Mechanical Science for Technicians. Volume 2

228 Pages·1988·6.858 MB·English

by  Ian McDonagh (Auth.)

Mechanical science
for technicians
volume 2
Ian McDonagh
Senior lecturer in engineering,
Wirral Metropolitan College
Edward Arnold
A member of the Hodder Headline Group
LONDON SYDNEY AUCKLAND
Edward Arnold is a division of Hodder Headline PLC
338 Euston Road, London NW1 3BH
© 1988 I McDonagh
First published in the United Kingdom 1984
9th impression 1995
British Library Cataloguing in Publication Data
McDonagh, Ian
Mechanical science for technicians.
Vol.2
1. Mechanics
I. Title
620.1 TA350
ISBN 0 7131 3445 3
All rights reserved. No part of this publication may be reproduced or
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Typeset in 10/11 IBM Press Roman by
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Printed and bound in the United Kingdom by
Athenaeum Press Ltd, Gateshead, Tyne & Wear
Preface
This book has been written to meet the requirements of the Business &
Technician Education Council (BTEC) standard unit Mechanical science IV
(U82/041). Its aims are (a) to introduce the student to the principles of
stress-and-strain analysis, (b) to extend the ability of the student to tackle
problems involving deflections of beams, and (c) to extend the student's
understanding of dynamics and vibrations.
SI units have been used throughout in the text, with the following prefer-
red multiples and submultiples:
Prefix Symbol Multiplication factor
giga G 109 = 1000000000
mega M 106 = 1000000
kilo k 103 =1000
milli m 10-3 = 0.001
micro μ 10~6 = 0.000001
Ian McDonagh
1 Principles of stress-and-strain
analysis
1.1 Classification of materials
Materials may be classified by their elastic properties, as follows:
a) Isotropie materials These are materials in which the elastic properties
are the same in all directions. Most metals are isotropic materials.
b) Orthotropic materials These are materials in which the elastic properties
vary with direction. For example, timber has two values for modulus of
elasticity, E - one value is for loads applied across the grain, while the
second values is for loads applied with the grain.
c) Non-isotropic materials These are materials which exhibit non-uniform
elastic properties throughout. Typical non-isotropic materials include
ceramics and glass. In non-isotropic materials it is not possible to predict
the effect of loading on the deformation of the material.
In considering the principles of stress-and-strain analysis, it will be assumed
that the materials are isotropic,
1.2 Relationship between stress and strain
As an elastic material is stretched, it will also contract in a direction perpen-
dicular to the applied force, as shown in fig. 1.1, inducing a compressive
strain in the material at right angles to the line of action of the tensile force.
This strain, known as the lateral strain, is directly proportional to the direct
or longitudinal strain produced by the force.
I Lateralstraine
F
Longitudinal
Fig. 1.1
Let ex = longitudinal strain and ey = lateral strain
then
or € = ve
y x
1
The constant v (nu) is known as Poisson's ratio,
e lateral strain
v
i.e. Poisson s ratio v = — =
e longitudinal strain
x
Since it is a ratio of like quantities, Poisson's ratio has no units. For metals,
Poisson's ratio can vary between 0.25 and 0.33.
Let o be the stress producing the longitudinal strain e in fig. 1.1 ; then
x x
where E is the modulus of elasticity or Young's modulus for the material.
The lateral strain e induced by o is
y x
The minus sign indicates that e is compressive.
y
€x = l(oX'vay) €χ=ϊ~(σχ+υσν)
\°v
ey = - (ay -υσχ) ■-Ε(σν + υοχ
(a) (b)
Fig. 1.2
For the stressed element shown in fig. 1.2(a), the total strain in the direc-
tion off oo iiss
xx
e = strain due to o — strain due to o
x x y
°y - 1 / Λ
i.e. e,, = —
-°y .. °χ = 1
also e = -j - v — = - (σ - vo)
y χ y
2
In fig. 1.2(b), σ is tensile and o is compressive:
χ y
£χ = — + v -f = - (σ + w )
χ y
°v °x 1 ,
and €3, = *- - v — = (o + vo)
y x
Example 1 Two mutually perpendicular stresses of magnitude 50 N/mm2
(tensile) and 35 N/mm2 (compressive) act at a point in an element. If
E = 200 kN/mm2 and v = 0.3, calculate the strains in the direction of the
stresses.
where o = 50 N/mm2 Oy = —35 N/mm2 (i.e. compressive)
x
1; = 0.3 and E = 200 x 103 N/mm2
e* = l- [50 N/mm2 - 0.3 x (-35 N/mm2)]
