Part 3 · Theory of stress

3.7 Principal stresses and the linear elastic law

Source: Not in lessons 6–8. Written for the exercise sheets for the first partial (posted 13 Oct 2026, “Teoria della deformazione, della tensione e legame costitutivo”), which ask for all of it

Objectives

Recipe. The linearised balance equations, and the inertia force they leave.
Theorem. A symmetric stress has three real principal stresses on three orthogonal directions, where there is no shear.
Recipe. Principal stresses and directions, and the type of stress state.
Recipe. Stress from strain with the isotropic linear elastic law, $S = λ (tr E) I + 2 μ E$ .

Warningbeyond the lectures, but on the exam

Lessons 6–8 stop at the second Piola–Kirchhoff tensor. The practice sheets for the first partial also ask for the linearised equations, principal stresses and stress from strain with a linear elastic law (the syllabus’s block 4, the constitutive law, which the lessons have not reached). Everything in this section is written for those sheets, from standard continuum mechanics, and is marked “beyond the lectures”. Carboni’s own treatment, when it comes, takes precedence.

The linearised equations

Proposition 3.7.1small displacement gradientbeyond the lectures

If the displacement gradient is small, $F = I + H \approx I$ and $det F \approx 1$ (§2.7 Infinitesimal strain). Then to first order $\overset{˘}{T} \approx T \approx S, \overset{ρ}{˘} \approx ρ, \overset{˘}{b} \approx b,$ and the difference between reference and current coordinates disappears. All three systems become $\frac{\partial S _{ij}}{\partial X _{j}} + b_{i} = ρ a_{i}, S = S^{T}, S m = g on \partial B_{f} .$

Proof

$S = det F F^{- 1} \overset{˘}{T} F^{- T}$ with $det F = 1 + O (ε)$ and $F^{- 1} = I - H + O (ε^{2})$ , so $S = \overset{˘}{T} + O (ε) \overset{˘}{T}$ : the three tensors differ by terms of the order of the strain times the stress, negligible against the stress itself. The same for the densities and body forces.

The sheets call $ρ a = \nabla \cdot S + b$ the inertia force per unit volume needed for the balance; in d’Alembert’s form (Definition 1.1.5) the inertia force is $- ρ a$ and the balance reads $\nabla \cdot S + b - ρ a = 0$ .

Principal stresses

The same spectral theorem that gave the principal stretches of the symmetric $C$ (Key observation 2.3.1) applies to any symmetric stress tensor, $\overset{˘}{T}$ or $S$ .

Theorem 3.7.2principal stressesbeyond the lectures

A symmetric stress tensor $\overset{˘}{T}$ has three real eigenvalues $σ_{1}, σ_{2}, σ_{3}$ , the principal stresses, and three mutually orthogonal unit eigenvectors $n_{1}, n_{2}, n_{3}$ , the principal directions of stress: $\overset{˘}{T} n_{k} = σ_{k} n_{k}, det (\overset{˘}{T} - σ I) = 0 ⟺ σ^{3} - I_{1} σ^{2} + I_{2} σ - I_{3} = 0,$ with the invariants $I_{1} = tr \overset{˘}{T}$ , $I_{2} = \frac{1}{2} [(tr \overset{˘}{T})^{2} - tr (\overset{˘}{T}^{2})]$ , $I_{3} = det \overset{˘}{T}$ .

On a face normal to $n_{k}$ the stress vector is $\overset{˘}{T} n_{k} = σ_{k} n_{k}$ : purely normal, no shear.
In the principal basis $\overset{˘}{T} = diag (σ_{1}, σ_{2}, σ_{3})$ .
The largest and smallest principal stresses are the largest and smallest normal stresses over all orientations.

Proof

The first three points are the spectral theorem for a real symmetric matrix and the definition of eigenvector. For the last: write a unit normal $\overset{˘}{n} = \sum a_{k} n_{k}$ with $\sum a_{k}^{2} = 1$ ; then $σ_{n} = \overset{˘}{n} \cdot \overset{˘}{T} \overset{˘}{n} = \sum_{k} σ_{k} a_{k}^{2}$ , a weighted average of the principal stresses, as for the stretches in Theorem 2.3.2.

Definition 3.7.3triaxial, cylindrical and spherical stressbeyond the lectures

Named by the multiplicity of the principal stresses, as for strain (Definition 2.3.5):

three distinct $σ_{k}$ : triaxial;
two equal: cylindrical (axisymmetric); every direction normal to the single one is principal;
three equal: spherical (hydrostatic), $\overset{˘}{T} = σ I$ ; every direction is principal and no face carries shear.

Recipe 3.7.4principal stresses and directionsbeyond the lectures

Look for a split. If row and column 3 are zero off the diagonal, $e_{3}$ is principal with $σ = T_{33}$ , and the other two come from the $2 \times 2$ block: $σ_{I, II} = \frac{T _{11} + T _{22}}{2} \pm (\frac{T _{11} - T _{22}}{2})^{2} + T_{12}^{2}, tan 2 θ = \frac{2 T _{12}}{T _{11} - T _{22}},$ with $θ$ the angle of a principal direction from $e_{1}$ in the $1$ – $2$ plane. (Any other zero row and column works the same way.)
Otherwise solve the cubic $σ^{3} - I_{1} σ^{2} + I_{2} σ - I_{3} = 0$ .
Directions. For each $σ_{k}$ solve $(\overset{˘}{T} - σ_{k} I) n = 0$ and normalise.
Check $\sum σ_{k} = I_{1}$ and $\prod σ_{k} = I_{3}$ . Name the state.

