Recent content by chowdhury

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    A FEM basis polynomial order and the differential equation order

    Is there a good rubric on how to choose the order of polynomial basis in an Finite element method, let's say generic FEM, and the order of the differential equation? For example, I have the following equation to be solved ## \frac{\partial }{\partial x} \left ( \epsilon \frac{\partial u_{x}...
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    A Gradient of higher rank tensor

    Does anybody know how to derivemy above three queries? Thanks.
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    A Gradient of higher rank tensor

    In this particular equation, I expect the result to be of rank ZERO. $$\nabla \cdot (\bf{\epsilon}^{S} \cdot \nabla \phi) = \nabla \cdot (\bf{e} : \nabla_{s} \bf{u}) = \textup{rank zero} $$
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    A Gradient of higher rank tensor

    I understand up to this, I cannot say further below.
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    A Gradient of higher rank tensor

    @anuttarasammyak : I am sorry, I am not clear how to proceed, better, I would seek @Orodruin as the Judge!
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    A Gradient of higher rank tensor

    @anuttarasammyak : yes in most cases, Query 1.) here is what I try to follow $$\nabla \cdot (\bf{\epsilon}^{S} \cdot \nabla \phi) = \nabla \cdot (\bf{e} : \nabla_{s} \bf{u}) $$ Now $$\nabla_{i} \epsilon_{ij}^{S} \nabla_{j} \phi = \nabla_{i} \epsilon_{ij}^{S} \phi_{,j} =...
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    A Gradient of higher rank tensor

    [FONT=-apple-system]## \nabla ## is the usual operator in maths [FONT=-apple-system]##\nabla_s## is the symmetric part of the gradient operator, which results in strain of the material [FONT=-apple-system]##e## is the piezoelectric coefficient and it is a third rank tensor...
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    A Gradient of higher rank tensor

    @anuttarasammyak : I will provide after the last #16 of mine. Thanks for being so patient.
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    A Gradient of higher rank tensor

    @Orodruin: @anuttarasammyak: I derived this quantity, following the book Acoustic Fields and Waves in Solids. Volume I Hardcover – April 20, 1973 $$ \bf{c^{eff}} = \bf{c^{E}} + e^{transpose} \cdot (\bf{(\epsilon^{S}})^{-1} \bf{e})$$ The notation my be incorrect, but the true nature is CORRECT...
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    A Gradient of higher rank tensor

    @Orodruin : I was told this book is the Bible, and there is no mistake in it. Without complaining, which is easy to rant to anybody in the world, I took an introspective view and try to educate myself.
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    A Gradient of higher rank tensor

    I applied the defintions, but could not find the reasonable notation, that is the problem.
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    A Gradient of higher rank tensor

    @Orodruin : I have noticed that in the field of piezoelectricity, "they" denote like this, Here is another book, https://www.amazon.com/dp/3540686800/?tag=pfamazon01-20 What is driving me crazy is this statement, from the end of this book Acoustic Fields and Waves in Solids. Volume I...
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    A Gradient of higher rank tensor

    @anuttarasammyak Here is the notation for double dot in the book, [FONT=-apple-system]
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    A Gradient of higher rank tensor

    @Orodruin : I understand the transpose operator appears in matrix operations, however, I have seen the transpose operation in a 3rd rank tensor in Wikipedia, as suggested by one of the moderators, @andrewkirk. What I am lacking how to represent them. Thanks for your clarification. How to say...
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    A Gradient of higher rank tensor

    @Orodruin : If there a way to thank you by offering and serving you a good coffee, I would be pleased to do so to you. But we live far apart, so Thank you is all I can say at this moment. I understand your suggestion on the recent book, but this is what I was given and for some reason, I have...