Part 2 · Theory of deformation

2.8 Worked problems: from a displacement to strain

Source: Carboni’s exercise session of lesson 5 (12 Oct 2026), pp. 1–9; the exercise sheets for the first partial (posted 13 Oct 2026), deformation parts

Objectives

Recipe. The whole chain for an exam problem: $U$ or $p$ → $F$ , $det F$ → $C$ , $E$ → stretches and shear → $E^{(1)}$ , $Δ$ , deviator → $W$ → principal strains.
Picture. When the displacement gradient is not small, the finite and the linear answers disagree, and why.

Lesson 5 was an exercise session: one problem taken through every tool of this part. Its steps 1–3 already appear as Example 2.1.9 and Example 2.2.13; here is the whole problem, then the exam-style problems of the practice sheets.

Recipe 2.8.1a deformation problem, start to finishbeyond the lectures

$p = X + U$ ; $F = \partial p / \partial X$ (row $i$ : derivatives of $P_{i}$ ), at the point. Admissible if $det F > 0$ (Recipe 2.1.8).
$H = F - I$ ; $C = F^{T} F$ ; $E = \frac{1}{2} (C - I)$ .
Stretches $ν_{m} = m^{T} Cm$ (check with $∣ Fm ∣$ ); shear $sin γ_{mn} = m^{T} Cn / (ν_{m} ν_{n})$ for $m ⊥ n$ ; current angle $\overset{˘}{θ}_{mn} = 90° - γ_{mn}$ .
Linear theory: $E^{(1)} = \frac{1}{2} (H + H^{T})$ ; $ε_{m} = m^{T} E^{(1)} m$ , $γ_{mn} = 2 m^{T} E^{(1)} n$ ; $Δ = tr E^{(1)}$ ; spherical part $\frac{Δ}{3} I$ , deviator $E^{(1)} - \frac{Δ}{3} I$ .
$W = \frac{1}{2} (H - H^{T})$ ; $ω = (W_{32}, W_{13}, W_{21})$ . A rigid rotation is present if $W \neq = 0$ .
Principal strains: eigenvalues and unit eigenvectors of $E^{(1)}$ (or of $C$ for the finite principal stretches); name the state.

The exercise session of lesson 5

Problem (notes L5, p. 1, in English on the sheet, headed “21/10/2019 Continuum Mechanics”). Given the displacement $U (X) = X_{1}^{2} e_{1} + (X_{2}^{2} + X_{3}^{2} - X_{3}) e_{2} + (X_{3}^{2} + X_{2}^{2} + X_{2}) e_{3},$ the point $A = (1, 1, 1)$ and the fibres $m = \frac{2}{2} (e_{2} + e_{1})$ , $n = \frac{2}{2} (e_{2} - e_{1})$ :

show that the transformation is an admissible deformation at $A$ ;
compute $F$ and $C$ at $A$ ;
find the stretches along $m$ and $n$ at $A$ and the shear strain between them;
compute $E^{(1)}$ at $A$ and from it the elongations along $m$ and $n$ and the shear strain between them, the volumetric dilatation, the deviatoric and spherical parts;
compute the infinitesimal rigid rotation tensor at $A$ , and say whether the displacement contains a rigid rotation;
(added on p. 8) find the principal strains and directions.

Steps 1–3 are Example 2.1.9 and Example 2.2.13: $F (A) = 300033013$ , $det F = 18$ , $C (A) = 9000181201210$ , $ν_{m} = ν_{n} = 27/2 = 3.674$ , $γ_{mn} = arcsin \frac{1}{3} = 19.47°$ .

Example 2.8.2step 4: infinitesimal strainnotes L5, pp. 5–7

$E_{ij}^{(1)} = \frac{1}{2} (\partial U_{i} / \partial X_{j} + \partial U_{j} / \partial X_{i})$ :

$E_{11}^{(1)} = \frac{1}{2} (2 \cdot 2 X_{1}) = 2 X_{1}$ , $E_{22}^{(1)} = 2 X_{2}$ , $E_{33}^{(1)} = 2 X_{3}$ ;
$E_{23}^{(1)} = E_{32}^{(1)} = \frac{1}{2} (2 X_{3} - 1 + 2 X_{2} + 1) = X_{2} + X_{3}$ ;
all the others are zero. $E^{(1)} (A) = 200022022 .$ Elongations and shear:
$ε_{m} = m^{T} E^{(1)} m = [2, 2, 2] \cdot [\frac{2}{2}, \frac{2}{2}, 0] = 1 + 1 = 2$ ;
$ε_{n} = [- 2, 2, 2] \cdot [- \frac{2}{2}, \frac{2}{2}, 0] = 1 + 1 = 2$ ;
$γ_{mn} = 2 m^{T} E^{(1)} n = [22, 22, 22] \cdot [- \frac{2}{2}, \frac{2}{2}, 0] = - 2 + 2 = 0$ .

