DYNAMICS OF ROTATING SYSTEMS questions Mechanical Principles

MP-4-1.pdf

Home >Engineering homework help >Mechanical Engineering homework help >DYNAMICS OF ROTATING SYSTEMS questions Mechanical Principles

MODULE TITLE : MECHANICAL PRINCIPLES

TOPIC TITLE : DYNAMICS OF ROTATING SYSTEMS

LESSON 1 : SINGLE AND MULTI-LINK MECHANISMS

MP - 4 - 1

Published by Teesside University Open Learning (Engineering)

School of Science & Engineering

Teesside University

Tees Valley, UK

TS1 3BA

+44 (0)1642 342740

retrieval system, or transmitted, in any form or by any means, electronic, mechanical,

photocopying, recording or otherwise without the prior permission

of the Copyright owner.

This book is sold subject to the condition that it shall not, by way of trade or

otherwise, be lent, re-sold, hired out or otherwise circulated without the publisher's

prior consent in any form of binding or cover other than that in which it is

published and without a similar condition including this

condition being imposed on the subsequent purchaser.

________________________________________________________________________________________

INTRODUCTION ________________________________________________________________________________________

Newton’s second law is the basic law in dynamics, and it relates force to

acceleration. When analysing a dynamic system, in which translational motion

occurs, we have to determine velocity and acceleration in order to calculate the

force or torque and power transmitted.

Dynamics of rotating systems deals with the study of operation principles and

balancing in some mechanisms widely used in engineering, such as slider-

crank, three/four bar and flywheel, etc. In this lesson, we will only consider

the first two mechanisms.

________________________________________________________________________________________

YOUR AIMS ________________________________________________________________________________________

After studying this lesson, you should be able to:

• understand how a mechanism works

• construct space , velocity vector and acceleration vector diagrams

• use these diagrams to determine the geometry, velocity and

acceleration in a mechanism

• calculate forces and torques acting on links in a mechanism.

________________________________________________________________________________________

STUDY ADVICE ________________________________________________________________________________________

To complete this lesson, you will need graph paper, ruler, compass, protractor

and set square.

Teesside University Open Learning (Engineering)

________________________________________________________________________________________

MECHANISMS ________________________________________________________________________________________

A mechanism can be defined as a device used to translate a movement or

displacement at one point to another point some distance away. A device used

to transmit a force or torque is termed a machine. Any machine can be

considered as a group of interconnected mechanisms which convert one type

of motion to a variety of other motions. These changes may be to convert

rotary motion to straight line motion or to convert reciprocal (back and forth)

motion to intermittent motion. They may also transform a fixed type of

motion, for example by magnifying a linear motion or by slowing down a

rotary motion.

Both machines and mechanisms consist of rigid bodies. The relative motions

among the rigid bodies are definite. However, machines transform energy to

do work, whereas mechanisms do not necessarily perform this function. The

term machinery generally means machines and mechanisms.

The various parts of a mechanism are called links. These may be items such as

levers, cranks, sliders, shafts, pulleys and bearings.

There are a number of different types of mechanism. FIGURES 1 to 3 show

some examples.

Teesside University Open Learning (Engineering)

LEVER

A lever is a rigid beam that can rotate about a fixed point called the fulcrum.

An effort applied to one end of the beam will cause a load to be moved at the

other. The force and the load move in opposite directions.

Levers are an essential part of many mechanisms. They can be used to change

the amount, the strength and the direction of movement. The position of the

force and the load are interchangeable and by moving them to different points

on the lever, different effects can be produced. Also, by moving the fulcrum

nearer to the load, you can lift a large load with only a little effort. This is

called mechanical advantage. FIGURE 1 shows an example of a lever.

FIG. 1 Lever

(a) Mechanism (b) Skeleton outline

Teesside University Open Learning (Engineering)

SLIDER-CRANK

A slider-crank mechanism changes rotary (crank) to reciprocal motion (piston)

or vice versa. In the car engine, the reciprocating motion of the piston caused

by gas pressure from exploding fuel is converted into rotary motion as the rod

moves the crankshaft around as illustrated in FIGURE 2.

An air compressor uses this principle in reverse – an electric motor turns the

crankshaft and the piston moves up and down to compress the air.

FIG. 2 A power cylinder (slider-crank) in a car engine

(a) Mechanism (b) Skeleton outline

Piston

Rod

Crank

Teesside University Open Learning (Engineering)

FOUR-BAR CHAIN

FIGURE 3 shows the basic four-bar chain used in a vehicle suspension unit. It

can be seen that this mechanism consists of four links with four turning joints.

The links of the four-bar chain may be moveable, like links 1, 2 and 4, or fixed

like link 3.

The four-bar mechanism has some special configurations created by making

one or more links infinite in length. The slider-crank mechanism shown in

FIGURE 2 can be treated as a four-bar chain with the slider replacing an

infinitely long output link.

FIG. 3 A vehicle suspension unit (four-bar chain)

By adjusting the length of the links and the joint points, you will find that

many mechanisms are made up of a series of four-bar chains.

(a) Mechanism (b) Skeleton outline

A B

D C

A B

Teesside University Open Learning (Engineering)

________________________________________________________________________________________

VELOCITY DIAGRAMS ________________________________________________________________________________________

The velocity of an object refers to the speed and direction in which it moves,

so velocity is a vector quantity and can be represented by a line drawn to scale

and in the correct direction. Velocity vectors are used when dealing with many

kinds of mechanism, such as turbine blades and the cranks and connecting rods

of reciprocating engines. They are used in conjunction with acceleration

diagrams to determine various forces acting on components and, although the

procedures are more complex, the basic principles are the same for both.

