Part 2 · Theory of deformation

2.3 Principal stretches and principal directions

Source: Carboni’s notes for lesson 3, Teoria della deformazione 2, pp. 2–6

Objectives

Theorem. The largest and smallest stretches at a point are the square roots of the eigenvalues of $C$ , along its eigenvectors.
Recipe. Principal stretches and directions: solve $det (C - λ I) = 0$ , then $(C - λ I) m = 0$ with $∣ m ∣ = 1$ .
Picture. A small sphere becomes an ellipsoid whose axes are the deformed principal directions.
Vocabulary. Triaxial, cylindrical and spherical states of deformation.

For the mechanics it matters to know, at one or more points of the body, the largest and smallest stretch and the directions along which they occur: a material breaks or yields where it is stretched most. In Figure 2.2.1 the stretch of the fibre changed as it turned; this section finds where it peaks.

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Figure 2.3.1 The stretch of a fibre against its angle. Turn

m

(left) or drag its point along the graph (below the readouts):

ν_{m}

moves between a maximum and a minimum, reached at two orthogonal directions, the principal directions

m_{1}, m_{2}

(dashed, orange). The fibre snaps to them. Their images are the axes of the ellipse into which the unit circle goes.

The extremal problem

Problem. Find the extremes of $ν_{m}^{2} = m^{T} C m$ over all unit vectors, that is under the constraint $m^{T} m = 1$ .

A constrained extremum calls for a Lagrange multiplier $λ$ , a number: look for the stationary points of $f_{m} = ν_{m}^{2} - λ (m^{T} m - 1) = m^{T} C m - λ (m^{T} m - 1) = m^{T} (C - λ I) m + λ .$

Set both derivatives to zero: $⎩ ⎨ ⎧ \frac{\partial f _{m}}{\partial m} = 2 (C - λ I) m = 0 \frac{\partial f _{m}}{\partial λ} = - (m^{T} m - 1) = 0 ⟹ {(C - λ I) m = 0 m^{T} m = 1 (characteristic equation) (normalisation)$

Why the derivative of m^T A m is 2Am

For a symmetric matrix $A$ (beyond the lectures), write the sums out: $m^{T} Am = \sum_{i} \sum_{j} A_{ij} m_{i} m_{j}$ . Differentiating by $m_{k}$ hits $m_{i}$ when $i = k$ and $m_{j}$ when $j = k$ : $\frac{\partial}{\partial m _{k}} i, j \sum A_{ij} m_{i} m_{j} = j \sum A_{kj} m_{j} + i \sum A_{ik} m_{i} = 2 j \sum A_{kj} m_{j} = 2 (Am)_{k},$ using $A_{ik} = A_{ki}$ . And $m^{T} m$ is the case $A = I$ , with derivative $2 m$ .

Warningslip in the notes

L3 p. 3 writes $\partial f_{m} / \partial m = (C - λ I) m$ , without the factor 2. Since it is set to zero, the resulting equation is the same.

Key observation 2.3.1extremal stretches are an eigenvalue problemnotes L3, p. 3

Extremal stretches are eigenvalue problems for $C$ .

Non-zero solutions $m$ exist only if $det (C - λ I) = 0$ : the characteristic equation of $C$ , a cubic in $λ$ .
$C$ is symmetric and positive definite (Proposition 2.2.3, Proposition 2.2.4). So its three eigenvalues $λ_{1}, λ_{2}, λ_{3}$ are real and positive, and there are always three orthonormal eigenvectors $m_{1}, m_{2}, m_{3}$ .

Theorem 2.3.2principal stretchesnotes L3, pp. 2–4

Let $λ_{k}$ , $m_{k}$ be the eigenvalues and unit eigenvectors of $C$ . Then $ν_{k}^{2} = m_{k}^{T} C m_{k} = λ_{k}, ν_{k} = λ_{k} .$ The $ν_{k}$ are the principal stretches and the $m_{k}$ the principal directions of stretch. The largest and smallest principal stretches are the largest and smallest stretches over all directions.

Proof

Eigenvalues are squared stretches. Multiply $(C - λ_{k} I) m_{k} = 0$ on the left by $m_{k}^{T}$ : $m_{k}^{T} C m_{k} - λ_{k} m_{k}^{T} m_{k} = 0 ⟹ ν_{k}^{2} = m_{k}^{T} C m_{k} = λ_{k} .$

They are the maximum and minimum (beyond the lectures). Write any unit $m = a_{1} m_{1} + a_{2} m_{2} + a_{3} m_{3}$ , with $a_{1}^{2} + a_{2}^{2} + a_{3}^{2} = 1$ because the $m_{k}$ are orthonormal. Then $Cm = λ_{1} a_{1} m_{1} + λ_{2} a_{2} m_{2} + λ_{3} a_{3} m_{3}$ , and $ν_{m}^{2} = m^{T} C m = λ_{1} a_{1}^{2} + λ_{2} a_{2}^{2} + λ_{3} a_{3}^{2},$ a weighted average of the eigenvalues, with weights $a_{k}^{2}$ adding up to 1. It is never above the largest eigenvalue nor below the smallest, and it reaches them along the matching eigenvectors.

