Part 2 · Theory of deformation

2.5 Change of volume and area

Source: Carboni’s notes for lesson 4, Teoria della deformazione 3, p. 1 (results), and lesson 6, Teoria della tensione, postulati di Cauchy, pp. 1–6 (proofs)

Objectives

Theorem. A small volume changes by the factor $det F$ .
Theorem. A small oriented area changes by Nanson’s formula, $d \overset{˘}{a} = det F F^{- T} d a$ .
Vocabulary. Volume dilatation, volumetric strain, incompressible, areal dilatation.
Picture. A normal is carried by $F^{- T}$ , not by $F$ : it does not move with the material.

Lesson 4 states these results; lesson 6 proves them, at the start of the theory of stress, because forces per unit volume and per unit area have to be carried between the two configurations (§3.1 Forces and mass in two configurations). This section gives each result with its proof.

The proofs use the permutation symbol $ϵ_{ijk}$ (Levi-Civita, or Ricci, symbol): a number for each triple of indices, $ϵ_{ijk} = ⎩ ⎨ ⎧ + 1 - 1 0 (ijk) an even permutation of (123) : 123, 231, 312 (ijk) an odd permutation : 132, 213, 321 an index repeats.$

Change of volume

Definition 2.5.1volume dilatationnotes L4, p. 1

The volume dilatation (dilatazione volumetrica) at a point is the ratio of a small deformed volume to the original one: $Λ = Δ V \to 0 lim \frac{Δ V ˘}{Δ V} = \frac{d V ˘}{d V} .$

Theorem 2.5.2change of volumenotes L6, pp. 1–4

$Λ = \frac{d V ˘}{d V} = det F .$

The proof writes a volume as a triple product, $det$ of three vectors, in index form. Here first with the sums written out, then with the summation convention.

Proof

A small parallelepiped. Take three small fibres $d X^{(1)}, d X^{(2)}, d X^{(3)}$ at $P$ (the superscript in brackets numbers the fibre; the subscript will be the component). Two of them span a face of area $d A$ and unit normal $m$ , and the volume is the face times the height along the third: $d A m = d X^{(1)} \times d X^{(2)}, d V = d A m \cdot d X^{(3)} = (d X^{(1)} \times d X^{(2)}) \cdot d X^{(3)} .$
The cross product in components. $d X^{(1)} \times d X^{(2)} = e_{1} d X_{1}^{(1)} d X_{1}^{(2)} e_{2} d X_{2}^{(1)} d X_{2}^{(2)} e_{3} d X_{3}^{(1)} d X_{3}^{(2)} = i = 1 \sum 3 j = 1 \sum 3 k = 1 \sum 3 ϵ_{ijk} e_{i} d X_{j}^{(1)} d X_{k}^{(2)},$ which, written out, is $e_{1} (d X_{2}^{(1)} d X_{3}^{(2)} - d X_{3}^{(1)} d X_{2}^{(2)}) + e_{2} (d X_{3}^{(1)} d X_{1}^{(2)} - d X_{1}^{(1)} d X_{3}^{(2)}) + e_{3} (d X_{1}^{(1)} d X_{2}^{(2)} - d X_{2}^{(1)} d X_{1}^{(2)}) .$ From here on, a repeated index is summed from 1 to 3 (Einstein’s convention), and the $\sum$ signs are dropped: $ϵ_{ijk} e_{i} d X_{j}^{(1)} d X_{k}^{(2)}$ means the triple sum above. Dotting with $d X^{(3)}$ picks the component $i$ : $d V = ϵ_{ijk} d X_{i}^{(3)} d X_{j}^{(1)} d X_{k}^{(2)} = ϵ_{L MN} d X_{M}^{(1)} d X_{N}^{(2)} d X_{L}^{(3)},$ the last form only renaming the summed indices.
The deformed parallelepiped. The deformed fibres are $d p^{(n)} = F d X^{(n)}$ , that is $d p_{j}^{(1)} = F_{j M} d X_{M}^{(1)}$ , $d p_{k}^{(2)} = F_{k N} d X_{N}^{(2)}$ , $d p_{i}^{(3)} = F_{i L} d X_{L}^{(3)}$ . The same formula gives their volume: $d \overset{˘}{V} = ϵ_{ijk} d p_{i}^{(3)} d p_{j}^{(1)} d p_{k}^{(2)} = ϵ_{ijk} F_{i L} F_{j M} F_{k N} d X_{M}^{(1)} d X_{N}^{(2)} d X_{L}^{(3)} .$
The key identity (boxed on L6 p. 4), the step the proof turns on: $ϵ_{ijk} F_{i L} F_{j M} F_{k N} = det F ϵ_{L MN} .$ For $(L, M, N) = (1, 2, 3)$ it reads $ϵ_{ijk} F_{i 1} F_{j 2} F_{k 3} = det F$ : the left side is the expansion of the determinant over permutations, one product of entries from different rows for each permutation, with its sign. Swapping two of $L, M, N$ swaps two columns of $F$ , which flips the sign of the determinant, as it flips $ϵ_{L MN}$ ; a repeated index gives two equal columns and zero, as $ϵ_{L MN}$ is (why: beyond the lectures).
Substitute: $d \overset{˘}{V} = det F ϵ_{L MN} d X_{M}^{(1)} d X_{N}^{(2)} d X_{L}^{(3)} = det F d V .$

