Part 3 · Kinematics

3.5 Homogeneous and rigid deformations

Source: Favata’s notes, Kinematics §1.5, p. 11–12 (homogeneous deformations, translation, rotation, rigid deformations (1.8), Exercises 1.4, 1.5)

Objectives

Recognise a homogeneous deformation, and what stays exact for it.
Prove that a deformation keeping all distances is a translation plus a rotation.
Recipe: $f (x) = f (x_{0}) + F (x - x_{0})$ ; $\nabla (a \cdot b) = (\nabla a)^{⊤} b + (\nabla b)^{⊤} a$ .
Picture: the linear guess and the exact image of a disc coincide, at every size.
Theorem: rigid deformations are rigid-body deformations.
Vocabulary: homogeneous, translation, rotation about a point, rigid.

Homogeneous deformations

Definition 3.5.1Homogeneous deformation

A deformation is homogeneous if $F$ is the same at every point. Then $f (x) = f (x_{0}) + F (x - x_{0}) \forall x, x_{0} .$

Why the formulabeyond the notes

$g (x) := f (x) - Fx$ has gradient $F - F = 0$ , so on a connected body it is constant: $g (x) = g (x_{0})$ . Rearranged, that is the formula.

Key observation 3.5.2Every part deforms in the same way

The expansion $f (x + h) = f (x) + Fh$ is exact, with no $o (∣ h ∣)$ term. So:

straight lines go to straight lines, planes to planes, parallel lines stay parallel;
a sphere goes to an ellipsoid (a circle to an ellipse);
the ratios $d ℓ / d ℓ_{0}$ , $d a / d a_{0}$ , $d v / d v_{0}$ hold for finite segments, faces and volumes.

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Figure 3.5.1 The probe of Figure 3.2.1 again. For the shear, the stretch and the rotation, the exact image of the disc and the linear guess coincide however large

∣ h ∣

is, and the arrows are the same wherever the probe goes: one

F

for the whole body. Switch to the bending to see a deformation that is not homogeneous.

Examples from the notes:

Translation: $f (x) = x + c$ , $c$ constant. $F = I$ , $u = c$ .
Rotation about $x_{0}$ : $f (x) = x_{0} + R (x - x_{0})$ with $R \in Rot$ (Definition 2.6.4). The body turns about the point $x_{0}$ as $R$ prescribes; $F = R$ . About $e_{3}$ through the origin: Exercise 3.5.1.

More (beyond the notes):

Example 3.5.3Stretch along the axesbeyond the notes

$f (x) = α_{1} x_{1} e_{1} + α_{2} x_{2} e_{2} + α_{3} x_{3} e_{3}$ , $α_{i} > 0$ : $F = \sum_{i} α_{i} e_{i} \otimes e_{i}$ . A unit cube becomes an $α_{1} \times α_{2} \times α_{3}$ box (Exercise 3.5.3).

Example 3.5.4Simple shearbeyond the notes

$f_{1} = x_{1} + γ x_{2}$ , $f_{2} = x_{2}$ , $f_{3} = x_{3}$ : $F = I + γ e_{1} \otimes e_{2}$ . Planes $x_{2} =$ const slide over one another like a deck of cards; $det F = 1$ (Exercise 3.4.3).

Rigid deformations

Definition 3.5.5Rigid deformation

A deformation is rigid if it keeps the distance between any two points: $∣ f (x) - f (z) ∣ = ∣ x - z ∣ \forall x, z \in Ω. (1.8)$

The tool for what follows is Exercise 3.5.2: $\nabla (a \cdot b) = (\nabla a)^{⊤} b + (\nabla b)^{⊤} a$ .

Theorem 3.5.6Rigid means translation plus rotation

A deformation is rigid iff $f (x) = translation f (x_{0}) - x_{0} + rotation about x_{0} x_{0} + Q (x - x_{0}), Q \in Rot constant .$

Proof

Square (1.8): $(f (x) - f (z)) \cdot (f (x) - f (z)) = (x - z) \cdot (x - z)$ for all $x, z$ .
Differentiate in $x$ , holding $z$ fixed, with Exercise 1.5 ( $a = b = f (x) - f (z)$ , gradient $F (x)$ ; on the right, gradient $I$ ): $2 F^{⊤} (x) (f (x) - f (z)) = 2 (x - z) .$
Differentiate in $z$ , holding $x$ fixed: $- 2 F^{⊤} (x) F (z) = - 2 I$ , so $F^{⊤} (x) F (z) = I \forall x, z .$
(Spelled out beyond the notes.) With $z = x$ , $F (x)$ is orthogonal (Theorem 2.6.2); then $F^{⊤} (x) = F (x)^{- 1}$ , and step 3 gives $F (z) = F (x)$ : $F$ is constant. By (1.1), $det F = + 1$ , so $F = Q \in Rot$ .
A constant $F$ is homogeneous: $f (x) = f (x_{0}) + Q (x - x_{0})$ ; add and subtract $x_{0}$ .

