Part 2 · Theory of deformation

2.1 Configurations and the deformation gradient

Source: Carboni’s notes for lesson 2, Teoria della deformazione 1, pp. 1–4

Objectives

Vocabulary. Reference and current configuration, deformation, displacement, fibre.
Theorem. Near a point, any deformation acts on fibres as a linear map: $\overset{˘}{P} \overset{˘}{Q} = F PQ$ .
Recipe. Compute $F$ from the deformation $p (X)$ or the displacement $U (X)$ , and check that the deformation is admissible.
Picture. $F$ turns a small square into a parallelogram.

The oscillator had one number, $U$ , for its deformation. A body has a deformation at every one of its points, and in every direction through each point. This part builds the tools that measure it, without any reference to forces: pure geometry.

The continuum and its configurations

A continuum is a model of a physical body in which its molecular structure is ignored: the body fills a region of space without gaps, and quantities such as displacement, density or internal force are smooth fields defined at every point of that region.

Fix an origin $O$ and an orthonormal basis ${e_{1}, e_{2}, e_{3}}$ : three unit vectors, mutually perpendicular. (Carboni’s handwriting writes the basis letter like $ℓ$ ; the typed sheets write $e$ .)

Definition 2.1.1reference and current configurationnotes L2, pp. 1–2

The reference configuration $B$ is the region the body occupies at the start. A point $P \in B$ has the position vector $X = X_{1} e_{1} + X_{2} e_{2} + X_{3} e_{3} .$
The current configuration $\overset{˘}{B}$ is the region the body occupies at time $t$ . The same material point is now at $\overset{˘}{P}$ , with the position vector $p (X, t)$ .

A breve ( $˘$ ) marks everything in the current configuration: $\overset{˘}{P}$ , $\overset{˘}{B}$ , later $\overset{˘}{V}$ , $\overset{˘}{θ}$ .

Definition 2.1.2deformation and displacementnotes L2, p. 2

The map $(X, t) ⟼ p (X, t), t \in [t_{0}, \infty),$ which sends each material point to where it is at time $t$ , is the deformation (or transformation). Followed in time it describes the motion of the body. The displacement is what is added to the starting position: $p (X, t) = X + U (X, t) ⟺ U (X, t) = p (X, t) - X .$

In components, $p = P_{1} e_{1} + P_{2} e_{2} + P_{3} e_{3}$ with $P_{1} = X_{1} + U_{1} (X, t), P_{2} = X_{2} + U_{2} (X, t), P_{3} = X_{3} + U_{3} (X, t) .$ $X$ is a label: it says which material point we mean. $p$ and $U$ are vector-valued functions of that label and of time.

Warningtwo different U

In Carboni’s notation the bold $U$ is the displacement here, and later (§2.4 Polar decomposition) the same letter is the right stretch tensor of the polar decomposition. The context always says which: the displacement is a vector field $U (X)$ , the stretch tensor a $3 \times 3$ symmetric matrix.

What a deformation must satisfy

Not every map $p$ describes a real body. The notes ask for three things.

Definition 2.1.3admissible deformationnotes L2, p. 2

A deformation is admissible if:

Regularity. Each $P_{i} (X, t)$ is continuous, and so are its derivatives $\partial P_{i} / \partial X_{j}$ for all $i, j$ .
Bijectivity (impenetrability). At each $t$ , $p$ is one-to-one and onto: one material point, one position, and back. So the inverse map exists. Two pieces of matter never end up in the same place.
Positive volume ratio. A small volume $d V$ around $P$ goes to a volume $d \overset{˘}{V}$ around $\overset{˘}{P}$ with $\frac{d V ˘}{d V} = det (\frac{\partial p}{\partial X}) > 0.$

Why the volume ratio is this determinant is proved in §2.5 Change of volume and area. Once $F$ is defined below, condition 3 reads simply $det F > 0$ .

Remarkbeyond the lectures

What the sign of $det F$ means.

