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Mechanical Engineering Principles
Fourth Edition
Why are competent engineers so vital?

Engineering is among the most important of all professions. It is the authors’ opinions that engineers save
more lives than medical doctors (physicians). For example, poor water, or the lack of it, is the second largest
cause of human death in the world, and if engineers are given the ‘tools’, they can solve this problem. The
largest cause of human death is caused by the malarial mosquito, and even death due to malaria can be de-
creased by engineers – by providing helicopters for spraying areas infected by the mosquito and making and
designing medical syringes and pills to protect people against catching all sorts of diseases. Most medicines
are produced by engineers! How does the engineer put 1 mg of ‘medicine’ precisely and individually into mil-
lions of pills, at an affordable price?
Moreover, one of the biggest contributions by humankind was the design of the agricultural tractor, which
was designed and built by engineers to increase food production many-fold, for a human population which
more-or-less quadruples every century! It is also interesting to note that the richest countries in the world are
very heavily industrialized. Engineers create wealth! Most other professions don’t!
Even in blue sky projects, engineers play a major role. For example, most rocket scientists are chartered
engineers or their equivalents and Americans call their chartered engineers (and their equivalents), scien-
tists. Astronomers are space scientists and not rocket scientists; they could not design a rocket to conquer
outer space. Even modern theoretical physicists are mainly interested in astronomy and cosmology and also
nuclear science. In general a theoretical physicist cannot, without special training, design a submarine struc-
ture to dive to the bottom of the Mariana Trench, which is 11.52 km or 7.16 miles deep, or design a very long
bridge, a tall city skyscraper or a rocket to conquer outer space. It may be shown that the load on a submarine
pressure hull of diameter 10 m and length 100 m is equivalent to carrying the total weight of about 7 million
London double-decker buses!
This book presents a solid foundation for the reader in mechanical engineering principles, on which s/he
can safely build tall buildings and long bridges that may last for a thousand years or more. It is the authors’
experience that it is most unwise to attempt to build such structures on shaky foundations; they may come
tumbling down – with disastrous consequences.
John Bird is the former Head of Applied Electronics in the Faculty of Technology at Highbury College, Ports-
mouth, UK. More recently, he has combined freelance lecturing at the University of Portsmouth with Examiner
responsibilities for Advanced Mathematics with City and Guilds and examining for the International Baccalaureate
Organisation. He is the author of over 135 textbooks on engineering and mathematical subjects with worldwide
sales of over one million copies. He is a chartered engineer, a chartered mathematician, a chartered scientist and a
Fellow of three professional institutions, and is currently lecturing at the Defence School of Marine Engineering in
the Defence College of Technical Training at H.M.S. Sultan, Gosport, Hampshire, UK, one of the largest technical
training establishments in Europe.
Carl Ross gained his first degree in Naval Architecture from King’s College, Durham University, his PhD in Struc-
tural Engineering from the Victoria University of Manchester, and was awarded his DSc in Ocean Engineering from
the CNAA, London. His research in the field of engineering led to advances in the design of submarine pressure
hulls. His publications and guest lectures to date exceed some 300 papers, books, etc., and he was Professor of
Structural Dynamics at the University of Portsmouth, UK.
Carl Ross’s websites have an enormous content on science, technology and education.
Use any search engine (such as Google):
carltfross YouTube or carl t f ross YouTube or YouTube carl t f ross
Some quotes from Albert Einstein (14 March 1879–18 April 1955)
‘Scientists investigate that which already is; Engineers create that which has never been’
‘Imagination is more important than knowledge. For knowledge is limited to all we now know and understand,
while imagination embraces the entire world, and all there ever will be to know and understand’
‘Everybody is a genius. But if you judge a fish by its ability to climb a tree, it will live its whole life believing
that it is stupid’
‘To stimulate creativity, one must develop the childlike inclination for play’
Mechanical Engineering Principles
Fourth Edition
John Bird BSc(Hons), CEng, CMath, CSci, FIMA, FIET, FCollT

Carl Ross BSc(Hons), PhD, DSc, CEng, FRINA, MSNAME
Fourth edition published 2020
by Routledge
2 Park Square, Milton Park, Abingdon, Oxon, OX14 4RN
and by Routledge
52 Vanderbilt Avenue, New York, NY 10017
Routledge is an imprint of the Taylor & Francis Group, an informa business
© 2020 John Bird and Carl Ross
The right of John Bird and Carl Ross to be identified as authors of this work has been asserted by them in accordance with sec-
tions 77 and 78 of the Copyright, Designs and Patents Act 1988.
All rights reserved. No part of this book may be reprinted or reproduced or utilised in any form or by any electronic, mechani-
cal, or other means, now known or hereafter invented, including photocopying and recording, or in any information storage or
retrieval system, without permission in writing from the publishers.
Trademark notice: Product or corporate names may be trademarks or registered trademarks, and are used only for identification
and explanation without intent to infringe.
First edition published by Elsevier 2002

Third edition published by Routledge 2015
British Library Cataloguing-in-Publication Data

A catalogue record for this book is available from the British Library
Library of Congress Cataloging-in-Publication Data

A catalog record has been requested for this book
ISBN: 978-0-367-25326-4 (hbk)

ISBN: 978-0-367-25324-0 (pbk)
ISBN: 978-0-429-28720-6 (ebk)
Typeset in Times New Roman

by Servis Filmsetting Ltd, Stockport, Cheshire
Visit the companion website: www.routledge.com/cw/bird

Contents
Prefaceix 3.9 Ductility, brittleness and malleability 57
3.10 Modulus of rigidity 57
3.11 Thermal strain 57
Part One Revision of Mathematics 1 3.12 Compound bars 58
1 Revisionary mathematics 3
4 Tensile testing 64
1.1 Introduction 3
4.1 The tensile test 64
1.2 Radians and degrees 4
4.2 Worked problems on tensile testing 66
1.3 Measurement of angles 4
4.3 Further worked problems on tensile testing 68
1.4 Triangle calculations 5
4.4 Proof stress 70
1.5 Brackets 8
1.6 Fractions 8
5 Forces acting at a point 72
1.7 Percentages 10
5.1 Scalar and vector quantities 72
1.8 Laws of indices 12
5.2 Centre of gravity and equilibrium 73
1.9 Simultaneous equations 14
5.3 Forces 73
5.4 The resultant of two coplanar forces 74
Revision Test 1 Revisionary mathematics 18
5.5 Triangle of forces method 75
5.6 The parallelogram of forces method 76
2 Further revisionary mathematics 20 5.7 Resultant of coplanar forces by
2.1 Units, prefixes and engineering notation 21 calculation77
2.2 Metric–US/Imperial conversions 24 5.8 Resultant of more than two coplanar forces 77
2.3 Straight line graphs 28 5.9 Coplanar forces in equilibrium 79
2.4 Gradients, intercepts and equation of a graph 30 5.10 Resolution of forces 81
2.5 Practical straight line graphs 32 5.11 Summary 84
2.6 Introduction to calculus 34
2.7 Basic differentiation revision 34 6 Simply supported beams 87
2.8 Revision of integration 36 6.1 The moment of a force 87
2.9 Definite integrals 38 6.2 Equilibrium and the principle of moments 88
2.10 Simple vector analysis 39 6.3 Simply supported beams having
point loads 90
Revision Test 2 Further revisionary mathematics 43 6.4 Simply supported beams with couples 94
Revision Test 3 Forces, tensile testing

and beams 98
Part Two Statics and Strength
of Materials 45
7 Forces in structures 99
3 The effects of forces on materials 47 7.1 Introduction 99
3.1 Introduction 48 7.2 Worked problems on mechanisms
3.2 Tensile force 48 and pin-jointed trusses 100
3.3 Compressive force 48 7.3 Graphical method 101
3.4 Shear force 48 7.4 Method of joints (a mathematical method) 105
3.5 Stress 49 7.5 The method of sections (a mathematical
3.6 Strain 50 method)110
3.7 Elasticity, limit of proportionality
and elastic limit 52 8 Bending moment and shear force diagrams 113
3.8 Hooke’s law 53 8.1 Bending moment (M)113
vi Contents
8.2 Shearing force (F)114 14 The matrix displacement method 178
8.3 Worked problems on bending 14.1 Introduction 179
moment and shearing force diagrams 114 14.2 The matrix displacement method 179
8.4 Uniformly distributed loads 123 14.3 The structural stiffness matrix (K)180
14.4 Elemental stiffness matrix
9 First and second moments of area 128 for a plane rod 181
9.1 Centroids 128 14.5 Slope-deflection equations 184
9.2 The first moment of area 129 14.6 Continuous beams 185
9.3 Centroid of area between a curve
14.7 Analysis of pin-jointed trusses on
and the x-axis129
SmartPhones, tablets and Microsoft
9.4 Centroid of area between a curve and computers191
the y-axis129
14.8 Analysis of continuous beams on
9.5 Worked problems on centroids of
SmartPhones, tablets and Microsoft
simple shapes 130
computers193
9.6 Further worked problems on centroids
14.9 Analysis of rigid-jointed plane
of simple shapes 131
frames on SmartPhones, tablets
9.7 Second moments of area of regular
and Microsoft computers 196
sections132
9.8 Second moment of area for ‘built-up’
sections139 Part Three Dynamics 199
Revision Test 4 Forces in structures, 15 Linear and angular motion 201
bending moment and shear 15.1 The radian 201
force diagrams, and second 15.2 Linear and angular velocity 201
moments of area 145 15.3 Linear and angular acceleration 203
15.4 Further equations of motion 204
10 Bending of beams 146 15.5 Relative velocity 206
10.1 Introduction 146
σ M E 16 Linear momentum and impulse 210
10.2 To prove that = = 147 16.1 Linear momentum 210
y I R
10.3 Worked problems on the bending 16.2 Impulse and impulsive forces 213
of beams 148
17 Force, mass and acceleration 218
11 Torque 152 17.1 Introduction 218
11.1 Couple and torque 152 17.2 Newton’s laws of motion 219
11.2 Work done and power transmitted 17.3 Centripetal acceleration 222
by a constant torque 153 17.4 Rotation of a rigid body about
11.3 Kinetic energy and moment of inertia 155 a fixed axis 223
11.4 Power transmission and efficiency 158 17.5 Moment of inertia (I)224
12 Twisting of shafts 162 18 Work, energy and power 227

τ T Gθ 18.1 Work 227
12.1 To prove that = = 162
r J L 18.2 Energy 231
12.2 Worked problems on the 18.3 Power 232
twisting of shafts 164 18.4 Potential and kinetic energy 235
18.5 Kinetic energy of rotation 238
Revision Test 5 Bending of beams, torque
and twisting of shafts 168 Revision Test 6 Linear and angular motion,
momentum and impulse,
force, m
ass and acceleration,
13 An introduction to matrix algebra 169 work, energy and power 241
13.2 Elementary matrix algebra 170 19 Friction 242
13.3 Addition and subtraction of matrices 171 19.1 Introduction to friction 242
13.4 Matrix multiplication 171 19.2 Coefficient of friction 243
13.5 Two by two determinants 173 19.3 Applications of friction 244
13.6 Three by three determinants 174
Contents vii
19.4 Friction on an inclined plane 245 24 Thermal expansion 296

19.5 Motion up a plane with the pulling 24.1 Introduction 296
force P parallel to the plane 245 24.2 Practical applications of thermal
19.6 Motion down a plane with the expansion297
pulling force P parallel to the plane 246 24.3 Expansion and contraction of water 297
19.7 Motion up a plane due to a horizontal 24.4 Coefficient of linear expansion 297
force P 246 24.5 Coefficient of superficial expansion 299
19.8 The efficiency of a screw jack 249 24.6 Coefficient of cubic expansion 300
20 Motion in a circle 253
20.1 Introduction 253 Revision Test 8 Heat energy and transfer,
20.2 Motion on a curved banked track 255 and thermal expansion 304
20.3 Conical pendulum 256
20.4 Motion in a vertical circle 258 25 Hydrostatics 305
20.5 Centrifugal clutch 260 25.1 Pressure 306
25.2 Density 307
21 Simple harmonic motion 262
25.3 Fluid pressure 309
21.1 Introduction to simple harmonic
25.4 Atmospheric pressure 310
motion (SHM) 262
25.5 Archimedes’ principle 311
21.2 The spring-mass system 263
25.6 Measurement of pressure 313
21.3 The simple pendulum 265
25.7 Barometers 313
21.4 The compound pendulum 266
25.8 Absolute and gauge pressure 315
21.5 Torsional vibrations 267
25.9 The manometer 315
22 Simple machines 269 25.10 The Bourdon pressure gauge 316
22.1 Machines 269 25.11 Vacuum gauges 317
22.2 Force ratio, movement ratio 25.12 Hydrostatic pressure on submerged
and efficiency 269 surfaces318
22.3 Pulleys 271 25.13 Hydrostatic thrust on curved surfaces 319
22.4 The screw-jack 273 25.14 Buoyancy 319
22.5 Gear trains 273 25.15 The stability of floating bodies 319
22.6 Levers 275
26 Fluid flow 326
26.1 Differential pressure flowmeters 326
Revision Test 7 Friction, motion in a circle,
simple harmonic motion and 26.2 Orifice plate 327
simple machines 279 26.3 Venturi tube 328
26.4 Flow nozzle 328
26.5 Pitot-static tube 328
26.6 Mechanical flowmeters 330
Part Four Heat Transfer and Fluid 26.7 Deflecting vane flowmeter 330
Mechanics 281 26.8 Turbine type meters 330
26.9 Float and tapered-tube meter 331
23 Heat energy and transfer 283 26.10 Electromagnetic flowmeter 331
26.11 Hot-wire anemometer 332
23.2 The measurement of temperature 285
26.12 Choice of flowmeter 332
23.3 Specific heat capacity 285
26.13 Equation of continuity 333
23.4 Change of state 287
26.14 Bernoulli’s equation 333
23.5 Latent heats of fusion and vaporisation 288
26.15 Impact of a jet on a stationary plate 335
23.6 A simple refrigerator 289
23.7 Conduction, convection and radiation 290 27 Ideal gas laws 338
23.8 Vacuum flask 290 27.1 Boyle’s law 338
23.9 Use of insulation in conserving fuel 291 27.2 Charles’ law 340
23.10 Thermal efficiency 291 27.3 The pressure or Gay-Lussac’s law 341
23.11 Calorific value and combustion 291 27.4 Dalton’s law of partial pressure 342
23.12 Heat exchangers 291 27.5 Characteristic gas equation 343
viii Contents
27.6 Worked problems on the characteristic
gas equation 343 Revision Test 9 Hydrostatics, fluid flow,
27.7 Further worked problems on the gas laws and temperature
characteristic gas equation 345 measurement 359
28 The measurement of temperature 349 A list of formulae for mechanical

