# 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:

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

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

Another pattern is used to calculate stresses:

$\tau&space;=\frac{M_{s}}{I_{s}}*\rho$

where:
$M_{s}$ – 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:

$Is=\frac{\pi&space;*r^{4}}{2}=\frac{\pi&space;*D^{4}}{32}$

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

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

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

$\varphi&space;=\frac{M_{s}*L}{G*I_{s}}$

where:
$M_{s}$ – 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.

Data:
$G&space;=&space;83,00&space;GPa&space;=&space;83&space;*&space;10^{6}&space;kPa$
$k_{s}&space;=&space;40,00MPa&space;=&space;40&space;*&space;10^{3}kPa$

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

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

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

1. Stress value in section 1-1

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

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

1. Torsion angle

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