Micromechanics of composites/Proof 11

Identity 1

Let $𝑨$ and $𝑩$ be two second order tensors. Show that

$𝑨:𝑩=({𝑨}^T\cdot 𝑩):1.$

Proof:

Using index notation,

$𝑨:𝑩=A_{ij}{B}_{ij}=A_{ji}^T{B}_{ij}=A_{ji}^T{B}_{ik}{\delta }_{jk}=[{𝑨}^T\cdot 𝑩{]}_{jk}{\delta }_{jk}=({𝑨}^T\cdot 𝑩):1.$

Hence,

$𝑨:𝑩=({𝑨}^T\cdot 𝑩):1◻$

Identity 2

Let $𝑨$ be a second order tensor and let $𝐚$ and $𝐛$ be two vectors. Show that

$𝑨:(𝐚\otimes 𝐛)=(𝑨\cdot 𝐛)\cdot 𝐚.$

Proof:

It is convenient to use index notation for this. We have

$𝑨:(𝐚\otimes 𝐛)=A_{ij}{a}_i{b}_j=(A_{ij}{b}_j)a_i=(𝑨\cdot 𝐛)\cdot 𝐚.$

Hence,

$𝑨:(𝐚\otimes 𝐛)=(𝑨\cdot 𝐛)\cdot 𝐚◻$

Identity 3

Let $𝑨$ and $𝑩$ be two second order tensors and let $𝐚$ and $𝐛$ be two vectors. Show that

$(𝑨\cdot 𝐚)\cdot (𝑩\cdot 𝐛)=({𝑨}^T\cdot 𝑩):(𝐚\otimes 𝐛).$

Proof:

Using index notation,

$(𝑨\cdot 𝐚)\cdot (𝑩\cdot 𝐛)=(A_{ij}{a}_j)(B_{ik}{b}_k)=(A_{ij}{B}_{ik})(a_j{b}_k)=(A_{ji}^T{B}_{ik})(a_j{b}_k)=({𝑨}^T\cdot 𝑩):(𝐚\otimes 𝐛).$

Hence,

$(𝑨\cdot 𝐚)\cdot (𝑩\cdot 𝐛)=({𝑨}^T\cdot 𝑩):(𝐚\otimes 𝐛)◻$

Identity 4

Let $𝑨$ be a second order tensors and let $𝐚$ and $𝐛$ be two vectors. Show that

$(𝑨\cdot 𝐚)\otimes 𝐛=𝑨\cdot (𝐚\otimes 𝐛)\text{and}𝐚\otimes (𝑨\cdot 𝐛)=[𝑨\cdot (𝐛\otimes 𝐚){]}^T=(𝐚\otimes 𝐛)\cdot {𝑨}^T.$

Proof:

For the first identity, using index notation, we have

$[(𝑨\cdot 𝐚)\otimes 𝐛{]}_{ik}=(A_{ij}{a}_j)b_k=A_{ij}(a_j{b}_k)=A_{ij}[𝐚\otimes 𝐛{]}_{jk}=𝑨\cdot (𝐚\otimes 𝐛).$

Hence,

$(𝑨\cdot 𝐚)\otimes 𝐛=𝑨\cdot (𝐚\otimes 𝐛)◻$

For the second identity, we have

$[𝐚\otimes (𝑨\cdot 𝐛){]}_{ij}=a_i(A_{jk}{b}_k)=(a_i{b}_k)A_{jk}=(a_i{b}_k)A_{kj}^T=[(𝐚\otimes 𝐛)\cdot {𝑨}^T{]}_{ij}.$

Therefore,

$𝐚\otimes (𝑨\cdot 𝐛)=(𝐚\otimes 𝐛)\cdot {𝑨}^T.$

Now, $𝐚\otimes 𝐛=[𝐛\otimes 𝐚{]}^T$ and $(𝑨\cdot 𝑩{)}^T={𝑩}^T\cdot {𝑨}^T$. Hence,

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

Therefore,

$𝐚\otimes (𝑨\cdot 𝐛)=[𝑨\cdot (𝐛\otimes 𝐚){]}^T=(𝐚\otimes 𝐛)\cdot {𝑨}^T◻$

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