28.1 Liquid-in-glass thermometer 349 engineering principles 360
28.2 Thermocouples 351
28.3 Resistance thermometers 352 Metric to Imperial conversions and vice versa 365
28.4 Thermistors 354 Greek alphabet 366
28.5 Pyrometers 354
28.6 Temperature indicating paints Glossary of terms 367
and crayons 356
28.7 Bimetallic thermometers 356 Answers to multiple-choice questions 372
28.8 Mercury-in-steel thermometer 356 Index 374
28.9 Gas thermometers 356
28.10 Choice of measuring devices 357
Preface
Mechanical Engineering Principles, Fourth Edition Part 1 Revision of Mathematics

aims to broaden the reader’s knowledge of the basic Part 2 Statics and Strength of Materials
principles that are fundamental to mechanical engineer- Part 3 Dynamics
ing design and the operation of mechanical systems. Part 4 Heat Transfer and Fluid Mechanics
Modern engineering systems and products still rely Mechanical Engineering Principles, Fourth Edition
upon static and dynamic principles to make them is suitable for the following:
work. Even systems that appear to be entirely elec-
tronic have a physical presence governed by the prin- (i) Undergraduate courses in Mechanical,
ciples of statics. Civil, Structural, Aeronautical & Marine
Engineering, together with Naval Architecture
In this fourth edition of Mechanical Engineering (ii) National Certificate/Diploma courses in
Principles, some further material has been added on Mechanical Engineering
factor of safety involved with tensile testing, thermal (iii) Any introductory/access/foundation course
efficiency and power rating of heat exchangers, den- involving Mechanical Engineering Principles
sity, and the importance of specific gravity and its re- at University, and Colleges of Further and
lationship to density. A new chapter has been added Higher education.
on the matrix displacement method, which is one of
the most powerful computer methods for analysing Although pre-requisites for the modules covered in this
complex structures and associated problems in con- book include Foundation Certificate/diploma, or simi-
tinuum mechanics; the method is used to analyse plane lar, in Mathematics and Science, each topic consid-
pin-jointed trusses and continuous beams. Guidance ered in the text is presented in a way that assumes
software that can be used via SmartPhones, iPads or that the reader has little previous knowledge of that
laptops is included. This chapter, although challeng- topic.
ing, is included to stimulate a student’s interest in the
Mechanical Engineering Principles, Fourth Edition
computing possibilities often needed in mechanical en-
contains some 415 worked problems, followed by
gineering. Since this new chapter involves matrices, a
over 750 further problems (all with answers). The
new prior chapter has been introduced explaining the
further problems are contained within some 160 Exer-
basic principles of matrix algebra, a topic not all stu-
cises; each Exercise follows on directly from the rel-
dents will have necessarily met at this stage of their
evant section of work, every few pages. In addition,
studies. Minor modifications and some further worked
the text contains 312 multiple-choice questions (all
problems have all been added to the text.
with answers), and 270 short answer questions, the
Thanks are due to Mr Kyrin Hay at the Defence Col- answers for which can be determined from the preced-
lege of Marine Engineering for his help and advice ing material in that particular chapter. Where at all pos-
with some of the new material. sible, the problems mirror practical situations found in
Full solutions are available to both students and mechanical engineering. Four hundred line diagrams
staff, and much more besides, from the website enhance the understanding of the theory.
www.routledge.com/cw/bird – see page x. At regular intervals throughout the text are some 9
For clarity, the text is divided into four sections, these Revision Tests to check understanding. For example,
being: Revision Test 1 covers material contained in Chapter 1,
x Preface
Test 2 covers the material in Chapter 2, Test 3 covers ‘Learning by Example’ is at the heart of Mechanical
the material in Chapters 3 to 6, and so on. No answers Engineering Principles, Fourth Edition.
are given for the questions in the Revision Tests, but an
Instructor’s guide has been produced giving full solu- JOHN BIRD
tions and a suggested marking scheme. The guide is Defence College of Technical Training,
available to lecturers/instructors via the website – see HMS Sultan, formerly
below. University of Portsmouth and
At the end of the text, a list of relevant formulae and Highbury College, Portsmouth
metric to imperial conversions is included for easy CARL ROSS
reference, together with the Greek alphabet and a former Professor, University of Portsmouth
glossary of terms.
From the website 6. Use any search engine (such as google)

The following support material is available for content on science, technology and
from www.routledge.com/cw/bird education:
carltfross YouTube or carl t f ross You Tube
For Students:
or YouTube carl t f ross
1.
Fully worked solutions to all 750 further
For Lecturers/Instructors:
questions contained in the 160 Practice
Exercises 1–6. As per students 1–6 above
2. A list of essential formulae 7. Full solutions and marking scheme for each
3. A full glossary of terms of the 9 Revision Tests; also, each test may
4. Multiple-choice questions be downloaded for distribution to students
5. Information on 20 famous engineers men- 8. All 400 illustrations used in the text may
tioned in the text be downloaded for use in PowerPoint
presentations
Part One
Revision of Mathematics
Chapter 1
Revisionary mathematics
Why it is important to understand: Revisionary mathematics
Mathematics is a vital tool for professional and chartered engineers. It is used in mechanical & manu-
facturing engineering, in electrical & electronic engineering, in civil & structural engineering, in naval
architecture & marine engineering and in aeronautical & rocket engineering. In these various branches
of engineering, it is very often much cheaper and safer to design your artefact with the aid of mathemat-
ics – rather than through guesswork. ‘Guesswork’ may be reasonably satisfactory if you are designing
an artefact similar to one that has already proven satisfactory; however, the classification societies will
usually require you to provide the calculations proving that the artefact is safe and sound. Moreover,
these calculations may not be readily available to you and you may have to provide fresh calculations,
to prove that your artefact is ‘roadworthy’. For example, if you design a tall building or a long bridge
by ‘guesswork’, and the building or bridge do not prove to be structurally reliable, it could cost you a
fortune to rectify the deficiencies. This cost may dwarf the initial estimate you made to construct these
artefacts, and cause you to go bankrupt. Thus, without mathematics, the prospective professional or
chartered engineer is very severely handicapped.
At the end of this chapter you should be able to:

• convert radians to degrees
• convert degrees to radians
• calculate sine, cosine and tangent for large and small angles
• calculate the sides of a right-angled triangle
• use Pythagoras’ theorem
• use the sine and cosine rules for acute-angled triangles
• expand equations containing brackets
• be familiar with summing vulgar fractions
• understand and perform calculations with percentages
• understand and use the laws of indices
• solve simple simultaneous equations
1.1 Introduction knowledge of mathematics. This chapter highlights

some areas of mathematics which will make the
As highlighted above, it is not possible to understand understanding of the engineering in the following
aspects of mechanical engineering without a good chapters a little easier.
Mechanical Engineering Principles, Bird and Ross, ISBN 9780367253264

4 Mechanical Engineering Principles
[
1.2 Radians and degrees �
Part One
(a)
rad or 0.7854 rad
4
There are 2π radians or 360° in a complete circle, thus:
�
(b) rad or 1.5708 rad
π radians = 180° from which, 2
2�
180° � (c) rad or 2.0944 rad
1 rad = or 1° = rad 3
� 180
where π = 3.14159265358979323846 .... to 20 decimal
(d) π rad or 3.1416 rad ]
places!
Problem 1. Convert the following angles to

degrees correct to 3 decimal places: 1.3 Measurement of angles
(a) 0.1 rad (b) 0.2 rad (c) 0.3 rad
Angles are measured starting from the horizontal ‘x’
axis, in an anticlockwise direction, as shown by θ1 to
180°
(a) 0.1 rad = 0.1 rad × = 5.730° θ4 in Figure 1.1. An angle can also be measured in a
� rad
clockwise direction, as shown by θ5 in Figure 1.1, but
180° in this case the angle has a negative sign before it. If,
(b) 0.2 rad = 0.2 rad × = 11.459°
� rad for example, θ4 = 300° then θ5 = – 60°.
180°
(c) 0.3 rad = 0.3 rad × = 17.189°
� rad
Problem 2. Convert the following angles to

radians correct to 4 decimal places:
(a) 5° (b) 10° (c) 30°
� rad �
(a) 5° = 5° × = rad = 0.0873 rad
180° 36
� rad �
(b) 10° = 10° × = rad = 0.1745 rad
180° 18
Figure 1.1
� rad �
(c) 30° = 30° × = rad = 0.5236 rad
180° 6 Problem 3. Use a calculator to determine the
cosine, sine and tangent of the following angles,
Now try the following Practice Exercise each measured anticlockwise from the horizontal
‘x’ axis, each correct to 4 decimal places:
(a) 30° (b) 120° (c) 250°
Practice Exercise 1 Radians and degrees
(d) 320° (e) 390° (f) 480°
1.
Convert the following angles to degrees
correct to 3 decimal places (where necessary):
(a) cos 30° = 0.8660 sin 30° = 0.5000
(a) 0.6 rad (b) 0.8 rad tan 30° = 0.5774
(c) 2 rad (d) 3.14159 rad
[(a) 34.377° (b) 45.837°

(c) 114.592° (d) 180° ] (b) cos 120° = – 0.5000
tan 120° = – 1.7321
sin 120° = 0.8660
2.
Convert the following angles to radians (c) cos 250° = – 0.3420 sin 250° = – 0.9397
correct to 4 decimal places: tan 250° = 2.7475
(a) 45° (b) 90°
(d) cos 320° = 0.7660 sin 320° = – 0.6428
(c) 120° (d) 180°
tan 320° = – 0.8391
Revisionary mathematics 5
(e) cos 390° = 0.8660 sin 390° = 0.5000 Now try the following Practice Exercise
tan 390° = 0.5774
Part One
(f) cos 480° = – 0.5000 sin 480° = 0.8660 Practice Exercise 2 Measurement of
tan 480° = – 1.7321 angles
1.
Find the cosine, sine and tangent of the
These angles are now drawn in Figure 1.2. Note that following angles, where appropriate each
cosine and sine always lie between –1 and +1 but correct to 4 decimal places:
that tangent can be >1 and <1 (a) 60° (b) 90° (c) 150°
(d) 180° (e) 210° (f) 270°
(g) 330° (h) – 30° (i) 420°
(j) 450° (k) 510°
[(a) 0.5, 0.8660, 1.7321

(b) 0, 1, ∞
(c) – 0.8660, 0.5, – 0.5774
(d) –1, 0, 0
(e) – 0.8660, – 0.5, 0.5774
(f) 0, –1, – ∞
(g) 0.8660, – 0.5000, – 0.5774
Figure 1.2 (h) 0.8660, – 0.5000, – 0.5774
(i) 0.5, 0.8660, 1.7321
Note from Figure 1.2 that θ = 30º is the same as (j) 0, 1, ∞
θ = 390º and so are their cosines, sines and tangents. (k) – 0.8660, 0.5, – 0.5774]
Similarly, note that θ = 120º is the same as θ = 480º
and so are their cosines, sines and tangents. Also, note
that θ = – 40º is the same as θ = + 320º and so are their
cosines, sines and tangents.
It is noted from above that 1.4 Triangle calculations
• in the first quadrant, i.e. where θ varies from 0º
to 90º, all (A) values of cosine, sine and tangent are (a) Sine, cosine and tangent
positive
bc ab
• in the second quadrant, i.e. where θ varies from From Figure 1.4, sin θ = cos θ =
90º to 180º, only values of sine (S) are positive ac ac
• in the third quadrant, i.e. where θ varies from bc
180º to 270º, only values of tangent (T) are positive tan θ =
ab
• in the fourth quadrant, i.e. where θ varies from
270º to 360º, only values of cosine (C) are positive.
These positive signs, A, S, T and C are shown in
Figure 1.3.
Figure 1.4
Problem 4. In Figure 1.4, if ab = 2 and ac = 3,

determine the angle θ.
Figure 1.3 It is convenient to use the expression for cos θ, since

‘ab’ and ‘ac’ are given.
ab 2 Now try the following Practice Exercise

Hence, cos θ = ac = 3 = 0.66667
Part One
Practice Exercise 3 Sines, cosines and

from which, θ = cos–1(0.66667) = 48.19º tangents, and
Pythagoras’ theorem
Problem 5. In Figure 1.4, if bc = 1.5 and
ac = 2.2, determine the angle θ. In Problems 1 to 5, refer to Figure 1.5.
1. If ab = 2.1 m and bc = 1.5 m, determine
It is convenient to use the expression for sin θ, since angle θ.[35.54°]
‘bc’ and ‘ac’ are given. 2. If ab = 2.3 m and ac = 5.0 m, determine
bc 1.5 angle θ.[62.61°]
Hence, sin θ = = = 0.68182
ac 2.2 3. If bc = 3.1 m and ac = 6.4 m, determine
angle θ.[28.97°]
from which, θ= sin–1(0.68182) = 42.99º
4. If ab = 5.7 cm and bc = 4.2 cm, determine
Problem 6. In Figure 1.4, if bc = 8 and ab = 1.3, the length ac [7.08 cm]
determine the angle θ. 5. If ab = 4.1 m and ac = 6.2 m, determine
length bc. [4.65 m]
It is convenient to use the expression for tan θ, since
‘bc’ and ‘ab’ are given.
bc 8 (c) The sine and cosine rules
Hence, tan θ = = = 6.1538 For the triangle ABC shown in Figure 1.6,
ab 1.3
from which, θ = tan–1(6.1538) = 80.77º
(b) Pythagoras’ theorem

Pythagoras’ theorem* states that:
(hypotenuse)2 = (adjacent side)2 + (opposite side)2
i.e. in the triangle of Figure 1.5,
ac2 = ab2 + bc2
Figure 1.5
Problem 7. In Figure 1.5, if ab = 5.1 m and

bc = 6.7 m, determine the length of the
hypotenuse, ac.
From Pythagoras, ac2 = ab2 + bc2

= 5.12 + 6.72 = 26.01 + 44.89
*Pythagoras of Samos (born approximately 570BC and
= 70.90 died around 495BC) was an Ionian Greek philosopher and
mathematician, best known for the Pythagorean Theorem. To
from which, ac = 70.90 = 8.42 m
find out more go to www.routledge.com/cw/bird
8.04 8.2
i.e. =
sin 70° sin B
Part One
from which, 8.04 sin B = 8.2 sin 70°
8.2sin 70°
and sin B = = 0.95839
8.04
−1
Figure 1.6 and B = sin (0.95839) = 73.41°
a b c Since A + B + C = 180°, then

the sine rule states: = =
sin A sin B sin C C = 180° – A – B = 180° – 70° – 73.41° = 36.59°
and the cosine rule states: a 2 = b 2 + c 2 − 2bc cos A

Now try the following Practice Exercise
Problem 8. In Figure 1.6, if a = 3 m, A = 20° and
B = 120°, determine lengths b and c and angle C. Practice Exercise 4 Sine and cosine rules
In Problems 1 to 4, refer to Figure 1.6.
a b
Using the sine rule, =
sin A sin B 1. If b = 6 m, c = 4 m and B = 100°, determine
angles A and C and length a.
3 b
i.e. = [A = 38.96°, C = 41.04°, a = 3.83 m]
sin 20° sin120°
2. If a = 15 m, c = 23 m and B = 67°, determine
3sin120° 3 × 0.8660 length b and angles A and C.
from which, b = =
sin 20° 0.3420 [b = 22.01 m, A = 38.86°, C = 74.14°]
= 7.596 m
3. If a = 4 m, b = 8 m and c = 6 m, determine
Angle, C = 180° – 20° – 120° = 40° angle A.[28.96°]
c a 4. If a = 10.0 cm, b = 8.0 cm and c = 7.0 cm,
Using the sine rule again gives: =
sin C sin A determine angles A, B and C.
a sin C 3 × sin 40° [A = 83.33°, B = 52.62°, C = 44.05°]
i.e. c = =
sin A sin 20° 5. In Figure 1.7, PR represents the inclined jib
= 5.638 m of a crane and is 10.0 m long. PQ is 4.0 m
long. Determine the inclination of the jib to the
Problem 9. In Figure 1.6, if b = 8.2 cm, c = 5.1 cm vertical (i.e. angle P) and the length of tie QR.
and A = 70°, determine the length a and angles B
and C.
From the cosine rule,

a 2 = b 2 + c 2 − 2bc cos A
= 8.2 2 + 5.12 − 2 × 8.2 × 5.1 × cos70°
= 67.24 + 26.01 – 2(8.2)(5.1)cos70°
= 64.643
Hence, length, a = 64.643 = 8.04 cm

Figure 1.7
a b
Using the sine rule: = [P = 39.73°, QR = 7.38 m]
sin A sin B
1.5 Brackets Expand the brackets in Problems 3 to 7.

