TABLE OF CONTENTS In this section we consider a particular solution for the Airy stress function that allows some important properties of displacement fields to be discussed, and relationships between solutions to be explored.
Choosing
gives rise to a particularly simple stress field given by ? rr =D(1+2log r), ? ??= D(3+2log r), ? r ?=0.
Of particular interest are the strain and displacement fields for this solution.
The strain tensor is computed from stress using the isotropic compliance tensor in terms of the Kolosov constant K, with 2 ? set to unity for simplicity. Application of the inc operator shows, however, that this strain is not compatible. Hence the application of IntegrateStrain runs into difficulties. If the steps of this procedure are formally completed, however, a displacement field can be found that contains terms depending on z. If these are discarded, a plane displacement field remains that in fact can be shown to give rise to the correct strain.
<b class="bold">Stress = AiryStress[D r^2 Log [r]]</b><b class="bold">Strain=DDot[IsotropicComplianceK[K],Stress];</b><b class="bold">Inc [Strain]</b><b class="bold">theta = IntegrateGrad[-Curl[Strain]];</b><b class="bold">omega =Table[Sum[Signature[{i, j, k}] theta [[k]],</b><b class="bold">{k, 3}], {i, 3}, {j, 3}];</b><b class="bold">Uint = Simplify[IntegrateGrad[Strain + omega]]</b><b class="bold">Eint = Simplify [(Grad[uint] + Transpose[Grad[uint]])/2]</b>The displacement field has the form
Of primary importance here is the fact that the tangential displacement component u ? contains the polar angle ? and therefore is multivalued, unless a branch cut is introduced. Taking the location...
