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Co-rotational formulation for trusses

Hi, I have been reading a chapter on Co-rotational Formulation for trusses in Non-Linear Finite Element Analysis. And, I am here with a question which a very basic question concerning linear analysis. 

Now, a co-rotational formulation for trusses (say) separates the rigid motion from the motion of the truss and what remains is strain causing the deformation of the truss element. My basic question is, that, while doing linear analysis, is it that, the displacements which we obtain in case of trusses or even beams include rigid body motion as well? 

In that case, if the displacements we obtain from linear analysis include rigid body motion as well, then, all our internal forces are incorrect as far as linear analysis is concerned? Because, if motion includes rigid body motion, then, the strain obtained includes rigid body motion and hence all internal forces are not in fact functions of deformation as they should be? Please help/clarify.

yawlou's picture

For a linear analysis to be valid, displacements and rotations must be small.  As long as this restriction is maintained the rigid body effects are negligible in a linear analysis.

 It is when displacements or rotations get large that rigid body effects causes errors if they are not properly accounted for.

 (I believe this is the correct answer.  If this is incorrect, somebody please do correct me.  But, I believe my response gets directly to the answer.)

 regards,

Louie

Thanks sir, that means, linear analysis includes rigid body motion as well (though their magnituse is small and negligible)?

yawlou's picture

Yes, although there may be more to it than just a simple yes.

Salwan

I'm just recalling the derivation of the stiffness matrix of a truss element - linear analysis:

we apply a unit displacement at node 1 and measure forces at nodes 1 and 2.These are AE/L and -AE/L respectively.Similarly, apply unit disp;acement at node 2 and measure forces at 1 and 2.

So element stiffness matrix is k11 = AE/L ; k12=k21= - AE/L; k22= AE/L.

Where are the rigid body displacements accomodated here?

Also, here we can say the element stiffness is a result of attaching a local element reference frame as in co-rotational?

Sir, also, see the attachment, co-rotational crisfield attached in the first post of this thread.

As I see, he has just taken a coordinate axis along the axis of the element , then, got element stiffness and then carried out transformation.

This is what we do in linear analysis as above.Then, how does linear analysis not include rigid body displacements?

yawlou's picture

It is difficult for me to answer your question succintly.  I would be interested to see someone more experienced then me answer your question.  Your questions are good questions.

The most direct answer to your question is as follows:

1.  a linear analysis is valid only for small displacements.

2.  if displacements get too large, in a linear analysis, the rigid body motions cause erroneous strains and hence erroneous stresses.

It is not that a linear analysis does not included rigid body motions, it is that a linear analysis does not either update the geometry or subtract rigid body motions to get the true strains.

You should analyze a cantilever beam using a linear analysis.  Then do the same beam with a corotational analysis.  Observe the difference when you cause large vertical displacements.  You will start to see the linear analysis diverge from what happens in reality at large displacements quite quickly.  (for example the tip displacement in the horizontal direction stays zero in a linear analysis, this is physically clearly wrong.)

 

 

Sir, my question is that in co-rotational analysis (derivation of crisfield also attached in the first post in this thread); we get element stiffness and then carry out transformation.

The same procedure is also carried out while analyssing a truss using linear analysis.I mean to say ; since, we have a coordinate system along the axis of the element both in co-rotational analysis as well as linear analysis;then;a linear analysis ALSO DOES NOT INCLUDE RIGID BODY DISPLACEMENTS AS CO-ROTATIONAL ANALYSIS.

yawlou's picture

Kajal,

What you have said above is incorrect I believe for the following reason.

1.  The linear analysis does do the same transformation, but it does it only once and does not update the angles for the cosine and sine terms as in a corotational analysis.

2.  The linear analysis is done in one step, whereas a corotational analysis is incremental due to the geometry changing and the need for updating the geometry during each step.

3.  Also the linear analysis does not check equilibrium at the end, whereas a corotational analysis does and potentially makes corrections by a newtonraphson procedure.

 The simple transformation in a linear analysis does not remove the rigid body motions. 

regards,

Louie

likask's picture

Recently I publish paper about co-rotational formulation applied to analysis of fracture in solids. Formulation is tailored to the problem, however simple example in section 7.1 shows what is difference between small and large/moderate rotations. 

http://dx.doi.org/10.1016/j.cma.2008.11.018

Regards,

Lukasz 

 

 

Sir:

Yes, I understand that linear analysis does the tarnsformation only once and also does not check equilibrium.And the result on the whole are not as realistic like non-linear-corotational.

