iMechanica - Comments for "Geometric non-linearity basic question"
https://imechanica.org/node/5462
Comments for "Geometric non-linearity basic question"enOn the choice of a measure of strain
https://imechanica.org/comment/10871#comment-10871
<a id="comment-10871"></a>
<p><em>In reply to <a href="https://imechanica.org/comment/10862#comment-10862">Hi Zhigang,
Thanks, your</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>
There seem to be two issues:
</p>
<ol><li>When is infinitesimal kinematics valid?</li>
<li>What is a good measure of strain?</li>
</ol><p>
The two issues are separate. I have talked about 1 in the previous comment. Regarding 2, we may note the following:
</p>
<ul><li>Indeed, as you have pointed out, the deformation gradient <strong>F</strong> is a product of stretch and rotation.
</li>
<li>This fact should be remembered if one uses deformation gradient as a measure of strain. For example, the energy density is invariant of rotation, and can only depend on <strong>F</strong> through the product <strong>F</strong>^T<strong>F</strong>.</li>
<li>By choosing <strong>F</strong> as a measure of strain, one can retain linear kinematics, and lump all nonlinear effects in the energy-density function. I take this approach in my <a href="http://imechanica.org/node/538">notes on the general theory of finite deformation</a>. I like this approach because all PDEs in this approach are linear, and the variational statement looks exactly the same as that in infinitesimal kinematics.</li>
<li>Of course, you can also regard the product <strong>F</strong>^T<strong>F</strong> as a measure of strain.</li>
</ul><p>The two approaches are fully equivalent. Consequently, <strong>the choice of measure of strain has no bearing on whether infinitesimal kinematics is valid for a particular problem</strong>. This is the point I wished to make. Sorry for being unclear last time.</p>
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</ul>Thu, 21 May 2009 10:08:00 +0000Zhigang Suocomment 10871 at https://imechanica.orgHi Zhigang,
Thanks, your
https://imechanica.org/comment/10862#comment-10862
<a id="comment-10862"></a>
<p><em>In reply to <a href="https://imechanica.org/comment/10859#comment-10859">Re: Confusion about two explanations on geometric nonlinearity</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>
Hi Zhigang,
</p>
<p>
Thanks, your explanation gives some more insight into the concept. However, I would like to have some clarification on the last point. How can we directly use the deformation gradient tensor as a measure of strain, since it includes the effect of rigid body rotations also? Won't the process of removing the rotations from it, will make the strain-displacement relations non-linear, unless we use a corotational formulation (and only in case of small-strain geometric non-linearity) as Peyman mentioned?
</p>
<p>
Thanks again,
</p>
<p>
Jayadeep
</p>
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</ul>Wed, 20 May 2009 04:39:47 +0000Jayadeep U. B.comment 10862 at https://imechanica.orgRe: Confusion about two explanations on geometric nonlinearity
https://imechanica.org/comment/10859#comment-10859
<a id="comment-10859"></a>
<p><em>In reply to <a href="https://imechanica.org/comment/10844#comment-10844">Confusion about two explanations on geometric nonlinearity</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>
Here are a few paragraphs from my <a href="/node/5065 ">lecture notes on finite deformation</a>. Hope they help.
</p>
<p>
When a structure deforms, each state of deformation must obey Newton’s law. This principle has often been violated in our past work. For example, in analyzing a truss, we have balanced forces in the deformed truss using the shape of the undeformed truss, neglecting the deformation.
</p>
<p>
This negligence is often justified on the ground that deformation in most engineering structures is small. You might think that a structure suffering a small strain, say less than 1%, entitles you to neglect the change in shape when you balance forces. A counter example, however, is familiar to you. When a column buckles, the strain in the column is indeed small, but you must enforce equilibrium in the deflected state of the column.
</p>
<p>
The essential point is this: <strong>we must enforce Newton’s law in every deformed state of a structure, and use this correct principle as a basis to justify any simplification</strong>. This consideration alone requires us to examine finite deformation, even when the strain is small everywhere in the structure.
