DYNAMICS OF ROTATING SYSTEMS questions Mechanical Principles

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MODULE TITLE : MECHANICAL PRINCIPLES

TOPIC TITLE : DYNAMICS OF ROTATING SYSTEMS

LESSON 3 : THE BALANCE OF MASSES ROTATING

IN SEVERAL PLANES

MP - 4 - 3

Published by Teesside University Open Learning (Engineering)

School of Science & Engineering

Teesside University

Tees Valley, UK

TS1 3BA

+44 (0)1642 342740

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________________________________________________________________________________________

INTRODUCTION ________________________________________________________________________________________

We can now extend the procedure learned in the previous lesson to balancing a

shaft which has several components attached along its length. Although we could

balance each component separately, we will find that we can achieve complete

balance by using only two added masses. Now, not only do the individual mass

centres have different planes of rotation located along the shaft, the radial lines

joining the axis of rotation to the mass centres will have different angles relative

to each other. This means that the moments of the centrifugal forces about any

point on the shaft axis will affect the shaft in different directions. We will,

therefore, have to extend our knowledge of the moments of forces as vector

quantities in order to evaluate the masses to be added to the system.

________________________________________________________________________________________

YOUR AIMS ________________________________________________________________________________________

When you have completed this lesson you should be able to:

• recognize that the moments of forces have direction as well as

magnitude and are therefore vector quantities

• construct moment vector diagrams as well as the force vector diagrams

• manipulate the diagrams to predict the required characteristics of the

counterbalances.

________________________________________________________________________________________

STUDY ADVICE ________________________________________________________________________________________

You will need graph paper, ruler, protractor and set square to complete this

lesson.

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________________________________________________________________________________________

COUNTERBALANCE NOT IN THE SAME PLANE ________________________________________________________________________________________

The question posed at the end of the previous lesson was to provoke you into

thinking about a rotating shaft which has eccentric masses distributed along the

length of the shaft, and to show you that two masses must be added to the

system to balance even a single unbalanced component, if its plane of rotation

cannot be used for the counterweight (see FIGURE 1).

Remember that the objective of balancing the system is to reduce the bearing

reactions to zero.

Since F2 is equal in magnitude

Now take moments about B. The sum of the moments must be zero as the

shaft is not turning about this point (remember we take anticlockwise moments

as positive).

∴ × + +( ) +

– . . .400 0 04 0 04 0 04

4000 0 04 0 04 0 04 0

48 16 0 08

× × +( ) =

∴ + × =

. – . .

– – .

A 0

Thuus

0 08 32

400

. –

–

A =

m r mc c cω 2 2400 10 100

100 4= × × =

∴

N or 00 N

mc = 0.4 kg

∴ = × × = 00 NF1 0 8 5

100 100 42.

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The –ve sign shows that the assumption that RA is upwards was in fact wrong

and that the direction should be downwards. So there is a bearing reaction of

400 N at A.

FIG. 1

Support bearings

Out of balance component

Planes of rotation

(a)

(b)

4cm 4cm 4cm

Bearing reaction

Shaft

The free body diagram

F1 = mrω 2

Centrifugal force from the

component B

RBRA F2 = mcrcω 2

Centrifugal force from the counterweight

0.8 kg 5 cm 10 cm 100 rad s–1

m r rc ω

= = = =

Counterweight

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As ΣF must also be zero then

The +ve sign shows that RB is upwards.

So we can't balance the shaft with just one counterweight, since both the forces

and the moments must be balanced. However, we can with two

counterweights.

Let us put a second counterweight next to the bearing at B just to illustrate

balancing. Furthermore, we will put it on the same side as the eccentric mass.

Now, mcA will not have the same value as previously, and so we have to

calculate it again. Assume that RA and RB will be zero, so that an equilibrium

will exist without needing bearing reactions (FIGURE 2(a)), i.e. Σ forces = 0 and Σ moments about B = 0.

Let us take moments first, see FIGURE 2(b).

The resulting equations will contain a common ω2 quantity in each term which can be omitted leaving mass × radius (mr) products instead of forces. It is convenient however to refer to the mass × radius products as forces (i.e. equal to mrω2 products where ω = 1).

