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The book I'm referring to says that a shaft with a circular cross section in pure torsion will have it's cross sections remain flat during the loading. That is, the cross sections won't deform, they will just 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 is that so?

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  • $\begingroup$ Check out lemon shafts used in pto $\endgroup$
    – Solar Mike
    Dec 7 '21 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
    Dec 7 '21 at 14:16
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We say and write what we observe. In the history, when an axisymmetric cross sectional beam (i.e. the beam's cross section still remains the same when rotated about its longitudinal centeroidal 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-axisymmteric cross sectional beam (i.e. the beam's cross section DOESN'T remain the same when rotated about its logitudinal centeroidal axis) was subjected to torsion, it was observed that the cross sections didn't remain plane/flat. Infact, they wrapped 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 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 cross section (above/below the neutral axis) being subjected to compression, and 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.

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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.

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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

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warping

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