone please guide me on these queries, so that it will be
usefull toall:
1. When we should go for Explicit Analysis(few thumb rules).
2. What is the difference between Implicit Dynamics and
ExplicitDynamics. (is Explicit analysis is prefered for all dynamic
problems??)
Waiting for the answer.......!
Sagar
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stiffness matrix in explicit methodSubmitted by julianxqwang on
Fri, 2011-08-05 02:23.
Hi Everyone,
I have been trying to use DYNA3D for dynamic analysis of very
flexible
structures. The discussion here is very instructive. I have one
question: in the
explicit method, is the tangent or the nonlinear stiffness
matrix explicitly
generated so that it can be output or not? If not, is there any
type of stiffness
matrix used in the solution process and can be output?
-
5/21/2014 Implicit and Explicit finite element method |
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Hope someone here could give me a response. Thanks!
Julian
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Stiffness matrix in explicit methodSubmitted by Nachiket Gokhale
on Fri, 2011-08-05 10:06.
No, typically tangent matrices are not explicitly generated. I'd
be surprised if
Dyna3D does this. -Nachiket
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Limitation of ABAQUS for explicit and implicitMethodSubmitted by
manojsinghkiran on Fri, 2012-02-17 07:32.
Hi everyone
I am doing resaech in Damahe Mechanics .
I want to Know , How the result very in Explicit and implicit
FEM in ABAQUS.
What is limitation to use ABAQUS at high velocity Impact .
With Regards
Manoj Kumar
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Staitc riks and arc lengthSubmitted by shijo_iitr on Wed,
2012-03-07 02:04.
can someone help me with static riks method and the associated
option of arc
length???
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explicit and implicit methodsSubmitted by mohamadzare on Thu,
2013-01-03 00:55.
as we know finite element method is a method forsolving
gifferential equations that governed tophysical problem. beyond
many of engineeringproblems, is a certain differential equation
governs that. for example consider heat transfer in a longrod that
governing equation is "Q/t=k*2 Q/x2" (0) that Q is temprature and t
is time and x is coordinatealong the rod. by using taylor
aproximation we can write:
2 Q/x2 =(Q(x+x)-2*Q(x)+Q(x-x))/x2 (1)
proof:Q(x+x)=Q(x)+Q'(x)*x+(Q''(x)/2)*x2
Q(x-x)=Q(x)-Q'(x)*x+(Q''(x)/2)*(-x)2
=Q(x)-Q'(x)*x+(Q''(x)/2)*x2
putting this data to equation (1), we can write: 2
Q/x2=Q''(x)
and Q/t=(Q(x,t+t)-Q(x,t))/t (2)
main difference between explicit and implicit method startin
here that in the explicit method we calculate equation(0)
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in time=t. by using explicit formulation we can write:
2 Q/x2
=(Q(x+x)t-2*Q(x)t+Q(x-x)t)/x2=(1/k)(Q(x,t+t)-Q(x,t))/t
lets we name Q(x,t)=Qmt and Q(x+x,t)=Qm+1 t Q(x,t+t)=Qm t+1 that
super script t refer time andsubscript m refer to segment that we
consider for solve theequation(m= a certain element in the rod that
is created bymeshing in the finite element method).
then we can write :
k*(Qm+1 t -2*Qmt + Qm-1 t )/x2= (Qm t+1 -Qmt )/ t (k*t/x2)*(Qm+1
t -2*Qmt + Qm-1 t) +Qmt=Qm t+1 (3)
that as see temprature at time t+1 is explicitly calculate,cause
we know all quantities at left side of equation (3) attime t(for
example we know the temprature at the end andstart of the rod at
time t=0 from boundary condition). asshow in here this is a
progressive procedure that dontsatisfy the equilibirium of
temperature in all of the rod, butsatisfy this in certain
level.
what about the implicit method?
in the implicit method the equation (0) is solved at timet+t.
let thinking about this.
we can write equation (1) at time t+t (or for
simplicityt+1):
k*(Qm+1 t+1 -2*Qmt+1 + Qm-1 t+1 )/x2= (Qm t+1 -Qmt )/ t
(k*t/x2)*(Qm+1 t+1 -2*Qmt+1 + Qm-1t+1) +Qmt=Qm t+1 (4)
as see in above equatin (4) for solving this equation andobtain
temperature at time t+1 at certain location, we needto know all
quantities at the left side of equation (4) intime t+1. cause we
dont know at first this quantities attime t+1 (our oubject really
is obtain this values at timet+1), we should solve a set of
equation to obtain solution.by enforcing boundary condition at the
end and start of therod, we can solve heat transfer equation. as
you see usingthe implicit method satisfy the heat equilibirium at
alllocation of the rod.since we must solve a set of equation inthis
method,the calculation time is more than explicitmethod,
particulary when our problem's domain is relativelylarge. really in
heat conduction problem, if we devide therod to n element, we
should solve n simultaneous equationto obtaing solution in each
time steps. if our problem be 2or 3 dimensional problem, and in
each nodes of the domainafter meshing, we have certain degrees of
freedom, theadvantages of using explicite metod apear. cause
forimplicit method we should solve (n=number of
nodes)*(dof)simultaneous equation for obtain solution at each
step.
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Explicit vs Implicit for material properties varying
withtimeSubmitted by BMEstudent on Tue, 2012-07-03 19:09.
Hi all,
I followed comments on this topic. I know about mathematical
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5/21/2014 Implicit and Explicit finite element method |
iMechanica
http://imechanica.org/node/5396 11/12
background of both methods and little bit and stability and
conditional
stability of these methods. I have a general question about
material
nonlinearity and choosing between one of these methods.
I'm modeling a viscoelastic material in Abaqus. As you know
this
type of material is history dependent material and based on this
fact my
supervisor says since in explicit method, we update material
properties in each
step directly from previous step, we should use explicit
procedure and not
implicit one. Mathematically I think, providing proper time
steps for both
methods, the results should not differ but I cannot answer
philosophically my
supervisor! What do you guys think?
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Crisfield's bookSubmitted by mahmood seraji on Wed, 2012-09-12
21:32.
Dear yawlou
Would you please give some more information about the introduced
book of
(Crisfield's book volume 1)?
Thanks
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Crisfield's booksSubmitted by yawlou on Wed, 2012-09-26
17:23.
The Crisfield books I mentioned are as follows:
M. A. Crisfield, Essentials, Volume 1, Non-LinearFinite Element
Analysis of Solids and Structures,Wiley, 1996
M. A. Crisfield, Advanced Topics, Volume 2, Non-LinearFinite
Element Analysis of Solids and Structures, Wiley,1997
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Thanks a lot dear yawlou.Submitted by mahmood seraji on Mon,
2012-10-15 23:28.
Thanks a lot dear yawlou.
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Arc length methodSubmitted by irushdie on Wed, 2012-09-26
00:11.
Can anyone please tell me the alogotithm for E Ramm's arc length
method. I
am using the following type of algorithm but am not getting the
correct result--
--
u{0}, lamda=0
Fext
n no of steps
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5/21/2014 Implicit and Explicit finite element method |
iMechanica
http://imechanica.org/node/5396 12/12
Kg (stiffness with u{0})
u1=(Kg)^-1 * Fext'
[L D]=ldl(Kg)
m=det(D)
arc_length=norm(u1)
lamda= 0.1*arc_length/sqrt(u1'*u1+1)
if m