Strain Measures in three dimensions

Initial orthonormal basis:

$({\boldsymbol {E}}_{1},{\boldsymbol {E}}_{2},{\boldsymbol {E}}_{3})$ Deformed orthonormal basis:

$(\mathbf {e} _{1},\mathbf {e} _{2},\mathbf {e} _{3})$ We assume that these coincide.

Motion

$\mathbf {x} ={\boldsymbol {\varphi }}(\mathbf {X} ,t)=\mathbf {x} (\mathbf {X} ,t)$ {\begin{aligned}{\boldsymbol {F}}&={\frac {\partial {\boldsymbol {\varphi }}}{\partial \mathbf {X} }}={\boldsymbol {\nabla }}_{o}{\boldsymbol {\varphi }}\\&={\frac {\partial \mathbf {x} }{\partial \mathbf {X} }}={\boldsymbol {\nabla }}_{X}{\boldsymbol {\varphi }}\end{aligned}} Effect of ${\boldsymbol {F}}$ :

{\begin{aligned}d\mathbf {x} _{1}&={\boldsymbol {F}}\bullet d\mathbf {X} _{1}~;&d\mathbf {x} _{2}&={\boldsymbol {F}}\bullet d\mathbf {X} _{2}\end{aligned}} ${\boldsymbol {F}}=F_{iJ}~\mathbf {e} _{i}\otimes {\boldsymbol {E}}_{J}$ Index notation:

$F_{iJ}={\frac {\partial x_{i}}{\partial X_{J}}}$ The determinant of the deformation gradient is usually denoted by $J$ and is a measure of the change in volume, i.e.,

$J=\det {\boldsymbol {F}}$ Push Forward and Pull Back

Forward Map:

$\mathbf {x} ={\boldsymbol {\varphi }}(\mathbf {X} ,t)$ ${\boldsymbol {F}}={\frac {\partial \mathbf {x} }{\partial \mathbf {X} }}={\boldsymbol {\nabla }}_{o}{\boldsymbol {\varphi }}$ ${\boldsymbol {F}}=\sum _{i,J=1}^{3}{\frac {\partial x_{i}}{\partial X_{J}}}~\mathbf {e} _{i}\otimes {\boldsymbol {E}}_{J}$ $d\mathbf {x} ={\boldsymbol {F}}\bullet d\mathbf {X} ={\boldsymbol {\varphi }}_{*}[d\mathbf {X} ]$ Push Forward operation:

${\boldsymbol {\varphi }}_{*}[\bullet ]$ • $d\mathbf {X}$ = material vector.
• $d\mathbf {x}$ = spatial vector.

Inverse map:

$\mathbf {X} ={\boldsymbol {\varphi }}^{-1}(\mathbf {x} ,t)$ ${\boldsymbol {F}}^{-1}={\frac {\partial \mathbf {X} }{\partial \mathbf {x} }}={\boldsymbol {\nabla }}{\boldsymbol {\varphi }}^{-1}$ ${\boldsymbol {F}}^{-1}=\sum _{i,J=1}^{3}{\frac {\partial X_{I}}{\partial x_{j}}}~{\boldsymbol {E}}_{I}\otimes \mathbf {e} _{j}$ $d\mathbf {X} ={\boldsymbol {F}}^{-1}\bullet d\mathbf {x} ={\boldsymbol {\varphi }}^{*}[d\mathbf {x} ]$ Pull Back operation:

${\boldsymbol {\varphi }}^{*}[\bullet ]$ • $d\mathbf {X}$ = material vector.
• $d\mathbf {x}$ = spatial vector.
Example

Motion:

{\begin{aligned}x_{1}&={\cfrac {1}{4}}(18+4X_{1}+6X_{2})\\x_{2}&={\cfrac {1}{4}}(14+6X_{2})\end{aligned}} $F_{ij}={\frac {\partial x_{i}}{\partial X_{j}}}\implies \mathbf {F} ={\frac {1}{2}}{\begin{bmatrix}2&3\\0&3\end{bmatrix}}$ $\mathbf {F} ^{-1}={\cfrac {1}{3}}{\begin{bmatrix}3&-3\\0&2\end{bmatrix}}$ Push Forward:

{\begin{aligned}{\boldsymbol {\varphi }}_{*}[{\boldsymbol {E}}_{1}]&=\mathbf {F} {\begin{bmatrix}1\\0\end{bmatrix}}={\begin{bmatrix}1\\0\end{bmatrix}}\\{\boldsymbol {\varphi }}_{*}[{\boldsymbol {E}}_{2}]&=\mathbf {F} {\begin{bmatrix}0\\1\end{bmatrix}}={\begin{bmatrix}1.5\\1.5\end{bmatrix}}\end{aligned}} Pull Back:

{\begin{aligned}{\boldsymbol {\varphi }}^{*}[\mathbf {e} _{1}]&=\mathbf {F} ^{-1}{\begin{bmatrix}1\\0\end{bmatrix}}={\begin{bmatrix}1\\0\end{bmatrix}}\\{\boldsymbol {\varphi }}^{*}[\mathbf {e} _{2}]&=\mathbf {F} ^{-1}{\begin{bmatrix}0\\1\end{bmatrix}}={\begin{bmatrix}-1\\2/3\end{bmatrix}}\end{aligned}} Cauchy-Green Deformation Tensors

Right Cauchy-Green Deformation Tensor

Recall:

$d\mathbf {x} _{1}={\boldsymbol {F}}\bullet d\mathbf {X} _{1}~;~~d\mathbf {x} _{2}={\boldsymbol {F}}\bullet d\mathbf {X} _{2}$ Therefore,

$d\mathbf {x} _{1}\bullet d\mathbf {x} _{2}=({\boldsymbol {F}}\bullet d\mathbf {X} _{1})\bullet ({\boldsymbol {F}}\bullet d\mathbf {X} _{2})$ Using index notation:

{\begin{aligned}d\mathbf {x} _{1}\bullet d\mathbf {x} _{2}&=(F_{ij}~dX_{j}^{1})(F_{ik}~dX_{k}^{2})\\&=dX_{j}^{1}~(F_{ij}~F_{ik})~dX_{k}^{2}\\&=d\mathbf {X} _{1}\bullet ({\boldsymbol {F}}^{T}\bullet {\boldsymbol {F}})\bullet d\mathbf {X} _{2}\\&=d\mathbf {X} _{1}\bullet {\boldsymbol {C}}\bullet d\mathbf {X} _{2}\end{aligned}} Right Cauchy-Green tensor:

${\boldsymbol {C}}={\boldsymbol {F}}^{T}\bullet {\boldsymbol {F}}$ Left Cauchy-Green Deformation Tensor

Recall:

$d\mathbf {X} _{1}={\boldsymbol {F}}^{-1}\bullet d\mathbf {x} _{1}~;~~d\mathbf {X} _{2}={\boldsymbol {F}}^{-1}\bullet d\mathbf {x} _{2}$ Therefore,

$d\mathbf {X} _{1}\bullet d\mathbf {X} _{2}=({\boldsymbol {F}}^{-1}\bullet d\mathbf {x} _{1})\bullet ({\boldsymbol {F}}^{-1}\bullet d\mathbf {x} _{2})$ Using index notation:

{\begin{aligned}d\mathbf {X} _{1}\bullet d\mathbf {X} _{2}&=(F_{ij}^{-1}~dx_{j}^{1})(F_{ik}^{-1}~dx_{k}^{2})\\&=dx_{j}^{1}~(F_{ij}^{-1}~F_{ik}^{-1})~dx_{k}^{2}\\&=d\mathbf {x} _{1}\bullet ({\boldsymbol {F}}^{-T}\bullet {\boldsymbol {F}}^{-1})\bullet d\mathbf {x} _{2}\\&=d\mathbf {x} _{1}\bullet ({\boldsymbol {F}}\bullet {\boldsymbol {F}}^{T})^{-1}\bullet d\mathbf {x} _{2}\\&=d\mathbf {x} _{1}\bullet \mathbf {b} ^{-1}\bullet d\mathbf {x} _{2}\end{aligned}} Left Cauchy-Green (Finger) tensor:

$\mathbf {b} ={\boldsymbol {F}}\bullet {\boldsymbol {F}}^{T}$ Strain Measures

Green (Lagrangian) Strain

{\begin{aligned}{\frac {1}{2}}(d\mathbf {x} _{1}\bullet d\mathbf {x} _{2}&-d\mathbf {X} _{1}\bullet d\mathbf {X} _{2})\\&={\frac {1}{2}}d\mathbf {X} _{1}\bullet ({\boldsymbol {C}}-{\boldsymbol {I}})\bullet d\mathbf {X} _{2}\\&=d\mathbf {X} _{1}\bullet {\boldsymbol {E}}\bullet d\mathbf {X} _{2}\end{aligned}} Green strain tensor:

