Multiplicative decomposition of the deformation gradient

Remarks

Example: let $F^p$ be the simple shear with rows $(1, \gamma, 0)$, $(0, 1, 0)$, $(0, 0, 1)$ and let $F^e = \operatorname{diag}(1 + e, 1, 1)$. Then $F = F^e F^p$ has rows $(1 + e, (1 + e)\gamma, 0)$, $(0, 1, 0)$, $(0, 0, 1)$, and $\det F^p = 1$ expresses that metal plasticity preserves volume. Non-example: the polar decomposition $F = RU$ is also a product, but $U$ is not a plastic part, since the stress depends on $U$. It is a purely kinematic split, not an elastic–plastic one. The fields $F^e$ and $F^p$ are in general not gradients of any map, so the intermediate configuration exists only locally. The split is unique only up to a rotation $Q$ of the intermediate configuration: $F^e Q^T$ and $Q F^p$ give the same $F$. For small displacement gradients, $F^e = I + H^e$ and $F^p = I + H^p$ give $F \approx I + H^e + H^p$, the additive split of small strain theory.