Strain and stress¶

Strain¶

The general Seth-Hill formula of strain measures in the Lagrangian and Eulerian coordinates are defined as [Ogd97]

(11)\begin{aligned} &\frac{1}{n} \left( \textbf{U}^n - \bm{I} \right), \quad &\frac{1}{n} \left( \textbf{v}^n - \bm{I} \right), \quad \text{if} \quad n \neq 0,\\ &\log \textbf{U}, \quad &\log \textbf{v}, \quad \text{if} \quad n = 0, \end{aligned}

where $$n$$ is a real number and $$\textbf{U}, \textbf{v}$$ are unique SPD right (or material) and left (or spatial) stretch tensors defined by the unique polar decomposition of the deformation gradient (8) $$\bm{F} = \textbf{R} \textbf{U} = \textbf{v} \textbf{R}$$ with $$\textbf{R}^T \textbf{R} = \bm{I}$$ as the rotation tensor. The eigenvalues of the symmetric material tensor $$\textbf{U}$$ are $$\lambda_i, \, i=1,2,3$$, called the principal stretches and their corresponding orthonormal eigenvectors $$\hat{\bm{N}}_i$$ called principal referential directions, i.e.

(12)$\textbf{U} \hat{\bm{N}}_i= \lambda_i \hat{\bm{N}}_i. \quad i=1,2,3$

From the polar decomposition we deduce the eigenvalues problems for symmetric spatial tensor $$\textbf{v}$$ as

(13)$\textbf{v} \hat{\bm{n}}_i= \lambda_i \hat{\bm{n}}_i. \quad i=1,2,3$

where $$\hat{\bm{n}}_i = \bm{R} \hat{\bm{N}}_i$$. For the special case $$n=2$$, we define right and left Cauchy-Green tensors

$\bm{C} = \bm{F}^T \bm{F} = \textbf{U}^2, \quad \bm{b} = \bm{F} \bm{F}^T = \textbf{v}^2$

and Green-Lagrange and Green-Euler strains with $$\bm H = \nabla_X \bm u$$

(14)$\bm{E} = \frac{1}{2} \left( \bm{C} - \bm{I} \right) = \frac{1}{2} \left( \bm{H} + \bm{H}^T + \bm{H}^T \bm{H} \right).$
(15)$\bm{e} = \frac{1}{2} \left( \bm{b} - \bm{I} \right) = \frac{1}{2} \left( \bm{H} + \bm{H}^T + \bm{H} \bm{H}^T \right).$

Clearly $$\textbf{U}$$, $$\bm{C}$$ and $$\bm{E}$$ have the same orthonormal eigenvectors $$\hat{\bm{N}}_i$$ and their eigenvalues are related through $$\lambda_i^C = \lambda_i^2 = 1 + 2\lambda_i^E$$ i.e.,

(16)\begin{aligned} \bm{C} \hat{\bm{N}}_i&= \lambda_i^C \hat{\bm{N}}_i = \lambda_i^2 \hat{\bm{N}}_i. \quad i=1,2,3 \\ \bm{E} \hat{\bm{N}}_i&= \lambda_i^E \hat{\bm{N}}_i = \frac{1}{2} \left( \lambda_i^2 - 1 \right)\hat{\bm{N}}_i. \quad i=1,2,3, \end{aligned}

Similarly the eigenvectors of $$\textbf{v}$$, $$\bm{b}$$ and $$\bm{e}$$ are $$\hat{\bm{n}}_i$$ and their eigenvalues satisfy $$\lambda_i^b = \lambda_i^2 = 1 + 2\lambda_i^e$$ i.e.,

(17)\begin{aligned} \bm{b} \hat{\bm{n}}_i&= \lambda_i^b \hat{\bm{n}}_i = \lambda_i^2 \hat{\bm{n}}_i. \quad i=1,2,3 \\ \bm{e} \hat{\bm{n}}_i&= \lambda_i^e \hat{\bm{n}}_i = \frac{1}{2} \left( \lambda_i^2 - 1 \right)\hat{\bm{n}}_i. \quad i=1,2,3, \end{aligned}

Stress¶

Consider a continuum $$\Omega$$ subjected to a deformation mapping $$\bm{\chi}$$ that results in the deformed configuration as shown in Initial (undeformed) and current (deformed) configurations of a continuum body.. Let the force vector on an elemental area $$da$$ with normal $$\bm n$$ in the deformed configuration be $$d \bm f$$. Suppose that the area element in the undeformed configuration that corresponds to $$da$$ is $$dA$$ with normal $$\bm N$$. The force $$d \bm f$$ can be expressed in terms of a stress vector $$\bm t$$ and $$\bm T$$ times the area in current and initial configurations, respectively as

(18)$d\bm f = \bm t d a = \bm T d A$

From Cauchy’s formula [Hol00], we have $$\bm t = \bm \sigma \cdot \bm n$$, where $$\bm \sigma$$ is the Cauchy stress tensor. In a similar fashion, we introduce a stress tensor $$\bm P$$, called the first Piola-Kirchhoff stress tensor, such that $$\bm T = \bm P \cdot \bm N$$. Then using (18) we can write

$\bm \sigma \cdot \bm n da = \bm P \cdot \bm N d A \quad \text{or} \quad \bm \sigma \cdot d \bm a = \bm P \cdot d \bm A,$

from (10) we arrive at

(19)$\bm P = J \bm \sigma \bm{F}^{-T}.$

The first Piola-Kirchhoff stress tensor, also referred to as the nominal stress tensor, or Lagrangian stress tensor, gives the current force per unit undeformed area.

In large deformation analysis, we introduce the second Piola-Kirchhoff stress tensor $$\bm S$$, associated with the force $$d \bm F$$ in the undeformed elemental area $$d \bm A$$ that corresponds to the force $$d \bm f$$ on the deformed elemental area $$d \bm a$$

(20)$d \bm F = \bm S \cdot d \bm A.$

Thus, the second Piola-Kirchhoff stress tensor gives the transformed current force per unit undeformed area. Similar to relationship between $$d\bm x$$ and $$d\bm X$$ (7), the force $$d \bm f$$ is related to the force $$d \bm F$$ as

$d \bm F = \bm{F}^{-1} d \bm{f} = \bm{F}^{-1} (\bm P \cdot d \bm A),$

Hence, the second Piola-Kirchhoff stress tensor is related to the first Piola-Kirchhoff stress tensor and Cauchy stress tensor as

(21)$\bm S = \bm{F}^{-1} \bm{P} = J \bm{F}^{-1} \bm{\sigma} \bm{F}^{-T}.$

Clearly, $$\bm S$$ is symmetric whenever $$\bm \sigma$$ is symmetric but $$\bm P$$ is not symmetric. We can defined the symmetric Kirchhoff stress tensor $$\bm \tau$$, in current configuration by pushing forward $$\bm S$$, as

(22)$\bm{\tau} = \bm F \bm S \bm F^T = J \bm \sigma = J \bm{P} \bm{F}^T.$