In this section, we study the construction of velocity diagrams, which will be

used for analysis of the motion of links in a mechanism. They need to be

drawn accurately and to a suitable scale.

VELOCITY VECTORS

Since velocity has magnitude and direction it can be represented by a vector,

i.e. a straight line. A velocity of 50 m s–1 due north would be represented as

shown in FIGURE 4.

FIG. 4 Vector diagram

50 m s–1

Teesside University Open Learning (Engineering)

Example 1

Calculate the vertical rate of climb and the horizontal forward velocity for a

rocket travelling at 300 km h–1 at an angle of 15° to the horizontal.

Solution

FIG. 5

From FIGURE 5, by calculation:

horizontal velocity component cos 15= =Vh 300 °°

forward velocity km h

–1

300 0 9659

289 8

eertical velocity component sin 15°

= =Vv 300

3300 0 2588

77 65

.rate of climb km h–1

300 km h–1

15°

Forward velocity Vh

Climbing velocity

Teesside University Open Learning (Engineering)

ABSOLUTE AND RELATIVE VELOCITY

An absolute velocity is the velocity of an object measured from a coordinate,

which is normally fixed on the ground or anything rigidly attached to the

ground and not moving. A relative velocity is the velocity of an object

measured relative to another object that is moving.

According to these definitions, actually, there is nothing ‘absolute’. All motion

is relative. A body is said to move if it changes its position in space relative to

some other body or fixed point. Velocities are commonly quoted relative to

the earth which is regarded as being ‘fixed’. This is obviously not true, as we

all know the earth is moving, but it provides a convenient datum from which to

measure motion.

Example 2

(a) If you are sitting in a stationary car A and another vehicle B passes at

40 km h–1, the velocity of car B relative to you is 40 km h–1. The velocity

of car A relative to a building by the road (or the absolute velocity of A)

is 0, since both car A and the building (the latter is fixed to the ground)

are not moving.

(b) If you are travelling at 40 km h–1 alongside another vehicle B, also

moving at 40 km h–1, in the same direction, the velocity of B relative to

you is zero, since you are both travelling at the same speed and in the

same direction. However, both you and vehicle B have the velocity of

40 km h–1 relative to the building.

you at 40 km h–1 moving in the opposite direction, the velocity of B

relative to A is 80 km h–1. Note that the velocity magnitude of car A and

B relative to the building is the same, which is 40 km h–1.

Teesside University Open Learning (Engineering)

TANGENTIAL VELOCITY

Consider the motion of link AB, which is pinned at A and revolving about A at

an angular velocity ω as shown in FIGURE 6. Then point B is rotating around A with a velocity of vBA. The direction of vBA is always tangential and hence

at 90° to the link. The magnitude of vBA is

vBA = rω

where vBA = the magnitude of velocity at B relative to point A (m s –1)

r = the length of link AB (m)

ω = the angular velocity (rad s–1).

FIG. 6

B r

ω v

Teesside University Open Learning (Engineering)

RADIAL VELOCITY

Consider a sliding link C, which can only slide on link AB, as shown in

FIGURE 7. The motion of link C along link AB is a radial motion relative to

A with a velocity of vCA radial. Since link AB is rotating about A, then link C

also has a tangential velocity vCA tangential at the same time.

FIG. 7

Note that the two relative velocities of sliding link C, vCA tangential and

vCA radial are normal to each other and occur simultaneously. The true velocity

of link C relative to A (or the absolute velocity since point A is fixed to the

ground) vCA can be determined using the diagram shown in FIGURE 8.

FIG. 8 Velocity vector diagram

v CA tangential

v CA radial

v CA

v CA tangential

v CA radial

Teesside University Open Learning (Engineering)

In a mathematical version, vCA is the vector sum of vCA tangential and

vCA radial, which is

Example 3

Refer to FIGURE 9. Slide C is moving along link AB with a velocity of

0.4 m s–1. Determine the velocity of C relative to A when it is 20 cm away

from A and link AB is rotating about A with an angular velocity of 8 rad s–1.

FIG. 9

C 20 cm

0.4 m s–1

8 rad s–1

v CA tangential

v v vCA CA CA= +tangential radial

Teesside University Open Learning (Engineering)

Solution

Since link AB is rotating about A, then slide C has a tangential velocity

vCA tangential relative to A. The magnitude of vCA tangential can be obtained by

Now, drawing the velocity vector diagram to scale (1 cm represents 0.2 m s–1),

we can find out the magnitude and direction of the velocity of C relative to A,

as shown in FIGURE 10.

FIG. 10

0.4 m s–1

1.6 m s–1

1.65 m s–1

v rCA tangential

–1 m s

= × ×

20 10 8

1 6

2–

Teesside University Open Learning (Engineering)

Looking at FIGURE 10, ab, bc and ca represent the magnitudes of velocities

vCA radial, vCA tangential and vCA, respectively, where 1cm represents 0.2 m s –1.

Thus

then

Also, the angle θ can be measured using a protractor, which gives θ = 14°.

In this particular case, we also can calculate vCA trigonometrically as

vCA tangential and vCA radial are normal to each other. Hence,

Since

then,

The following examples show the method of drawing velocity vector diagrams

and how to use them to determine velocities of the links in the slide-crank and

four-bar systems.

v v vCA CA CA= +

= +

2 2

2 21 6 0 4

tangential radial

. .

