Part 3 · Theory of stress

3.8 Worked problems: stress

Source: The exercise sheets for the first partial (posted 13 Oct 2026), stress parts: Esercitazione, Esercizi per I esonero bis, Problem 2, and the Simulazione per I esonero, items h)–i)

Objectives

Recipe. An exam stress problem start to finish: balance at a point, principal stresses, stress from strain, and the three stress tensors from one another.

The practice sheets for the first partial combine everything in this part with the linear elastic law of §3.7 Principal stresses and the linear elastic law. The solutions were worked for this book: the sheets give only the questions.

Recipe 3.8.1a stress problem, start to finishbeyond the lectures

Balance at a point. Divergence along the rows, $(\nabla \cdot S)_{i} = \partial S_{ij} / \partial x_{j}$ ; then $ρ a = \nabla \cdot S + b$ (Proposition 3.7.1), or $b = - \nabla \cdot S$ at rest.
Principal stresses: Recipe 3.7.4.
Stress from strain: Recipe 3.7.7.
The other tensors: $T = FS$ , $\overset{˘}{T} = \frac{1}{d e t F} T F^{T}$ (check it is symmetric).
Stress on a plane: Recipe 3.3.5.

Exercises

Exercise 3.8.1

Exercises for the first partial (bis), Problem 2. With the data of Problem 1 (Exercise 2.8.1), the body is homogeneous, isotropic and linear elastic, with $E = 5$ , $ν = \frac{1}{4}$ , so $μ = λ = 2$ . At $P = (0, 0, - 1)$ : (1) compute $S$ from the constitutive law, assuming the linear law between $S$ and $E$ holds for large strains too; (2) compute the first Piola–Kirchhoff tensor $T$ and the Cauchy stress $\overset{˘}{T}$ ; (3) compute the principal stresses and directions of $S$ ; (4) compute the Cauchy stress vector on the plane with normal $\overset{˘}{n} = \frac{1}{2} (1, 0, 1)$ .

(The sheet calls $S$ the “tensore delle tensioni nominali”; in lesson 8 “nominal” is the first Piola–Kirchhoff $T$ .)

Solution

From Problem 1: $F (P) = 120 - 1 10 001$ , $det F = 3$ , $E = 2 \frac{1}{2} 0 \frac{1}{2} \frac{1}{2} 0 000$ , $tr E = \frac{5}{2}$ . Check the constants: $λ = \frac{5 \cdot \frac{1}{4}}{\frac{5}{4} \cdot \frac{1}{2}} = 2$ , $μ = \frac{5}{2 \cdot \frac{5}{4}} = 2$ .

$S = 2 \cdot \frac{5}{2} I + 4 E = 1320270005$ .
$T = FS = 11280 - 5 110 005$ (not symmetric), and $\overset{˘}{T} = \frac{1}{3} T F^{T} = \frac{1}{3} 1617017670005 = 5.333 5.667 0 5.667 22.333 0 00 1.667,$ symmetric, as it must be.
$e_{3}$ is principal with $σ = 5$ . The $1$ – $2$ block: $σ = 10 \pm 9 + 4 = 10 \pm 13$ .
- $σ_{I} = 10 + 13 = 13.606$ , along $(0.9571, 0.2898, 0)$ ;
- $σ_{II} = 10 - 13 = 6.394$ , along $(- 0.2898, 0.9571, 0)$ ;
- $σ_{III} = 5$ , along $(0, 0, 1)$ .
$tan 2 θ = \frac{4}{6}$ , so the in-plane directions are $e_{1}, e_{2}$ turned by $θ = 16.85°$ about $e_{3}$ . Triaxial.
$\overset{˘}{t}_{n} = \overset{˘}{T} \overset{˘}{n} = \frac{1}{3 2} (16, 17, 5) = (3.7712, 4.0069, 1.1785)$ .
- Normal stress $σ_{n} = \overset{˘}{n} \cdot \overset{˘}{t}_{n} = \frac{16 + 5}{6} = \frac{7}{2}$ : tension.
- Shear $τ = ∣ \overset{˘}{t}_{n} ∣^{2} - σ_{n}^{2} = \frac{570}{18} - \frac{49}{4} = 4.4064$ .

Exercise 3.8.2

Esercitazione (exercise sheet). In the linearised theory, the second Piola–Kirchhoff tensor and the body force are $S = x_{2} sin x_{1} 0 - x_{1} x_{2}^{2} x_{3} 0 x_{1}^{2} 0 - x_{1} x_{2}^{2} x_{3} 0 - x_{3}, b = x_{1}^{2} e_{1} + x_{2} e_{2} + x_{1}^{3} e_{3}, A = (π, 1, 1) .$ (a) Compute the inertia force per unit volume that satisfies the balance of linear momentum at $A$ . (b) Compute the principal stresses and directions of $S$ at $A$ . © With the Lamé constants $λ = 153250$ MPa, $μ = 78947$ MPa, the point $B = (1, 2, 3)$ and the infinitesimal strain $E^{(1)} = α x_{2}^{3} 0 - x_{1} x_{2}^{2} x_{3} 0 x_{1}^{2} 0 - x_{1} x_{2}^{2} x_{3} 0 - x_{3},$ compute $S$ at $B$ with the linear elastic law.

