When we twist a shaft/beam/rod, shearing stresses are produced. The angle by which the end of the shaft rotates can be found as:

$${\displaystyle \theta ={\frac {TL}{GJ}}}$$

where $T$ is torque, $L$ is the length, $G$ the shear modulus, and $J$ a torsion constant depending on the geometry of the cross section. For circular cross section, $J$ equals the second moment of area. For non-circular shafts there is an effect called "warping" present where the cross sections don't remain planar during twisting; if we had a beam with a rectangular cross section and held it radially away from us and twisted it, the cross sections would bend towards us/away from us. Do I understand warping correctly?

But shafts with circular cross sections do not experience this. Planar cross sections remain planar, and deform only in 2D. Why is this?

  • $\begingroup$ Without digging too much into the mechanics, I might start by looking into the distribution of shear stress along the perimeter of the shape. For a circle section, the shear stress is uniform all around the perimeter. For angular sections the distance to the neutral axis for shear is changing and thus the shear stress is change. The must be some coupling of this in plane shear stress with the out of plane deformation. Something interesting is happening on a 3d stress cube this much is for sure. $\endgroup$
    – ShadowMan
    Commented Nov 15, 2019 at 20:59

2 Answers 2


Let's imagine a solid cylinder attached to a fix support at one end and free at other. We twist it at the other end by a torque.

It will rotate about it's axis without warping, ( small angles condition).

Now we cut the cylinder with laser longitudinally in 4 cuts into 8 triangle sections, radiating from the center of the circle, so if you look at the cross section it look like a circle with 4 diagonal lines at 45 degrees.

Now if we twist this cylinder with the same torque we see the triangles wind around each other like a cable and the end of them is not a plane any more. it looks like a concrete drill bit. The circular triangles have warped and if you measure the strain half of each triangle is shortened and half elongated.

The reason is the strain energy stored into this axial deformation is more than what would be if there was only shear strain.

Now if you weld the seams back together you deny the triangles the opportunity to wind about. So the plane of the rotated cylinder remains plane and the strain stored is just shear.


The reason is symmetry.

For a circular rod under torsion, if you rotate the rod through any angle the warped shape has to look the same, because there is no "special" point around the circumference where you can start measuring angles from.

So, if the rod warps, the warped shape can only vary with the radius across the section.

Similarly, the warping must be the same at every cross section along the length of the rod.

Now consider what the two ends of the rod might look like. Suppose you twist the rod with a clockwise torque, and one end of the rod warps to bulge outwards. The other end must therefore be bulging inwards. Now suppose you turn the rod round, end-to-end. The torque on the rod is still in the same clockwise direction, but now the inward and outward bulges are reversed.

There is no physical reason why the rod should bulge one way and not the other since it is perfectly symmetrical and uniform. So the only possible "warped shape" that satisfies all these conditions is no warping at all.

Of course you can reach the same conclusion using 3D continuum mechanics, but learning how to deduce things using symmetry instead of solving systems of partial differential equations is a valuable practical tool in many situations!


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