2
$\begingroup$

What is the difference of engineering and true stress? Also, why does the engineering stress decrease after the ultimate tensile strength point?

$\endgroup$

1 Answer 1

1
$\begingroup$

In tensile testing, Stress is usually measured indirectly by measurement of the applied force over strain $\epsilon$.

If you divide that force $F$ by the cross-section of your specimen at the start of testing, $A_0$, you gain a value $\sigma_e$ with the dimension of a stress. $$\sigma_e=\frac{F}{A_0}$$ This is the engineering stress.

If, instead of $A_0$, you use the cross section corresponing with the elongation of the Probe, $A$, you get the true stress $\sigma$. $$\sigma=\frac{F}{A}$$

$\epsilon$ is defined by $$\epsilon=\frac{\Delta L}{L_0}$$ with $\Delta L$ being the Elongation and $L_0$ being the starting length. Because the solid material of the specimen is incompressible, its Volume $V$ has to stay constant in spite of strain. $$V=L_0*A_0=\int _0^L A(x)dx=const.$$

As a result of an increase in $L$ with constant $V$, A is changing throughout the whole Experiment. Even before reaching ultimate tensile strength, $\sigma$ differs from $\sigma_e$.

For many ductile materials we see the development of a constriction at a random point on the specimen. The reason for this is that the minimum-energy direction of plastic deformation is in a direction at an angle of 45° in relation to the direction of $F$. Now, at this constriction point, $A$ is drastically reduced which results in a large $\sigma$. $\sigma_e$, on the other hand, usually declines, due to the reduction in necessary $F$. This is the behaviour you've described in your question.

Engineering Stress is a measure for the applied force during tensile testing, rather than the actual stress. However, for many applications with elastic behaviour, it is deemed "close enough". The main reasons for this are the increased cost in having $A$ measured additionally and that the remaining uncertainty of testing can be higher than the difference between $\sigma$ and $\sigma_e$ (before reaching ultimate tensile strength).

In forming technology, though, true stress is very relevant, since the main interest is plastic behaviour, where larger differences between engineering stress and true stress appear.

$\endgroup$
2
  • $\begingroup$ Except that solids are not incompressible and the volume changes under stress (See bulk modulus). The fact is, that it is nearly incompressible as extension in one direction results in contraction in other directions, but $$\frac{\Delta V}{V} \approx \frac{\sigma}{E} (1-2 \nu) $$ $\endgroup$ Dec 29, 2016 at 15:41
  • $\begingroup$ @ja72 Initially, I was going to write nearly incompressible but I decided to leave it out in order to avoid confusion. $\endgroup$
    – John Steed
    Dec 29, 2016 at 18:23

This site is temporarily in read-only mode and not accepting new answers.

Not the answer you're looking for? Browse other questions tagged .