As you said, linear analysis does transformation once- a single step analysis- but in that single step the rigid body motion is NOT included.Suppose, if I had a queston that , one has to 'separate' the rigid body component from the deformation obtained from linear analysis, then can I say that , deformaton (obtained in single step) does not include rigid body component?

 

yawlou's picture

Kajal,

To answer your question, NO.

 Prove to yourself that a linear analysis DOES include rigid body motion.

1.  Do a linear analysis on a truss that is 20 feet long.

2.  Make the joints at 5 feet on center.

3.  At the left end make the truss haved pinned supported joints so the truss is like a cantilever.

4.  At 10 feet put a vertical point load(do not put loads anywhere else).

 

5.  When the truss deflects, all truss members to the right of the point load should have no stress in them.  The free truss joints will have displacements.  Even the truss members to the right of the point load will have displacements.  Hence, those members to the right of the point load will have moved rigidly.

Therefore, a linear analysis does include rigid body motion.

If a linear analysis did not include rigid body motion, then there would be no need to remove rigid body motions during each step in an incremental corotational analysis.  I'm not sure why you keep insisting that a linear analysis does not include rigid body motion.  It does include rigid body motion.  Perhaps I do not understand your question.

 regards,

 

Louie

 

 

Dear Sir,

Yes, I now understand from the points 1 through 5 above  with reference to the example you've cited that linear analysis DOES include rigid body motons.

You also said,

"---If a linear analysis did not include rigid body motion, then there would be no need to remove rigid body motions during each step in an incremental corotational analysis.--"

Can you tell me in which step we remove the rigid body motions in co-rotational analysis?

You can mention with reference to your paper below or with reference to Crisfields derivation attached at the first thread of this post:

http://people.wallawalla.edu/~louie.yaw/Co-rotational_docs/2Dcorot_truss.pdf

Kajal

yawlou's picture

Kajal,

 

Equation 3 in my truss paper calculates the local axial displacement only.  L is based on global displacements which includes rigid body displacements in addition to structural deformations. Calculating d=L-Lo determines the local strain causing deformations only.  Hence, d does not included the rigid body displacements or rotations. Hence, equation 3 is where the rigid body displacements are removed.

Incidentally, calculating d as (L^2-Lo^2)/(L+Lo)  is a better conditioned formula as indicated by Crisfield.

regards,

Louie

Sir:

Sorry for replying late as I was unwell from past 3 days.

Can we also say that since in the stiffness matrix formulation, we have considered the change in angle 'theta' the rigid body rotations are accounted for/eliminated?

Because, in ther formulations (i.e. Total Lagrangian) we start with strain (Engineering strain = Final length - Original length / Original Length) but never accounted for change in theta?

Kajal

Peyman Khosravi's picture

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Hi;

First I should say that I enjoyed this discussion and I should thank Louie
for his explanations. Here is my answer based on what I learned in the past;

Rigid body (RB) motion in stress analysis and finite element can only be
understood through equilibrium. Let's put it this way: what is really the definition
of small deflection and large deflection analysis? How small is small enough to
do the small deflection analysis? The whole corotational analysis and problems
that we encounter with them during corotational analysis and convergence are
related to this point.

Small deformation analysis (SDA) and large deformation analysis (LDA) are
different in the way we satisfy equilibrium. If in solving our problem we
consider only one geometry, it is SDA and equilibrium is satisfied for
that geometry. If that geometry is the undeformed geometry, forces found in the
analysis are in equilibrium only on that geometry. Deflections also are not
real because we didn’t consider the real deformed geometry in our analysis. If
defections are small we can say OK they are at least some approximation of the
real deflections otherwise they are totally wrong. An example is a truss
element that has a large rigid body rotation. We can have a RB rotation by
rotating a horizontal truss element about one end (let’s say by 45 degrees) See
Eq. 17 of Louie’s PDF file mentioned above for element stiffness matrix and you
will see that you get some forces in the element. Those forces are not real
(otherwise they should be zero) and don’t satisfy equilibrium on the deformed
geometry, however they satisfy equilibrium on the undeformed geometry.

