iMechanica - Comments for "Stresses and Strains"
https://imechanica.org/node/2663
Comments for "Stresses and Strains"enjoint material interface
https://imechanica.org/comment/9594#comment-9594
<a id="comment-9594"></a>
<p><em>In reply to <a href="https://imechanica.org/node/2663">Stresses and Strains</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>
In joint material interface between elastic and elastic-plastic material, how can we show the continuity of displacement equation theoretically?It must be continuous and compatible of deformation in practical case.Could anyone help me to understand this answer?
</p>
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</ul>Mon, 19 Jan 2009 04:13:23 +0000nadimcomment 9594 at https://imechanica.orgjump conditions
https://imechanica.org/comment/6506#comment-6506
<a id="comment-6506"></a>
<p><em>In reply to <a href="https://imechanica.org/node/2663">Stresses and Strains</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 Kamyar:</p>
<p>First of all, I don't know what you mean exactly by "Traction stresss" here. Let me assume you mean traction.</p>
<p>I think the best way of looking at problems like this is to think of<br />
balance laws on measure-zero sets (in 3D a surface, for example) and<br />
kinematics of deformation and not to make any general rules. </p>
<p>An interface means a surface on which deformation gradient (or in the<br />
linearized approximation, strain tensor) has a jump discontinuity. But<br />
then the question is weather deformation gradient can have an arbitrary<br />
jump. The answer is no and the jump in deformation gradient has to<br />
satisfy the Hadamard's compatibility equations, i.e. [|F|]= a tensor<br />
product N, where N is unit normal of the undeformed interface at a<br />
given point X on the undeformed interface and "a" is an arbitrary<br />
vector. If you assume that there are no defects like a crack, etc. that<br />
would mean that deformation mapping is single-valued or displacement<br />
field (in the linearized approximation) is continuous. F maps a tangent<br />
vector in the undeformed configuration to a deformed tangent vector<br />
(you could think of these tangent vectors as line elements with<br />
infinitesimal lengths). Given a tangent vector in the undeformed<br />
interface, both F^- and F^+ (deformation gradients on both sides of the<br />
interface) must map it to the same deformed tangent map otherwise the<br />
deformation mapping cannot be single-valued (slip). This would then<br />
give Hadamard's compatibility conditions. In the linearized theory, you<br />
know that displacements are different on both sides of the interface<br />
but displacement field is continuous on the interface (displacement<br />
jump is the zero vector) and hence you can conclude that partial<br />
derivatives of displacements taken in the interface should agree too,<br />
i.e. those components of strain that do not involve partial derivatives<br />
with respect to normal to the interface should be continuous. Other<br />
components are, in general, discontinuous.</p>
<p>Continuity of tractions results from localization of balance of linear<br />
momentum on the interface in the case of stationary interfaces (I don't<br />
like the term "Newton's third law"). Of course, when an interface is<br />
dynamically moving, e.g. a shock wave or a failure front, traction is<br />
not continuous. Instead, jump in traction is proportional to the normal<br />
velocity of the interface.</p>
<p>Arash</p>
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</ul>Tue, 05 Feb 2008 23:07:50 +0000arash_yavaricomment 6506 at https://imechanica.orgBecause traction strains involve d(displacement)/dr terms
https://imechanica.org/comment/6495#comment-6495
<a id="comment-6495"></a>
<p><em>In reply to <a href="https://imechanica.org/node/2663">Stresses and Strains</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 Kamyar and Ying,
</p>
<p>
You have brought up a very interesting problem and I would like to join the discussion here.
</p>
<p>
First let me rephrase Kamyar's problem. You are talking about a <em>cylindrical</em> inclusion with radius R embeded in a matrix whose outer radius goes to infinite. In the following we make use of the conventional cylindrical coordinate to discuss this problem. To distinguish the inclusion and the matrix we use "1" to denote the inclusion and "2" to denote the matrix. By "antiplane problem" you mean we apply shear stresses <em>σzr</em> at the ends of this composite. Maybe it's easier to imagine if the matrix outer radius is finite (set to be <em>Ro</em>) because we also need to apply shear stress at the outer circumference surface of the matrix, pointing to <em><strong>z</strong></em> direction, i.e. the other pair of <em>σzr</em>.
</p>
<p>
Kamyar's question is that on the inclusion/matrix interface, i.e. at r=R, why εzr1 ≠ εzr2 but εzθ1 = εzθ2?
</p>
<p>
First, as pointed out by Ying, we all agree that at r=R, u1=u2, v1=v2, w1=w2.
