iMechanica - Comments for "Deformation theories"
https://imechanica.org/node/5510
Comments for "Deformation theories"enThank you Peyman
https://imechanica.org/comment/11130#comment-11130
<a id="comment-11130"></a>
<p><em>In reply to <a href="https://imechanica.org/node/5510">Deformation theories</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>
Many thanks for your explanation.
</p>
<p>
You told that for thin beam we can neglect the shear deformation, but in my problem, since the loading is only an end moment to a cantilever beam, is it not obvious that the beam will be free from shear deformation even if it is not a thin beam?
</p>
<p>
I am taking a 3 node isoparametric beam element for this problem with a convected natural coordinate system. The virual work form is linearized by using Total Lagrangian formulation. Now, my question is, what will be the Green-Lagrange strain components. I am taking it as <span>{E</span><span>11 E22}</span>. Is it ok?
</p>
<p>
</p>
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</ul>Fri, 05 Jun 2009 09:25:59 +0000surajitdascomment 11130 at https://imechanica.orgAnswer to surajitdas' question
https://imechanica.org/comment/11126#comment-11126
<a id="comment-11126"></a>
<p><em>In reply to <a href="https://imechanica.org/node/5510">Deformation theories</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>
Yes you can neglect shear deformation if the beam is thin, however you should note that in this case pure strain is also small since moset of the deformations of the beam elements are rigid body motion and rotation, and the length of the beam does not change that much, and if the beam is thin the bending strain can also stay in the elastic range. You can measure the strain in rotated coordinates after removing rigid body motions. this means you have to calculate the pure strains by measuring how much each piece of beam is stretched and how much each each end of it is rotated WITH RESPECT TO A LINE THAT CONNECTS THE TWO NEW POSITIONS OF ITS END POINTS.
</p>
<p>
Peyman
</p>
<p>
</p>
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</ul>Fri, 05 Jun 2009 01:02:36 +0000Peyman Khosravicomment 11126 at https://imechanica.orgStrain components
https://imechanica.org/comment/11093#comment-11093
<a id="comment-11093"></a>
<p><em>In reply to <a href="https://imechanica.org/node/5510">Deformation theories</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 am trying a problem for a cantilever beam in X-Y plane with an end moment considering Large displacement, large rotation, large strain. My questions are:
</p>
<p>
1) Since an end moment is being applied can i neglect shear deformation?
</p>
<p>
2) In this case what should be the components of the inceremntal strain matrices.
</p>
<p>
Please Help
</p>
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</ul>Wed, 03 Jun 2009 06:19:52 +0000surajitdascomment 11093 at https://imechanica.orgLogarithmic strain
https://imechanica.org/comment/11043#comment-11043
<a id="comment-11043"></a>
<p><em>In reply to <a href="https://imechanica.org/node/5510">Deformation theories</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>
Assume a bar with initial length 100 cm which stretches by 1 cm and becomes 101 cm:
</p>
<p>
engineering strain=(101-100)/100= 0.01
</p>
<p>
logarithmic strain=ln(101/100)= 0.00995
</p>
<p>
Yes, chapter 4 of the volume 1 is enough to learn this topic;
</p>
<p>
Regards
</p>
<p>
Peyman Khosravi
</p>
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</ul>Sun, 31 May 2009 02:05:02 +0000Peyman Khosravicomment 11043 at https://imechanica.orgLogarithmic strains
https://imechanica.org/comment/11042#comment-11042
<a id="comment-11042"></a>
<p><em>In reply to <a href="https://imechanica.org/node/5510">Deformation theories</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>
thank you peymann--
</p>
<p>
1) As you mentioned about logarithmic strain above, can you give a numerical example to show that engineering strain will be different from logarithmic strain--extremely grateful---
</p>
<p>
2)as suggested will read further on other strain measures--will the chapter 4 of volume 1 of crisfield's book of basic continuum mechanics suffice for the purpose or more detailed books are required?