1 65

0 4 1 6

0 25

= °

m s

tan

–1

vCA

= = ≅2 cm, bc cm and ca cm

8 8 26.

ddial –1

tangential –1 m s m s= =0 4 1 6. , .vCA ,,

and m s–1vC ≅ 1 65

Teesside University Open Learning (Engineering)

Example 4

For the slide-crank mechanism shown in FIGURE 11, determine the angular

velocity of link BC when the angle ∠BAC = 60°.

FIG. 11

Solution

Link AB is rotating about A with the angular velocity of 40 rad s–1, then the

magnitude of velocity B relative to A, vBA, is obtained by

Since vBA is the tangential velocity, its vector is at right angles to link AB.

v lBA AB AB=

= × ×

200 10 40

3–

m s–1

60°

200 mm 500 mm

ω = 40 rad s–1

Teesside University Open Learning (Engineering)

From the question, we know that link AC and slide D are jointed at point C.

The latter is moving in the horizontal direction, so point C does the same.

Hence, joint C is moving relatively to A in the horizontal direction with the

velocity vCA, and also moving relatively to B in the direction normal to link BC

with the velocity vCB simultaneously, as shown in FIGURE 12. The magnitude

of vCB can be calculated by

where ωCB is the angular velocity of link BC.

FIG. 12

Now, we can draw the velocity vector diagram with a scale of 1 cm

representing 1 m s–1 on graph paper using the following steps:

(i) draw line ef at an angle of 60° to the hor izontal, as shown in

FIGURE 13(a);

(ii) starting with point b on ef, draw line bg at right angles to ef as shown in

FIGURE 13(b);

60°

ω = 40 rad s–1

ω CB

v CBv

v CA

v lCB CB CB= ω

Teesside University Open Learning (Engineering)

(iii) find point c on line bg, which is 8 cm away from point b, as shown in

FIGURE 13(c);

(iv) starting with point c, draw the horizontal line cd. Lines cd and ef meet at

point a as shown in FIGURE 13 (d).

Then, we have triangle ∆abc, which is the velocity vector diagram for this example. From this diagram we can find out the velocity magnitudes of vBA,

vCA and vCB by measuring the length of each side of this triangle.

You can mark velocities vBA, vCA and vCB, on the diagram with a proper

direction as shown in FIG 13 (e). Now, we have:

bc = 8 cm vBA = 8 m s –1

ca = 9.24 cm vCA = 9.24 m s –1

ab = 4.62 cm vCB = 4.62 m s –1

Since vCB = lCBωCB and lCB = 500 mm = 0.5 m, then,

The angular velocity of link BC is 9.24 rad s–1.

ω CB CB

l =

4 62 0 5

9 24

. .

. rad s–1

Teesside University Open Learning (Engineering)

FIG. 13

Teesside University Open Learning (Engineering)

(a)

60°

(b)

(c)

8 cm

60°

FIG. 13 (continued)

60°

(e)

c a

d v

v BA

v CB

60°

(d)

c a

Teesside University Open Learning (Engineering)

Example 5

For the four-bar chain shown in FIGURE 14, determine the angular velocity of

link CD and the velocity of point P relative to point B.

FIG. 14

Solution

Look at FIGURE 14, since link AB is rotating about A with the angular

velocity of 4 rad s–1, then the magnitude of the velocity of B relative to A, vBA,

is obtained by

v lBA AB AB=

= ×

0 5 4

m s–1

0.5 m

0.75 m

0.6 m 0.6 m

0.9 m D

4 rad s–1

Teesside University Open Learning (Engineering)

Joint point C has two velocities, vCD and vCB. vCD is the velocity of C relative

to D (or the absolute velocity since D is fixed to the ground), and is at right

angles to CD. The magnitude of vCD can be determined by

vCB is the velocity of C relative to B, which is at right angles to BC, with

magnitude of

From the question, we know that P is the mid-point of CB. Hence, the velocity

direction of vPB is at right angles to CB as well. The magnitude of vPB is

FIG. 15

4 rad s–1

v BA

v PB

v CB

ω CD

ω CB

v CD

v lPB PB CB= ω . . . . . . . . . . . . . . . . . . . . . . (3)

v lCB CB CB= ω . . . . . . . . . . . . . . . . . . . . . . (2)

v lCD CD CD= ω . . . . . . . . . . . . . . . . . . . . . . (1)

Teesside University Open Learning (Engineering)

First, we should draw a space diagram to scale of the mechanism, say 1 cm

represents 0.1 m. Referring to FIGURE 16, the steps used in drawing the

space diagram are:

(i) draw a 9 cm horizontal line AD;

(ii) draw arc EF centred at A with a radius of 5 cm;

(iii) draw arc GH centred at D with a radius of 7.5 cm;

(iv) draw arc JK centred at B on EF (B is any point on EF) with a radius of

12 cm. JK and GH meet at point C;

(v) connect BC;

(vi) find the middle point of BC and mark as point P.

Then, the space diagram of this four-bar chain mechanism in the scale of 1 cm

to 0.1 m has been completed, in which AD = 9 cm, AB = 5 cm,

BC = 12 cm, and CD = 7.5 cm.

FIG. 16

G C H

5 cm

12 cm

7.5 cm

9 cm

Teesside University Open Learning (Engineering)

Now, following the steps below, we can draw velocity vector diagram shown

in FIGURE 17 with an appropriate scale, say 1 cm represents 0.5 m s–1:

(i) starting with point a, draw line ab with the length of 4 cm, which is at

right angles to AB;

(ii) draw line de passing through point b and normal to BC;

(iii) starting with point a draw line af, which is at right angles to CD; lines af

and de meet at point c.

