Introduction
The
traction vector, \({\bf T}\), is simply the internal force vector
on a cross-section divided by that cross-section's area.
\[
{\bf T} = { {\bf F}_{\text{internal}} \over \text{Area} }
\]
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 in the x-direction. So
\[
{\bf F}_{\text{internal}} = 4,000 \, {\bf i} \, \text{N}
\]
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.
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
\({\bf F} = 4,679 \, {\bf i} + 7,486 \, {\bf j} + 3,440 \, {\bf k} \, \text{N}\).
Forces on Cross Sections
Before closing, recall one more time that the force on a cross-section is
\[
{\bf F} \; = \; \int {\bf T} \, dA \; = \; \int \boldsymbol{\sigma} \cdot {\bf n} \, dA
\]
Both forms turn up often in literature.
