Applied Mechanics for Engineers

236 Pages·1966·3.399 MB·English

by  C. B. Smith, N. Hiller and G. E. Walker (Auth.)

Applied Mechanics
for Engineers
VOLUME 1
by
C. B. SMITH
B.Sc.(Eng.), C.Eng., A.M.I.Mech.E., A.M.I.C.E.
Senior Lecturer in Mechanical Engineering,
Norwich City College
PERGAMON PRESS
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Copyright © 1966 Pergamon Press Ltd.
First edition 1966
Library of Congress Catalog Card No. 66-16883
Printed in Great Britain by Bell and Bain Ltd,, Glasgow
This book is sold subject to the condition
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of without the publisher's consent,
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Preface
THE idea of this book is to provide an introduction to mechanics
applied to engineering and an attempt has been made to support
the theory with a large number of worked examples. The theory
in each section has been built up as far as possible from first
principles and it is hoped the reader will work in this way and will
not try to rely on memory for a large number of formulae. In
practice it is found that each problem is a new problem often
requiring a new approach and this is where the student who relies
on his memory for a standard formula breaks down and is unable
to cope with the new situation.
In the first volume the work covered corresponds to the first
year of the Ordinary National Certificate in Engineering. Since
the syllabus of the 01 course varies from one college to another it
will be found that in some cases sections of the work covered are
not required and in other cases there may be something missing,
but it is hoped that the gaps left will be very small.
As far as possible the book has been written to conform with
Letter Symbols, Signs and Abbreviations laid down in British
Standard 1991: 1957.
The calculations in this book have all been made with the
assistance of a slide rule and it is hoped that the reader will have a
slide rule and will make full use of it. Anyone who has not already
obtained a slide rule is recommended to purchase one with a
log-log scale on it for the solution of problems of the form PV n.
Whilst very little use will be found for the log-log scale in the 01
course, it will be found that in later years more and more problems
will occur in the solution of which the log-log scale is of great help.
The Author would like to thank all his colleagues who have
ix
A*
χ PREFACE
offered helpful advice and also those who read through the script
and suggested certain modifications.
The tables on pp. 218-227 are reproduced from Schofield,
Technical Tables for Schools and Colleges (Pergamon Press), and
the steam table on pp. 228 is reproduced by permission from the
Abridged Callendar Steam Tables (London, Edward Arnold
(Publishers) Ltd.). The Author would like to thank Edward
Arnold (Publishers) Ltd., for their assistance.
CHAPTER 1
Forces and Moments
Force
A force may be defined as any action which alters or tends to
alter a body's state of rest or of uniform motion in a straight line.
It may be quite obvious that a body which is at rest will not move
unless acted on by a force, but it must also be realized that a body
which is moving in a straight line at a certain velocity will con-
tinue to move in that direction with a constant velocity if it is not
acted on by some force.
Characteristics of a Force
A force has three characteristics:
(1) Magnitude.
(2) Line of action, which indicates where the force acts.
(3) Direction along the line of action.
Vector
The above characteristics of a force may be represented by a
straight line known as a vector. The magnitude of the force will be
indicated by the length of the line to some scale, the line of action
by the position of the line, and the direction by the direction in
which the line is drawn.
EXAMPLE 1.1. Suppose that the garden roller shown in Fig. 1.1
weighs 1 cwt, i.e. 112 lbf, and requires a force of 50 lbf to pull it
along in the direction shown.
To represent the forces acting on the roller, that is, the force
l
2 APPLIED MECHANICS FOR ENGINEERS
W = 112 lbf due to the weight and the pull F = 50 lbf necessary
to move it along, a vector or force diagram may be drawn as shown
in Fig. 1.2. In this diagram a vertical line of length equivalent to
112 lbf is drawn to represent the weight W and a further line of
length 50 lbf is added in the direction of the pull F. Then due to
the weight W and the force F, the force acting on the roller is the
vector sum of W and F as shown by the dotted line.
α
A
FIG. 1.1. Garden Roller FIG. 1.2. Force Diagram
Bow's Notation
When drawing vector diagrams it is usual to use a system of
lettering known as Bow's notation. To use this system the spaces
between the various forces are lettered and each force is identified
by the letters in the spaces either side of the force. In the case of
the garden roller shown in Fig. 1.1, the spaces are lettered A, B, C,
as shown. The force Wis identified as AB and is represented in the
force diagram by ab, and the force F is identified as BC and
represented by be. The sum of the two forces is represented by the
vector ac. When drawing the vector diagram it does not matter
whether the forces are taken in a clockwise or anticlockwise
direction, but one direction must be decided on and used through-
out the drawing. In the example shown the direction chosen is
FORCES AND MOMENTS 3
anticlockwise, and after drawing the vector ab vertically down-
wards to represent the weight W, the force F is now added by
drawing the vector be in the direction of the force F. The vector
sum of FT and Fis then shown by ac and this equivalent force will
