Part 2 · Theory of deformation

2.7 Infinitesimal strain

Source: Carboni’s notes for lesson 4, Teoria della deformazione 3, pp. 3–6

Objectives

Vocabulary. Infinitesimal strain tensor $E^{(1)}$ , volumetric dilatation $Δ$ , spherical and deviatoric parts, infinitesimal rotation tensor $W$ .
Picture. The diagonal of $E^{(1)}$ holds elongations, the off-diagonal half the shear angles.
Recipe. Elongation, shear and volume change from $E^{(1)}$ : $ε_{m} = m^{T} E^{(1)} m$ , $γ_{mn} = 2 m^{T} E^{(1)} n$ , $Δ = tr E^{(1)}$ .
Theorem. Every displacement gradient is a strain plus a rotation: $H = E^{(1)} + W$ .

Most engineering deformations are small: a steel beam stretches by a thousandth. Then the quadratic term of $E$ (Theorem 2.6.6) is a thousandth of a thousandth, and dropping it makes everything linear. This section is that linearised (infinitesimal) theory, and its limits.

Linearisation

Let the displacement gradient be small: $H = O (ε)$ , with $ε$ a small number (its entries are of the order of $ε$ ). Then $H^{T} H = O (ε^{2})$ is negligible, and $E$ reduces to its linear part.

Definition 2.7.1infinitesimal strain tensorEssentialnotes L4, p. 3

$E^{(1)} = \frac{1}{2} (H + H^{T}) = \frac{1}{2} [(\nabla U)^{T} + \nabla U], E_{ij}^{(1)} = \frac{1}{2} (\frac{\partial U _{i}}{\partial X _{j}} + \frac{\partial U _{j}}{\partial X _{i}})$ is the infinitesimal (linearised) Green–Lagrange strain tensor: the symmetric part of the displacement gradient. It approximates the finite one, $E = E^{(1)} + O (ε^{2}), E^{(1)} = O (ε) .$

Warningslips in the notes

L4 p. 3 calls $E^{(1)}$ “il tensore di Green-Lagrange delle tensioni infinitesime”: it is the tensor of infinitesimal strains (deformazioni); stress (tensione) comes later. The same page writes $E := E^{(1)} + O (ε^{2})$ with a definition sign: it is an approximation, not a definition. On p. 4 the component formula is written $E_{ij}$ , without the $^{(1)}$ .

What the components mean

For small strains $ν_{j} ≃ 1$ and $γ_{ij} ≃ 0$ . Linearise the entries of $E$ (Key observation 2.6.7):

Diagonal. With $ν_{j} - 1 = ε_{j}$ (the elongation) and $ν_{j} + 1 = ε_{j} + 2$ : $\frac{ν _{j}^{2} - 1}{2} = \frac{( ν _{j} - 1 ) ( ν _{j} + 1 )}{2} = \frac{2 ε _{j} + ε _{j}^{2}}{2} ≃ ε_{j} .$
Off-diagonal. With $ν_{i} ν_{j} ≃ 1$ and $sin γ_{ij} ≃ γ_{ij}$ : $\frac{ν _{i} ν _{j} sin γ _{ij}}{2} ≃ \frac{γ _{ij}}{2} .$

Key observation 2.7.2elongations and half-shearsnotes L4, p. 4

$E^{(1)} = ε_{1} sym \frac{γ _{12}}{2} ε_{2} \frac{γ _{13}}{2} \frac{γ _{23}}{2} ε_{3}$ The diagonal holds elongations, the off-diagonal holds half the shear angles.

Proposition 2.7.3elongation and shear in any directionnotes L4, p. 4

For unit fibres $m$ , and $n ⊥ m$ , $ε_{m} = m^{T} E^{(1)} m, γ_{mn} = 2 m^{T} E^{(1)} n,$ to first order in $ε$ .

Proofbeyond the lectures

Use $1 + x = 1 + \frac{x}{2} + O (x^{2})$ and $\frac{1}{1 + x} = 1 - x + O (x^{2})$ on Proposition 2.6.4:

$ε_{m} = 2 m^{T} Em + 1 - 1 = m^{T} Em + O (ε^{2}) = m^{T} E^{(1)} m + O (ε^{2})$ .
$sin γ_{mn} = 2 m^{T} En (1 + O (ε)) = 2 m^{T} E^{(1)} n + O (ε^{2})$ , and $sin γ = γ + O (γ^{3})$ .

