iMechanica - Comments for "Co-rotational formulation for trusses"
https://imechanica.org/node/7020
Comments for "Co-rotational formulation for trusses"enIf this is your
https://imechanica.org/comment/12864#comment-12864
<a id="comment-12864"></a>
<p><em>In reply to <a href="https://imechanica.org/node/7020">Co-rotational formulation for trusses</a></em></p>
<div class="field field-name-comment-body field-type-text-long field-label-hidden"><div class="field-items"><div class="field-item even"><p>
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.
</p>
<p>
Peyman
</p>
</div></div></div><ul class="links inline"><li class="comment_forbidden first last"><span><a href="/user/login?destination=node/7020%23comment-form">Log in</a> or <a href="/user/register?destination=node/7020%23comment-form">register</a> to post comments</span></li>
</ul>Sat, 21 Nov 2009 00:50:58 +0000Peyman Khosravicomment 12864 at https://imechanica.orgPeyman and Louie please comment
https://imechanica.org/comment/12855#comment-12855
<a id="comment-12855"></a>
<p><em>In reply to <a href="https://imechanica.org/node/7020">Co-rotational formulation for trusses</a></em></p>
<div class="field field-name-comment-body field-type-text-long field-label-hidden"><div class="field-items"><div class="field-item even"><p>
Peyman and Louie:
</p>
<p>
Can we say, what really makes the difference in co-rotational is:
</p>
<p>
1)Length is updated following every iteration whereas in TL, we use the initial geometry each time
</p>
<p>
2)The angle made by truss element is updated each tim based on the updated length
</p>
<p>
3)Rigid body motions are removed by equation 3 of Louie's document
</p>
<p>
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.
</p>
<p>
Am I right?Please add a review/comments.
</p>
<p>
Kajal
</p>
</div></div></div><ul class="links inline"><li class="comment_forbidden first last"><span><a href="/user/login?destination=node/7020%23comment-form">Log in</a> or <a href="/user/register?destination=node/7020%23comment-form">register</a> to post comments</span></li>
</ul>Thu, 19 Nov 2009 23:27:29 +0000kajalschopracomment 12855 at https://imechanica.orgMaterial Non linearity in 2D truss
https://imechanica.org/comment/12846#comment-12846
<a id="comment-12846"></a>
<p><em>In reply to <a href="https://imechanica.org/node/7020">Co-rotational formulation for trusses</a></em></p>
<div class="field field-name-comment-body field-type-text-long field-label-hidden"><div class="field-items"><div class="field-item even"><p>
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.
</p>
<p>
</p>
<p>
How do i attach files in imechanica, i don find any attachment tags
</p>
<p>
By the way</p>
<p>
I am Karthic, a PhD student from Nanyang Technological university,Singapore
</p>
</div></div></div><ul class="links inline"><li class="comment_forbidden first last"><span><a href="/user/login?destination=node/7020%23comment-form">Log in</a> or <a href="/user/register?destination=node/7020%23comment-form">register</a> to post comments</span></li>
</ul>Thu, 19 Nov 2009 05:04:25 +0000karthic_newbeecomment 12846 at https://imechanica.orgok peyman and louie---
https://imechanica.org/comment/12842#comment-12842
<a id="comment-12842"></a>
<p><em>In reply to <a href="https://imechanica.org/node/7020">Co-rotational formulation for trusses</a></em></p>
<div class="field field-name-comment-body field-type-text-long field-label-hidden"><div class="field-items"><div class="field-item even"><p>
ok, peyman;I will re-read and come back.
</p>
<p>
Many many thanks to Louie and Peyman for answering patiently.I apologise for asking too much.
</p>
<p>
However,the thread is not done yet and I will come back after re-reading.Please keep helping.
</p>
<p>
Thanks again
</p>
</div></div></div><ul class="links inline"><li class="comment_forbidden first last"><span><a href="/user/login?destination=node/7020%23comment-form">Log in</a> or <a href="/user/register?destination=node/7020%23comment-form">register</a> to post comments</span></li>
</ul>Wed, 18 Nov 2009 23:38:37 +0000kajalschopracomment 12842 at https://imechanica.orgMore on corotational
https://imechanica.org/comment/12835#comment-12835
<a id="comment-12835"></a>
<p><em>In reply to <a href="https://imechanica.org/node/7020">Co-rotational formulation for trusses</a></em></p>
<div class="field field-name-comment-body field-type-text-long field-label-hidden"><div class="field-items"><div class="field-item even"><p>
Kajal;
</p>
<p>
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.
</p>
<p>
Peyman
</p>
</div></div></div><ul class="links inline"><li class="comment_forbidden first last"><span><a href="/user/login?destination=node/7020%23comment-form">Log in</a> or <a href="/user/register?destination=node/7020%23comment-form">register</a> to post comments</span></li>
</ul>Wed, 18 Nov 2009 15:32:25 +0000Peyman Khosravicomment 12835 at https://imechanica.orgalso--peyman and louie
https://imechanica.org/comment/12831#comment-12831
<a id="comment-12831"></a>
<p><em>In reply to <a href="https://imechanica.org/node/7020">Co-rotational formulation for trusses</a></em></p>
<div class="field field-name-comment-body field-type-text-long field-label-hidden"><div class="field-items"><div class="field-item even"><p>
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.
</p>
<p>
</p>
<p>
Kajal
</p>
</div></div></div><ul class="links inline"><li class="comment_forbidden first last"><span><a href="/user/login?destination=node/7020%23comment-form">Log in</a> or <a href="/user/register?destination=node/7020%23comment-form">register</a> to post comments</span></li>
</ul>Wed, 18 Nov 2009 09:33:31 +0000bruno-pagecomment 12831 at https://imechanica.orgPeyman and Louie
https://imechanica.org/comment/12830#comment-12830
<a id="comment-12830"></a>
<p><em>In reply to <a href="https://imechanica.org/node/7020">Co-rotational formulation for trusses</a></em></p>
<div class="field field-name-comment-body field-type-text-long field-label-hidden"><div class="field-items"><div class="field-item even"><p>
Peyman and Louie:
</p>
<p>
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- <em>co rotational crisfield.pdf</em>).Right?
</p>
<p>
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- <em>Deep_Truss_element_using_Greens strain.pdf</em>)Right?
</p>
<p>
3) What is the difference beween the two?
</p>
<p><span>Why does 2 not remove rigid body translations and rotations when 1 does (As you say)?</span><span>Peyman, can you please explain in this perspective?</span> </p>
<p>
</p>
</div></div></div><ul class="links inline"><li class="comment_forbidden first last"><span><a href="/user/login?destination=node/7020%23comment-form">Log in</a> or <a href="/user/register?destination=node/7020%23comment-form">register</a> to post comments</span></li>
</ul>Wed, 18 Nov 2009 09:25:01 +0000bruno-pagecomment 12830 at https://imechanica.orgequation 3 removes rigid body rotations and translations!
https://imechanica.org/comment/12825#comment-12825
<a id="comment-12825"></a>
<p><em>In reply to <a href="https://imechanica.org/node/7020">Co-rotational formulation for trusses</a></em></p>
<div class="field field-name-comment-body field-type-text-long field-label-hidden"><div class="field-items"><div class="field-item even"><p>
Kajal,
</p>
<p>
As I mentioned previously above.
</p>
<p>
In the corotational truss formulation rigid body rotations and translations are removed when we use equation (3) in my pdf file.
</p>
<p>
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.
</p>
<p>
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.
</p>
<p>
regards,
</p>
<p>
Louie
</p>
</div></div></div><ul class="links inline"><li class="comment_forbidden first last"><span><a href="/user/login?destination=node/7020%23comment-form">Log in</a> or <a href="/user/register?destination=node/7020%23comment-form">register</a> to post comments</span></li>
</ul>Wed, 18 Nov 2009 01:55:47 +0000yawloucomment 12825 at https://imechanica.orgKajal;
Rigid body is
https://imechanica.org/comment/12823#comment-12823
<a id="comment-12823"></a>
<p><em>In reply to <a href="https://imechanica.org/node/7020">Co-rotational formulation for trusses</a></em></p>
<div class="field field-name-comment-body field-type-text-long field-label-hidden"><div class="field-items"><div class="field-item even"><p>
Kajal;
</p>
<p>
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.
</p>
<p>
Peyman
</p>
</div></div></div><ul class="links inline"><li class="comment_forbidden first last"><span><a href="/user/login?destination=node/7020%23comment-form">Log in</a> or <a href="/user/register?destination=node/7020%23comment-form">register</a> to post comments</span></li>
</ul>Tue, 17 Nov 2009 21:48:43 +0000Peyman Khosravicomment 12823 at https://imechanica.orgPeyman,
https://imechanica.org/comment/12814#comment-12814
<a id="comment-12814"></a>
<p><em>In reply to <a href="https://imechanica.org/node/7020">Co-rotational formulation for trusses</a></em></p>
<div class="field field-name-comment-body field-type-text-long field-label-hidden"><div class="field-items"><div class="field-item even"><p>
Peyman,
</p>
<p>
Ok:
</p>
<p>
1)Can you tell me which step in the derivation removes the RB motions, both:translations and rotations?
</p>
<p>
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?
</p>
<p>
3)Suppose, if I do not want to remove RB rotations then what would be the difference in the derivation?
</p>
<p>
</p>
<p>
kajal
</p>
</div></div></div><ul class="links inline"><li class="comment_forbidden first last"><span><a href="/user/login?destination=node/7020%23comment-form">Log in</a> or <a href="/user/register?destination=node/7020%23comment-form">register</a> to post comments</span></li>
</ul>Tue, 17 Nov 2009 02:32:57 +0000kajalschopracomment 12814 at https://imechanica.orgKajal; Honestly I still
https://imechanica.org/comment/12813#comment-12813
<a id="comment-12813"></a>
<p><em>In reply to <a href="https://imechanica.org/node/7020">Co-rotational formulation for trusses</a></em></p>
<div class="field field-name-comment-body field-type-text-long field-label-hidden"><div class="field-items"><div class="field-item even"><p>
Kajal;
</p>
<p>
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)"?
</p>
<p>
Peyman
</p>
</div></div></div><ul class="links inline"><li class="comment_forbidden first last"><span><a href="/user/login?destination=node/7020%23comment-form">Log in</a> or <a href="/user/register?destination=node/7020%23comment-form">register</a> to post comments</span></li>
</ul>Tue, 17 Nov 2009 02:07:40 +0000Peyman Khosravicomment 12813 at https://imechanica.orgI'll try and explain my question clearly--
https://imechanica.org/comment/12807#comment-12807
<a id="comment-12807"></a>
<p><em>In reply to <a href="https://imechanica.org/node/7020">Co-rotational formulation for trusses</a></em></p>
<div class="field field-name-comment-body field-type-text-long field-label-hidden"><div class="field-items"><div class="field-item even"><p>
Thanks Louie,I will go through Peymans response minutely and carefully as you've suggested.
</p>
<p>
Peyman:
</p>
<p>
1)Look at Crisfield'd derivation of corotational formulation attached at the first post of this thread.
</p>
<p>
2)We first get strain along the axis of the element.
</p>
<p>
3)We then get internal force vector corresponding to the axis of the element.
</p>
<p>
4)Next, we transform this internal force vector to global axis by multiplying with transformation matrix.
</p>
<p>
5)Next, we get stiffness matrix by differentiating the transformed internal force vector.
</p>
<p>
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.
</p>
<p>
My question is:
</p>
<p>
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?
</p>
<p>
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?
</p>
<p>
Respects,
</p>
<p>
Kajal
</p>
<p>
</p>
</div></div></div><ul class="links inline"><li class="comment_forbidden first last"><span><a href="/user/login?destination=node/7020%23comment-form">Log in</a> or <a href="/user/register?destination=node/7020%23comment-form">register</a> to post comments</span></li>
</ul>Mon, 16 Nov 2009 07:41:53 +0000kajalschopracomment 12807 at https://imechanica.orggood questions difficult to answer
https://imechanica.org/comment/12806#comment-12806
<a id="comment-12806"></a>
<p><em>In reply to <a href="https://imechanica.org/node/7020">Co-rotational formulation for trusses</a></em></p>
<div class="field field-name-comment-body field-type-text-long field-label-hidden"><div class="field-items"><div class="field-item even"><p>
Kajal,
</p>
<p>
You are asking good questions Kajal. This topic is a bit difficult to explain. I still struggle with some of these same questions.
</p>
<p>
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.
