Tensile Test

The Specimen

Standardization

Specimens are standardized to allow for consistent results across labs.

Specimens are said to be proportional when gauge length $L_0$ is proportional to the square root of the area of the cross-section $A_0$.

Circular Cross-Section

$$\begin{array}{rl}& L_0=5d\text{ Rounded to nearest 5mm}\\ & L_0+\frac{d}{2}<L_C<L_0+2d\end{array}$$

Rectangular Cross-Section

$$\begin{array}{rl}& L_0=5.65\sqrt{A_0}\text{ Rounded to nearest 5mm}\\ & L_0+1.5\sqrt{A_0}\le L_C\le L_0+2.5\sqrt{A_0}\end{array}$$

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The Setup

The specimen is slowly pulled, measuring the elongation $\Delta L=L-L_0$.

Loading rate matters!

We therefore have the following standard rates for steel and aluminium respectively (units in $Nm{m}^{-2}{s}^{-1}$):

$$\begin{array}{rl}& 6\le \frac{d\sigma }{dt}\le 30\\ & 2\le \frac{d\sigma }{dt}\le 10\end{array}$$

Note that in the test, we have stress in one direction, and ideally, this should be the maximum principal stress ${\sigma }_0$, therefore giving the stress tensor:

$$[\sigma ]=\left[\begin{array}{cccc}{\sigma }_1& 0& 0& \\ 0& 0& 0& \\ 0& 0& 0& \end{array}\right]$$

If we want to test out maximum tangential stress, we apply the tension at a $45^{\circ }$ angle with slip planes, leading to:

$${\tau }_{\text{max}}=\frac{{\sigma }_1-{\sigma }_3}{2}$$

This is the maximum shear stress given by the fractured plane. This formula can be visualized using Mohr’s Circle: the highest point on the graph is the top of the circle spanning from the minimum stress to the maximum stress.

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Results

From any point we can get strain percent:

$$\epsilon =\frac{\Delta L}{L_0}=\frac{L_i-L_0}{L_0}$$

We can also get poisson’s ratio by using:

$${\epsilon }_x={\epsilon }_y=-\nu {\epsilon }_z$$

For isotropic materials.