Part 3 · Theory of stress

3.2 Cauchy's postulates and lemma

Source: Carboni’s notes for lesson 6, p. 8, and lesson 7, Lemma, teorema fondamentale, equazioni di bilancio, pp. 1–3

Objectives

Vocabulary. Cut, stress vector (real or Cauchy stress) $\overset{˘}{t}_{n}$ , contact couple.
Theorem. Cauchy’s two postulates: the force per unit area across a cut has a limit, the couple per unit area does not survive.
Theorem. Cauchy’s lemma: $\overset{˘}{t}_{- n} = - \overset{˘}{t}_{n}$ , action and reaction across a cut.
Picture. Squash a small cylinder across the cut: only its two faces are left.

Inside a loaded body every part pulls or pushes on its neighbours. To see those forces, cut the body in thought: on each side of the cut, the force that the other side used to exert is now an external force. Cauchy’s postulates say what such a force looks like near a point.

The stress vector

Cut the current configuration $\overset{˘}{B}$ through a point $\overset{˘}{P}$ by a plane with unit normal $\overset{˘}{n}$ . On a small element $Δ \overset{˘}{A}$ of the cut, the material on the side $\overset{˘}{n}$ points to exerts on the other side a force $Δ \overset{˘}{F}$ (lesson 6, p. 8).

Definition 3.2.1stress vectornotes L6, p. 8

First postulate (axiom) of Cauchy. The contact force per unit area has a limit as the element shrinks: $\overset{˘}{t} = Δ \overset{˘}{A} \to 0 lim \frac{Δ F ˘}{Δ A ˘} .$ $\overset{˘}{t}$ is the stress vector (real stress, or Cauchy stress) at $\overset{˘}{P}$ on that cut, and $\overset{˘}{t} d \overset{˘}{A}$ the internal contact force (forza di interazione) on a small element. The limit depends on the point and on the orientation of the cut, so it is written $\overset{˘}{t}_{n}$ (or $\overset{˘}{t}_{\overset{n}{˘}}$ ). It is a vector, in general not along $\overset{˘}{n}$ .

Definition 3.2.2second postulate: no point couplesnotes L6, p. 8; L7, p. 1

Second postulate. The contact couple per unit area vanishes in the limit: $Δ \overset{˘}{A} \to 0 lim \frac{Δ M ˘}{Δ A ˘} = 0 .$ There are no local (point) couples: the interactions across a cut are forces only.

Remarkbeyond the lectures

Continua where the second postulate fails, with couple stresses, exist (some granular and microstructured materials are modelled this way); they are not in this course. §3.4 The balance equations shows that this postulate is what makes the stress tensor symmetric.

Warningslip in the notes

On L7 p. 1 the second postulate’s limit is written $\overset{˘}{A} \to 0$ ; it is $Δ \overset{˘}{A} \to 0$ , as on L6 p. 8.

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Figure 3.2.1 A cut through a point, in the plane. Turn the cut by dragging its pink normal

\overset{˘}{n}

. The stress vector

\overset{˘}{t}_{n}

acts on the grey side; its normal and shear parts are dashed. The other side of the cut feels the opposite vector (dashed teal): that is Cauchy’s lemma, below. Change the stress state in the controls.

Cauchy’s lemma

Theorem 3.2.3Cauchy’s lemmanotes L7, pp. 2–3

The stress vectors on the two faces of a cut are equal and opposite: $\overset{˘}{t}_{- n} = - \overset{˘}{t}_{n} .$ It is action and reaction for the two sides of the cut, derived from the balance of linear momentum.

Proof

In the current configuration take a small cylinder at $\overset{˘}{P}$ with axis $\overset{˘}{n}$ , radius $d \overset{r}{˘}$ and height $d \overset{˘}{h}$ : $d \overset{˘}{A} = π d \overset{r}{˘}^{2}, d \overset{˘}{V} = d \overset{˘}{A} d \overset{˘}{h} = π d \overset{r}{˘}^{2} d \overset{˘}{h} .$ The forces on it are:

