1
$\begingroup$

Lets I have a material with ultimate tensile strength $S_{UTS}$.

Lets say $S_{UTS}$ is produced by strain $\epsilon_{UTS}$

My question is, if the material is strained such that it achieves $\epsilon_{UTS}$ over may days if not weeks of time, will it simply plastically deform or will it break?

EDIT: the material is a metal

$\endgroup$

4 Answers 4

2
$\begingroup$

I think its the other way around. I.e. the tensile testing takes place in an quasi static (relatively slow rate). This means that you shouldn't expect (usually) larger strains to failure.

However, when you are starting to compress time (i.e. do things faster), then you have adiabatic processes, inertial effects and/or stress waves that take place and lead to (usually) higher failure stresses and lower failure strains.

As I tried to stress, to my knowledge this is what usually happens. There might be exceptions for specific cases of materials (auxetic or hyperelastic) that behave in a different way. However, for most structural engineering materials that is not the case.


This is purely a conjecture (based on the different strain rate law that exist), however I feel that the change has to do with the log of the time.

Lets assume a tensile test would last about 100 sec to failure so the log10 should be about 2. Dropping to 10 seconds is a large change in terms of logarithm (log10 is 1). However, going the other way (e.g. to log10 3) means increasing from 1 min 40 seconds to approximately 16 min 40 sec. Going even higher to log10=4 would mean about 3 hours.

$\endgroup$
2
$\begingroup$

This is not the answer but to remind you that you need to identify the material of concern (as not all materials have the same failure mode), and the UTS on the curve (whether it is the engineering stress-strain curve or the true stress-strain curve). The best bet is predicting its behavior from its own stress-strain curve.

enter image description here enter image description here enter image description here

$\endgroup$
1
$\begingroup$

Generally not a metal; possibly with a polymer. If the metal is tested above its recrystallization temperature, such as lead at room temperature, possibly recrystallization of strained material with time, might permit higher stress.

$\endgroup$
1
$\begingroup$

In most engineering materials, the duration of applying tension is not going to affect the stress-strain curve.

The plastic range you mention will lead to strain hardening regardless of the time rate and the stress increases due to the narrowing of the specimen till the specimen gets to ultimate tensile strength.

There are differences in the curve though due to material ductility or brittleness not the time of application of tension.

Here is a quote from Wiki.

The second stage is the strain hardening region. This region starts as the stress goes beyond the yielding point, reaching a maximum at the ultimate strength point, which is the maximal stress that can be sustained and is called the ultimate tensile strength (UTS). In this region, the stress mainly increases as the material elongates, except that for some materials such as steel, there is a nearly flat region at the beginning. The stress of the flat region is defined as the lower yield point (LYP) and results from the formation and propagation of Lüders bands. Explicitly, heterogeneous plastic deformation forms bands at the upper yield strength and these bands carrying with deformation spread along the sample at the lower yield strength. After the sample is again uniformly deformed, the increase of stress with the progress of extension results from work strengthening, that is, dense dislocations induced by plastic deformation hampers the further motion of dislocations. To overcome these obstacles, a higher resolved shear stress should be applied. As the strain accumulates, work strengthening gets reinforced, until the stress reaches the ultimate tensile strength.

The third stage is the necking region. Beyond tensile strength, a neck forms where the local cross-sectional area becomes significantly smaller than the average. The necking deformation is heterogeneous and will reinforce itself as the stress concentrates more at small section. Such positive feedback leads to quick development of necking and leads to fracture. Note that though the pulling force is decreasing, the work strengthening is still progressing, that is, the true stress keeps growing but the engineering stress decreases because the shrinking section area is not considered. This region ends up with the fracture. After fracture, percent elongation and reduction in section area can be calculated.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.