The principal directions stay orthogonal

Proposition 2.3.3principal directions undergo no shearnotes L3, p. 4

The images $\overset{˘}{m}_{k}$ of the principal directions are orthogonal too: $\overset{˘}{m}_{i}^{T} \overset{˘}{m}_{j} = 0$ for $i \neq = j$ . Principal directions undergo no shear.

Proof

The $m_{k}$ are orthogonal, $m_{i}^{T} m_{j} = 0$ . With $ν_{i} = F m_{i} = ν_{i} \overset{˘}{m}_{i}$ : $ν_{i} ν_{j} \overset{˘}{m}_{i}^{T} \overset{˘}{m}_{j} = ν_{i}^{T} ν_{j} = (F m_{i})^{T} (F m_{j}) = m_{i}^{T} C m_{j} = λ_{j} m_{i}^{T} m_{j} = 0.$ Stretches are positive, so $\overset{˘}{m}_{i}^{T} \overset{˘}{m}_{j} = 0$ . In the language of Theorem 2.2.9, $sin γ_{ij} = m_{i}^{T} C m_{j} / (ν_{i} ν_{j}) = 0$ .

Key observation 2.3.4sphere to ellipsoidnotes L3, p. 4

A small sphere around $P$ becomes an ellipsoid around $\overset{˘}{P}$ . Its semi-axes point along $\overset{˘}{m}_{1}, \overset{˘}{m}_{2}, \overset{˘}{m}_{3}$ , with lengths proportional to $ν_{1}, ν_{2}, ν_{3}$ . In the basis ${m_{1}, m_{2}, m_{3}}$ , $C$ is diagonal: $C = ν_{1}^{2} 00 0 ν_{2}^{2} 0 00 ν_{3}^{2} .$

Why a sphere becomes an ellipsoidbeyond the lectures

A point $X$ of the unit sphere has $X = a_{1} m_{1} + a_{2} m_{2} + a_{3} m_{3}$ with $\sum a_{k}^{2} = 1$ . Its image is $FX = a_{1} ν_{1} \overset{˘}{m}_{1} + a_{2} ν_{2} \overset{˘}{m}_{2} + a_{3} ν_{3} \overset{˘}{m}_{3}$ . In the orthonormal frame $\overset{˘}{m}_{k}$ its coordinates are $y_{k} = a_{k} ν_{k}$ , and they satisfy $\frac{y _{1}^{2}}{ν _{1}^{2}} + \frac{y _{2}^{2}}{ν _{2}^{2}} + \frac{y _{3}^{2}}{ν _{3}^{2}} = 1,$ the equation of an ellipsoid with semi-axes $ν_{1}, ν_{2}, ν_{3}$ along $\overset{˘}{m}_{1}, \overset{˘}{m}_{2}, \overset{˘}{m}_{3}$ .

Types of deformation state

The multiplicity of the eigenvalues names the state of deformation at the point.

Definition 2.3.5triaxial, cylindrical and spherical deformationnotes L3, pp. 3–6

Triaxial: $λ_{1} \neq = λ_{2} \neq = λ_{3}$ . Three distinct principal stretches and three principal directions, fixed up to sign. A sphere becomes a general ellipsoid.
Cylindrical: $λ_{1} = λ_{2} \neq = λ_{3}$ . $λ_{3}$ has its own direction $m_{3}$ . The double eigenvalue has a whole plane of eigenvectors (its eigenspace), orthogonal to $m_{3}$ : any orthonormal pair $m_{1}, m_{2}$ in that plane will do, and every direction in it has the stretch $ν = ν_{1} = ν_{2}$ . A cylinder with axis along $m_{3}$ is mapped into a cylinder.
Spherical (isotropic): $λ_{1} = λ_{2} = λ_{3}$ , so $ν_{1} = ν_{2} = ν_{3} = ν$ . Every direction is principal, any orthonormal triad will do, and a sphere is mapped into a sphere of radius $ν$ times the original.