400 400 400 0

400

– – + =

= +

B N

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FIG. 2

The mass × radius product of the eccentric mass is:

F = × = ↑0 8 5. 4 kg cm

mcA

Planes of rotation of the masses distributed along the shaft

Eccentric mass

First counterweight A on opposite side of the shaft

(a) The rotating system

8 cm 4 cm

FcA

mcB

Second counterweight on the same side of

the shaft

FcB

0.8 kg 5 cm rB = 10 cm

m r rA

= = =

(b) The free body diagram

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The mass × radius product of the first counterweight (A) is:

The mass × radius product of the second counterweight (B) is:

Take moments of mass about point B (as unknown FcB has no moment about

this point).

ΣMB = 0

ΣF = 0

Hence, mcA = = 1.2 kg and mcB = = 0.8 kg.

We didn't need any bearing reactions, therefore, to maintain equilibrium of the

shaft.

8 10

12 10

4 12 0– + =

∴ = ↑

cB 8 kg cm

– 4 8 4 4 0

4 12 4

× +( ) + × =

∴ = ×

= ↓

kg cm for equuilibrium of moments

F m rcB cB B= × ↑

F m rcA cA A= × ↓

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FIG. 3

For FIGURE 3, calculate the masses required to balance the shaft if planes A and B,

on either side of the eccentric mass, have to be used.

...................................................................................................................................................

________________________________________________________________________________________

r Shaft Eccentric

mass

0.8 kg 5 cm 5 cm 10 cm

6 cm

4 cm

m r rA rB

= = = =

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Solution

The free body diagram is drawn. Both of the counterbalance forces are placed in the

opposite sense to the out of balance force.

The sign of the following results will tell us if we are correct.

For equilibrium of the shaft ΣF = 0 and ΣM B

= 0 about B.

Taking moments about B:

ΣM B

= 0

(–ve moment from F as it tends to turn shaft clockwise about B)

But F = 0.8 × 5 = 4 kg cm

ΣF = 0

F F FA B– – = 0

∴ =

1 6 kg cm

10 16F

–

+ × +( ) × =F FA 6 4 4 0–

6 cm 4 cm B

FA = mArA

FB = mBrB

F = mr

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(F A

is –ve as we have chosen the counterweight force directions as downwards).

We can now calculate the required masses.

THE MOMENT OF A FORCE AS A VECTOR

In the question on page 7 you will have probably assumed that the radii of the

balancing masses were parallel to the radius of the eccentric mass but on the

opposite side of the shaft as we did in the solution above. So looking end on

we would see:

Eccentric mass

Shaft

F m r F r

A A A A A

= = =

∴ = =

where 1 kg cm and cm

6 5

1 6 5

00.32 kg

where 2 kg cm and cm

F m r F rB B B B B= = =

∴

.4 10

mmB = = 2 4 10 .

0.24 kg.

∴ =

2 4 kg cm

4 1 6– . –

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Is there any reason why we couldn't incline the radii of the counterbalances in

the way shown below?

To effectively answer this question we need to realize that the moment of a

force, or the 'turning' effect of the force, is a vector quantity M and has the

usual properties of vectors. This means that we can represent the moment of a

force by a line of appropriate length, drawn in the direction of the moment and

so we can add moments together graphically, in the same way as we sum

forces, velocities, etc. The magnitude of the moment we already know (it's the

product of the force and its moment arm), but what will its direction be?

The illustration in FIGURE 4 shows a force (applied by the hand) turning a

body (a nut on a right-hand screw thread) with the moment arm being the

spanner length. The nut will rotate about the axis of the screw thread X – X′.

Eccentric mass

mA mB

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FIG. 4

This axis is the line of action of the moment vector. The nut will move along

the thread towards X′, this gives the sense of the vector. This is known as the corkscrew, i.e. clockwise screws into the page and anticlockwise screws out of

the page.

FIGURE 5 shows the force F and its associated moment Mo acting on an axis

through a point O. The magnitude of

Notice that we need all three coordinates x, y and z to describe the directions of

F, d and Mo, with the origin located at O. This is to emphasise the three

dimensional nature of forces and moments in general.

FIG. 5

O d

M d Fo = × Nm

Force

Moment arm Axis

X′

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A handy memory aid to find the direction of the moment is illustrated in

Figure 6. Think of a corkscrew being turned.