In geometrically nonlinear analysis we are trying to satisfy equilibrium on
the deformed configuration. This means that we have TWO geometries from the beginning.
You may say OK I know that, but the reality is more complicated than this
because sometimes deformed and undeformed geometries are so close together that
one doesn’t see the necessity for that consideration. Here is more explanation.

Consider a corotational analysis for a general element. The method is
basically to put a copy of the element on its final position and then remove
the rigid body (RB) motion and measure the remaining deformation. One may say at
this stage that the force of the element is [K_linear]*{remaining deformation}.
And also for the next iteration when he needs the tangent stiffeness matrix, he
may add the element stiffness to the conventional stress stiffness matrix (that
is available in any FEM book and is a function of the element internal force)
to get the tangent stiffness matrix. Right? Wrong! The reason is that these stiffness
matrix equations satisfy equilibrium on the undeformed geometry (i.e.
the copy element) and not on the deformed geometry (even though they are very close
together since we have put them over each other). Why is it so important to
satisfy the equilibrium on final geometry? Because otherwise we will have
convergence problem during iterations. Having this problem was the origin of
inventing projector matrix to REALLY PURIFY those seemingly pure deformations
such that they are in equilibrium on the final geometry. (refer to Rankin’s
papers)

In finite element analysis we usually formulate the elements in a way that
they produce no forces under RB motion. However one can easily check that it is
not the case for RB rotations (as we checked before). Bottom line is that if you
ask why it is not working the answer is because this hasn’t been designed to
work for any geometry. SDA only works if you don’t care about the final shape
and satisfying equilibrium on the final shape. What you get from SDA may be
said that includes rigid body motion to some degree, but it is not the correct
deformation. So the RB motion that is found in SDA is also not correct.

Peyman  

Peyman,thanks for putting your poin of view.However, my question is, that,whether, if in the stiffness matrix formulation, which we get by adfifferemtiating the internal force vector (qi), the change in qi accounts for both change due to dsiplacement as well as the angle 'theta', then is it that the rigid body displacements have been eliminated?

kajal

Peyman Khosravi's picture

Kajal;

I don't understand your question. In linear analysis everything is based on UNDEFROMED geometry so even if we talk abaout any change of displacement, those are WITH THE ASSUMTION THAT THE WHOLE THING IS SMALL. so there is no point to talk about RB motion in linear analysis when deformations are large. in this case all the answer is already simply wrong because we assumed they are small but they are not.

peyman

yawlou's picture

Kajal,

You are asking good questions Kajal.  This topic is a bit difficult to explain.  I still struggle with some of these same questions.

I'm glad that Peyman has added his perspectives on the discussion.  He has brought up some very important additional ways to look at the issues.  I particularly like several things that he has pointed out.

1.  It is helpful to examine small displacement analysis and large displacement analysis by looking at them from an equilibrium point of view.

2.  It is difficult to say that rigid body displacements are included or not included in a small displacement analysis, because a small displacement analysis has some inherent flaws in it.  (But, often it is sufficiently accurate for engineering purposes so long as displacements are small.)

It is good that more than one way of trying to explain it has been presented here.  It will be beneficial for you to examine Peyman's comments carefully.

regards,

Louie

 

Thanks Louie,I will go through Peymans response minutely and carefully as you've suggested.

Peyman:

1)Look at Crisfield'd derivation of corotational formulation attached at the first post of this thread.

2)We first get strain along the axis of the element.

3)We then get internal force vector corresponding to the axis of the element.

4)Next, we transform this internal force vector to global axis by multiplying with transformation matrix.

5)Next, we get stiffness matrix by differentiating the transformed internal force vector.

6)Now, when we differentiate this internal force vector, we differentiate partially with respect to the 'theta'- which is incorporated in the transformation matrix and displacement 'p' along the axis.

My question is:

A)We had never before carried out the differentiation of internal force vector with respect to 'theta'- that means in corotational when we do so, the change in stiffness due to 'theta' is accounted and hence the rigid body displacement are removed from the results?Because, surely, it is this differentiation with respect to 'theta' is making all the difference which was never done before.I'm trying to figure out how?

B)Can you answer the question with this perspective-i.e.differentiationg internal force vector with respect to 'theta'-brings about what so special in co-rotational?