{\begin{aligned}{\boldsymbol {E}}&={\frac {1}{2}}({\boldsymbol {C}}-{\boldsymbol {I}})\\&={\frac {1}{2}}({\boldsymbol {F}}^{T}\bullet {\boldsymbol {F}}-{\boldsymbol {I}})\\&={\frac {1}{2}}\left[{\boldsymbol {\nabla }}_{o}\mathbf {u} +({\boldsymbol {\nabla }}_{o}\mathbf {u} )^{T}+{\boldsymbol {\nabla }}_{o}\mathbf {u} \bullet ({\boldsymbol {\nabla _{o}\mathbf {u} )^{T}}}\right]\end{aligned}} Index notation:

{\begin{aligned}E_{ij}&={\frac {1}{2}}(F_{ki}~F_{kj}-\delta _{ij})\\&={\frac {1}{2}}\left({\frac {\partial u_{i}}{\partial X_{j}}}+{\frac {\partial u_{j}}{\partial X_{i}}}+{\frac {\partial u_{k}}{\partial X_{i}}}{\frac {\partial u_{k}}{\partial X_{j}}}\right)\end{aligned}} Almansi (Eulerian) Strain

{\begin{aligned}{\frac {1}{2}}(d\mathbf {x} _{1}\bullet d\mathbf {x} _{2}&-d\mathbf {X} _{1}\bullet d\mathbf {X} _{2})\\&={\frac {1}{2}}d\mathbf {x} _{1}\bullet ({\boldsymbol {I}}-\mathbf {b} ^{-1})\bullet d\mathbf {x} _{2}\\&=d\mathbf {x} _{1}\bullet \mathbf {e} \bullet d\mathbf {x} _{2}\end{aligned}} Almansi strain tensor:

{\begin{aligned}\mathbf {e} &={\frac {1}{2}}({\boldsymbol {I}}-\mathbf {b} ^{-1})\\&={\frac {1}{2}}({\boldsymbol {I}}-{\boldsymbol {F}}^{-T}\bullet {\boldsymbol {F}}^{-1})\end{aligned}} Index notation:

{\begin{aligned}e_{ij}&={\frac {1}{2}}(\delta _{ij}-F_{ki}^{-1}~F_{kj}^{-1})\end{aligned}} Push Forward and Pull Back

Recall:

$d\mathbf {x} _{1}\bullet \mathbf {e} \bullet d\mathbf {x} _{2}=d\mathbf {X} _{1}\bullet {\boldsymbol {E}}\bullet d\mathbf {X} _{2}$ Now,

{\begin{aligned}d\mathbf {x} _{1}\bullet \mathbf {e} \bullet d\mathbf {x} _{2}&=({\boldsymbol {F}}\bullet d\mathbf {X} _{1})\bullet \mathbf {e} \bullet ({\boldsymbol {F}}\bullet d\mathbf {X} _{2})\\&=d\mathbf {X} _{1}\bullet ({\boldsymbol {F}}^{T}\bullet \mathbf {e} \bullet {\boldsymbol {F}})\bullet d\mathbf {X} _{2}\\&=d\mathbf {X} _{1}\bullet {\boldsymbol {E}}\bullet d\mathbf {X} _{2}\end{aligned}} Therefore,

{\begin{aligned}{\boldsymbol {E}}&={\boldsymbol {F}}^{T}\bullet \mathbf {e} \bullet {\boldsymbol {F}}\\\implies \mathbf {e} &={\boldsymbol {F}}^{-T}\bullet {\boldsymbol {E}}\bullet {\boldsymbol {F}}^{-1}\end{aligned}} Push Forward:

$\mathbf {e} ={\boldsymbol {\varphi }}_{*}[{\boldsymbol {E}}]={\boldsymbol {F}}^{-T}\bullet {\boldsymbol {E}}\bullet {\boldsymbol {F}}^{-1}$ Pull Back:

${\boldsymbol {E}}={\boldsymbol {\varphi }}^{*}[\mathbf {e} ]={\boldsymbol {F}}^{T}\bullet \mathbf {e} \bullet {\boldsymbol {F}}$ Some useful results

Derivative of J with respect to the deformation gradient

We often need to compute the derivative of $J=\det {\boldsymbol {F}}$ with respect to the deformation gradient ${\boldsymbol {F}}$ . From tensor calculus we have, for any second order tensor ${\boldsymbol {A}}$ ${\cfrac {\partial }{\partial {\boldsymbol {A}}}}(\det {\boldsymbol {A}})=\det {\boldsymbol {A}}~{\boldsymbol {A}}^{-T}$ Therefore,

 ${\cfrac {\partial J}{\partial {\boldsymbol {F}}}}=J~{\boldsymbol {F}}^{-T}$ Derivative of J with respect to the right Cauchy-Green deformation tensor

The derivative of J with respect to the right Cauchy-Green deformation tensor (${\boldsymbol {C}}$ ) is also often encountered in continuum mechanics.