FIG. 6

Notice that the direction of the moment is 90° anticlockwise from that of the

force when viewed along the z-axis from the positive direction.

Equivalent Systems

We can, now, replace any force F acting at one point on a body, such as the

shaft we are trying to balance, by the force F and its moment Mo at any other

chosen point O on the body.

For example, returning to the system shown in FIGURE 4, the effect of the

force on the nut is a force F equal to the force on the end of the spanner plus

the moment Mo of magnitude d × F Nm.

Into page

M moment vector

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FIG. 7

FIGURE 7 shows the equivalent force and moment lying on an imaginary

reference plane located at O and situated at right angles to the axis of the

spanner, labelled as the x-direction in the figure.

Turning point O

Reference plane

Mo = dF

This system can be

replaced by

this system at the point O

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For each of the four cases of a centrifugal force acting on a shaft at the location shown,

choose one of the options which represents the effects of the force transferred to

point O, drawn on the reference plane to the scales 1 cm = 1 kg cm for force and

1 cm = 5 kg cm2 for the moment. Note that there are two perpendicular lines drawn

for each option, one will be the centrifugal force, the other the moment.

(1)

F = 4 kg cm

O z

x y

5 c m

Shaft Reference plane

(2)

F = 5 kg cm

O z

4 c m

(3)

F = 3 kg cm

O z

5 c m

Sh aft

45°

(4)

F = 3 kg cm

O z

5 c m

225°Sh aft

Sh aft

x x

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1. ......................................................................................................................

2. ......................................................................................................................

3. ......................................................................................................................

4. ......................................................................................................................

(a) (b)

(c)

(d)

(e) (f)

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Solution

1. Option (d), not (f) or (c).

2. Option (c) as (f) has incorrect lengths for F and M.

3. Option (a).

4. Option (a) also, although the lines represent different vectors to those in 3.

Option (e) would have been the response to 4, if F had been on the opposite side of O.

3 kg cm 3 cm to the direction

3 kg

≡ ° +

× =

⏐135

5 15

ccm 3 cm

to direction

2 ≡ ° +

⏐45 y

225° + 90° = 315°

225°

3 kg cm 3 cm to direction

3 5 kg cm 32

≡ ° +

× = ≡

⏐45

ccm to direction⏐135° + y

3 cm

5 kg cm 5 cm in the direction

5 20 kg cm 5 c2

≡ +

× = ≡

4 mm in the direction.– y

5 kg cm 5 cm in the direction

5 20 kg cm 5 c2

≡ +

× = ≡

4 mm in the direction.+ y

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Solution

Returning to the question on page 7, let's see if we can find a solution with the

counterbalances as suggested on page 9.

Referring to FIGURE 3, let us put the balancing mass at A at an angle θA to the line of the radius of the eccentric mass and then transfer the effects of both

masses mA and m to point B.

• The effects of counterweight A at point B and sketched on the reference

plane through B, which is also the plane of rotation of counterweight B,

are:

the centrifugal force FA and the moment of this force MA shown by

the dashed line.

• The effects of the eccentric mass m are:

centrifugal force F and the moment of this force M again shown by a

dashed line.

• The effects of the balance mass B is just the centrifugal force FB shown

on the reference plane. It has no moment at B.

Reference plane

θΑ

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• For equilibrium of the shaft under the action of the forces and moments

acting at B (which is equivalent to the original force system), the vector

sum of the forces must equal zero, i.e. the polygon of the forces drawn on

the reference plane must close, written symbolically as ΣF = 0.

This is possible, but the vector sum of the moments must also be zero or

ΣM = 0.

You can see that this is only possible if MA (the moment caused by the

balancing mass at A) is made equal in magnitude and direction, but of opposite

sense, to M. The angle, therefore, must be determined.

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________________________________________________________________________________________

SEVERAL MASSES REVOLVING IN DIFFERENT PLANES ________________________________________________________________________________________

FIG. 8

FIGURE 8 shows a system of masses m1, m2 and m3 revolving at radii r1, r2 and r3. The angular locations of the arms, or radii, are shown in the end view,

measured from the first radius.