FIG. 17

Then, we have triangle ∆abc, in which ab, bc and ac represent the velocity magnitudes of vBA, vCB and vCD, respectively. You can mark these velocities

on the diagram with a proper direction. Now, we have:

ab = 4 cm vBA = 2 m s –1

bc = 2 cm vCB = 1 m s –1

ac = 4.3 cm vCD = 2.15 m s –1

C ab = 4 cm

v BA v

v CD

Teesside University Open Learning (Engineering)

Applying equation (1), vCD = lCDωCD, the angular velocity of link CD is calculated by

Applying equation (2), vCB = lCBωCB, the angular velocity of link BC can be determined by

The magnitude of vPB is obtained using equation (3):

v lPB PB CB=

= ×

0 6 0 83

0 5

. .

. rad s–1

ω CB CB

l =

1 1 2

0 83

. rad s–1

ω CD CD

l =

2 15 0 75

2 87

. rad s–1

Teesside University Open Learning (Engineering)

________________________________________________________________________________________

ACCELERATION DIAGRAMS ________________________________________________________________________________________

It is important to determine the acceleration of links since acceleration

produces inertia forces in the link that stress the component parts of the

mechanism.

Similar to velocity, acceleration is also a vector quantity and can be determined

by means of an acceleration vector diagram. Accelerations may be relative or

absolute in the same way as described for velocity.

When a body is subject to more than one acceleration, then the resultant

acceleration is determined by using a parallelogram of vectors. In this section,

we will consider two forms of acceleration occurring in the rotating motion of

links, centripetal (or radial) and tangential, because the gravitational

acceleration due to the weight of links can normally be neglected.

According to our previous study, we already know that when an object rotates

about a centre with radius r it has a tangential velocity v due to the angular

velocity ω. When the velocity changes, either the magnitude or the direction or both, there will be acceleration.

Teesside University Open Learning (Engineering)

CENTRIPETAL ACCELERATION

In a rotating motion, the tangential velocity v of a body always changes

direction, which continually makes it accelerate towards the centre even

though this body never moves any closer to the centre. This acceleration is

called centripetal acceleration or radial acceleration ar. The magnitude of ar can be determined by:

Since v = ωr, then,

where ar = the centripetal acceleration (m s –2)

ω = the angular velocity (rad s–1) r = the radius of the rotating motion (m)

v = the tangential velocity (m s–1)

Note that the direction of the vector ar starts at the end of the link and is

always towards the centre of rotation as shown in FIGURE 18.

FIG. 18 The formation of resultant acceleration a

a t

a r

a v

rr =

2 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5( )

a rr = ω 2 . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4( )

Teesside University Open Learning (Engineering)

TANGENTIAL ACCELERATION

Tangential acceleration at only occurs if an object has an angular acceleration

due to the change of angular velocity ω. The direction of the tangential acceleration vector at is always parallel to the tangential velocity vector v, and

perpendicular to the radius vector of the circular motion as shown in

FIGURE 18. The magnitude of at is obtained by:

Again, the example given below will show you how to draw an acceleration

vector diagram and its application.

Example 6

Using the conditions given in Example 4 shown in FIGURE 11 (reproduced

below), find the acceleration of slide D and the acceleration of link CB.

FIG. 11 (Reproduced)

60°

200 mm 500 mm

ω = 40 rad s–1

a v

tt = =

( )d d

d . . . . . . . . . . . . .

ω .. . . . . . . 6( )

Teesside University Open Learning (Engineering)

Solution

As mentioned in Example 4, slide D has the same motion as joint C.

Therefore, if we can determine the kinematic property of link BC, the problem

will be solved.

From Example 4, we already know:

vBA = 8 m s –1, vCA = 9.24 m s

–1, vCB = 4.62 m s –1 and ωCB = 9.24 rad s–1

The directions of velocity vectors vBA, vCA and vCB are shown in FIGURE 12.

FIG. 12 (Reproduced)

Then, the magnitude of the centripetal acceleration of joint B, (aBA)r, is

obtained by applying equation (4):

a lBA r AB AB( ) =

= × ×

ω 2

3 2200 10 40

320

–

m s–2

60°

ω = 40 rad s–1

ω CB

v CBv

v CA

Teesside University Open Learning (Engineering)

There is no tangential acceleration for link AB as ωAB = constant.

The true or resultant acceleration of joint C is formed by the centripetal

acceleration relative to point B, (aCB)r, and the tangential acceleration relative

to point B, (aCB)t.

Again, applying equation (4), we can calculate the magnitude of the centripetal

acceleration of joint C relative to B, (aCB)r, which is

At this stage, the magnitude of the tangential acceleration of C relative to B,

(aCB)t, and the true acceleration of C or D relative to A, aCA, are unknown.

However, it is sure that (aCB)t is along the line at right angles to link CB, and

aCA is along line AC (horizontal line). FIGURE 19 shows the acceleration

vectors related to link CB.

FIG. 19

60°

ω AB

ω CB

(a CB

) r

(a CB

) t

a CA

(a BA

) r

a lCB r CB CB( ) =

= × ×

ω 2

3 2500 10 9 24

42 69

– .