act in direction ac.
Types of Forces
When defining a force it is usual to state whether the force is
tensile or compressive. In the case of the roller shown in
Fig. 1.1. the pull F necessary to move the roller is a tensile force
acting in the direction shown whereas the force W due to the
weight of the roller sets up a state of compression between the
roller and the ground.
Coplanar forces all lie in a single plane. Most of the work
covered in this book will deal with coplanar forces.
Concurrent forces have lines of action all passing through a
single point. It will be shown later on page 7 that three
coplanar forces in equilibrium must be concurrent.
Reaction. To every force there is an equal and opposite
reaction.
EXAMPLE 1.2. Suppose that in a tug-of-war team each of the six
members of the team at one end of the rope exerts a pull of 2 cwtf
on the rope; then the total pull exerted by the team on the rope
will bea6x2 = 12 cwtf which will set up a tensile force of 12 cwtf
in the central part of the rope.
To balance this pull, the other end of the rope must be anchored
either by the pull of the other team or by some other fixing to
produce an equal and opposite force of 12 cwtf.
If the other team is not able to produce a static pull of 12 cwtf,
then there will still be a tensile force of 12 cwtf in the rope and this
will overcome the static resistance of the other team and cause
movement.
4 APPLIED MECHANICS FOR ENGINEERS
Moment
The moment of a force about any point is the product of the
magnitude of the force multiplied by the perpendicular distance
between the line of action of the force and the point considered.
EXAMPLE 1.3. If a horizontal lever 3 ft long is mounted on a
shaft at one end and is subject to a vertical force of 10 lbf at the
FIG. 1.4.
other end as shown in Fig. 1.3, then the 10 lbf force exerts a
turning moment or " torque " of 10 lbf χ 3 ft = 30 lbf/ft on the
shaft.
Care must be taken to make sure that the distance measured, i.e.
the radius at which the force acts, is measured perpendicular to the
line of action of the force. If in the above example the force acts
at 30° to the vertical as shown in Fig. 1.4, then the perpendicular
distance between the line of action of the force and the axis of the
shaft is 3 ft χ cos 30° = 3x0-866 = 2-598 ft and the turning
moment or torque is now 10 lbf χ 2-598 ft = 25-98 lbf/ft.
FORCES AND MOMENTS 5
Couple
When two equal and opposite parallel forces act on a body, then,
since the forces are equal and opposite, the total force acting on
the body is zero but if moments are taken about any point in the
line of action of one of the forces or about any other point, it will
be found that the body is subject to a turning moment equal to one
of the forces multiplied by the perpendicular distance between the
lines of action of the two forces. Such a moment produced by the
action of two equal and opposite forces is called a " couple ".
8 lbf
8 lbf
FIG. 1.5.
EXAMPLE 1.4. In the case of the lever shown in Fig. 1.5, which is
mounted on a shaft at its midpoint and is subject to two forces each
of 8 lbf acting at 2 ft from the shaft, it will be found that since the
forces are equal and opposite, there is no force on the shaft, but
that there is a couple or turning moment of
81bfx2ft+81bfx2ft = 81bfx4ft = 32 lbf/ft
acting on the shaft.
Resultant Force
A single force which will have the same effect as a number of
separate forces is known as a resultant force.
In the case of the tug-of-war team, the resultant force due to
the six 2 cwtf forces exerted by the members of the team is
6x2 cwtf = 12 cwtf.
In the case of the roller shown in Fig. 1.1, the resultant force due
6 APPLIED MECHANICS FOR ENGINEERS
to the pull F lbf and the weight W lbf of the roller is the vector ac
acting in the direction ac and, if the roller is stationary, then the
ground will produce a reaction ca equal and opposite to this
resultant.
Component Force
Any force can be considered as being the sum of a number of
component forces of which the vector sum will have the same effect
as the original force.
When dealing with component forces, it is usual to choose two
perpendicular directions, say ox and oy and to resolve each force
in these directions.
A force acting in a direction at angle θ to the ox direction has
a component F = F cos θ in the direction ox and a component
x
F = F sin θ in the oy direction.
y
If the force F on the roller shown in Fig. 1.1 is at an angle θ
to the horizontal, then it will have a component F cos θ in the
horizontal direction, i.e. parallel to the ground which will be the
effective force pulling the roller along and it will have a vertical
component F sin θ which will tend to lift the roller off the ground
and consequently will reduce the compressive force between the
roller and the ground due to the weight of the roller. Considera-
tion of these two component effects will show why it is necessary
to alter the angle θ for different ground conditions. If the surface
of the ground is hard, there will be little tendency for the roller to
sink into the ground and a relatively small force at a small angle θ
will pull the roller along. If, however, the ground is soft, then the
angle Θ must be made large so that the force Fhas a relatively large
vertical component which will prevent the roller from sinking into
the ground.
In the case of the lever shown in Fig. 1.4, the 10 lbf has a
component 10 cos 30° = 8-66 lbf in direction perpendicular to the
lever and this will produce a moment of 8-66 χ 3 = 25-98 lbf about
the shaft. The component 10 sin 30° = 5 lbf exerts no turning
moment on the shaft.