Part 3 · Kinematics

3.4 Change in area and volume

Source: Favata’s notes, Kinematics §1.4, p. 9–11 (cofactor (1.5)–(1.6), Nanson’s formula (1.7), change in volume, Exercises 1.2, 1.3)

Objectives

Compute how a small oriented area and a small volume change.
Understand why normals do not follow $F$ , and what they follow instead.
Recipe: Nanson’s formula $\frac{d a}{d a _{0}} n_{cur} = F^{*} n_{ref}$ ; $d v / d v_{0} = det F$ .
Picture: a face seen edge-on, its material normal fibre, and its actual new normal.
Theorem: $Av \times Aw = A^{*} (v \times w)$ .
Vocabulary: cofactor, Nanson’s formula, isochoric.

How a cross product transforms

A small parallelogram on the infinitesimal sides $a$ , $b$ at $x_{0}$ has $n_{ref} = \frac{a \times b}{∣ a \times b ∣}, d a_{0} = ∣ a \times b ∣.$ After the deformation its sides are $Fa$ , $Fb$ , so $d a = ∣ Fa \times Fb ∣$ and $n_{cur} = Fa \times Fb / d a$ . The question is how $F$ acts on a cross product.

Theorem 3.4.1Cofactor

For invertible $A$ , $Av \times Aw = A^{*} (v \times w) \forall v, w, A^{*} := (det A) A^{- ⊤} . (1.6)$ $A^{*}$ is the cofactor of $A$ : the tensor that maps cross products, hence oriented areas.

Proof

The determinant multiplied out, $Au \cdot (Av \times Aw) = (det A) u \cdot (v \times w)$ , holds for all $u, v, w$ : if they are dependent, so are their images, and both sides vanish. Set $\overset{ˉ}{u} := Au$ , so $u = A^{- 1} \overset{ˉ}{u}$ : $\overset{ˉ}{u} \cdot (Av \times Aw) = (det A) A^{- 1} \overset{ˉ}{u} \cdot (v \times w) = \overset{ˉ}{u} \cdot (det A) A^{- ⊤} (v \times w),$ moving $A^{- 1}$ across the scalar product. Two vectors with the same scalar product against every $\overset{ˉ}{u}$ are equal.

The name comes from the matrix: the entries of $[A^{*}]$ are the cofactors, signed $2 \times 2$ minors, of $[A]$ (beyond the notes). With Exercise 2.6.3 it explains why the cross product follows rotations: for $R \in Rot$ , $R^{*} = R^{- ⊤} = R$ .

Nanson’s formula

Apply (1.6) with $A = F$ ; since $a \times b = d a_{0} n_{ref}$ , $Fa \times Fb = d a_{0} F^{*} n_{ref} .$ Its length is $d a$ , its direction $n_{cur}$ :

Theorem 3.4.2Nanson’s formula

$\frac{d a}{d a _{0}} n_{cur} = F^{*} n_{ref}, \frac{d a}{d a _{0}} = ∣ F^{*} n_{ref} ∣, n_{cur} = \frac{F ^{*} n _{ref}}{∣ F ^{*} n _{ref} ∣} . (1.7)$

One vector, $F^{*} n_{ref}$ , carries both: its direction is the new normal, its length the area ratio.

Key observation 3.4.3Normals go with F⁻ᵀ, not with F

A normal is whatever is orthogonal to every tangent $t$ of the surface. Tangents are material fibres, so they become $Ft$ , and $F^{- ⊤} n_{ref} \cdot Ft = n_{ref} \cdot F^{- 1} Ft = n_{ref} \cdot t = 0.$ $F^{- ⊤} n_{ref}$ is orthogonal to all the deformed tangents. The material fibre that was normal to the face, $F n_{ref}$ , generally is not.

Loading the demo…

Figure 3.4.1 A face seen edge-on: the grey line in the reference, the teal line after the deformation. Drag its normal

n_{ref}

around. The dashed arrow is

F n_{ref}

, the fibre that used to be normal; the solid one is

F^{*} n_{ref}

, which stays at right angles to the face and whose length is the area ratio. Under the shear, turn

n_{ref}

e_{2}

: the faces slide on themselves, nothing changes. Turn it to

e_{1}

: the two arrows part.

