17.1.1 Tensor Definition

When defining the state of strain at a point it is meaningful to define the strain tensor in terms of the deformation (or displacement) tensor

$$e_{ij}= \left( \begin{array}{ccc} e_{xx} & e_{xy} & e_{xz} \\... ...artial y} & \frac{\partial w }{\partial z} \end{array}\right).$$ (17.5)

In general the deformation tensor is composed of a strain tensor and a rotation tensor as follows:

 

$$\epsilon_{ij} + w_{ij},$$ e_ij =

$$\frac{1}{2}(e_{ij}+e_{ij}) + \frac{1}{2}(e_{ij}-e_{ij}).$$ e_ij = (17.6)

Thus from the tensor description the strain is a symmetric second rank tensor of the following form:

$$\epsilon_{ij}= \left( \begin{array}{ccc} \epsilon_{xx} & \eps... ...artial z}) & \frac{\partial w}{\partial z} \end{array}\right).$$ (17.7)

The strain tensor defined in equation () is of the same form as the stress tensor defined in equation (). Therefore, as described by equations (), () and () for stress the strain tensor can also be decompose d into a volumetric component, dilation, and a component associated with a change of shape. Also a similar triaxial state of strain can be described in terms of the principal strains, thus allowing allowing an effective strain to be defined. This is particularly useful in strain hardening as described in the previous section, when an effective plastic strain $\epsilon_{{\it eff}}^p$ is often required and can be defined in terms of the invariant of the strain tensor as follows:

$$\epsilon_{{\it eff}}^p = \sqrt{\frac{2}{3}} \{ \epsilon_{ij}^p \epsilon_{ij}^p \}^{\frac{1}{2}}.$$ (17.8)

The strain tensor component of the deformation tensor is associated constitutively with the stress tensor as follows:

$$\sigma_{ij} = C_{ijkl} \epsilon_{kl}$$ (17.9)

where C_ijkl is the fourth rank tensor of elastic constants. As the stress and strain tensors are symmetric and the material can be assumed to be isotropic and homogenous, the independent components of the tensor of elastic constants can be reduced significantly. This allows equation () to be simplified to

$$\sigma_{ij} = 2\mu \epsilon_{ij} + \lambda \epsilon_{kk} \delta_{ij}$$ (17.10)

where $\mu$ and $\lambda$ are the Lamé constants, which can be defined in terms of the Youngs modulus E and the Poisson ratio $\nu$ as

$$\begin{eqnarray*}\mu & = & \frac{E}{2(1+\nu)}, \\ \lambda & = & \frac{\nu E}{(1+\nu)(1-2\nu)}. \end{eqnarray*}$$

The Lamé constant $\mu$ is equivalent to the shear modulus G, there is no direct physical equivalent for the Lamé constant $\lambda$. The constitutive relationship can be decomposed into deviatoric and hydrostatic components, respectively,

$$\begin{eqnarray*}s_{ij} & = & \frac{E}{1+\nu} {\epsilon_{ij}}' = 2G{\epsilon_{ij... ...igma_{ii} & = & \frac{E}{1-2\nu}\epsilon_{kk} = 3K\epsilon_{kk}. \end{eqnarray*}$$

where ${\epsilon_{ij}}'$ is the deviatoric strain and K is the bulk modulus.