</p>
<p>
1. It is helpful to examine small displacement analysis and large displacement analysis by looking at them from an equilibrium point of view.
</p>
<p>
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.)
</p>
<p>
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.
</p>
<p>
regards,
</p>
<p>
Louie
</p>
<p>
</p>
</div></div></div><ul class="links inline"><li class="comment_forbidden first last"><span><a href="/user/login?destination=node/7020%23comment-form">Log in</a> or <a href="/user/register?destination=node/7020%23comment-form">register</a> to post comments</span></li>
</ul>Mon, 16 Nov 2009 05:55:01 +0000yawloucomment 12806 at https://imechanica.orgKajal;
I don't understand
https://imechanica.org/comment/12805#comment-12805
<a id="comment-12805"></a>
<p><em>In reply to <a href="https://imechanica.org/node/7020">Co-rotational formulation for trusses</a></em></p>
<div class="field field-name-comment-body field-type-text-long field-label-hidden"><div class="field-items"><div class="field-item even"><p>
Kajal;
</p>
<p>
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.
</p>
<p>
peyman
</p>
</div></div></div><ul class="links inline"><li class="comment_forbidden first last"><span><a href="/user/login?destination=node/7020%23comment-form">Log in</a> or <a href="/user/register?destination=node/7020%23comment-form">register</a> to post comments</span></li>
</ul>Mon, 16 Nov 2009 01:54:17 +0000Peyman Khosravicomment 12805 at https://imechanica.orgpeyman, my question is--
https://imechanica.org/comment/12804#comment-12804
<a id="comment-12804"></a>
<p><em>In reply to <a href="https://imechanica.org/node/7020">Co-rotational formulation for trusses</a></em></p>
<div class="field field-name-comment-body field-type-text-long field-label-hidden"><div class="field-items"><div class="field-item even"><p>
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?
</p>
<p>
kajal
</p>
</div></div></div><ul class="links inline"><li class="comment_forbidden first last"><span><a href="/user/login?destination=node/7020%23comment-form">Log in</a> or <a href="/user/register?destination=node/7020%23comment-form">register</a> to post comments</span></li>
</ul>Sun, 15 Nov 2009 23:50:22 +0000kajalschopracomment 12804 at https://imechanica.orgRigid body motion in Corotational analysis
https://imechanica.org/comment/12803#comment-12803
<a id="comment-12803"></a>
<p><em>In reply to <a href="https://imechanica.org/node/7020">Co-rotational formulation for trusses</a></em></p>
<div class="field field-name-comment-body field-type-text-long field-label-hidden"><div class="field-items"><div class="field-item even"><p>Normal<br />
0</p>
<p>MicrosoftInternetExplorer4</p>
<p>/* Style Definitions */<br />
table.MsoNormalTable<br />
{mso-style-name:"Table Normal";<br />
mso-tstyle-rowband-size:0;<br />
mso-tstyle-colband-size:0;<br />
mso-style-noshow:yes;<br />
mso-style-parent:"";<br />
mso-padding-alt:0cm 5.4pt 0cm 5.4pt;<br />
mso-para-margin:0cm;<br />
mso-para-margin-bottom:.0001pt;<br />
mso-pagination:widow-orphan;<br />
font-size:10.0pt;<br />
font-family:"Times New Roman";}</p>
<p>
Hi;
</p>
<p>
First I should say that I enjoyed this discussion and I should thank Louie<br />
for his explanations. Here is my answer based on what I learned in the past;
</p>
<p>
Rigid body (RB) motion in stress analysis and finite element can only be<br />
understood through equilibrium. Let's put it this way: what is really the definition<br />
of small deflection and large deflection analysis? How small is small enough to<br />
do the small deflection analysis? The whole corotational analysis and problems<br />
that we encounter with them during corotational analysis and convergence are<br />
related to this point.