Warningslip in the notes

L4 p. 4 writes $γ_{mn} = 2 m^{T} E n$ , without the $^{(1)}$ . The formula is the linearised one, so the tensor is $E^{(1)}$ ; with the finite $E$ the exact relation is the $sin γ_{mn}$ formula of Proposition 2.6.4.

Loading the demo…

Figure 2.7.1 Exact against linear. The slider scales the displacement gradient

H

. For small

ε

the solid (exact) and dashed (linear) curves of the elongation

ε_{m}

, the shear

γ_{mn}

and the area change coincide; for large

ε

they part. The preset “rotation” is the most striking: the exact measures stay zero (a rotation does not deform), the linear ones do not.

Spherical and deviatoric parts

Definition 2.7.4spherical and deviatoric partsnotes L4, p. 5

Split $E^{(1)}$ as $E^{(1)} = \frac{Δ}{3} I + E^{(1)}^{'}, Δ = tr E^{(1)} = E_{11}^{(1)} + E_{22}^{(1)} + E_{33}^{(1)}, E^{(1)}^{'} = E^{(1)} - \frac{Δ}{3} I .$

$\frac{Δ}{3} I$ is the spherical part: it causes only changes of volume.
$E^{(1)}^{'}$ is the deviatoric part: it causes only changes of shape.

Theorem 2.7.5the trace is the volume changebeyond the lectures

To first order, the relative change of volume is the trace of the infinitesimal strain: $\frac{d V ˘ - d V}{d V} ≃ Δ = ε_{1} + ε_{2} + ε_{3},$ the volumetric dilatation. The deviator has zero trace, so it changes no volume (to first order); the spherical part has the same elongation $Δ/3$ in every direction and no shear, so it changes no shape.

Proof

$det (I + H) = 1 + tr H + O (ε^{2})$ : expanding the determinant, the only product of entries without two off-diagonal factors is the product of the diagonal, $(1 + H_{11}) (1 + H_{22}) (1 + H_{33}) = 1 + H_{11} + H_{22} + H_{33} + O (ε^{2})$ ; every other term has at least two off-diagonal entries of $H$ .
$tr H = tr E^{(1)}$ , because the antisymmetric part of $H$ has zeros on the diagonal.
So $Λ - 1 = det F - 1 ≃ Δ$ , using Theorem 2.5.2.
The deviator: $tr E^{(1)}^{'} = Δ - 3 \cdot Δ/3 = 0$ .
The spherical part: $m^{T} \frac{Δ}{3} Im = Δ/3$ for every unit $m$ , and $2 m^{T} \frac{Δ}{3} In = 0$ for $m ⊥ n$ .

Infinitesimal rigid motion and the rotation tensor

If the deformation is only a rigid motion of the body, the strain vanishes: $E^{(1)} = \frac{1}{2} (H + H^{T}) = 0 ⟹ \nabla U = - (\nabla U)^{T} :$ the displacement gradient is antisymmetric (the notes add $+ O (ε)$ : this holds to first order).

Expand $U$ about a point $X_{0}$ to first order: $U (X) = U (X_{0}) + Θ (X - X_{0}),$ with $Θ$ the antisymmetric gradient. This is the general form of an infinitesimal rigid displacement: $U (X_{0})$ is a translation, $Θ (X - X_{0})$ a small rotation about $X_{0}$ .

Definition 2.7.6infinitesimal rotation tensornotes L4, pp. 5–6

$W = \frac{1}{2} (H - H^{T}) = \frac{1}{2} [(\nabla U)^{T} - \nabla U], W_{ij} = \frac{1}{2} (\frac{\partial U _{i}}{\partial X _{j}} - \frac{\partial U _{j}}{\partial X _{i}})$ is the infinitesimal rigid rotation tensor (the notes also write it $Θ$ ): the antisymmetric part of the displacement gradient.