</p>
<p>
</p>
<p>
I should add a point. It is sometimes stated that the significance of finite deformation may be judged if we need to use nonlinear strain-displacement relation. I believe this statement is meaningless. The form of strain-displacement relation depends what measure of strain you use. If you happen to use the deformation gradient as a measure of strain, then the strain-displacement relation is always linear, no matter how large deformation is.
</p>
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</ul>Wed, 20 May 2009 01:53:02 +0000Zhigang Suocomment 10859 at https://imechanica.orgIs a bar under large axial strain geometrically nonlinear?
https://imechanica.org/comment/10858#comment-10858
<a id="comment-10858"></a>
<p><em>In reply to <a href="https://imechanica.org/node/5462">Geometric non-linearity basic question</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.
</p>
<p>
See <a href="http://en.wikipedia.org/wiki/Infinitesimal_strain_theory#Infinitesimal_strain_tensor">http://en.wikipedia.org/wiki/Infinitesimal_strain_theory#Infinitesimal_strain_tensor</a>
</p>
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</ul>Wed, 20 May 2009 01:26:51 +0000Biswajit Banerjeecomment 10858 at https://imechanica.orgHi Jayadeep;
I am not
https://imechanica.org/comment/10857#comment-10857
<a id="comment-10857"></a>
<p><em>In reply to <a href="https://imechanica.org/node/5462">Geometric non-linearity basic question</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>
Hi Jayadeep;
</p>
<p>
I am not sure if every large strain problem is actually a geometrically nonlinear problem. For example, is a bar under large axial strain a geometrically nonlinear problem?
</p>
<p>
I think we can say a geometrically nonlinear problem is the one that has large deflections, plus large or small rotations. there is no clear line here but to me, any problem in which the deformed shape is so far from undeformed shape is geometrically nonlinear.
</p>
<p>
Thanks
</p>
<p>
Peyman khosravi
</p>
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</ul>Wed, 20 May 2009 00:56:52 +0000Peyman Khosravicomment 10857 at https://imechanica.orgThanks for putting all the aspects into proper perspective
https://imechanica.org/comment/10855#comment-10855
<a id="comment-10855"></a>
<p><em>In reply to <a href="https://imechanica.org/node/5462">Geometric non-linearity basic question</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>
Hi Peyman,
</p>
<p>
Thanks for that nice explanation. I had come across rotational effects and other concepts you mentioned in your comment, but didn't pursue them enough to develop a proper understanding. Thanks again for putting the different aspects of geometric non-linearity in proper perspective.
</p>
<p>
Just one more question in this connection: can I make a statement that "the geometric non-linearity is either due to large strains or finite rotations or both"?
</p>
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</ul>Tue, 19 May 2009 16:54:26 +0000Jayadeep U. B.comment 10855 at https://imechanica.orgLarge strain and small strain problems
https://imechanica.org/comment/10849#comment-10849
<a id="comment-10849"></a>
<p><em>In reply to <a href="https://imechanica.org/node/5462">Geometric non-linearity basic question</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>
Hi Jayadeep;
</p>
<p>
This is a good question. Actually geometric nonlinearity means when change of the shape of the structure under the load makes the linear analysis invalid (since in linear analysis we refer everything to the undeformed configuration). This may happen in different cases and ofcourse we may use different solutions for it. In all of these cases you can say that the linear definitions of strains are not correct any more and we need to add some nonlinear (or higher order terms) to it. One example is a piece of rubber under the load. It deforms largly and then the small-deformation definitions of strains ex,ey,exy are not valid for that. Another example is a fishing rod which has large deformations and so again analysis based on small deformation is not valid for that.
</p>
<p>
However, there is a basic difference between these two examples. A rubber deforms largly under the load and it also developes large strains. A fishing rod deflects largly under the tip load but it doesn't have large strains. As a result, people usually devide geometric nonlinearity to two categories: small strain, and large strain problems.
