BalancingBalancing 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 CollegeDate 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.

Contents• Balancing of Shafts

• First Step

• Second Step

• Third Step

• Fourth Step

• Fifth Step

• Sixth Step

• Credits

File Name Unit Outcome Key Words

Balancing of shafts 1.4 Beams, balancing, rotation, mass, stress, shafts

These files support the Edexcel HN unit – Mechanical Principles (NQF L4)

For further information regarding unit outcomes go to Edexcel.org.uk/ HN/ Engineering / Specifications

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.

When viewed from end D, the mass at C is 120o 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.

The first step, is to sketch the layout of the shaft with relative dimensions

0.6

0.4

0.2

A

B

C

D

120O

B 2 kg

C 2.5 kg

R = 60 mm

R = 50 mm

Plane m (Kg)

r (m)

mr (Kgm)

x(m)

mrx(Kgm2)

A mA rA mA rA 0 0

B 2 0.05 0.1 0.2 0.02

C 2.5 0.06 0.15 0.4 0.06

D mD rD mD rD 0.6 0.6 mD rD

Second step: Produce a table

Third step: Construct the mrx diagram or polygon, using a scale of 1mm = 0.001 Kgm2

120o

79o

0.02 Kgm2

0.06 Kgm2

0.6 mD rD

From the mrx diagram measure the moment which represents 0.6 mD rD

0.6 mD rD = 53 x 0.001 Kgm2 = 0.053 Kgm2

mD rD = 0.053 / 0.6 = 0.0883 Kgm

The angle is also measured anticlockwise from B is 79o

Fourth step: The mass - radius polygon can now be constructed. This is known as the mr diagram

d

A =134o

0.15kgm

b

a

c

0.1Kgm

0.0883kgm

mA rA

Scale used 1mm = 0.001 Kgm2

Fifth step: The vector has to be measured at this point; its value is 30.1mm

mA rA = 30.1 x 0.002kgm = 0.0602 Kgm

The angle =134o measured anticlockwise from B

The centripetal force F = mr2

Where = 210 rpm x 2 rad/rev = 22 rad/s

60(sec/min)

At bearing A

FA = 0.0602 Kgm x (22 rad/s)2 = 29.1 N

RA = - FA = -29.1N

The minus sign indicating that the direction is opposite to the direction FA

Sixth step: Produce a diagram showing the direction of bearing reactions

B

RA

RB

CFA

FB

134O

79O

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.

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