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Stress-strain relation material property

  1. Jul 20, 2006 #1
    when we normally see the "stress -strain" curve is having stress on y axis and strain on x axis. that mean stress is the
    response of the structure and strain is the input

    can i say the following

    stress -strain curve,stress in the x axis and strain in the y axis. because when i apply a force F through a material of cross section and then the strain occurs.
    Last edited by a moderator: Jul 20, 2006
  2. jcsd
  3. Jul 20, 2006 #2
    Stress strain relation. Is it a material property . Or is it the property of
    the specimen on which the tension test is done.

    i have a rod fix it at one end and pull the other end by distance x. So the
    specimen has a restoring force f. is this force acts throughout all the cross sections of the specimen.

    why do we divide the change in length/specimen length and call it as strain and

    why do we divide the restoring force/area and call it as stress

    and finally

    why do we find the relation between strain and stress and call it as material property . why can't we call it as the specimen's property
  4. Jul 20, 2006 #3


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    Although I can see how you came up with that idea, there is nothing more fundamental about strain as opposed to stress. They are both reactions of a material to an applied force and shouldn't be looked at as if one is a variable of the other. Nevertheless, we obviously have equations which equate one to the other.
    Stress and strain could be graphed inversely as you suggest (ie: stress on x, strain on y). Not sure what you're getting at here.
  5. Jul 20, 2006 #4


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    Yes, stress and strain is a property of both the specimen and the material. Note that most metals have some kind of heat history as well as chemical make up, such that even two samples with the same chemical makeup can have very different material properties (except modulus).
    These are things we want to know. They give us the information needed to design something to. It wouldn't make sense for example, to apply a load and measure the change in length divided by the force since such a measurement, although it would be useful for the specimen, would need to be altered for use on any other specimen. Knowing the modulus of elasticity or the yield strength of a material are fundamental properties of any identical material.

    Note the relation between stress and strain gives us modulus of elasticity. We're not interested in the strain of a specimen exactly, we're interested in the identical properties of a given material.
  6. Jul 20, 2006 #5


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    Stricly speaking, no.

    The stress that appears in the stress-strain curve is meaningful only if we it is a material property. Thus, we are plotting the stress developed in the specimen in response to the applied stress. But there is no good way to measure the induced crystalline stress. So, the only way to do this, is to ensure that the induced stress is equal to the applied stress, and measure the applied stress (which is easy to do). However, this equality holds only when the specimen is in equilibrium, or more accurately, a sufficiently good pseudo-equilibrium, meaning that the strain-rate must be fairly small.

    So, with this new picture, we understand the stress-strain curve to represent the nature of the stress induced in a material when it is subjected to different strains.

    The reason for using stress and strain as opposed to say, force and elongation is precisely that you want to study material properties and not specimen properties. In reality, however, this is only approximately true (that the stress-strain relationship is independent of geometry) - or else we wouldn't need rigorous ASTM guidelines for measuring say, the flexural modulus of SS 304.
  7. Jul 20, 2006 #6


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    Thermomechanics doesn't impose that many restrictions on how to understand and define stress & strain (or the direction you wish to view them from). Pretty much just obey the fundamental laws of thermodynamics and in process properties such as being work/power conjugate will result. Much of the make-up of continuum mechanics arises from such thermomechanical principles (internal & state variables, state functions, normality, .... along the general fundamental ones & balance laws of course) and history how the whole "thing" has been build up - naturally for example typical strain tensors arise quite understandably when studying deformation, the story behind the kinetic (constitutive) relationship and definition of stress ("need" for stress :biggrin: ) being a tad more complex, but a complete 'defined' construct anyways (perhaps a better word would be somewhere between defined and derived).
  8. Aug 7, 2006 #7
    i understand it like this. any material when subjected to loading will
    reorganise itself in to a new shape and then apply a counteracting load
    this reorganisation is called strain

    and the response is the stress
  9. Aug 7, 2006 #8


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    When an object is loaded, there is a force imbalance on the atoms of the material. The loading force exceeds the restoring electrostatic forces resulting in relative motion (ie:strain). As the strain increases, the restoring force also increases till it equals the applied force. At this point we have equilibrium. We now record the strain in the member and the stress required to attain equilibrium, and we have a data point for our stress-strain curve.

    Think about the stress as coming from the developed restoring force rather than the applied force. Think about the harmonic oscillator, as the basis for this. For any displacement x, of the oscillator, we can calculate the restoring force, [itex]F(x) = -m\omega ^2x [/itex]. That makes it easier to distinguish the dependent and independent variables.
  10. Aug 25, 2006 #9
    i understood the stress part. I a axial loaded member how come a shear stress occur?
  11. Aug 25, 2006 #10
    Both the state of stress and strain in a solid are described bysecond-rank symmetric Cartesian tensors whose 9 (6 independent) components vary with the coordinate axes that are chosen and (maybe) also from place to place in the solid.

    If an axial or tensile load is applied to a "member", say to a uniform rod, along whose axis a coordinate (say z of a Cartesian system) is chosen to lie, the stress tensor will have only one component, namely the zz component and it won't vary from place to place. Nice and simple.

    But if the coordinate axes are chosen to be at say at 45 degrees to the rod's axis, the state of stress referred to these axes will include shear stresses as well as "normal" (like xx,yy or zz component) stresses, although again they won't vary from place to place. More complicated.

    The state of applied stress is uniquely determined by the loads that are applied to the member. The way you describe the stress can be simple (as in the first case above) -- or complicated (the second case) if you happen to choose axes that don't match the symmetry of the member that is stressed.

    Strain is more complicated. And there are also frozen-in internal stresses to be considered. Solids are in practice full of them. But all this is another long story!
  12. Sep 8, 2006 #11
    strain energy

    why do we study strain energy and strain energy density in a mechanics course. what is the application?
  13. Sep 8, 2006 #12


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    There are applications in strength theories. For example, the 4th theory of strength (Beltrami, Haigh) states that two elements consisting of the same material, indipendent of the stress state, are in the same condition if their specific potential energies of strain are equal.
    In this case, the specific potential energy of strain serves as a connection between a single-axial and a three-axial stress state, where, for the single axial state it is given with [tex]u=\frac{\sigma \epsilon}{2}[/tex] (1), and for the three-axial state [tex]u=\sum_{i=1}^3 \frac{\sigma_{i} \epsilon_{i}}{2}[/tex] (2). (note: sigma represents stress, and epsilon represents strain) If you use Hooke's Law ( [tex]E=\frac{\sigma}{\epsilon}[/tex] ) on (1) and (2), and if you express (1) as [tex]\sigma_{equivalent}=\sqrt{2Eu}[/tex], and put (2) expressed with u into (1), you get an equivalent stress equation according to the 4th strength theory [tex]\sigma_{equivalent}=\sqrt{\sigma_{1}^2+\sigma_{2}^2+\sigma_{3}^2-2\nu(\sigma_{1}\sigma_{2}+\sigma_{2}\sigma_{3}+\sigma_{3}\sigma_{1})}[/tex]. ([tex]\nu[/tex] is Poisson's number)Why would we use this? Well, when we test materials in the lab, we actually run single-axial tests. So, for a three-axial stress state, we can find it's equivalent single-axial stress. Sadly, this theory isn't applied very often because the lack of experimental proof, but I hope it was illustrative.
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