Volumetric dilatation $Δ = tr E^{(1)} = 2 + 2 + 2 = 6$ . Deviatoric and spherical parts: $E^{(1)}^{'} = E^{(1)} - \frac{Δ}{3} I = 000002020, \frac{Δ}{3} I = 200020002, E^{(1)} = E^{(1)}^{'} + \frac{Δ}{3} I .$

Warningslips in the worked sheet

L5 p. 5 ends the $E_{21}^{(1)}$ line with “ $E_{13} (A) = 0$ ”: it is $E_{21} (A)$ . On p. 6 the first product for $ε_{n}$ writes the column $n$ without its minus sign; the next line uses $(- \frac{2}{2}, \frac{2}{2}, 0)$ , as it should.

Remarkbeyond the lectures

The linear theory is far off here. The displacement gradient at $A$ has entries 2 and 3, nowhere near small, so $E^{(1)}$ is only a formal answer:

elongation: exact $ν_{m} - 1 = 2.674$ , against $ε_{m} = 2$ ;
shear: exact $19.5°$ , against $γ_{mn}^{(1)} = 0$ ;
volume: exact ratio $det F = 18$ , against $1 + Δ = 7$ .

The exam asks for both kinds of quantity; just do not expect them to agree for a large deformation (Figure 2.7.1 shows how they part).

Example 2.8.3step 5: rigid rotationnotes L5, pp. 7–8

$W_{ij} = \frac{1}{2} (\partial U_{i} / \partial X_{j} - \partial U_{j} / \partial X_{i})$ :

the diagonal is zero, and so are $W_{12}, W_{13}, W_{21}, W_{31}$ ;
$W_{23} = \frac{1}{2} (2 X_{3} - 1 - 2 X_{2} - 1) = X_{3} - X_{2} - 1$ , so $W_{23} (A) = - 1$ ;
$W_{32} = \frac{1}{2} (2 X_{2} + 1 - 2 X_{3} + 1) = X_{2} - X_{3} + 1$ , so $W_{32} (A) = 1$ . $W (A) = 000001 0 - 1 0 = 000 00 ω_{1} 0 - ω_{1} 0, ω_{1} = 1.$ $W \neq = 0$ : the displacement contains a rigid rotation, about $e_{1}$ .

Warningslips in the worked sheet

L5 p. 7 simplifies $W_{23}$ to $\frac{2 ( X _{2} - X _{3} ) - 2}{2}$ : it is $\frac{2 ( X _{3} - X _{2} ) - 2}{2}$ . On p. 8, $W_{32}$ starts $\frac{1}{2} (2 X_{3} + 1 - \dots)$ : the first term is $2 X_{2}$ . Both slips vanish at $A$ , where $X_{2} = X_{3}$ , so the values $- 1$ and $1$ are right.

Example 2.8.4step 6: principal strainsnotes L5, pp. 8–9

$det (E^{(1)} - λ I) = det 2 - λ 00 0 2 - λ 2 02 2 - λ = (2 - λ) [(2 - λ)^{2} - 4] = 0.$

One root: $λ_{1} = 2$ .
The bracket: $λ^{2} - 4 λ = λ (λ - 4) = 0$ , so $λ_{2} = 0$ and $λ_{3} = 4$ .