</p>
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</ul>Sat, 30 May 2009 22:41:55 +0000kajalschopracomment 11042 at https://imechanica.orgDifferent measures of strain
https://imechanica.org/comment/11039#comment-11039
<a id="comment-11039"></a>
<p><em>In reply to <a href="https://imechanica.org/node/5510">Deformation theories</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;<br />
Behind each definition of strain there is a mathematical reason. Here is one example. We define the engineering strain as (delta L)/L0. It may seem fine. however, a closer look may lead us to say it is more exact to calculate the final strain as the integral of all the small strains during this stretching process. and at each moment, we use the current length. so we can say: e=int(dL/L) calculated from L0 to L. This is actually ln(L/L0) and is called logarithmic or "true" strain. Other definitions of strain have reasons behind them (even if they don't have an understandable physical meaning) and have been defined because we needed them. It is beyond this limited post and you should refer to the continuum mechanics books. But I can give you one example. Green strain (also called green lagrange strain) is the one we need in total and updated lagrangian methods. Its definition is with respect to the initial configuration and this makes it suitable for that method. Please refer to the references for more explanations.<br />
Second question: Sorry. I meant rigid body large rotations. Linear analysis has problem with that and if you are getting large rotations after doing a linear analysis you should then re-examine the problem with nonlinear analysis.<br />
Third question: you know the new position of the end points. Points 1 and 2 are going to the new positions 1’ and 2’. Your new x axis which was already from 1 to 2 in now from 1’ to 2’. Pick up the old configuration, put it on the new one with their x axes coinciding and match the points 1 and 1’. Measure the distance between 2 and 2’. This is actually the change of length in rotated coordinate system. Divide it by initial length. You get the strain.<br />
Peyman Khosravi</p>
<p>
</p>
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</ul>Sat, 30 May 2009 18:35:15 +0000Peyman Khosravicomment 11039 at https://imechanica.orgThanks Peymann for your
https://imechanica.org/comment/11038#comment-11038
<a id="comment-11038"></a>
<p><em>In reply to <a href="https://imechanica.org/node/5510">Deformation theories</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 Peymann for your reply.You said that:
</p>
<p>
1)In regard to my first question, you mentioned different measures of strain-as Engineering strain, Logarithmic strain and Green's strain-but why do we have such measures of strain at all? Is the simple Engineering strain not enough?What is that leads to study different measures of strain and DOES EACH OF THEM HAVE A PHYSICAL MEANING?
</p>
<p>
2)You said that rotated strain means getting rid of rigid body displacements-but when we do a lienar FE analysis of trusses-using truss elements thus- we do not have rigid body displacements in the results at all? Do we?
</p>
<p>
3)And in continuation of 2, after a certain laod level in geometric non-linear analysis, how much do we know that the element/structure has undergne rigid body displacement-to get a new coordinate system in accordance to the actual deformation than including rigid body displacements?
</p>
<p>
</p>
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</ul>Sat, 30 May 2009 17:39:30 +0000kajalschopracomment 11038 at https://imechanica.orgRotated strains
https://imechanica.org/comment/11036#comment-11036
<a id="comment-11036"></a>
<p><em>In reply to <a href="https://imechanica.org/node/5510">Deformation theories</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 Kajal;<br />
First of all I have to talk about the strain, so for now forget the coordinate systems and just imagine a simple horizontal bar element with initial length L0 which is stretched to new length L. And we also assume that the change of length is small.<br />
There are different definitions (measures) for the strain. Each has a different name. Think of them as different units for measuring something. The most traditional one is engineering strain defined as:<br />
e= (L-L0)/L0<br />
Other definitions also exist, such as Green’s strain:<br />
e=(L^2-L0^2)/(2L0^2)<br />
or logarithmic strain:<br />
e=ln(L/L0)<br />
or Almansi’s strain:<br />
e=(L^2-L0^2)/(2L^2)</p>
<p>What you should note is that these are just different measures of strain. The important thing is that when you use one of these strain definitions in calculating the work done by the internal force (1/2 integral of the (strain)*(stress) over the volume) you should use also its matching stress definition to end up with the correct result, i.e. for each strain definition there is a stress definition which matches that.<br />
