Micromechanics of composites/Proof 3

Surface-volume integral relation 1

Let $\mathrm{\Omega }$ be a body and let $\partial \mathrm{\Omega }$ be its surface. Let $𝐧$ be the normal to the surface. Let $𝐯$ be a vector field on $\mathrm{\Omega }$ and let $𝑺$ be a second-order tensor field on $\mathrm{\Omega }$. Show that

$\underset{\partial \mathrm{\Omega }}{\int }𝐯\otimes ({𝑺}^T\cdot 𝐧)\text{dA}=\underset{\mathrm{\Omega }}{\int }[\mathrm{\nabla }𝐯\cdot 𝑺+𝐯\otimes (\mathrm{\nabla }\bullet {𝑺}^T)]\text{dV}.$

Proof:

Recall the relation

$\mathrm{\nabla }\bullet [(𝐯\cdot 𝐚)(𝑺\cdot 𝐛)]=𝐚\cdot [\{\mathrm{\nabla }𝐯\cdot 𝑺+𝐯\otimes (\mathrm{\nabla }\bullet {𝑺}^T)\}\cdot 𝐛].$

Integrating over the volume, we have

$\underset{\mathrm{\Omega }}{\int }\mathrm{\nabla }\bullet [(𝐯\cdot 𝐚)(𝑺\cdot 𝐛)]\text{dV}=\underset{\mathrm{\Omega }}{\int }𝐚\cdot [\{\mathrm{\nabla }𝐯\cdot 𝑺+𝐯\otimes (\mathrm{\nabla }\bullet {𝑺}^T)\}\cdot 𝐛\text{dV}.$

Since $𝐚$ and $𝐛$ are constant, we have

$\underset{\mathrm{\Omega }}{\int }\mathrm{\nabla }\bullet [(𝐯\cdot 𝐚)(𝑺\cdot 𝐛)]\text{dV}=𝐚\cdot [\{\underset{\mathrm{\Omega }}{\int }[\mathrm{\nabla }𝐯\cdot 𝑺+𝐯\otimes (\mathrm{\nabla }\bullet {𝑺}^T)]\text{dV}\}\cdot 𝐛].$

From the divergence theorem,

$\underset{\mathrm{\Omega }}{\int }\mathrm{\nabla }\bullet 𝐮\text{dV}=\underset{\partial \mathrm{\Omega }}{\int }𝐮\cdot 𝐧\text{dA}$

we get

$\underset{\mathrm{\Omega }}{\int }\mathrm{\nabla }\bullet [(𝐯\cdot 𝐚)(𝑺\cdot 𝐛)]\text{dV}=\underset{\partial \mathrm{\Omega }}{\int }[(𝐯\cdot 𝐚)(𝑺\cdot 𝐛)]\cdot 𝐧\text{dA}.$

Using the relation

$[(𝐯\bullet 𝐚)(𝑺\bullet 𝐛)]\cdot 𝐧=𝐚\cdot [\{𝐯\otimes ({𝑺}^T\bullet 𝐧)\}\cdot 𝐛]$

we get

$\underset{\mathrm{\Omega }}{\int }\mathrm{\nabla }\bullet [(𝐯\cdot 𝐚)(𝑺\cdot 𝐛)]\text{dV}=\underset{\partial \mathrm{\Omega }}{\int }𝐚\cdot [\{𝐯\otimes ({𝑺}^T\bullet 𝐧)\}\cdot 𝐛]\text{dA}.$

Since $𝐚$ and $𝐛$ are constant, we have

$\underset{\mathrm{\Omega }}{\int }\mathrm{\nabla }\bullet [(𝐯\cdot 𝐚)(𝑺\cdot 𝐛)]\text{dV}=𝐚\cdot [\{\underset{\partial \mathrm{\Omega }}{\int }𝐯\otimes ({𝑺}^T\bullet 𝐧)\text{dA}\}\cdot 𝐛].$

Therefore,

$𝐚\cdot [\{\underset{\partial \mathrm{\Omega }}{\int }𝐯\otimes ({𝑺}^T\bullet 𝐧)\text{dA}\}\cdot 𝐛]=𝐚\cdot [\{\underset{\mathrm{\Omega }}{\int }[\mathrm{\nabla }𝐯\cdot 𝑺+𝐯\otimes (\mathrm{\nabla }\bullet {𝑺}^T)]\text{dV}\}\cdot 𝐛].$

Since $𝐚$ and $𝐛$ are arbitrary, we have

$\underset{\partial \mathrm{\Omega }}{\int }𝐯\otimes ({𝑺}^T\bullet 𝐧)\text{dA}=\underset{\mathrm{\Omega }}{\int }[\mathrm{\nabla }𝐯\cdot 𝑺+𝐯\otimes (\mathrm{\nabla }\bullet {𝑺}^T)]\text{dV}◻$

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