Warning

The notes print $n_{cur}$ with $Fa \times Fa$ on p. 10; it is $Fa \times Fb$ . And $F^{*}$ is defined only for invertible $F$ , which (1.1) guarantees.

Nanson’s formula is what later moves tractions and stresses between the configurations (the Piola–Kirchhoff stresses).

Change in volume

Four points $x_{0}$ , $x_{0} + a$ , $x_{0} + b$ , $x_{0} + c$ span a small parallelepiped of volume $d v_{0} = ∣ a \cdot b \times c ∣$ ; after the deformation, $d v = ∣ Fa \cdot Fb \times Fc ∣$ .

Theorem 3.4.4Change in volume

$\frac{d v}{d v _{0}} = \frac{∣ Fa \cdot Fb \times Fc ∣}{∣ a \cdot b \times c ∣} = det F .$

Without the absolute values the ratio is $det F$ by definition; with them it is $∣ det F ∣ = det F$ by (1.1).

Definition 3.4.5Isochoricbeyond the notes

A deformation with $det F = 1$ everywhere is isochoric: it keeps volumes. It is the kinematics of incompressible materials.

beyond the notes

Most books write $J := det F$ . The volume of a deformed part $P$ is the sum of the small boxes, $vol f (P) = \int_{P} det F d v_{0}$ .

Loading the demo…

Figure 3.4.2 A small cube of the body, and its image under

F

, drawn as a 3D map of the columns

F e_{j}

. It starts as the simple shear, which leans the cube without changing its volume:

det F = 1

. Drag the columns: the volume ratio is the readout’s

det

, and pushing a column through the plane of the other two is what (1.1) forbids.

Everything from C

Since $det C = det F^{⊤} det F = (det F)^{2}$ , volumes follow from $C$ ; so do areas (Exercise 3.4.1). Lengths, angles, areas and volumes all follow from $C$ .

Exercises

Exercise 3.4.1notes p. 10 (Exercise 1.2)

Prove that the change in area can be expressed through $C$ .

Solution

$∣ F^{*} n_{ref} ∣^{2} = (det F)^{2} F^{- ⊤} n_{ref} \cdot F^{- ⊤} n_{ref} = (det F)^{2} F^{- 1} F^{- ⊤} n_{ref} \cdot n_{ref}$ , and $F^{- 1} F^{- ⊤} = (F^{⊤} F)^{- 1} = C^{- 1}$ , $(det F)^{2} = det C$ : $\frac{d a}{d a _{0}} = det C C^{- 1} n_{ref} \cdot n_{ref} .$

Exercise 3.4.2notes p. 11 (Exercise 1.3)

Prove that the change in volume can be expressed through $C$ .

Solution

$det C = (det F)^{2}$ and $det F > 0$ , so $d v / d v_{0} = det C$ .

Exercise 3.4.3beyond the notespractice (K.1)

For the simple shear $F = I + γ e_{1} \otimes e_{2}$ , find $det F$ , and the area change and new normal of the faces with $n_{ref} = e_{2}$ and $n_{ref} = e_{1}$ . (Lengths and angles: Exercise 3.3.1.)

Solution

$det F = 1$ : isochoric. $F^{- 1} = I - γ e_{1} \otimes e_{2}$ (Exercise 2.5.2), so $F^{*} = F^{- ⊤} = I - γ e_{2} \otimes e_{1}$ .

$n_{ref} = e_{2}$ : $F^{*} e_{2} = e_{2}$ . Area and normal unchanged: the horizontal planes slide on themselves.
$n_{ref} = e_{1}$ : $F^{*} e_{1} = e_{1} - γ e_{2}$ , so $d a / d a_{0} = 1 + γ^{2}$ and $n_{cur} = (e_{1} - γ e_{2}) / 1 + γ^{2}$ .
Check directly: the face’s sides $e_{2}$ , $e_{3}$ become $γ e_{1} + e_{2}$ and $e_{3}$ , and $(γ e_{1} + e_{2}) \times e_{3} = e_{1} - γ e_{2}$ .
$F e_{1} = e_{1}$ is not the new normal.