Part One
3. 2(x – 2y + 3) [2x – 4y + 6]

The use of brackets, which are used in many engi-
neering equations, is explained through the following 4. (3x – 4y) + 3(y – z) – (z – 4x)
worked problems. [7x – y – 4z]
5. 2x + [y – (2x + y)][0]
Problem 10. Expand the bracket to determine A,
given A = a(b + c + d) 6. 24a – [2{3(5a – b) – 2(a + 2b)} + 3b]
[11b – 2a]
Multiplying each term in the bracket by ‘a’ gives:
7. ab[c + d – e(f – g + h{i + j})]
A = a(b + c + d) = ab + ac + ad
[abc + abd – abef + abeg – abehi – abehj]
Problem 11. Expand the brackets to determine A,
given A = a[b(c + d) – e(f – g)]
When there is more than one set of brackets the 1.6 Fractions
innermost brackets are multiplied out first. Hence,
A = a[b(c + d) – e(f – g)] = a[bc + bd – ef + eg] 2
An example of a fraction is where the top line, i.e.
(Note that –e × –g = +eg) 3
the 2, is referred to as the numerator and the bottom
Now multiplying each term in the square brackets by line, i.e. the 3, is referred to as the denominator.
‘a’ gives: A proper fraction is one where the numera-
tor is smaller than the denominator, examples being
A = abc + abd – aef + aeg
2 1 3 5
, , , , and so on.
Problem 12. Expand the brackets to determine A, 3 2 8 16
An improper fraction is one where the denomi-
given A = a[b(c + d – e) – f (g – h{j – k})]
nator is smaller than the numerator, examples being
The inner brackets are determined first, hence 3 2 8 16
, , , , and so on.
2 1 3 5
A = a[b(c + d – e) – f (g – h{j – k})] Addition of fractions is demonstrated in the follow-
= a[b(c + d – e) – f (g – hj + hk)] ing worked problems.
= a[bc + bd – be – fg + fhj – fhk]
i.e. A = abc + abd – abe – afg + afhj – afhk 1 1
Problem 14. Evaluate A, given A = +
2 3
Problem 13. Evaluate A, given
A = 2[3(6 – 1) – 4(7{2 + 5} – 6)] The lowest common denominator of the two denomi-
nators 2 and 3 is 6, i.e. 6 is the lowest number that both
A = 2[3(6 – 1) – 4(7{2 + 5} – 6)] 2 and 3 will divide into.
= 2[3(6 – 1) – 4(7 × 7 – 6)] 1 3 1 2 1 1
Then = and = i.e. both and have the
= 2[3 × 5 – 4 × 43] 2 6 3 6 2 3
common denominator, namely 6.
= 2[15 – 172] = 2[– 157] = – 314
The two fractions can therefore be added as:
1 1 3 2 3+ 2 5
A= + = + = =
Now try the following Practice Exercise 2 3 6 6 6 6
Practice Exercise 5 Brackets 2 3

Problem 15. Evaluate A, given A = +
In Problems 1 and 2, evaluate A. 3 4
1. A = 3(2 + 1 + 4) [21] A common denominator can be obtained by multiply-
2. A = 4[5(2 + 1) – 3(6 – 7)] [72] ing the two denominators together, i.e. the common
denominator is 3 × 4 = 12.
The two fractions can now be made equivalent, i.e.

2 8 3 9 1 2
Problem 18. Evaluate
Part One
= and = +
3 12 4 12 4 3
so that they can be easily added together, as follows:

2 3 8 9 8 + 9 17 (i) Press function
A= + = + = =
3 4 12 12 12 12
(ii) Type in 1
2 3 5
i.e. A= + = 1 (iii) Press ↓ on the cursor key and type in 4
3 4 12
1
(iv) appears on the screen
1 2 3 4
Problem 16. Evaluate A, given A = + +
6 7 2 (v) Press → on the cursor key and type in +

A suitable common denominator can be obtained by (vi) Press function
multiplying 6 × 7 = 42, because all three denominators
divide exactly into 42. (vii) Type in 2
1 7 2 12 3 63 (viii) Press ↓ on the cursor key and type in 3
Thus, = , = and =
6 42 7 42 2 42 (ix) Press → on the cursor key
1 2 3 11
Hence, A = + + (x) Press = and the answer appears
6 7 2 12
7 12 63 7 + 12 + 63 82 41 (xi) Press S ⇔ D function and the fraction changes
= + + = = =
to a decimal 0.9166666....
42 42 42 42 42 21
1 2 3 20 1 2 11
i.e. A = + + =1 Thus, + = = 0.9167 as a decimal, correct to
6 7 2 21 4 3 12
4 decimal places.
Problem 17. Determine A as a single fraction,
1 2
given A = + It is also possible to deal with mixed numbers on the
x y calculator.

A common denominator can be obtained by multiply- Press Shift then the function and appears

ing the two denominators together, i.e. xy
1 3
1 y 2 2x Problem 19. Evaluate 5 − 3
Thus, = and = 5 4
x xy y xy
1 2 y 2x
Hence, A = + = + (i) Press Shift then the function and
x y xy xy
appears on the screen
y + 2x
i.e. A= (ii) Type in 5 then → on the cursor key
xy
(iii)Type in 1 and ↓ on the cursor key
Note that addition, subtraction, multiplication and divi- 1
(iv) Type in 5 and 5 appears on the screen
sion of fractions may be determined using a calculator 5
(for example, the CASIO fx-83ES or fx-991ES). (v) Press → on the cursor key

Locate the and functions on your calculator (vi) Type in – and then press Shift then the

(the latter function is a shift function found above 1
function and 5 – appears on the screen
5

the function) and then check the following worked (vii) Type in 3 then → on the cursor key

problems. (viii) Type in 3 and ↓ on the cursor key
1 3 of commercial life, as well as in engineering. Interest

(ix) Type in 4 and 5 − 3 appears on the screen rates, sale reductions, pay rises, exams and VAT are all
Part One
5 4
29 examples where percentages are used.
(x) Press = and the answer appears Percentages are fractions having 100 as their
20
denominator.
(xi) Press S ⇔ D function and the fraction changes 40
to a decimal 1.45 For example, the fraction is written as 40% and is
100
1 3 29 9 read as ‘forty per cent’.
Thus, 5 − 3 = =1 = 1.45 as a decimal. The easiest way to understand percentages is to go
5 4 20 20
through some worked examples.
Now try the following Practice Exercise Problem 20. Express 0.275 as a percentage.
Practice Exercise 6 Fractions 0.275 = 0.275 × 100% = 27.5%

In Problems 1 to 3, evaluate the given fractions.
Problem 21. Express 17.5% as a decimal number.
1.
1
3
+
1
4
[ ]
7
12 17.5% =
17.5
= 0.175
[ ]
1 1 9 100
2. +
5 4 20 5
[ ]
1 1 1 7 Problem 22. Express as a percentage.
3. + − 8
6 2 5 15
5 5 500
In Problems 4 and 5, use a calculator to evaluate = × 100% = % = 62.5%
the given expressions. 8 8 8
4.
1 3 8
– ×
3 4 21
[ ]
1
21
Problem 23. In two successive tests a student
gains marks of 57/79 and 49/67. Is the second mark
[ ]
3 4 2 4 9 better or worse than the first?
5. × – ÷ –
4 5 3 9 10
3 5 1 57 = 57 × 100% 5700
57/79 = = %
6. Evaluate + − as a decimal, correct to 79 79 79
8 6 2
4 decimal places. [ 17
24
= 0.7083 ] = 72.15% correct to 2 decimal places.
49/67 =
49 49
= × 100% =
4900
%
8 2 67 67 67
7. Evaluate 8 ÷ 2 as a mixed number.
9 3
[ ] 1 = 73.13% correct to 2 decimal places.
3
3 Hence, the second test is marginally better than the
1 1 7 first test.
8. Evaluate 3 × 1 − 1 as a decimal, cor-
5 3 10
This question demonstrates how much easier it is to
rect to 3 decimal places. [2.567]
compare two fractions when they are expressed as per-
2 3
9. Determine + as a single fraction. centages.
x y
[ 3x + 2 y
xy ] Problem 24. Express 75% as a fraction.
75 3
75% = =
100 4
75
1.7 Percentages The fraction is reduced to its simplest form
100
by cancelling, i.e. dividing numerator and denominator
Percentages are used to give a common standard. The by 25.
use of percentages is very common in many aspects
Problem 25. Express 37.5% as a fraction. Problem 30. A drilling speed should be set to
Part One
400 rev/min. The nearest speed available on the
37.5 machine is 412 rev/min. Calculate the percentage
37.5% = over-speed.
100
375 % over-speed
= by multiplying numerator and
1000 available speed − correct speed
denominator by 10 = × 100%
correct speed
15
= by dividing numerator and 412 − 400
40 = × 100%
denominator by 25 400
3 12
= by dividing numerator and
8 = 400 × 100% = 3%
denominator by 5
Problem 26. Find 27% of £65. Now try the following Practice Exercise
27 Practice Exercise 7 Percentages

27% of £65 = × 65 = £17.55 by calculator
100
In Problems 1 and 2, express the given numbers
as percentages.
Problem 27. A 160 GB iPod is advertised as
costing £190 excluding VAT. If VAT is added at 1. 0.057 [5.7%]
20%, what will be the total cost of the iPod?
2. 0.374 [37.4%]
20 3. Express 20% as a decimal number [0.20]
VAT = 20% of £190 = × 190 = £38
100
11
4. Express as a percentage [68.75%]
Total cost of iPod = £190 + £38 = £228 16
A quicker method to determine the total cost is: 5
1.20 × £190 = £228 5.
Express as a percentage, correct to 3
13
decimal places [38.462%]
Problem 28. Express 23 cm as a percentage of
72 cm, correct to the nearest 1%. 6. Place the following in order of size, the
smallest first, expressing each as percent-
23 ages, correct to 1 decimal place:
23 cm as a percentage of 72 cm =
× 100% 12 9 5 6
72 (a) (b) (c) (d)
= 31.94444...% 21 17 9 11
= 3 2% correct
to the nearest
[(b) 52.9%, (d) 54.5%,
(c) 55.6%, (a) 57.1% ]
1% 7. Express 65% as a fraction in its simplest
Problem 29. A box of screws increases in price

form [ ]
13
20
from £45 to £52. Calculate the percentage change 8. Calculate 43.6% of 50 kg [21.8 kg]
in cost, correct to 3 significant figures.
9. Determine 36% of 27 m [9.72 m]
new value − original value 10. Calculate correct to 4 significant figures:
% change = × 100%
original value (a) 18% of 2758 tonnes
52 − 45 7 (b) 47% of 18.42 grams
= × 100% = × 100
45 45 (c) 147% of 14.1 seconds
[(a) 496.4 t (b) 8.657 g (c) 20.73 s]
= 15.6% = percentage change in cost
Law 3: When a number which is raised to a power is
11. Express: (a) 140 kg as a percentage of 1 t raised to a further power, the indices are multiplied.
Part One
(b) 47 s as a percentage of 5 min (c) 13.4 cm 3

as a percentage of 2.5 m For example, ( 22 ) = 22× 3 = 26
[(a) 14% (b) 15.67% (c) 5.36%]
2
12. A computer is advertised on the internet at
and ( 34 ) = 34 × 2 = 38
£520, exclusive of VAT. If VAT is payable More generally, (am)n = amn
at 20%, what is the total cost of the com-
3
puter? [£624] For example, ( d 2 ) = d 2× 3 = d 6
13.
Express 325 mm as a percentage of 867
mm, correct to 2 decimal places. Law 4: When a number has an index of 0, its value is 1.
[37.49%]
For example, 30 = 1
14. hen signing a new contract, a Premiership
W
and 170 = 1
footballer’s pay increases from £15,500 to
£21,500 per week. Calculate the percentage More generally, a0 = 1
pay increase, correct to 3 significant figures.
[38.7%] Law 5: A number raised to a negative power is the
15. A metal rod 1.80 m long is heated and its reciprocal of that number raised to a positive power.
length expands by 48.6 mm. Calculate the 1 1
percentage increase in length. [2.7%] For example, 3–4 = and = 23
34 2 −3
1
More generally, a–n =
an
1.8 Laws of indices 1
For example, a −2 =
The manipulation of indices, powers and roots is a cru- a2
cial underlying skill needed in algebra. Law 6: When a number is raised to a fractional
Law 1: When multiplying two or more numbers power the denominator of the fraction is the root
having the same base, the indices are added. of the number and the numerator is the power.
2
3
For example, 2 2 × 2 3 = 2 2+ 3 = 2 5 For example, 83 = 8 2 = (2)2 = 4
5 4 × 5 2 × 5 3 = 5 4+ 2 + 3 = 5 9 1
and 2
and 25 2 =
251 = 251
More generally, am × an = am+n 2
= ± 5 (Note that ≡ )
For example, a3 × a4 = a 3+ 4 = a7 m
n
More generally, an = am
Law 2: When dividing two numbers having the
same base, the index in the denominator is subtract- 4
3
ed from the index in the numerator. For example, x3 = x4
25
For example, = 2 5− 3 = 2 2 Problem 31. Evaluate in index form 53 × 5 × 52
23
78
and = 7 8− 5 = 7 3 53 × 5 × 52 = 53 × 51 × 52 (Note that 5 means 51)
75
= 53+1+ 2 = 56 from law 1
am
More generally, = am–n
an
c5 35
For example, = c 5− 2 = c 3 Problem 32. Evaluate
c2 34
35 (a) 41/2 = 4 = ±2
From law 2: = 35− 4 = 31 = 3
Part One
34 (b) 163/4 = 4
16 3 = (2)3 = 8
24 (Note that it does not matter whether the 4th root
Problem 33. Evaluate of 16 is found first or 16 cubed is found first; the
24
same answer will result.)
24
4 = 2 4 − 4 from law 2 (c) 272/3 = 3
27 2 = (3)2 = 9
2
= 2 0 = 1 from law 4 1 1 1 1
(d) 9–1/2 = = = = ±
Any number raised to the power of zero equals 1 91/ 2 9 ± 3 3
3 × 32 Problem 39. Simplify a 2b 3c × ab 2c 5

34
a 2b 3c × ab 2c 5 = a 2 × b 3 × c × a × b 2 × c 5
3 × 32 31 × 32 31+ 2 33
=
= = 4 = a 2 × b 3 × c1 × a1 × b 2 × c 5
34 34 34 3
Grouping together like terms gives:
= 33− 4 = 3−1 from laws 1 and 2
a 2 × a 1 × b 3 × b 2 × c1 × c 5
1
= from law 5 Using law 1 of indices gives:
3
10 3 × 10 2 a 2+1 × b 3+ 2 × c1+5 = a 3 × b5 × c 6
10 8 i.e. a 2b 3c × ab 2c 5 = a 3 b 5 c 6
10 3 × 10 2 10 3+ 2 10 5
= = from law 1 x 5 y 2z
10 8 10 8 10 8 Problem 40. Simplify
x2y z3
= 10 5−8 = 10 −3 from law 2
1 1 x 5 y 2z x 5 × y 2 × z x5 y 2 z
= = from law 5 2 3
=
2 3 = 2× 1× 3
10 + 3 1000 x yz x × y×z x y z
10 3 × 10 2 1 = x 5− 2 × y 2 −1 × z 1− 3 by law 2
Hence, = 10 −3 = = 0.001
10 8 1000 3
or x y
3 −2
= x 3 × y 1 × z −2 = x y z
Problem 36. Simplify: (a) (23)4 (b) (32)5 z2
expressing the answers in index form.
Now try the following Practice Exercise
From law 3: (a) (23)4 = 23×4 = 212
(b) (32)5 = 32×5 = 310 Practice Exercise 8 Laws of indices
In Problems 1 to 18, evaluate without the aid of
(10 2 ) 3 a calculator:
10 4 × 10 2
1. Evaluate 2 2 × 2 × 2 4 [2 7 = 128]
(10 2 ) 3 10 (2 × 3) 2. Evaluate 35 × 33 × 3 in index form
From laws 1, 2, and 3: =
10 4 × 10 2 10 (4 + 2) [ 39 ]
10 6 27
= 6 = 106–6 3. Evaluate [2 4 = 16]
10 23
= 100 = 1
4. Evaluate
33
35
[ 3−2 =
1
32
=
1
9 ]
Problem 38. Evaluate: (a) 41/2 (b) 163/4
(c) 272/3 (d) 9–1/2 5. Evaluate 7 0 [1]
4
23 × 2 × 26
( x 3 ) [x12]
Part One
6. Evaluate [2 3 = 8] 27.

