Strain tensor

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The strain tensor, ε, is a symmetric tensor used to quantify the strain of an object undergoing a small 3-dimensional deformation:

the diagonal coefficients ε_ii are the relative change in length in the direction of the i direction (along the x_i-axis) ;
the other terms ε_ij = 1/2 γ_ij (i ≠ j) are the shear strains, i.e. half the variation of the right angle (assuming a small cube of matter before deformation).

The deformation of an object is defined by a tensor field, i.e., this strain tensor is defined for every point of the object. This field is linked to the field of the stress tensor by the generalized Hooke's law.

In case of small deformations, the strain tensor is the Green tensor or Cauchy's infinitesimal strain tensor, defined by the equation:

$\varepsilon_{ij} = {1 \over 2} \left ({\part u_i \over \part x_j} + {\part u_j \over \part x_i}\right )$

Where u represents the displacement field of the object's configuration (i.e., the difference between the object's configuration and its natural state). This is the 'symmetric part' of the Jacobian matrix. The 'antisymmetric part' is called the small rotation tensor.

For large (finite) deformations see Finite Deformation Tensors.

1 Demonstration in simple cases
- 1.1 One-dimensional elongation
- 1.2 Pure shear strain
2 Relative variation of the volume
3 Derivation of the strain tensor
4 See also
5 References

[edit] Demonstration in simple cases

[edit] One-dimensional elongation

When the [AB] segment, parallel to the x₁-axis, is deformed to become the [A'B' ] segment, the deformation being also parallel to x₁

the ε₁₁ strain is (expressed in algebraic length):

$\varepsilon_{11} = \frac{\Delta l}{l_0} = \frac{\overline{A'B'}-\overline{AB}}{\overline{AB}}$

Considering that

$\overline{AA'} = u_1(A)$ and $\overline{BB'} = u_1(B)$

the strain is

$\varepsilon_{11} = \frac{\overline{AB} + \overline{BB'}-\overline{A'A}}{\overline{AB}} - 1$

$\varepsilon_{11} = \frac{u_1(B)-u_1(A) + \overline{AB}}{\overline{AB}} - 1$

The series expansion of u₁ is

$u_1(B) \simeq u_1(A) + \frac{\partial u_1}{\partial x_1} \cdot \overline{AB}$

and thus

$\varepsilon_{11} = \frac{\partial u_1}{\partial x_1}$

And in general

$\varepsilon_{ii} = \frac{\partial u_i}{\partial x_i} = \frac{1}{2} \left ( \frac{\partial u_i}{\partial x_i} + \frac{\partial u_i}{\partial x_i} \right )$

[edit] Pure shear strain

Let us now consider a pure shear strain. An ABCD square, where [AB] is parallel to x₁ and [AD] is parallel to x₂, is transformed into a AB'C'D' rhombus, symmetric to the first bisecting line.

The tangent of the γ angle is:

$\tan(\gamma) = \frac{\overline{BB'}}{\overline{AB}}$

for small deformations,

$\tan(\gamma) \simeq \gamma$

and

$\overline{BB'} = u_2(B) \simeq u_2(A) + \frac{\partial u_2}{\partial x_1} \cdot \overline{AB}$

and u₂(A) = 0. Thus,

$\gamma \simeq \frac{\partial u_2}{\partial x_1}$

Considering now the [AD] segment:

$\gamma \simeq \frac{\partial u_1}{\partial x_2}$

and thus

$\gamma = \frac{1}{2} \gamma_{12} = \varepsilon_{12} = \frac{1}{2} \left ( \frac{\partial u_1}{\partial x_2} + \frac{\partial u_2}{\partial x_1} \right )$

where γ₁₂ is the engineering strain, which is equal to 2γ.

It is interesting to use the average because the formula is still valid when the rhombus rotates; in such a case, there are two different angles $\gamma_B = \widehat{B'AB}$ and $\gamma_D = \widehat{D'AD}$ and the formula allows for neglecting the variation of angle due to rigid-body motion (which gives no contribution to the strain).

[edit] Relative variation of the volume

The dilatation (the relative variation of the volume) δ = ΔV/V₀, is the trace of the tensor:

$\delta=\frac{\Delta V}{V_0} = \varepsilon_{11} + \varepsilon_{22} + \varepsilon_{33}$

Actually, if we consider a cube with an edge length a, it is a quasi-cube after the deformation (the variations of the angles do not change the volume) with the dimensions $a \cdot (1 + \varepsilon_{11}) \times a \cdot (1 + \varepsilon_{22}) \times a \cdot (1 + \varepsilon_{33})$ and V₀ = a³, thus

$\frac{\Delta V}{V_0} = \frac{\left ( 1 + \varepsilon_{11} + \varepsilon_{22} + \varepsilon_{33} + \varepsilon_{11} \cdot \varepsilon_{22} + \varepsilon_{11} \cdot \varepsilon_{33}+ \varepsilon_{22} \cdot \varepsilon_{33} + \varepsilon_{11} \cdot \varepsilon_{22} \cdot \varepsilon_{33} \right ) \cdot a^3 - a^3}{a^3}$

as we consider small deformations,

$1 \gg \varepsilon_{ii} \gg \varepsilon_{ii} \cdot \varepsilon_{jj} \gg \varepsilon_{11} \cdot \varepsilon_{22} \cdot \varepsilon_{33}$

therefore the formula.

Real variation of volume (top) and the approximated one (bottom): the green drawing shows the estimated volume and the orange drawing the neglected volume

In case of pure shear, we can see that there is no change of the volume.

[edit] Derivation of the strain tensor

(Symon (1971) Ch. 10)

Let the position of a point in a material be specified by a vector $\mathbf{x}$ with components $x i$ . Let the point then move a small distance to a new position specified by a vector with components

$x'_i=x_i+u_i(\mathbf{x})\,$

where $u i$ is a vector function of $\mathbf{x}$ . Let $x i + d x i$ be a point close to $x i$ . After the motion, it will be in a new position given by:

$x'_i+dx'_i = x_i+dx_i + u_i(\mathbf{x}\!+\!\mathbf{dx})\,$

Since the $u i$ are small, we may approximate them by the first two terms in their Taylor series

$x'_i+dx'_i \approx x_i+dx_i + u_i + (\partial_j u_i)\,dx_j$

where we have used $\partial_j$ to represent $\partial/\partial x_j$ and we have used Einstein notation in which repeated indices in a product are assumed to be summed (i.e. index j in this case). $\partial_j u_i$ is just the Jacobian matrix of the $u i$ function. If we represent the unit matrix by $δ i j$ then the above equation may be written:

$x'_i+dx'_i = x'_i + (\delta_{ij}+\partial_j u_i)\,dx_j$

It is seen that the final term (the displacement matrix) specifies the infinitesimal change in the position ( $d x' i$ ) of the nearby particle. If the $u i$ are constants, the displacement matrix will be the unit matrix, and the resulting displacement will simply be a rigid translation. Any matrix may be written as the sum of an antisymmetric matrix and a symmetric matrix. Writing the diplacement matrix (in parentheses in the above equation) in this manner yields:

$\delta_{ij}+\frac{1}{2}(\partial_j u_i-\partial_i u_j) +\frac{1}{2}(\partial_j u_i+\partial_i u_j)$

The first two terms are the unit matrix and the antisymmetric part of the displacement matrix. These are the first two terms in the Taylor series of a rigid rotation about the translated point $x' i$ . They constitute an infinitesimal rotation and therefore do not represent a deformation of the material. It is the second, symmetric matrix which represents the deformation of the material and this is just the strain tensor $\varepsilon_{ij}$ :