This page reviews the familiar stress tensor.
Stress is always simply \(Force / Area\),
but some complexity does arrise because the relative orientation
of the force vector to the surface normal dictates the type
of stress. When the force vector is normal to the surface,
as shown to the right, the stress is called normal stress and
represented by \(\sigma\).
When the force vector is parallel to the surface, the stress is called
shear stress and represented by \(\tau\). When the
force vector is somewhere in between, then
its normal and parallel components are used as follows.
This page is a near-duplicate of the earlier stress.html page in the
Introductory Mechanics section.
If you have read that page, then this one can be skipped.
The earlier page served as the complete discussion of stress because it was in the
Introductory Mechanics section (keyword here being
Introductory). This time, this page is only the introduction to a full chapter
on the subject. The reason being that, this time, we will worry about the complications
that arise when large deformations and rotations are present.
For example, when large deformations are present, does one use the initial or deformed area
to calculate stress? And if a part rotates 90° such that a force originally in the x-direction
ends up acting in the y-direction, then should the corresponding stress be \(\sigma_{xx}\) or
\(\sigma_{yy}\)?
Component Definitions
Many components are needed to capture a complete stress state.
Consider the object in the 2-D example here that is being pulled in
simple tension, though not in a direction parallel to any global axis.
Standard practice is to (virtually) cut it perpendicular to the global
axes as shown. This first cut results in an area with unit normal
parallel to the global x-axis. The force on this area contains both
normal and parallel components. The stresses are defined as
\[
\sigma_{xx} = {F_x \over A_x} \qquad \text{and} \qquad \tau_{xy} = {F_y \over A_x}
\]
Note how the two subscripts on the stress variables match those on the force
and area components with one subscript coming from each.
Alternately, one could (virtually) cut the object horizontally to produce a
surface with an outward normal in the y-direction. This leads to
\[
\sigma_{yy} = {F_y \over A_y} \qquad \text{and} \qquad \tau_{yx} = {F_x \over A_y}
\]
If a numerical example were worked out, one would notice an amazing result.
It is that \(\tau_{xy} = \tau_{yx}\). This will always be true
in order to maintain rotational equilibrium. This is discussed
in more detail next.
Equilibrium
The complete (2D) stress state at a point is shown below. The key difference
between the left and right figures is the shear stresses. But they will be discussed
later.
First, let's look at the normal stresses, \(\sigma_{xx}\) and \(\sigma_{yy}\).
Note how the x-normal stress, \(\sigma_{xx}\), is present on
both the left and right sides of each square in order to maintain horizontal
equilibrium. These x-normal stresses represent tension because they point out of the
square. Tensile normal stresses have positive values, and compressive normal stresses
have negative values.
The y-normal stresses, \(\sigma_{yy}\), are also present on two surfaces, top and bottom,
in order to maintain vertical equilibrium. Like \(\sigma_{xx}\), \(\sigma_{yy}\) is also
drawn to represent tension, which is positive.
The difference between the left and right pictures is that \(\tau_{yx}\) in the
left figure is replaced by \(\tau_{xy}\) in the right figure.
The left figure contains two shear stress values, \(\tau_{xy}\), which rotates
the square counter-clockwise, and \(\tau_{yx}\), which rotates the square clockwise.
But if the two shear values are not equal, then the square will not be in rotational
equilibrium. The only way to maintain rotational equilibrium is for \(\tau_{xy}\)
to be equal to \(\tau_{yx}\). So there is no need to have two separate variables.
The right figure contains only one, \(\tau_{xy}\).
2-D Notation
Stress is in fact a tensor. Why? Because it obeys standard coordinate
transformation principles of tensors. This alone appears to be enough
to make it so. It can be written in any of several
different forms as follows. They are all identical.
\[
\boldsymbol{\sigma} =
\left[ \matrix{
\sigma_{11} & \sigma_{12} \\
\sigma_{21} & \sigma_{22} }
\right]
=
\left[ \matrix{
\sigma_{xx} & \sigma_{xy} \\
\sigma_{yx} & \sigma_{yy} }
\right]
=
\left[ \matrix{
\sigma_{xx} & \tau_{xy} \\
\tau_{yx} & \sigma_{yy} }
\right]
\]
But since \(\tau_{xy} = \tau_{yx}\), all the tensors can also be written as
\[
\boldsymbol{\sigma} =
\left[ \matrix{
\sigma_{11} & \sigma_{12} \\
\sigma_{12} & \sigma_{22} }
\right]
=
\left[ \matrix{
\sigma_{xx} & \sigma_{xy} \\
\sigma_{xy} & \sigma_{yy} }
\right]
=
\left[ \matrix{
\sigma_{xx} & \tau_{xy} \\
\tau_{xy} & \sigma_{yy} }
\right]
\]
Setting \(\tau_{xy} = \tau_{yx}\) has the effect of making (requiring in fact)
the stress tensors symmetric.