(L6 p. 4 relabels the summed indices cyclically before step 5, pairing $L$ with $d X^{(1)}$ . A triple product is unchanged by a cyclic permutation of its factors, so this is the same $d V$ .)

Warningslips in the notes

In lesson 6’s index computations: on p. 2 the $e_{1}$ term of the cross product starts $d X_{1}^{(1)} d X_{3}^{(2)}$ (it is $d X_{2}^{(1)} d X_{3}^{(2)}$ ; p. 3 has it right, with the 2 overwritten in blue); on pp. 3–5 the third factor of $F$ is written $F_{km}$ next to $d X_{n}^{(2)}$ . It must be $F_{kn}$ , with the same index as the fibre component it multiplies. Lesson 6 also writes the reference fibres $d x^{(n)}$ ; here they are $d X^{(n)}$ , as in lesson 2.

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Figure 2.5.1 A unit cube (grey) and the parallelepiped on the columns

ν_{1}, ν_{2}, ν_{3}

F

. Its volume is

det F = ν_{1} ν_{2} ν_{3}

; a rotation does not change it. Make the product of the three stretches equal to 1 (for example 2, 1, 0.5) for a deformation that keeps volume.

Definition 2.5.3volumetric strain, incompressiblenotes L4, p. 1

The volumetric strain (deformazione volumetrica) is the relative change of volume: $ε_{V} = \frac{d V ˘ - d V}{d V} = Λ - 1 = det F - 1.$ A deformation with $Λ = det F = 1$ , that is $ε_{V} = 0$ , keeps volume; a body for which this holds at every point, for every deformation it can undergo, is incompressible.

Remarkbeyond the lectures

Since $det F = det R det U = det U = ν_{1} ν_{2} ν_{3}$ (Theorem 2.4.1), the volume ratio is the product of the principal stretches: the ellipsoid of semi-axes $ν_{k}$ has $ν_{1} ν_{2} ν_{3}$ times the volume of the unit sphere. And $det C = (det F)^{2}$ .

Change of area

A small surface element of area $d A$ and unit normal $m$ is the vector $d a = d A m$ . After the deformation it is $d \overset{˘}{a} = d \overset{˘}{A} \overset{˘}{m}$ .

Definition 2.5.4areal dilatationnotes L4, p. 1

The areal dilatation (dilatazione areale o superficiale) of an element with normal $m$ is $Λ_{m} = Δ A \to 0 lim \frac{∣Δ a ˘ ∣}{∣Δ a ∣} = \frac{d A ˘}{d A} .$ It depends on the orientation $m$ of the element, as a stretch depends on the direction of a fibre.

Theorem 2.5.5Nanson’s formulanotes L6, pp. 5–6

$d \overset{˘}{a} = det F (F^{- 1})^{T} d a, Λ_{m} = det F m^{T} (F^{- 1} F^{- T}) m .$ The vector $λ_{m} = d \overset{˘}{a} /∣ d a ∣ = Λ_{m} \overset{˘}{m} = det F F^{- T} m$ is the areal stretch vector (vettore di stretch areolare).

( $F^{- T}$ is short for $(F^{- 1})^{T} = (F^{T})^{- 1}$ . The name “Nanson’s formula” is not in the notes.)