Theorem 2.7.7strain plus rotationnotes L4, p. 6

Every displacement gradient splits into its symmetric and antisymmetric parts: $H = \frac{1}{2} (H + H^{T}) + \frac{1}{2} (H - H^{T}) = E^{(1)} + W,$ and on a fibre, $Hm = deformation E^{(1)} m + rigid motion Wm .$

In Figure 2.7.1, the orange dashed square is $I + E^{(1)}$ and the teal dotted one $I + W$ : for small $ε$ the violet square is, to first order, the first deformed by the second’s turn.

Proposition 2.7.8W as a rotation vectorbeyond the lectures

An antisymmetric $3 \times 3$ matrix has only three independent entries: $W = 0 ω_{3} - ω_{2} - ω_{3} 0 ω_{1} ω_{2} - ω_{1} 0, Wv = ω \times v .$ So $W$ turns the neighbourhood by the small angle $∣ ω ∣$ about the axis $ω = (W_{32}, W_{13}, W_{21})$ .

Proof

Write out $ω \times v = (ω_{2} v_{3} - ω_{3} v_{2}, ω_{3} v_{1} - ω_{1} v_{3}, ω_{1} v_{2} - ω_{2} v_{1})$ and compare it row by row with $Wv$ .

Warningthe limit of the linear theorybeyond the lectures

$E^{(1)}$ vanishes for infinitesimal rotations only. For a finite rotation $F = R$ , $E = 0$ exactly (Exercise 2.6.2) but $E^{(1)} \neq = 0$ : the neglected $H^{T} H$ is not small. The linear theory needs small strains and small rotations.

Exercises

Exercise 2.7.1

$H = 1 0^{- 3} 300 21 - 2 04 - 1$ , uniform. Find $E^{(1)}$ , $W$ , $ω$ and $Δ$ . For $m = \frac{1}{2} (1, 1, 0)$ and $n = \frac{1}{2} (- 1, 1, 0)$ , compare $ε_{m}$ , $γ_{mn}$ and the volume change with the exact values.

Solution

$E^{(1)} = 1 0^{- 3} 310111 01 - 1$ , $W = 1 0^{- 3} 0 - 1 0 10 - 3 030$ , so $ω = (W_{32}, W_{13}, W_{21}) = 1 0^{- 3} (- 3, 0, - 1)$ .
$Δ = (3 + 1 - 1) \times 1 0^{- 3} = 3 \times 1 0^{- 3}$ .
$ε_{m} = \frac{1}{2} (3 + 1 + 2 \cdot 1) \times 1 0^{- 3} = 3.000 \times 1 0^{- 3}$ ; exact $ν_{m} - 1 = 3.003 \times 1 0^{- 3}$ .
$γ_{mn} = 2 m^{T} E^{(1)} n = (1 - 3) \times 1 0^{- 3} = - 2.000 \times 1 0^{- 3}$ ; exact $- 1.992 \times 1 0^{- 3}$ .
Volume: $1 + Δ = 1.003000$ ; exact $det F = 1.003007$ .
The differences are $O (1 0^{- 6})$ , the size of $H^{T} H$ .

Exercise 2.7.2

For $F = R$ , the $90°$ rotation about $e_{3}$ , compute $C$ , $E$ and $E^{(1)}$ . Which one says correctly that nothing deforms?

Solution

$F = 010 - 1 00 001$ : $C = I$ and $E = 0$ , correct.
$H = R - I$ , so $E^{(1)} = diag (- 1, - 1, 0)$ : it reports a 100% shortening along $e_{1}$ and $e_{2}$ that is not there.
Here $H^{T} H = diag (2, 2, 0)$ is not negligible, and it exactly cancels $H + H^{T}$ .

Exercise 2.7.3

For simple shear $H = 000 k 00 000$ , split $H$ into $E^{(1)} + W$ . What is the linear shear angle $γ_{12}$ , and the rotation? Compare with the exact $γ_{12} = arctan k$ for $k = \frac{1}{2}$ .

Solution

$E^{(1)} = 0 k /2 0 k /2 00 000$ , $W = 0 - k /2 0 k /2 00 000$ .
$γ_{12} = 2 E_{12}^{(1)} = k$ ; $ω = (0, 0, - k /2)$ : a clockwise turn by $k /2$ .
For $k = \frac{1}{2}$ : linear $0.5$ rad, exact $arctan \frac{1}{2} = 0.4636$ rad, about 8% apart. Simple shear is a stretch along one diagonal, a shortening along the other, and a rotation.