. m s–2

Teesside University Open Learning (Engineering)

This time, we need help from the space diagram of the mechanism in order to

decide the directions of the accelerations of (aCB)r and (aCB)t. Following the

steps below, this diagram (shown in FIGURE 20) can be easily drawn:

(i) draw a horizontal line AD;

(ii) draw line AB with length of 2 cm representing 200 mm, which is at 60° to

line AD;

(iii) draw arc EF centred at B with a radius of 5 cm. Arc EF and line AD meet

at point C.

Now, we have triangle ABC, which is the space diagram of the slide-crank

mechanism.

FIG. 20

60° A

C D

F 5 cm

2 cm

Teesside University Open Learning (Engineering)

Referring to FIGURE 21, the stage by stage construction of the acceleration

vector diagram with an appropriate scale (say 1 cm representing 40 m s–2) is as

follows:

(i) starting with point a (or B), draw line ab along CB produced with the

length of 1.1 cm (approx.) representing 42.69 m s–2, the magnitude of the

centripetal acceleration of point C relative to B, (aCB)r;

(ii) again, starting with point a, draw vector ad along AB produced with the

length of 8 cm representing 320 m s–2, the magnitude of the centripetal

acceleration of point B relative to A, (aBA)r;

(iii) starting with point b, draw line bf that is at right angles to ab;

(iv) starting with point d, draw a horizontal line to meet line bf at c.

Then, we have quadrangle abcd, which is the acceleration vector diagram for

link BC, in which ab, ad, bc and cd represent the magnitudes of the

accelerations of (aCB)r, (aBA)r, (aCB)t and aCA, respectively, as follows:

ab = 1.1 cm (aCB)r = 42.69 m s –2

ad = 8 cm (aBA)r = 320 m s –2

bc = 7.0 cm (aCB)t = 280 m s –2

cd = 2.6 cm (aCA)t = 104 m s –2

Thus, slide D has an acceleration of 104 m s–2.

Teesside University Open Learning (Engineering)

FIG. 21

From our previous study, we already know the relationship between the linear

and angular acceleration is

In this case, we have

a lCB t CB CB( ) = α

a l= α

c d

8 cm

1.1 cm

a CA

(a CB

) r

(a CB

) t

(a BA

) r

60°

60° C

Teesside University Open Learning (Engineering)

Then, the angular acceleration of link CB is:

Example 7

Find the angular acceleration of the link CD for the case shown in FIGURE 22.

FIG. 22

D 60° ω

AB = 500 rad s–1

70° 250 mm

200 mm

100 mm

α CB CB t

l =

( )

= ×

280 500 10

560

3–

rad s–2

Teesside University Open Learning (Engineering)

Solution

Firstly, we should draw the space diagram shown in FIGURE 23 with an

appropriate scale, say 1 cm represents 50 mm, by following the steps below:

(i) draw the horizontal line AD with the length of 5 cm;

(ii) starting with point D, draw line DC with the length of 4 cm at the angle of

70° to DA;

(iii) starting with point A, draw line AB with the length of 2 cm with angle

∠BAD equal to 60°;

(iv) connect CB.

Now we have the quadrangle ABCD, which is the space diagram of the four-

bar chain mechanism in this example. Measuring we find that CB = 3.32 cm,

which gives the length of link CB as 166 mm.

FIG. 23

D 60°70°

Teesside University Open Learning (Engineering)

We can only calculate the magnitude of velocity B relative to A, vBA, which is:

Following the steps below, the velocity vector diagram shown in FIGURE 24

can be constructed with the scale of 1 cm to 10 m s–1:

(i) starting with point a (point B in the space diagram), draw line ab with the

length of 5 cm (representing 50 m s–1). The line ab is at right angles to

AB;

(ii) starting with point b, draw line bf that is at right angles to line CB;

(iii) starting with point a, draw line ac at right angles to CD, to meet bf at c.

Now, we have triangle ∆abc. After marking the velocities vBA, vBC and vCD with proper direction, we find that:

ab = 5 cm, vBA = 50 m s –1

bc = 4.2 cm, vBC = 42 m s –1

ac = 1.9 cm, vCD = 19 m s –1

v lBA BA BA=

= × ×

500 100 10

3–

m s–1

Teesside University Open Learning (Engineering)

FIG. 24

Next, we should calculate all magnitudes of the accelerations possible.

The magnitude of the centripetal acceleration of B relative A is:

The magnitude of the centripetal acceleration of C relative B is:

The magnitude of the centripetal acceleration of C relative D is:

At this stage, any tangential accelerations are unknown, but they can be

determined by means of the acceleration vector diagram. Note that there is no

tangential acceleration of B relative to A as link BA is rotating at a constant

angular velocity.

a v

lCD r CD

CD ( ) = =

× =

2 2

19 200 10

1805 –

m s–2

a v

lCB r CB

CB ( ) = =

× =

2 2

42 166 10

10 627 –

m s–2

a v

lBA r BA

BA ( ) = =

× =

2 2

50 100 10

25 000 –

m s–2

60°70°

v BA

v BCv

Teesside University Open Learning (Engineering)

Referring to FIGURE 25, the stage by stage construction of this diagram with

an appropriate scale, say 1 cm represents 2500 m s–2, is done as follows:

(i) starting with point a (point B in the space diagram) and along line BA,

draw line ab with the length of 10 cm;

(ii) starting with point b, draw line bc with the length of 4.46 cm that is

parallel to BC;

(iii) starting with point c, draw line de that shows the direction of the

tangential acceleration vector of C relative to B, so de should be at right

angles to line CB;

(iv) starting with point a, draw line af with the length of 0.72 cm. Line af is

parallel to line CD;

(v) starting with point f, draw line fg that is at right angles to CD and meets

de at point h.

