Part 3 · Theory of stress

3.3 The Cauchy–Poisson theorem and the stress tensor

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

Objectives

Theorem. Cauchy–Poisson: the stress vectors on three orthogonal faces give the one on any face, $\overset{˘}{t}_{n} = \overset{˘}{t}_{j} \overset{n}{˘}_{j}$ .
Vocabulary. Cauchy stress tensor $\overset{˘}{T}$ , normal stress, shear stress.
Picture. Cauchy’s tetrahedron; the columns of $\overset{˘}{T}$ are the stress vectors on the coordinate faces.
Recipe. The stress vector on a plane, and its normal and shear parts: $\overset{˘}{t}_{n} = \overset{˘}{T} \overset{˘}{n}$ , $σ_{n} = \overset{˘}{n} \cdot \overset{˘}{T} \overset{˘}{n}$ , $τ = ∣ \overset{˘}{t}_{n} ∣^{2} - σ_{n}^{2}$ .

At a point there are infinitely many cuts, one for each normal $\overset{˘}{n}$ , and a stress vector on each. This section shows they are not independent: three of them determine all the others, linearly. That linear map is the stress tensor.

The theorem

Write $\overset{˘}{t}_{j}$ for the stress vector on the face with normal $e_{j}$ ( $j = 1, 2, 3$ ), and $\overset{˘}{n} = \overset{n}{˘}_{1} e_{1} + \overset{n}{˘}_{2} e_{2} + \overset{n}{˘}_{3} e_{3}$ .

Theorem 3.3.1Cauchy–Poisson theoremnotes L7, pp. 3–4

The stress vector on the cut with normal $\overset{˘}{n}$ is $\overset{˘}{t}_{n} = \overset{˘}{t}_{1} \overset{n}{˘}_{1} + \overset{˘}{t}_{2} \overset{n}{˘}_{2} + \overset{˘}{t}_{3} \overset{n}{˘}_{3} = j = 1 \sum 3 \overset{˘}{t}_{j} \overset{n}{˘}_{j} .$ Also called Cauchy’s fundamental theorem.

Proof

Take a small tetrahedron at the point: three faces with outward normals $- e_{1}, - e_{2}, - e_{3}$ and areas $d \overset{˘}{A}_{1}, d \overset{˘}{A}_{2}, d \overset{˘}{A}_{3}$ , and an inclined face with outward normal $\overset{˘}{n}$ and area $d \overset{˘}{A}_{n}$ . Balance of linear momentum: $\overset{˘}{t}_{n} d \overset{˘}{A}_{n} + \overset{˘}{t}_{- e_{1}} d \overset{˘}{A}_{1} + \overset{˘}{t}_{- e_{2}} d \overset{˘}{A}_{2} + \overset{˘}{t}_{- e_{3}} d \overset{˘}{A}_{3} + \overset{˘}{b} d \overset{˘}{V} = \frac{d}{d t} (\overset{ρ}{˘} d \overset{˘}{V} \overset{˘}{v}) .$ If $h$ is the size of the tetrahedron, the areas go like $h^{2}$ and the volume like $h^{3}$ : the body force and inertia terms are infinitesimals of higher order and drop out as the tetrahedron shrinks. Divide by $d \overset{˘}{A}_{n}$ : $\overset{˘}{t}_{n} + \overset{˘}{t}_{- e_{1}} \frac{d A ˘ _{1}}{d A ˘ _{n}} + \overset{˘}{t}_{- e_{2}} \frac{d A ˘ _{2}}{d A ˘ _{n}} + \overset{˘}{t}_{- e_{3}} \frac{d A ˘ _{3}}{d A ˘ _{n}} = 0 .$ Two facts finish it:

$d \overset{˘}{A}_{j} / d \overset{˘}{A}_{n} = \overset{n}{˘}_{j}$ . Each coordinate face is the projection of the inclined face on the plane normal to $e_{j}$ , and projecting multiplies an area by the cosine of the angle between the normals, $\overset{˘}{n} \cdot e_{j} = \overset{n}{˘}_{j}$ (why: beyond the lectures).
By the lemma (Theorem 3.2.3), $\overset{˘}{t}_{- e_{j}} = - \overset{˘}{t}_{e_{j}} = - \overset{˘}{t}_{j}$ .

So $\overset{˘}{t}_{n} - \overset{˘}{t}_{j} \overset{n}{˘}_{j} = 0$ (summed over $j$ ).

Warningslips in the notes

The force labels in the tetrahedron figure on L7 p. 3 do not match its faces: one coordinate face’s force is written with $d \overset{˘}{A}_{n}$ , another with $d A^{3}$ for $d \overset{˘}{A}_{2}$ . The equation on p. 4 has the right pairing, $\overset{˘}{t}_{- e_{j}} d \overset{˘}{A}_{j}$ . The line after dividing by $d \overset{˘}{A}_{n}$ has no “ $= 0$ ”.

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Figure 3.3.1 Cauchy’s tetrahedron in a uniform stress. Drag the pink point to tilt the inclined face. Each stress vector is drawn at its face’s centre, scaled by the face’s area, so the arrows are forces: they add up to zero (readout), and the ratios of the areas are the components of

\overset{˘}{n}

. The size slider shows the volume shrinking faster than the areas.

The Cauchy stress tensor

The theorem says $\overset{˘}{t}_{n}$ depends linearly on $\overset{˘}{n}$ : it is a matrix times $\overset{˘}{n}$ .

Definition 3.3.2Cauchy stress tensorEssentialnotes L7, p. 5

The Cauchy stress tensor has the stress vectors on the coordinate faces as its columns: $\overset{˘}{T} = [\overset{˘}{t}_{1} \overset{˘}{t}_{2} \overset{˘}{t}_{3}] = \overset{σ}{˘}_{11} \overset{σ}{˘}_{21} \overset{σ}{˘}_{31} \overset{σ}{˘}_{12} \overset{σ}{˘}_{22} \overset{σ}{˘}_{32} \overset{σ}{˘}_{13} \overset{σ}{˘}_{23} \overset{σ}{˘}_{33}, \overset{˘}{t}_{n} = \overset{˘}{T} \overset{˘}{n}, \overset{˘}{t}_{j} = \overset{˘}{T}_{ij} e_{i} .$

In components, $\overset{˘}{t}_{n 1} \overset{˘}{t}_{n 2} \overset{˘}{t}_{n 3} = \overset{˘}{T}_{11} \overset{˘}{T}_{21} \overset{˘}{T}_{31} \overset{˘}{T}_{12} \overset{˘}{T}_{22} \overset{˘}{T}_{32} \overset{˘}{T}_{13} \overset{˘}{T}_{23} \overset{˘}{T}_{33} \overset{n}{˘}_{1} \overset{n}{˘}_{2} \overset{n}{˘}_{3}, \overset{˘}{t}_{ni} = j = 1 \sum 3 \overset{˘}{T}_{ij} \overset{n}{˘}_{j} .$

Key observation 3.3.3reading the indicesnotes L7, p. 5

$\overset{˘}{T}_{ij} = e_{i} \cdot \overset{˘}{t}_{j}$ is the $i$ -th component of the stress vector on the face with normal $e_{j}$ : the second index names the face, the first the direction. Lesson 7 calls $\overset{˘}{t}_{n}$ the Cauchy stress vector on $\overset{˘}{n}$ and $\overset{˘}{T}$ the Cauchy tensor of true stress (“del vero stress”).

The same structure as the deformation gradient (Key observation 2.2.7): there the columns of $F$ were the images of the basis fibres, here the columns of $\overset{˘}{T}$ are the stress vectors on the basis faces.

Remark

Use the current normal. Beside a sketch of the tetrahedron, L7 p. 7 has three red notes: $\overset{˘}{t} = \overset{n}{˘}_{j} \overset{˘}{t}_{j}$ , then “ $\overset{˘}{n} \neq = n_{j} e_{j}$ ” with the equals sign struck through, then $t_{n} = Tn$ . The middle one is unclear. A likely reading: the current normal $\overset{˘}{n}$ is not the reference normal $n = n_{j} e_{j}$ , so the theorem must be used with the components of $\overset{˘}{n}$ . The reference normal comes back with the Piola–Kirchhoff tensor in §3.5 The problem in two configurations, and the first Piola–Kirchhoff tensor.

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Figure 3.3.2 Cauchy–Poisson in the plane. The dashed red and green arrows are the stress vectors

\overset{˘}{t}_{1}

and

\overset{˘}{t}_{2}

on the coordinate faces (the columns of

\overset{˘}{T}

). Weighted by

\overset{n}{˘}_{1}

and

\overset{n}{˘}_{2}

they add up, head to tail, to the stress vector

\overset{˘}{t}_{n}

on the cut. Turn the cut.