27
7.
6
Evaluate 10 × 10 [10 2 = 100]
10 5
28.
−3
( y2 ) [ y −6 or y16 ]
2
8. Evaluate 10 ÷ 10 [10 3 = 1000]
4 29. (t × t 3) [t 8]
3 4 −2
9. Evaluate 10 × 10 30. (c −7 ) [ c14]
10 9
[ 10 −2 =
1
10 2
=
1
100
= 0.01 ] 31.
 a 2 3
 
5
a 
[ a −9 or
a9
1
]
10. Evaluate 56 × 52 ÷ 57 [5]  1 4
11. Evaluate (72)3 in index form [7 6 ]
32.  3
b  [ b12
1
or b −12 ]
12. Evaluate (33)2[36 = 729]  b 2 −2
33.   [b10]
7
b 
37 × 34
13. Evaluate in index form [36 ]
35 34.
1
( )
s3
3
[ 1
s9
or s −9 ]
(9 × 32 ) 3
14. Evaluate in index form [ 34 ]
(3 × 27) 2 35. p 3qr 2 × p 2q 5r × pqr 2 [ p 6q 7 r 5 ]
(16 × 4) 2
15. Evaluate
(2 × 8) 3
[1] 36.
x3 y 2z
x5 y z 3
[ x −2 y z −2 or
y
x2 z2
]
5−2
16. Evaluate [52 = 25]
5−4
17. Evaluate
32 × 3−4
33
[ 3−5 =
35
1
=
1
243 ] 1.9 Simultaneous equations
The solution of simultaneous equations is demonstrat-

7 2 × 7 −3
18. Evaluate [7 2 = 49] ed in the following worked problems.
7 × 7 −4
In Problems 19 to 36, simplify the following, Problem 41. If 6 apples and 2 pears cost £1.80
giving each answer as a power: and 8 apples and 6 pears cost £2.90, calculate how
much an apple and a pear each cost.
19. z 2 × z 6 [ z 8 ]
20. a × a 2 × a 5 [ a 8] Let an apple = A and a pear = P, then:
21. n8 × n −5 [ n 3] 6A + 2P = 180 (1)
22. b 4 × b 7 [ b11] 8A + 6P = 290 (2)
23. b 2 ÷ b5 b −3 or
1
b3
[ ] From equation (1), 6A = 180 – 2P
and
180 − 2 P
A= = 30 – 0.3333P(3)
24. c 5 × c 3 ÷ c 4 [c 4 ] 6
m5 × m 6 From equation (2), 8A = 290 – 6P
25. [m 4]
m 4 × m3 290 − 6 P
and A= = 36.25 – 0.75P(4)
8
26.
( x 2 )( x)
x6
[ x −3 or
1
x3 ] Equating (3) and (4) gives:
30 – 0.3333P = 36.25 – 0.75P
i.e. 0.75P – 0.3333P = 36.25 – 30 and 6s + 12b = 18 (2)
Part One
and 0.4167P = 6.25 Multiplying equation (1) by 3 gives:
6.25 6s + 60b = 30 (3)

and P= = 15
0.4167 Equation (3) – equation (2) gives: 48b = 12
12
Substituting in (3) gives: A = 30 – 0.3333(15) from which, b= = 0.25
48
= 30 – 5 = 25
Hence, an apple costs 25p and a pear costs 15p Substituting in (1) gives: 2s + 20(0.25) = 10
The above method of solving simultaneous equations i.e. 2s = 10 – 20(0.25)
is called the substitution method.
i.e. 2s = 5
Problem 42. If 6 bananas and 5 peaches cost 5
£3.45 and 4 bananas and 8 peaches cost £4.40, and s = = 2.5
2
calculate how much a banana and a peach each
cost. Therefore, a spanner costs £2.50 and a bolt costs
£0.25 or 25p
Let a banana = B and a peach = P, then:
6B + 5P = 345 (1) Now try the following Practice Exercises
4B + 8P = 440 (2)
Multiplying equation (1) by 2 gives: Practice Exercise 9 Simultaneous
equations
12B + 10P = 690 (3)
1. If 5 apples and 3 bananas cost £1.45 and
Multiplying equation (2) by 3 gives: 4 apples and 6 bananas cost £2.42, deter-
mine how much an apple and a banana each
12B + 24P = 1320 (4) cost. [apple = 8p, banana = 35p]
Equation (4) – equation (3) gives: 14P = 630 2. If 7 apples and 4 oranges cost £2.64 and
3 apples and 3 oranges cost £1.35, determine
630 how much an apple and an orange each cost.
from which, P= = 45
14 [apple = 28p, orange = 17p]
Substituting in (1) gives: 6B + 5(45) = 345
3.
Three new cars and four new vans sup-
i.e. 6B = 345 – 5(45) plied to a dealer together cost £93000, and
five new cars and two new vans of the same
i.e. 6B = 120 models cost £99000. Find the respective
costs of a car and a van.
120
and B= = 20 [car = £15000, van = £12000]
6
4.
In a system of forces, the relationship
Hence, a banana costs 20p and a peach costs 45p between two forces F1 and F2 is given by:
The above method of solving simultaneous equations
is called the elimination method. 5F1 + 3F2 = – 6
3F1 + 5F2 = – 18
Problem 43. If 20 bolts and 2 spanners cost £10, Solve for F1 and F2
and 6 spanners and 12 bolts cost £18, how much [F1 = 1.5, F2 = – 4.5]
does a spanner and a bolt cost?
5. Solve the simultaneous equations:
a+b=7
Let s = a spanner and b = a bolt. a – b = 3 [a = 5, b = 2]
Therefore, 2s + 20b = 10 (1)
6. Solve the simultaneous equations:

Part One
8a – 3b = 51
3a + 4b = 14 [a = 6, b = – 1]
Practice Exercise 10 Multiple-choice Figure 1.9

questions on
revisionary 8. In the triangle ABC shown in Figure 1.10,
mathematics side ‘a’ is equal to:
(Answers on page 372) (a) 61.27 mm
1. 73º is equivalent to: (b) 86.58 mm
(a) 23.24 rad (b) 1.274 rad (c) 96.41 mm
(c) 0.406 rad (d) 4183 rad (d) 54.58 mm
2. 0.52 radians is equivalent to:
(a) 93.6º (b) 0.0091º
(c) 1.63º (d) 29.79º
3. 3π/4 radians is equivalent to:
(a) 135º (b) 270º
(c) 45º (d) 67.5º
4. In the right-angled triangle ABC shown in
Figure 1.8, sine A is given by:
(a) b/a (b) c/b Figure 1.10
(c) b/c (d) a/b
9. In the triangle ABC shown in Figure 1.10,
angle B is equal to:
(a) 0.386º (b) 22.69º
(c) 74.71º (d) 23.58º
Figure 1.8 10. Removing the brackets from the expres-

sion: a[b + 2c – d{(e – f) – g(m – n)}] gives:
5. In the right-angled triangle ABC shown in
Figure 1.8, cosine C is given by: (a) ab + 2ac – ade – adf + adgm – adgn
(a) a/b (b) c/b (b) ab + 2ac – ade – adf – adgm – adgn
(c) a/c (d) b/a (c) ab + 2ac – ade + adf + adgm – adgn
6. In the right-angled triangle ABC shown in (d) ab + 2ac – ade – adf + adgm + adgn
Figure 1.8, tangent A is given by:
5 1 2
(a) b/c (b) a/c 11. + − is equal to:
(c) a/b (d) c/a 6 5 3
7. In the right-angled triangle PQR shown in 1 11

(a) (b)
Figure 1.9, angle R is equal to: 2 30
1 7
(a) 41.41º (b) 48.59º (c) − (d) 1
(c) 36.87º (d) 53.13º 2 10
1 2 2 1 is equal to: (16 × 4) 2
Part One
12. 1 +1 ÷ 2 − 18. The engineering expression is
3 3 3 3 (8 × 2) 4
2 19 equal to:
(a) 1 (b)
7 24 (a) 4 (b) 2 −4
1 5 1
(c) 2 (d) 1 (c) 2 (d) 1
21 8 2
3 3 1 2
13. ÷ 1 is equal to: 19. (16
−
4 – 27
−
3) is equal to:
4 4
3 9 7
(a) (b) 1 (a) (b) – 7
7 16 18
5 1 8 1
(c) 1 (d) 2 (c) 1 (d) – 8
16 2 9 2
14.
11 mm expressed as a percentage of 20.
The solution of the simultaneous equa-
41 mm is: tions 3a – 2b = 13 and 2a + 5b = – 4
(a) 2.68, correct to 3 significant figures is:
(b) 2.6, correct to 2 significant figures (a) a = – 2, b = 3
(c) 26.83, correct to 2 decimal places (b) a = 1, b = – 5
(d) 0.2682, correct to 4 decimal places (c) a = 3, b = – 2
2 −3 (d) a = – 7, b = 2
15. The value of − 1 is equal to:
2 −4
(a) 1 (b) 2
1 1
(c) –
2
(d)
2
References
4
16. In an engineering equation 3 = 1 . The There are many aspects of mathematics needed in engineering
3r 9 studies; a few have been covered in this chapter. For
value of r is: further engineering mathematics, see the following
(a) – 6 (b) 2 references:
(c) 6 (d) – 2 [1] BIRD J. O. Basic Engineering Mathematics 7th Edition,
Taylor & Francis, 2017.
3
− [2] BIRD J. O. Engineering Mathematics 8th Edition, Taylor
17. 16 4 is equal to: & Francis, 2017.
1
(a) 8 (b) –
23
1
(c) 4 (d)
8
For fully worked solutions to each of the problems in Practice Exercises 1 to 10 in this chapter,
go to the website:
www.routledge.com/cw/bird
Revision Test 1 Revisionary mathematics

Part One
This Revision Test covers the material contained in Chapter 1. The marks for each question are shown in brackets
at the end of each question.
1. Convert, correct to 2 decimal places: A triangular plot of land ABC is shown in
5.
Figure RT1.4. Solve the triangle and determine its
(a) 76.8° to radians
area
(b) 1.724 radians to degrees (4)
2. In triangle JKL in Figure RT1.1, find:
(a) length KJ correct to 3 significant figures
(b) sin L and tan K, each correct to 3 decimal places
Figure RT1.4
(9)
6. Figure RT1.5 shows a roof truss PQR with rafter

PQ = 3 m. Calculate the length of (a) the roof rise
PP′ (b) rafter PR, and (c) the roof span QR. Find
Figure RT1.1 also (d) the cross-sectional area of the roof truss
(4)
3. In triangle PQR in Figure RT1.2, find angle P in
decimal form, correct to 2 decimal places
Figure RT1.5
(10)
Figure RT1.2
olve triangle ABC given b = 10 cm, c = 15 cm
7. S
(2) and ∠A = 60°.(7)
4. In triangle ABC in Figure RT1.3, find lengths AB 8. Remove the brackets and simplify
and AC, correct to 2 decimal places 2(3x – 2y) – (4y – 3x)(3)
9. Remove the brackets and simplify
10a – [3(2a – b) – 4(b –a) + 5b](4)
10. Determine, correct to 2 decimal places, 57% of
17.64 g(2)
11.
Express 54.7 mm as a percentage of 1.15 m,
correct to 3 significant figures. (3)
12. Simplify:
Figure RT1.3 3 7 5 1 5
(a) − (b) 1 − 2 + 3 (8)
(4) 4 15 8 3 6
13. Use a calculator to evaluate: 2

(23 × 16)
Part One
7 3 3 (b) 3
(a) 1 × × 3
9 8 5 (8 × 2)
2 1
(b) 6 ÷ 1  1 −1
3 3 (c)  2  (7)
4 
1 1 2
(c) 1 × 2 ÷ (10)
3 5 5
17. Evaluate:
14. Evaluate:  2 −2 2
  −
(a) 3 × 2 3 × 2 2 1  3 9
(a) (27) − (b) (5)
1 3  2 2
(b) 49 2 (4)  
 3
15. Evaluate: 18. Solve the simultaneous equations:
27 10 4 × 10 × 10 5 (a) 2 x + y = 6
(a) (b) (4)
22 10 6 × 10 2 5x – y = 22
16. Evaluate: (b) 4 x – 3y = 11
23 ×2× 22 3x + 5y = 30 (10)
(a)
24
For lecturers/instructors/teachers, fully worked solutions to each of the problems in Revision Test 1,
together with a full marking scheme, are available at the website:
www.routledge.com/cw/bird
Chapter 2
Further revisionary
mathematics
Why it is important to understand: Further revisionary mathematics
In engineering there are many different quantities to get used to, and hence many units to become familiar
with. For example, force is measured in newtons, electric current is measured in amperes and pressure is
measured in pascals. Sometimes the units of these quantities are either very large or very small and hence
prefixes are used. For example, 1000 pascals may be written as 103 Pa which is written as 1 kPa in prefix
form, the k being accepted as a symbol to represent 1000 or 103. Studying, or working, in an engineering
discipline, you very quickly become familiar with the standard units of measurement, the prefixes used and
engineering notation. An electronic calculator is extremely helpful with engineering notation.
Most countries have used the metric system of units for many years; however, there are other coun-
tries, such as the USA, who still use the imperial system. Hence, metric to imperial unit conversions,
and vice versa, are internationally important and are contained in this chapter.
Graphs have a wide range of applications in engineering and in physical sciences because of their
inherent simplicity. A graph can be used to represent almost any physical situation involving discrete
objects and the relationship among them. If two quantities are directly proportional and one is plotted
against the other, a straight line is produced. Examples of this include an applied force on the end of a
spring plotted against spring extension, the speed of a flywheel plotted against time, and strain in a wire
plotted against stress (Hooke’s law). In engineering, the straight line graph is the most basic graph to
draw and evaluate.
There are many practical situations engineers have to analyse which involve quantities that are vary-
ing. Typical examples include the stress in a loaded beam, the temperature of an industrial chemical, the
rate at which the speed of a vehicle is increasing or decreasing, the current in an electrical circuit or the
torque on a turbine blade. Differential calculus, or differentiation, is a mathematical technique for analys-
ing the way in which functions change. This chapter explains how to differentiate the five most common
functions. Engineering is all about problem solving and many problems in engineering can be solved using
calculus. Physicists, chemists, engineers, and many other scientific and technical specialists use calculus in
their everyday work; it is a technique of fundamental importance. Integration has numerous applications
in engineering and science and some typical examples include determining areas, mean and r.m.s. values,
volumes of solids of revolution, centroids, second moments of area, and differential equations. Standard
integrals are covered in this chapter, and for any further studies in engineering, differential and integral
calculus are unavoidable.
Mechanical Engineering Principles, Bird and Ross, ISBN 9780367253264