The polygon abchf is the acceleration vector diagram for the four-bar chain

mechanism. The length of line fh, 14.2 cm, represents the tangential

acceleration magnitude of C relative to D, which is 35 500 m s–2. Therefore,

the angular acceleration of the link CD, αCD can be determined by:

then

From the study of this example, it is found that the link CD has angular

acceleration although the link BA rotates at a constant speed.

a l

CD t CD

CD CD

( ) =

= = ×

35 500 200 10

177 50 3–

00 rad s–2

Teesside University Open Learning (Engineering)

FIG. 25

Teesside University Open Learning (Engineering)

f (a

CD ) r

(a CD

) t

(a BA

) r

(a CB

) r

(a CB

) t

D 60°70°

POWER TRANSMISSION AND EFFICIENCY

The reason for finding the accelerations of links in the mechanism is to

calculate forces or torques exerted on the links, in order to determine the power

transmission and the efficiency of the mechanism.

When an object of mass m is in linear motion with an acceleration of a, such as

slide D in Examples 4 and 6, according to Newton’s second law the inertial

force acting on it is

In a rotating system, it has been shown that power transmitted P is the product

of torque T and angular velocity ω:

And, torque T is the product of the moment of inertia I and angular

acceleration α:

The efficiency η is defined as the ratio of the output power Pout to the input power Pin:

η = P

P out

T I= α

P T= ω

F ma=

Teesside University Open Learning (Engineering)

Example 7

Using the conditions and results of Examples 4 and 6, now consider the slide D

as a piston having a mass of 0.5 kg and a diameter of 80 mm, which

reciprocates in a horizontal cylinder as shown in FIGURE 26. The pressure in

the cylinder is 1 MPa. Assuming that the links BA and BC have negligible

inertia and friction, calculate the effective turning moment (or torque) acting

on the crank BA.

FIG. 26

60°

ω AB

= 40 rad s–1

200 mm 500 mm 1 MPa

Teesside University Open Learning (Engineering)

Solution

The effective turning moment acting on the crank BA can be obtained by

FIG. 27

Based on the space diagram (scale: 1 cm represents 100 mm), it can be found

that θ = 20° and the angle ∠CBE = 10°, as shown in FIGURE 27. Thus, the effective force F is calculated by

FBC is the force acting on link CB by the piston due to the pressure in the

cylinder, which can be determined by the analysis as follows.

For the piston D, the net force in the moving direction is

where p = the pressure in the cylinder = 1 MPa = 1 × 106 Pa

A = the cross-section area of the cylinder

F pA F maD BC D= =– cos 20°

F FBC= cos 10°

60° A

C E

5 cm2 cm

30°

F BC

10°

20°

T FlBA=

Teesside University Open Learning (Engineering)

m = the mass of the piston = 0.5 kg

aD = the acceleration of the piston, from Example 6,

Thus, the force acting on D by link BC, FBC, is calculated by

Therefore, the effective turning moment acting on the crank BA can be

obtained from:

T Fl F lBA BC BA= =

= × × ×

cos 10°

cos 10° 5297 200 10 3–

== 1043 3. Nm

F pA ma

BC D=

= × × × ×

–

. – .–

cos 20°

co 1 10 5 03 10 0 5 1046 3

ss 20°

N= 5297

aD = 104 m s –2

A d= = ×( ) = ×π π 4 4

80 10 5 03 102 3 2 3– –. m2

Teesside University Open Learning (Engineering)

________________________________________________________________________________________

SELF-ASSESSMENT QUESTION ________________________________________________________________________________________

1. A horizontal single cylinder reciprocating engine is shown in FIGURE 28.

The crank BA rotates at ωBA = 300 rad s–1. The lengths of the crank BA and the rod BC are 50 mm and 200 mm respectively. The piston E has a

mass of 0.4 kg and a diameter of 60 mm. The pressure in the cylinder is

15 bar.

Assuming the inertia of the crank and the rod and the friction in the

mechanism can be ignored, calculate:

(a) the inertial force acting on the piston

(b) the effective turning torque acting on the crank BA.

FIG. 28

ω BA

C 30°

Teesside University Open Learning (Engineering)

________________________________________________________________________________________

NOTES ________________________________________________________________________________________

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

Teesside University Open Learning (Engineering)

________________________________________________________________________________________

SOLUTION TO SELF-ASSESSMENT QUESTION ________________________________________________________________________________________

1. (a) Firstly, draw the space diagram to scale, as shown in FIGURE 29

(scale: 1 cm represents 20 mm), and find lCA = 242 mm, ∠BCA = 7°.

Mark F and FCB on the diagram and find ∠CBD = 53°.

F is the force to produce the effective torque on the crank BA, which

Note that

FIG. 29

53°

30°C A

BF CB

D 7°

10 cm

12.1 cm

2.5 cm

T F l

F F

cos 53°

Teesside University Open Learning (Engineering)

The magnitude of velocity B relative to A,

Next, draw the velocity vector diagram as shown in FIGURE 30

(scale: 1 cm represents 2 m s–1). Then find that:

ab = 7.5 cm vBA = 15 m s –1

bc = 6 cm vBC = 12 m s –1

ca = 4.5 cm vCA = 9 m s –1

Now calculate all the accelerations possible.