</p>
<p>
Small deformation analysis (SDA) and large deformation analysis (LDA) are<br />
different in the way we satisfy equilibrium. If in solving our problem we<br />
consider only one geometry, it is SDA and equilibrium is satisfied for<br />
that geometry. If that geometry is the undeformed geometry, forces found in the<br />
analysis are in equilibrium only on that geometry. Deflections also are not<br />
real because we didn’t consider the real deformed geometry in our analysis. If<br />
defections are small we can say OK they are at least some approximation of the<br />
real deflections otherwise they are totally wrong. An example is a truss<br />
element that has a large rigid body rotation. We can have a RB rotation by<br />
rotating a horizontal truss element about one end (let’s say by 45 degrees) See<br />
Eq. 17 of Louie’s PDF file mentioned above for element stiffness matrix and you<br />
will see that you get some forces in the element. Those forces are not real<br />
(otherwise they should be zero) and don’t satisfy equilibrium on the deformed<br />
geometry, however they satisfy equilibrium on the undeformed geometry.
</p>
<p>
In geometrically nonlinear analysis we are trying to satisfy equilibrium on<br />
the deformed configuration. This means that we have TWO geometries from the beginning.<br />
You may say OK I know that, but the reality is more complicated than this<br />
because sometimes deformed and undeformed geometries are so close together that<br />
one doesn’t see the necessity for that consideration. Here is more explanation.
</p>
<p>
Consider a corotational analysis for a general element. The method is<br />
basically to put a copy of the element on its final position and then remove<br />
the rigid body (RB) motion and measure the remaining deformation. One may say at<br />
this stage that the force of the element is [K_linear]*{remaining deformation}.<br />
And also for the next iteration when he needs the tangent stiffeness matrix, he<br />
may add the element stiffness to the conventional stress stiffness matrix (that<br />
is available in any FEM book and is a function of the element internal force)<br />
to get the tangent stiffness matrix. Right? Wrong! The reason is that these stiffness<br />
matrix equations satisfy equilibrium on the undeformed geometry (i.e.<br />
the copy element) and not on the deformed geometry (even though they are very close<br />
together since we have put them over each other). Why is it so important to<br />
satisfy the equilibrium on final geometry? Because otherwise we will have<br />
convergence problem during iterations. Having this problem was the origin of<br />
inventing projector matrix to REALLY PURIFY those seemingly pure deformations<br />
such that they are in equilibrium on the final geometry. (refer to Rankin’s<br />
papers)
</p>
<p>
In finite element analysis we usually formulate the elements in a way that<br />
they produce no forces under RB motion. However one can easily check that it is<br />
not the case for RB rotations (as we checked before). Bottom line is that if you<br />
ask why it is not working the answer is because this hasn’t been designed to<br />
work for any geometry. SDA only works if you don’t care about the final shape<br />
and satisfying equilibrium on the final shape. What you get from SDA may be<br />
said that includes rigid body motion to some degree, but it is not the correct<br />
deformation. So the RB motion that is found in SDA is also not correct.
</p>
<p>
Peyman<span> </span>
</p>
</div></div></div><ul class="links inline"><li class="comment_forbidden first last"><span><a href="/user/login?destination=node/7020%23comment-form">Log in</a> or <a href="/user/register?destination=node/7020%23comment-form">register</a> to post comments</span></li>
</ul>Sun, 15 Nov 2009 17:41:46 +0000Peyman Khosravicomment 12803 at https://imechanica.orgCan we say--
https://imechanica.org/comment/12799#comment-12799
<a id="comment-12799"></a>
<p><em>In reply to <a href="https://imechanica.org/node/7020">Co-rotational formulation for trusses</a></em></p>
<div class="field field-name-comment-body field-type-text-long field-label-hidden"><div class="field-items"><div class="field-item even"><p>
Sir:
</p>
<p>
Sorry for replying late as I was unwell from past 3 days.
</p>
<p>
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?
</p>
<p>
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?
</p>
<p>
Kajal
</p>
</div></div></div><ul class="links inline"><li class="comment_forbidden first last"><span><a href="/user/login?destination=node/7020%23comment-form">Log in</a> or <a href="/user/register?destination=node/7020%23comment-form">register</a> to post comments</span></li>
</ul>Sun, 15 Nov 2009 00:41:28 +0000kajalschopracomment 12799 at https://imechanica.orgequation (3)
https://imechanica.org/comment/12747#comment-12747
<a id="comment-12747"></a>
<p><em>In reply to <a href="https://imechanica.org/comment/12746#comment-12746">Many thanks Sir yes--</a></em></p>
<div class="field field-name-comment-body field-type-text-long field-label-hidden"><div class="field-items"><div class="field-item even"><p>
Kajal,
</p>
<p>
</p>
<p>
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.
