Strain Tensor

The strain tensor ε is a symmetric tensor used to quantify the strain of an object undergoing a 3-dimensional deformation:
  • the diagonal coefficients εii are the relative change in length in the direction of the i direction (along the xi-axis) ;
  • the other terms εij (i ≠ j) are the γ, 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 stress tensor by the generalized Hooke's law. In case of small deformations, the strain tensor is the Green 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.

Demonstration in simple cases

One-dimensional elongation

When the AB segment, parallel to the x1-axis, is deformed to become the segment, the deformation being also parallel to x''1 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{A'A} + \overline{AB} + \overline{BB'}}{\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 )

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 = \varepsilon_{12} = \frac{1}{2} \left ( \frac{\partial u_1}{\partial x_2} + \frac{\partial u_2}{\partial x_1} \right )
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} et \gamma_D = \widehat{D'AD}.

Relative variation of the volume

The relative variation of the volume ΔV/V0 is the trace of the tensor :
\frac{\Delta V}{V_0} = \varepsilon_{11} + \varepsilon_{22} + \varepsilon_{33}
Actually, if we considere 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 considere small deformations,
1 >> \varepsilon_{ii} >> \varepsilon_{ii} \cdot \varepsilon_{jj} >> \varepsilon_{11} \cdot \varepsilon_{22} \cdot \varepsilon_{33}
therefore the formula. 300px
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.

 

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