Linear Elastic Solid. Hookean Material

For a linear elastic solid or a Hookean material, it is assumed that the Cauchy stress is a linear function of the infinitesimal strain. That is,

$$\begin{split}&T_{11} = C_{1111}e_{11} + C_{1112}e_{12} + \cdo... ...11}e_{11} + C_{3312}e_{12} + \cdots C_{3333}e_{33}. \end{split}$$ (5.2.1)

We will assume that $T_{ij}$ is symmetric and since $e_{AB}$ is also symmetric, the above six equations can be written as

$$T_{ij} = C_{ijkl} e_{kl}$$ (5.2.2)

in which

$$C_{ijkl} = C_{jikl} = C_{ijlk}.$$ (5.2.3)

Since $T_{ij}$ and $e_{ij}$ are components of second order tensors, $C_{ijkl}$ are components of a fourth-order tensor. It is known as the elasticity tensor. It is this tensor which characterizes the mechanical properties of a particular anisotropic Hookean elastic solid. The anisotropy of the material is represented by the fact that the components of $C_{ijkl}$ are in general different for different choices of coordinate axes. If the body is homogeneous, that is, the mechanical properties are the same for every particle of the body, then $C_{ijkl}$ are constants (i.e. independent of position). We shall only study homogeneous bodies.

Because of the symmetry relations (5.2.3), the fourth order tensor $C_{ijkl}$ has 36 independent components. Thus, for a linear anisotropic elastic material, we need no more than 36 material constants to specify its mechanical properties.

It follows from (5.2.1) that whenever $e_{ij} = 0,\ T_{ij} = 0$. Thus, in the reference configuration, there is no stress. This implies that there are no initial stresses present.

A material is said to be isotropic if its mechanical properties can be described without reference to direction. That is, the components of the elasticity tensor $C_{ijkl}$ remain the same regardless of how the rectangular Cartesian coordinate axes are rotated. In other words,

$$C_{ijkl} = C^\prime_{ijkl}$$ (5.2.4)

under all orthogonal transformations of coordinate axes. A tensor having the same components with respect to every orthonormal basis is known as an isotropic tensor. An example of an isotropic tensor is $\delta_{ij}$. It is obvious that the following three fourth-order tensors are isotropic:

$$A_{ijkl} = \delta_{ij}\delta_{kl},\ B_{ijkl} = \delta_{ik}\delta_{jl},\ D_{ijkl} = \delta_{il} \delta_{jk}.$$

In fact, it can be shown that any fourth order isotropic tensor can be represented as a linear combination of the above three tensors. Thus, for an isotropic, linear elastic material, the stress-strain law (5.2.2) can be written as

$$T_{ij} = (\lambda A_{ijkl} + \mu B_{ijkl} + \beta D_{ijkl})e_{kl}.$$

In order for the symmetry relation (5.2.3)$_2$ to hold, $\beta = \mu$. Thus

$$T_{ij} =\ (\lambda\delta_{ij}\delta_{kl} + \mu\delta_{ik}\delta_{jl} + \mu\delta_{il}\delta_{jk})e_{kl},$$

$$=\ \lambda e_{kk} \delta_{ij} + 2\mu e_{ij}.$$ (5.2.5)

In the preceding equations $\lambda$ and $\mu$ are constants. Equation (5.2.5) is the constitutive equation for a linear elastic isotropic material. The two material constants $\lambda$ and $\mu$ are known as Lame's constants. Since $e_{ij}$ are dimensionless, $\lambda$ and $\mu$ are of the same dimensions as the stress tensor, force per unit area. For a given material, the values of Lamé's constants are to be determined from suitable experiments.

We now write (5.2.5) in a form usually studied in the Mechanics of Deformable Bodies course. Taking the trace of both sides of (5.2.5) we obtain

$$T_{ii} = (3\lambda + 2\mu )e_{kk}.$$ (5.2.6)

Assuming that

$$(3\lambda + 2\mu ) \ne 0,\ \mu \ne 0,$$ (5.2.7)

we get

$$e_{kk} = \frac{1}{(3\lambda + 2\mu)} T_{kk},$$

and hence

$$e_{ij} = \frac{1}{2\mu}\left[T_{ij} - \frac{\lambda}{3\lambda + 2\mu} T_{kk}\delta_{ij}\right]\ .$$ (5.2.8)

To get a physical interpretation of Lamé's constants in terms of Young's modulus $E$ and Poisson's ratio $\nu$, we note that, in a simple tension test, with the axial load $P$ applied along $X_1$-axis,

$$T_{ij} = \frac{P}{A}\delta_{i1}\delta_{j1}\ ,$$

where $A$ is the area of cross-section of the prismatic body. Also

$$e_{11} = \frac{T_{11}}{E},\ \nu = -\frac{e_{22}}{e_{11}}\ .$$ (5.2.9)

Substituting for $e_{11}$ and $e_{22}$ from (5.2.8) into (5.2.9) and assuming that $(\lambda + \mu ) \ne 0$ we arrive at

$$E = \frac{\mu (3\lambda + 2\mu )}{\lambda + \mu},\ \nu = \frac{\lambda}{2(\lambda + \mu )}\ .$$ (5.2.10)

The elimination of $\lambda$ from these two equations gives

$$\mu = \frac{E}{2(1 + \nu )}\ .$$ (5.2.11)

In order for a tensile load to produce extension in the direction of loading, it is clear from (5.2.9)$_1$ that $E$ must be $> 0$.