Eigenvectors, each normalised:

$λ_{1} = 2$ : $(E^{(1)} - 2 I) x_{1} = 0$ gives $0 = 0$ , $2 x_{3} = 0$ , $2 x_{2} = 0$ , so $x_{1} = (1, 0, 0)$ .
$λ_{2} = 0$ : $2 x_{1} = 0$ and $2 x_{2} + 2 x_{3} = 0$ , so $x_{1} = 0$ , $x_{2} = q$ , $x_{3} = - q$ ; normalising, $2 q^{2} = 1$ gives $q = \frac{2}{2}$ , and $x_{2} = (0, \frac{2}{2}, - \frac{2}{2})$ .
$λ_{3} = 4$ : $x_{1} = 0$ and $- 2 x_{2} + 2 x_{3} = 0$ , so $x_{3} = (0, \frac{2}{2}, \frac{2}{2})$ .

Three distinct principal strains: a triaxial state.

Warningslips in the worked sheet

On L5 p. 9: for $λ_{1} = 2$ the system reads $4 x_{3} = 0$ , $4 x_{2} = 0$ (the entries of $E^{(1)} - 2 I$ are 2; the conclusion is the same); for $λ_{3} = 4$ the matrix is written $E^{(1)} + 2 I$ , but it is $E^{(1)} - 4 I$ , which is what the system below it solves.

Exercises

Exercise 2.8.1

Exercises for the first partial (bis), Problem 1. Given $p (X) = (X_{1} + X_{2} X_{3}) e_{1} + (X_{2} - 2 X_{3} X_{1}) e_{2} + (X_{3} + X_{1} X_{2}) e_{3}$ and $P = (0, 0, - 1)$ : (1) compute $F$ and check that $p$ is admissible at $P$ ; (2) compute $H$ at $P$ ; (3) compute $C$ and $E$ at $P$ ; (4) compute the stretches along $m = \frac{1}{2} (1, 1, 0)$ and $n = \frac{1}{2} (- 1, 1, 0)$ and the shear angle between them; (5) compute $E^{(1)}$ and $W$ at $P$ .

Solution

$F (P) = 120 - 1 10 001$ , $det F = 3 > 0$ : admissible (Exercise 2.1.1).
$H = F - I = 020 - 1 00 000$ .
$C = 510120001$ , $E = 2 \frac{1}{2} 0 \frac{1}{2} \frac{1}{2} 0 000$ .
$ν_{m} = \frac{3 2}{2} = 2.1213$ , $ν_{n} = 1.5811$ , $sin γ_{mn} = - 1/ 5$ , $γ_{mn} = - 26.57°$ : the fibres open to $116.57°$ (Exercise 2.2.1).
$E^{(1)} = \frac{1}{2} (H + H^{T}) = 0 \frac{1}{2} 0 \frac{1}{2} 00 000$ , $W = \frac{1}{2} (H - H^{T}) = 0 \frac{3}{2} 0 - \frac{3}{2} 00 000$ : a rotation about $e_{3}$ with $ω_{3} = \frac{3}{2}$ .

Again the gradient is not small. The linear theory gives $ε_{m} = \frac{1}{2}$ (exact $ν_{m} - 1 = 1.12$ ) and $γ_{mn}^{(1)} = 0$ (exact $- 26.6°$ ).

Exercise 2.8.2

Simulation of the first partial, items a)–g) (a 2-hour, 30-point mock exam). Given $U (X) = - X_{1} X_{2} sin X_{1} e_{1} - X_{2} X_{3} e^{- 2 X_{2}} e_{2} + X_{3} cos X_{3} e_{3},$ the fibres $m = \frac{1}{3} (e_{1} + e_{2} + e_{3})$ and $n = \frac{1}{2} (- e_{1} + e_{2})$ and the point $A = (π /2, 0, π /2)$ : a) compute $F$ , $C$ and $E$ at $A$ (4 pt); b) show that $p (X)$ describes an admissible deformation (2 pt); c) compute the stretches along $m$ and $n$ and the shear strain between them (2 pt); d) compute the angle between $m$ and $n$ in the current configuration and its change from the reference one (2 pt); e) in the infinitesimal theory, compute $E^{(1)}$ at $A$ , its spherical (volumetric dilatation) and deviatoric parts (4 pt); f) compute $W$ at $A$ (2 pt); g) compute the principal strains and directions of $E^{(1)}$ at $A$ and name the type of strain state (5 pt).

Solution

a) $H_{ij} = \partial U_{i} / \partial X_{j}$ : $H (X) = - X_{2} (X_{1} cos X_{1} + sin X_{1}) 00 - X_{1} sin X_{1} X_{3} (2 X_{2} - 1) e^{- 2 X_{2}} 0 0 - X_{2} e^{- 2 X_{2}} cos X_{3} - X_{3} sin X_{3}, H (A) = 000 - \frac{π}{2} - \frac{π}{2} 0 00 - \frac{π}{2} .$ $F = I + H = 100 - \frac{π}{2} 1 - \frac{π}{2} 0 00 1 - \frac{π}{2} \approx 100 - 1.5708 - 0.5708 0 00 - 0.5708$ $C = 1 - \frac{π}{2} 0 - \frac{π}{2} 1 - π + \frac{π ^{2}}{2} 0 00 (1 - \frac{π}{2})^{2} \approx 1 - 1.5708 0 - 1.5708 2.7932 0 00 0.3258, E \approx 0 - 0.7854 0 - 0.7854 0.8966 0 00 - 0.3371$ b) $det F (A) = (1 - \frac{π}{2})^{2} = 0.3258 > 0$ : admissible at $A$ . Two diagonal entries of $F$ are negative, but their product is positive. The volume around $A$ shrinks to a third.

c) $Fm = \frac{1 - π /2}{3} (1, 1, 1)$ , so $ν_{m} = ∣1 - \frac{π}{2} ∣ = 0.5708$ . $ν_{n}^{2} = \frac{1}{2} (C_{11} + C_{22} - 2 C_{12}) = 1 + \frac{π ^{2}}{4}$ , $ν_{n} = 1.8621$ . $m ⊥ n$ in the reference configuration, and $sin γ_{mn} = m^{T} Cn / (ν_{m} ν_{n}) = 0.6888$ , so $γ_{mn} = 0.7598$ rad $= 43.53°$ .

d) $\overset{˘}{θ}_{mn} = 90° - 43.53° = 46.47°$ : the right angle has closed by $43.53°$ .

e) $E^{(1)} = 0 - \frac{π}{4} 0 - \frac{π}{4} - \frac{π}{2} 0 00 - \frac{π}{2}$ ; $Δ = - π$ ; spherical part $- \frac{π}{3} I$ ; deviator $\frac{π}{3} - \frac{π}{4} 0 - \frac{π}{4} - \frac{π}{6} 0 00 - \frac{π}{6}$ (trace zero). The linear $Δ = - 3.14$ is meaningless as a volume change; the exact one is $det F - 1 = - 0.674$ .

f) $W = 0 \frac{π}{4} 0 - \frac{π}{4} 00 000$ : a rotation about $e_{3}$ with $ω_{3} = \frac{π}{4}$ .

g) $e_{3}$ is an eigenvector with $ε = - \frac{π}{2}$ . The $1$ – $2$ block is $\frac{π}{4} [0 - 1 - 1 - 2]$ , with eigenvalues $\frac{π}{4} μ$ where $μ^{2} + 2 μ - 1 = 0$ , $μ = - 1 \pm 2$ :

$ε_{I} = \frac{π}{4} (2 - 1) = 0.3253$ , along $(0.9239, - 0.3827, 0)$ ;
$ε_{II} = - \frac{π}{2} = - 1.5708$ , along $(0, 0, 1)$ ;
$ε_{III} = - \frac{π}{4} (1 + 2) = - 1.8961$ , along $(0.3827, 0.9239, 0)$ .

The in-plane principal directions are $e_{1}, e_{2}$ turned by $- 22.5°$ about $e_{3}$ . Three distinct values: a triaxial strain state.

Exercise 2.8.3

A displacement $U = α (X_{2} e_{1} + X_{1} e_{2})$ , with $α = 1 0^{- 3}$ . Find $E^{(1)}$ , $W$ , the principal strains and directions, and the change of the right angle between $e_{1}$ and $e_{2}$ . Is there a rotation?

Solution

$H = α 010100000$ , symmetric: $E^{(1)} = H$ , $W = 0$ , no rotation.
$γ_{12} = 2 E_{12}^{(1)} = 2 α$ : the right angle closes by $2 \times 1 0^{- 3}$ rad.
Principal strains $α$ , $- α$ , $0$ , along $\frac{1}{2} (1, 1, 0)$ , $\frac{1}{2} (- 1, 1, 0)$ , $e_{3}$ : a “pure shear” is a stretch along one diagonal and an equal shortening along the other. $Δ = 0$ : no volume change.