Micromechanics of composites/Proof 4: Difference between revisions

Latest revision as of 22:27, 25 July 2017

Question

Let $\partial Ω$ be a surface. Let $𝐱$ be the position vector of points on the surface and let $𝐭$ be a vector field that are defined on $\partial Ω$ . If

\int_{\partial Ω} 𝐱 \times 𝐭 dA = 𝟎

show that

\int_{\partial Ω} 𝐱 \otimes 𝐭 dA = \int_{\partial Ω} 𝐭 \otimes 𝐱 dA .

Proof

If we assume a Cartesian basis, we can write the given relation in index notation as

\int_{\partial Ω} e_{i j k} x_{j} t_{k} dA = 0

where $e_{i j k}$ is the Levi-Civita (permutation) symbol. Since $e_{i j k}$ is does not depend upon the position we can write

e_{i j k} \int_{\partial Ω} x_{j} t_{k} dA = 0 .

Define

A_{j k} : = \int_{\partial Ω} x_{j} t_{k} dA .

Then,

e_{i j k} A_{j k} = 0 .

Expanding, we get

e_{i 11} A_{11} + e_{i 12} A_{12} + e_{i 13} A_{13} + e_{i 21} A_{21} + e_{i 22} A_{22} + e_{i 23} A_{23} + e_{i 31} A_{31} + e_{i 32} A_{32} + e_{i 33} A_{33} = 0 i = 1, 2, 3 .

Expanding further, we get three equations

\begin{matrix} e_{112} A_{12} + e_{113} A_{13} + e_{121} A_{21} + e_{123} A_{23} + e_{131} A_{31} + e_{132} A_{32} & = 0 \\ e_{212} A_{12} + e_{213} A_{13} + e_{221} A_{21} + e_{223} A_{23} + e_{231} A_{31} + e_{232} A_{32} & = 0 \\ e_{312} A_{12} + e_{313} A_{13} + e_{321} A_{21} + e_{323} A_{23} + e_{331} A_{31} + e_{332} A_{32} & = 0 \end{matrix}

or,

\begin{matrix} e_{123} A_{23} + e_{132} A_{32} & = 0 \\ e_{213} A_{13} + e_{231} A_{31} & = 0 \\ e_{312} A_{12} + e_{321} A_{21} & = 0 \end{matrix}

or,

A_{23} = A_{32}; A_{13} = A_{31}; A_{12} = A_{21} .

Therefore,

\int_{\partial Ω} x_{2} t_{3} dA = \int_{\partial Ω} x_{3} t_{2} dA = \int_{\partial Ω} t_{2} x_{3} dA; \int_{\partial Ω} x_{1} t_{3} dA = \int_{\partial Ω} x_{3} t_{1} dA = \int_{\partial Ω} t_{1} x_{3} dA; \int_{\partial Ω} x_{1} t_{2} dA = \int_{\partial Ω} x_{2} t_{1} dA = \int_{\partial Ω} t_{1} x_{2} dA .

Also, by symmetry,

\int_{\partial Ω} x_{1} t_{1} dA = \int_{\partial Ω} t_{1} x_{1} dA; \int_{\partial Ω} x_{2} t_{2} dA = \int_{\partial Ω} t_{2} x_{2} dA; \int_{\partial Ω} x_{3} t_{3} dA = \int_{\partial Ω} t_{3} x_{3} dA .

Therefore we may write,

\int_{\partial Ω} x_{j} t_{k} dA = \int_{\partial Ω} t_{j} x_{k} dA .

Reverting back to direct tensor notation, we get

\int_{\partial Ω} 𝐱 \otimes 𝐭 dA = \int_{\partial Ω} 𝐭 \otimes 𝐱 dA .

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Micromechanics of composites/Proof 4: Difference between revisions

Latest revision as of 22:27, 25 July 2017

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