Part 2 · Theory of deformation

2.2 Stretch, shear and the Cauchy–Green tensor

Source: Carboni’s notes for lesson 2, Teoria della deformazione 1, pp. 4–10; lesson 3, pp. 1–2; lesson 4, p. 2

Objectives

Vocabulary. Stretch $ν_{m}$ , stretch vector $ν_{m}$ , shear angle $γ_{mn}$ , Cauchy–Green tensor $C$ .
Recipe. The stretch of any fibre and the shear between two orthogonal fibres, from $C$ .
Picture. The columns of $F$ are the images of the basis fibres; the entries of $C$ are their squared lengths and the angles between them.
Theorem. $C$ is symmetric and positive definite, and $C = I$ exactly when nothing changes length.

$F$ says where each fibre goes, but it mixes two things: how the neighbourhood deforms (lengths and angles change) and how it moves rigidly (it may also turn). This section extracts the deformation: one number for the change of length of a fibre, one for the change of angle between two, and one tensor, $C$ , that holds them all.

Stretch

Definition 2.2.1stretchEssentialnotes L2, p. 4

The stretch of a fibre is how many times longer it has become: $ν = \frac{∣ P ˘ Q ˘ ∣}{∣ PQ ∣} .$

$ν ⎩ ⎨ ⎧ > 1 = 1 < 1 the fibre lengthens (allungamento) its length is unchanged it shortens (accorciamento)$

Write it with $F$ , using $∣ a ∣ = a^{T} a$ for the length of a column vector:

$∣ \overset{˘}{P} \overset{˘}{Q} ∣ = (F PQ)^{T} (F PQ) = PQ^{T} F^{T} F PQ$ , and $∣ PQ ∣ = PQ^{T} PQ$ .
So $ν = \frac{PQ ^{T} F ^{T} F PQ}{∣ PQ ∣ ∣ PQ ∣}$ .
Divide each $PQ$ by its length: what is left is the unit vector along the fibre, $m = PQ /∣ PQ ∣$ .

Definition 2.2.2Cauchy–Green tensorEssentialnotes L2, p. 5

$C = F^{T} F$ is the Cauchy–Green deformation tensor (many texts call it the right Cauchy–Green tensor). The stretch of the fibre along the unit vector $m$ is $ν_{m} = m^{T} C m .$

The stretch depends on the point (through $C$ ) and on the direction $m$ : it is a property of a fibre, not of the point alone. In Figure 2.2.1 turn the fibre $m$ and watch $ν_{m}$ change.

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Figure 2.2.1 The stretch of a fibre. Turn

m

on the left; on the right its image

ν_{m} = Fm

has length

ν_{m}

, computed both as

m^{T} Cm

and as

∣ Fm ∣

. Try the rotation:

F

changes,

C = I

does not, and every stretch is 1.

Properties of C

Proposition 2.2.3C is symmetricnotes L3, p. 2

$C^{T} = C$ .

Proof

$C^{T} = (F^{T} F)^{T} = F^{T} (F^{T})^{T} = F^{T} F = C$ , using $(AB)^{T} = B^{T} A^{T}$ .

Proposition 2.2.4C is positive definitebeyond the lectures

$m^{T} C m > 0$ for every $m \neq = 0$ : no fibre collapses to zero length.

Proof

$m^{T} Cm = (Fm)^{T} (Fm) = ∣ Fm ∣^{2} \geq 0$ . It is zero only if $Fm = 0$ ; since $det F > 0$ , $F$ is invertible and then $m = F^{- 1} 0 = 0$ .

When nothing changes length

Proposition 2.2.5C = Inotes L2, p. 5

$ν_{m} = 1 for every direction m ⟺ C = I .$

Proof

$\Leftarrow$ : $m^{T} Im = m^{T} m = 1$ for a unit vector.

$\Rightarrow$ (beyond the lectures): let $A = C - I$ , symmetric, with $m^{T} Am = 0$ for every unit $m$ .

$m = e_{i}$ gives $A_{ii} = 0$ .
$m = (e_{i} + e_{j}) / 2$ gives $\frac{1}{2} (A_{ii} + A_{jj} + 2 A_{ij}) = 0$ , so $A_{ij} = 0$ .

So $A = 0$ .

Warningslip in the notes

L2 p. 5 writes $m^{T} C m = 1 ⟺ C = I$ without saying for which $m$ . One fibre of unit stretch does not make $C = I$ (in Figure 2.2.1, simple shear leaves $e_{1}$ with $ν = 1$ ); the equivalence needs every direction $m$ .

