Search Continuum Mechanics Website

\[ {\bf T} = { {\bf F} \over \text{Area} } \]

So \({\bf T}\) has units of stress, like

\[ {\bf F} = 4,000 \, {\bf i} \, \text{N} \]

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.

\[ \sigma = {\bf T} \cdot {\bf n} \qquad \text{and} \qquad \tau = {\bf T} \cdot {\bf s} \]

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.

\[ {\bf T} = {1 \over \text{462 mm}^2} \;\; \text{4,000}\,{\bf i}\,\text{N} = 8.66 \, {\bf i} \, \text{MPa} \]

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

\[ \tau = {\bf T} \cdot {\bf s} = (8.66, \; 0, \; 0) \cdot (-\sin 30^\circ, \; \cos 30^\circ, \; 0) = -4.33 \, \text{MPa} \]

\[ \sigma_{xx} \, A \, \cos \theta + \tau_{xy} \, A \, \sin \theta = T_x \, A \] \[ \tau_{xy} \, A \, \cos \theta + \sigma_{yy} \, A \, \sin \theta = T_y \, A \]

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.

\[ \boldsymbol{\sigma} = \left[ \matrix{ 50 & 10 & 30 \\ 10 & 95 & 20 \\ 30 & 20 & 15 } \right] \]

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

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

\[ \boldsymbol{\sigma} = \left[ \matrix{ 50 & 10 & 30 \\ 10 & 95 & 20 \\ 30 & 20 & 15 } \right] \]

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

\[ \begin{eqnarray} {\bf n} \times {\bf s} \; & = & \; (0.400 \, {\bf i} + 0.600 \, {\bf j} + 0.693 \, {\bf k}) \times (-0.832 \, {\bf i} + 0.555 \, {\bf j} + 0.000 \, {\bf k}) \\ \\ & = & -0.385 \, {\bf i} - 0.576 \, {\bf j} + 0.721 \, {\bf k} \\ \end{eqnarray} \]

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

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.

\[ \tau \; = \; {\bf s} \cdot \boldsymbol{\sigma} \cdot {\bf n} \; = \; \boldsymbol{\sigma} : ({\bf s} \otimes {\bf n}) \]

(The same could also be done to compute the normal stress as well.)

This diadic product for shear arises so often in metal plasticity that it is represented by the single letter \({\bf p}\), and named the Schmidt tensor after the engineer who studied metal plasticity in the early 1900's.

\[ {\bf p} \; = \; {\bf s} \otimes {\bf n} \; = \; \left[ \matrix{ s_1 n_1 & s_1 n_2 & s_1 n_3 \\ s_2 n_1 & s_2 n_2 & s_2 n_3 \\ s_3 n_1 & s_3 n_2 & s_3 n_3 } \right] \]

\[ {\bf F} \; = \; \int {\bf T} \, dA \; = \; \int \boldsymbol{\sigma} \cdot {\bf n} \, dA \]

Both forms turn up often in literature.

Also, please consider visiting an advertiser on this page. Doing so helps generate revenue to support this website.

Bob McGinty

bmcginty@gmail.com

Click here to see a sample page in each of the two formats.

Click here to see a sample page in each of the two formats.

Copyright © 2012 by Bob McGinty