</p>
<p>
In small strain geometric nonlinearity (which usually happens in thin structures such as this fishing rod or a thin plate) we usually have large nodal rotations as well. Since most part of the deformation in these cases is the rigid body motion of elements (in FE analysis) we can remove this part and then use small strain formulations without any need to reformulate the strain (since the remaining part of the strain is small). This is the basic idea behind corotational nonlinear analysis.
</p>
<p>
Peyman Khosravi
</p>
<p>
</p>
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</ul>Mon, 18 May 2009 16:26:26 +0000Peyman Khosravicomment 10849 at https://imechanica.orgConfusion about two explanations on geometric nonlinearity
https://imechanica.org/comment/10844#comment-10844
<a id="comment-10844"></a>
<p><em>In reply to <a href="https://imechanica.org/node/5462">Geometric non-linearity basic question</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>
Hi all,
</p>
<p>
In solid mechanics books, geometric non-linearity is explained as a result of the non-linear strain displacement relations, when the deformations are larege (like the non-linear terms in the Green-Lagrange strain tensor). However, ANSYS software manuals explain the geometric non-linearity using the "fishing rod" example, where the deformation appears to affect the loading. Are these two explanations equivalent? This point has been troubling me for quite some time, since the first explanation uses local parameters (strain and displacement), while the second one uses global parameters (load and overall deformation). Thanks in advance for giving some clarity on the matter.
</p>
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</ul>Mon, 18 May 2009 11:20:44 +0000Jayadeep U. B.comment 10844 at https://imechanica.orgGeometric nonlinearity
https://imechanica.org/comment/10841#comment-10841
<a id="comment-10841"></a>
<p><em>In reply to <a href="https://imechanica.org/node/5462">Geometric non-linearity basic question</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>
Hi;
</p>
<p>
When I said linear analysis, I meant non-iterative solution, i.e. solving d=inv(K)*F only once and then stop.
</p>
<p>
Regarding Eq. 1.56 there is a typo. Pi(p) should be changed to qi(p) and so: qi(p+deltap)=qi(p)+dqi(p)/dp*delta p
</p>
<p>
This simply means that change of q due to a small change in p is equal to its value at p plus its derivative multiplied by the small change in p (simple definition of partial derivative). I think you had problem because of the typo.
</p>
<p>
Thanks
</p>
<p>
Peyman Khosravi
</p>
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</ul>Mon, 18 May 2009 03:32:29 +0000Peyman Khosravicomment 10841 at https://imechanica.orgDiscussion continues-thanks to Peymann
https://imechanica.org/comment/10840#comment-10840
<a id="comment-10840"></a>
<p><em>In reply to <a href="https://imechanica.org/node/5462">Geometric non-linearity basic question</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>
Hi,
</p>
<p>
I have 2 questions:
</p>
<p>
First one related to what you have said above:
</p>
<p>
1) You said when a bar is straight it's lateral stiffness is zero-fine.In the book by Crisfield which we've been discussing, in the example concerned - page 3, it is observed that stiffness is falling with every incremental laod applied- also bar is seflecting more with every incremental laod, then, how can you say:
</p>
<p>
"--However when the bar starts to move, it gains stiffness but this is not detectabe by linear analysis--"??
</p>
<p>
2)This is not related to your answer above, but, related to Crisfield's book:
</p>
<p>
In eqn 1.56, crisfield states that:
</p>
<p>
"--In order to produce an incremental solution procedure, the internal force qi corresponding to the displacement p can be expanded by means of a truncated Taylor series following which he gives the expression 1.56-can you kindly tell me the meaning/interpretation of this expression 1.56? Some insight into Taylor series expansion would really be helpful
</p>
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</ul>Sun, 17 May 2009 23:40:55 +0000kajalschopracomment 10840 at https://imechanica.orgAnswer to your questions on geometrical nonlinearity
https://imechanica.org/comment/10833#comment-10833
<a id="comment-10833"></a>
<p><em>In reply to <a href="https://imechanica.org/node/5462">Geometric non-linearity basic question</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>
Hi
</p>
<p>
If you mean that with z=0 (i.e. when the bar is initially horizontal before applying the load) there is no initaial stiffness, you are right. A bar has zero stiffness in the lateral direction. You can check it by modeling a bar (not a beam) and do the LINEAR analysis by any software. You will see that it gives you an unstability error. However when the bar starts to move, it gains stiffness but this is not detectabe by linear analysis. That is why we need to do nonlinear analysis to see the actual change in the bahavior of the bar as it moves.