To calculate the derivative of $J=\det {\boldsymbol {F}}$ with respect to ${\boldsymbol {C}}$ , we recall that (for any second order tensor ${\boldsymbol {T}}$ )

${\frac {\partial {\boldsymbol {C}}}{\partial {\boldsymbol {F}}}}:{\boldsymbol {T}}={\frac {\partial }{\partial {\boldsymbol {F}}}}({\boldsymbol {F}}^{T}\cdot {\boldsymbol {F}}):{\boldsymbol {T}}=({\boldsymbol {\mathsf {I}}}^{T}:{\boldsymbol {T}})\cdot {\boldsymbol {F}}+{\boldsymbol {F}}^{T}\cdot ({\boldsymbol {\mathsf {I}}}:{\boldsymbol {T}})={\boldsymbol {T}}^{T}\cdot {\boldsymbol {F}}+{\boldsymbol {F}}^{T}\cdot {\boldsymbol {T}}$ Also,

${\frac {\partial J}{\partial {\boldsymbol {F}}}}:{\boldsymbol {T}}={\frac {\partial J}{\partial {\boldsymbol {C}}}}:({\frac {\partial {\boldsymbol {C}}}{\partial {\boldsymbol {F}}}}:{\boldsymbol {T}})={\frac {\partial J}{\partial {\boldsymbol {C}}}}:({\boldsymbol {T}}^{T}\cdot {\boldsymbol {F}}+{\boldsymbol {F}}^{T}\cdot {\boldsymbol {T}})=\left[{\boldsymbol {F}}\cdot {\frac {\partial J}{\partial {\boldsymbol {C}}}}\right]:{\boldsymbol {T}}+\left[{\boldsymbol {F}}\cdot \left({\frac {\partial J}{\partial {\boldsymbol {C}}}}\right)^{T}\right]:{\boldsymbol {T}}$ From the symmetry of ${\boldsymbol {C}}$ we have

${\frac {\partial J}{\partial {\boldsymbol {C}}}}=\left({\frac {\partial J}{\partial {\boldsymbol {C}}}}\right)^{T}$ Therefore, involving the arbitrariness of ${\boldsymbol {T}}$ , we have

${\frac {\partial J}{\partial {\boldsymbol {F}}}}=2~{\boldsymbol {F}}\cdot {\frac {\partial J}{\partial {\boldsymbol {C}}}}$ Hence,

${\frac {\partial J}{\partial {\boldsymbol {C}}}}={\frac {1}{2}}~{\boldsymbol {F}}^{-1}\cdot {\frac {\partial J}{\partial {\boldsymbol {F}}}}~.$ Also recall that

${\frac {\partial J}{\partial {\boldsymbol {F}}}}=J~{\boldsymbol {F}}^{-T}$ Therefore,

 ${\frac {\partial J}{\partial {\boldsymbol {C}}}}={\frac {1}{2}}~J~{\boldsymbol {F}}^{-1}\cdot {\boldsymbol {F}}^{-T}={\cfrac {J}{2}}~{\boldsymbol {C}}^{-1}$ In index notation,

 ${\frac {\partial J}{\partial C_{IJ}}}={\cfrac {J}{2}}~C_{IJ}^{-1}$ Derivative of the inverse of the right Cauchy-Green tensor

Another result that is often useful is that for the derivative of the inverse of the right Cauchy-Green tensor (${\boldsymbol {C}}$ ).

Recall that, for a second order tensor ${\boldsymbol {A}}$ ,

${\frac {\partial {\boldsymbol {A}}^{-1}}{\partial {\boldsymbol {A}}}}:{\boldsymbol {T}}=-{\boldsymbol {A}}^{-1}\cdot {\boldsymbol {T}}\cdot {\boldsymbol {A}}^{-1}$ In index notation

${\frac {\partial A_{ij}^{-1}}{\partial A_{kl}}}~T_{kl}=B_{ijkl}~T_{kl}=-A_{ik}^{-1}~T_{kl}~A_{lj}^{-1}$ or,

${\frac {\partial A_{ij}^{-1}}{\partial A_{kl}}}=B_{ijkl}=-A_{ik}^{-1}~A_{lj}^{-1}$ Using this formula and noting that since ${\boldsymbol {C}}$ is a symmetric second order tensor, the derivative of its inverse is a symmetric fourth order tensor we have

 ${\frac {\partial C_{IJ}^{-1}}{\partial C_{KL}}}=-{\frac {1}{2}}~(C_{IK}^{-1}~C_{JL}^{-1}+C_{JK}^{-1}~C_{IL}^{-1})$ 