Micromechanics of composites/Proof 1

Tensor-vector identity - 1

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

Proof:

Using the identity $𝐚\cdot ({𝑨}^T\cdot 𝐛)=𝐛\cdot (𝑨\cdot 𝐚)$ we have

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

Also, using the definition $(𝐮\otimes 𝐯)\cdot 𝐚=(𝐚\cdot 𝐯)𝐮$ we have

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

Therefore,

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

Using the identity $𝐚\cdot ({𝑨}^T\cdot 𝐛)=𝐛\cdot (𝑨\cdot 𝐚)$ we have

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

Finally, using the relation $(𝐮\otimes 𝐯{)}^T=𝐯\otimes 𝐮$, we get

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

Hence,

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