$det F \neq = 0$ : no neighbourhood is squashed to zero volume, and $p$ can be inverted near each point.
$det F > 0$ : orientation is kept. A right-handed triad of fibres stays right-handed: a body cannot be turned inside out, which a mirror image would do ( $det F < 0$ ).

Try the presets “turned inside out” and “collapsed” in Figure 2.1.1.

The deformation gradient

How does the deformation act near a point? Take a second point $Q$ close to $P$ , at $X + d X$ .

Definition 2.1.4fibrenotes L2, p. 3

The short segment $PQ = d X$ from $P$ to a nearby point $Q$ is a fibre: a small material line through $P$ . The deformation takes it to $\overset{˘}{P} \overset{˘}{Q}$ , the segment between the new positions $\overset{˘}{P}$ and $\overset{˘}{Q}$ .

Theorem 2.1.5the deformation gradientnotes L2, pp. 3–4

To first order in the length of the fibre, $\overset{˘}{P} \overset{˘}{Q} = F PQ, F = \frac{\partial p}{\partial X}, F_{ij} = \frac{\partial P _{i}}{\partial X _{j}} .$

Proof

Subtract the images of the two ends: $\overset{˘}{P} \overset{˘}{Q} = p (X + d X, t) - p (X, t) .$
Taylor-expand the first term about $X$ : $p (X + d X, t) = p (X, t) + \frac{\partial p}{\partial X}_{(X, t)} d X + O (∣ d X ∣^{2}) .$
The $p (X, t)$ terms cancel. The fibre is short, so drop the higher-order terms: $\overset{˘}{P} \overset{˘}{Q} = \frac{\partial p}{\partial X} d X$ .

Component by component, the same steps read $\overset{˘}{PQ}_{i} = P_{i} (X_{j} + d X_{j}, t) - P_{i} (X_{j}, t) = \frac{\partial P _{i}}{\partial X _{1}} d X_{1} + \frac{\partial P _{i}}{\partial X _{2}} d X_{2} + \frac{\partial P _{i}}{\partial X _{3}} d X_{3},$ that is, as a matrix product, $\overset{˘}{PQ}_{1} \overset{˘}{PQ}_{2} \overset{˘}{PQ}_{3} = \frac{\partial P _{1}}{\partial X _{1}} \frac{\partial P _{2}}{\partial X _{1}} \frac{\partial P _{3}}{\partial X _{1}} \frac{\partial P _{1}}{\partial X _{2}} \frac{\partial P _{2}}{\partial X _{2}} \frac{\partial P _{3}}{\partial X _{2}} \frac{\partial P _{1}}{\partial X _{3}} \frac{\partial P _{2}}{\partial X _{3}} \frac{\partial P _{3}}{\partial X _{3}} d X_{1} d X_{2} d X_{3} .$

Definition 2.1.6deformation gradientEssentialnotes L2, p. 4

The matrix $F = \partial p / \partial X$ , with entries $F_{ij} = \partial P_{i} / \partial X_{j}$ (row $i$ : which component of $p$ ; column $j$ : which coordinate it is differentiated by), is the deformation gradient. It maps each fibre at $P$ to its image: $\overset{˘}{P} \overset{˘}{Q} = F PQ$ .

Key observation 2.1.7deformation is locally linearnotes L2, p. 4

Near a point, any deformation acts on fibres as a linear map, and $F$ is that map. However curved the deformation is on the scale of the body, a small enough neighbourhood sees only $F$ : straight fibres stay straight, parallel fibres stay parallel, a small square becomes a parallelogram and a small circle an ellipse.

$F$ changes from point to point and in time: it is a tensor field $F (X, t)$ . Studying it at one point is studying the deformation locally.

Loading the demo…

Figure 2.1.1 A plane

F

acting near a point. Left, the reference neighbourhood: a unit square on

e_{1}, e_{2}

and a fibre

m

you can turn. Right, the current one: drag the tips of

ν_{1}

and

ν_{2}

to change

F

. The columns of

F

are where

e_{1}

and

e_{2}

go, and every other fibre follows by linearity.

