Micromechanics of composites/Proof 12

Question

Let $𝑨$ and $𝑩$ be two second-order tensor fields. Let the average of any second-order tensor field ($𝑺$) over the region $\mathrm{\Omega }$ (of volume $V$) be defined as

$⟨𝑺⟩:=\frac{1}{V}\underset{\mathrm{\Omega }}{\int }𝑺\text{dV}.$

Show that

$⟨𝑨\cdot 𝑩⟩-⟨𝑨⟩\cdot ⟨𝑩⟩=⟨[𝑨-⟨𝑨⟩]\cdot [𝑩-⟨𝑩⟩]⟩.$

Proof

Expanding out the right hand side, we have

$\begin{array}{cc}⟨[𝑨-⟨𝑨⟩]\cdot [𝑩-⟨𝑩⟩]⟩& =\frac{1}{V}\underset{\mathrm{\Omega }}{\int }[𝑨-⟨𝑨⟩]\cdot [𝑩-⟨𝑩⟩]\text{dV}\\ & =\frac{1}{V}\underset{\mathrm{\Omega }}{\int }[𝑨\cdot 𝑩-⟨𝑨⟩\cdot 𝑩-𝑨\cdot ⟨𝑩⟩+⟨𝑨⟩\cdot ⟨𝑩⟩]\text{dV}\\ & =\frac{1}{V}\underset{\mathrm{\Omega }}{\int }𝑨\cdot 𝑩\text{dV}-\frac{1}{V}\underset{\mathrm{\Omega }}{\int }⟨𝑨⟩\cdot 𝑩\text{dV}-\frac{1}{V}\underset{\mathrm{\Omega }}{\int }𝑨\cdot ⟨𝑩⟩\text{dV}+\frac{1}{V}\underset{\mathrm{\Omega }}{\int }⟨𝑨⟩\cdot ⟨𝑩⟩\text{dV}.\end{array}$

Now $⟨𝑨⟩$ and $⟨𝑩⟩$ are constants with respect to the integration. Hence,

$\begin{array}{cc}⟨[𝑨-⟨𝑨⟩]\cdot [𝑩-⟨𝑩⟩]⟩& =\frac{1}{V}\underset{\mathrm{\Omega }}{\int }𝑨\cdot 𝑩\text{dV}-⟨𝑨⟩\cdot \left(\frac{1}{V}\underset{\mathrm{\Omega }}{\int }𝑩\text{dV}\right)-\left(\frac{1}{V}\underset{\mathrm{\Omega }}{\int }𝑨\text{dV}\right)\cdot ⟨𝑩⟩+⟨𝑨⟩\cdot ⟨𝑩⟩\left(\frac{1}{V}\underset{\mathrm{\Omega }}{\int }\text{dV}\right)\\ & =⟨𝑨\cdot 𝑩⟩-⟨𝑨⟩\cdot ⟨𝑩⟩-⟨𝑨⟩\cdot ⟨𝑩⟩+⟨𝑨⟩\cdot ⟨𝑩⟩\\ & =⟨𝑨\cdot 𝑩⟩-⟨𝑨⟩\cdot ⟨𝑩⟩.\end{array}$

Therefore,

$⟨𝑨\cdot 𝑩⟩-⟨𝑨⟩\cdot ⟨𝑩⟩=⟨[𝑨-⟨𝑨⟩]\cdot [𝑩-⟨𝑩⟩]⟩◻$

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