Respects,

Kajal

 

Peyman Khosravi's picture

Kajal;

Honestly I still don't understand what the question is. This derivation has teta because it has been transformed to the global coordinate system. Can you explain how you conclude from "change of stiffness due to teta is accounted", that "RB rotations has been removed (or considered)"? 

Peyman

Peyman,

Ok:

1)Can you tell me which step in the derivation removes the RB motions, both:translations and rotations?

2)WE have the internal force vector (qi) and then we obtain the change in qi (i.e. the stiffness matri)due to change in 'theta' and due to change in displacement 'p', right?

3)Suppose, if I do not want to remove RB rotations then what would be the difference in the derivation?

 

kajal

Peyman Khosravi's picture

Kajal;

Rigid body is removed in another step during corotational. Refer to chapter 7 or refer to the Louei's PDF file. removing rigid body rotation and translation is a different story.

Peyman

yawlou's picture

Kajal,

As I mentioned previously above.

 In the corotational truss formulation rigid body rotations and translations are removed when we use equation (3) in my pdf file.

In equation (3) we get L by using the current joint coordinates (after the structure has rotated and translated) and Lo is based on the original coodinates of the truss joints.  Hence, d=L-Lo effectively removes all the rigid body translation and rotation that happened and you end up with just the pure deformation along the axis of the truss member.

I'm not sure why you are trying to find removal of rigid body displacements and rotations somewhere else in the formulation.  It happens in equation (3) for trusses.

regards,

Louie

Peyman and Louie:

1) Louie-as you say that rigid body rotations are removed in equation 3 of your pdf and not anywhere else in the formulation or the code.Suppose, I'm using Crisfield's derivation it would mean that we are removing the rigid body translations and roattions in equation  3.123(Attached at the first post of this thread- co rotational crisfield.pdf).Right?

2) Now, suppose I am doing a Total LAgrangian formulation using Greens strain for trusses-refer chapter 3 of Crisfield equation 3.54 (which is the similar step) ,also attached at the first post of this thread, (attached- Deep_Truss_element_using_Greens strain.pdf)Right?

3) What is the difference beween the two?

Why does 2 not remove rigid body translations and rotations when 1 does (As you say)?Peyman, can you please explain in this perspective?

 

Peyman and Louie, in the atatched pdf's please look at the equations mentioned above,please neglect any comments or notes ajoining the equation(s)-these were just for my math derivations.My querstions are clearly descrined in 1,2,3 above.

 

Kajal

Peyman Khosravi's picture

Kajal;

The difference between TL(total lagrangian) and CR (corotational) is that in TL everything is done from the beginning to the end in the global coordinate system with respect to the initial geometry. TL works usually with Green's strain as well. The story is not just removing RB motion which of course can be done easily in both methods. In CR we switch between local and global axe frequently and we use a shadow element to calculate the strain in the local system. In TL stiffness and force and ... are calculated from the beginning to the end in the global system but in CR they are calculated in local and global by using shadow element. Before asking how this is done please read Louie's PDF carefully and be patient. Louie has made a lot of effort in preparing that excelent document and you should really read that first. not everything has benn explaned in Crisfield in the best way.

Peyman

ok, peyman;I will re-read and come back.

Many many thanks to Louie and Peyman for answering patiently.I apologise for asking too much.

However,the thread is not done yet and I will come back after re-reading.Please keep helping.

Thanks again

I have a query on material non linearity of 2D truss, i posed this question to Prof.Louie. i want to attach that file here where i have solved it and having some problems. I need to check up with u ppl whether my solution way is correct or not.

 

How do i attach files in imechanica, i don find any attachment tags

 By the way

I am Karthic, a PhD student from Nanyang Technological university,Singapore

Peyman and Louie:

Can we say, what really makes the difference in co-rotational is:

1)Length is updated following every iteration whereas in TL, we use the initial geometry each time

2)The angle made by truss element is updated each tim based on the updated length

3)Rigid body motions are removed by equation 3 of Louie's document

4)Though in TL we have a similar equation 3 (strain = change in length / org length) it is not so realsitic because we have everything in terms of initial geometry- so removal of the rigid body motions cannot be really demarkated there.

Am I right?Please add a review/comments.

Kajal

Peyman Khosravi's picture

If this is your interpretation, that is fine. Let's not discuss more on that and move on. You will see, as you study more, that things will become more clear.

Peyman

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