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Pipe Stress engineers seem to only be concerened with lengthwise expansion. Why is this?

When something expands due to temperature increase, does poisson's ratio apply? Or is poisson's ratio only due to an external force being applied?

Mike K
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  • do you have a link to support your statement? – aaaaa says reinstate Monica Apr 08 '17 at 06:03
  • no. I went on an interview for finite element analysis of refinery pipes today. We spent hours looking at a program that visually shows the expansion of pipes. All we discussed was the problems that length increase causes. – Mike K Apr 08 '17 at 06:08
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    If a pipe has a diameter of 0.1 metres and a length of 100 metres and their percentage increase in size due to a temperature rise is the same which is going to increase in length more? What will be the consequences of each of these expansion? – Farcher Apr 08 '17 at 07:12
  • @Farcher or if you consider railroad track beam, which change size with temperature flactuations – aaaaa says reinstate Monica Apr 08 '17 at 17:11

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Lengthwise expansion is much more important than lateral expansion because pipes are far longer than they are wide. But even if lateral expansion is negligible, it still generally occurs, mediated by the coefficient of thermal expansion and possibly also the Poisson effect, depending on the constraints. A comparison of two examples should make this clear.

The temperature-dependent generalized Hooke's Law for a solid isotropic material can be written as $$\epsilon_{ij}=\frac{1+\nu}{E}\sigma_{ij}+\left(\alpha\Delta T-\frac{\nu}{E}\sigma_{kk}\right)\delta_{ij}$$

which is just a shorthand way of writing $$\epsilon_{11}=\frac{1}{E}\sigma_{11}-\frac{\nu}{E}\sigma_{22}-\frac{\nu}{E}\sigma_{33}+\alpha\Delta T$$ $$\epsilon_{22}=\frac{1}{E}\sigma_{22}-\frac{\nu}{E}\sigma_{11}-\frac{\nu}{E}\sigma_{33}+\alpha\Delta T$$ and $$\epsilon_{33}=\frac{1}{E}\sigma_{33}-\frac{\nu}{E}\sigma_{11}-\frac{\nu}{E}\sigma_{22}+\alpha\Delta T$$ plus three constitutive equations for shear strain.

In the lateral direction, with no stresses applied, we have $$\epsilon_{22}=\alpha\Delta T$$

which doesn't contain Poisson's ratio. So as you might intuit, the strain in every direction is equal (because Nature doesn't have a preferred direction) and originates solely from thermal expansion. (Of course, since the deformation is proportional to both the strain and the geometry, the lateral deformation might still be negligible relative to the axial deformation.)

For a rod constrained in the lengthwise direction only, $\epsilon_{11}=0$, and we can solve for $\sigma_{11}$ to obtain $$\epsilon_{22}=(1+\nu)\alpha\Delta T$$ In this case, Poisson's ratio enters the expression because of the axial constraint.

Because the cross section of a pipe is mostly empty space (i.e., the thin-wall assumption), the lateral strain of the pipe as a whole is going to correspond much more closely to the first $\epsilon_{22}$ expression than the second. In other words, the constraint causes the walls to get thicker than they would otherwise be, but the width of the entire pipe cross section is relatively insensitive to $\nu$.

In fact, the thermal expansion of the width of the entire pipe corresponds to the circumferential expansion of the pipe walls (imagine cutting the pipe down its length on one side and flattening out the circular cross section to form a plate that is stress-free on its top and bottom sides and on two edges).

For very thick walls, however, the pipe will start to resemble a solid material, and Poisson's effect may no longer be negligible.

Chemomechanics
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