</p>
<p>
Incidentally, calculating d as (L^2-Lo^2)/(L+Lo) is a better conditioned formula as indicated by Crisfield.
</p>
<p>
regards,
</p>
<p>
Louie
</p>
</div></div></div><ul class="links inline"><li class="comment_forbidden first last"><span><a href="/user/login?destination=node/7020%23comment-form">Log in</a> or <a href="/user/register?destination=node/7020%23comment-form">register</a> to post comments</span></li>
</ul>Sun, 08 Nov 2009 23:09:44 +0000yawloucomment 12747 at https://imechanica.orgMany thanks Sir yes--
https://imechanica.org/comment/12746#comment-12746
<a id="comment-12746"></a>
<p><em>In reply to <a href="https://imechanica.org/comment/12730#comment-12730">Linear analysis DOES include rigid body motion</a></em></p>
<div class="field field-name-comment-body field-type-text-long field-label-hidden"><div class="field-items"><div class="field-item even"><p>
Dear Sir,
</p>
<p>
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.
</p>
<p>
You also said,
</p>
<p>
"---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.--"
</p>
<p>
Can you tell me in which step we remove the rigid body motions in co-rotational analysis?
</p>
<p>
You can mention with reference to your paper below or with reference to Crisfields derivation attached at the first thread of this post:
</p>
<p>
<a href="http://people.wallawalla.edu/~louie.yaw/Co-rotational_docs/2Dcorot_truss.pdf">http://people.wallawalla.edu/~louie.yaw/Co-rotational_docs/2Dcorot_truss.pdf</a>
</p>
<p>
Kajal
</p>
</div></div></div><ul class="links inline"><li class="comment_forbidden first last"><span><a href="/user/login?destination=node/7020%23comment-form">Log in</a> or <a href="/user/register?destination=node/7020%23comment-form">register</a> to post comments</span></li>
</ul>Sun, 08 Nov 2009 22:52:57 +0000kajalschopracomment 12746 at https://imechanica.orgLinear analysis DOES include rigid body motion
https://imechanica.org/comment/12730#comment-12730
<a id="comment-12730"></a>
<p><em>In reply to <a href="https://imechanica.org/node/7020">Co-rotational formulation for trusses</a></em></p>
<div class="field field-name-comment-body field-type-text-long field-label-hidden"><div class="field-items"><div class="field-item even"><p>
Kajal,
</p>
<p>
To answer your question, NO.
</p>
<p>
Prove to yourself that a linear analysis DOES include rigid body motion.
</p>
<p>
1. Do a linear analysis on a truss that is 20 feet long.
</p>
<p>
2. Make the joints at 5 feet on center.
</p>
<p>
3. At the left end make the truss haved pinned supported joints so the truss is like a cantilever.
</p>
<p>
4. At 10 feet put a vertical point load(do not put loads anywhere else).
</p>
<p>
</p>
<p>
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.
</p>
<p>
Therefore, a linear analysis does include rigid body motion.
</p>
<p>
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.
</p>
<p>
regards,
</p>
<p>
</p>
<p>
Louie
</p>
<p>
</p>
<p>
</p>
</div></div></div><ul class="links inline"><li class="comment_forbidden first last"><span><a href="/user/login?destination=node/7020%23comment-form">Log in</a> or <a href="/user/register?destination=node/7020%23comment-form">register</a> to post comments</span></li>
</ul>Fri, 06 Nov 2009 05:29:34 +0000yawloucomment 12730 at https://imechanica.orgSir:
Yes, I understand
https://imechanica.org/comment/12708#comment-12708
<a id="comment-12708"></a>
<p><em>In reply to <a href="https://imechanica.org/node/7020">Co-rotational formulation for trusses</a></em></p>
<div class="field field-name-comment-body field-type-text-long field-label-hidden"><div class="field-items"><div class="field-item even"><p>
Sir:
</p>
<p>
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.
</p>
<p>
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?