Proof

Let the element be spanned by two fibres: $d a = d X^{(1)} \times d X^{(2)}$ and $d \overset{˘}{a} = d p^{(1)} \times d p^{(2)}$ .
In components, with $d p_{j}^{(1)} = F_{j M} d X_{M}^{(1)}$ and $d p_{k}^{(2)} = F_{k N} d X_{N}^{(2)}$ : $(d \overset{˘}{a})_{i} = ϵ_{ijk} d p_{j}^{(1)} d p_{k}^{(2)} = ϵ_{ijk} F_{j M} F_{k N} d X_{M}^{(1)} d X_{N}^{(2)} .$
The identity used (boxed on L6 p. 5) is $ϵ_{r jk} F_{j M} F_{k N} = det F (F^{- 1})_{L r} ϵ_{L MN} .$ To get it (beyond the lectures), multiply the volume identity $ϵ_{ijk} F_{i L} F_{j M} F_{k N} = det F ϵ_{L MN}$ by $(F^{- 1})_{L r}$ and sum over $L$ . On the left, $F_{i L} (F^{- 1})_{L r} = (F F^{- 1})_{i r} = δ_{i r}$ (1 if $i = r$ , 0 otherwise), which turns $ϵ_{ijk}$ into $ϵ_{r jk}$ .
Substitute, and recognise $ϵ_{L MN} d X_{M}^{(1)} d X_{N}^{(2)} = (d a)_{L}$ from step 2 of the volume proof: $(d \overset{˘}{a})_{r} = det F (F^{- 1})_{L r} (d a)_{L} = det F [(F^{- 1})^{T} d a]_{r} .$
Divide by $∣ d a ∣ = d A$ : $λ_{m} = det F F^{- T} m$ .
Its length is $Λ_{m}$ : $Λ_{m} = ∣ λ_{m} ∣ = det F (F^{- T} m)^{T} (F^{- T} m) = det F m^{T} F^{- 1} F^{- T} m .$

Key observation 2.5.6normals do not move with the materialbeyond the lectures

The new normal $\overset{˘}{m}$ is along $F^{- T} m$ , in general not along $Fm$ . A fibre that was perpendicular to the surface does not stay perpendicular to it: under a shear the surface tilts one way and the fibre another. This is why the stress section needs Nanson’s formula.

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Figure 2.5.2 Nanson’s formula in the plane, where an “area” element is a short segment (violet) with normal

m

. Turn the normal on the left. On the right the segment is deformed with the material, and its new normal is along

λ_{m} = det F F^{- T} m

, of length

Λ_{m}

, equal to the deformed segment’s length. The dashed

Fm

is where the material fibre along the old normal went: not normal any more.

Exercises

Exercise 2.5.1

A bar is pulled along $e_{1}$ and narrows sideways: $F = diag (2, 0.8, 0.8)$ . Find $Λ$ , and $Λ_{m}$ for elements with normal $e_{1}$ , $e_{2}$ and $\frac{1}{2} (1, 1, 0)$ . Is the deformed normal of the last one still along $(1, 1, 0)$ ?

Solution

$Λ = det F = 1.28$ , and $F^{- T} = diag (\frac{1}{2}, \frac{5}{4}, \frac{5}{4})$ .
Normal $e_{1}$ (the cross-section): $λ = 1.28 (\frac{1}{2}, 0, 0)$ , so $Λ_{e_{1}} = 0.64 = 0.8 \cdot 0.8$ . The section shrinks with the two lateral stretches.
Normal $e_{2}$ : $Λ_{e_{2}} = 1.28 \cdot \frac{5}{4} = 1.6 = 2 \cdot 0.8$ .
Normal $\frac{1}{2} (1, 1, 0)$ : $λ = 1.28 \cdot \frac{1}{2} (\frac{1}{2}, \frac{5}{4}, 0) = (0.4525, 1.1314, 0)$ , so $Λ_{m} = 1.2185$ .
The new normal is along $(0.4525, 1.1314, 0)$ : not along $(1, 1, 0)$ , and not along $Fm \propto (2, 0.8, 0)$ either.

Exercise 2.5.2

For simple shear $F = 100 k 10 001$ , show that the volume is kept, and find the areal dilatation and the new normal of the face whose normal was $e_{1}$ .

Solution

$det F = 1$ .
$F^{- 1} = 100 - k 10 001$ , $F^{- T} = 1 - k 0 010001$ .
$λ_{e_{1}} = F^{- T} e_{1} = (1, - k, 0)$ , so $Λ = 1 + k^{2}$ and $\overset{˘}{m} = (1, - k, 0) / 1 + k^{2}$ : the vertical face has tilted and grown, while the horizontal face ( $m = e_{2}$ , $λ = (0, 1, 0)$ ) keeps its area and normal.

Exercise 2.5.3

At the point $A$ of the exercise session $det F = 18$ . What is the volumetric strain? Which is the deformed volume of a cube of side 1 mm there?

Solution

$ε_{V} = det F - 1 = 17$ : the volume grows by 1700%. A 1 mm³ cube becomes a parallelepiped of 18 mm³ (to first order, for a cube small enough that $F$ is uniform over it).