iMechanica - Comments for "Question about Anisotropic Plasticity models (Hill48)"
https://imechanica.org/node/20181
Comments for "Question about Anisotropic Plasticity models (Hill48)"enOk, I think I got it. So if I
https://imechanica.org/comment/28544#comment-28544
<a id="comment-28544"></a>
<p><em>In reply to <a href="https://imechanica.org/comment/28543#comment-28543">Dear Aleksander, </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, I think I got it. So if I understand you correctly, both Hill48 and von Mises criteria are based on critical value of distortional energy. In case of von Mises criterion, the limit energy is the same for every direction (isotropic criterion) - as a result the deviatoric projection of the criterion is circle. However, when we consider Hill48 model, the energy threshold is dependent on the direction of loading (anisotropy). As a result, the deviatoric projection is an ellipse. The material coordinate system is aligned with major/minor axis of said ellipse and so the quadratic form can be defined with diagonal matrix, correct? The last think that comes to my mind would be to ask if material directionw (min/max strength of material - according to Hill) are always aligned with the direction of manufacturing? I assume we are discussing materials that can be reasonably well modelled with Hill criterion, e.g. thin sheets of metal. </p>
<p>Once again thank you very much, your insight was really helpful.</p>
<p>Aleksander</p>
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</ul>Fri, 19 Aug 2016 13:51:47 +0000Aleksander Marekcomment 28544 at https://imechanica.orgDear Aleksander,
https://imechanica.org/comment/28543#comment-28543
<a id="comment-28543"></a>
<p><em>In reply to <a href="https://imechanica.org/comment/28542#comment-28542">Dear Stefan,</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>Hill's quadratic is analogous to an elastic strain energy (or its conjugate, in stress space), with the additional feature of incompressibility. The "stiffness" matrix is its Hessian. This is in the "diagonal" form in the symmetry frame. As we know from linear algebra, any quadratic can be diagonalized by performing appropriate (orthogonal) transformations. With one major difference: here the transformations are those induced in the stress space by orthogonal transformations in the coordinate space (the 3D space) - the usual transformation formulas for the stress components under a change of coordinates. </p>
<p>Good luck,<br />Stefan</p>
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</ul>Thu, 18 Aug 2016 16:53:58 +0000Stefan C. Soarecomment 28543 at https://imechanica.orgDear Stefan,
https://imechanica.org/comment/28542#comment-28542
<a id="comment-28542"></a>
<p><em>In reply to <a href="https://imechanica.org/node/20181">Question about Anisotropic Plasticity models (Hill48)</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 Stefan,</p>
<p>Thank you for outlining those principles for me, mostly it is with agreement with my intuition, so that's good. However, could you elaborate on point 1. - 'diagonal' form of criterion? When I was asked about that while ago by my supervisor, that was exactly the point I wanted to make ('I assume that similarly to case of orthotropic elasticity there must be some frame that leads to 'principal' formulation') but I was unsure where would it manifest in plasticity. In case of elasticity when principal stiffness axes are found, the stiffness matrix indeed takes diagonal form (e.g. bending of beam in principal axes only leads to deflection in direction of loading). In case of orthotropic elasticity, as I understand, principal axes are defined so that normal strains do not produce shearing stresses, am I right? How would it translate to plasticity? Considering effective yield stress as:<br />σeff=σ'Pσ<br />does only material axes produce matrix P that has 0 terms corresponding to normal-shear interaction terms? Or I think equivalently you can say that only in this frame principal stress direction is aligned with principal axes of the strain-increment (which is also true for unique angle given by Hill in his book).</p>
<p>2) Clear on that, thanks.</p>
<p>3) Yes, I meant objectivity, I confused the terms. And as you discussed in point 1, the criterion is objective, thanks.</p>
<p>As for the book, I've just borrowed it from library and mostly it covers what Hill wrote in his 1947 paper introducing the Hill48 model, however, the book gives explicit formula for yield stress in any direction, given the yield stresses in material axes which is very useful for me. Still, I'd like to find something more on that 'diagonal' form of plasticity in material axes (I understand the point and I totally agree, I just need to convince my supervisor).</p>
<p>Cheers,</p>
<p>Aleksander</p>
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</ul>Thu, 18 Aug 2016 10:36:24 +0000Aleksander Marekcomment 28542 at https://imechanica.orgDear Aleksander,
https://imechanica.org/comment/28540#comment-28540
<a id="comment-28540"></a>
<p><em>In reply to <a href="https://imechanica.org/node/20181">Question about Anisotropic Plasticity models (Hill48)</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>Let us recall first that anisotropy means variation with respect to directions in space, or orientation. Therefore, describing this variation requires a reference frame, usually referred to as "material" frame. Material properties are "objective" (let us assume that) and thus one could pick any material frame to describe them. However, as you also mentioned in your post, symmetries, if present, lead to the selection of a natural "symmetry" frame, one in which special directions or planes allow for simpler mathematical descriptions.</p>
<p>Taking Hill's quadratic as example:</p>
<p>1) It is a "scalar" orthotropic function with one tensor argument. It acquires the simplest ("diagonal") form in the unique (up to a symmetry transformation) frame associated with the three orthogonal symmetry planes. Of course, once the parameters of Hill'48 are known in this particular frame one can express it with respect to any other frame. With additional ("apparent") parameters ! Thus, only with respect to the symmetry frame the set of parameters is "optimal" (translating in a minimal set of experiments).</p>
<p>2) Regarding uniaxial tests: Restricting further our discussion to the case of orthotropic sheet metal, and aligning the material frame along the symmetry axes, it should be obvious that one needs to perform such tests only for directions in the first quadrant. Thus a 120-degs test is identical to a 60-degs one.</p>
<p>3) Invariance with respect to coordinate systems is a matter of "objectivity". Thermodynamics is concerned with dissipation. This has to be positive, always. This means that Hill's quadratic has to be positive definite to be thermodynamically admissible (a condition it satisfies most of the time, since experimental data "suggests" convex shapes). </p>
<p>The best reference on Hill'48 is... Hill's book, The mathematical theory of plasticity. </p>
<p>Best <br />Stefan</p>
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</ul>Thu, 18 Aug 2016 06:32:32 +0000Stefan C. Soarecomment 28540 at https://imechanica.org