Part 2 · Theory of deformation

2.6 The Green–Lagrange tensor

Source: Carboni’s notes for lesson 4, Teoria della deformazione 3, pp. 1–3

Objectives

Vocabulary. Strain measure, elongation $ε_{m}$ , displacement gradient $H$ , Green–Lagrange tensor $E$ .
Recipe. $E$ from $F$ , or from the displacement gradient: $E = \frac{1}{2} (H + H^{T}) + \frac{1}{2} H^{T} H$ .
Theorem. Stretch and shear in terms of $E$ .

$C$ measures deformation, but it is the identity, not zero, when nothing deforms. A strain measure should vanish when the deformation does, so that “small strain” means “small numbers”. This section shifts $C$ to get one.

A measure that vanishes when nothing deforms

The undeformed neighbourhood is the case $C = I$ (Proposition 2.2.5). Then $C = I ⟹ U = I (C = U^{2}, U positive definite) ⟹ F = RU = R .$ Conversely, if $F = R$ then $C = R^{T} R = R^{- 1} R = I$ .

Key observation 2.6.1rigid exactly when C = Inotes L4, p. 1

$F$ is a pure rotation exactly when $C = I$ . Subtracting $I$ from $C$ gives a measure that is zero in that case, and only then.

Definition

Definition 2.6.2Green–Lagrange tensorEssentialnotes L4, p. 2

$E = \frac{1}{2} (C - I) = \frac{1}{2} (F^{T} F - I) ⟺ C = 2 E + I$ is the Green–Lagrange (finite) strain tensor. Like $C$ it is symmetric; it is $0$ exactly when the neighbourhood moves rigidly.

The factor $\frac{1}{2}$ makes $E$ agree, for small deformations, with the elongations and half-shears of §2.7 Infinitesimal strain.

Definition 2.6.3elongationnotes L4, p. 2

The elongation of the fibre along $m$ is $ε_{m} = ν_{m} - 1$ : positive when the fibre lengthens, negative when it shortens, zero when its length is unchanged.

Proposition 2.6.4stretch and shear from Enotes L4, p. 2

$ε_{m} = 2 m^{T} E m + 1 - 1, sin γ_{mn} = \frac{2 m ^{T} E n}{( 1 + ε _{m} ) ( 1 + ε _{n} )} (m ⊥ n) .$

Proof

Substitute $C = 2 E + I$ in Definition 2.2.2: $ν_{m} = m^{T} (2 E) m + m^{T} Im = 2 m^{T} E m + 1$ , and subtract 1.
For orthogonal fibres, $m^{T} Cn = 2 m^{T} E n + m^{T} n = 2 m^{T} E n$ , since $m^{T} n = 0$ (beyond the lectures). And $ν_{m} ν_{n} = (1 + ε_{m}) (1 + ε_{n})$ . Put both into Theorem 2.2.9.

Warningslip in the notes

L4 p. 2 writes $ν_{m} = m^{T} (2 E) m^{T} + \dots$ : the second $m$ has a stray transpose.

E in terms of the displacement

Most problems give the displacement $U (X)$ . Since $p = X + U$ , $F = \frac{\partial p}{\partial X} = I + \frac{\partial U}{\partial X} .$

Definition 2.6.5displacement gradientnotes L4, p. 2

$H = \frac{\partial U}{\partial X} = (\nabla U)^{T}, H_{ij} = \frac{\partial U _{i}}{\partial X _{j}}, F = I + H .$ In this course $(\nabla U)_{ij} = \partial U_{j} / \partial X_{i}$ , hence the transpose: $H$ has the components of $U$ along its rows, like $F$ .

Theorem 2.6.6finite strain from the displacementnotes L4, p. 3

$E = \frac{1}{2} (H + H^{T}) + \frac{1}{2} H^{T} H = \frac{1}{2} [(\nabla U)^{T} + \nabla U] + \frac{1}{2} \nabla U (\nabla U)^{T} .$ This is exact, with no assumption on the size of the deformation: the tensor of finite strains.

Proof

$E = \frac{1}{2} [(I + H)^{T} (I + H) - I] = \frac{1}{2} (I + H + H^{T} + H^{T} H - I) = \frac{1}{2} (H + H^{T}) + \frac{1}{2} H^{T} H .$ With $H = (\nabla U)^{T}$ , $H^{T} = \nabla U$ and $H^{T} H = \nabla U (\nabla U)^{T}$ .

In components (beyond the lectures), with sums written out: $E_{ij} = \frac{1}{2} (\frac{\partial U _{i}}{\partial X _{j}} + \frac{\partial U _{j}}{\partial X _{i}}) + \frac{1}{2} k = 1 \sum 3 \frac{\partial U _{k}}{\partial X _{i}} \frac{\partial U _{k}}{\partial X _{j}} .$ The first part is linear in the displacement, the second quadratic: the next section drops it.

Components of E

From the components of $C$ (Key observation 2.2.11), since $I$ only touches the diagonal:

Key observation 2.6.7what the entries of E meannotes L4, p. 3

$E = \frac{ν _{1}^{2} - 1}{2} sym \frac{ν _{1} ν _{2} sin γ _{12}}{2} \frac{ν _{2}^{2} - 1}{2} \frac{ν _{1} ν _{3} sin γ _{13}}{2} \frac{ν _{2} ν _{3} sin γ _{23}}{2} \frac{ν _{3}^{2} - 1}{2}$

Example 2.6.8finite strain at a pointbeyond the lectures

$F (P) = 120 - 1 10 001$ (Exercise 2.1.1).

$H = F - I = 020 - 1 00 000$ .
$C = 510120001$ , so $E = \frac{1}{2} (C - I) = 2 \frac{1}{2} 0 \frac{1}{2} \frac{1}{2} 0 000$ .
Check with Theorem 2.6.6: $\frac{1}{2} (H + H^{T}) = 0 \frac{1}{2} 0 \frac{1}{2} 00 000$ and $\frac{1}{2} H^{T} H = \frac{1}{2} 400010000$ ; their sum is $E$ . ✓
Read the entries: $E_{11} = 2 = (ν_{1}^{2} - 1) /2$ gives $ν_{1} = 5$ , the length of the first column $(1, 2, 0)$ of $F$ . ✓

Exercises

Exercise 2.6.1

For $F = diag (2, 0.8, 0.8)$ , compute $E$ and the elongations of the three basis fibres.

Solution

$C = diag (4, 0.64, 0.64)$ , $E = diag (1.5, - 0.18, - 0.18)$ . Elongations $ε = ν - 1$ : $1$ , $- 0.2$ , $- 0.2$ (check with $2 E_{ii} + 1 - 1$ : $4 - 1 = 1$ , $0.64 - 1 = - 0.2$ ).

Exercise 2.6.2

Show that a rigid rotation $F = R$ gives $E = 0$ although $H = R - I \neq = 0$ . Which term of Theorem 2.6.6 cancels which?

Solution

$C = R^{T} R = I$ , so $E = 0$ . In Theorem 2.6.6, $\frac{1}{2} (H + H^{T}) = \frac{1}{2} (R + R^{T}) - I$ and $\frac{1}{2} H^{T} H = \frac{1}{2} (R^{T} R - R - R^{T} + I) = I - \frac{1}{2} (R + R^{T})$ : the quadratic part exactly cancels the linear one. Dropping the quadratic part, as the next section does, breaks this.