Strength of Materials/Torsion

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Circular Shaft under Torsion

Torsion acting on a long bar tends to twist it in the direction of the torque. The analysis is performed on a section perpendicular to the axis, and the sum of the internal resisting torque is set equal to the external torque acting on the system. The following assumptions will be made for the bars studied in this chapter:

  1. Plane sections remain plane after the torque is applied.
  2. The shear strain γ varies linearly in the radial direction.
  3. The material is linearly elastic, so that Hooke's law applies.

Torsion FormulaEdit

We want to find the maximum shear stress τmax which occurs in a circular shaft of radius c due to the application of a torque T. Using the assumptions above, we have, at any point r inside the shaft, the shear stress is τr = r/c τmax.

∫τrdA r = T

∫ r2/c τmax dA = T

τmax/c∫r2 dA = T

Now, we know,

J = ∫ r2 dA

is the polar moment of inertia of the cross sectional area..

Thus, the maximum shear stress

τmax = Tc/J

The above equation is called the torsion formula.

Now, for a solid circular shaft, we have,

J = π/32(d)4

Further, for any point at distance r from the center of the shaft, we have, the shear stress τ is given by

τ = Tr/J This formula cannot be correct. Dimensional analysis:

 - τ must be in N/mm2
 - Tr is in N x mm
 - J is in mm4
 - The result of the equation is in N x mm / mm4 = N / mm3
 - Instead of J, I think we should use Zp.  This would give the correct result in dimensional analysis.  See here (make a correct address) .

We only consider the torsional loading of simple circular shafts in this analysis, ie cylinders or non-eccentric tubes without splits. Circular shafts are most commonly used as torque carrying members in machinery with rotating parts (like drive shafts of motors). This is fortuitous, as the analysis of non circular members under torsion is not simple to perform analytically.

Angle of TwistEdit


The above image shows the twist of a shaft acted upon by a torque T at one end. We know that the shear angle γ is given by

γ = τ/G

For a shaft of radius c, we have

φ c = γ L

where L is the length of the shaft. Now, τ is given by

τ = Tc/J

so that

φ = TL/GJ

The angle of twist φ for a circular shaft acted upon by a torque Tx at a point x along its axis is given by:


where Jx is the moment at section x. Note that you can use the torsion formula for shafts with slowly varying area as long as they are circular.