Part 3 · Kinematics

3.2 The deformation gradient

Source: Favata’s notes, Kinematics §1.2–1.3, p. 5–8 (deformation gradient, det F > 0, Remark 1.1, displacement, Remark 1.2, Exercise 1.1)

Objectives

Understand $F$ as the linear map that best approximates the deformation near a point.
Understand why $det F > 0$ is required everywhere.
Recipe: compute $[F]$ as the Jacobian matrix $f_{i, j}$ ; $H = F - I$ .
Pictures: a small disc and its image, closer and closer to an ellipse; a bar whose points must keep their order.
Vocabulary: two-point tensor, displacement, displacement gradient, admissible.
Essential vocabulary: deformation gradient.

Zooming in

The notes assume $f$ and $f^{- 1}$ continuous and differentiable. This leaves out cracks and the other pathologies of fracture mechanics, which the course does not treat.

To see what $f$ does near $x$ , probe a direction: $h = ε e$ , with $e$ a unit vector, and let $ε \to 0$ .

Definition 3.2.1Deformation gradientEssential

The deformation gradient at $x$ is the tensor $F (x) = \nabla f (x)$ with $F (x) e := ε \to 0 lim \frac{f ( x + ε e ) - f ( x )}{ε} \forall e,$ equivalently $f (x + h) - f (x) = F (x) h + o (∣ h ∣) .$

Key observation 3.2.2What F does

Up to terms small compared with $∣ h ∣$ , the small material segment from $x$ to $x + h$ is carried to $F (x) h$ . $F (x)$ acts on infinitesimal material fibres, not on positions: translating the whole body changes $f$ but not $F$ .

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Figure 3.2.1 Drag the probe

x

over the body on the left. On the right, the exact image of the small disc around it (violet) against the linear guess

f (x) + F (x) h

(orange, dashed). In the controls, shrink

∣ h ∣

: the two curves merge, which is what “best linear approximation” means. The arrows are the columns of

[F (x)]

. Pick the fold and move the probe to the sides: where the body is shaded red,

det F \leq 0

Recipe 3.2.3Computing F

In a Cartesian basis, $F e_{j} = \partial f / \partial x_{j} = f_{, j}$ (the limit with $e = e_{j}$ is a partial derivative), so $F = f_{, j} \otimes e_{j}, F_{ij} = f_{i, j}, [F] = f_{1, 1} f_{2, 1} f_{3, 1} f_{1, 2} f_{2, 2} f_{3, 2} f_{1, 3} f_{2, 3} f_{3, 3} .$ Row $i$ holds the derivatives of the component $f_{i}$ ; column $j$ is the image of a fibre along $e_{j}$ . It is the Jacobian matrix of $f$ .

Why the Cartesian form holdsbeyond the notes

With $I = e_{j} \otimes e_{j}$ and $A (a \otimes b) = Aa \otimes b$ (Proposition 2.2.7): $F = FI = F e_{j} \otimes e_{j} = f_{, j} \otimes e_{j}$ . Components: $F_{ij} = F e_{j} \cdot e_{i} = f_{, j} \cdot e_{i} = f_{i, j}$ .

Warning

The comma means a derivative: $f_{i, j} = \partial f_{i} / \partial x_{j}$ . The first index is the component, the second the variable. Swapping them gives $[F]^{⊤}$ , a different tensor.

det F > 0

Definition 3.2.4Admissible deformation

The notes require $det F (x) > 0 \forall x \in Ω. (1.1)$

Recall that the determinant multiplies signed volumes. With $A = F$ and an infinitesimal parallelepiped on $u, v, w$ :

$det F = 0$ : the small box is crushed to a sheet, a segment or a point. Excluded.
$det F < 0$ : the orientation of the triple flips, so matter has passed through itself. Excluded.

Consequences (beyond the notes): $F (x)$ is invertible at every point (Theorem 2.5.2), and by the chain rule on $f^{- 1} (f (x)) = x$ , the gradient of $f^{- 1}$ at $f (x)$ is $F (x)^{- 1}$ .

