# Would poisson's ratio also result in stresses in lateral directions?

This is kind of like a basic question which students might ask in their bachelor's level of education, which I want to refresh my knowledge about. So if I apply a load in a single longitudinal direction, say $$X$$ direction onto a material having a poisson's ratio of say 0.3, I know that it will also result in strains along the other lateral directions i.e. $$Y$$ and $$Z$$. But will it also create stresses in those directions i.e. $$\sigma_Y$$ and $$\sigma_Z$$? (assume no longitudinal force is applied in the $$Y$$ and $$Z$$ directions).

• Stresses are related to forces. All internal forces in the material must balance the external forces. Therefore, for homogeneous deformations, there cannot be unbalanced stresses in Y and Z. Jul 28 '21 at 23:20
• @BiswajitBanerjee, then it means either there are forces or there aren't any in the directions of Y and Z. Since I am not applying any contraint in the Y and Z directions, this makes me believe that I don't have any resistance to deformation in Y and Z directions. Hence there are no forces at all in these directions. Which basically implies that there are no stresses in these directions. Jul 29 '21 at 11:12
• That's correct. Jul 30 '21 at 4:31

It should be obvious that your examples will have stress components in the X and Y directions (assuming the structure is in the XY plane and the thickness is in the Z direction.)

Think about the boundary conditions around the curved edge. The stress component normal to the boundary must equal the external force normal to the boundary, which is zero.

So unless you think all the stress components at the curved boundary are zero (which doesn't make any sense - if it was true, cutting a hole in a plate would not cause a stress concentration around the hole but a reduction in stress around the hole!) there must be direct stress components in both X and Y, and also shear stress in the XY plane.

The previous two paragraphs have nothing to do with Poisson's ratio.

If Poisson's ratio is non-zero and there is a non-uniform stress field in the XY plane, there will be also a non-uniform strain field in the XY plane. Therefore (from Poisson's ratio) there will be non-uniform strains in the Z direction.

Therefore the top and bottom surfaces of the plate are no longer flat because the thickness is not uniform. The same argument as before shows there must be a stress component in the Z direction.

For a "thin" plate the stresses in the Z direction are small and can usually be ignored in practice, but if you make a detailed 3D model you will find they are not exactly zero.

IMHO there is not an easy way to answer this. The short way to answer is that if there is strain there is stress (and vice versa). However there are a few exceptions.

The most prominent example are thermally induced strains (that is something that I intended to write in the initial answer but I missed out). Thermally induced strains in an unconstrained object would not necessarily induce stresses (or so I understand and I would be happy to hear counterarguments).

Another example, I can think of, is that in continuum mechanics you can assume that a thin plate with forces acting parallel to the direction is in plane stress. e.g.

Figure 1: examples of plain stress (source: Haad)

However, what that really means is that the out of plane stress are negligible compared to the in plane stresses.

So, the way I interpret it, is that stresses develop in all direction when there is a strain involved, however, the exact magnitude is a matter of boundary constraints and loading.

• can you check out the comment I made under my post? When I replied to Biswajit Banerjee. Jul 29 '21 at 16:48

Not if the loading in the x-direction is uniform and causes uniform compression deformation in the x-direction.

It will just cause strain in the y and z-direction.

But if there is any constrain on the part not allowing it to expand laterally then there will be lateral stresses.

Here is a Link to article. in Wikiwand on the subject.

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