Before you read this, I suggest you read post 21.21.
If you’ve read post 2.21, you may wonder why anyone would be interested in torsion of a cylinder. The reason is that long, thin cylinders (rods) are used as shafts to transfer rotational motion in many different types of machines. The drivetrain of a car (the system that transfers rotation of the engine to rotation of the wheels) contains several rotating shafts – for example, the driveshaft and axles. A propellor shaft transfers rotation motion from an engine to a propellor in a boat or plane.
Now let’s think about the power transmitted by a rotating shaft. Power is the rate of doing work. For a constant force, F, moving an object in the direction in which it acts, we can define work as
W = Fr
where r is the distance moved. Then power, P, is given by differentiation of W with respect to time, t, so that, for a constant force
P = F(dr/dt).
In a rotating system, according to post 17.39, the power is then given by
P = T(dθ/dt)
where T is the torque generated by the engine and θ is the angle that it turns, so that dθ/dt is the angular speed, ω. So, we could write that
P = T(dθ/dt) = Tω = 2πfT (1)
where f is the frequency of rotation of the shaft.
According to equation 20 of post 21.21, the maximum shear stress in the shaft is
τmax = Tro/J (2)
where ro is the external radius of the shaft. J is the polar second moment of area of the cross section of the shaft. Then
τmax = (Pro)/(ωJ) or ro = (τmaxωJ)/P (3).
In principle, we can use equation 3 to calculate the external radius of the shaft we need to use to drive a machine at an angular speed ω with an engine of power (sometimes called its “brake horsepower”) P.
Let’s see how we can do this for a solid shaft. Then, according to post 21.21
J = (πro4/2) (4).
Then, from equations 3 and 4
ro = (πro4τmaxω)/(2P) or ro = [2P/(πτmaxω)]1/3 (5)
Equation 5 allows us to calculate the radius of shaft we need to use with an engine of power P rotating with an angular speed ω. The value of τmax that we need to use, to ensure that the shaft is not damaged by torsion, is discussed in post 21.21.
If we want to reduce its mass, we could use a hollow shaft – a tube of internal radius ri and external radius ro. Its strength would be slightly reduced but this could be compensated for by a small increase in ro. The reason is that the stress in a solid shaft is zero at its centre and increases with increasing radius, as explained in post 21.21. Now J depends on both ri and ro (see post 21.21), so there are many combinations of values that we can use to design a rotating shaft that will not be damaged by the required values of P and ω.
In post 17.25, we saw that explaining science is often a process of diminishing deception – we give a simple explanation of an idea before thinking about it more deeply, at a later stage of the learning process. This post shows another way in which we solve problems by building on simple ideas. In post 21.21, we analysed torsion of a static cylinder – not itself an important or very interesting topic. But, in this post, we have seen that we can build on the results of this analysis to design an important component of many machines.