Another stress state, called simple shear, is the one for which all stress components except one pair of off-diagonal elements vanish. In particular, we choose $T_{12} = T_{21} \ne 0$. Equation (5.2.8) gives

$$e_{12} = \frac{T_{12}}{2\mu}\ .$$ (5.2.12)

In order for a simple shear force on $X_2$ plane to produce a sliding of the $X_2$ plane in the direction of the applied load, $e_{12}$ and $T_{12}$ must be of the same sign. For that to be true, $\mu$ must be $> 0$.

For $E$ and $\mu$ to be positive, it follows from (5.2.11) that $-1<\nu$. Now consider a stress state called, hydrostatic stress, for which $T_{ij} = -p\delta_{ij}$. For this case, equation (5.2.8) gives

$$e_{kk} = - \frac{1}{\left(\lambda + \frac{2}{3}\mu\right)}p\ .$$ (5.2.13)

Since $e_{kk} = \frac{dv-dV}{dV}$, therefore, in order for the hydrostatic pressure to produce a decrease in volume, we must have

$$\lambda + (2/3)\mu > 0\ .$$ (5.2.14)

Substituting in this equation from (5.2.10), we obtain

$$\frac{2\mu}{3}\left(\frac{1 + \nu}{1 - 2\nu}\right)>0\ .$$ (5.2.15)

On the assumption that $\mu > 0$, $\nu > -1$, the inequality (5.2.15) implies that $\nu$ must be $\le 1/2$. Rewriting (5.2.13) as

$$e_{kk} = -\frac{3}{2\mu}\frac{1 - 2\nu}{1 + \nu} p,$$ (5.2.16)

we see that for $\nu = 1/2,\ e_{kk} = 0$. That is, there is no change in volume. Since for incompressible materials $e_{kk} = 0$, therefore, $\nu = 1/2$ for incompressible materials.

Equations (5.2.10), (5.2.11) and (5.2.5) yield

$$T_{ij} = \frac{2\mu\nu}{1-2\nu} e_{kk}\delta_{ij} + 2 \mu e_{ij}\ .$$ (5.2.17)

For an incompressible material, the first term on the right-hand side of (5.2.17) is of the form % and hence is indeterminate. It is usually denoted by $-p\delta_{ij}$ and the constitutive relation for an incompressible linear elastic material becomes

$$T_{ij} = -p\delta_{ij} + 2\mu e_{ij}\ .$$ (5.2.18)

Here $p$ is called the hydrostatic pressure, and it can not be determined from the strain field. However, whenever surface tractions are prescribed on at least a part of the boundary, $p$ can be uniquely determined.

Example:

a)

For an isotropic Hookean material, show that the principal axes of stress and strain coincide.

b)

Find a relation between the principal values of stress and strain.

Solution: Note that eigenvectors of $T_{ij}$ and $e_{ij}$ are the principal axes of stress and strain respectively.

Let $\mathbf{n}^{(1)}$ be an eigenvector of $e_{ij}$ and $e^{(1)}$ be the corresponding eigenvalue. That is

$$e_{ij}n^{(1)}_j = e^{(1)}n^{(1)}_i\ .$$

By Hooke's law we have

$$T_{ij}n^{(1)}_j =\ (\lambda e_{kk}\delta_{ij} + 2\mu e_{ij})n^{(1)}_j\ ,$$

$$=\ \lambda e_{kk}n^{(1)}_i + 2\mu e_{ij} n^{(1)}_j\ ,$$

$$=\ \lambda e_{kk} n^{(1)}_i + 2\mu e^{(1)}n^{(1)}_i\ ,$$

$$=\ (\lambda e_{kk} + 2\mu e^{(1)})n^{(1)}_i\ . \ast$$

Therefore, $\mathbf{n}^{(1)}$ is also an eigenvector of $T_{ij}$ and the corresponding eigenvalue is $\lambda e_{kk} + 2\mu e^{(1)}$.

b) It is clear from eqn. ($\ast$) that the eigenvalue $T^{(1)}$ of $T_{ij}$ corresponding to the eigenvector $\mathbf{n}^{(1)}$ is $\lambda e_{kk} + 2\mu e^{(1)}$. Since $e_{kk} = e^{(1)} + e^{(2)} + e^{(3)}$, therefore,

$$T^{(1)} = \lambda (e^{(1)} + e^{(2)} + e^{(3)}) + 2\mu e^{(1)}\ .$$

Similarly,

$$T^{(2)} = \lambda (e^{(1)} + e^{(2)} + e^{(3)}) + 2\mu e^{(2)},$$

$$T^{(3)} = \lambda (e^{(1)} + e^{(2)} + e^{(3)}) + 2\mu e^{(3)}\ .$$

Exercise: (Recall the exercise given on page 2-11). For $W = \mu e_{ij}e_{ij} + \frac{\lambda}{2} (e_{kk})^2$, show that $\frac{\partial W}{\partial e_{ij}} = T_{ij}$.

Hint: It is advisable to do the problem long-hand. That is, first expand the given expression for $W$. Substitute $e_{21} = e_{12} = (e_{12} + e_{21})/2$ etc. in it and then carry out the differentiation with respect to $e_{11},\ e_{22},\ e_{12}$ etc.

$W$ is called the stored energy function or the strain energy density.