Rationale For Plain Strain Assumption in Long or Infinite Bars

Question

When plane strain is discussed, there are two examples given, both of which involve two dimensional boundary stress conditions. One is a constrained object that cannot move along the third axis. From Johnston and Beer:

This occurs in a plate subjected to uniformly distributed loads along its edges and restrained from expanding or contracting laterally by smooth, rigid, and fixed supports

The other is the infinite bar (or very long bar far from the ends). Again from Johnston and Beer:

It is also found in a bar of infinite length subjected to uniformly distributed loads on its sides, because by reason of symmetry, the elements located in a transverse plane cannot move out of that plane. This idealized model shows that a long bar subjected to uniformly distributed transverse loads is in a state of plane strain in any given transverse section that is not located too close to either end of the bar.

Another example can be seen on Wikipedia. As you can see, this appeals to the long / infinite nature of the bar. It does not appeal to something like friction along the bar length.

The first one is obvious. The object is constrained by a fixture to prevent third axis movement, and so it cannot. Since the boundary stresses imposed on the object are two dimensional, there is no reason to break the symmetry. In this case, the third axis of stress to resist the Poisson ratio effect is imposed by the fixture.

The second is not so obvious. The same third axis stress is still required to satisfy the Poisson ratio requirement, but where it originates from is not obvious.

Personally, I don't accept the hand wavy appeal to "symmetry" due to the bar being very long (nearly infinite). If transverse compression is imposed on an infinite bar, it seems no less reasonable to me to assume positive longitudinal strain (due to the Poisson ratio) rather than negative longitudinal stress (again due to the Poisson ratio), as is being claimed by all the textbooks.

The situation is even worse for the long, but finite bar. In this case, there is an end, even if it is far relative to the transverse dimensions. Since (near the center), we have a longitudinal stress (by the plane strain assumption) and at the ends, we do not have longitudinal stress, there must be a transition from no longitudinal stress to the constant value needed to satisfy the Poisson ratio. Thus, there must be some portion near the ends of the bar that has an axial normal stress gradient. There is really only one way this can happen according to the equilibrium stress differential equations. There needs to be a longitudinal shear stress gradient that offsets the longitudinal normal stress gradient.

To me, it seems more intuitive that if the ends of the bar have a zero stress condition, then the whole length of the bar should have zero longitudinal stress. This would require the bar to have some longitudinal strain, as required by the Poisson ratio effect.

Does anyone know of an analytic solution to this problem, or at least have an intuitive explanation for the plane strain condition?

Please see some other (inadequate) answers from over the years:

Physics Stack Exchange (this one appeals to friction from the application of the boundary condition)

Research Gate (This one just repeats the statement without justifying it)

Answer 1

The constraint holding the axial strain to nearly zero doesn't need to be applied by an external factor such as a clamp or friction. It can be applied by the object itself.

Consider a long bar or rod with lateral tensile stress applied to the top and lateral compressive stress applied to the bottom. Because of Poisson effects, one side would tend to exhibit axial elongation, and the other side would tend to exhibit axial contraction.

If we conceptually split the bar lengthwise to produce two unbonded halves subject to these tendencies, because of the long length, one half would elongate appreciably, and the other would contract appreciably. Small strains have led to large displacements—unacceptable opposing displacements, in fact.

Because we don't have two unattached halves—we have one bar that needs to retain its single ends of approximately the original cross-sectional shape. We thus idealize the axial strain as being exactly zero, at least away from the ends, for ease of analysis. This constitutes an internal constraint that generally makes the system stiffer.

Of course, if this idealization isn't appropriate for your case(s) of interest, don't use it!

Answer 2

Any nonuniform axial strain distribution leads to absurd distortions of the cross-section in case of infinite bar, so the axial strain distribution in infinitely long bar must be uniform. Plane strain is usually defined with one normal strain component being zero, although there is also a variation where this strain is not zero, but just uniform (special case of even more general "generalised plane strain").