1
$\begingroup$

The book I'm referring to says that a shaft with a circular cross-section in pure torsion will have its cross-sections flat during the loading. The cross sections won't deform; they will rotate.

It then says that this is not the case for non-circular cross-sections. In a non-circular c/s shaft, the c/s distort and are not flat during the loading.

I wanted to know why that is so.

$\endgroup$
2
  • $\begingroup$ Check out lemon shafts used in pto $\endgroup$
    – Solar Mike
    Commented Dec 7, 2021 at 9:49
  • $\begingroup$ Same reason we don't use square wheels. Relative to adjacent cross sections, a rotated circle is the same. A rotated square is not. $\endgroup$
    – DKNguyen
    Commented Dec 7, 2021 at 14:16

3 Answers 3

3
$\begingroup$

We say and write what we observe. In history, when an axisymmetric cross-sectional beam (i.e. the beam's cross-section still remains the same when rotated about its longitudinal centroidal axis) was subjected to torsion, the cross-sections of the beam remained plane/flat. This is true for both, solid beam and hollow beam. However, when a non-axisymmetric cross-sectional beam (i.e. the beam's cross-section DOESN'T remain the same when rotated about its longitudinal centroidal axis) was subjected to torsion, it was observed that the cross-sections didn't remain plane/flat. In fact, they warped when subjected to twisting. The general equation for calculating stresses due to pure torsion for a solid circular beam is shown below:

enter image description here

where T: Torque applied, r: distance from the centroid, J: Polar moment of inertia. Now, the derivation of this equation involves an assumption that the plane section must remain plane, i.e. the cross-section cannot warp. Therefore, this equation cannot be applied to any other non-axisymmetric cross-sectional beams.

Warping basically refers to one-half of the cross section (above/below the neutral axis) being subjected to compression, and the other half to tension. This behavior is what we usually see in a beam subjected to bending, however, this behavior is also observed in non-axisymmetric cross-sections subjected to torsion only. Calculating shear stresses for a non-circular cross-section is somewhat more complex and complicated than the circular ones.

$\endgroup$
1
  • $\begingroup$ No, J is not the polar moment of inertia, it is the torsional constant. For axisymmetric circular shapes it happens to be the same number as the polar moment of inertia. $\endgroup$ Commented Jun 10 at 21:55
3
$\begingroup$

not so much a formal answer (more like an answer for the intuition), is that during torsion on non circular shafts, the stresses tend to deform the material in such a way that the cross-section resembles more that of a circular.

In order for that to happen , material needs to be deformed/pulled from adjacent section. The end result is that it contracts.

$\endgroup$
3
$\begingroup$

A non-circular section under the torsion deforms in both St Venant shear and also by warping.

For example, an I beam deforms both by warping and shear deformation. Warping creates tension and compression in flanges. so the plane of the flange does not remain flat anymore.

source

warping of open section

.

warping

$\endgroup$

Not the answer you're looking for? Browse other questions tagged or ask your own question.