# Nonlinear finite elements/Homework 7/Hints
## Hints 1: Index notation

Index notation:
    $\sigma _{ij}=2\mu ~\varepsilon _{ij}+\lambda ~\varepsilon _{kk}~\delta _{ij}~.$
If $j=i$
    ${\begin{aligned}\sigma _{ii}&=2\mu ~\varepsilon _{ii}+\lambda ~\varepsilon _{kk}~\delta _{ii}\\&=2\mu ~\varepsilon _{kk}+3\lambda ~\varepsilon _{kk}\\&=(2\mu +3\lambda )~\varepsilon _{kk}\end{aligned}}$
    $\sigma _{kk}=(2\mu +3\lambda )~\varepsilon _{kk}$
Dummy indices are replaceable.

## Hint 2: Index notation

Index notation:
    $\sigma _{ij}=2\mu ~\varepsilon _{ij}+\lambda ~\varepsilon _{kk}~\delta _{ij}~.$
Multiply by $\delta _{ij}$:
    ${\begin{aligned}\sigma _{ij}~\delta _{ij}&=2\mu ~\varepsilon _{ij}~\delta _{ij}+\lambda ~\varepsilon _{kk}~\delta _{ij}~\delta _{ij}\\\implies \sigma _{jj}&=2\mu ~\varepsilon _{ii}+\lambda ~\varepsilon _{kk}~\delta _{ii}\\\implies \sigma _{kk}&=2\mu ~\varepsilon _{kk}+3\lambda ~\varepsilon _{kk}\\\implies \sigma _{kk}&=(2\mu +3\lambda )~\varepsilon _{kk}\end{aligned}}$
Multiplication by $\delta _{ij}$ leads to replacement of one index.
    $A_{ij}~\delta _{kl}=?\qquad A_{ij}~\delta {jl}=?$

## Hint 3: Index notation

Index notation:
    ${\begin{aligned}{\boldsymbol {\sigma }}&=\sigma _{ij}~\mathbf {e} _{i}\otimes \mathbf {e} _{j}\\{\boldsymbol {\varepsilon }}&=\varepsilon _{ij}~\mathbf {e} _{i}\otimes \mathbf {e} _{j}\end{aligned}}$
From the definition of dyadic product, we can show
    ${\begin{aligned}(\mathbf {a} \otimes \mathbf {b} ):(\mathbf {u} \otimes \mathbf {v} )&=(\mathbf {a} \bullet \mathbf {u} )(\mathbf {b} \bullet \mathbf {v} )\end{aligned}}$
Contraction gives:
    ${\begin{aligned}{\boldsymbol {\sigma }}:{\boldsymbol {\varepsilon }}&=(\sigma _{ij}~\mathbf {e} _{i}\otimes \mathbf {e} _{j}):(\varepsilon _{kl}~\mathbf {e} _{k}\otimes \mathbf {e} _{l})\\&=\sigma _{ij}~\varepsilon _{kl}~(\mathbf {e} _{i}\otimes \mathbf {e} _{j}):(\mathbf {e} _{k}\otimes \mathbf {e} _{l})\\&=\sigma _{ij}~\varepsilon _{kl}~(\mathbf {e} _{i}\bullet \mathbf {e} _{k})(\mathbf {e} _{j}\bullet \mathbf {e} _{l})\\&=\sigma _{ij}~\varepsilon _{kl}~\delta _{ik}~\delta _{jl}\\&=\sigma _{ij}~\varepsilon _{ij}\end{aligned}}$

## Hint 4: Tensor product

Index notation:
    ${\begin{aligned}{\boldsymbol {\varepsilon }}&=\varepsilon _{ij}~\mathbf {e} _{i}\otimes \mathbf {e} _{j}\\{\boldsymbol {\mathsf {C}}}&=C_{ijkl}~\mathbf {e} _{i}\otimes \mathbf {e} _{j}\otimes \mathbf {e} _{k}\otimes \mathbf {e} _{l}\end{aligned}}$
Definition of dyadics products:
    ${\begin{aligned}(\mathbf {a} \bullet \mathbf {b} )\otimes \mathbf {x} &=(\mathbf {b} \bullet \mathbf {x} )\mathbf {a} \\(\mathbf {a} \bullet \mathbf {b} \otimes \mathbf {c} )\otimes \mathbf {x} &=(\mathbf {c} \bullet \mathbf {x} )(\mathbf {a} \otimes \mathbf {b} )\\(\mathbf {a} \bullet \mathbf {b} \otimes \mathbf {c} \otimes \mathbf {d} )\otimes \mathbf {x} &=(\mathbf {d} \bullet \mathbf {x} )(\mathbf {a} \otimes \mathbf {b} \otimes \mathbf {c} )\end{aligned}}$
We can show that
    ${\begin{aligned}(\mathbf {a} \otimes \mathbf {b} \otimes \mathbf {c} \otimes \mathbf {d} ):(\mathbf {u} \otimes \mathbf {v} )&=((\mathbf {a} \bullet \mathbf {b} \otimes \mathbf {c} \otimes \mathbf {d} )\otimes \mathbf {v} )\bullet \mathbf {u} \end{aligned}}$
Contraction gives:
    ${\begin{aligned}{\boldsymbol {\mathsf {C}}}:{\boldsymbol {\varepsilon }}&=(C_{ijkl}~\mathbf {e} _{i}\otimes \mathbf {e} _{j}\otimes \mathbf {e} _{k}\otimes \mathbf {e} _{l}):(\varepsilon _{mn}~\mathbf {e} _{m}\otimes \mathbf {e} _{n})\\&=C_{ijkl}~\varepsilon _{mn}~(\mathbf {e} _{i}\otimes \mathbf {e} _{j}\otimes \mathbf {e} _{k}\otimes \mathbf {e} _{l}):(\mathbf {e} _{m}\otimes \mathbf {e} _{n})\\&=C_{ijkl}~\varepsilon _{mn}~((\mathbf {e} _{i}\bullet \mathbf {e} _{i}\otimes \mathbf {e} _{k}\otimes \mathbf {e} _{l})\otimes \mathbf {e} _{n})\bullet \mathbf {e} _{m}\\&=C_{ijkl}~\varepsilon _{mn}~(\mathbf {e} _{l}\bullet \mathbf {e} _{n})(\mathbf {e} _{i}\otimes \mathbf {e} _{j}\otimes \mathbf {e} _{k})\bullet \mathbf {e} _{m}\\&=C_{ijkl}~\varepsilon _{mn}~\delta _{ln}(\mathbf {e} _{k}\bullet \mathbf {e} _{m})(\mathbf {e} _{i}\otimes \mathbf {e} _{j})=C_{ijkl}~\varepsilon _{mn}~\delta _{ln}~\delta _{km}~\mathbf {e} _{i}\otimes \mathbf {e} _{j}\\&=C_{ijkl}~\varepsilon _{kl}\end{aligned}}$