Loading the demo…

Figure 3.7.1 Principal stresses in the plane. Turn the cut: on the dashed orange directions the stress vector is purely normal (the cut snaps to them). On the right, each cut is a point

(σ_{n}, τ)

; as the normal turns by

θ

the point runs round Mohr’s circle by

2 θ

, and the principal stresses are where it crosses the axis. Mohr’s circle is not in the course: it is here as a picture of the formula in step 1 of the recipe (centre

\frac{T _{11} + T _{22}}{2}

, radius the square root).

Example 3.7.5a stress state with one known directionbeyond the lectures

$\overset{˘}{T} = 50300 30 - 20 0 0010$ MPa.

$e_{3}$ is principal: $σ = 10$ MPa.
Block: $\frac{50 - 20}{2} \pm 3 5^{2} + 3 0^{2} = 15 \pm 46.10$ : $σ_{I} = 61.10$ MPa, $σ_{III} = - 31.10$ MPa.
$tan 2 θ = \frac{60}{70}$ , $θ = 20.30°$ : $n_{I} = (0.9379, 0.3469, 0)$ , $n_{III} = (- 0.3469, 0.9379, 0)$ .
Check: $I_{1} = 40 = 61.10 + 10 - 31.10$ ; $I_{3} = det \overset{˘}{T} = 10 \cdot (- 1000 - 900) = - 19000 = 61.10 \cdot 10 \cdot (- 31.10)$ . Triaxial.

The linear elastic law

Strain and stress have been built separately; a constitutive law links them, as $F_{e l} = K U$ did for the spring (Key observation 1.1.3).

Definition 3.7.6isotropic linear elastic lawbeyond the lectures

A homogeneous, isotropic, linear elastic material has $S = λ (tr E) I + 2 μ E, S_{ij} = λ E_{kk} δ_{ij} + 2 μ E_{ij},$ with $λ$ , $μ$ the Lamé constants ( $μ$ is the shear modulus) and $δ_{ij}$ 1 if $i = j$ , 0 otherwise. In terms of Young’s modulus $E$ and Poisson’s ratio $ν$ : $λ = \frac{E ν}{( 1 + ν ) ( 1 - 2 ν )}, μ = \frac{E}{2 ( 1 + ν )} .$

In the linear theory $E$ is $E^{(1)}$ and $S \approx \overset{˘}{T}$ : this is Hooke’s law.
With finite strains the sheets keep the same law between $S$ and the Green–Lagrange $E$ (a Saint Venant–Kirchhoff material), and then go to $T = FS$ and $\overset{˘}{T} = \frac{1}{d e t F} T F^{T}$ .
The two are paired because both are symmetric and refer to the reference configuration (Key observation 3.6.4).

Remarkbeyond the lectures

Reading the law. The spherical part of the strain gives a spherical stress: $tr S = (3 λ + 2 μ) tr E$ , so a pure volume change is resisted by the bulk modulus $λ + \frac{2}{3} μ$ . The deviatoric part gives a deviatoric stress, $2 μ$ times it: a pure change of shape is resisted by $μ$ . In particular $S_{12} = 2 μ E_{12} = μ γ_{12}$ in the linear theory: shear stress is the shear modulus times the shear angle.

Recipe 3.7.7stress from strainbeyond the lectures

$E$ at the point (finite: $\frac{1}{2} (C - I)$ ; linear: $E^{(1)}$ , as the problem says).
$λ$ , $μ$ from $E$ , $ν$ if needed.
$S = λ (tr E) I + 2 μ E$ : add $λ tr E$ to each diagonal entry of $2 μ E$ .
If asked: $T = FS$ , $\overset{˘}{T} = \frac{1}{d e t F} T F^{T}$ .

Exercises

Exercise 3.7.1

Steel has $E = 210$ GPa and $ν = 0.3$ . Find $λ$ and $μ$ . A small strain $E^{(1)} = 1 0^{- 3} diag (1, - 0.3, - 0.3)$ is applied: find $S$ . What kind of test is it?

Solution

$λ = \frac{210 \cdot 0.3}{1.3 \cdot 0.4} = 121.2$ GPa, $μ = \frac{210}{2.6} = 80.77$ GPa.
$tr E^{(1)} = 0.4 \times 1 0^{- 3}$ , so $λ tr E^{(1)} = 48.46$ MPa.
$S_{11} = 48.46 + 2 \cdot 80.77 \cdot 1 = 210.0$ MPa; $S_{22} = S_{33} = 48.46 - 2 \cdot 80.77 \cdot 0.3 = 0.0$ MPa.
A uniaxial tension test: the lateral contraction $- ν ε$ is exactly what makes the lateral stresses vanish, and $S_{11} = E ε_{1}$ .

Exercise 3.7.2

Find the principal stresses and directions of $\overset{˘}{T} = 00 - π 0 π^{2} 0 - π 0 - 1$ , and name the state.

Solution

$e_{2}$ is principal: $σ = π^{2} = 9.8696$ .
The $1$ – $3$ block $[0 - π - π - 1]$ : $σ^{2} + σ - π^{2} = 0$ , so $σ = \frac{- 1 \pm 1 + 4 π ^{2}}{2}$ : $2.6811$ and $- 3.6811$ .
Directions: for $2.6811$ , $- 2.6811 n_{1} - π n_{3} = 0$ , so $n \propto (- π, 0, 2.6811)$ , normalised $(- 0.7606, 0, 0.6492)$ ; for $- 3.6811$ , $(0.6492, 0, 0.7606)$ .
Check: $9.8696 + 2.6811 - 3.6811 = π^{2} - 1 = tr \overset{˘}{T}$ . Triaxial.