We need to find the masses and angular locations of the mass radii of the two

balancing masses, mA and mB, rotating in planes A and B, which will give

complete balance to the shaft. The plane of rotation of mB is shown in the

figure and will be used as a reference plane. The locations of all of the masses,

including A and B, along the shaft are given.

Remember that the centrifugal force from each and every mass can be

replaced by an equal and parallel force acting through a chosen point, together

2End view

2 3

1A B

L3L1

r2 L2

Reference plane through B

r A

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with the moment of the force trying to turn the shaft about an axis through the

point, at right angles to the line of action of the force.

These effects from all of the forces are evaluated at the same point on the

shaft. B has been chosen in the illustration shown in FIGURE 8. We shall see

why shortly.

FIGURE 9 illustrates the forces and moments referred to point B on the shaft.

The force and moment diagrams are shown separately for clarity. Although

the effects of the counterweights at A and B are indicated, we don't know the

magnitudes or the direction of either centrifugal forces FA and FB or of the

moment MA.

FIG. 9

Notice that we do know the moment from counterweight B, it is zero. It has

no moment about the chosen point B.

For the shaft to be in equilibrium, there must be no net moment at this chosen

point nor must there be a net force.

So, ΣMB = 0 and ΣF = 0, which will ensure that there are no loads on the support bearings caused by the eccentricity of the masses. We will develop

M3 = m3r3L3

F3 = m3r3

F2 = m2r2

F1 = m1r1

FB FA

Shaft

M2 = m2r2L2

M1 = m1r1L1

Shaft

Force Moments

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this technique of evaluating the balancing masses required by working through

a numerical example, in the form of a guided activity.

Try the example yourself. We will then go through each step in the solution

(we call this technique the solution algorithm).

FIGURE 10 shows the location, along the shaft, of the planes of rotation of three

eccentric masses m1, m2 and m3, and the two balancing masses mA and mB. The end

view shows the angular locations of the radii of the masses. The table gives the known

values of the masses and their radii. The question continues overleaf.

FIG. 10

1 60°

100°

m1 = 0.4 kg

m2 = 0.8 kg

m3 = 0.5 kg

r1 = 5 cm

r2= 5 cm

r3 = 9 cm

m1 mA m2 mB m3

10 10 10 10

mA = ? rA = 8 cm

mB = ? rB = 8 cm

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Use worksheet 1 to record your calculations and to draw the force and moment vector

polygons.

• Calculate the reduced centrifugal force caused by each eccentric mass. Record in

the appropriate cell in the table, e.g. for m1: Force = m1 ×× r1 = 0.4 ×× 5 = 2 kg cm.

• Choose one of the planes of rotation to be the reference plane. In this case I have

chosen the plane of rotation of mB. I could have chosen A. Why?

• Determine the moment arm of each force from this chosen point B, e.g. for F1 the moment arm L1 = 10 + 10 + 10 = 30 cm. Notice the moment arm for FB is

zero.

• Calculate the reduced moment of each centrifugal force about B.

For example, M1 = L1 ×× F1 = L1 ×× m1r1 = 30 ×× 2 = 60 kg cm2.

You can check your answers by following the steps laid out in the Solution beginning

on page 24.

...................................................................................................................................................

________________________________________________________________________________________

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Worksheet 1

z60°

100°

Item

0.4

0.8

0.5

100

2.0

mrL

F M

O1 Moment polygon

Force polygon

(kg) (cm) (°) (kg cm2)(cm)(kg cm)

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Solution

• Draw the vector sum of M1, M2 and M3 in that order. Start M1 from O1 and use a scale

of 1 cm = 10 kg cm2. You should end up at point 'a' on the worksheet and your vector

diagram should be this shape.

As the sum of all the moments of the forces about B must equal zero, we must make

the moment of the counterweight at A have a vector of such a magnitude and direction,

that when added to the polygon, starting at 'a' it ends on O1. So join 'a' to O1, to get this

equilibriant, as it is known. Measure the length of aO1, using the scale to get the value

of MA, and then calculate mArA and so mA. Measure the direction from M1 and record.

• We now have a value for the centrifugal force from counterweight A and its direction.

You should have FA = mArA ≈ 4.4 kg cm at an angle of 174° measured from the direction of r1.

