Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Questions regarding Tensile Testing of metal

  1. Jul 24, 2012 #1
    Can anyone explain me how the load is applied to a metal specimen during tensile test?
    I am confused because I read in my textbook that after the ultimate stress is achieved and if the load is continued, then the metal specimen fails. But from the stress-strain diagram, we can see that the stress is decreased (the y axis) after ultimate tensile strength. Why is it decreased if we maintain the same stress?
    To be clear my question is that is the load applied in Universal Testing Machine is constant or not? I mean is the load uniform over time?
    thanks in advance.
  2. jcsd
  3. Jul 24, 2012 #2
    Well you certainly are confused because you have said to opposite things as highlighted above.

    No matter it is an easy thing to understand.

    No machine is designed to apply a stress. It is designed to apply a load (force), usually by a hydraulic mechanism.

    This load is shown on a dial or electronic readout. The readout may be calibrated in stress units but that then leads to only being true for a particular size of specimen so force units are more usual.

    The hydraulic pump increases the force on the specimen from zero at a specific rate of increase, by pulling on the ends on the specimen.

    At this happens the specimen stretches longitudinally in the direction of the load
    and also becomes thinner in cross section (have you heard of poisson?).

    Eventually we find that the stretching of the specimen continues but our force increase mechanism can no longer increase the force - in fact the load dial drops back a bit.
    We have just past the ultimate stength.

    At this time the specimen will have developed a very pronounced 'neck' which is a thinning of the cross section. Continued pulling will cause the specimen to break at this neck.

    If you measure the cross sectional area of the neck and divide the failure load by this area you will find that the stress did not fall after all.

    You should look up or ask your tutor about the difference between 'engineering stress' and 'true stress'.
  4. Jul 24, 2012 #3
    Now it's clear. Thank you very much.
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook