Part 3 · Theory of stress

3.1 Forces and mass in two configurations

Source: Carboni’s notes for lesson 6, Teoria della tensione, postulati di Cauchy, pp. 1 and 6–7

Objectives

Vocabulary. Body forces, surface forces, internal contact forces; nominal and real forces; nominal and real density.
Recipe. Carry a force density between the configurations: $b = det F \overset{˘}{b}$ , $f = Λ_{m} \overset{˘}{f}$ , $ρ = det F \overset{ρ}{˘}$ .
Theorem. Conservation of mass.

The theory of deformation measured shape. The theory of stress measures forces, and the first question is: per unit of what? A body has two descriptions, the reference configuration $B$ and the current one $\overset{˘}{B}$ (Definition 2.1.1), and a force spread over a volume or a surface can be divided by the volume or area it has in either. This is why lesson 6 proved the change of volume and area (§2.5 Change of volume and area) first.

External and internal forces

A body in the current configuration $\overset{˘}{B}$ is loaded by:

Definition 3.1.1body, surface and contact forcesnotes L6, p. 1

Body forces (forze di volume) act on every small volume: $\overset{˘}{b} d \overset{˘}{V}$ in the current configuration, $b d V$ in the reference one. Gravity is the usual example. Units: N/m³.
Surface forces (forze di superficie) act on the part $\partial B_{f}$ of the boundary where forces are prescribed: $\overset{˘}{f} d \overset{˘}{A}$ , or $f d A$ . Units: N/m².
Internal contact forces: cut the body by a surface through a point; the two sides push and pull on each other across it, with a force $\overset{˘}{t} d \overset{˘}{A}$ on each small element of the cut. These are the subject of the rest of this part.

Lesson 6 also draws a couple $\overset{˘}{M}$ acting on the cut; Cauchy’s second postulate (§3.2 Cauchy's postulates and lemma) says it vanishes per unit area.

Nominal and real forces

A force is the same force whichever configuration it is measured in. Only what it is divided by changes. So the force on a small volume, and on a small surface element, must come out equal in both descriptions (L6 p. 6 calls this the “conservation” of the forces).

Proposition 3.1.2nominal and real forcesnotes L6, p. 6

$b d V = \overset{˘}{b} d \overset{˘}{V} ⟹ b = \frac{d V ˘}{d V} \overset{˘}{b} = det F \overset{˘}{b},$ $f d A = \overset{˘}{f} d \overset{˘}{A} ⟹ f = \frac{d A ˘}{d A} \overset{˘}{f} = Λ_{m} \overset{˘}{f},$ with $Λ_{m} = det F m^{T} F^{- 1} F^{- T} m$ the areal dilatation of the element, whose reference normal is $m$ (Theorem 2.5.5).

$\overset{˘}{b}$ , $\overset{˘}{f}$ are the real forces: per unit of current volume or area, the force the material actually feels.
$b$ , $f$ are the nominal forces: the same force per unit of reference volume or area.

(Lesson 6 writes the areal dilatation as $λ$ ; the theory of deformation called it $Λ_{m}$ , and this book keeps one name.) Each is the limit of a force over the volume or area it acts on (L6 p. 7): $b (X) = Δ V \to 0 lim \frac{Δ F _{b}}{Δ V}, f (X) = Δ A \to 0 lim \frac{Δ F _{f}}{Δ A},$ with $Δ F_{b}$ , $Δ F_{f}$ the resultant force on the small volume or area: a force here, not the deformation gradient (the notes write the numerators as $Δ b$ and $Δ f$ ).

Example 3.1.3a compressed part under gravitybeyond the lectures

A steel part of density $ρ = 7800$ kg/m³ is compressed to $det F = 0.98$ , and $g = 9.81$ m/s².

The nominal body force of gravity is $ρ g = 76.52$ kN/m³: the weight of a piece of material divided by its reference volume. It does not change with the deformation.
The real one is $\overset{˘}{b} = b / det F = 78.08$ kN/m³: the same weight in a smaller volume.

Conservation of mass

Definition 3.1.4nominal and real densitynotes L6, p. 7

The nominal density is mass per unit reference volume, the real density mass per unit current volume: $ρ (X) = Δ V \to 0 lim \frac{Δ m ( X )}{Δ V}, \overset{ρ}{˘} (p (X, t)) = Δ \overset{˘}{V} \to 0 lim \frac{Δ m ˘ ( p ( X , t ) )}{Δ V ˘} .$

Theorem 3.1.5conservation of massnotes L6, p. 7

A small piece of the body keeps its mass while it deforms: $ρ d V = \overset{ρ}{˘} d \overset{˘}{V} ⟹ ρ = \frac{d V ˘}{d V} \overset{ρ}{˘} = det F \overset{ρ}{˘} .$

Mass is measured in kg; force is mass times acceleration, $1$ N $= 1$ kg·m/s².

Warningslip in the notes

The real density’s limit is written $Δ V \to 0$ . It is $Δ \overset{˘}{V} \to 0$ , the volume in the denominator.

quantity	nominal (reference)	real (current)	relation
body force	$b$ , per $d V$	$\overset{˘}{b}$ , per $d \overset{˘}{V}$	$b = det F \overset{˘}{b}$
surface force	$f$ , per $d A$	$\overset{˘}{f}$ , per $d \overset{˘}{A}$	$f = Λ_{m} \overset{˘}{f}$
density	$ρ$ , per $d V$	$\overset{ρ}{˘}$ , per $d \overset{˘}{V}$	$ρ = det F \overset{ρ}{˘}$

Exercises

Exercise 3.1.1

A rubber block of density $ρ = 1100$ kg/m³ is stretched to $F = diag (2, 0.8, 0.8)$ . Find the real density and the real body force of gravity.

Solution

$det F = 1.28$ , so $\overset{ρ}{˘} = 1100/1.28 = 859.4$ kg/m³ and $\overset{˘}{b} = \overset{ρ}{˘} g = 8.43$ kN/m³, against the nominal $ρ g = 10.79$ kN/m³.

Exercise 3.1.2

In the same block, a pressure of $p = 10$ kPa (real, per unit current area) acts on the face whose reference normal is $e_{1}$ . What is the nominal surface force, and the total force on a face of reference area 4 cm²?

Solution

The face’s areal dilatation is $Λ_{e_{1}} = 0.8 \cdot 0.8 = 0.64$ (Exercise 2.5.1).
Nominal: $f = Λ \overset{˘}{f} = 6.4$ kPa.
Force: $6.4 kPa \cdot 4 cm^{2} = 2.56$ N, the same as $10 kPa \cdot (0.64 \cdot 4) cm^{2}$ .