= MA

Moment polygon 1 cm = 10 kg cm2

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• Now draw the sum of the vectors F1, F2, F3 and FA starting at O2, using a scale of

1 cm = 1 kg cm. You should end up at point 'b' on the worksheet and your vector

diagram shape should be of the form shown in this sketch.

• Again the closing line bO2 is the equilibriant, in this case the centrifugal force

generated by the counterweight B.

Measure the length and direction of bO2. Use the scale of this force diagram to

evaluate FB and then mB, so completing the table.

Force polygon 1cm = 1kg cm

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z60°

100°

Item

0.4

0.8

0.5

0.55

1.06

100

174

278

2.0

4.0

4.5

4.43

8.5

mrL

88.5

F M

O1 Moment polygon 1cm = 10 kg cm2

Force polygon 1cm = 1 kg cm

(kg) (cm) (°) (kg cm2)(cm)(kg cm)

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________________________________________________________________________________________

MODIFICATION OF THE MOMENT VECTOR ________________________________________________________________________________________

You may have found using the right-hand screw rule to find the directions of

the moments of the centrifugal forces to be prone to error (and tiring on the

wrist). We can simplify that procedure. To see how it works, take a piece of

tracing paper and trace the moment polygon, remembering to put the

directional arrows on the moment vectors. Rotate the tracing about O1 by 90°

in the anticlockwise direction. You should get:

Now compare the directions of the moments produced by the forces. You

should conclude that:

(i) M1 is parallel to F1 but of opposite direction, as are M2 and MA to F2 and

FA respectively, i.e. all the forces to the left of the chosen point for which

the moment arms measured from that point to the force, are in the

negative x direction.

M3 M1

Rotated moment polygon 1 cm = 10 kg cm2

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(ii) M3 is parallel to F3 and in the same direction. Notice that this moment

arm L3 is measured in the positive direction. Their positions relative to

each other have not changed.

We can incorporate this idea into our solution strategy by setting the moment

arms of forces to the left of the chosen point negative in our table and those

forces to the right, positive. On calculating the moments of F1 and F2, we

will retain the negative sign, denoting that the moments produced by those

forces will be drawn in the opposite direction to the force in the moment

polygon, shown in FIGURE 14.

How would the table of values have looked if we had solved the previous example using

the rotated moment polygon from the start.

...................................................................................................................................................

________________________________________________________________________________________

Item

m r θ mr L mrL (kg) (cm) (°) (kg cm2)(cm)(kg cm)

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Solution

Example 1

The following example illustrates the use of the modified moment diagram.

Check each step and item in the table.

Take extra care in checking the directions of the moments with respect to the

directions of the centrifugal forces determined by the sign of the moment arm.

B has been used as the point chosen to refer the forces and moments.

A shaft carries four components of mass m1 = 20 kg, m2 = 30 kg,

m3 = 24 kg and m4 = 26 kg, rotating with their mass centres at 9 cm, 7 cm,

10 cm and 12 cm respectively from the axis of rotation. They are spaced at

intervals of 10 cm along the shaft with the angular locations of m2, m3 and m4 at, respectively, 45°, 75° and 135° measured anticlockwise from m1.

Determine the masses, and angular locations measured from m1, of the two

counterweights required to completely balance the shaft, if they rotate at radii

of 24 cm.

The first mass mA is located midway between masses m1 and m2 whilst the

second, mB, is placed midway between masses m3 and m4.

Item

0.4

0.8

0.5

0.55

1.06

100

174

278

2.0

4.0

4.5

4.43

8.5

–30

–10

+10

–20

mrL

–60

–40

+45

–88.5

F M

(kg) (cm) (°) (kg cm2)(cm)(kg cm)

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Worksheet 2

1. Draw the configuration diagram with the radii of the masses at their given angular position.

Construct the table with the known quamtities. Calculate the reduced force and moments, take care over the sign of the moment arms L. So F = mr and M = mrL.

Draw the moment diagram from a suitable pole O

1 with

a convenient scale.

Close the moment diagram with the broken outline. This is the moment which counterweight A must produce to balance the moments of forces taken about point B, ∑MB = 0.

Measure M A to

obtain its magnitude and direction. Fill in the approriate cells, notably F

A which is

needed for the next step.

Close the force diagram by the broken line, measure the length and angle and complete the highlighted cells.