FIG. 30

ω BC

C A

a ωBA

vBA vBC

vCA

v lBA BA BA=

= × ×

300 50 10

3–

m s–1

Teesside University Open Learning (Engineering)

Point B only has a centripetal acceleration relative to A, (aBA)r, as the

crank rotates at a constant angular velocity ωBA. The magnitude of (aBA)r is

The magnitude of the centripetal acceleration of B relative to C,

(aBC)r, can be obtained by

The tangential acceleration of B relative to C is unknown at this

stage, but the direction must be at right angles to BC.

Then, we can draw the acceleration vector diagram as shown in

FIGURE 31 (scale: 1 cm represents 450 m s–2).

a v

lBC r BC

BC ( ) = =

2 2

12 200 10

720

–

m s–2

a v

lBA r BA

BA ( ) = =

2 2

15 50 10

4500

–

m s–2

Teesside University Open Learning (Engineering)

FIG. 31

ω B

b(a C

B ) r

(a B

A ) r

a C A

4 5 0 0 m

s– 2

(a B

C ) t

7 2 0 m

s– 2

Teesside University Open Learning (Engineering)

From FIGURE 31 we can find that

ab = 10 cm (aBA)r = 4500 m s –2

bc = 1.6 cm (aBC)r = 720 m s –2

cd = 4.8 cm (aBC)t = 2160 m s –2

ad = 9.6 cm aCA = 4320 m s –2

For the piston D, the inertial force is maCA, which is

(b) The forces on the piston D as a free body are shown in FIGURE 32.

FIG. 32

In the horizontal direction, the force balance fomula is

then F P ma

BA CA=

= × × × ×( )

–

––

cos 7°

15 10 4

60 10 17285 3 2π

oos 7°

N= 2532 0.

P F ma P pABA CA– cos 7° and = =

7° F

a CA

0 4 4320 1728. × = N

Teesside University Open Learning (Engineering)

The effective turning torque acting on the crank BA is

T Fl F lBA BA BA= =

= × × ×

cos 53° .

cos 53°2532 0 50. 110

76 2

3–

.= Nm

Teesside University Open Learning (Engineering)

________________________________________________________________________________________

SUMMARY ________________________________________________________________________________________

The kinematics of single and multi-link mechanisms in rotating systems is a

complicated topic, which involves the construction and study of space

diagrams, velocity vector diagrams and acceleration diagrams. Without help

from these diagrams, some engineering problems would be difficult to solve.

In the next lesson, we will look at the balancing of single plane and multi-

plane rotating mass systems.

Teesside University Open Learning (Engineering)