</p>
<p>
Then let's list all the strain-displacment equations in cylindrical coordinates. We keep consistent with Kamyar's division of strains:</p>
<p>
<strong>Traction strains</strong> (If we make an imaginary cut at the interface, those are the strains that will be exposed. Because the interface normal is in <em><strong>r</strong></em> direction, all strains of this kind should include an r in their subscripts):
</p>
<p>
εr=∂u/∂r
</p>
<p>εzr=[∂w/∂r+∂u/∂z]/2</p>
<p>
εrθ=[(1/r)∂u/∂θ+∂v/∂r-v/r]/2
</p>
<p>
Non-traction strains (all the rest strains)
</p>
<p>
εθ=(∂v/∂θ+u)/r
</p>
<p>
εz=∂w/∂z
</p>
<p>
εθz=[∂v/∂z+(1/r)∂w/∂θ]/2
</p>
<p>
Looking carefully, we can discover that all traction strains involve a differential term with respect to r while all non-traction terms only include derivatives with respect to <em>z</em> or <em>θ</em>. Why does this matter? Well, remember the difference between strain and displacement is an integral over a length. While the integral limits in z or θ directions are consistent for both the inclusion and the matrix, the starting and ending limits of the r-direction integral is very different. For the cylindrical inclusion, to find the interface displacements we need to integrate the traction strains over [0, R], but [R,Ro] for the matrix. Therefore, because all the displacements pairs are equal at the interface, any strain involving derivatives with respect to r cannot be equal at the interface; but any strain doesn't involve d(displacement)/dr should be equal at r=R.</p>
<p>
A final notice is that we need to be careful when we try to generalize the conclusions from this problem. As you know, the strict boundary conditions at interfaces should be tractions equal and displacements equal. Regarding to the strains, we really need to examine the strain-displacment equations along with the problem geometry. Therefore, it is not always true that non-traction strains are cotinuous at the interface.</p>
<p>
</p>
<p>
</p>
<p>
</p>
<p>
</p>
<p>
</p>
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</ul>Tue, 05 Feb 2008 04:33:43 +0000Nanshu Lucomment 6495 at https://imechanica.orgcontinuity of derivatives
https://imechanica.org/comment/6494#comment-6494
<a id="comment-6494"></a>
<p><em>In reply to <a href="https://imechanica.org/comment/6492#comment-6492">Strains</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 Kamyar,
</p>
<p>
Partial derivatives are nothing but infinitesimal changes of a function along a particular direction. Since the displacement is continuous across the interface, the change of the function as you move within the plane of interface is also uniquely defined, i.e. continuous across the interface. However, when you move away from the interface, the change need not to be unique. Therefore, the partial derivatives on the tangent plane of the interface are continuous, while those on a direction out of the tangent planes need not be continuous.
</p>
<p>
Wei
</p>
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</ul>Tue, 05 Feb 2008 04:33:01 +0000Wei Hongcomment 6494 at https://imechanica.orgThe displacement determiend
https://imechanica.org/comment/6493#comment-6493
<a id="comment-6493"></a>
<p><em>In reply to <a href="https://imechanica.org/node/2663">Stresses and Strains</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 class="MsoNormal">
<span>Dear Kamyar,</span>
</p>
<p><span>You know, the strain is continuous at interface, which is basically determined by the compatible of the displacement at the interface. Here, you problem is out. For the Cartesian Coordinate (like X and Y), the displacement at the interface should be compatible for u^1=u^2 and v^1=v^2. As the epsilonx=d(u)/d(x) and epsilony=d(v)/d(y), it obviously that the epsilonx^1= epsilonx^2 and epsilony^1= epsilony^2. However, it dose not exist in the polar coordination. As the epsilon r=d(u)/d(r), but epsilon theta = (d(v)/d(theta))/r + u/r, I think the epsilon theta^1= epsilon theta^2. You could get it by yourself. </span></p>
<p class="MsoNormal">
<span>I hope it could help you.</span>
</p>
<p>
Ying Li Department of Engineering Mechanics Tsinghua University Beijing, 100084, P. R. CHINA
</p>
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</ul>Mon, 04 Feb 2008 14:54:59 +0000Ying Licomment 6493 at https://imechanica.orgStrains
https://imechanica.org/comment/6492#comment-6492
<a id="comment-6492"></a>
<p><em>In reply to <a href="https://imechanica.org/node/2663">Stresses and Strains</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 Ying
</p>
<p>
Thank you for your answer. Yes, the displacement should be continuous at the interface, but why some components of its partial derivative are continuous and some of them are discontinuous? Please note that the traction strains (I mean the strains related to traction stresses) are discontious. Suppose we have an antiplane problem and an inclusion (r<R) embeded in an infinite medium. In this case \sigma_{zr}^{(1)}=\sigma_{zr}^{(2)} on r=R but \epsilon_{zr}^{(1)} != \epsilon_{zr}^{(2)} while \epsilon_{z\theta}^{(1)}=\epsilon_{z\theta}^{(2)} at the interface.
</p>
<p>
Could you please explain more?
</p>
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</ul>Mon, 04 Feb 2008 12:04:55 +0000Kamyar M Davoudicomment 6492 at https://imechanica.orgWhat are non-traction strains?
https://imechanica.org/comment/6491#comment-6491
<a id="comment-6491"></a>
<p><em>In reply to <a href="https://imechanica.org/node/2663">Stresses and Strains</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>What are non-traction strains?</p>
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</ul>Mon, 04 Feb 2008 04:27:15 +0000Amit.Ranadecomment 6491 at https://imechanica.orgthe compatible of the deformation
https://imechanica.org/comment/6490#comment-6490
<a id="comment-6490"></a>
<p><em>In reply to <a href="https://imechanica.org/node/2663">Stresses and Strains</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 Kamyar,
</p>
<p>
The non-traction strains are continuous because the deformation of the structures could be compatible. That is to say, the dispalcement at the interface should be continuous.
</p>
<p>
Ying Li Department of Engineering Mechanics Tsinghua University Beijing, 100084, P. R. CHINA
</p>
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</ul>Mon, 04 Feb 2008 03:40:55 +0000Ying Licomment 6490 at https://imechanica.org