3.2.3 Stress Tensor

When a solid is strained by external forces the positions between atoms change and it is in a non-equilibrium state. As a result some internal forces appear. These forces tend to return the strained solid to its equilibrium state. The induced forces are called the internal stresses. If a solid is not strained, the internal stresses vanish.

Considering some volume of a solid the total force applied to it can be written as an integral over the same volume:

$$\int\vec{F}\,dV,$$ (3.21)

where $\vec{F}$ is the force applied to a unit volume of the solid. As the forces between internal parts of the volume are balanced, only external parts contribute. Further the macroscopic principle^3.2 means that the total force can only be expressed as a surface integral.

Therefore each of the three components of the total force (3.21) should be a surface integral. This fact implies the that vector $\vec{F}$ must be the divergence of a tensor of rank two. Denoting this tensor as $\sigma_{ik},\thickspace i,k=1,2,3$, $\vec{F}$ can be expressed as:

$$F_{i}=\frac{\partial\sigma_{ik}}{\partial x_{k}}.$$ (3.22)

From this expression the following formula is obtained:

$$\int F_{i}\,dV=\oint\sigma_{ik}\,df_{k},$$ (3.23)

which is only valid for the weak strain condition^3.3.

The tensor $\sigma_{ik}$ is called the stress tensor. As it can be seen from (3.23), $\sigma_{ik}df_{k}$ is the $i$-th component of the force applied to the surface element $d\vec{f}$. Choosing a surface element in the $x,y$; $y,z$ and $x,z$ planes it can be seen that the component $\sigma_{ik}$ of the stress tensor is the $i$-th component of the force applied to a unit area perpendicular to the $x_{k}$ axis. For example, the normal component of the force applied to a unit area perpendicular to $x$ is equal to $\sigma_{xx}$ and the forces tangential to it are equal to $\sigma_{yx}$ and $\sigma_{zx}$ as it is shown in Fig. 3.6.

Figure 3.6: The stress tensor components in terms of the applied forces.

It is well known that the moment of force can be written as an antisymmetric tensor of rank two^3.4. Thus the total moment of force applied to the volume under consideration is:

$$M_{ik}=\int(F_{i}x_{k}-F_{k}x_{i})\,dV.$$ (3.24)

Like the total force the moment of force also must be expressed through a surface integral. Substituting (3.22) into (3.24) gives:

$$M_{ik}=\oint(\sigma_{il}x_{k}-\sigma_{kl}x_{i})df_{l}+\int(\sigma_{ki}-\sigma_{ik})dV.$$ (3.25)

This expression represents a surface integral only if the following equality is satisfied:

$$\sigma_{ik}=\sigma_{ki},$$ (3.26)

which removes the volume integral in (3.25). Hence an important property of the stress tensor is its symmetry. In principle the integrand in the volume integral in (3.25) can be a complete divergence of a tensor of rank three which is antisymmetric in the first two indices. This tensor must be expressed through derivatives like $\partial u_{i}/\partial x_{k}$. Thus the stress tensor would contain terms of higher order derivatives of the displacement vector $\vec{u}$. Within this work such terms are considered negligible. Hence the symmetry of the stress tensor is again justified.

It should be noted that the stress tensor can be transformed into diagonal form even without this approximation. The point is that its definition (3.22) is not unique. Any transformation

$$\widetilde{\sigma}_{ik}-\sigma_{ik}=\frac{\partial\chi_{ikl}}{\partial x_{l}},$$ (3.27)

where the tensor $\chi_{ikl}$ is antisymmetric with respect to the last two indices, does not change the force $\vec{F}$. S. Smirnov: