iMechanica - fundamental solution
https://imechanica.org/taxonomy/term/5141
enFundamental Solutions and Dual Boundary Element Method for Crack Problems in Plane Cosserat Elasticity
https://imechanica.org/node/17204
<div class="field field-name-taxonomy-vocabulary-6 field-type-taxonomy-term-reference field-label-hidden"><div class="field-items"><div class="field-item even"><a href="/taxonomy/term/76">research</a></div></div></div><div class="field field-name-taxonomy-vocabulary-8 field-type-taxonomy-term-reference field-label-hidden"><div class="field-items"><div class="field-item even"><a href="/taxonomy/term/10095">cosserat</a></div><div class="field-item odd"><a href="/taxonomy/term/607">boundary element method</a></div><div class="field-item even"><a href="/taxonomy/term/2608">BEM</a></div><div class="field-item odd"><a href="/taxonomy/term/5141">fundamental solution</a></div><div class="field-item even"><a href="/taxonomy/term/420">cracks</a></div><div class="field-item odd"><a href="/taxonomy/term/10096">crack in Cosserat materials</a></div><div class="field-item even"><a href="/taxonomy/term/10097">Cosserat elasticity</a></div><div class="field-item odd"><a href="/taxonomy/term/10098">micropolar</a></div><div class="field-item even"><a href="/taxonomy/term/10099">Griffith crack</a></div></div></div><div class="field field-name-body field-type-text-with-summary field-label-hidden"><div class="field-items"><div class="field-item even"><p><a href="http://orbilu.uni.lu/handle/10993/17878">http://orbilu.uni.lu/handle/10993/17878</a></p>
<p><a href="http://hdl.handle.net/10993/17878">http://hdl.handle.net/10993/17878</a></p>
<p>In this paper, both singular and hypersingular fundamental solutions of plane Cosserat elasticity are derived and given in a ready-to-use form. The hypersingular fundamental solutions allow to formulate the analogue of Somigliana stress identity, which can be used to obtain the stress and couple stress fields inside the domain from the boundary values of the displacements, microrotation and stress and couple stress tractions. Using these newly derived fundamental solutions, the boundary integral equations of both types are formulated and solved by the boundary element method. Simultaneous use of both types of the equations (approach known as the dual BEM) allows to treat problems where parts of the boundary are overlapping, such as crack problems, and to do this for general geometry and loading conditions. The high accuracy of the boundary element method for both types of equations is demonstrated for a number of benchmark problems, including a Griffith crack problem and a plate with an edge crack. The detailed comparison of the BEM-results and the analytical solution for a Griffith crack is given, particularly, in terms of stress and couple stress intensity factors, as well as the crack opening displacements and microrotations on the crack faces. A modified method for computing the couple stress intensity factors is also proposed and evaluated. Finally, the asymptotic behavior of the solution to the Cosserat crack problems, in the vicinity of the crack tip is analyzed.</p>
</div></div></div>Sat, 20 Sep 2014 08:03:24 +0000Stephane Bordas17204 at https://imechanica.orghttps://imechanica.org/node/17204#commentshttps://imechanica.org/crss/node/17204A boundary element formulation problem
https://imechanica.org/node/8151
<div class="field field-name-taxonomy-vocabulary-8 field-type-taxonomy-term-reference field-label-hidden"><div class="field-items"><div class="field-item even"><a href="/taxonomy/term/607">boundary element method</a></div><div class="field-item odd"><a href="/taxonomy/term/3058">anisotropic</a></div><div class="field-item even"><a href="/taxonomy/term/5141">fundamental solution</a></div><div class="field-item odd"><a href="/taxonomy/term/5142">laplacian</a></div><div class="field-item even"><a href="/taxonomy/term/5143">biharmonic</a></div><div class="field-item odd"><a href="/taxonomy/term/5144">green's function</a></div></div></div><div class="field field-name-body field-type-text-with-summary field-label-hidden"><div class="field-items"><div class="field-item even"><p>I am working on some boundary integral equation formulation and I am currently stuck with some mathematics. I wish anyone can help me out with this. </p>
<p> </p>
<p>I have an anisotropic (sometimes called generalized) biharmonic differential operator which takes the form </p>
<p>L = k11 D1^4 + k12 D1^2 D2^2 + k22 D2^4 </p>
<p>where D1 = d/dx, D2 = d/dy, my problem is two dimensional.</p>
<p>I need to find a fundamental solution (Green's function) for this operator, that is </p>
<p>L(u) = -delta</p>
<p>where delta is Dirac delta function.</p>
<p>I thought of two solutions but they both seem to fail, first, i thought of making some coordinate transformation so that the coefficients of the terms of the operator become the same, and hence, use the traditional biharmonic Green's function (1/8/pi r^2 ln(r) ), however, my proposed transformation takes the form </p>
<p>xi = a11 x + a12 y</p>
<p>eta = a21 x + a22 y</p>
<p>there a11, a12, a21 and a22 are to be determined according to the above requirement (all operator terms have equal coefficients), however, using this transformation got me more terms in the final operator form (terms involving D1^3 D2 and D1 D2^3)</p>
<p>My second solution was to decompose the anisotropic biharmonic operator into two anisotropic laplacian operators and solving two anisotropic laplacian equations instead of one, I succeeded in the decomposition and I've solved the first anisotropic laplacian equation, however, I cannot solve the seconf one. the solution I got was the following</p>
<p>L = (a1 D1^2 + a2 D2^2)*(b1 D1^2 + b2 D2^2)</p>
<p>let L1 = (a1 D1^2 + a2 D2^2) </p>
<p>and</p>
<p>L2 = (b1 D1^2 + b2 D2^2)</p>
<p>now, L1(L2(u)) = -delta</p>
<p>let L2(u) = v</p>
<p>then L1(v) = -delta</p>
<p>this gives v = -1/sqrt(a1*a2)*ln(r')</p>
<p>where r' = sqrt(x^2/a1 + y^2/a2)</p>
<p>now, we have L2(u) = v</p>
<p>thus L2(u) = -1/2/sqrt(a1*a2)*ln(x^2/a1 + y^2/a2)</p>
<p>which gives (b1 D1^2 + b2 D2^2)(u) = -1/2/sqrt(a1*a2)*ln(x^2/a1 + y^2/a2)</p>
<p>I am stuck at this point, I have no clue how to solve this</p>
<p>any ideas ??</p>
<p> </p>
<p>thanks</p>
<p> </p>
<p>Ahmed </p>
</div></div></div>Wed, 05 May 2010 21:07:58 +0000ahmed.hussein8151 at https://imechanica.orghttps://imechanica.org/node/8151#commentshttps://imechanica.org/crss/node/8151