Further revisionary mathematics 21
Vectors are an important part of the language of science, mathematics and engineering. They are
used to discuss multivariable calculus, electrical circuits with oscillating currents, stress and strain in
structures and materials, and flows of atmospheres and fluids, and they have many other applications.
Resolving a vector into components is a precursor to computing things with or about a vector quantity.
Because position, velocity, acceleration, force, momentum and angular momentum are all vector quan-
tities, resolving vectors into components is a most important skill required in any engineering studies.
At the end of this chapter you should be able to:

• state the seven SI units
• understand derived units
• recognise common engineering units
• understand common prefixes used in engineering
• use engineering notation and prefix form with engineering units
• understand and calculate metric to imperial conversions and vice versa
Part One
• understand rectangular axes, scales and co-ordinates
• plot given co-ordinates and draw the best straight line graph
• determine the gradient and vertical-axis intercept of a straight line graph
• state the equation of a straight line graph
• plot straight line graphs involving practical engineering examples
• state that calculus comprises two parts – differential and integral calculus
• differentiate y = axn by the general rule
• differentiate sine, cosine, exponential and logarithmic functions
• understand that integration is the reverse process of differentiation
• determine integrals of the form axn where n is fractional, zero, or a positive or negative integer
• integrate standard functions – cos ax, sin ax, eax, 1
• evaluate definite integrals x
• evaluate 2 by 2 and 3 by 3 determinants
• determine scalar (or dot) products of two vectors
• determine vector (or cross) products of two vectors
2.1 Units, prefixes and engineering SI units

notation The system of units used in engineering and science
is the Systeme Internationale d’Unites (International
Of considerable importance in engineering is knowl- System of Units), usually abbreviated to SI units, and
edge of units of engineering quantities, the prefixes is based on the metric system. This was introduced in
used with units, and engineering notation. 1960 and is now adopted by the majority of countries
We need to know, for example, that as the official system of measurement.
The basic seven units used in the SI system are listed
80 kN = 80 × 103N which means 80,000
on page 22 with their symbols.
newtons
There are, of course, many other units than the seven
and 25 mJ = 25 × 10–3J which means 0.025 joules shown. These other units are called derived units and
are defined in terms of the standard units listed.
and 50 nF = 50 × 10–9F which means 0.000000050
For example, speed is measured in metres per second,
farads
therefore using two of the standard units, i.e. metres
This is explained in this chapter. and seconds.
Quantity Unit Symbol Capacitance farad F

Length metre m (1 m = 100 cm Electrical ohm Ω
= 1000 mm) resistance
Mass kilogram kg (1 kg = 1000 g) Inductance henry H
Time second s Moment of force newton metre Nm
Electric ampere A Stress pascal Pa
current Torque newton metre Nm
Thermo- kelvin K (K = °C + 273) Momentum kilogram metre per kg m/s
dynamic second
temperature
Luminous candela cd
Common prefixes
intensity
SI units may be made larger or smaller by using pre-
Amount of mole mol fixes which denote multiplication or division by a par-
substance ticular amount.
The most common multiples are listed below. Knowl-
Part One
edge of indices is needed since all of the prefixes are

Some derived units are given special names.
powers of 10 with indices that are a multiple of 3.
For example, force = mass × acceleration, has units of
kilogram metre per second squared, which uses three of
the base units, i.e. kilograms, metres and seconds. The Prefix Name Meaning
unit of kg m/s2 is given the special name of a newton. T tera multiply i.e. × 1,000,000,000,000
Below is a list of some quantities and their units that by 1012
are common in engineering. G giga multiply i.e. × 1,000,000,000
by 109
M mega multiply i.e. × 1,000,000
Quantity Unit Symbol by 106
Length metre m k kilo multiply i.e. × 1,000
Area square metre m2 by 103
Volume cubic metre m3 m milli multiply 1 1
i.e. × 3 = = 0.001
by 10–3 10 1000
Mass kilogram kg
µ micro multiply 1 1
Time second s i.e. × 6 =
by 10–6 10 1,000,000
Electric current ampere A = 0.000001
Speed, velocity metre per second m/s n nano multiply 1 1
Acceleration metre per second m/s2 by 10–9 i.e. × 9 =
10 1,000,000,000
squared = 0.000 000 001
Density kilogram per cubic kg/m3 p pico multiply 1 1
metre by 10–12 i.e. × 1012 1,000,000,000,000
=
Temperature kelvin or Celsius K or °C = 0.000 000 000 001
Angle radian or degree rad or °
Angular velocity radian per second rad/s Here are some examples of prefixes used with engi-
Frequency hertz Hz neering units.
A frequency of 15 GHz means 15 × 109 Hz, which
Force newton N is 15,000,000,000 hertz*, i.e. 15 gigahertz is written
Pressure pascal Pa as 15 GHz and is equal to 15 thousand million hertz.
Energy, work joule J Instead of writing 15,000,000,000 hertz, it is much
Power watt W neater, takes up less space and prevents errors caused
by having so many zeros, to write the frequency as
Electric potential volt V 15 GHz.
A voltage of 40 MV means 40 × 106 V, which is For example, a force of 43,645 N can be re-written as
40,000,000 volts, i.e. 40 megavolts is written as 40 MV 43.645 × 103 N and from the list of prefixes can then be
and is equal to 40 million volts. expressed as 43.645 kN. Thus,
12 12 43,645 N ≡ 43.645 kN
Energy of 12 mJ means 12 × 10–3 J or 3 J or J,
10 1000
which is 0.012 J, i.e. 12 millijoules is written as 12 mJ To help further, on your calculator is an ‘ENG’ button.
and is equal to 12 thousandths of a joule. Enter the number 43,645 into your calculator and then
150 press ‘=’
A time of 150 ns means 150 × 10–9 s or s, which is
109 Now press the ‘ENG’ button and the answer is
0.000 000 150 s, i.e. 150 nanoseconds is written as 150 ns 43.645 × 103 .
and is equal to 150 thousand millionths of a second. We then have to appreciate that 103 is the prefix ‘kilo’
A force of 20 kN means 20 × 103 N, which is 20,000 giving 43,645 N ≡ 43.645 kN
newtons, i.e. 20 kilonewtons is written as 20 kN and is In another example, let a current be 0.0745 A
equal to 20 thousand newtons. Enter 0.0745 into your calculator. Press ‘=’
Engineering notation Now press ‘ENG’ and the answer is 74.5 × 10−3 .
Engineering notation is a number multiplied by a power We then have to appreciate that 10–3 is the prefix ‘milli’
Part One
of 10 that is always a multiple of 3. giving 0.0745 A ≡ 74.5 mA
For example,
43,645 = 43.645 × 103 in engineering notation and
0.0534 = 53.4 × 10−3 in engineering notation Problem 1. Express the following in engineering
notation and in prefix form:
In the list of engineering prefixes on page 22 it is ap- (a) 300,000 W (b) 0.000068 H
parent that all prefixes involve powers of 10 that are
multiples of 3. (a) Enter 300,000 into the calculator. Press ‘=’
Now press ‘ENG’ and the answer is 300 × 103
From the table of prefixes on page 22, 103 cor-
responds to kilo.
Hence, 300,000 W = 300 × 103 W in engineering
notation
= 300 kW in prefix form
(b) Enter 0.000068 into the calculator. Press ‘=’
Now press ‘ENG’ and the answer is 68 × 10−6
From the table of prefixes on page 22, 10−6 cor-
responds to micro.
Hence, 0.000068 H = 68 × 10−6 H in engineering
notation
= 68 µH in prefix form
Problem 2. Express the following in engineering

notation and in prefix form:
(a) 42 × 105 Ω (b) 47 × 10−10 F
(a) Enter 42 × 105 into the calculator. Press ‘=’

*Heinrich Rudolf Hertz (22 February 1857–1 January 1894)
was the first person to prove the existence of electromagnetic Now press ‘ENG’ and the answer is 4.2 × 106
waves. The scientific unit of frequency was named hertz in his From the table of prefixes on page 22, 106 cor-
honour. To find out more go to www.routledge.com/cw/bird
responds to mega.
Hence, 42 × 105 Ω = 4.2 × 106 Ω in engineering
notation 9. 35,000,000,000 Hz [35 GHz]
= 4.2 MΩ in prefix form 10. 1.5 × 10–11 F [15 pF]
47 11. 0.000017 A [17 µA]

(b) Enter 47 ÷ 1010 = into the cal-
12. 46200 Ω [46.2 kΩ]
culator. Press ‘=’ 10,000,000,000
Now press ‘ENG’ and the answer is 4.7 × 10–9 13. Rewrite 0.003 mA in µA [3 µA]
From the table of prefixes on page 22, 10–9 cor- 14. Rewrite 2025 kHz as MHz [2.025 MHz]
responds to nano. 15. Rewrite 5 × 104 N in kN [50 kN]
Hence, 47 ÷ 10–10 F = 4.7 × 10–9 F in engineering 16. Rewrite 300 pF in nF [0.3 nF]
notation
17. Rewrite 6250 cm in metres [62.50 m]
= 4.7 nF in prefix form
18. Rewrite 34.6 g in kg [0.0346 kg]
Problem 3. Rewrite (a) 14,700 mm in metres 19. The tensile stress acting on a rod is 5600000
(b) 276 cm in metres (c) 3.375 kg in grams Pa. Write this value in engineering notation.
Part One
[5.6 × 106 Pa = 5.6 MPa]

(a) 1 m = 1000 mm 20. The expansion of a rod is 0.0043 m. Write
1 1 this in engineering notation.
Hence, 1 mm = = 3 = 10−3 m
1000 10
Hence, 14,700 mm = 14,700 × 10–3 m = 14.7 m [4.3 × 10–3 m = 4.3 mm]
1 1
(b) 1 m = 100 cm hence 1 cm = = = 10−2 m
100 102
2.2 Metric–US/Imperial conversions
Hence, 276 cm = 276 × 10–2 m = 2.76 m
(c) 1 kg = 1000 g = 103 g The Imperial System (which uses yards, feet, inches,
Hence, 3.375 kg = 3.375 × 103 g = 3375 g etc. to measure length) was developed over hundreds of
years in the UK, then the French developed the Metric
System (metres) in 1670, which soon spread through
Now try the following Practice Exercise Europe, even to England itself in 1960. But the USA
and a few other countries still prefer feet and inches.
When converting from metric to imperial units,
Practice Exercise 11 Engineering notation or vice versa, one of the following tables (2.1 to 2.8)
In Problems 1 to 12, express in engineering nota- should help.
tion in prefix form: Table 2.1 Metric to Imperial length
1. 60,000 Pa [60 kPa] Metric US or Imperial
2. 0.00015 W [150 µW or 0.15 mW] 1 millimetre, mm 0.03937 inch
3. 5 × 107 V [50 MV] 1 centimetre, cm = 10 mm 0.3937 inch
1 metre, m = 100 cm 1.0936 yard
4. 5.5 × 10–8 F [55 nF]
1 kilometre, km = 1000 m 0.6214 mile
5. 100,000 N [100 kN]
6. 0.00054 A [0.54 mA or 540 µA] Problem 4. Calculate the number of inches in
7. 15 × 105 Ω [1.5 MΩ] 350 mm, correct to 2 decimal places.
8. 225 × 10–4 V [22.5 mV] 350 mm = 350 × 0.03937 inches = 13.78 inches from
Table 2.1
(b) 5.2 nautical miles = 5.2 × 1.852 km

Problem 5. Calculate the number of inches in
= 9.630 km from Table 2.2
52 cm, correct to 4 significant figures.
52 cm = 52 × 0.3937 inches = 20.47 inches from Table 2.1 Table 2.3 Metric to Imperial area
Metric US or Imperial
Problem 6. Calculate the number of yards in
1 cm2 = 100 mm2 0.1550 in2
74 m, correct to 2 decimal places.
1 m2 = 10,000 cm2 1.1960 yd2
74 m = 74 × 1.0936 yards = 80.93 yds from Table 2.1 1 hectare, ha = 10,000 m2 2.4711 acres
1 km2 = 100 ha 0.3861 mile2
Problem 7. Calculate the number of miles in
12.5 km, correct to 3 significant figures.
Problem 13. Calculate the number of square
12.5 km = 12.5 × 0.6214 miles = 7.77 miles from inches in 47cm2, correct to 4 significant figures.
Table 2.1
47cm2 = 47 × 0.1550 in2 = 7.285 in2 from Table 2.3
Table 2.2 Imperial to Metric length Problem 14. Calculate the number of square
Part One
US or Imperial Metric yards in 20 m2, correct to 2 decimal places.
1 inch, in 2.54 cm 20 m2 = 20 × 1.1960 yd2 = 23.92 yd2 from Table 2.3
1 foot, ft = 12 in 0.3048 m
Problem 15. Calculate the number of acres in 23
1 yard, yd = 3 ft 0.9144 m
hectares of land, correct to 2 decimal places.
1 mile = 1760 yd 1.6093 km
1 nautical mile = 2025.4 yd 1.852 km 23 hectares = 23 × 2.4711 acres = 56.84 acres from
Table 2.3
Problem 8. Calculate the number of centimetres
Problem 16. Calculate the number of square miles
in 35 inches, correct to 1 decimal place.
in a field of 15 km2 area, correct to 2 decimal places.
35 inches = 35 × 2.54 cm = 88.9 cm from Table 2.2
15 km2= 15 × 0.3861 mile2 = 5.79 mile2 from Table 2.3
Problem 9. Calculate the number of metres in 66
inches, correct to 2 decimal places. Table 2.4 Imperial to Metric area
US or Imperial Metric
66 66
66 inches = feet = × 0.3048 m = 1.68 m from 1 in2 6.4516 cm2
12 12 Table 2.2 1 ft2 = 144 in2 0.0929 m2
Problem 10. Calculate the number of metres in 1 yd2 = 9 ft2 0.8361 m2
50 yards, correct to 2 decimal places. 1 acre = 4840 yd2 4046.9 m2
1 mile2 = 640 acres 2.59 km2
50 yards = 50 × 0.9144 m = 45.72 m from Table 2.2
Problem 17. Calculate the number of square cen-
Problem 11. Calculate the number of kilometres timetres in 17.5 in2, correct to the nearest square
in 7.2 miles, correct to 2 decimal places. centimetre.
7.2 miles = 7.2 × 1.6093 km = 11.59 km from Table 2.2 17.5 in2 = 17.5 × 6.4516 cm2 = 113 cm2 from Table 2.4
Problem 12. Calculate the number of (a) yards Problem 18. Calculate the number of square
(b) kilometres in 5.2 nautical miles. metres in 205 ft2, correct to 2 decimal places.
(a) 5.2 nautical miles = 5.2 × 2025.4 yards 205 ft2 = 205 × 0.0929 m2 = 19.04 m2 from Table 2.4
= 10532 yards from Table 2.2
Problem 19. Calculate the number of square metres Problem 25. Calculate the number of cubic cen-
in 11.2 acres, correct to the nearest square metre. timetres in 3.75 in3, correct to 2 decimal places.
11.2 acres = 11.2 × 4046.9 m2 = 45325 m2 from 3.75 in3 = 3.75 × 16.387 cm3 = 61.45 cm3 from Table 2.6
Table 2.4
Problem 26. Calculate the number of cubic me-
Problem 20. Calculate the number of square tres in 210 ft3, correct to 3 significant figures.
kilometres in 12.6 mile2, correct to 2 decimal places.
210 ft3 = 210 × 0.02832 m3 = 5.95 m3 from Table 2.6
12.6 mile2 = 12.6 × 2.59 km2 = 32.63 km2 from
Table 2.4 Problem 27. Calculate the number of litres in
4.32 US pints, correct to 3 decimal places.
Table 2.5 Metric to Imperial volume/capacity
4.32 US pints = 4.32 × 0.4732 litres = 2.044 litres from
Metric US or Imperial
Table 2.6
1 cm3 0.0610 in3
Problem 28. Calculate the number of litres in
Part One
1 dm3 = 1000 cm3 0.0353 ft3