Lengths alone fix $C$ . Angles then come for free, as the next sections show: the back page When nothing deforms collects every way of saying “nothing deforms”.

Stretch vectors

Take three fibres along the basis directions, $PQ_{j} = ∣ PQ_{j} ∣ e_{j}$ . Their images per unit original length are $\frac{P ˘ Q ˘ _{j}}{∣ PQ _{j} ∣} = \frac{F e _{j} ∣ PQ _{j} ∣}{∣ PQ _{j} ∣} = F e_{j} .$

Definition 2.2.6stretch vectornotes L2, pp. 6–7

The stretch vector of direction $j$ is $ν_{j} = F e_{j} = F_{1 j} F_{2 j} F_{3 j} = ν_{j} \overset{˘}{m}_{j},$

its length $∣ ν_{j} ∣ = ∣ \overset{˘}{P} \overset{˘}{Q}_{j} ∣/∣ PQ_{j} ∣ = ν_{j}$ is the stretch of that fibre;
its direction $\overset{˘}{m}_{j}$ is where the fibre points in the current configuration.

More generally $ν_{m} = Fm = ν_{m} \overset{˘}{m}$ for any unit $m$ .

Key observation 2.2.7the columns of Fnotes L2, p. 7

The columns of $F$ are the images of the basis vectors: $F = [ν_{1} ν_{2} ν_{3}] .$ For any direction $m = m_{1} e_{1} + m_{2} e_{2} + m_{3} e_{3}$ , linearity gives $ν_{m} = Fm = m_{1} ν_{1} + m_{2} ν_{2} + m_{3} ν_{3} .$

This is exactly what the handles of Figure 2.1.1 are: dragging $ν_{1}$ rewrites the first column of $F$ .

Warningslip in the notes

L2 p. 6 writes $∣ ν_{j} ∣ = ∣ \overset{˘}{P} \overset{˘}{Q}_{j} ∣/∣ PQ ∣$ : the denominator is $∣ PQ_{j} ∣$ , the fibre along $e_{j}$ . The figure on the same page labels the deformed points $Q_{1}, Q_{2}, Q_{3}$ without breves.

The shear angle

Fibres also change the angle between them. The notes sketch a square grid of fibres sheared into parallelograms: a cell $P, Q_{1}, Q, Q_{2}$ with a right angle at $P$ becomes $\overset{˘}{P}, \overset{˘}{Q}_{1}, \overset{˘}{Q}, \overset{˘}{Q}_{2}$ with an angle $\overset{˘}{θ}$ at $\overset{˘}{P}$ , while its sides keep their length (L2 p. 7). Shear is independent of stretch: an angle can change while no fibre along the sides changes length.

Take two fibres from $P$ , $PM$ and $PN$ , with unit vectors $m = PM /∣ PM ∣$ and $n = PN /∣ PN ∣$ , orthogonal in the reference configuration. After the deformation they make the angle $\overset{˘}{θ}_{mn}$ .

Definition 2.2.8shear anglenotes L2, p. 7

The shear angle (scorrimento angolare, shear strain) between two fibres that were orthogonal is how much the right angle has closed: $γ_{mn} = \frac{π}{2} - \overset{˘}{θ}_{mn} .$

Remarkbeyond the lectures

With this sign convention $γ_{mn} > 0$ when the fibres close up and $γ_{mn} < 0$ when they open.
$γ$ is an angle, measured in radians.

Theorem 2.2.9shear from Cnotes L2, pp. 8–9; L4, p. 2

For two unit fibres $m ⊥ n$ , $sin γ_{mn} = cos \overset{˘}{θ}_{mn} = \frac{m ^{T} C n}{ν _{m} ν _{n}} = \frac{m ^{T} C n}{m ^{T} C m n ^{T} C n} .$