Item

17.5

22.7

135

195

296

180

210

240

312

420

545

–25

–15

–5

–20

mrL

–4500

–3150

–1200

1560

–8400

F M

Draw the force diagram to a suitable scale and from a convenient pole O

2 .

Moment polygon 1cm = 800 kg cm2

Force polygon 1cm = 50 kg cm

4 (kg) (cm) (°) (kg cm2)(cm)(kg cm)

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Now you tackle the same problem (Example 1 on page 29) but solve by taking

moments about point A. You should achieve the same results of course.

Use Worksheet 3 for the moment diagram.

...................................................................................................................................................

________________________________________________________________________________________

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Worksheet 3

Item

180

210

240

312

mrL

F M

Moment polygon 1cm = 800 kg cm2

Force polygon 1cm = 800 kg cm2

4 3

m θ (kg) (cm) (°) (kg cm2)(cm)(kg cm)

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Solution

Item

17.5

22.6

195

296

180

210

240

312

420

542

–5

mrL

–900

1050

3600

7800

10840

F M

4 3

Moment diagram 1cm = 800 kg cm2

Force polygon 1cm = 50 kg cm2

(kg) (cm) (°) (kg cm2)(cm)(kg cm)

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________________________________________________________________________________________

SELF-ASSESSMENT QUESTIONS ________________________________________________________________________________________

1. A turbine rotor, rotating at an angular velocity of 300 rad s–1, is supported

in bearings A and B, a distance of 3.65 m apart. The rotor is found to be

out of balance to the extent of the equivalent of two unbalanced masses of

1.6 kg and 2 kg at radii of 0.45 m and 0.6 m respectively, and are located

between A and B at distances of 0.5 m and 2.75 m respectively from A.

The angle between the radii is 135° measure in CCW direction from the

radius r1, or the first mass.

Evaluate the magnitude and direction of the reaction force on the shaft by

each of the bearings, A and B, when r1 makes an angle of 0° with the

horizontal.

2. Determine for the previous example, the mass and angular location of

each of the two counterweights required to completely balance the turbine

rotor if the first can be located, between A and B, 2.1 m from B, and the

other immediately adjacent to B. Both counterweights will rotate at a

radius of 0.5 m.

3. A shaft carries four masses, m1 = 20 kg, m2 = 30 kg, m3 = 24 kg and

m4 = 26 kg rotating at radii 9, 7, 10 and 12 cm, respectively, from the

axis of rotation and with angular displacements, measured anticlockwise

when viewed from the left-hand end of the shaft, of 0°, 45°, 120° and

210° respectively. The masses rotate in planes spaced at 10 cm intervals

along the shaft with the plane of rotation of m1 at the left-hand end of the

shaft.

Determine the masses of the two counterweights, A and B, rotating with

24 cm radii if the plane of rotation of A is located midway between planes

1 and 2, and the plane of B is located midway between planes 3 and 4.

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4. A shaft carries four masses A, B, C and D along its length and in that

order. The masses of B and C are 20 kg and 10 kg respectively and both

are at a radius of 0.1 m, while A and D are both at a radius of 0.2 m. The

angle between the radii of C and B is 100° and that of B from A is 190° –

both angles being measured in the same direction. The planes containing

A and B are 0.25 m apart and those containing B and C are 0.5 m apart.

The shaft is to be in complete balance.

Determine:

(a) the magnitudes of the masses A and D

(b) the distance between the planes C and D

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________________________________________________________________________________________

ANSWERS TO SELF-ASSESSMENT QUESTIONS ________________________________________________________________________________________

1. For equilibrium of the shaft, the vector sum of all of the forces acting on

the shaft, including the bearing reactions, equals zero, i.e. ΣF = 0. If F1 = centrifugal force from m1, F2 = centrifugal force from m2 and RA and RB are the bearing reactions then:

Also for equilibrium, the sum of the moments of those forces taken about

a convenient point on the shaft equals zero, i.e.

Choose the point on the centreline of the shaft at B as the convenient

point because the bearing reaction at B has no moment about this point.

Use the tabular method to keep control of the calculations even though

the cells under m and r for items A and B are not applicable in this case.

The cells for mr are used for recording the bearing reactions.

