## Axial Stress and Strain

June 27, 2015

Published in Reference, Strength of Materials

## Normal Stress

Stress characterizes the internal forces within a body, where forces act between one particle and the next within the crystal structure. A stressed body can be studied by visualizing the load path through the part. The definition of a tensile stress depends only on force and area, and as such it represents a load density within the material. Consider a round bar as shown below, which has a concentrated tensile load P acting on small regions at the center of both end faces.

Since the load originates and terminates at small localized regions, the stress field is non-uniform throughout the length of the bar, where it is more concentrated at the points of load application and wider at the midspan of the bar, and might look something like the sketch shown below.

If we take a cut at section B-B, we find that the reaction force R over this area must be equal to the applied load P. This implies that the sum of all the internal forces across the section B-B must be equal to the external load P. The tensile stress computed using statics can only give us an average value of stress:

\[ \sigma_{ave} = \frac{P}{A} \]

This stress value is an average value and resembles the stress in the bar at regions not near the ends (approximately equal to or greater than the width of the member, per Saint-Venant principle). If we divide the cut surface B-B into lots of very small areas, the stress at a point Q can be found by taking the limit of the force over the area at Q as the area goes to 0:

\[ \sigma_Q = \lim_{\Delta A \to 0}\frac{\Delta R}{\Delta A} \]

If the force acts in the direction of the bar but on a sloped plane, normal and shear stresses can be found on the surface by:

\[ P_N = P\cos{\theta} \]

\[ P_T = P\sin{\theta} \]

\[ \sigma_N = \frac{P_N}{A_{\theta}} = \frac{P\cos{\theta}}{\frac{A}{\cos{\theta}}} = \frac{P}{A}cos^2{\theta} \]

\[ \tau_N = \frac{P_T}{A_{\theta}} = \frac{P\sin{\theta}}{\frac{A}{\cos{\theta}}} = \frac{P}{A}\sin{\theta}\cos{\theta} \]

Plotting the normal and shear components versus face rotation angle, as a multiple of P/A:

## Normal Strain

Strain characterizes the deformation of a body under load. For axial loading of a bar with constant cross section, strain is defined by the ratio of the deformation or extension of the bar to the original length:

\[ \epsilon = \frac{L-L_0}{L_0} = \frac{\delta}{L_0} \]

For a bar of constant cross section, the extension or change in length under axial loading can be found by:

\[ \delta = \epsilon L \]

\[ E = \frac{\sigma}{\epsilon} \]

\[ \delta = \frac{\sigma}{E}L_0 \]

\[ \delta = \frac{P/A}{E}L_0 = \frac{PL_0}{AE} \]

For a bar of non-constant cross section, the total deformation can be found by dividing up the length into infinitely small elements of width dx, then integrating over the length of the bar to find the total deformation. For example, a bar with a tapered shape as shown below.

At any point the infinitesimal strain is given by:

\[ d\delta = \frac{Pdx}{AE} \]

The total strain for the whole bar is then the sum of these small strains:

\[ \delta_{tot} = \int_0^L d\delta \]

Therefore,

\[ \delta_{tot} = \int_0^L \frac{P(x)}{A(x)E(x)} dx = \frac{P}{E}\int_0^L\frac{1}{A(x)}dx \]

\[ A(x) = \pi y^2(x) = \pi(mx+b)^2 \]

\[ \delta_{tot} = \frac{P}{E\pi} \int_0^L \frac{1}{(mx+b)^2} dx \]

\[ \epsilon_{tot} = \frac{\delta_{tot}}{L_0} \]