Proof

The scalar product of the deformed fibres is $\overset{˘}{P} \overset{˘}{M} \cdot \overset{˘}{P} \overset{˘}{N} = ∣ \overset{˘}{P} \overset{˘}{M} ∣ ∣ \overset{˘}{P} \overset{˘}{N} ∣ cos \overset{˘}{θ}_{mn} .$
Solve for the cosine, writing the lengths as stretch times original length, $∣ \overset{˘}{P} \overset{˘}{M} ∣ = ν_{m} ∣ PM ∣$ and $∣ \overset{˘}{P} \overset{˘}{N} ∣ = ν_{n} ∣ PN ∣$ : $cos \overset{˘}{θ}_{mn} = \frac{P ˘ M ˘ ^{T} P ˘ N ˘}{ν _{m} ∣ PM ∣ ν _{n} ∣ PN ∣} .$
Replace each deformed fibre with $F$ times the original one: $cos \overset{˘}{θ}_{mn} = \frac{( F PM ) ^{T} ( F PN )}{ν _{m} ν _{n} ∣ PM ∣ ∣ PN ∣} = \frac{PM ^{T} F ^{T} F PN}{ν _{m} ν _{n} ∣ PM ∣ ∣ PN ∣} = \frac{m ^{T} C n}{ν _{m} ν _{n}} .$
Finally $cos \overset{˘}{θ}_{mn} = cos (π /2 - γ_{mn}) = sin γ_{mn}$ .

Key observation 2.2.10C holds every length and angle

$C$ alone gives every length and every angle change at the point: the stretch of any fibre, and the shear between any two orthogonal fibres. It is the metric of the deformed neighbourhood, read in reference coordinates. A rigid rotation, which changes $F$ , leaves $C$ alone.

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Figure 2.2.2 The shear between two fibres.

m

and

n

are orthogonal on the left; turn them. On the right, the angle between their images is

\overset{˘}{θ}_{mn} = 90° - γ_{mn}

. Under “unequal stretches”,

e_{1}

and

e_{2}

stay orthogonal but the diagonals do not: shear depends on the pair of fibres.

The components of C

Take the fibres along the basis, $m = e_{i}$ and $n = e_{j}$ with $i \neq = j$ , and write $γ_{ij}$ for their shear angle. $e_{i}^{T} C e_{j}$ picks out the entry $C_{ij}$ , so:

stretch: $ν_{i} = e_{i}^{T} C e_{i} = C_{ii}$ , so $C_{ii} = ν_{i}^{2}$ ;
shear: $sin γ_{ij} = \frac{e _{i}^{T} C e _{j}}{ν _{i} ν _{j}} = \frac{C _{ij}}{ν _{i} ν _{j}}$ , so $C_{ij} = ν_{i} ν_{j} sin γ_{ij}$ .

Key observation 2.2.11what the entries of C meannotes L2, p. 10; L3, p. 1

$C = ν_{1}^{2} sym ν_{1} ν_{2} sin γ_{12} ν_{2}^{2} ν_{1} ν_{3} sin γ_{13} ν_{2} ν_{3} sin γ_{23} ν_{3}^{2}$

Diagonal: the squared stretches of the three basis fibres.
Off-diagonal: the shear between pairs of basis fibres, weighted by their stretches.

Why: C is a table of scalar productsbeyond the lectures

$C_{ij} = (F^{T} F)_{ij} = ν_{i} \cdot ν_{j}$ : the entry $(i, j)$ of $C$ is the scalar product of the $i$ -th and $j$ -th columns of $F$ . On the diagonal that is $∣ ν_{i} ∣^{2} = ν_{i}^{2}$ ; off the diagonal it is $∣ ν_{i} ∣∣ ν_{j} ∣ cos \overset{˘}{θ}_{ij} = ν_{i} ν_{j} sin γ_{ij}$ .

Proposition 2.2.12C = I is a rigid motionbeyond the lectures

If $C = I$ , then $F^{T} F = I$ : $F$ is an orthogonal matrix. For any orthonormal $m, n$ , $ν_{m} = m^{T} m = 1, sin γ_{mn} = \frac{m ^{T} n}{ν _{m} ν _{n}} = 0.$ Lengths are kept and orthogonal fibres stay orthogonal: the neighbourhood only translates and rotates.

Example 2.2.13the exercise session, step 3notes L5, pp. 3–4

From Example 2.1.9, at $A$ , $C (A) = F^{T} F = 300031033300033013 = 9000181201210,$ with the fibres $m = \frac{2}{2} (e_{1} + e_{2})$ and $n = \frac{2}{2} (e_{2} - e_{1})$ , orthogonal.