Normal and shear components

On an element with normal $\overset{˘}{n}$ , take two unit vectors $\overset{˘}{ℓ}$ , $\overset{˘}{m}$ in its plane. The stress vector splits into:

Definition 3.3.4normal and shear stressnotes L7, p. 6

$\overset{σ}{˘}_{nn} = \overset{˘}{n} \cdot \overset{˘}{t}_{n} (normal stress), \overset{σ}{˘}_{ℓ n} = \overset{˘}{ℓ} \cdot \overset{˘}{t}_{n}, \overset{σ}{˘}_{mn} = \overset{˘}{m} \cdot \overset{˘}{t}_{n} (shear, or tangential, stresses) .$ For the coordinate faces this is how to read the matrix: the diagonal entries are normal stresses, the others shear stresses, and the notes rename them so: $\overset{˘}{T} = \overset{σ}{˘}_{11} \overset{τ}{˘}_{21} \overset{τ}{˘}_{31} \overset{τ}{˘}_{12} \overset{σ}{˘}_{22} \overset{τ}{˘}_{32} \overset{τ}{˘}_{13} \overset{τ}{˘}_{23} \overset{σ}{˘}_{33} .$

Recipe 3.3.5the stress vector on a planebeyond the lectures

Normalise the normal: $\overset{˘}{n} = v /∣ v ∣$ .
$\overset{˘}{t}_{n} = \overset{˘}{T} \overset{˘}{n}$ (row $i$ of $\overset{˘}{T}$ dotted with $\overset{˘}{n}$ gives component $i$ ).
Normal stress $σ_{n} = \overset{˘}{n} \cdot \overset{˘}{t}_{n} = \overset{˘}{n}^{T} \overset{˘}{T} \overset{˘}{n}$ ; $σ_{n} > 0$ is tension, $σ_{n} < 0$ compression.
Shear: the vector $\overset{˘}{t}_{n} - σ_{n} \overset{˘}{n}$ lies in the plane; its size is $τ = ∣ \overset{˘}{t}_{n} ∣^{2} - σ_{n}^{2}$ .

Example 3.3.6stress on an inclined planebeyond the lectures

$\overset{˘}{T} = 50300 30 - 20 0 0010$ MPa, $\overset{˘}{n} = \frac{1}{3} (1, 1, 1)$ .

$\overset{˘}{t}_{n} = \frac{1}{3} (50 + 30, 30 - 20, 10) = \frac{1}{3} (80, 10, 10) = (46.19, 5.77, 5.77)$ MPa, $∣ \overset{˘}{t}_{n} ∣ = 46.90$ MPa.
$σ_{n} = \frac{1}{3} (80 + 10 + 10) = 33.33$ MPa: tension.
$τ = 46.9 0^{2} - 33.3 3^{2} = 33.00$ MPa. The shear vector $\overset{˘}{t}_{n} - σ_{n} \overset{˘}{n} = (26.94, - 13.47, - 13.47)$ MPa lies in the plane: its components add to zero, so it is orthogonal to $(1, 1, 1)$ .

Exercises

Exercise 3.3.1

At a point $\overset{˘}{T} = 1000 - 40 000 - 40 060$ MPa. Find the stress vector, the normal stress and the shear stress on the plane with normal $\overset{˘}{n} = \frac{1}{2} (1, 0, 1)$ .

Solution

$\overset{˘}{t}_{n} = \frac{1}{2} (100 - 40, 0, - 40 + 60) = \frac{1}{2} (60, 0, 20) = (42.43, 0, 14.14)$ MPa.
$σ_{n} = \frac{1}{2} (60 + 20) = 40$ MPa (tension).
$∣ \overset{˘}{t}_{n} ∣^{2} = \frac{1}{2} (3600 + 400) = 2000$ , so $τ = 2000 - 1600 = 20$ MPa.

Exercise 3.3.2

Which stress state has, on every plane through the point, a stress vector along the plane’s normal? Show it with the theorem.

Solution

$\overset{˘}{T} = σ I$ (hydrostatic): then $\overset{˘}{t}_{n} = σ \overset{˘}{n}$ for every $\overset{˘}{n}$ , with no shear. Conversely, if every $\overset{˘}{n}$ is mapped to a multiple of itself, every vector is an eigenvector of $\overset{˘}{T}$ , which forces $\overset{˘}{T} = σ I$ .

Exercise 3.3.3

On the coordinate faces of a point, $\overset{˘}{t}_{1} = (2, 1, 0)$ , $\overset{˘}{t}_{2} = (1, - 1, 0)$ , $\overset{˘}{t}_{3} = (0, 0, 0)$ . Write $\overset{˘}{T}$ and find the stress vector on the cut with normal $(cos 30°, sin 30°, 0)$ .

Solution

$\overset{˘}{T} = [\overset{˘}{t}_{1} ∣ \overset{˘}{t}_{2} ∣ \overset{˘}{t}_{3}] = 210 1 - 1 0 000$ . Then $\overset{˘}{t}_{n} = \overset{˘}{t}_{1} cos 30° + \overset{˘}{t}_{2} sin 30° = (2 \cdot 0.866 + 0.5, 0.866 - 0.5, 0) = (2.232, 0.366, 0)$ . It is the default state of Figure 3.3.2 with $θ = 30°$ .