Solution

(a) The divergence runs along the rows:

row 1: $\partial_{1} (x_{2} sin x_{1}) + \partial_{3} (- x_{1} x_{2}^{2} x_{3}) = x_{2} cos x_{1} - x_{1} x_{2}^{2}$ ;
row 2: $\partial_{2} (x_{1}^{2}) = 0$ ;
row 3: $\partial_{1} (- x_{1} x_{2}^{2} x_{3}) + \partial_{3} (- x_{3}) = - x_{2}^{2} x_{3} - 1$ .

At $A$ : $\nabla \cdot S = (- 1 - π, 0, - 2)$ and $b = (π^{2}, 1, π^{3})$ , so $ρ a = \nabla \cdot S + b = (π^{2} - π - 1, 1, π^{3} - 2) = (5.7280, 1, 29.0063) .$ The inertia force in d’Alembert’s sense is $- ρ a$ .

(b) $S (A) = 00 - π 0 π^{2} 0 - π 0 - 1$ : this is Exercise 3.7.2. $σ_{I} = π^{2} = 9.8696$ along $(0, 1, 0)$ ; $σ_{II} = 2.6811$ along $(- 0.7606, 0, 0.6492)$ ; $σ_{III} = - 3.6811$ along $(0.6492, 0, 0.7606)$ . Triaxial.

© At $B$ : $E^{(1)} = α 80 - 12 010 - 12 0 - 3$ , $tr E^{(1)} = 6 α$ . Then $S_{ij} = 6 α λ δ_{ij} + 2 μ E_{ij}$ , with $6 λ = 919500$ MPa and $2 μ = 157894$ MPa: $S (B) = α 21826520 - 1894728 010773940 - 1894728 0445818 MPa .$ (The constants are steel’s: $E = 210$ GPa, $ν = 0.33$ .)

Exercise 3.8.3

Simulation of the first partial, items h)–i). With the data of items a)–g) (Exercise 2.8.2): (h) with the Lamé constants $λ = 2$ GPa and $μ = 1$ GPa, compute $S$ , $T$ and $\overset{˘}{T}$ at $A$ , starting from the finite Green–Lagrange tensor $E$ (4 pt); (i) compute the principal stresses and directions of the Cauchy stress tensor and name the type of stress state (5 pt).

(Item i) writes “tensore della tensione di Cauchy $T$ ” without the breve; the words say Cauchy, so it is solved for $\overset{˘}{T}$ .)

Solution

From a): $F = 100 - \frac{π}{2} 1 - \frac{π}{2} 0 00 1 - \frac{π}{2}$ , $det F = (1 - \frac{π}{2})^{2} = 0.3258$ , $E = 0 - \frac{π}{4} 0 - \frac{π}{4} \frac{π ^{2}}{4} - \frac{π}{2} 0 00 \frac{π ^{2}}{8} - \frac{π}{2}$ .

(h) $tr E = \frac{3 π ^{2}}{8} - π = 0.5595$ . With $S = 2 tr E I + 2 E$ (GPa): $S = \frac{π ( 3 π - 8 )}{4} - \frac{π}{2} 0 - \frac{π}{2} \frac{π ( 5 π - 12 )}{4} 0 00 π (π - 3) = 1.1190 - 1.5708 0 - 1.5708 2.9122 0 00 0.4448 GPa$ $T = FS = 3.5864 0.8966 0 - 6.1453 - 1.6623 0 00 - 0.2539 GPa, \overset{˘}{T} = \frac{1}{det F} T F^{T} = 40.6357 10.7662 0 10.7662 2.9122 0 00 0.4448 GPa .$ $\overset{˘}{T}_{11}$ is large because $det F$ is small: the same force sits on a much smaller current area.

(i) $e_{3}$ is principal with $σ = π (π - 3) = 0.4448$ GPa. The $1$ – $2$ block gives the other two:

$σ_{I} = 43.492$ GPa, along $(0.9666, 0.2564, 0)$ ;
$σ_{II} = 0.4448$ GPa, along $(0, 0, 1)$ ;
$σ_{III} = 0.0559$ GPa, along $(0.2564, - 0.9666, 0)$ .

Three distinct values: a triaxial stress state. The in-plane directions are $e_{1}, e_{2}$ turned by $14.86°$ about $e_{3}$ . For comparison, $S$ has principal values $3.824$ , $0.445$ , $0.207$ GPa: the three tensors differ a lot here, because the strain is far from small.

Exercise 3.8.4

A thin plate is in plane stress, $\overset{˘}{T} = 80 - 30 0 - 30 200 000$ MPa. Find the principal stresses, the angle of the principal directions, and the largest shear stress on a plane containing $e_{3}$ .

Solution

$e_{3}$ : $σ = 0$ .
Block: $50 \pm 3 0^{2} + 3 0^{2} = 50 \pm 42.43$ : $σ_{I} = 92.43$ MPa, $σ_{II} = 7.57$ MPa.
$tan 2 θ = \frac{- 60}{60} = - 1$ , $θ = - 22.5°$ : $n_{I} = (cos 22.5°, - sin 22.5°, 0)$ .
The largest in-plane shear is the radius of Mohr’s circle, $42.43$ MPa, on planes at $45°$ to the principal ones (Figure 3.7.1).