</p>
<p>
</p>
</div></div></div><ul class="links inline"><li class="comment_forbidden first last"><span><a href="/user/login?destination=node/7020%23comment-form">Log in</a> or <a href="/user/register?destination=node/7020%23comment-form">register</a> to post comments</span></li>
</ul>Tue, 03 Nov 2009 19:57:08 +0000kajalschopracomment 12708 at https://imechanica.orgi do not believe that is correct
https://imechanica.org/comment/12706#comment-12706
<a id="comment-12706"></a>
<p><em>In reply to <a href="https://imechanica.org/comment/12703#comment-12703">Question is--</a></em></p>
<div class="field field-name-comment-body field-type-text-long field-label-hidden"><div class="field-items"><div class="field-item even"><p>
Kajal,
</p>
<p>
What you have said above is incorrect I believe for the following reason.
</p>
<p>
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.
</p>
<p>
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.
</p>
<p>
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.
</p>
<p>
The simple transformation in a linear analysis does not remove the rigid body motions.
</p>
<p>
regards,
</p>
<p>
Louie
</p>
</div></div></div><ul class="links inline"><li class="comment_forbidden first last"><span><a href="/user/login?destination=node/7020%23comment-form">Log in</a> or <a href="/user/register?destination=node/7020%23comment-form">register</a> to post comments</span></li>
</ul>Tue, 03 Nov 2009 18:42:13 +0000yawloucomment 12706 at https://imechanica.orgRecently I publish paper
https://imechanica.org/comment/12705#comment-12705
<a id="comment-12705"></a>
<p><em>In reply to <a href="https://imechanica.org/node/7020">Co-rotational formulation for trusses</a></em></p>
<div class="field field-name-comment-body field-type-text-long field-label-hidden"><div class="field-items"><div class="field-item even"><p>
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.
</p>
<p>
<a href="http://dx.doi.org/10.1016/j.cma.2008.11.018">http://dx.doi.org/10.1016/j.cma.2008.11.018</a>
</p>
<p>
Regards,
</p>
<p>
Lukasz
</p>
<p>
</p>
<p>
</p>
</div></div></div><ul class="links inline"><li class="comment_forbidden first last"><span><a href="/user/login?destination=node/7020%23comment-form">Log in</a> or <a href="/user/register?destination=node/7020%23comment-form">register</a> to post comments</span></li>
</ul>Tue, 03 Nov 2009 17:05:36 +0000likaskcomment 12705 at https://imechanica.orgQuestion is--
https://imechanica.org/comment/12703#comment-12703
<a id="comment-12703"></a>
<p><em>In reply to <a href="https://imechanica.org/node/7020">Co-rotational formulation for trusses</a></em></p>
<div class="field field-name-comment-body field-type-text-long field-label-hidden"><div class="field-items"><div class="field-item even"><p>
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.
</p>
<p>
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.
</p>
</div></div></div><ul class="links inline"><li class="comment_forbidden first last"><span><a href="/user/login?destination=node/7020%23comment-form">Log in</a> or <a href="/user/register?destination=node/7020%23comment-form">register</a> to post comments</span></li>
</ul>Tue, 03 Nov 2009 13:16:30 +0000kajalschopracomment 12703 at https://imechanica.orgnot easy to answer succintly
https://imechanica.org/comment/12702#comment-12702
<a id="comment-12702"></a>
<p><em>In reply to <a href="https://imechanica.org/node/7020">Co-rotational formulation for trusses</a></em></p>
<div class="field field-name-comment-body field-type-text-long field-label-hidden"><div class="field-items"><div class="field-item even"><p>
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.
</p>
<p>
The most direct answer to your question is as follows:
</p>
<p>
1. a linear analysis is valid only for small displacements.
</p>
<p>
2. if displacements get too large, in a linear analysis, the rigid body motions cause erroneous strains and hence erroneous stresses.
</p>
<p>
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.
</p>
<p>
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.)
</p>
<p>
</p>
<p>
</p>
</div></div></div><ul class="links inline"><li class="comment_forbidden first last"><span><a href="/user/login?destination=node/7020%23comment-form">Log in</a> or <a href="/user/register?destination=node/7020%23comment-form">register</a> to post comments</span></li>
</ul>Tue, 03 Nov 2009 01:33:36 +0000yawloucomment 12702 at https://imechanica.orgreply
https://imechanica.org/comment/12695#comment-12695
<a id="comment-12695"></a>
<p><em>In reply to <a href="https://imechanica.org/comment/12692#comment-12692">Thanks sir, that means,</a></em></p>
<div class="field field-name-comment-body field-type-text-long field-label-hidden"><div class="field-items"><div class="field-item even"><p>
Yes, although there may be more to it than just a simple yes.
</p>
</div></div></div><ul class="links inline"><li class="comment_forbidden first last"><span><a href="/user/login?destination=node/7020%23comment-form">Log in</a> or <a href="/user/register?destination=node/7020%23comment-form">register</a> to post comments</span></li>
</ul>Tue, 03 Nov 2009 01:20:55 +0000yawloucomment 12695 at https://imechanica.orgSir, also, see the
https://imechanica.org/comment/12701#comment-12701
<a id="comment-12701"></a>
<p><em>In reply to <a href="https://imechanica.org/node/7020">Co-rotational formulation for trusses</a></em></p>
<div class="field field-name-comment-body field-type-text-long field-label-hidden"><div class="field-items"><div class="field-item even"><p>
Sir, also, see the attachment, co-rotational crisfield attached in the first post of this thread.
</p>
<p>
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.
</p>
<p>
This is what we do in linear analysis as above.Then, how does linear analysis not include rigid body displacements?
</p>
</div></div></div><ul class="links inline"><li class="comment_forbidden first last"><span><a href="/user/login?destination=node/7020%23comment-form">Log in</a> or <a href="/user/register?destination=node/7020%23comment-form">register</a> to post comments</span></li>
</ul>Tue, 03 Nov 2009 00:33:26 +0000kajalschopracomment 12701 at https://imechanica.orgI'm just recalling the
https://imechanica.org/comment/12700#comment-12700
<a id="comment-12700"></a>
<p><em>In reply to <a href="https://imechanica.org/node/7020">Co-rotational formulation for trusses</a></em></p>
<div class="field field-name-comment-body field-type-text-long field-label-hidden"><div class="field-items"><div class="field-item even"><p>
I'm just recalling the derivation of the stiffness matrix of a truss element - linear analysis:
</p>
<p>
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.
</p>
<p>
So element stiffness matrix is k11 = AE/L ; k12=k21= - AE/L; k22= AE/L.
</p>
<p>
Where are the rigid body displacements accomodated here?
</p>
<p>
Also, here we can say the element stiffness is a result of attaching a local element reference frame as in co-rotational?
</p>
</div></div></div><ul class="links inline"><li class="comment_forbidden first last"><span><a href="/user/login?destination=node/7020%23comment-form">Log in</a> or <a href="/user/register?destination=node/7020%23comment-form">register</a> to post comments</span></li>
</ul>Mon, 02 Nov 2009 22:45:40 +0000bruno-pagecomment 12700 at https://imechanica.orgok, please whats ur question??
https://imechanica.org/comment/12693#comment-12693
<a id="comment-12693"></a>
<p><em>In reply to <a href="https://imechanica.org/node/7020">Co-rotational formulation for trusses</a></em></p>
<div class="field field-name-comment-body field-type-text-long field-label-hidden"><div class="field-items"><div class="field-item even"><p>Salwan</p>
</div></div></div><ul class="links inline"><li class="comment_forbidden first last"><span><a href="/user/login?destination=node/7020%23comment-form">Log in</a> or <a href="/user/register?destination=node/7020%23comment-form">register</a> to post comments</span></li>
</ul>Mon, 02 Nov 2009 14:32:18 +0000Mscappmechcomment 12693 at https://imechanica.orgThanks sir, that means,
https://imechanica.org/comment/12692#comment-12692
<a id="comment-12692"></a>
<p><em>In reply to <a href="https://imechanica.org/node/7020">Co-rotational formulation for trusses</a></em></p>
<div class="field field-name-comment-body field-type-text-long field-label-hidden"><div class="field-items"><div class="field-item even"><p>
Thanks sir, that means, linear analysis includes rigid body motion as well (though their magnituse is small and negligible)?
</p>
</div></div></div><ul class="links inline"><li class="comment_forbidden first last"><span><a href="/user/login?destination=node/7020%23comment-form">Log in</a> or <a href="/user/register?destination=node/7020%23comment-form">register</a> to post comments</span></li>
</ul>Mon, 02 Nov 2009 08:12:27 +0000bruno-pagecomment 12692 at https://imechanica.orgError | iMechanica