<< /ASCII85EncodePages false /AllowTransparency false /AutoPositionEPSFiles true /AutoRotatePages /None /Binding /Left /CalGrayProfile (Dot Gain 20%) /CalRGBProfile (sRGB IEC61966-2.1) /CalCMYKProfile (U.S. Web Coated \050SWOP\051 v2) /sRGBProfile (sRGB IEC61966-2.1) /CannotEmbedFontPolicy /Error /CompatibilityLevel 1.4 /CompressObjects /Tags /CompressPages true /ConvertImagesToIndexed true /PassThroughJPEGImages true /CreateJDFFile false /CreateJobTicket false /DefaultRenderingIntent /Default /DetectBlends true /ColorConversionStrategy /LeaveColorUnchanged /DoThumbnails false /EmbedAllFonts true /EmbedJobOptions true /DSCReportingLevel 0 /SyntheticBoldness 1.00 /EmitDSCWarnings false /EndPage -1 /ImageMemory 1048576 /LockDistillerParams false /MaxSubsetPct 100 /Optimize true /OPM 1 /ParseDSCComments true /ParseDSCCommentsForDocInfo true /PreserveCopyPage true /PreserveEPSInfo true /PreserveHalftoneInfo false /PreserveOPIComments false /PreserveOverprintSettings true /StartPage 1 /SubsetFonts true /TransferFunctionInfo /Apply /UCRandBGInfo /Preserve /UsePrologue false /ColorSettingsFile () /AlwaysEmbed [ true ] /NeverEmbed [ true ] /AntiAliasColorImages false /DownsampleColorImages true /ColorImageDownsampleType /Bicubic /ColorImageResolution 300 /ColorImageDepth -1 /ColorImageDownsampleThreshold 1.50000 /EncodeColorImages true /ColorImageFilter /DCTEncode /AutoFilterColorImages true /ColorImageAutoFilterStrategy /JPEG /ColorACSImageDict << /QFactor 0.15 /HSamples [1 1 1 1] /VSamples [1 1 1 1] >> /ColorImageDict << /QFactor 0.15 /HSamples [1 1 1 1] /VSamples [1 1 1 1] >> /JPEG2000ColorACSImageDict << /TileWidth 256 /TileHeight 256 /Quality 30 >> /JPEG2000ColorImageDict << /TileWidth 256 /TileHeight 256 /Quality 30 >> /AntiAliasGrayImages false /DownsampleGrayImages true /GrayImageDownsampleType /Bicubic /GrayImageResolution 300 /GrayImageDepth -1 /GrayImageDownsampleThreshold 1.50000 /EncodeGrayImages true /GrayImageFilter /DCTEncode /AutoFilterGrayImages true /GrayImageAutoFilterStrategy /JPEG /GrayACSImageDict << /QFactor 0.15 /HSamples [1 1 1 1] /VSamples [1 1 1 1] >> /GrayImageDict << /QFactor 0.15 /HSamples [1 1 1 1] /VSamples [1 1 1 1] >> /JPEG2000GrayACSImageDict << /TileWidth 256 /TileHeight 256 /Quality 30 >> /JPEG2000GrayImageDict << /TileWidth 256 /TileHeight 256 /Quality 30 >> /AntiAliasMonoImages false /DownsampleMonoImages true /MonoImageDownsampleType /Bicubic /MonoImageResolution 1200 /MonoImageDepth -1 /MonoImageDownsampleThreshold 1.50000 /EncodeMonoImages true /MonoImageFilter /CCITTFaxEncode /MonoImageDict << /K -1 >> /AllowPSXObjects false /PDFX1aCheck false /PDFX3Check false /PDFXCompliantPDFOnly false /PDFXNoTrimBoxError true /PDFXTrimBoxToMediaBoxOffset [ 0.00000 0.00000 0.00000 0.00000 ] /PDFXSetBleedBoxToMediaBox true /PDFXBleedBoxToTrimBoxOffset [ 0.00000 0.00000 0.00000 0.00000 ] /PDFXOutputIntentProfile () /PDFXOutputCondition () /PDFXRegistryName (http://www.color.org) /PDFXTrapped /Unknown /Description << /ENU (Use these settings to create PDF documents with higher image resolution for high quality pre-press printing. The PDF documents can be opened with Acrobat and Reader 5.0 and later. These settings require font embedding.) /JPN <FEFF3053306e8a2d5b9a306f30019ad889e350cf5ea6753b50cf3092542b308030d730ea30d730ec30b9537052377528306e00200050004400460020658766f830924f5c62103059308b3068304d306b4f7f75283057307e305930023053306e8a2d5b9a30674f5c62103057305f00200050004400460020658766f8306f0020004100630072006f0062006100740020304a30883073002000520065006100640065007200200035002e003000204ee5964d30678868793a3067304d307e305930023053306e8a2d5b9a306b306f30d530a930f330c8306e57cb30818fbc307f304c5fc59808306730593002> /FRA <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> /DEU <FEFF00560065007200770065006e00640065006e0020005300690065002000640069006500730065002000450069006e007300740065006c006c0075006e00670065006e0020007a0075006d002000450072007300740065006c006c0065006e00200076006f006e0020005000440046002d0044006f006b0075006d0065006e00740065006e0020006d00690074002000650069006e006500720020006800f60068006500720065006e002000420069006c0064006100750066006c00f600730075006e0067002c00200075006d002000650069006e00650020007100750061006c00690074006100740069007600200068006f006300680077006500720074006900670065002000410075007300670061006200650020006600fc0072002000640069006500200044007200750063006b0076006f0072007300740075006600650020007a0075002000650072007a00690065006c0065006e002e00200044006900650020005000440046002d0044006f006b0075006d0065006e007400650020006b00f6006e006e0065006e0020006d006900740020004100630072006f0062006100740020006f0064006500720020006d00690074002000640065006d002000520065006100640065007200200035002e003000200075006e00640020006800f600680065007200200067006500f600660066006e00650074002000770065007200640065006e002e00200042006500690020006400690065007300650072002000450069006e007300740065006c006c0075006e00670020006900730074002000650069006e00650020005300630068007200690066007400650069006e00620065007400740075006e00670020006500720066006f0072006400650072006c006900630068002e> /PTB <FEFF005500740069006c0069007a006500200065007300740061007300200063006f006e00660069006700750072006100e700f5006500730020007000610072006100200063007200690061007200200064006f00630075006d0065006e0074006f0073002000500044004600200063006f006d00200075006d00610020007200650073006f006c007500e700e3006f00200064006500200069006d006100670065006d0020007300750070006500720069006f0072002000700061007200610020006f006200740065007200200075006d00610020007100750061006c0069006400610064006500200064006500200069006d0070007200650073007300e3006f0020006d0065006c0068006f0072002e0020004f007300200064006f00630075006d0065006e0074006f0073002000500044004600200070006f00640065006d0020007300650072002000610062006500720074006f007300200063006f006d0020006f0020004100630072006f006200610074002c002000520065006100640065007200200035002e00300020006500200070006f00730074006500720069006f0072002e00200045007300740061007300200063006f006e00660069006700750072006100e700f50065007300200072006500710075006500720065006d00200069006e0063006f00720070006f0072006100e700e3006f00200064006500200066006f006e00740065002e> /DAN <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> /NLD <FEFF004700650062007200750069006b002000640065007a006500200069006e007300740065006c006c0069006e00670065006e0020006f006d0020005000440046002d0064006f00630075006d0065006e00740065006e0020007400650020006d0061006b0065006e0020006d00650074002000650065006e00200068006f00670065002000610066006200650065006c00640069006e00670073007200650073006f006c007500740069006500200076006f006f0072002000610066006400720075006b006b0065006e0020006d0065007400200068006f006700650020006b00770061006c0069007400650069007400200069006e002000650065006e002000700072006500700072006500730073002d006f006d0067006500760069006e0067002e0020004400650020005000440046002d0064006f00630075006d0065006e00740065006e0020006b0075006e006e0065006e00200077006f007200640065006e002000670065006f00700065006e00640020006d006500740020004100630072006f00620061007400200065006e002000520065006100640065007200200035002e003000200065006e00200068006f006700650072002e002000420069006a002000640065007a006500200069006e007300740065006c006c0069006e00670020006d006f006500740065006e00200066006f006e007400730020007a0069006a006e00200069006e006700650073006c006f00740065006e002e> /ESP <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> /SUO <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> /ITA <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> /NOR <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> /SVE <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> >> >> setdistillerparams << /HWResolution [2400 2400] /PageSize [612.000 792.000] >> setpagedevice