RemarkF is a two-point tensor (Remark 1.1)

With separate spaces for the two configurations, $F$ maps $V_{ref}$ into $V_{cur}$ . Strictly it is not a second-order tensor, which maps a space into itself, but a two-point tensor (double vector). Its input leg lives in the reference configuration, its output leg in the current one. That is why $F^{⊤} F$ , which goes $V_{ref} \to V_{cur} \to V_{ref}$ , is a genuine tensor on the reference configuration (§3.3 Change in length and angle).

Displacement

Definition 3.2.5Displacement and displacement gradient

$u (x) := f (x) - x, H := \nabla u = F - I . (1.3)$

The second because $\nabla x = I$ ( $x_{i, j} = δ_{ij}$ ). In a Cartesian basis $H = u_{, j} \otimes e_{j}$ , $H_{ij} = u_{i, j}$ .

No deformation, $f (x) = x$ : $u = 0$ , $H = 0$ , $F = I$ .
$H$ measures how far $F$ is from the identity. “Small $H$ ” is where the infinitesimal theory starts.

The 1D picture

Remark 1.2: condition (1.1) is clearest in one dimension. Take a continuum along $e$ , with $x = x e$ , $f (x) = f (x) e$ , $u (x) = u (x) e$ .

$[F]$ is the $1 \times 1$ matrix $f^{'}$ , so (1.1) says $f^{'} > 0$ : the deformation must be increasing.
With $u = f - x$ , $f^{'} = 1 + u^{'}$ , and the condition reads $u^{'} (x) > - 1$ .
On a segment $[0, L]$ , integrating gives $u (L) > u (0) - L$ : the deformed length $f (L) - f (0) = L + u (L) - u (0)$ is positive.

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Figure 3.2.2 Drag the nodes of the graph of

u

. Below, the bar before and after: its points are numbered. Wherever the graph falls more steeply than the dotted slope

- 1

f^{'} = 1 + u^{'} \leq 0

, the piece turns red and its points come out in the wrong order: matter has passed through itself. Try to make

u (L) - u (0) < - L

Key observation 3.2.6Order is orientation

In 1D, $f^{'} > 0$ says that points keep their order along $e$ . If $B$ were to the right of $A$ before and to the left after, the segment from $A$ to $B$ would have turned around, and somewhere in between two pieces of matter would have crossed. The inequality is strict: $u (L) - u (0) = - L$ would put both ends in one place.

Exercises

Exercise 3.2.1notes p. 6 (Exercise 1.1)

Find $F$ for $f (x) = x + α x_{2} e_{2} + β x_{2} x_{3} e_{3}$ .

Solution

$f_{1} = x_{1}$ , $f_{2} = (1 + α) x_{2}$ , $f_{3} = x_{3} + β x_{2} x_{3}$ , so $[F] = 100 0 1 + α β x_{3} 00 1 + β x_{2},$ $F = f_{, j} \otimes e_{j} = e_{1} \otimes e_{1} + (1 + α) e_{2} \otimes e_{2} + β x_{3} e_{3} \otimes e_{2} + (1 + β x_{2}) e_{3} \otimes e_{3} .$

Exercise 3.2.2beyond the notes

For the deformation of Exercise 3.2.1, find $det F$ and say where it is admissible.

Solution

$[F]$ is lower triangular, so $det F = (1 + α) (1 + β x_{2})$ . Condition (1.1) asks this to be positive at every point of $Ω$ : for instance $α > - 1$ and $1 + β x_{2} > 0$ throughout the body. For a body with $∣ x_{2} ∣ < 1$ and $α > - 1$ , any $∣ β ∣ < 1$ will do.

Exercise 3.2.3beyond the notes

In 1D on $[0, 1]$ , is $u (x) = - \frac{3}{2} x^{2}$ admissible? Where does it fail, and what happens to the points there?

Solution

$f^{'} (x) = 1 + u^{'} (x) = 1 - 3 x$ , negative for $x > \frac{1}{3}$ . On $(\frac{1}{3}, 1]$ the order is reversed: $f (\frac{1}{3}) = \frac{1}{3} - \frac{1}{6} = \frac{1}{6}$ , $f (1) = 1 - \frac{3}{2} = - \frac{1}{2}$ , so the right end lands to the left of the point $x = \frac{1}{3}$ . Not admissible; and indeed $u (1) - u (0) = - \frac{3}{2} < - 1 = - L$ .