For the answer to the second part of your question, now consider the simple definition of engineering strain e= (L-L0)/L0. It simply means change of the length divided by the initial length. Let’s say we want to use this definition for a truss element. Assume that this element has rotated, moved, and stretched and finally reached to its final configuration. The final strain in the element is only a function of its final length. We don’t care how much it has rotated or moved as rigid body movement. We just measure its final length and we find how much its length has changed. So the general rule is to somehow eliminate all extra rigid body motions or rotations and look at the bar element as if it is a horizontal bar stretched to its new length. The way we do this is by assuming that a coordinate system is attached to the bar and moves and rotates with it. So rotated strain simply means the strain in the local coordinate system of the element which is attached to it. It means the strain after getting rid of all extra rigid body movements which produce no strain.<br />
Shallow truss element is formulated by approximating for small angles. It is not general and get rid of it. Try to learn the rotated strains because that is the real thing and is what you should always use. It is simple and exact.</p>
<p>Peyman Khosravi</p>
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</ul>Sat, 30 May 2009 06:22:58 +0000Peyman Khosravicomment 11036 at https://imechanica.orgthank you again peymann
https://imechanica.org/comment/11031#comment-11031
<a id="comment-11031"></a>
<p><em>In reply to <a href="https://imechanica.org/node/5510">Deformation theories</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>
<font>thank you very much peymann.Basically, I am reading crisfield's book for non-linear finite element analysis.Can you kindly help me here:</font>
</p>
<p>
<font>1)What is the difference between a shallow truss problem (discussed in chapters 1 ans 2) and the type of problem in chapter 3?</font>
</p>
<p>
<font>2)chapter 3 talks of strain mesures- Green's strain,rotated engineering strain and rotated log strain.The book says we use rotated coordinate system in chapter 3-what is meant by this rotated coordinate system?</font>
</p>
<p>
<font>Sorry, if I am asking a very fundamental question-actually am self sudying the course but at the same time determined and interested-plz help!</font>
</p>
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</ul>Fri, 29 May 2009 22:25:08 +0000kajalschopracomment 11031 at https://imechanica.orgWhy nonlinear analysis
https://imechanica.org/comment/11022#comment-11022
<a id="comment-11022"></a>
<p><em>In reply to <a href="https://imechanica.org/node/5510">Deformation theories</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>
It is a good question. The answer is this: what we have here is a special case of problem which we can solve it analytically. Not every problem is like that. Actually the conventional linear structural analysis also fails in this case. If you model this beam and solve it with a linear finite element analysis software you will see that the vertical deflection at the tip is not correct. the correct value should be zero since the beam has been rolled to a circle and the tip is touching the end point but linear structural analysis doesn't show that. Of course you can find an ANALYTICAL equation which gives you the exact position of the tip at every moment but that is not conventional linear structural analysis.
</p>
<p>
to understand it better consider this. imagine a cantilever beam with end shear load at the tip. linear structural analysis finds zero horizontal deflection at the tip which is wrong. geometrically nonlinear analysis finds the correct answer.
</p>
<p>
Using geometrically nonlinear analysis you are able to find the exact geometry of the beam at each load level.
</p>
<p>
Peyman Khosravi
</p>
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</ul>Fri, 29 May 2009 01:50:49 +0000Peyman Khosravicomment 11022 at https://imechanica.orgthank you peymann for the
https://imechanica.org/comment/11017#comment-11017
<a id="comment-11017"></a>
<p><em>In reply to <a href="https://imechanica.org/node/5510">Deformation theories</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>
thank you peymann for the response.IF we solve such a problem using conventional linear analysis- we will get the same results (final length = 2*pi*R) at some magnitude of moment?
</p>
<p>
IF so, what is it that causes different/more realistic results solving as geometric non-linear problem?
</p>
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</ul>Thu, 28 May 2009 19:35:35 +0000kajalschopracomment 11017 at https://imechanica.orgHi Jayadep,
Since
https://imechanica.org/comment/11012#comment-11012
<a id="comment-11012"></a>
<p><em>In reply to <a href="https://imechanica.org/comment/10947#comment-10947">Practical significance</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 Jayadep,
</p>
<p>
Since we can only observe the motion of deformable materials, the translation, rotation and strain of materials are just defined based on our imaged theories. Hence, I think these conditions are possible deformations.