1 m3 = 1000 dm3 1.3080 yd3 8.62 US gallons, correct to 2 decimal places.
1 litre = 1 dm3 = 1000 cm3 2.113 fluid pt = 1.7598 pt 8.62 US gallons = 8.62 × 3.7854 litres
= 32.63 litres from Table 2.6
Problem 21. Calculate the number of cubic
inches in 123.5 cm3, correct to 2 decimal places.
Table 2.7 Metric to Imperial mass
123.5cm3 = 123.5 × 0.0610 in3 = 7.53 in3 from Metric US or Imperial
Table 2.5
1 g = 1000 mg 0.0353 oz
Problem 22. Calculate the number of cubic feet 1 kg = 1000 g 2.2046 lb
in 144 dm3, correct to 3 decimal places. 1 tonne, t = 1000 kg 1.1023 short ton
1 tonne, t = 1000 kg 0.9842 long ton
144 dm3 = 144 × 0.0353 ft3 = 5.083 ft3 from Table 2.5
Problem 23. Calculate the number of cubic yards The British ton is the long ton, which is 2240
in 5.75 m3, correct to 4 significant figures. pounds, and the US ton is the short ton which is
2000 pounds.
5.75 m3 = 5.75 × 1.3080 yd3 = 7.521 yd3 from Table 2.5
Problem 29. Calculate the number of ounces in a
Problem 24. Calculate the number of US fluid mass of 1346 g, correct to 2 decimal places.
pints in 6.34 litres of oil, correct to 1 decimal place.
1346 g = 1346 × 0.0353 oz = 47.51 oz from Table 2.7
6.34 litre = 6.34 × 2.113 US fluid pints = 13.4 US fluid
pints from Table 2.5 Problem 30. Calculate the mass, in pounds, in a
210.4 kg mass, correct to 4 significant figures.
Table 2.6 Imperial to Metric volume/capacity
210.4 kg = 210.4 × 2.2046 lb = 463.8 lb from Table 2.7
1 in3 16.387 cm3 Problem 31. Calculate the number of short tons
1 ft3 0.02832 m3 in 5000 kg, correct to 2 decimal places.
1 US fl oz = 1.0408 UK fl oz 0.0296 litres
1 US pint (16 fl oz) = 0.8327 UK pt 0.4732 litres 5000 kg = 5 t = 5 × 1.1023 short tons
= 5.51 short tons from Table 2.7
1 US gal (231 in3) = 0.8327 UK gal 3.7854 litres
Table 2.8 Imperial to Metric mass Now try the following Practice Exercise
1 oz = 437.5 grain 28.35 g Practice Exercise 12 Metric/imperial
1 lb = 16 oz 0.4536 kg conversions
1 stone = 14 lb 6.3503 kg In the following Problems, use the metric/impe-
1 hundredweight, cwt = 112 lb 50.802 kg rial conversions in Tables 2.1 to 2.8
1 short ton 0.9072 tonne 1. Calculate the number of inches in 476 mm,
1 long ton 1.0160 tonne correct to 2 decimal places [18.74 in]
2. Calculate the number of inches in 209 cm,
Problem 32. Calculate the number of grams in correct to 4 significant figures [82.28 in]
5.63 oz, correct to 4 significant figures.
3. Calculate the number of yards in 34.7 m,
5.63 oz = 5.63 × 28.35 g = 159.6 g from Table 2.8 correct to 2 decimal places [37.95 yd]
4. Calculate the number of miles in 29.55 km,
Problem 33. Calculate the number of kilograms
correct to 2 decimal places
Part One
in 75 oz, correct to 3 decimal places.
[18.36 miles]
75 75 5. Calculate the number of centimetres in
75 oz = lb = × 0.4536 kg = 2.126 kg from
16 16 16.4 inches, correct to 2 decimal places
Table 2.8
[41.66 cm]
6. Calculate the number of metres in 78
Problem 34. Convert 3.25 cwt into (a) pounds
inches, correct to 2 decimal places
(b) kilograms.
[1.98 m]
(a) 3.25 cwt = 3.25 × 112 lb = 364 lb from Table 2.8 7. Calculate the number of metres in 15.7
(b) 3.25 cwt = 3.25 × 50.802 kg = 165.1 kg from yards, correct to 2 decimal places
Table 2.8 [14.36 m]
8. Calculate the number of kilometres in 3.67
Temperature miles, correct to 2 decimal places
To convert from Celsius to Fahrenheit, first multiply by [5.91 km]
9/5, then add 32. 9. Calculate the number of (a) yards (b) kilo-
To convert from Fahrenheit to Celsius, first subtract 32, metres in 11.23 nautical miles
then multiply by 5/9. [(a) 22,745 yd (b) 20.81 km]
10. Calculate the number of square inches in
Problem 35. Convert 35ºC to degrees Fahrenheit. 62.5 cm2, correct to 4 significant figures
9 [9.688 in2]
F = 9 C + 32 hence 35ºC = (35) + 32 = 63 + 32 = 95ºF
5 5 11. Calculate the number of square yards in
15.2 m2, correct to 2 decimal places
Problem 36. Convert 113ºF to degrees Celsius. [18.18 yd2]
9 12. Calculate the number of acres in 12.5

C= 32) hence 113ºF = 5 (113 − 32) = 5 (81)
(F+−32
C hectares, correct to 2 decimal places
5 9 9
= 45ºC [30.89 acres]
13. Calculate the number of square miles in
A summary of metric to imperial conversions, and vice 56.7 km2, correct to 2 decimal places
versa, is shown on page 365. [21.89 mile2]
14. Calculate the number of square centi- 29. Calculate the number of grams in 7.78 oz,
metres in 6.37 in2, correct to the nearest correct to 4 significant figures [220.6 g]
square centimetre [41 cm2]
30. Calculate the number of kilograms in
15. Calculate the number of square metres in 57.5 oz, correct to 3 decimal places
308.6 ft2, correct to 2 decimal places [1.630 kg]
[28.67 m2]
31. Convert 2.5 cwt into (a) pounds (b) kilograms
16. Calculate the number of square metres [(a) 280 lb (b) 127.2 kg]
in 2.5 acres, correct to the nearest square
metre [10117 m2] 32. Convert 55ºC to degrees Fahrenheit [131ºF]
17. Calculate the number of square kilometres 33. Convert 167ºF to degrees Celsius [75 ºC]
in 21.3 mile2, correct to 2 decimal places
[55.17 km2]
18. Calculate the number of cubic inches in 2.3 Straight line graphs
200.7 cm3, correct to 2 decimal places
A graph is a visual representation of information, show-
[12.24 in3]
Part One
ing how one quantity varies with another related quantity.

19. Calculate the number of cubic feet in The most common method of showing a relationship
214.5 dm3, correct to 3 decimal places between two sets of data is to use a pair of reference
[7.572 ft3] axes – these are two lines drawn at right angles to each
20. Calculate the number of cubic yards in other (often called Cartesian or rectangular axes), as
13.45 m3, correct to 4 significant figures shown in Figure 2.1.
[17.59 yd3] y
21. Calculate the number of US fluid pints in 4

B (4, 3)
15 litres, correct to 1 decimal place 3
[31.7 fluid pints] 2 A (3, 2)
Origin
22. Calculate the number of cubic centimetres 1
in 2.15 in3, correct to 2 decimal places
[35.23 cm3] 4 3 2 1 0 1 2 3 4 x
1
23. Calculate the number of cubic metres in 2
175 ft3, correct to 4 significant figures C (3, 2)
[4.956 m3] 3
4
24. Calculate the number of litres in 7.75 US
pints, correct to 3 decimal places
[3.667 litres] Figure 2.1
25. Calculate the number of litres in 12.5 US The horizontal axis is labelled the x-axis, and the verti-
gallons, correct to 2 decimal places cal axis is labelled the y-axis.
[47.32 litres] The point where x is 0 and y is 0 is called the origin. x
values have scales that are positive to the right of the
26. Calculate the number of ounces in 980 g,
origin and negative to the left.
correct to 2 decimal places [34.59 oz]
y values have scales that are positive up from the origin
27. Calculate the mass, in pounds, in 55 kg, and negative down from the origin.
correct to 4 significant figures [121.3 lb] Co-ordinates are written with brackets and a comma
in between two numbers.
28. Calculate the number of short tons in
For example, point A is shown with co-ordinates (3, 2)
4000 kg, correct to 3 decimal places
and is located by starting at the origin and moving 3
[4.409 short tons]
units in the positive x direction (i.e. to the right) and
then 2 units in the positive y direction (i.e. up).
When co-ordinates are stated, the first number is al- Similarly, determine the force applied when the
ways the x value, and the second number is always the load is zero. It should be close to 11 N.
y value. Where the straight line crosses the vertical axis is
Also in Figure 2.1, point B has co-ordinates (– 4, 3) and called the vertical-axis intercept. So in this case,
point C has co-ordinates (–3, –2) the vertical-axis intercept = 11 N at co-ordinates
The following table gives the force F newtons which, (0, 11)
when applied to a lifting machine, overcomes a corre-
The graph you have drawn should look something like
sponding load of L newtons.
Figure 2.2 shown below.
F (newtons) 19 35 50 93 125 147
Graph of F against L
L (newtons) 40 120 230 410 540 680
160
150
1. Plot L horizontally and F vertically. 140
2. Scales are normally chosen such that the graph oc- 130
120
cupies as much space as possible on the graph pa-
110
per. So in this case, the following scales are chosen: 100
F (newtons)
Part One
90
Horizontal axis (i.e. L): 1 cm = 50 N
80
Vertical axis (i.e. F): 1 cm = 10 N 70
60
3. Draw the axes and label them L (newtons) for the 50
horizontal axis and F (newtons) for the vertical 40
axis. 30
20
4. Label the origin as 0. 10
5. Write on the horizontal scaling at 100, 200, 300,
0 100 200 300 400 500 600 700 800
and so on, every 2 cm. L (newtons)
6. Write on the vertical scaling at 10, 20, 30, and so
on, every 1 cm. Figure 2.2
7. Plot on the graph the co-ordinates (40, 19),

(120, 35), (230, 50), (410, 93), (540, 125) and In another example, let the relationship between two
(680, 147) marking each with a cross or a dot. variables x and y be y = 3x + 2
8. Using a ruler, draw the best straight line through When x = 0, y = 3 × 0 + 2 = 0 + 2 = 2
the points. You will notice that not all of the points
When x = 1, y = 3 × 1 + 2 = 3 + 2 = 5
lie exactly on a straight line. This is quite normal
with experimental values. In a practical situation When x = 2, y = 3 × 2 + 2 = 6 + 2 = 8, and so on.
it would be surprising if all of the points lay ex-
The co-ordinates (0, 2), (1, 5) and (2, 8) have been
actly on a straight line.
produced and are plotted, with others, as shown in
9. Extend the straight line at each end. Figure 2.3.
10. From the graph, determine the force applied when
the load is 325 N. It should be close to 75 N. This y
process of finding an equivalent value within the
8
given data is called interpolation.
6
Similarly, determine the load that a force of 45 N y 3x 2
4
will overcome. It should be close to 170 N.
2
11. From the graph, determine the force needed to over-
come a 750 N load. It should be close to 161 N. This 1 0 1 2 x
process of finding an equivalent value outside the
given data is called extrapolation. To extrapolate
we need to have extended the straight line drawn. Figure 2.3
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To summarize flying when a cross wind is blowing, it will be said
that the direction of actual travel will not be the direction indicated by
the axis of the airplane; and that therefore while in a picture of the
situation the airplane appears to skid sideways along the whole
course it must be borne in mind that actually there is no skidding
whatever but the air is meeting the airplane in normal manner. The
situation is analogous to that of a fly going from one side to the other
of the cabin of a moving ship, where the actual course through space
of the fly is an apparent skid, due to the resultant of its own and the
ship’s movement.
Variation of Velocity and Direction With Height

(25 miles per hour wind)
Height in feet At surface 500 1000 2000 3000 4000 5000
Velocity change in per cent 100 135 172 188 196 200 200
Clockwise deviation in degrees 0 5 10 16 19 20 21
Effect of Wind on Radius of Action.—Not only is the direction of