That the columns of $F$ are the images of the basis vectors is the subject of Stretch vectors in the next section.

Warningslip in the notes

L2 p. 3 writes the second term of the difference as $P_{i} (X_{j})$ , without the $t$ ; it is $P_{i} (X_{j}, t)$ .

The displacement gradient

Often the data is the displacement $U$ rather than $p$ . Since $p = X + U$ and $\partial X / \partial X = I$ (the identity matrix), $F = I + H, H = \frac{\partial U}{\partial X}, H_{ij} = \frac{\partial U _{i}}{\partial X _{j}} .$ $H$ is the displacement gradient. Lesson 4 introduces it (Definition 2.6.5); it is used here already because it is the quickest way to $F$ from a displacement.

Recipe 2.1.8deformation gradient at a pointbeyond the lectures

Write the components $P_{i} (X) = X_{i} + U_{i} (X)$ (or differentiate $U$ and add $I$ at the end).
Differentiate: row $i$ of $F$ holds $\partial P_{i} / \partial X_{1}$ , $\partial P_{i} / \partial X_{2}$ , $\partial P_{i} / \partial X_{3}$ .
Evaluate at the point.
Admissible there? Compute $det F$ and check it is positive.

Example 2.1.9the exercise session, steps 1–2notes L5, pp. 2–3

$U (X) = X_{1}^{2} e_{1} + (X_{2}^{2} + X_{3}^{2} - X_{3}) e_{2} + (X_{3}^{2} + X_{2}^{2} + X_{2}) e_{3}$ , at $A = (1, 1, 1)$ .

$P_{1} = X_{1}^{2} + X_{1}$ , $P_{2} = X_{2}^{2} + X_{3}^{2} - X_{3} + X_{2}$ , $P_{3} = X_{3}^{2} + X_{2}^{2} + X_{2} + X_{3}$ .
Differentiate: $F (X) = 2 X_{1} + 1 00 0 2 X_{2} + 1 2 X_{2} + 1 0 2 X_{3} - 1 2 X_{3} + 1 .$
At $A$ : $F (A) = 300033013 .$
$det F (A) = 3 (3 \cdot 3 - 1 \cdot 3) = 18 > 0$ : admissible at $A$ . The volume around $A$ grows 18 times.

Exercises

Exercise 2.1.1

$p (X) = (X_{1} + X_{2} X_{3}) e_{1} + (X_{2} - 2 X_{3} X_{1}) e_{2} + (X_{3} + X_{1} X_{2}) e_{3}$ . Compute $F (X)$ , then $F$ and $det F$ at $P = (0, 0, - 1)$ . Is the deformation admissible there? (Exercises for the first partial, bis, Problem 1.)

Solution

$F (X) = 1 - 2 X_{3} X_{2} X_{3} 1 X_{1} X_{2} - 2 X_{1} 1, F (P) = 120 - 1 10 001 .$ $det F (P) = 1 \cdot 1 - (- 1) \cdot 2 = 3 > 0$ : admissible at $P$ .

Exercise 2.1.2

For simple shear, $p = (X_{1} + k X_{2}) e_{1} + X_{2} e_{2} + X_{3} e_{3}$ with $k$ a number, find $F$ , $H$ and $det F$ . What happens to a unit square on $e_{1}, e_{2}$ ?

Solution

$F = 100 k 10 001$ , $H = 000 k 00 000$ , $det F = 1$ .
$e_{1}$ stays put, $e_{2}$ goes to $(k, 1, 0)$ : the square becomes a parallelogram of the same base and height, so the same area. Layers parallel to $e_{1}$ slide over each other, like a pack of cards.

Exercise 2.1.3

Is $p = (2 X_{1}, - X_{2}, X_{3})$ an admissible deformation? What does it do?

Solution

$F = diag (2, - 1, 1)$ , $det F = - 2 < 0$ : not admissible. It stretches along $e_{1}$ and reflects across the plane $X_{2} = 0$ , turning the body inside out, which no real body can do.