</p>
<p>
</p>
<p>
Shunlai
</p>
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</ul>Thu, 28 May 2009 07:41:55 +0000Shunlai.Zangcomment 11012 at https://imechanica.orgKajal;
You can easily
https://imechanica.org/comment/11010#comment-11010
<a id="comment-11010"></a>
<p><em>In reply to <a href="https://imechanica.org/node/5510">Deformation theories</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 can easily show (by using your equations and considering that for a full circle from a beam with initail length L we have: L=2*pi*R) that strain=pi*t/L .
</p>
<p>
if we consider strains up to 0.005 as small (which is usually assumed) you can see that for a strip of plate with 1mm thickness and 70 cm length, the strain is small even when we roll it into a full circle. The goal of this example is to show that even in case of very large deflections we may have small pure strains and here small means in the range that for example we can use conventional small strain finite element methods (ofcourse with a rotated coordinate system like corotational).
</p>
<p>
Peyman Khosravi
</p>
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</ul>Thu, 28 May 2009 01:04:37 +0000Peyman Khosravicomment 11010 at https://imechanica.orgHi Sugeng;
Here we assume
https://imechanica.org/comment/11009#comment-11009
<a id="comment-11009"></a>
<p><em>In reply to <a href="https://imechanica.org/comment/10944#comment-10944">Re:Thank you Peymann and Shunlai but</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 Sugeng;
</p>
<p>
Here we assume that wecan neglect the shear deformation (because the beam is very thin). This matches with Euler's beam theory however you have to consider a rotated coordinate system for each beam segment. This can be done easily with corotational analysis.
</p>
<p>
Peyman Khosravi
</p>
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</ul>Thu, 28 May 2009 00:52:22 +0000Peyman Khosravicomment 11009 at https://imechanica.orgsmall strain large rotation-peymann
https://imechanica.org/comment/10948#comment-10948
<a id="comment-10948"></a>
<p><em>In reply to <a href="https://imechanica.org/node/5510">Deformation theories</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>
<font>thanks peymann--yes i do understand the plastic example.But considering the enormous rotation (As well as translation) that occurs , can you numerically show that strain is not large- using the equations pertaining to bending theory above where strain is y /R?you can show example of the plastic sheet itself (also proving numerically its plastic limit/ yield point is not reached)</font>
</p>
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</ul>Wed, 27 May 2009 07:42:39 +0000kajalschopracomment 10948 at https://imechanica.orgPractical significance
https://imechanica.org/comment/10947#comment-10947
<a id="comment-10947"></a>
<p><em>In reply to <a href="https://imechanica.org/comment/10932#comment-10932">The concepts of deformation</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>
Can you please comment on the practical significance of items (1), (5), (6) & (7) above?
</p>
<p>
Thanks,
</p>
<p>
Jayadeep
</p>
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</ul>Wed, 27 May 2009 05:19:13 +0000Jayadeep U. B.comment 10947 at https://imechanica.orgRe:Thank you Peymann and Shunlai but
https://imechanica.org/comment/10944#comment-10944
<a id="comment-10944"></a>
<p><em>In reply to <a href="https://imechanica.org/comment/10937#comment-10937">Thank you Peymann and Shunlai but</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>
Are you sure that Euler beam theory considering large displacement and rotation? I doubt it
</p>
<p>
You can produce a circle from cantilever beam only if you apply constant moment at free beam tip or follower forces which always perpendicular to beam length for instance.
</p>
<p>
Small strain regime means that values from all of strain definition (log, green lagrange, etc) almost coincident within that regime
</p>
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</ul>Wed, 27 May 2009 01:38:00 +0000Sugeng Waluyocomment 10944 at https://imechanica.orgMy example was for a very thin beam...
https://imechanica.org/comment/10943#comment-10943
<a id="comment-10943"></a>
<p><em>In reply to <a href="https://imechanica.org/node/5510">Deformation theories</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 Kajal;
</p>
<p>
If we make a beam very thin, then when we see that when we relaese it after deforming it, it goes back to its initial shape, like a spring. My example was for a quite thin and long beam. I wanted to show that it is possible to have very large deformations and rotations but small strain. And here small strain means in the range of elastic behavior without going into plasticity. Your calculations are correct though. you have definitely seen paper-thin sheets of plastic that we can deform them like a circle and they stay in their elastic region (small strain) and go back to their initial shape like a spring.