Conversely (beyond the notes), $∣ f (x) - f (z) ∣ = ∣ Q (x - z) ∣ = ∣ x - z ∣$ , since orthogonal tensors keep lengths.

A rigid deformation is a rigid-body deformation: the composition of a translation and a rotation.

Key observation 3.5.7F is not a strain measurebeyond the notes

A rigid rotation has $F = Q \neq = I$ , yet nothing is strained: $C = Q^{⊤} Q = I$ , every fibre keeps its length and every angle is kept. Conversely, $C = I$ everywhere on a connected body makes the deformation rigid (Gurtin). That is why strain is measured through $C$ , not $F$ .

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Figure 3.5.2 A rigid rotation: the square turns, the circle lands on itself, every fibre has stretch

1

and the angle between the two fibres is kept.

[F]

is far from

I

[C]

is exactly

I

. Compare “stretch and rotation”, which is not rigid.

Exercises

Exercise 3.5.1notes p. 11 (Exercise 1.4)

In a Cartesian frame ${e_{i}}$ , find the effect on the basis vectors of a rotation by $θ$ about the $e_{3}$ axis through the origin.

Solution

$f (x) = Rx$ , homogeneous with $F = R$ as in Example 2.6.5: $R e_{1} = cos θ e_{1} + sin θ e_{2}, R e_{2} = - sin θ e_{1} + cos θ e_{2}, R e_{3} = e_{3},$ the columns of $[R]$ . The notes print $R e_{1}$ with $- sin θ e_{2}$ ; the matrix gives $+$ .

Exercise 3.5.2notes p. 12 (Exercise 1.5)

Let $a (x)$ and $b (x)$ be vector fields. Find the gradient of $a (x) \cdot b (x)$ .

Solution

$(a_{i} b_{i})_{, j} = a_{i, j} b_{i} + a_{i} b_{i, j}$ , and $a_{i, j} b_{i} = ((\nabla a)^{⊤} b)_{j}$ , so $\nabla (a \cdot b) = (\nabla a)^{⊤} b + (\nabla b)^{⊤} a .$

Exercise 3.5.3beyond the notespractice (K.2)

For the stretch along the axes $F = \sum_{i} α_{i} e_{i} \otimes e_{i}$ , find $C$ , $d v / d v_{0}$ , and the area ratio of the face normal to $e_{3}$ . Then specialise to an incompressible uniaxial stretch $λ$ along $e_{1}$ .

Solution

$C = \sum_{i} α_{i}^{2} e_{i} \otimes e_{i}$ ; $d v / d v_{0} = α_{1} α_{2} α_{3}$ .
$F^{*} = det F F^{- ⊤} = α_{2} α_{3} e_{1} \otimes e_{1} + α_{1} α_{3} e_{2} \otimes e_{2} + α_{1} α_{2} e_{3} \otimes e_{3}$ , so $F^{*} e_{3} = α_{1} α_{2} e_{3}$ : the face keeps its normal, and its area scales by the product of its two side stretches.
Incompressible uniaxial: $α_{1} = λ$ , $α_{1} α_{2} α_{3} = 1$ , $α_{2} = α_{3}$ give $α_{2} = α_{3} = 1/ λ$ . Pulling lengthens along $e_{1}$ and thins the cross-section.

Exercise 3.5.4beyond the notes

Is $f (x) = Rx + c$ with $[R] = 010 - 1 00 001$ rigid? About which point does it rotate, when $c = 2 e_{1}$ ?

Solution

$R$ is the rotation by $9 0^{\circ}$ about $e_{3}$ , so it is rigid. A point $x_{0}$ in the 1–2 plane is fixed when $R x_{0} + 2 e_{1} = x_{0}$ : $(- y + 2, x) = (x, y)$ , so $x = y = 1$ . The deformation is the rotation by $9 0^{\circ}$ about the axis through $e_{1} + e_{2}$ parallel to $e_{3}$ : $f (x) = x_{0} + R (x - x_{0})$ with $x_{0} = e_{1} + e_{2}$ .