X 200 x 103 N/mm2
= 3.025 x 10~4 (i.e. tensile)
Also, € = - (o - ι>σ)
y y χ
1 - (-35 N/mm2 - 0.3 x 50 N/mm2)
200 x 103 N/mm2
= -2.5 x 10~4 (i.e. compressive)
i.e. the strains are 3.025 x 10~4 (tensile) and 2.5 x 10~4 (compressive).
Example 2 The maximum and minimum mutually perpendicular strains in
a stressed component were found to be 8.5 x 10"4 and 1.5 x 10"4, both
tensile. If E = 200 GN/m2 and v = 0.25, calculate the maximum and mini-
mum stresses.
(0
and GO
Multiplying equation (i) by E and equation (ii) by Ev and adding the results,
3
(e + ve)E
o x y
x
l-*2
_ (ß+ve)E
Also y x
1 -jr
where e =8.5x 10~4 e = 1.5 x 10"4 E = 200 x 109 N/mm2
x v
and i; = 0.25
(8.5+0.25 x 1.5) x ΚΓ" χ 200 x 10' N/m2
°x =
1 -0.252
= 189.3 x 106 N/m2 or 189.3 N/mm2
(1.5 + 0.25 x 8.5) x 10~4 x 200 x 109
=
and
1 - 0.252
= 77.3 x 106 N/m2 or 77.3 N/mm2
i.e. the maximum and minimum stresses are respectively 189.3 N/mm2 and
77.3 N/mm2, both tensile.
1.3 Volumetric strain
Volumetric strain is defined as change in volume (δ V) per unit original
volume (V)
change in volume
i.e. volumetric strain =
original volume
The symbol used for volumetric strain is e,
v
bV
Λ e = —
v
V
1.4 Relationship between linear strain and volumetric strain
The cube of side length / shown in fig. 1.3 is subjected to three mutually
perpendicular stresses, o, o, and o, producing strains e, e, and e
x y z x y z
respectively,
°x σν Oz
i.e. e= — - v - ^ -v —
x
E E E
= - [o - v{p + σ )]
x y ζ
E
4
ko {e = ^ (o - υ (σ +σ))
y y y χ ζ
σ,ίβ,)
σχ<€χ=£ Κ-υ(σ + σ))
κ ζ
azie7 = f Κ-υ(σ +σ ))
χ κ
Cube of side
length /
K<€K>
Fig. 1.3
and e = - [σ - v(p + σ^)]
z ζ x
Before stressing, the volume of the cube is Vs I3. After stressing, the
length of each side will increase by an amount equal to length of side x strain,
i.e. volume after stressing = (/ + le) (/ + le) (/ + le)
x y z
= /3(l+e +e +e )
x y 2
(ignoring small-second order terms in e2, i.e. e2 etc.)
x
Λ change in volume, δ V = /3(1 + e + e + e ) - /3
x y z
and vo1lu met.nc .s.trai n e = —δκ = '30+e*+ev+ez-/3-)
v
V I3
Or € = € + € + €
v x y z
w/z/c/i it is useful to remember.
For the special case where o = o - o = σ,
x y z
1
e = e = e = e = - [σ - (σ + o)v]
x y z
or e = - (1 - 2v)
E
and volumetric strain e = 3e = 3 x linear strain
v
which it is useful to remember.
5
Example A short column of rectangular cross-section 90 mm x 150 mm and
1 m long is subjected to an axial compressive force of 800kN. If the modulus
of elasticity is 150GN/m2 and Poisson's ratio is 0.28, calculate the change in
volume and state whether it is an increase or a decrease.
Let e be the longitudinal compressive strain due to the axial force and
x
e = € be the mutually perpendicular lateral strains, which will both be
y z
tensile. Then
° F
€χ " E ~ ÂË
where F = 800 x 103 N A = 0.09m x 0.15m = 0.0135m2
and £=150xl09N/m2
800 x 103 N
0.0135m2 xl50xl09N/m2
= —3.95 x 10~4 (i.e. compressive)
€ = e = -ve
y z x
where v = 0.28
·*· e = e = -0.28 x (-3.95 x 10"4) = 1.1 x 10"4
y z
Volumetric strain e = e + e + e
v x y z
= (-3.95+ 1.1 +1.1) xlO"4
= -1.75 xlO"4
change in volume, δ V = e x V
v
where V = lmx 0.0135m2 = 0.0135m3
Λ 3V = -1.75 x 10-4 x 0.0135m3
= -2.36xl0-6m3
i.e. the volume is decreased by 2.36 x 10~6 m3.
1.5 Bulk modulus of elasticity
The cube shown in fig. 1.4 is subjected to three mutually perpendicular com-
pressive stresses of equal magnitude σ. In the direction of each stress,
o
linear strain e = (1 - 2*>)
E
(the minus sign indicating that the strain is compressive).
o
Volumetric strain e = 3e = - 3 - (1 — 2v)
v
6
Fig. 1.4
The term 3(1 - 2v)/E is called the compressibility factor (k), and the
reciprocal of this value is known as the compressibility modulus or bulk
modulus of elasticity, K.
1 E
i.e. K =
k 3(1 - 2v)
The bulk modulus of elasticity is also defined as stress per unit volumetric
strain,
stress
i.e. bulk modulus of elasticity =
volumetric strain
Referring to fig. 1.4,
-o
-3σ(1 - 2v)/E
or K 1.1
3(1 -2v)
The units for bulk modulus of elasticity are newtons per square metre (N/m2).
Referring to equation 1.1, as the value of Poisson's ratio v approaches 0.5,
the value of K approaches infinity and, for values of v greater than 0.5, A'
will be a negative value. For any material, the bulk modulus of elasticity is a
positive value, therefore, for any material the value of Poisson 's ratio v is less
than 0.5.
7