## Hint 5 : Tensor product

Tensor Product of two tensors:
    ${\begin{aligned}{\boldsymbol {A}}&=A_{ij}~\mathbf {e} _{i}\otimes \mathbf {e} _{j}\\{\boldsymbol {B}}&=B_{kl}~\mathbf {e} _{k}\otimes \mathbf {e} _{l}\end{aligned}}$
Tensor product:
    ${\begin{aligned}{\boldsymbol {A}}\otimes {\boldsymbol {B}}&=(A_{ij}~\mathbf {e} _{i}\otimes \mathbf {e} _{j})\otimes (B_{kl}~\mathbf {e} _{k}\otimes \mathbf {e} _{l})\\&=A_{ij}B_{kl}\mathbf {e} _{i}\otimes \mathbf {e} _{j}\otimes \mathbf {e} _{k}\otimes \mathbf {e} _{l}\end{aligned}}$

## Hint 6: Vector transformations

Change of basis: Vector transformation rule
    $v_{i}^{'}=L_{ij}v_{j}$
$L_{ij}$ are the direction cosines.
    ${\begin{aligned}L_{11}&=\mathbf {e} _{1}^{'}\bullet \mathbf {e} _{1};&L_{12}&=\mathbf {e} _{1}^{'}\bullet \mathbf {e} _{2};&L_{13}&=\mathbf {e} _{1}^{'}\bullet \mathbf {e} _{3}\\L_{21}&=\mathbf {e} _{2}^{'}\bullet \mathbf {e} _{1};&L_{22}&=\mathbf {e} _{2}^{'}\bullet \mathbf {e} _{2};&L_{23}&=\mathbf {e} _{2}^{'}\bullet \mathbf {e} _{3}\\L_{31}&=\mathbf {e} _{3}^{'}\bullet \mathbf {e} _{1};&L_{32}&=\mathbf {e} _{3}^{'}\bullet \mathbf {e} _{2};&L_{33}&=\mathbf {e} _{3}^{'}\bullet \mathbf {e} _{3}\end{aligned}}$
In matrix form
    $\mathbf {v} ^{'}=\mathbf {L} ~\mathbf {v} ;~~\mathbf {v} =\mathbf {L} ^{T}~\mathbf {v} ^{'};~~\mathbf {L} \mathbf {L} ^{T}=\mathbf {I} \implies \mathbf {L} ^{T}=\mathbf {L} ^{-1}$
Other common form: Vector transformation rule
    $v_{i}^{'}=Q_{ji}v_{j}$
    ${\begin{aligned}Q_{11}&=\mathbf {e} _{1}\bullet \mathbf {e} _{1}^{'};&Q_{12}&=\mathbf {e} _{1}\bullet \mathbf {e} _{2}^{'};&Q_{13}&=\mathbf {e} _{1}\bullet \mathbf {e} _{3}^{'}\\Q_{21}&=\mathbf {e} _{2}\bullet \mathbf {e} _{1}^{'};&Q_{22}&=\mathbf {e} _{2}\bullet \mathbf {e} _{2}^{'};&Q_{23}&=\mathbf {e} _{2}\bullet \mathbf {e} _{3}^{'}\\Q_{31}&=\mathbf {e} _{3}\bullet \mathbf {e} _{1}^{'};&Q_{32}&=\mathbf {e} _{3}\bullet \mathbf {e} _{2}^{'};&Q_{33}&=\mathbf {e} _{3}\bullet \mathbf {e} _{3}^{'}\end{aligned}}$
In matrix form
    $\mathbf {v} ^{'}=\mathbf {Q} ^{T}~\mathbf {v} ;~~\mathbf {v} =\mathbf {Q} ~\mathbf {v} ^{'};~~\mathbf {Q} \mathbf {Q} ^{T}=\mathbf {I} \implies \mathbf {Q} ^{T}=\mathbf {Q} ^{-1}$

## Hint 7: Tensor transformations

Change of basis: Tensor transformation rule
    $T_{ij}^{'}=L_{ip}L_{jq}T_{pq}$
where $L_{ij}$ are the direction cosines.
In matrix form,
    $\mathbf {T} ^{'}=\mathbf {L} \mathbf {T} \mathbf {L} ^{T}$
Other common form
    $T_{ij}^{'}=Q_{pi}Q_{qj}T_{pq}$
In matrix form,
    $\mathbf {T} ^{'}=\mathbf {Q} ^{T}\mathbf {T} \mathbf {Q}$