top face (outward normal $\overset{˘}{n}$ ): $\overset{˘}{t}_{n} d \overset{˘}{A}$ ;
bottom face (outward normal $- \overset{˘}{n}$ ): $\overset{˘}{t}_{- n} d \overset{˘}{A}$ ;
the lateral surface: the element at angle $\overset{φ}{˘}$ has area $d \overset{r}{˘} d \overset{φ}{˘} d \overset{˘}{h}$ , so the total is $\int_{0}^{2 π} \overset{˘}{t}_{ℓ} d \overset{r}{˘} d \overset{˘}{h} d \overset{φ}{˘} = d \overset{r}{˘} d \overset{˘}{h} \int_{0}^{2 π} \overset{˘}{t}_{ℓ} d \overset{φ}{˘} = 2 π d \overset{r}{˘} d \overset{˘}{h} \overset{˘}{t}_{ℓ m},$ with $\overset{˘}{t}_{ℓ m}$ the mean of $\overset{˘}{t}_{ℓ}$ around the mantle (the notes’ subscript is hard to read; “mean” is what the step means);
the body force $\overset{˘}{b} d \overset{˘}{V}$ .

Newton’s second law (balance of linear momentum) for the cylinder, with $\overset{˘}{v}$ its velocity: $\overset{˘}{t}_{n} d \overset{˘}{A} + \overset{˘}{t}_{- n} d \overset{˘}{A} + 2 π d \overset{r}{˘} d \overset{˘}{h} \overset{˘}{t}_{ℓ m} + \overset{˘}{b} d \overset{˘}{V} = \frac{d}{d t} (\overset{ρ}{˘} d \overset{˘}{V} \overset{˘}{v}) .$ Substitute $d \overset{˘}{A}$ and $d \overset{˘}{V}$ , and let the height go to zero first, $d \overset{˘}{h} \to 0$ : $d \overset{˘}{h} \to 0 lim [π d \overset{r}{˘}^{2} (\overset{˘}{t}_{n} + \overset{˘}{t}_{- n}) + 2 π d \overset{r}{˘} d \overset{˘}{h} \overset{˘}{t}_{ℓ m} + \overset{˘}{b} π d \overset{r}{˘}^{2} d \overset{˘}{h}] = d \overset{˘}{h} \to 0 lim \frac{d}{d t} (\overset{ρ}{˘} π d \overset{r}{˘}^{2} d \overset{˘}{h} \overset{˘}{v}) .$ Every term with $d \overset{˘}{h}$ vanishes: the mantle, the body force and the inertia. What is left is $π d \overset{r}{˘}^{2} (\overset{˘}{t}_{n} + \overset{˘}{t}_{- n}) = 0 ⟹ \overset{˘}{t}_{- n} = - \overset{˘}{t}_{n} .$

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Figure 3.2.2 The cylinder of the proof, side on. Squash it: the mantle (blue) and the body force (red) scale with the height and vanish; the forces on the two faces (teal) do not, so they must balance each other.

Exercises

Exercise 3.2.1

At a point, the stress vector on the cut with normal $\overset{˘}{n} = e_{1}$ is $\overset{˘}{t}_{1} = (50, 30, 0)$ MPa. What is the stress vector on the same cut seen from the other side? Which part of each is normal and which shear?

Solution

By the lemma, $\overset{˘}{t}_{- e_{1}} = - \overset{˘}{t}_{1} = (- 50, - 30, 0)$ MPa. On the first side the normal part is $\overset{˘}{t}_{1} \cdot e_{1} = 50$ MPa (tension: it pulls the other side) and the shear part $30$ MPa along $e_{2}$ . On the other side the normal is $- e_{1}$ , and $\overset{˘}{t}_{- e_{1}} \cdot (- e_{1}) = 50$ MPa: the same tension, with the shear reversed.

Exercise 3.2.2

In the proof of the lemma, why must the height of the cylinder go to zero before its radius? What would go wrong the other way round?

Solution

With $d \overset{˘}{h} \to 0$ at fixed $d \overset{r}{˘}$ , the mantle term $2 π d \overset{r}{˘} d \overset{˘}{h}$ and the volume terms $π d \overset{r}{˘}^{2} d \overset{˘}{h}$ vanish while the faces $π d \overset{r}{˘}^{2}$ stay: the balance isolates the two faces. Shrinking the radius first at fixed height, the faces ( $\propto d \overset{r}{˘}^{2}$ ) would vanish faster than the mantle ( $\propto d \overset{r}{˘}$ ), and the balance would say something about the mantle instead.