</p>
<p>
Regards
</p>
<p>
Peyman
</p>
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</ul>Wed, 27 May 2009 00:11:12 +0000Peyman Khosravicomment 10943 at https://imechanica.orgThank you Peymann and Shunlai but
https://imechanica.org/comment/10937#comment-10937
<a id="comment-10937"></a>
<p><em>In reply to <a href="https://imechanica.org/node/5510">Deformation theories</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>
Peymann, you said,in the cantilever example:
</p>
<p>
"----But you should note that even in this case, the strain in the beam (both axial and shear strains) are again small (because the beam is thin) and we don't deal with a large strain problem"
</p>
<p>
But, as per the bending equation- Euler Bernouli hypothesis:
</p>
<p>
M / I = f / y = E / R
</p>
<p>
where:
</p>
<p>
M = bending moment
</p>
<p>
I = moment of ienrtia
</p>
<p>
f = bending (Axial) stress
</p>
<p>
E = modulus of elasticity
</p>
<p>
R = radius of curvature (that is the radius of the circle in the form of which the beam deflects)
</p>
<p>
y = distance of any fiber from neutral axis
</p>
<p>
Now:
</p>
<p>
M / I = f / E = y / R
</p>
<p>
That is:
</p>
<p>
M /I = epsilon (strain) = y /R
</p>
<p>
so:
</p>
<p>
as "M" increases , R decreases and hence epsilon increases
</p>
<p>
How can we say that strain is not large in the cantilver example you gave?
</p>
<p>
as M is very large, R wou;ld be less, hence epsilon the strain would be large
</p>
<p>
As beam is thin, agreed, y is not so large but we cannot neglect decreases in radius hence increase in starin due to large "M"?Can we?
</p>
<p>
So, how can we say that the catilver exapmle you gave is a small starin problem even though the beam is thin?
</p>
<p>
</p>
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</ul>Tue, 26 May 2009 18:31:53 +0000kajalschopracomment 10937 at https://imechanica.orgThe concepts of deformation
https://imechanica.org/comment/10932#comment-10932
<a id="comment-10932"></a>
<p><em>In reply to <a href="https://imechanica.org/node/5510">Deformation theories</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>
Here are some concepts on deformation.
</p>
<p>
1. large deformation. If we consider this concept from displacement, rotation and strain aspects. The below conditions could be called as large deformation.
</p>
<p>
(1) large displacement, small rotation, small strain;
</p>
<p>
(2) large displacement, small rotation, large strain;
</p>
<p>
(3) large displacement, large rotation, small strain;
</p>
<p>
(4) large displacement, large rotation, large strain;
</p>
<p>
(5) small displacement, small rotation, large strain;
</p>
<p>
(6) small displacement, large rotation, small strain;
</p>
<p>
(7) small displacement, large rotation, large strain;
</p>
<p>
2. small deformation, which is only correspodent to small displacement, small rotation, small strain.
</p>
<p>
<br />
From the above, you can see the shear strain and linear strain have no relations to the rotation of material.
</p>
<p>
</p>
<p>
Shunlai
</p>
<p>
</p>
<p>
</p>
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</ul>Tue, 26 May 2009 14:04:33 +0000Shunlai.Zangcomment 10932 at https://imechanica.orgRotations in deformation theory
https://imechanica.org/comment/10923#comment-10923
<a id="comment-10923"></a>
<p><em>In reply to <a href="https://imechanica.org/node/5510">Deformation theories</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>
Rotation means the rotation of a node in space with respect to its initial position. To make it simple, let's consider a cantilever beam with an end moment. For small amount of moment, the beam has small deformation and we can consider it a small deformation problem. However, when we increase the moment, the beam starts to deflect more and eventually we may make a full circle out of it. In this case each node on the beam has a large rotation and so we cannot consider it a small deformation problem. But you should note that even in this case, the strain in the beam (both axial and shear strains) are again small (because the beam is thin) and we don't deal with a large strain problem.
</p>
<p>
Peyman Khosravi
</p>
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</ul>Mon, 25 May 2009 23:44:08 +0000Peyman Khosravicomment 10923 at https://imechanica.org