Tips and tricks

Choosing the right mounting position

Question:

"If a rotor has two standard shaft ends - what can possibly go wrong"? The question alone contains an indication of where the problem may lie - the shaft ends may be skew.

Example:

Consider, for example, a rotor with a mass of 50 kg. Adhering to the quality grades specified in ISO 1940, you find that you have to achieve a permissible tolerance of Uperm = 250 g·mm in the plane of the center of gravity, equivalent to a permissible residual eccentricity eperm. of 5 µm.

Next, the question arises whether – for the sake of convenience – you can support the rotor 5 mm to the side of the plane of the service bearings. In your measurement room, you find that there is an in-phase eccentricity error of 6 µm (peak-to-peak) at the desired mounting positions for balancing A-A, relative to the subsequent operational bearing positions B-B, which corresponds to an eccentricity eA/B = 3 µm between the two axes of rotation defined by the mounting positions A-A and B-B.

(cf. Figure 1)

Is this minor eccentricity sufficient to jeopardize the balancing tolerance?

As we found, the axis of rotation of the rotor when mounted in your balancing machine is offset by eA/B = 3 µm from the the subsequent axis of rotation in service. For the sake of simplicity, it is assumed here that both shaft ends have the same eccentricity error in the same direction and that the only unbalance present in the rotor acts in the centre of gravity plane.


Fig.  1
Mounting in the service bearings results in the axis of rotation B-B. The center of gravity however, is not situated on the axis B-B - this introduces unbalance.

What happens next?
A further assumption being made here is that a residual unbalance of 250 g·mm can be achieved quickly in one balancing step. This unbalance corresponds exactly to the permissible residual unbalance. The centre of gravity will be 5µm away from the axis of rotation A-A. (Fig. 2 b). In its operating condition, however, the rotor turns about an axis offset by eA/B = 3 µm. This means that the center of gravity, which was situated quite close to the axis of rotation A-A on the balancing machine, also shifts by the same 3 µm during operation. It will move on a circular orbit around the final axis of rotation, i.e. the operational axis of rotation B-B. The entire mass of the rotor is acting as unbalance mass at this radius off set by 3 µm!

How high is the resulting unbalance?
The 3 µm offset of the axis of rotation results in a change in unbalance of : UA/B = m · eA/B = 50 kg · 3 µm = 150 g·mm. In the worst case assumed here, the residual rotor unbalance adds vectorially to this value, resulting in a total unbalance of 250 g·mm + 150 g·mm = 400 g·mm (Fig. 2c). Many users attempt to overcome this problem by specifying even tighter tolerances – e.g. 100 g·mm –. However, this will result in the balancing procedure becoming more time-consuming and therefore more costly. Conclusion: The best solution in these cases would be to mount the rotor on the balancing machine at its subsequent in-service bearing positions (Figs.3 a + b). If this is difficult to achieve on the balancing machine, we will be able to point out various possibilities and solutions available for your problem.

Remark

For the sake of greater clarity a number of simplifications have been made here. More detailed information can be obtained in our balancing seminars. We will also be happy to discuss your specific problems in greater detail during a personal meeting or at a seminar.


Fig. 2
Correct mounting on the balancing machine ensures that the axis of rotation A-A coincides with B-B. Due to the action of unbalance, the centre if gravity is not situated on the axis of rotation.

Fig. 3
Balancing ensures that the centre of gravity is situated close to the axis B-B, both on the balancing machine and in subsequent operation.

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