flight altered by the wind but also the radius of action from a
standpoint of gasoline capacity is altered. In the above machine the
gasoline capacity is sufficient for 3½ hr. of flight. How far can it go
across country and return before the gasoline is used up? Always
allow ½ hr. gasoline for climbing and for margin; this leaves 3 hr.,
which at 75 miles an hour is 225 miles, or 112 miles out and 112
miles back. Now suppose that a flight is to be made across country
directly in the teeth of a 40-mile wind; the radius of flight will be
altered as indicated by the following calculation: Speed outward is
obviously 75 minus 40 or 35 miles per hour. Speed on the return trip
is obviously 75 plus 40 or 115 miles per hour—3.29 times as fast—
and occupying a time which may be designated by the letter x. The
time on the outward trip may be designated by 3.29x, a total time of
x + 3.29x which we know equals 180 min. before the gas runs out.
Solve the equation x + 3.29x = 180 and we find that x is equal to 42
min., that is, the return trip requires 42 min., and the outward trip
requires 138 min. The distance covered on the outward trip is then
138/ of 35, which equals 80.5 miles. The radius is then reduced
60
from 112 miles to 80.5 miles.
In cases where the wind is not parallel to the line of flight the
actual velocity of course can not be obtained by adding up the
airplane and wind velocities, but must be obtained by the graphical
method mentioned above; thenceforward the calculation is the same.
Effect of Height.—Of course if one has to fly in the teeth of a wind
and can choose one’s own altitude, it is desirable to fly low where
the head wind has its smaller velocity, and when flying with the
following wind to rise to considerable altitudes. The proper height at
which to fly will be about 1500 to 3000 ft., for cross-country trips over
ordinary country; but may be increased when the wind is unsteady or
decreased when there are low-lying clouds. The steadiness as well
as the speed of the wind increases with the height. The character of
the country should be carefully investigated from the profile maps
before starting; all hilly parts should be marked on the map as a
warning against landing. Contour is not readily distinguished from a
height of 2000 ft. and for this reason points may be indicated on the
map where poor landing places make it desirable to fly high. The
character of the country or the scarcity of landing places may make it
advisable to fly at high altitudes for the following reasons: (1) in case
of engine failure a good margin of height is necessary to provide
length of glide to reach distant landing places; (2) there is then plenty
of space for righting the airplane in case of bumps, side slips, etc.;
(3) eddies or local currents due to inequalities of the ground do not
exist to great heights; (4) landmarks can be better distinguished from
high altitudes because the vision is better (however, one must never
trust to landmarks only in navigating but should constantly use a
compass if only as a check, and especially in passing through
clouds). Having selected in advance the proper height to use during
the trip climb to this height in circles; note the direction of wind drift
meanwhile to check up your estimate. Pass directly over the point of
departure and when over it point the nose of the airplane for a
moment directly toward the desired objective (which can be done
with the aid of the magnetic compass); select some distant object
which is dead ahead, and therefore directly in the course; then head
the nose of the machine up into the wind just enough so that the
direction of movement will be straight toward this distant object. The
direction of the nose of the machine thus set by a method distinct
from the graphical method above mentioned should exactly
correspond, however, with the calculated direction; and thus a
means of checking is obtained.
Effect of Fog.—The effect of fog upon navigating an airplane is
that it prevents the use of landmarks in aiding the pilot; also that it
upsets the pilot’s sense of level. These two effects are, of course,
independent of the fact that proper landing places are obscured, with
resultant peril in case of engine failure. Therefore, a fog should be
avoided whenever possible; when one comes up, the airplane
should descend, and should never attempt to get above it, as in
certain localities it may turn out to be a ground fog. If the fog is very
bad, land at the earliest opportunity. It is on account of fog that the
pilot avoids river valleys where frequently there is a haze from the
ground up to a height of 700 ft., preventing the view of proper
landing places in case of necessity.
Effect of Clouds on Navigation.—Flying in or above the clouds
is a similar case, inasmuch as landmarks can not be seen. It is not
wise to go above the clouds when on the sea coast, as offshore
winds may, unknown to the pilot, carry him out to sea; and any flight
over the sea which is to a distance greater than the safe return
gliding distance is, of course, perilous.
Navigation by Means of the Drift Indicator.—The drift indicator
is an instrument for determining directly the side drift of an airplane.
It enables the pilot by looking through a telescope at the ground to
determine exactly what his direction of motion is with relation to the
ground. The telescope is mounted vertically and is rotatable about its
own axis; it has a cross-hair which appears in the field of view during
the pilot’s observation of the ground. As the airplane speeds
overhead objects on the ground will appear through the telescope to
slip backward in the given direction; and when accustomed to the
use of this instrument the pilot can rotate the telescope until the
cross-hair is exactly parallel to the apparent line of motion of objects
on the ground. The telescope cross-hair is parallel to the axis of the
airplane normally and the scale attached to the telescope will in this
case read zero. When the pilot rotates the telescope so that the
cross-hair becomes parallel to the relative backward motion of the
ground the scale will read something different from zero and will give
the angle between the actual line of motion and the axis of the
airplane.
Such a drift indicator is, of course, useful only when the ground is
visible. The pilot knowing the angle between the airplane axis and
the line of motion and therefore knowing the deviation between the
supposed course and the actual course is able to make corrections
and steer the machine in its proper direction. This may be done by
altering the “lubber-line” or his compass just enough to offset the
side drift of the machine; after which the desired course may be
followed by simply keeping to the proper compass bearing. An
instrument has been devised wherein the rotation of the drift-
indicator telescope simultaneously alters the lubber-line zero. The
operator then has merely to take an occasional observation of the
apparent drift line of the ground, which observation automatically
shifts the lubber-line and navigation proceeds as if there were no
side wind blowing whatever. Knowing the angle between the
direction of movement and the airplane axis, the pilot may then
compute the speed of motion in a manner analogous to the graphical
method previously mentioned; or he can make use of a chart for the
determination of this speed.
Navigation over Water.—In flying over water the presence of
waves is a valuable guide to the aviator, for he knows that these
waves extend in a direction normal to the wind. Moreover, he knows
that the velocity of the waves bears some relation to the velocity of
the wind. In order to estimate the velocity of the waves it is only
necessary to know their wave length, that is, the distance between
two consecutive wave crests. The rule is that for a wave length of 10
ft. the velocity is 10 miles per hour, and will vary as the square root
of this wave length; that is, if the wave length is half, the velocity will
be 10 divided by the square root of 2, or 7.1 miles per hour.
CHAPTER VI
THE RIGGING OF AIRPLANES
Object.—The object of this chapter is to teach the elementary
principles of correct rigging. It is not expected that the student will
become an expert mechanic, but with this treatment as a basis and
through practice he will be able to judge whether or not a machine is
correctly and safely rigged. In other words, he will not have to
depend on someone else’s judgment as to whether panels, wires,
controls, struts, etc., of a machine are in good order, but he will be
able to observe understandingly that they are. If the engine goes
wrong he can land, if the rigging goes wrong he is in great difficulty.
Moreover, if the rigging is wrong, speed is lessened and the stability
is uncertain.
The first thing to be learned in rigging is a knowledge of the
peculiar terms which have come into use in aeronautics defining
different parts of the machines. Our present list of terms is derived,
partly from French, partly from English, and partly from American
terms. Thus different names may refer to the same part.
NOMENCLATURE
1. Tractor.—An airplane that is pulled through the air by a propeller situated in

front of the machine, is called a tractor.
2. Pusher.—If the propeller is back of the main lifting planes the machine is
called a pusher.
3. Fuselage or Body.—The main body of the airplane in which the pilot sits and
to which the landing gear, motor, controls, and sustaining surfaces are fixed. A
small body, especially in pusher types of machines, is called a Nacelle.
4. Cockpit.—The openings and space in the fuselage where pilot or observer
sits.
5. Stream-line Body.—The shape of a body or part which permits a regular flow
of air around and along it with the least resistance, in other words with minimum,
obstruction and eddying.
6. Fairing.—Building up a member or part of the plane with a false piece that it
may have a stream-line body.
7. Wings, Planes, Panels.—The main supporting surfaces of an airplane are
called wings, although the terms planes and panels are probably as frequently
used and even preferred by many. The term panel refers properly to a section of
the wings with the included struts and wires. The small panel directly above the
body is called the engine section panel or the center panel, while the panels to the
right and left of the body or fuselage are called the main panels. The main panels
are the right and left panels as seen from the seat. Each main panel may be
subdivided into the inner wing bay, the outer wing bay, and the overhang.
8. Landing Gear, Chassis or Undercarriage.—The wheels and the struts and
wires by which they are attached to the fuselage.
9. Horizontal Stabilizer or Horizontal Fin.—The horizontal fixed tail plane.
10. Vertical Stabilizer or Vertical Fin.—The small vertical fixed plane in front of
the rudder.
11. Rudder.—The hinged surface used to control the direction of the aircraft in
the horizontal plane. As with a boat, for steering or “yawing” or changing its
direction of travel.
12. Elevator or Flap; Flippers.—A hinged horizontal surface for controlling the
airplane up and down, usually attached to the fixed tail plane; for pitching the
machine or “nosing up” and “nosing down.”
13. Tail or “Empennages.”—A general name sometimes applied to the tail
surfaces of a machine.
14. Mast or Cabane.—The small vertical strut on top of the upper plane used
for bracing the overhang.
15. Ailerons.—Movable auxiliary surfaces used for the control of rolling or
banking motion. Other definitions are that they are for the lateral control or for
maintaining equilibrium. When they are a part of the upper plane they are
sometimes called wing flaps.
16. Landing Wires or Ground Wires (Single).—The single wires which support
the weight of the panels when landing or on the ground.
17. Flying Wires, or Load Wires (Double).—The wires which support the body
or fuselage from the planes when in flight.
18. Drift Wires.—The horizontal wires which lead from the nose of the fuselage
to the wings and thus keep them from collapsing backward. For the same reason
the wings have interior drift wires.
19. Diagonal Wires.—Any inclined bracing wires.
20. Skids.—(a) Tail Skid.—The flexible support under the tail of the machine.
(b) Wing Skid.—The protection under the outer edge of the lower wing.
(c) Chassis Skids.—Skids sometimes placed in front of the landing gear.
21. Horns, or Control Braces.—The steel struts on the controls to which the
control wires are attached.
22. Struts; Wing Struts.—The vertical members of the wing trusses of a
biplane, used to take pressure or compression, whereas the wires of the trusses
are used to take pull or tension. There are also fuselage struts and chassis struts.
23. Spar or Wing Bars.—The longitudinal members of the interior wing
framework.
24. Rib (Wing).—The members of the interior wing framework transverse to the
spars.
25. The Longerons or Longitudinals.—The fore and aft or lengthwise
members of the framing of the fuselage, usually continuous across a number of
points of support.
26. Engine (Right and Left Hand).—In the ordinary tractor machine, when
viewed from the pilot’s seat a right-handed engine revolves clockwise and right-
handed.
27. Propeller.—
28. Pitch (Propeller).—The distance forward that the propeller would travel in
one revolution, if there were no slip, that is, if it were moving in a thread cut at the
same inclination as the blade. Pitch angle refers to the angle of inclination of the
propeller blade.
29. Slip.—Slip is the difference between the actual travel forward of a screw
propeller in one revolution and its pitch.
30. Dope.—A general term applied to the material used in treating the cloth
surface of airplane members to increase strength, produce tautness, and act as a
filler to maintain air and moisture tightness. Usually of the cellulose type.
31. Controls.—Since there are three axes or main directions about which an
airplane may turn or rotate it follows that three controlling devices are required.
These are: (1) the elevator for pitching; (2) the rudder for steering or yawing; (3)
the ailerons for lateral, rolling or banking control.
The term controls is a general term used to distinguish the means provided for
operating the devices used to control speed, direction of flight and attitude of the
aircraft.
32. Cotter Pins.—Must be on every nut.
33. Castelled Nuts.—Admit cotter pins.
34. Turnbuckles.—Must be well and evenly threaded and locked with safety
wires.
35. Safety Wires.—For locking turnbuckles and hinge pins.
36. Shackle and Pin.—
37. Hinge Connections.—
38. Leading Edge or Entering Edge.—The front edge of a plane.
39. Trailing Edge.—The rear edge of a plane.
40. Stagger.—The horizontal distance that the entering edge of the upper wing
of a biplane is ahead of the entering edge of the lower wing.
41. Dihedral Angle.—A term used to denote that the wings are arranged to
incline slightly upward from the body toward their tips. The angle made with the
horizontal by this inclination of the wing is called the dihedral angle.
42. Angle of Incidence.—The angle at which a wing is inclined to the line of
flight.
43. Decalage.—Difference in angle of incidence between any two distinct
aerofoils on an airplane.
44. Chord.—Distance between the entering edge and trailing edge of a wing
measured on a straight line touching front and rear bottom points of a wing.
45. Camber.—The depth of the curve given to a sustaining surface such as a
wing. Thus it will be observed that the planes are not straight in cross-section but
are concave slightly upward. The depth of this concavity is the camber. Another
way of expressing this is that camber is the greatest distance between the surface
of a wing and its chord line.
46. Gap.—The distance between the lower and upper wings of a biplane.
47. Spread.—The distance over all from one wing tip to the other wing tip.
48. Aerofoil.—A general name applied to any wing or lifting surface of an
airplane.
49. Deadhead Resistance.—Each part of an airplane against which the wind
strikes offers a resistance against being moved through the air. This is called the
deadhead resistance or the parasite resistance. It is for the purpose of lessening
this resistance that the parts of a machine are stream-lined. Remember that force
or power must be applied to overcome this resistance and the lessening of such
resistance decreases the power necessary. A parallel illustration is to think of the
power necessary to push a board sideways through water.
50. Drift.—When the air strikes the inclined wing of an airplane its force has two
components. One part called the lift (see 52) acts up and tends to lift the machine.
The other part, called drift, tends to push the machine backward. This drift must
also be overcome by applying power enough to drive the machine forward.
51. Total Resistance.—Sometimes called drag. (49) Deadhead resistance
added to (50) drift, gives the total forces opposing the forward movement of the
airplane. This is called the total resistance and is overcome by the thrust of the
propeller.
52. Lift.—(See 50). The upward or vertical part of the air pressure acting against
the wings, and which is utilized to lift or support the airplane.
53. Center of Gravity.—The point of balance of an airplane which may be
otherwise defined as the point through which the mass of an airplane acts. If the
weight is too far forward the machine is nose-heavy. If the weight is too far behind
the center of lift the machine is tail-heavy.
54. Aspect Ratio.—The ratio of span to chord of a wing or any other aerofoil.
55. Gliding Angle (Volplane).—The angle made to the horizontal by the flight
path of an airplane with the engine shut off; e.g., an airplane is 1000 ft. high, when
its engine fails. Suppose its gliding angle is 1 in 6. Therefore, in still air it can glide
6000 ft. forward. The general term glide refers to flying without power.
56. The Angle of Best Climb.—The steepest angle at which an airplane can
climb.
57. Stability.—The property of an airplane to maintain its direction and to return
easily to its equilibrium or balance with a minimum of oscillation. This is sometimes
called dynamical stability. An airplane may have (first) inherent stability, which is
the stability due to the arrangement and disposition of its fixed parts. It may also
have stability with regard to any one of the three directions in which it may move.
These are named as follows: (1) directional stability, with reference to the vertical
axis; (2) lateral stability with reference to the longitudinal (or fore and aft) axis; (3)
longitudinal stability, stability with reference to the lateral (or thwartship) axis.
58. Flying Position.—Refers to the position of the fuselage when flying. With
the Curtiss JN-4 machines in this position the top longerons are horizontal and
level both ways. The engine bearers are also level, and the wings have an angle of
incidence of 2°.
59. Capacity.—The weight an airplane will carry in excess of the dead load
(dead load includes structure power plant and essential accessories).
60. Flight Path.—The path of the center of gravity of an aircraft with reference
to the air.
61. Stalling.—A term describing the condition of an airplane which from any
cause has lost the relative speed necessary for support and controlling, and
referring particularly to angles of incidence greater than the critical angle.
62. Sweepback.—The horizontal angle (if any) that the leading edge of a
machine makes with the crosswise or lateral axis of an airplane.
63. Nose Dive or Vol-pique.—A dangerously steep descent, head on.
CHAPTER VII
MATERIALS OF CONSTRUCTION
The materials of construction for airplanes should be of such
material, size and form as to combine greatest strength and least
weight. With metal parts in particular it may be necessary to
substitute less strong material for the sake of getting non-corrosive
qualities, ability to withstand bending, ductility or ease of bending,
etc. With wood, absence of warping is important as well. The
materials which are considered are the following: wood, steel,
including wires; special metals as aluminum, brass, monel metal,
copper, etc., and also linen and dope.
Strength of Materials.—It is important in a general way to
understand the terms used in speaking of strength of materials. Thus
we may have strength in tension, strength in compression, or
strength in shearing, bending and torsion. Some material fitted to
take tension will not take compression, as for example wire; some
material, as bolts, are suited to take shear, etc.
In general all material for airplanes has been carefully tested and
no excess material is used above that necessary to give the machine
the necessary strength.
Tension.—This means the strength of a material which enables it
to withstand a pull. Thus wires are used where strength of this kind is
required.
Compression.—This refers to strength against a pressure. Wire
has no strength for this purpose, and wood or sometimes steel is
used.
Shearing.—Refers to strength against cutting off sideways. Thus
the pull on an eyebolt tends to shear the eyebolt, or the side pull on
any bolt or pin tends to shear the pin.
Bending.—In bending material the fibres on the outside tend to
pull apart; those on the inside tend to go together. Thus on the
outside we have tension, and on the inside compression. Along the
center line there is neither tension or compression, it is the “neutral
axis.”
Torsion.—Torsion is a twisting force, such as an engine propeller
shaft receives.
Testing for Strength.—If a wire is an inch square in cross-section
and breaks when a load of 150,000 lb. is hung on it, we say that the
strength of the wire is 150,000 per square inch. Smaller wires
equally strong have a strength of 150,000 lb. per square inch also,
but they in themselves will not support a load of 150,000 lb. but only
the fraction of that, according to the fraction of a square inch
represented by their cross-section.
In the same way, a square inch of wood under a compressive load
may break at 5000 lb. If, however, the piece of wood is long in
proportion to its thickness, it will bend easily and support much less
weight. For example, a perfectly straight walking cane could perhaps
have a ton weight put on it without breaking but if the cane were not
set squarely or if it started to bend it would immediately break under
the load.
These cases illustrate the importance of having struts perfectly
straight, not too spindling and evenly bedded in their sockets. Some
training machines are built with a factor of safety of 12. That is to
say, the breaking strength of any part is twelve times the ordinary
load or stress under which the piece is placed. It should be
remembered, however, that under any unusual condition in the air,
such as banking, etc., extra strains are placed on the parts and the
factor of safety is much less than 12. Factor of safety of 12 thus does
not mean exactly what it does in other engineering work, where
allowances are made for severe conditions. The so-called factor of
safety of 12 in airplane work is probably no greater than a factor of
safety of 2 or 3 in regular engineering work.
There are three all-important features in the flying machine
construction, viz., lightness, strength and extreme rigidity. Spruce is
the wood generally used for parts when lightness is desired more
than strength, oak, ash, hickory and maple are all stronger, but they
are also considerably heavier, and where the saving of weight is
essential, the difference is largely in favor of the spruce. This will be
seen in the following condensed table of U. S. Government
Specifications.
Modulus of rupture,
Weight per cubic foot, Compression strength,
Wood pounds per square
pounds (15% moisture) pounds per square inch
inch
Hickory 50 16,300 7,300
White
46 12,000 5,900
Oak
Ash 40 12,700 6,000
Walnut 38 11,900 6,100
Spruce 27 7,900 4,300
White
29 7,600 4,800
Pine
A frequently asked question is: “Why is not aluminum or some

similar metal, substituted for wood?” Wood, particularly spruce, is
preferred because, weight considered, it is much stronger than
aluminum, and this is the lightest of all metals. In this connection the
following table will be of interest.
Weight in cubic Tensile strength per Compression strength

