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Mechanical Engineering Principles
Fourth Edition
‘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
and by Routledge
52 Vanderbilt Avenue, New York, NY 10017
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.
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.
[
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!
� 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
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°
Figure 1.4
Figure 1.5
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°
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
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
10 Mechanical Engineering Principles
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.
[ ]
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
Revisionary mathematics 11
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
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
Revisionary mathematics 13
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
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:
Problem 37. Evaluate
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]
14 Mechanical Engineering Principles
4
23 × 2 × 26
( x 3 ) [x12]
Part One
( )
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
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
Revisionary mathematics 15
Part One
and 0.4167P = 6.25 Multiplying equation (1) by 3 gives:
8a – 3b = 51
3a + 4b = 14 [a = 6, b = – 1]
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
18 Mechanical Engineering Principles
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)
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
Revisionary mathematics 19
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.
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.
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
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
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
8. 225 × 10–4 V [22.5 mV] 350 mm = 350 × 0.03937 inches = 13.78 inches from
Table 2.1
Further revisionary mathematics 25
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
26 Mechanical Engineering Principles
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
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
US or Imperial Metric
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
Further revisionary mathematics 27
Table 2.8 Imperial to Metric mass Now try the following Practice Exercise
US or Imperial Metric
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]
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
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).
Further revisionary mathematics 29
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
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
FOOTNOTES: