The requirement to provide balance for rotating systems is a vital component in ensuring long, reliable service. This document describes a graphical method that can be used to determine the out of balance forces of such a system and the correct size and position of the balance weight required to do this. A worked example of the method is shown.

1. Balancing Balancing of shafts HNC In Engineering – Mechanical Science Edexcel HN Unit: Engineering Science (NQF L4) © Leicester College 2009. This work is licensed under a Creative Commons Attribution 2.0 License . Author: Leicester College Date created: Date revised: 2009 Abstract: The requirement to provide balance for rotating systems is a vital component in ensuring long, reliable service. This document describes a graphical method that can be used to determine the out of balance forces of such a system and the correct size and position of the balance weight required to do this. A worked example of the method is shown.

3. Balancing of Shafts A shaft ABCD is 0.6m long and rotates in bearings at A and D which allow the ends of the shaft freedom of direction (i.e. the shaft is simply supported). Masses of 2kg and 2.5kg located at radii 50mm and 60mm are attached to the shaft at B and C.

4. When viewed from end D, the mass at C is 120 o clockwise from the mass at B. If AB = BC = CD = 0.2m The shaft rotates at 210 rev/min. Calculate the magnitude and direction of the dynamic reactions at the bearings .

5. The first step, is to sketch the layout of the shaft with relative dimensions 0.6 0.4 0.2 A B C D

6. 120 O B 2 kg C 2.5 kg R = 60 mm R = 50 mm

7. Second step: Produce a table Plane m (Kg) r (m) mr (Kgm) x (m) mr x (Kgm 2 ) A m A r A m A r A 0 0 B 2 0.05 0.1 0.2 0.02 C 2.5 0.06 0.15 0.4 0.06 D m D r D m D r D 0.6 0.6 m D r D

8. Third step: Construct the mr x diagram or polygon, using a scale of 1mm = 0.001 Kgm 2 120 o 79 o 0.02 Kgm 2 0.06 Kgm 2 0.6 m D r D

9. From the mr x diagram measure the moment which represents 0.6 m D r D 0.6 m D r D = 53 x 0.001 Kgm 2 = 0.053 Kgm 2 m D r D = 0.053 / 0.6 = 0.0883 Kgm The angle is also measured anticlockwise from B is 79 o

10. Fourth step: The mass - radius polygon can now be constructed. This is known as the mr diagram d  A =134 o 0.15kgm b a c 0.1Kgm 0.0883kgm m A r A Scale used 1mm = 0.001 Kgm 2

11. Fifth step: The vector has to be measured at this point; its value is 30.1mm m A r A = 30.1 x 0.002kgm = 0.0602 Kgm The angle =134 o measured anticlockwise from B The centripetal force F = mr  2 Where  = 210 rpm x 2  rad/rev = 22 rad/s 60(sec/min)

12. At bearing A F A = 0.0602 Kgm x (22 rad/s) 2 = 29.1 N R A = - F A = -29.1N The minus sign indicating that the direction is opposite to the direction F A

13. Sixth step: Produce a diagram showing the direction of bearing reactions B R A R B C F A F B 134 O 79 O

14. This resource was created Leicester College and released as an open educational resource through the Open Engineering Resources project of the Higher Education Academy Engineering Subject Centre. The Open Engineering Resources project was funded by HEFCE and part of the JISC/HE Academy UKOER programme. © 2009 Leicester College This work is licensed under a Creative Commons Attribution 2.0 License. The JISC logo is licensed under the terms of the Creative Commons Attribution-Non-Commercial-No Derivative Works 2.0 UK: England & Wales Licence. All reproductions must comply with the terms of that licence. The HEA logo is owned by the Higher Education Academy Limited may be freely distributed and copied for educational purposes only, provided that appropriate acknowledgement is given to the Higher Education Academy as the copyright holder and original publisher. The Leicester College name and logo is owned by the College and should not be produced without the express permission of the College.