Material
feet, pounds sq. in pounds per sq. in pounds
Spruce 27 7,900 4,300
Aluminum 162 15,000 12,000
Brass (sheet) 510 20,000 12,000
Steel (tool) 490 100,000 60,000
Nickel steel 480 100,000[1]
Copper (sheet) 548 30,000 40,000
Tobin bronze
... 80,000
(Turnbuckles)
Monel metal 540 90,000 30,000
Wood.—Present practice in airplane construction is to use wood
for practically all framing, in other words, for all parts which take
pressure or compression. Although wood is not as strong for its size
as steel and therefore offers more air resistance for the same
strength yet the fact that frame parts must not be too spindling, in
other words, that they must have a certain thickness in proportion to
their unsupported length, has led to the use of wood in spite of the
greater strength of steel. Some airplanes, however, as the
Sturtevant, are constructed with practically a steel framing.
It should be borne in mind that any piece or kind of wood will not
answer for framing, and more especially for repair parts. There is a
tremendous difference in the strength and suitability among different
woods for the work. For instance, a piece of wood of cross or
irregular grain, one with knots, or even one which has been bored or
cut or bruised on the outside, may have only half or less the strength
of the original piece. Air drying doubles the strength of green wood,
proper oven drying is better yet.
Notice how the ends of each piece are ferruled, usually with
copper or tin. This is to prevent the bolt pulling out with the grain of
the wood, and also prevents splitting and end checking and gives a
uniform base on which the pressure comes.
It is generally advised not to paint wood as it tends to conceal
defects from inspection. So varnish only.
Wrapping wooden members with linen or cord tightly and doping
this, both to make waterproof and to still further tighten, increases
the resistance to splitting. The absence of warping tendencies
determine often what wood to choose.
The selection of lumber and detection of flaws is a matter of
experience and should be cultivated. It is, however, nothing more
than the extension of the knowledge that leads a man to pick out a
good baseball bat.
Woods.—1. Spruce.—Should be clear, straight-grained, smooth
and free from knot holes and sap pockets, and carefully kiln-dried or
seasoned. It is about the lightest and for its weight the strongest
wood used. It is ordinarily used as a material for spars, struts,
landing gear, etc., as it has a proper combination of flexibility,
lightness and strength.
2. White Pine.—A very light wood used for wing ribs, and small
struts.
3. Ash.—Springy, strong in tension, hard and tough, but is
considerably heavier than spruce. Used for longerons, rudder post,
etc.
4. Maple.—Used for small wood details, as for blocks connecting
rib pieces across a spar or for spacers in a built-up rib.
5. Hard Pine.—Tough and uniform and recommended for long
pieces, such as the wooden braces in the wings.
6. Walnut, Mahogany, Quarter-sawed Oak.—The strength,
uniformity, hardness and finishing qualities make these woods
favorites for propeller construction.
7. Cedar Wood.—Is used occasionally for fuselage coverings or
for hull planking in hydroplanes as it is light, uniform and easily
worked. Veneers, or cross-glued thin layers of wood, are sometimes
used for coverings.
Laminated or built-up wooden members have been much used for
framing and for ribs and spars. The engine bearers are always of
wood on account of vibration and are also laminated. In lamination
the wooden strut is built up of several pieces of wood carefully glued
together. The grains of the different layers run in different directions,
consequently a stronger and more uniform stick often is secured.
The objection to laminated pieces comes from the weather causing
ungluing. Laminated pieces should be wrapped in linen or paper and
freshened with paint or varnish from time to time.
Forms.—Attention should be called to the hollowed form of many
of the wooden members. In any beam or strut, material at the center
of the cross-section is of far less value in taking the load than the
material away from the center. Therefore, to secure greatest strength
with least weight, it is permissible to lighten wooden members if
done understandingly.
Steel.—There is a tremendous difference in the strength, wearing
and other desirable qualities among different steels and irons. For
airplane work none but the best qualities are allowed. For this reason
the use of ordinary iron bolts (as stove bolts) or metal fastenings or
wire not standardized and of known qualities should not be
permitted. The airplane is no stronger than its weakest fitting. This
does not mean that the hardest and strongest steel must necessarily
be used, as ease of working and freedom from brittleness may be
just as important qualities, but the steel on all metal fittings should be
of high-grade uniform stock. A ductile, not too easily bent, mild
carbon steel is usually recommended for all steel plate, clips,
sockets and other metal parts. If any parts are required to be
tempered or hardened it must be remembered that they become
brittle and can not afterward be bent without annealing or softening.
Tool or drill steel is a name given to uniform or rather reliable grades
of steel adapted to heat treatment as tempering or annealing. Often
the bolts, clips, nuts, pins, devices and other fittings are of special
heat-treated nickel steel which must not be heated locally for
bending or for attachment. Such work seriously weakens the steel.
The steel is often copper-or nickel-plated and enamelled to prevent
rusting. Do not forget that the proper material may be twice as strong
as other material which looks the same but which has not received
special treatment.
Wires.—Only the highest grade of steel wire, strand and cord is
allowable. Manufacturers, as Roebling of Trenton, N. J., manufacture
special aviator wire and cord, which is given the highest possible
combination of strength and toughness, combined with ability to
withstand bending, etc. Steel wire ropes for airplane work are divided
into three classes as follows:
1. The solid wire = 1 wire (as piano-wire grade) and known as
aviation wire.
2. The strand stay, consisting either of 7 or 19 wires stranded
together and known as “aviator strand.” Flying and landing wires on
Curtiss.
3. Cord or Rope Stay.—Seven strands twisted together forming a
rope, each strand being of 7 or 19 wires and known to trade as
aviator cord. The wires are either tinned or galvanized as protection
against rust, etc. Ordinarily galvanizing is used, but hard wires and
very small wires are injured by the heat of galvanizing and they are
therefore tinned.
No. 1. The single wire is the strongest for its weight. Single wires
will not coil easily without kinking and are easily injured by a blow,
therefore their use is confined to the protected parts of the machine
such as brace wires in the fuselage and in the wings.
The strand stay (No. 2) of 7 or 19 wires is generally used for
tension wires, as it is more elastic (can be bent around smaller
curve) without injury, as the flying and landing wires on the Curtiss.
The smaller strands usually have 7 wires, the larger ones 19 wires.
No. 3. The Tinned Aviator Cord.—The 7 by 19 cord is used for
stays on foreign machines. It is 1¾ times as elastic as a solid wire of
the same material. On the Curtiss it is used for control wires. For
steering gear and controls extra flexible aviator cord is also
recommended. This has a cotton center which gives extra flexibility
and is used for steering gear and controls. It is 2¼ times as elastic
as a single wire.
Although wire strands or cords are not quite as strong for the
same size as a single wire they are preferred for general work, being
easier to handle and because a single weak spot in one wire does
not seriously injure the whole strand.
Especial care is necessary to avoid using common steel wires, or
strands which have a frayed or broken wire, or wire that has been
kinked and then straightened or wire that has been locally heated or
wire that has been bruised. All these factors weaken steel rope much
more than is supposed ordinarily.
Wire Fastening or Terminal Connections.—Wire terminals are
of four classes:
1. Ferrule and dip in solder, then bend back the end. With or
without thimble; used on single wires or on strand; 50 to 94 per cent.
as strong as the wire.
2. Thimble and End Splicing.—The splice must be long and
complete. Used on cable; 80 to 85 per cent. as strong as the strand;
breaks at last tuck in the splice.
3. Socket.—Nearly 100 per cent. strong.
4. End Wrap and Solder.—Simple and serviceable; not used for
hard wire.
Present practice is rather toward elimination of acid and solder,
imperfect bends, flattening of cable on bends, and toward care in
avoiding all injury as kinking to wire, strand and cord due to unskillful
handling of material in the field.
Other Metals.—Other metals as aluminum, brass, bronze, copper,
monel metal (copper and nickel) are used for certain airplane fittings
for the reasons of lightness, non-corrosive qualities, or ease of
bending, etc. The trouble with these metals is that they are not
uniform and reliable in strength and in an important part the great
strength combined with minimum weight given by steel is not
equalled by any of these metals. Aluminum is used on the engine
hood and also for control levers and for the backs of the seats. In
other words, for parts and castings which require light metal
construction, but which are under no particular stress. Tin and
copper are used for ferrules of wire joints and for tankage. Copper or
brass wire are used for safety wires. Special Tobin bronze is used for
turnbuckles as the part must not only be strong but free from any
tendency to rust. Monel metal (nickel 60 per cent., copper 35 per
cent., iron 5 per cent.) is strong and has the special property of being
acid- and rust-resisting. It has been used for metal fittings and even
for wires and for the water jacket of the motor. Until more strength
tests show greater uniformity of strength, it is to be recommended
with caution.
In dealing with metals like steel, it should be remembered that they
are subject to crystallization and fatigue.
Repeated jarring may cause a bar of steel to break easily at a
particular point, when the metal is said to have crystallized there.
Fatigue of a metal may be defined as loss of springiness which
may come from repeated bending and which lessens the strength of
metal. Above all, however, corrosion of steel must be guarded
against.
The above points should be clear, as in airplane work you are
dealing with a structure which is safe with perfect materials and
workmanship. The factor of safety, however, is not great enough to
permit carelessness, or defective material.
Linen.—The almost universal wing covering is fine, unbleached
Irish linen, stretched rather loosely on the wing frames and then
treated with dope.
The linen used weighs 3¾ to 4¾ oz. per square yard, and should
have a strength with the length of the cloth or “warp” of at least 60 lb.
per inch of width. The strength in this direction is slightly greater than
that taken crosswise of the cloth or on the filler or weft. There is a
gain of strength and tautness by varnishing or “doping.”
In general, it is desirable to have wing material which will not sag
easily and have the fabric yield rather than break. This often reduces
stress and saves complete failure.
Dope.—The linen must be coated with a more or less waterproof
dope. Some form of cellulose acetate or nitrate with more or less
softening material is used and to these some suitable solvent as
acetone is added.
The cellulose acetate or nitrate in the dope acts as a waterproof
sizing, shrinks the cloth tight, and prevents it from changing in
tightness due to moisture. Spar varnish protects this layer from
peeling and makes the wing more waterproof. In service, varnish or
dope must be applied every few weeks.
The U. S. Army practice calls for four coats of cellulose nitrate
dope followed by two coats of spar varnish to prevent inflammability.
Cellulose nitrate is more elastic and durable than the acetate but is
also more inflammable.
Commercial dopes with various desirable properties are: Cellon,
Novavia, Emaillite, Cavaro, Titanine, etc.
FOOTNOTES:
[1] But has very high elastic limit.

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An Introduction to Mechanical Engineering (4th Ed.) 4th

Why are competent engineers so vital?

John Bird BSc(Hons), CEng, CMath, CSci, FIMA, FIET, FCollT

Routledge is an imprint of the Taylor & Francis Group, an informa business

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First edition published by Elsevier 2002

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Visit the companion website: www.routledge.com/cw/bird

Revision Test 3 Forces, tensile testing

12 Twisting of shafts 162 18 Work, energy and power 227

19.4 Friction on an inclined plane 245 24 Thermal expansion 296

28 The measurement of temperature 349 A list of formulae for mechanical

Mechanical Engineering Principles, Fourth Edition Part 1 Revision of Mathematics

From the website 6. Use any search engine (such as google)

At the end of this chapter you should be able to:

1.1 Introduction knowledge of mathematics. This chapter highlights

Mechanical Engineering Principles, Bird and Ross, ISBN 9780367253264

Problem 1. Convert the following angles to

Problem 2. Convert the following angles to

[(a) 34.377° (b) 45.837°

[(a) 0.5, 0.8660, 1.7321

Problem 4. In Figure 1.4, if ab = 2 and ac = 3,

Figure 1.3 It is convenient to use the expression for cos θ, since

ab 2 Now try the following Practice Exercise

Practice Exercise 3 Sines, cosines and

(b) Pythagoras’ theorem

Problem 7. In Figure 1.5, if ab = 5.1 m and

From Pythagoras, ac2 = ab2 + bc2

a b c Since A + B + C = 180°, then

and the cosine rule states: a 2 = b 2 + c 2 − 2bc cos A

From the cosine rule,

Hence, length, a = 64.643 = 8.04 cm

1.5 Brackets Expand the brackets in Problems 3 to 7.

3. 2(x – 2y + 3) [2x – 4y + 6]

Practice Exercise 5 Brackets 2 3

The two fractions can now be made equivalent, i.e.

1 3 of commercial life, as well as in engineering. Interest

Practice Exercise 6 Fractions 0.275 = 0.275 × 100% = 27.5%

27 Practice Exercise 7 Percentages

Problem 29. A box of screws increases in price

(b) 47 s as a percentage of 5 min (c) 13.4 cm 3

3 × 32 Problem 39. Simplify a 2b 3c × ab 2c 5

6. Evaluate [2 3 = 8] 27.

The solution of simultaneous equations is demonstrat-

i.e. 0.75P – 0.3333P = 36.25 – 30 and 6s + 12b = 18 (2)

Revision Test 3 Forces, tensile testing

12 Twisting of shafts 162 18 Work, energy and power 227

19.4 Friction on an inclined plane 245 24 Thermal expansion 296

28 The measurement of temperature 349 A list of formulae for mechanical

From the website 6. Use any search engine (such as google)

[(a) 0.5, 0.8660, 1.7321

Practice Exercise 3 Sines, cosines and

3. 2(x – 2y + 3) [2x – 4y + 6]

6. Evaluate [2 3 = 8] 27.

i.e. 0.75P – 0.3333P = 36.25 – 30 and 6s + 12b = 18 (2)

6.25 6s + 60b = 30 (3)

6. Solve the simultaneous equations:

Practice Exercise 10 Multiple-choice Figure 1.9

Figure 1.8 10. Removing the brackets from the expres-

7. In the right-angled triangle PQR shown in 1 11

6. Figure RT1.5 shows a roof truss PQR with rafter

13. Use a calculator to evaluate: 2

47 11. 0.000017 A [17 µA]

(b) 5.2 nautical miles = 5.2 × 1.852 km