Introduction
The
traction vector, \({\bf T}\), is simply the force vector
on a cross-section divided by that cross-section's area.
\[
{\bf T} = { {\bf F} \over \text{Area} }
\]
So \({\bf T}\) has units of stress, like
MPa, but it is absolutely
a vector, not a stress tensor. So all the usual rules for vectors
apply to it. For example, dot products, cross products, and coordinate
transforms can be applied.
Calculating a Traction Vector
The object below has a 400 mm
^{2} cross sectional area and is being
pulled in tension by a 4,000 N force (red) in the x-direction.
So the (arbitrarily chosen) rightward pointing internal
force vector (blue) is
\[
{\bf F} = 4,000 \, {\bf i} \, \text{N}
\]
It is cut (virtually) so the traction vector is
\[
{\bf T} = \left( 1 \over \text{400 mm}^2 \right) \text{4,000}\,{\bf i}\,\text{N} = 10.0 \, {\bf i} \, \text{MPa}
\]
Note the units are MPa now rather than N.
This time, the object is cut (virtually again) at a 30° angle.
The applicable area in this case is
\[
A = {\text{400 mm}^2 \over \cos(30^\circ) } = \text{462 mm}^2
\]
And the traction vector is
\[
{\bf T} = \left( 1 \over \text{462 mm}^2 \right) \text{4,000}\,{\bf i}\,\text{N} = 8.66 \, {\bf i} \, \text{MPa}
\]
Note the direction of the traction vector is always the same as the internal force vector.
Only its magnitude changes with cut angle.
Normal and Shear Stresses
Normal and shear stresses are simply the components of the traction vector
that are normal and parallel to the area's surface as shown in the figure.
Using \({\bf n}\) for the unit normal vector to the surface, and
\({\bf s}\) for the unit vector parallel to it, means that
\[
\sigma = {\bf T} \cdot {\bf n} \qquad \text{and} \qquad \tau = {\bf T} \cdot {\bf s}
\]
It's very important to recognize that \(\sigma\) and \(\tau\) here are
each scalar values, not full tensors. This is the natural result of
the dot product operations involving \({\bf T}\), \({\bf n}\), and
\({\bf s}\). (Dot products produce scalar results.)
The normal and shear stress values here are scalars rather than tensors because
they are only two individual components of the full stress tensor.
Also, note that in 3-D, there are in fact an infinite number of \({\bf s}\)
vectors parallel to the surface, each having a different component in-and-out
of the page, so to speak. This is why it is common to specify one
parallel to the page and a second perpendicular to it.
Normal and Shear Stress from a Traction Vector
Recall that the traction vector from the above example was
\[
{\bf T} = {1 \over \text{462 mm}^2} \;\; \text{4,000}\,{\bf i}\,\text{N} = 8.66 \, {\bf i} \, \text{MPa}
\]
The unit normal to the surface is
\[
{\bf n} = (\cos 30^\circ, \; \sin 30^\circ, \; 0)
\]
So the normal stress on the surface is
\[
\sigma = {\bf T} \cdot {\bf n} = (8.66, \; 0, \; 0) \cdot (\cos 30^\circ, \; \sin 30^\circ, \; 0) = 7.5 \, \text{MPa}
\]
The vector parallel to the surface is
\[
{\bf s} = (-\sin 30^\circ, \; \cos 30^\circ, \; 0)
\]
The shear stress on the surface is
\[
\tau = {\bf T} \cdot {\bf s} = (8.66, \; 0, \; 0) \cdot (-\sin 30^\circ, \; \cos 30^\circ, \; 0) = -4.33 \, \text{MPa}
\]
Stress Tensors and Traction Vectors
The relationship between the traction vector and stress state at a
point results directly from setting the sum of
forces on an object equal to zero, i.e., imposing equilibrium.
\[
\sigma_{xx} \, A \, \cos \theta + \tau_{xy} \, A \, \sin \theta = T_x \, A
\]
\[
\tau_{xy} \, A \, \cos \theta + \sigma_{yy} \, A \, \sin \theta = T_y \, A
\]
The area, \(A\), cancels out of both sides leaving
\[
\sigma_{xx} \, \cos \theta + \tau_{xy} \, \sin \theta = T_x
\]
\[
\tau_{xy} \, \cos \theta + \sigma_{yy} \, \sin \theta = T_y
\]
but \(\cos \theta\) and \(\sin \theta\) are the
components of the unit normal to the surface,
\({\bf n} = (\cos \theta, \sin \theta)\), that
\({\bf T}\) is acting on.
