Free twisting of round bars

In this guide I will explain what it means to free twisting bars, give all the necessary formulas for calculating a bar with a circular or annular cross-section and at the same time I will carry out calculations for an exemplary task.

Free twisting – it is a torsion, in which the deplanations (warping) of neighboring cross-sections are free, that is, they do not press themselves. There is no normal stress.

In order for the rod to undergo free twisting, the need for a suitable shape, a proper method of fixing and proper application of loads.

Before starting to calculate an example bar, I will give all the necessary patterns. Let’s start with the formula for determining the diameter of the rod, it looks like this:

D=\sqrt[3]{\frac{16*M_{s}}{\pi *k_{s}}}

M_{s} – torque moment
k_{s} – allowable stresses
\pi – 3.14 …

Another pattern is used to calculate stresses:

\tau =\frac{M_{s}}{I_{s}}*\rho

M_{s} – torque moment
I_{s} – moment of inertia when turning. Note, I will also place patterns for this because they depend on the section. Another is for circular and another for annular.

Formula for the moment of inertia when turning for a circular cross-section:

Is=\frac{\pi *r^{4}}{2}=\frac{\pi *D^{4}}{32}

Formula for the moment of inertia when screwing for the ring section:

Is=\frac{\pi }{2}*(R^{4}-r^{4})

The last pattern has been left. This is a formula for the angle of twisting the rod, it looks like this:

\varphi =\frac{M_{s}*L}{G*I_{s}}

M_{s} – torque moment
L – length
I_{s} – moment of inertia when turning

So much theory, now it’s time for some counting. I will choose the diameter of the rod, calculate the stresses in section 1-1 and the total angle of torsion on the cross-section below.
It is a steel rod with circular cross-section, fixed along its length.

G = 83,00 GPa = 83 * 10^{6} kPa
k_{s} = 40,00MPa = 40 * 10^{3}kPa


We start calculating the given bar according to the following algorithm.

  1. Designation of Mr
    -M_{r} + M_{1} = 0
    M_{r} = 1,50kNm
  1. Determining the torque graph (internal forces)The entire length of the rod
    M_{s} = 1,50kNm
  1. Diameter determination
    D_{1} = \sqrt[3]{\frac{16,00 * 1,50}{(\Pi * 40,00 * 10,00^3)}} = 0,0576m = 57,60mm

    We \; \; accept\; \; D_{1} = 60,00mm = 0,06m

  1. Stress value in section 1-1

I_{s} = (\frac{\Pi * 0,06 ^ 4}{32,00}) = 1,27 * 10^{-6} m^{4}



T_{1-1} =\frac{(1,50 * 0,03)}{(1,27 * 10^{-6})}= 35,43 MPa



  1. Torsion angle

\varphi = \frac{1,50 * 0,40}{83,00 * 10^{6} * 1,27 * 10^{-6}} = 0,0057 \; \; radian.