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Suppose I pull a bar at both ends and there’s no movement in the bar. Now strain in this case is 0. But I know the stress is nonzero (right?) However strain is linearly related to stress, implying stress must be 0. Could someone point out what my misunderstanding is?

And let’s assume this is a simple material, and we aren’t concerning temperature or gravity. Make the system as simple as possible, as I’m taking a my first continuum mechanics course so I’m not super understanding with everything. I’d like to understand how strain can be 0, but stress is non-zero despite the linear relationship between the two.

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Thermal expansion of a constrained object can produce nonzero stress with zero strain, for example. This occurs because generalized Hooke’s Law contains a thermal expansion term. But pulling a stable unconstrained and initially unloaded solid will always produce a nonzero strain, as the elastic moduli are always positive.

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  • $\begingroup$ Huh really. I guess for metals, the Young’s modulus should be huge…which makes sense as in my exercise problems it’s usually written in terms of gigapascels. Thank you for clearing that up. Just for clarification, if the strain truly was 0, then there’d be zero stress, right? $\endgroup$
    – Spencer Kraisler
    Commented Nov 26, 2021 at 19:39
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    $\begingroup$ Yes, in a single direction, at steady state, barring any other effects (thermal expansion, piezoelectricity, magnetostriction, etc.). $\endgroup$
    – Chemomechanics
    Commented Nov 26, 2021 at 19:49
  • $\begingroup$ The reason for the uniaxial restriction is that I can pull on a material to cancel out Poisson contraction from loading of one or both other axes and obtain zero strain even with nonzero stress. $\endgroup$
    – Chemomechanics
    Commented Nov 27, 2021 at 0:53
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Suppose I pull a bar at both ends and there’s no movement in the bar. Now strain in this case is 0. But I know the stress is nonzero (right?)

You simply didn't pull hard enough for the stress to cause measurable strain for the given material modulus of elasticity.

I’d like to understand how strain can be 0, but stress is non-zero despite the linear relationship between the two.

If a bar is placed between two fixed members and is heated, it will be prevented from expanding axially that would otherwise occur due to thermal expansion. This creates an internal thermally induced axial stress with no axial strain. For a discussion, see this https://www.engineeringtoolbox.com/stress-restricting-thermal-expansion-d_1756.html

Hope this helps.

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  • $\begingroup$ That helps a lot. If I apply a force or stress to the ends, then there must be non-zero strain. I guess that explains why the Young’s modulus for metals are usually in terms of gigapascels. $\endgroup$
    – Spencer Kraisler
    Commented Nov 26, 2021 at 19:44
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    $\begingroup$ Exactly. Would you expect to see strain for a steel bar if you pulled on it with your hands? Maybe if you are Superman! $\endgroup$
    – Bob D
    Commented Nov 26, 2021 at 19:45
  • $\begingroup$ I guess this explains why we don’t write equations in terms of strain, but in stress, and why we measure stress but not strain in lab settings. As we’d be dealing with very very tiny strains. $\endgroup$
    – Spencer Kraisler
    Commented Nov 26, 2021 at 19:47
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    $\begingroup$ @SpencerKraisler But we do measure strain in lab settings. How else can we determine Young's Modulus for materials? It's just that you need tensile test equipment capable of applying a sufficient axial load to produce measurable strain. $\endgroup$
    – Bob D
    Commented Nov 26, 2021 at 19:58
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    $\begingroup$ First of all strain is units/units. You are describing the total deformation. Though I am not conversant in tensile testing, it is my understanding that a tensile tester can go as high as 600 kN. ResearchGate.net specifies dogbone specimen is 20 mm$^2$ by 50 mm long in middle. Carbon steel has an E of about 21 x 10$^{10}$ N/m$^2$. So if I did my math right, that would be a strain of about 0.145 mm/mm which for an original length of 50 mm would would theoretically give a maximum deformation of about 7.25 mm, unless that is beyond the elastic limit. $\endgroup$
    – Bob D
    Commented Nov 26, 2021 at 20:32

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