Replacing the \(\cos \theta\) and \(\sin \theta\) with \(n_x\) and \(n_y\) gives
\[
\sigma_{xx} \, n_x + \tau_{xy} \, n_y = T_x
\]
\[
\tau_{xy} \, n_x + \sigma_{yy} \, n_y = T_y
\]
Both equations can be summarized as
\[
{\bf T} = \boldsymbol{\sigma} \cdot {\bf n}
\]
or in tensor notation as
\[
T_i = \sigma_{ij} \, n_j
\]
The above equations are very useful, compact, matrix and tensor notation
representations of the equilibrium equations. The full equations, in 3-D,
are
\[
\sigma_{xx} \, n_x + \tau_{xy} \, n_y + \tau_{xz} \, n_z = T_x
\]
\[
\tau_{yx} \, n_x + \sigma_{yy} \, n_y + \tau_{yz} \, n_z = T_y
\]
\[
\tau_{zx} \, n_x + \tau_{zy} \, n_y + \sigma_{zz} \, n_z = T_z
\]
The tensor notation term, \(\sigma_{ij} \, n_j\), leads to
nine separate stress components. For example, both \(\sigma_{xz}\)
and \(\sigma_{zx}\) are present above, and both are always equal.
This is in fact common in all equations involving stress and strain.
Traction Vector from Stress Tensor
Given the stress tensor (in MPa)
\[
\boldsymbol{\sigma} =
\left[ \matrix{
50 & 10 & 30 \\
10 & 95 & 20 \\
30 & 20 & 15 }
\right]
\]
Calculate the traction vector on a surface with unit normal
\({\bf n} = (0.400, \, 0.600, \, 0.693)\).
\[
\left\{ \matrix{
T_x \\ T_y \\ T_z }
\right\}
=
\left[ \matrix{
50 & 10 & 30 \\
10 & 95 & 20 \\
30 & 20 & 15 }
\right]
\left\{ \matrix{
0.400 \\ 0.600 \\ 0.693 }
\right\}
=
\left\{ \matrix{
46.79 \\ 74.86 \\ 34.40 }
\right\}
\]
So \({\bf T} = 46.79 \, {\bf i} + 74.86 \, {\bf j} + 34.40 \, {\bf k} \, \text{MPa}\).
If the area is 100 mm
^{2}, then the force on it would be
\({\bf F} = 4,679 \, {\bf i} + 7,486 \, {\bf j} + 3,440 \, {\bf k} \, \text{N}\).
Stress Transforms
This section introduces an aspect of coordinate transformations of stress tensors
that is a subset of the general case, which comes later. It does so by
combining different equations involving the traction vector.
Recall that the normal and shear stresses on a surface are related
to the traction vector by
\[
\sigma = {\bf T} \cdot {\bf n} \qquad \text{and} \qquad \tau = {\bf T} \cdot {\bf s}
\]
Recall that the normal and shear stresses here are just scalar quantities on the surface,
not a full stress tensor.
But we also saw that the traction vector is related to the full stress tensor by
\[
{\bf T} = \boldsymbol{\sigma} \cdot {\bf n}
\]
Substituting this equation for \({\bf T}\) into the above ones gives
\[
\sigma = {\bf n} \cdot \boldsymbol{\sigma} \cdot {\bf n} \qquad \text{and} \qquad
\tau = {\bf s} \cdot \boldsymbol{\sigma} \cdot {\bf n}
\]
In tensor notation, the equations are
\[
\sigma = \sigma_{ij} \, n_i \, n_j \qquad \text{and} \qquad
\tau = \sigma_{ij} \, s_i \, n_j
\]
These represent very useful relationships between the stress tensor in the global
coordinate system and the normal and shear stress components at
any other
orientation.