</p>
<p>
Now about your other question. You are somehow right. A better explanation is as this. If a bar is under axial tension load, it is harder to deflect it laterally. If it is under axial compression load, it is easier. So there is something in the axial load which affects the lateral stiffeness. When a bar which is under axial load deflects laterally, its length increases, and so an extra amount of axial strain happens which is called higher order strain. For a horizontal bar with length L if one end moves laterally by a SMALL amount of V, we can show that its length increases almost by V^2/(2*L) and so a higher order strain equal to V^2/(2*L^2) happens. Now if the bar is initally under the axial load N , a work equal to (volume)*(stress)*(strain)=(AL)*(N/A)*(V^2/2L^2) or (1/2)(N/L)V^2 is associated with this axial load during lateral deformation. Since work is always equal to (1/2)(stiffness)(deflection^2) we can say that an extra amount of stiffness equal to N/L exists which is due to the axial load. Ofcourse if N is tensional this extra stiffness is positive and makes the lateral movement more difficult and vise versa.
</p>
<p>
So we can say that stress stiffness (which is the one that if you add it to the usuall stiffness you get the tangent stiffness) is associated with the work of the axial load (or in better words membrane forces, for example in plate structures) through lateral displacements.
</p>
<p>
A good book is also the finite element book written by Cook which you may want to have a look at it.
</p>
<p>
Peyman Khosravi
</p>
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</ul>Sun, 17 May 2009 04:41:37 +0000Peyman Khosravicomment 10833 at https://imechanica.organother question for Peyman et all
https://imechanica.org/comment/10832#comment-10832
<a id="comment-10832"></a>
<p><em>In reply to <a href="https://imechanica.org/node/5462">Geometric non-linearity basic question</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>
But Peyman--if z = 0 --that is bar is initially straight, then the expression:
</p>
<p>
Kt = EA/l (z/l)^2 + EA/l (2zw + w^2/ L^2) + N/L
</p>
<p>
is invalid?
</p>
<p>
Does it mean the expression is valid only inclined bars?
</p>
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</ul>Sat, 16 May 2009 21:26:49 +0000kajalschopracomment 10832 at https://imechanica.orgThank you Peymann
https://imechanica.org/comment/10831#comment-10831
<a id="comment-10831"></a>
<p><em>In reply to <a href="https://imechanica.org/node/5462">Geometric non-linearity basic question</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 again Peymann.Can you answer my second question above?Am I correct?Again, restating below:
</p>
<p>
"-----2.Now, am I right if I say that, geometric stiffness of any laod carrying element can be taken into account if the effect of internal force on the stiffness of the structural element/load carrying member is taken into account throughout its loading range? "
</p>
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</ul>Sat, 16 May 2009 18:25:54 +0000kajalschopracomment 10831 at https://imechanica.orgGeometric non-linearity
https://imechanica.org/comment/10829#comment-10829
<a id="comment-10829"></a>
<p><em>In reply to <a href="https://imechanica.org/node/5462">Geometric non-linearity basic question</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>
Hi;
</p>
<p>
You are making a mistake. Here, z shows the initial configuration of the bar. it means when no load is applied the bar is there. so the bar is initailly inclined. as you know, the linear stiffness is a function of the current geometry. so in this case you simply have a bar element which has an angle with horizon and its stiffness matrix is only related to z.
</p>
<p>
when the load is applied, the bar moves from z to the final configuration. as the angle is changed, at each position, the tangent stiffness matrix is calculated based on its current geometry (i.e. z+w) and also since it has now some axial force, we have to also add the effect of internal load of the bar.
</p>
<p>
Peyman Khosravi
</p>
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</ul>Sat, 16 May 2009 01:26:27 +0000Peyman Khosravicomment 10829 at https://imechanica.org