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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 reads

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

Login or register to post comments

goodSubmitted by safaei on Mon, 2009-11-09 08:04.

thanks


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?


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


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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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


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




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


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



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


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


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?



Respects,

kajal


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




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


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?



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?


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


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


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?



Hope someone here could give me a response. Thanks!

Julian


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


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


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???


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)



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.


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



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?


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


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


Thanks a lot dear yawlou.Submitted by mahmood seraji on Mon, 2012-10-15 23:28.

Thanks a lot dear yawlou.


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



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

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