Part 3 · Kinematics

3.6 Strain measures

Source: Favata’s notes, Kinematics §1.6, p. 12–13 (spectral form of C, right stretch tensor); the polar decomposition and the Green–Lagrange strain are beyond the notes so far

Objectives

Find the three fibres through a point that are stretched the most and the least, and stay orthogonal.
Split any deformation gradient into a pure stretch followed by a rotation.
Recipe: $C = \sum_{i} λ_{i} c_{i} \otimes c_{i}$ , $U = \sum_{i} λ_{i} c_{i} \otimes c_{i}$ , $R = F U^{- 1}$ .
Pictures: the principal directions and their images; $F = RU$ played as stretch, then turn.
Theorem: the polar decomposition.
Vocabulary: principal stretch, principal direction, right stretch tensor, polar decomposition, Green–Lagrange strain.

Principal directions

$C$ is symmetric and positive definite (Theorem 3.3.3). The spectral theorem and Theorem 2.7.9 give $C = i = 1 \sum 3 λ_{i} c_{i} \otimes c_{i}, λ_{i} > 0,$ with ${c_{i}}$ an orthonormal basis of eigenvectors of $C$ .

Warning

Here the $λ_{i}$ are the eigenvalues of $C$ , not stretches. The stretches are their square roots.

Definition 3.6.1Principal stretches and directionsbeyond the notes

The fibre along $c_{i}$ has stretch $C c_{i} \cdot c_{i} = λ_{i}$ : the principal stretch. $c_{i}$ is the principal direction.

Key observation 3.6.2Three fibres that stay orthogonalbeyond the notes

$F c_{i} \cdot F c_{j} = C c_{i} \cdot c_{j} = λ_{i} δ_{ij}$ : through every point there are three orthogonal fibres that are still orthogonal after the deformation. They are the axes of the ellipsoid that a small sphere becomes.
For any fibre $t = t_{i} c_{i}$ , $(d ℓ / d ℓ_{0})^{2} = \sum_{i} λ_{i} t_{i}^{2}$ , a weighted average of the $λ_{i}$ . The largest and smallest stretches at a point are along principal directions.

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Figure 3.6.1 Turn the fibre

t

and watch its stretch. It snaps when it lies on a principal direction, and the direction stays drawn with its image. Find both: they are at right angles before the deformation, and their images, the axes of the ellipse, are at right angles after.

The right stretch tensor

Definition 3.6.3Right stretch tensor

$U := i = 1 \sum 3 λ_{i} c_{i} \otimes c_{i}, U^{2} = C .$

$U$ is the square root of $C$ : symmetric, positive definite, and the only such tensor whose square is $C$ (Gurtin). The notes call its properties obvious; they are proved in Positive definite tensors and square roots.

Key observation 3.6.4U is pure stretchbeyond the notes

$U$ stretches each principal direction $c_{i}$ by $λ_{i}$ and does not turn it: three extensions along three orthogonal directions, nothing rigid.
It changes lengths exactly as $F$ does: $∣ Ut ∣^{2} = U^{2} t \cdot t = Ct \cdot t = ∣ Ft ∣^{2}$ .
So $F$ and $U$ can differ only by something that keeps lengths: a rotation.

Examples: for a stretch along the axes, $U = F$ ; for a rigid deformation, $U = I$ .

The notes break off here; what follows is the standard continuation (Gurtin), in the notes’ notation.

Polar decomposition

Theorem 3.6.5Polar decompositionbeyond the notes

Every $F$ with $det F > 0$ is uniquely $F = RU, U = F^{⊤} F, R := F U^{- 1} \in Rot .$

Proofbeyond the notes

$R$ is orthogonal: $R^{⊤} R = U^{- 1} F^{⊤} F U^{- 1} = U^{- 1} U^{2} U^{- 1} = I$ , using $U^{- ⊤} = U^{- 1}$ .
It is a rotation: $det R = det F / det U > 0$ .
Unique: if $F = R^{'} U^{'}$ with $R^{'}$ orthogonal and $U^{'}$ symmetric positive definite, then $C = U^{'} R^{' ⊤} R^{'} U^{'} = U^{'2}$ , so $U^{'} = U$ by uniqueness of the square root, and then $R^{'} = R$ .

Key observation 3.6.6Stretch, then turnbeyond the notes

Locally, every deformation is a pure stretch along the principal directions ( $U$ ), followed by a rotation ( $R$ ). The rigid information that $C$ discards is exactly $R$ .

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Figure 3.6.2 Play

F = RU

with the slider in the controls. The first half applies

U

: the principal directions are stretched and not turned, and the circle becomes the final ellipse in the wrong orientation. The second half applies

R

, which turns it rigidly into place. For the rigid rotation the first half does nothing.

beyond the notes

The same split in the other order: $F = VR$ , with the left stretch tensor $V := RU R^{⊤}$ and the left Cauchy–Green tensor $B := F F^{⊤} = V^{2}$ : turn first, then stretch along the turned directions $R c_{i}$ . $V$ and $B$ are tensors on the current configuration.

Green–Lagrange strain

Definition 3.6.7Green–Lagrange strainbeyond the notes

$E := \frac{1}{2} (C - I) = \frac{1}{2} (H + H^{⊤} + H^{⊤} H) .$

The second form: $C = (I + H)^{⊤} (I + H) = I + H + H^{⊤} + H^{⊤} H$ .
$E = 0$ exactly for rigid deformations (Key observation 3.5.7): what a strain measure should do.
Length change: $\frac{1}{2} ((d ℓ / d ℓ_{0})^{2} - 1) = Et \cdot t$ .
For small $H$ , dropping $H^{⊤} H$ leaves $sym H$ , the infinitesimal strain (Definition 2.3.1); its skew partner $skw H$ is the infinitesimal rotation.

Warning

The notes print “strain gradient” on p. 12; they mean the deformation gradient.

Exercises

Exercise 3.6.1beyond the notes

For the stretch along the axes of Exercise 3.5.3, find the principal stretches and directions, $U$ and $R$ .

Solution

$C = \sum_{i} α_{i}^{2} e_{i} \otimes e_{i}$ is already in spectral form: $c_{i} = e_{i}$ , principal stretches $α_{i}$ . $U = \sum_{i} α_{i} e_{i} \otimes e_{i} = F$ , so $R = I$ : no rotation.

Exercise 3.6.2beyond the notes

For the plane simple shear with $γ = \frac{3}{2}$ , $[F] = [10 \frac{3}{2} 1]$ , find the principal stretches.

Solution

$[C] = [1 \frac{3}{2} \frac{3}{2} \frac{13}{4}]$ , $tr = \frac{17}{4}$ , $det = \frac{13}{4} - \frac{9}{4} = 1$ . $λ^{2} - \frac{17}{4} λ + 1 = 0$ gives $λ = 4$ and $λ = \frac{1}{4}$ . Principal stretches $2$ and $\frac{1}{2}$ : their product is $det F = 1$ , as it must be for an isochoric deformation. The direction of stretch $2$ solves $(C - 4 I) c = 0$ : $- 3 c_{1} + \frac{3}{2} c_{2} = 0$ , so $c_{1} = (1, 2) / 5$ .

Exercise 3.6.3beyond the notes

Show that $F$ and $U$ have the same determinant, and that $B = F F^{⊤}$ has the same eigenvalues as $C$ .

Solution

$det F = det R det U = det U$ . $B = RUU R^{⊤} = RC R^{⊤}$ ; by Exercise 2.7.3 it has the eigenvalues of $C$ , with eigenvectors $R c_{i}$ .