Shafts are important parts of modern steam turbines. The manufacture of such shafts which are \(10\,\mathrm{m}\) in length and \(0.5\,\mathrm{m}\) in diameter is a complex technological problem. The shaft of a powerful turbine can withstand a load of about 200 ton and rotate with a speed of 3000 rpm.

At first glance, it might seem that such a shaft should be exceptionally hard and durable. This, however, is not so. At tens of thousands of revolutions per minute, a rigidly fastened and unbendable shaft will inevitably break, no matter how strong it may be.

It isn’t difficult to see why rigid shafts are unsuitable. No matter how precisely engineers work, they cannot avoid at least a slight asymmetry in the wheel of a turbine. Enormous centrifugal forces arise when such a wheel rotates; recall that their magnitudes are proportional to the square of the rotational speed. If they are not exactly balanced, the shaft will start “beating” against the ball bearings (for the unbalanced centrifugal forces “rotate” together with the machine), break them and smash the turbine.

At one time, this phenomenon created an unsurmountable obstacle to the increase in the rotational speed of a turbine. A way out of the situation was found at the last turn of the century. The flexible shaft was introduced into the technology of turbine construction.

In order to understand the idea behind this remarkable invention, we must compute the total effect of the centrifugal forces. But how can these forces be added? It turns out that the resultant of all the centrifugal forces acts at the centre of gravity of the shaft and has the same magnitude as if the entire mass of the wheel of the turbine were concentrated at the centre of gravity.

Let us denote the distance from the centre of gravity of the wheel of the turbine to its axis, distinct from zero because of a slight asymmetry in the wheel, by \(a\). During rotation, centrifugal forces will act on the shaft which will bend. Denote the displacement of the shaft by \(l\).

Let us compute this magnitude. We know the formula for centrifugal force (see here). This force is proportional to the distance from the centre of gravity to the axis, which is now \(a+l\), and is equal to \(4 \pi^{2} n^{2} M (a + l)\), where \(n\) is the number of revolutions per minute, and \(M\) is the mass of the rotating parts. The centrifugal force is balanced by the elastic force, which is proportional to the magnitude of the displacement of the shaft and is equal to \(kl\), where the coefficient \(k\) characterizes the rigidity of the shaft. Thus: \[kl= 4 \pi^{2} n^{2} M (a+l)\] whence \[l= \dfrac{a}{k/4 \pi^{2} n^{2} M - 1}\] Judging by this formula, fast rotations are no problem for a flexible shaft, For very large (even infinitely large) values of \(n\), the deflection \(l\) of the shaft does not grow without bound. The value of \(k/4 \pi^{2} n^{2} M\) figuring in our last formula tends to zero, and the deflection \(l\) of the shaft becomes equal in magnitude to the asymmetry, but opposite in sign.

This computational result implies that, for fast rotations, the asymmetrical wheel, instead of smashing the shaft, bends it in such a way as to cancel the effect of asymmetry. The bending shaft centres the rotating parts, transfers the centre of gravity to the axis of rotation by means of its deformation, and thus nullifies the action of the centrifugal force.

The flexibility of the shaft is by no means a drawback; on the contrary, it is a necessary condition for stability. As a matter of fact, it is necessary for stability that the shaft bend by a distance of the order of a without breaking.

An attentive reader may have noticed an error in the reasoning employed. If we displace a shaft “centring” during fast rotations from the position of equilibrium we have found and consider only centrifugal and elastic forces, it is easy to see that this equilibrium is unstable. It turns out, however, that Coriolis forces save the situation and make this equilibrium quite stable.

A turbine starts turning slowly. At first, when \(n\) is very small, the fraction \(k/4\pi^{2}n^{2}M\) will be great. As long as this fraction is greater than unity with increasing \(n\), the deflection of the shaft will have the same sign as that of the original displacement of the centre of gravity of the wheel. Therefore, at the beginning of the motion the bending shaft does not centre the wheel, but, on the contrary, increases the total displacement of the centre of gravity by means of its deformation, and hence also the centrifugal force. To the degree that \(n\) increases (with the condition \(k/4\pi^{2}n^{2}M > 1\) preserved), the displacement grows and, finally, the critical moment is reached.

The denominator of our formula for the displacement \(l\) vanishes when \(k/4\pi^{2}n^{2}M = 1\), and so the deflection of the shaft formally becomes infinitely large. The shaft will break at such a speed of rotation. In starting a turbine this moment must be passed very quickly; it is necessary to slip by the critical number of revolutions per minute and pass over to a much faster motion of the turbine for which the phenomenon of self-centring described above will begin.

But what is this critical moment? We can rewrite its condition in the following form: \[4 \pi^{2} \dfrac{M}{k} = \dfrac{M}{n^{2}}\] Or, expressing the number of revolutions per minute in terms of the period of rotation by means of the relation \(n = 1/ T\) and extracting square roots, we can rewrite it as follows: \[T = 2 \pi \, \sqrt{\dfrac{M}{k}}\] But what kind of quantity have we obtained in the right-hand side of the equality? Our formula looks rather familiar. Turning Section: Force and Potential Energy in Oscillations, we see that the period of free vibration of the wheel on the shaft figures in our right-hand side. The period \(2 \pi \sqrt{ M/k}\) is that with which the wheel of a turbine of mass \(M\) would vibrate on a shaft of rigidity \(k\) if we were to deflect the wheel to one side, so that it might vibrate by itself.

Thus, the dangerous instant is when the rotational period of the wheel of the turbine coincides with the period of free vibration of the system turbine-shaft. Resonance is responsible for the existence of a critical number of revolutions per minute.