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Figure 2.3.2 A unit sphere (grey, dashed) and the ellipsoid it becomes under

F

. Orange: the principal directions

m_{k}

(dashed) and the semi-axes

ν_{k} \overset{˘}{m}_{k}

. Set two stretches equal (they snap) to get the cylindrical case, three for the spherical one. Drag the background to turn it, or press spin.

Warningslip in the notes

L3 p. 5 says “gli autovettori $m_{1}$ e $m_{3}$ ” can be any orthogonal pair spanning the plane orthogonal to $m_{3}$ . The pair is $m_{1}, m_{2}$ .

Recipe 2.3.6principal stretches and directionsbeyond the lectures

Compute $C = F^{T} F$ .
Look for a direction already principal. If a row and column of $C$ are zero off the diagonal (say the third), then $e_{3}$ is a principal direction with $λ = C_{33}$ , and only a $2 \times 2$ problem is left.
Characteristic equation $det (C - λ I) = 0$ ; its roots are $λ_{1}, λ_{2}, λ_{3}$ (check: their sum is $tr C$ , their product $det C = (det F)^{2}$ ).
Eigenvectors. For each $λ_{k}$ , solve $(C - λ_{k} I) m = 0$ : two independent equations, one free component. Normalise to $∣ m_{k} ∣ = 1$ .
Stretches $ν_{k} = λ_{k}$ . Name the state from how many are equal.

Example 2.3.7simple shearbeyond the lectures

$F = 100 \frac{1}{2} 10 001$ , so $C = 1 \frac{1}{2} 0 \frac{1}{2} \frac{5}{4} 0 001$ .

The third row and column are zero off the diagonal: $e_{3}$ is principal, $λ = 1$ .
The $2 \times 2$ block: $det [1 - λ \frac{1}{2} \frac{1}{2} \frac{5}{4} - λ] = λ^{2} - \frac{9}{4} λ + 1 = 0$ , so $λ = \frac{9 \pm 17}{8}$ : $1.6404$ and $0.6096$ .
For $λ = 1.6404$ : $(1 - 1.6404) m_{1} + \frac{1}{2} m_{2} = 0$ , so $m \propto (1, 1.2808, 0)$ , normalised $(0.6154, 0.7882, 0)$ , about $52°$ from $e_{1}$ . The other in-plane direction is orthogonal to it.
Principal stretches $1.2808$ , $1$ , $0.7808$ ; their product is $1 = det F$ . Three distinct values: triaxial, even though $det F = 1$ . Compare with the “simple shear” preset of Figure 2.3.1, where the shear is twice as large.

Exercises

Exercise 2.3.1

For $F = 120 - 1 10 001$ (Exercise 2.1.1), find the principal stretches and directions. Name the state.

Solution

$C = 510120001$ ; $e_{3}$ is principal with $λ = 1$ .
Block: $λ^{2} - 7 λ + 9 = 0$ , $λ = \frac{7 \pm 13}{2}$ : $5.3028$ and $1.6972$ .
For $5.3028$ : $(5 - 5.3028) m_{1} + m_{2} = 0$ , $m \propto (1, 0.3028, 0)$ , normalised $(0.9571, 0.2898, 0)$ ; the other is $(- 0.2898, 0.9571, 0)$ .
$ν_{k} = 2.3028, 1.3028, 1$ . Product $3 = det F$ . Triaxial.

Exercise 2.3.2

At the point $A$ of the exercise session, $C (A) = 9000181201210$ (Example 2.2.13). Find the principal stretches and check them against $det F (A) = 18$ .

Solution

$e_{1}$ is principal: $λ = 9$ , $ν = 3$ .
Block: $λ^{2} - 28 λ + 36 = 0$ , $λ = 14 \pm 160$ : $26.649$ and $1.351$ .
$ν_{k} = 5.162, 3, 1.162$ . Product $5.162 \cdot 3 \cdot 1.162 = 18.0$ . ✓
The largest stretch, $5.16$ , is along $(0, 0.81, 0.58)$ : solve $(18 - 26.649) m_{2} + 12 m_{3} = 0$ , so $m \propto (0, 1, 0.7208)$ .

Exercise 2.3.3

A deformation has $C = diag (4, 4, 1)$ in the basis $e_{k}$ . What kind of state is it? Give two different valid choices of principal directions.

Solution

Cylindrical: $λ_{1} = λ_{2} = 4$ ( $ν = 2$ ), $λ_{3} = 1$ ( $ν = 1$ ) along $e_{3}$ . Any orthonormal pair in the $e_{1} e_{2}$ plane is principal: $e_{1}, e_{2}$ , or $\frac{1}{2} (1, 1, 0), \frac{1}{2} (- 1, 1, 0)$ . Every fibre in that plane is doubled in length.