Stretch along $m$ . $Cm = \frac{2}{2} (9, 18, 12)$ , so $ν_{m}^{2} = m^{T} Cm = \frac{9}{2} + \frac{18}{2} = \frac{27}{2}$ and $ν_{m} = 27/2 = 3.674$ . Check with the stretch vector: $ν_{m} = Fm = \frac{3 2}{2} (1, 1, 1)$ , of length $3 \cdot \frac{9}{2} = 27/2$ .
Stretch along $n$ . $Cn = \frac{2}{2} (- 9, 18, 12)$ , so $ν_{n}^{2} = \frac{9}{2} + \frac{18}{2} = \frac{27}{2}$ , $ν_{n} = 3.674$ .
Shear. $C_{mn} = m^{T} Cn = - \frac{9}{2} + \frac{18}{2} = \frac{9}{2}$ , so $sin γ_{mn} = \frac{9/2}{27/2} = \frac{1}{3}, γ_{mn} = arcsin \frac{1}{3} = 19.47° (0.3398 rad) .$
Check with the stretch vectors: $ν_{n} = Fn = \frac{3 2}{2} (- 1, 1, 1)$ and $cos \overset{˘}{θ}_{mn} = \frac{ν _{m} \cdot ν _{n}}{∣ ν _{m} ∣∣ ν _{n} ∣} = \frac{- \frac{9}{2} + \frac{9}{2} + \frac{9}{2}}{27/2} = \frac{1}{3}, \overset{˘}{θ}_{mn} = 70.53° = 90° - γ_{mn} .$

Warningslips in the worked sheet

On L5 p. 4: the $ν_{n}^{2}$ line writes its first term as $- \frac{9}{2}$ (it is $+ \frac{9}{2}$ , since $(- \frac{9 2}{2}) (- \frac{2}{2}) > 0$ ; the total $\frac{27}{2}$ is right); $C_{mn}$ is written $m^{T} C n^{T}$ (no transpose on $n$ ); and the angles are worked from the rounded $cos \overset{˘}{θ} = 0.334$ , giving $19.52°$ and $70.49°$ instead of the exact $19.47°$ and $70.53°$ .

Exercises

Exercise 2.2.1

With $F (P)$ from Exercise 2.1.1, compute $C$ at $P$ , the stretches along $m = \frac{1}{2} (1, 1, 0)$ and $n = \frac{1}{2} (- 1, 1, 0)$ , and the shear angle between them. (Exercises for the first partial, bis, Problem 1.)

Solution

$C = F^{T} F = 510120001$ .
$ν_{m}^{2} = \frac{1}{2} (5 + 2 + 2 \cdot 1) = \frac{9}{2}$ , so $ν_{m} = \frac{3 2}{2} = 2.1213$ .
$ν_{n}^{2} = \frac{1}{2} (5 + 2 - 2 \cdot 1) = \frac{5}{2}$ , so $ν_{n} = 1.5811$ .
$m^{T} Cn = \frac{1}{2} (- 5 + 1 - 1 + 2) = - \frac{3}{2}$ , so $sin γ_{mn} = \frac{- 3/2}{2.1213 \cdot 1.5811} = - \frac{1}{5}$ and $γ_{mn} = - 26.57°$ .
The fibres open to $\overset{˘}{θ}_{mn} = 116.57°$ : a negative shear angle.

Exercise 2.2.2

Simple shear $F = 100 \frac{1}{2} 10 001$ . Find $C$ , the stretches along $e_{1}$ , $e_{2}$ and $\frac{1}{2} (1, 1, 0)$ , and the shear angle $γ_{12}$ .

Solution

$C = 1 \frac{1}{2} 0 \frac{1}{2} \frac{5}{4} 0 001$ .
$ν_{1} = 1$ , $ν_{2} = \frac{5}{2} = 1.1180$ , and along the diagonal $ν^{2} = \frac{1}{2} (1 + \frac{5}{4} + 1) = \frac{13}{8}$ , $ν = 1.2748$ .
$sin γ_{12} = \frac{1/2}{1 \cdot 1.1180} = 0.4472$ , so $γ_{12} = 26.57° = arctan \frac{1}{2}$ : the vertical fibre tilts by $arctan \frac{1}{2}$ .

Exercise 2.2.3

Show that the rotation $F = cos α sin α 0 - sin α cos α 0 001$ has $C = I$ . What are the stretch vectors?

Solution

$C = F^{T} F$ has entries $cos^{2} α + sin^{2} α = 1$ on the diagonal and $- cos α sin α + sin α cos α = 0$ off it. The stretch vectors are the columns $(cos α, sin α, 0)$ , $(- sin α, cos α, 0)$ , $(0, 0, 1)$ : unit length, mutually orthogonal, the basis turned by $α$ .