Part 3 · Kinematics

3.3 Change in length and angle

Source: Favata’s notes, Kinematics §1.4, p. 8–9 (change in length, right Cauchy–Green tensor, Remark 1.3, footnote 3)

Objectives

Compute how much a small fibre is stretched, and how much an angle between two fibres changes.
Understand why $C = F^{⊤} F$ holds all of this, with less information than $F$ .
Recipe: $d ℓ / d ℓ_{0} = ∣ Ft ∣ = Ct \cdot t$ ; read stretches and shears off $[C]$ .
Pictures: a fibre turned around the unit circle and its stretch; two fibres and the angle between them.
Theorem: $C$ is symmetric and positive definite.
Vocabulary: stretch, shear.
Essential vocabulary: right Cauchy–Green tensor.

The stretch of a fibre

Take two nearby points $x_{0}$ and $x$ of the reference configuration: $x - x_{0} = d ℓ_{0} t, d ℓ_{0} = ∣ x - x_{0} ∣, t a unit vector .$ Their images bound a segment of length $d ℓ := ∣ f (x) - f (x_{0}) ∣$ . By the definition of $F$ , $f (x) - f (x_{0}) = d ℓ_{0} Ft + o (d ℓ_{0}),$ and taking lengths (a vector plus something $o (d ℓ_{0})$ changes length by $o (d ℓ_{0})$ ), $d ℓ = d ℓ_{0} ∣ Ft ∣ + o (d ℓ_{0})$ .

Theorem 3.3.1Change in length

For an infinitesimal fibre along the unit vector $t$ , $\frac{d ℓ}{d ℓ _{0}} = ∣ Ft ∣. (1.4)$

The ratio is the stretch of the fibre (the name is beyond the notes): $> 1$ it lengthens, $< 1$ it shortens, $= 1$ it keeps its length. In 1D it is $f^{'}$ .

Loading the demo…

Figure 3.3.1 Reference (grey, dashed) and current (teal) around a point, drawn over each other. Drag the fibre

t

around the unit circle: its image

Ft

ends on the ellipse, and its length is the stretch. Look for the longest and the shortest fibre; the readout says when you are on one. For the rigid rotation every fibre has stretch

1

, although

F \neq = I

The right Cauchy–Green tensor

Move $F$ across the scalar product (Definition 2.2.1): $∣ Ft ∣ = Ft \cdot Ft = F^{⊤} F t \cdot t .$

Definition 3.3.2Right Cauchy–Green tensorEssential

$C := F^{⊤} F, \frac{d ℓ}{d ℓ _{0}} = Ct \cdot t .$

Theorem 3.3.3C is symmetric and positive definite

$C^{⊤} = C$ (Remark 1.3);
$Ca \cdot a > 0$ for every $a \neq = 0$ (footnote 3).

Proof

$(F^{⊤} F)^{⊤} = F^{⊤} (F^{⊤})^{⊤} = F^{⊤} F$ (Theorem 2.2.6).
$Ca \cdot a = Fa \cdot Fa = ∣ Fa ∣^{2} \geq 0$ , with equality iff $Fa = 0$ . Since $det F > 0$ , $Fa = 0$ only for $a = 0$ (Theorem 2.5.2, condition 2).

Key observation 3.3.4C forgets the rotation

$F$ has 9 components, $C$ only 6. Length changes need only $C$ , so $F$ carries more than they need. The extra is the rigid part of the deformation: a rotation changes $F$ and leaves $C = I$ (§3.5 Homogeneous and rigid deformations, Theorem 3.6.5).

Angles

Two fibres $t_{1}$ , $t_{2}$ end up at an angle $ϑ$ with (beyond the notes) $cos ϑ = \frac{F t _{1} \cdot F t _{2}}{∣ F t _{1} ∣∣ F t _{2} ∣} = \frac{C t _{1} \cdot t _{2}}{C t _{1} \cdot t _{1} C t _{2} \cdot t _{2}} .$

Recipe 3.3.5Reading Cbeyond the notes

$C_{ij} = F e_{i} \cdot F e_{j}$ , the scalar products of the columns of $[F]$ .

Diagonal: $C_{11} = ∣ F e_{1} ∣^{2}$ , the squared stretch of the fibre along $e_{1}$ ; likewise $C_{22}$ , $C_{33}$ .
Off-diagonal: $C_{12} = ∣ F e_{1} ∣∣ F e_{2} ∣ cos ϑ_{12}$ , with $ϑ_{12}$ the angle between the deformed fibres. Nonzero means the right angle between $e_{1}$ and $e_{2}$ has changed: shear.

Loading the demo…

Figure 3.3.2 Two fibres,

t

and a second one. Start from

e_{1}

and

e_{2}

, at right angles: under the shear their images are not, and the readout gives the new angle through

C

. Now hunt for two fibres that stay at right angles after the deformation. There always is a pair; §3.6 Strain measures says why.

Key observation 3.3.6C is the local change of shape

Lengths and angles of all fibres at a point follow from $C$ alone. Together they are the local change of shape; areas and volumes will follow from $C$ too.

Exercises

Exercise 3.3.1beyond the notes

For the simple shear $F = I + γ e_{1} \otimes e_{2}$ find $C$ , the stretches of $e_{1}$ , $e_{2}$ and of the diagonals $(e_{1} \pm e_{2}) / 2$ , and the angle between the deformed $e_{1}$ , $e_{2}$ .

Solution

$C = (I + γ e_{2} \otimes e_{1}) (I + γ e_{1} \otimes e_{2}) = I + γ (e_{1} \otimes e_{2} + e_{2} \otimes e_{1}) + γ^{2} e_{2} \otimes e_{2}$ : $[C] = 1 γ 0 γ 1 + γ^{2} 0 001 .$

$e_{1}$ : $C_{11} = 1$ ; fibres along the shear keep their length.
$e_{2}$ : $1 + γ^{2}$ ; the vertical fibre tilts and lengthens.
$(e_{1} \pm e_{2}) / 2$ : $(C_{11} \pm 2 C_{12} + C_{22}) /2 = 1 \pm γ + γ^{2} /2$ : for $γ > 0$ one diagonal lengthens, the other shortens.
$cos ϑ_{12} = C_{12} / C_{11} C_{22} = γ / 1 + γ^{2}$ : the right angle closes.

Exercise 3.3.2beyond the notespractice (K.3)

For the deformation of Exercise 3.2.1, find $C$ and the stretch of a fibre along $e_{2}$ .

Solution

Columns of $[F]$ : $F e_{1} = e_{1}$ , $F e_{2} = (1 + α) e_{2} + β x_{3} e_{3}$ , $F e_{3} = (1 + β x_{2}) e_{3}$ . With $C_{ij} = F e_{i} \cdot F e_{j}$ : $[C] = 100 0 (1 + α)^{2} + β^{2} x_{3}^{2} β x_{3} (1 + β x_{2}) 0 β x_{3} (1 + β x_{2}) (1 + β x_{2})^{2} .$ The stretch along $e_{2}$ is $C_{22} = (1 + α)^{2} + β^{2} x_{3}^{2}$ , and varies from point to point: the deformation is not homogeneous.

Exercise 3.3.3beyond the notes

Show that a fibre’s stretch is $1$ for every $t$ iff $C = I$ .

Solution

If $C = I$ , $Ct \cdot t = 1$ . Conversely, $Ct \cdot t = 1 = It \cdot t$ for all unit $t$ gives $(C - I) u \cdot u = 0$ for all $u$ . A symmetric tensor with this property is zero: by the spectral theorem, its eigenvalues are $(C - I) a_{i} \cdot a_{i} = 0$ .