iMechanica - Comments for "Two-point Tensors"
https://imechanica.org/node/7131
Comments for "Two-point Tensors"enTwo-point tensor, deformation gradient, rotation, deformation
https://imechanica.org/comment/25317#comment-25317
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<p><em>In reply to <a href="https://imechanica.org/node/7131">Two-point Tensors</a></em></p>
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I have described these ideas with some care in <a href="http://imechanica.org/node/538">my class notes on finite deformation</a>. Here is a list of points:
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<strong>Deformation gradient </strong>
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<ul><li>To understand these ideas, all you need is a homogeneous deformation of a body, in a three-dimensional space, from a reference state to a current state.</li>
<li>Consider a set of material particles that forms a segment of a straight line in the space when the body is in the reference state. Because the deformation is homegeneous, the same set of material particles forms a segment of another straight line in the space when the body is in the current state. </li>
<li>In the reference state, the segment is an element <strong>Y</strong> in a vector space U. That is, a linear combination of two segments in U is yet another segment in U, as defined by the usual addition of vectors. </li>
<li>In the current state, the segement is an element <strong>y</strong> in a vector space V. Note that U and V are two distinct vector spaces: we will not form linear combination of an element in U and an element in V. </li>
<li>The deformation gradeint <strong>F</strong> is a linear map that maps an element in U to an element in V, <strong>y</strong> = <strong>FY</strong>. The two elements represent the same set of material particles, respectively, in the reference state and in the current state. </li>
<li>The two vector spaces U and V are distinct. We can choose different bases for the two vector spaces. By definition, the deformation grdient is a two-point tensor. (We define a two-point tensor as a linear map that maps an element in one vector space to an element in a different vector space.)</li>
</ul><p><strong>Geometric representation of the deformation gradient</strong></p>
<ul><li>Consider another set of material particles that forms a unit cube in the reference state. Because the deformation is homogeneous, the same set of material particles forms a parallelepiped in the current state.</li>
<li>We choose the three edges of the unit cube as the basis for both vector spaces.</li>
<li>Relative to this basis, we represent each straight segement <strong>Y </strong>in the reference state by a column, and represent the same set of material particles <strong>y </strong>in the current state by another column. We represent the deformation gradient <strong>F</strong> by a matrix.</li>
<li>Each column of the deformation gradeint <strong>F </strong>reprentsents a vector, corresponding to an edge of the parallelepiped.</li>
</ul><p><span><strong>Green deformation tensor</strong></span></p>
<ul><li><span>The state of matter does not change if the body undergoes a rigid-body rotation in the current state.</span></li>
<li><span>Thus, to characterize deformation without ridid-body rotation, we need to describe the size and the shape of the parallelepiped, but not the orientation of the parallelepiped.</span></li>
<li><span>The size and the shape of the parallelepiped are fully determined by six quantities: the lengths of the three edges and the angles between the three edges. </span><span>The six quantities do not form a tensor. </span></li>
<li><span>The inner products of the edges of the parallelepiped contain the same information of the lengths and angles. Furthermore, the six inner products do form a tensor if we write these inner products as </span><strong>F</strong>T<strong>F</strong><span>. This tensor is the Green deformation tensor.</span></li>
</ul><p><span>My </span><a href="http://imechanica.org/node/538">class notes on finite deformation</a> develop these ideas, and also contain the development of the nominal stress tensor.</p>
<p><span>Also see a thread on </span><a href="http://www.imechanica.org/node/15843">linear algebra, tensors, and linear maps between linear spaces</a> <span>. </span></p>
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</ul>Sat, 28 Dec 2013 16:38:00 +0000Zhigang Suocomment 25317 at https://imechanica.orgDear Wei,
I have a
https://imechanica.org/comment/13584#comment-13584
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<p><em>In reply to <a href="https://imechanica.org/node/7131">Two-point Tensors</a></em></p>
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Dear Wei,
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I have a question regarding to what you said.
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If we consider a body just did rigid body rotation, then the deformation gradient F=R, R is the rotation. If we suppose there is only one frame (Euleran frame), then R is not a two-point tensor, R=RijEiEj, both Ei and Ej are in the same fame
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however, if we still assume there exist a Lagrangian and a Eulerian frame, then F=R=Rij*ei*Ej is a two-point tensor?
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Is this correct based on what you said?
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</ul>Fri, 19 Feb 2010 21:22:16 +0000billowrivercomment 13584 at https://imechanica.orgRe: two-point tensor and Lagrange description
https://imechanica.org/comment/12879#comment-12879
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<p><em>In reply to <a href="https://imechanica.org/node/7131">Two-point Tensors</a></em></p>
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There are two separate issues here: two-point tensor and Lagrange desciption. While two-point tensor is not a must, a Lagrange system is usually required to properly describe a solid. A Lagrange system, in contrast to an Euler system, traces material points by its original (reference) position rather than its current position (so that a cooridnate system moves during deformation). The former is often adopted in solid mechanics while the later in fluid mechanics, because of the physics of the two types of materials.
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Solids (especially elastic solids) differ from fluids in that they have "memory": a solid knows its original state to some extent, while a fluid only cares about its current state (with some exceptions). Therefore, to fully describe a solid, we need to specify a reference state, and measure the difference between its current state and the reference. A two-point tensor, the deformation gradient, is naturally involved to bridge the two states. Two point tensors can be avoided by introducing the right Cauchy-Green deformation tensor F'F (or Green strain), and the second PK stress, with both "legs" in the reference state. Actually, these tensors will remove the rigid-body rotations, which are included in F.
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However, it is imposible to describe the deformation state of an elastic solid with Eulerian tensors only, defined in the current state. You need to tell the material where it was.
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</ul>Sun, 22 Nov 2009 19:52:36 +0000Wei Hongcomment 12879 at https://imechanica.org