Thursday, February 12, 2009

When the basis for r(0) is {ijk} then partials wrt k and partials wrt "s" have the same meaning as r(s)=sk. Therefore D1(ijk)r(ijk)-->r(0) and D2(ijk)r(ijk)-->D1(s)r(s). If this theory is sound then the planned "pure" shear test should work.

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