The cells are filled in as far as possible (notice the moment arm for

reaction B is zero) as shown on worksheet 4, e.g. Item 1 is the first mass

m1 = 1.5 kg rotating at a radius r1 = 0.45 m making an angle of 0° to

the datum at this instant.

ΣM

M M M M

+ + + =

0or 1 2 A B

F F R R

F r

1 + + + =

2 A B

where

and

1 1 1

2 2 2

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The magnitude of the centrifugal force F1 is

(use this form of the forces and moments until the final step).

This mass rotates at a distance of –3.15 m from B, the plane of

rotation being to the left of the moment axis, giving a moment of

m1r1L1 = 0.675 × –3.15 = –2.13 kg m2 about the point B. The direction of moment M1 will be represented by a line drawn parallel to

force F1, but directed in the opposite sense owing to the –ve sign. The

angular locations of the radii have been drawn in part (a) of the figure on

the worksheet, as the directions of r1 and r2 are the directions of the

centrifugal forces F1 and F2, i.e. from the shaft outwards.

The remaining steps of the solution are as follows.

• Draw the moment diagram to a scale of 1 cm = 0.3 kg m2 starting

from Pole O1, Worksheet 4(c). The broken line is the moment which

must be applied to the shaft to maintain equilibrium. In this case it is

provided by the bearing reaction at A.

Measure the length as 5.2 cm which represents a moment of

mA = 0.3 × 5.2 = 1.56 kg m2 which must be in the opposite direction to RA due to the –ve moment arm.

∴ × ( ) =

and m

Thus

R L

– .

–

1 56

3 65

11 56 3 65

0 427

. – .

.= kg m

F m r1 1 1 1 5 0 45

0 675

= × ×( ) = ×

ω 2

kg m

. .

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Record in the table.

The true value is multiplied by

The direction of MA is transferred to Figure 17(a), remembering again

that RA is in the opposite direction. Measure the angle of RA from radius

r1, as 209°. Record in the table.

• We can now evaluate RB by drawing the vector sum of all of the

forces acting on the shaft, which must close as ΣF = 0.

From the table F2 (= m2r2) = 1.2 kg m which is represented by a

line 1.2/0.2 = 6 cm in length drawn parallel to the radius r2 starting

from the head of vector F1. We end at the head of vector RA, so the

broken line is the force required on the shaft to maintain equilibrium

and is provided by the bearing reaction at B. Measure it as 4.25 cm,

thus RB = 4.25 × 0.2 = 0.85 kg m in reduced form. Record actual value of

Transfer the direction of RB to Worksheet 4(d) and measure its angle

from r1 as 311.5°.

• Thus R

= °

38 4 209

76 5 311 5

. .

⏐

RB = × = × =0 85 0 85 300 76 5 2 2. . .ω kN.

ω 2 2

A kN

= ×

∴ =

300

9 10

38 46

R . .

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Worksheet 4

RA RB

1cm = 0.2 kN

1cm = 0.3 kg cm2

0.5m

1 2

2.75m 3.65m

(b)

A B

(a)

Item

1.5

2.0

0.45

0.6

135

209

311.5

0.675

1.2

0.427

0.85

–3.15

–0.9

–3.65

mrL

–2.13

–1.08

–1.56

F M

M2 MA

(c)

(d)

135°

311.5°

209°

(kg) (cm) (°) (kg cm2)(cm)(kg cm)

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2. In this case we are adding mass, the counterweights, to the system to

eradicate the bearing reaction forces, by replacing them with centrifugal

forces from the counterweights. Let the mass of the first counterweight

be mC and that of the second be mB as it is located at B. A table is again

used to control the calculations. Choose the point B about which to take

moments. Then for equilibrium

The moment diagram is exactly the same as in the solution to Question 1.

The broken line representing the moment required about B to maintain

equilibrium is now provided by the centrifugal force from the counterweight

at C, thus relieving the bearing at A. Now as plane C is a little closer to B

than A was, then the force needed to give that moment is different.

at the angle 209° from r1. Thus

For equilibrium, ΣF = 0, so F1 + F2 + FC + FB = 0. Draw a new force diagram, as FC, although parallel to RA in the previous solution, is

of greater magnitude. The broken line, closing the polygon of forces, is

the force required from the counterweight at B, measured at 4.7 cm

equates to 0.94 kg m2. The direction is transferred to Figure (a) and

measured at 327° from r1. Mass mB is calculated as 0.94/0.5 = 1.88 kg.