Stress Transform Example
Recall the above stress tensor
\[
\boldsymbol{\sigma} =
\left[ \matrix{
50 & 10 & 30 \\
10 & 95 & 20 \\
30 & 20 & 15 }
\right]
\]
We had calculated the traction vector on a surface with unit normal
\({\bf n} = (0.400, \, 0.600, \, 0.693)\). This time, calculate
the normal and shear stresses on this surface.
The normal stress on the surface is
\[
\begin{eqnarray}
\sigma
& \; = \; &
\matrix{
\left\{ 0.400 \;\; 0.600 \;\; 0.693 \right\} \\
\\
\\
}
\left[ \matrix{
50 & 10 & 30 \\
10 & 95 & 20 \\
30 & 20 & 15 }
\right]
\left\{ \matrix{
0.400 \\ 0.600 \\ 0.693 }
\right\}
\\
\\
& = &
87.47 \text{ MPa}
\end{eqnarray}
\]
In order to compute a shear stress, we first need a specific one of an infinite
number of unit vectors parallel to the surface. Let's choose
\({\bf s} = (-0.832, \, 0.555, \, 0.000)\). A dot product will verify that
this vector is perpendicular to \({\bf n}\).
\[
\begin{eqnarray}
\tau
& \; = \; &
\matrix{
\left\{ -0.832 \;\; 0.555 \;\; 0.000 \right\} \\
\\
\\
}
\left[ \matrix{
50 & 10 & 30 \\
10 & 95 & 20 \\
30 & 20 & 15 }
\right]
\left\{ \matrix{
0.400 \\ 0.600 \\ 0.693 }
\right\}
\\
\\
& = &
2.62 \text{ MPa}
\end{eqnarray}
\]
So there is very little shear on this face
in the given s direction.
But this doesn't mean that there is no shear on the face at all. To see this, choose
a second direction parallel to the surface and perpendicular to the first
s.
Obtain this by crossing the unit normal vector with the first tangential vector.
\[
\begin{eqnarray}
{\bf n} \times {\bf s} \; & = & \; (0.400 \, {\bf i} + 0.600 \, {\bf j} + 0.693 \, {\bf k})
\times (-0.832 \, {\bf i} + 0.555 \, {\bf j} + 0.000 \, {\bf k}) \\
\\
& = &
-0.385 \, {\bf i} - 0.576 \, {\bf j} + 0.721 \, {\bf k} \\
\end{eqnarray}
\]
So the shear in the direction perpendicular to the first is
\[
\begin{eqnarray}
\tau
& \; = \; &
\matrix{
\left\{ \text{-}0.385 \;\; \text{-}0.576 \;\; 0.721 \right\} \\
\\
\\
}
\left[ \matrix{
50 & 10 & 30 \\
10 & 95 & 20 \\
30 & 20 & 15 }
\right]
\left\{ \matrix{
0.400 \\ 0.600 \\ 0.693 }
\right\}
\\
\\
& = &
-36.33 \text{ MPa}
\end{eqnarray}
\]
So there is a good bit of shear stress in this perpendicular direction.
And the negative value indicates that it is in the direction
opposite of the
s direction.
Transformation Tip
This transformation "trick" could be used to compute the normal and shear
stresses on all six faces of a cube at any random orientation, and in
the process, perform a complete coordinate transformation of a stress tensor.
But it's actually easier to do
\( \boldsymbol{\sigma}' = {\bf Q} \cdot \boldsymbol{\sigma} \cdot {\bf Q}^T\)
just as is the case for strain tensors.
But the opposite is also true. The stress tensor can be replaced with
the strain tensor to obtain
\[
\epsilon_{\text{normal}} \; = \; {\bf n} \cdot \boldsymbol{\epsilon} \cdot {\bf n} \qquad \text{and} \qquad
\gamma / 2 \; = \; {\bf s} \cdot \boldsymbol{\epsilon} \cdot {\bf n}
\]
Or in tensor notation as
\[
\epsilon_{\text{normal}} \; = \; \epsilon_{ij} \, n_i \, n_j \qquad \text{and} \qquad
\gamma / 2 \; = \; \epsilon_{ij} \, s_i \, n_j
\]
This works because since both stress and strain are tensors, then any
math operation that applies to one also applies to the other.
For a quick math review, note that \(s_i \, n_j\) in the above equations
can be interpreted as a diadic product of the two vectors, \({\bf s} \otimes {\bf n}\).
And then this result is "