This information is provided as a complimentary service to Granta users and visitors to our website.
Find out more about our materials information management and reference data products 


Torsion of Shafts

A torque, T, applied to the ends of an isotropic bar of uniform section, and acting in the plane normal to the axis of the bar, produces an angle of twist q. The twist is related to the torque by the first equation on Fig. A5, in which G is the shear modulus. For round bars and tubes of circular section, the factor K is equal to J, the polar moment of inertia of the section, defined in Appendix A1. For any other section shape K is less than J. Values of K are given in Fig. A1.

If the bar ceases to deform elastically, it is said to have failed. This will happen if the maximum surface stress exceeds either the yield strength sy of the material or the stress at which it fractures. For circular sections, the shear stress at any point a distance r from the axis of rotation is

[Equation]

The maximum shear stress, tmax, and the maximum tensile stress, smax, are at the surface and have the values

[Equation]

If tmax exceeds sy/2 (using a Tresca yield criterion), or if smax exceeds the modulus of rupture (MOR), the bar fails, as shown on Fig. A5. The maximum surface stress for the solid ellipsoidal, square, rectangular and triangular sections is at the points on the surface closest to the centroid of the section (the mid-points of the longer sides). It can be estimated approximately by inscribing the largest circle which can be contained within the section and calculating the surface stress for a circular bar of that diameter. More complex section-shapes require special consideration and, if thin, may additionally fail by buckling. Helical springs are a special case of torsional deformation. The extension of a helical spring of n turns of radius R, under a force F, and the failure force Fcrit, are given on Fig. A5.

Opens in a new window

Figure A5 Torsion of Shafts