Thus counterweight C is a mass of 1.49 kg rotating at a radius of 0.5 m at

an angle of 209° from r1 located at a distance of 2.1 m from B.

Counterweight B is a mass of 1.88 kg rotating at a radius of 0.5 m at an

angle of 327° from r1 and located at B.

mC = = – .

– . .

0 7436 0 5

1 49 kg

FC = = – . – .

. 1 56 2 1

0 743 kg m

M M M1 2 C+ + = 0

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Worksheet 5

mC mB

1cm = 0.2 kg m

0.5m

1 2

2.75m 3.65m

(b) (a)

Item

1.5

2.0

1.49

1.88

0.45

0.6

0.5

135

209

327

0.675

1.2

0.743

0.94

–3.15

–0.9

–2.1

mrL

–2.13

–1.08

–1.56

F M

(c)

2.1m

rB θB

135°

θC

(kg) (cm) (°) (kg cm2)(cm)(kg cm)

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3. Worksheet 6 (Solution to question 3)

Item

17.8

120

210

209

180

210

240

312

428

–25

–15

–5

–20

mrL 100

–45

–31.5

–12

+15.6

–85.6

F M

1cm = 75 kg cm

1cm = 800 kg cm2

(kg) (cm) (°) (kg cm2)

(cm)(kg cm)

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4. A difficult problem as the information is given in a complex form. It

seems to be a matter of rearranging the masses to give complete balance,

although we can look upon masses A and D to be the balancing masses.

The angular location of A is given (which we usually have to determine)

while the location of D along the shaft is not given (which it usually is).

We usually take moments about the location on the shaft of one of the

masses. This has to be A as we don't know where D is.

The initial table looks like this.

TABLE 1

We should measure all angles from the radius of A as we are given them

(except D).

ΣMA = 0

• We need to draw the moment diagram to scale, 1 cm = 0.1 kg m2.

The closing line then gives us the mrL value for 'D', i.e. 0.85 kg m2

and its direction 75° to the horizontal but as we don't know L for D

we can't find FD (= mr) from MD.

Item

0.2

0.1

0.2

190

290

+0.25

+0.75

mrL

0.5

0.75

F M

(kg) (cm) (°) (kg cm2)(cm)(kg cm)

Teesside University Open Learning (Engineering)

ΣF = 0

• We also know the direction of A. So we start the summation of the

forces with FB (= mBrB) of 2 kg m to the scale shown, add on FC of

1 kg m to scale then add on FD in direction only (as we don't know

its length).

Now we know that when FA is added, this final vector should end

where we started the summation, i.e. at O2. Therefore we draw a line

through 'O2' in the known direction of A, i.e. 0°. The intersection of

this line and the line representing FD in direction gives the magnitude

of FD and FA.

Intersection

FA Direction of FA

75°

Direction only

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Worksheet 7

Scale 1cm = 0.25 kg m

0.25 ?

A B

0.5 mB

190°

mC 100°

Item

6.2

6.8

0.2

0.1

0.2

190

290

1.24

1.36

+0.25

+0.75

+0.61

mrL

0.5

0.75

0.83

F M

MDMC

FD FC Scale 1cm = 0.1 kg m2

(kg) (m) (°) (kg m2)(m)(kg m)

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________________________________________________________________________________________

SUMMARY ________________________________________________________________________________________

• Two masses must be added to the shaft to completely balance the forces

and the moments caused by eccentric masses distributed along the shaft.

• Choosing the moment axis at the point of attachment of one of the

unknown counterbalances eliminates the moment of this force from the

moment equation.

• Using the moment of a force as a vector quantity in the same direction as

the force, if the moment arm is measured to the right from the moment

axis to the point of application of the force. Then the moment equation,

ΣM = 0, can be solved graphically to give the moment caused by one of the counterbalances.

• The other is found from the graphical solution to the force equation,

ΣF = Σmr = 0.

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<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> >> >> setdistillerparams << /HWResolution [2400 2400] /PageSize [612.000 792.000] >> setpagedevice