ENGINEERING PHYSICS

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EMAT10005 Engineering Physics 1D EMAT10005
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ENGINEERING PHYSICS
1 INTRODUCTION
1.1 What is Engineering Physics?
Physics is the science of observing the world in which we live and developing models to explain our
observations from a small set of basic principles.
Engineering is the application of these scientific principles for useful purposes.
The Engineering Physics unit provides the foundation of the basic physics needed for mechanical,
civil and materials engineering units in later parts of the degree programme.
1.2 Structure of Unit
Semester I covers
 mechanics of materials (behaviour of bodies subjected to loading)
 statics (study of forces acting on bodies in equilibrium)
Semester II covers
 kinematics (study of motion of a body, ignoring the forces involved)
 dynamics (study of motion of a body, considering the forces involved)
1.3 Problems
A good understanding of engineering physics is essential before embarking on specialist areas of
engineering of practical importance (such as aerodynamics, fluid mechanics and stress analysis). The
real test of understanding is the ability to apply the principles covered in the unit to practical problems.
Many worked examples will be covered in the lectures. It is essential that you tackle all the problem
sheets set over the academic year. Worked solutions will be made available on the www so that you
can check your solutions and ensure your understanding of the material. I caution you VERY
STRONGLY against the common belief that merely reading through the solutions and
convincing yourself that you by and large understand them is an adequate substitute for the
mental exercise of struggling to find solutions to the exercises unaided. It might be compared to
the belief that the best way of preparing to run a marathon is to watch videos of previous
marathon runners doing their thing.
The following textbooks contain most of the material covered in the unit and may be useful for
background reading. Many of the diagrams in these notes have been reproduced or adapted from these
sources.
Mechanics of Materials (5th Edn) by Gere, J M, Brooks/Cole, 2001.
Engineering Mechanics II: Dynamics (4th edn) by Merriam, J L, & Kraige, L G, John Wiley 1998
The following book is an incredibly useful reference work for the values of all sorts of physical
properties of common (and not so common) materials.
Tables of Physical and Chemical Constants, Kaye, G W C, Laby, T H, et al, 16
th
Edn), Longman,
1995
[Other books may also be recommended through the course for background reading or extra
examples.]
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SEMESTER 1 MECHANICS OF MATERIALS

The main objective of mechanics of materials is to calculate the stresses, strains and displacements of
structures due to loading. This is essential for the design of safe, economic structures, e.g. aircraft,
bridges, buildings, dams, pressure vessels, rotating machinery, robotic devices, and so on.
1 Concepts & Definitions
It is useful to begin by defining the following quantities.
1.1 Prismatic bar
A prismatic bar is a straight member of uniform cross-sectional area, A

Figure 1.1 - Cross-sections of typical prismatic bars
1.2 Axial force
An axial force, P, is a load applied in direction parallel to axis of member (unit = N)
Load P
L

Cross sectional area, A

Figure 1.2 - Extension of bar under load. Stress
A
P
, strain
L


1.3 Normal stress
The normal stress, , is the force per unit area perpendicular to a surface,
A
P
 (unit = N/m
2
= Pa)
1.4 Tensile stress
The Tensile stress is the force due to stretching (positive)
1.5 Compressive stress
The compressive stress is the force due to compression (negative)
1.6 Stress concentration
Localised stresses can be significantly greater than the average stress in the structure. This effect is
known as stress concentration and is an important design consideration. Remember the Comet!
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Stress in a uniform bar or plate Stress concentration around a hole

Figure 1.3 – Stress concentration
1.7 Bearing stress
The bearing stress, 
b
is the average stress between a bolt and a plate.
td
P
b
, where t = plate
thickness and d = bolt diameter

Figure 1.4 - Stress on bolt in single shear
1.8 Normal strain
The normal strain, , is the fractional change in length in the axial direction,
L

 where
 = displacement, L = natural length. ( > 0 for extension,  < 0 for shortening).
Example 1 A force of 100N acts over an area of 10mm
2
. Calculate the stress.
Example 2 A bar of length 1m is stretched a distance of 1mm. Calculate the strain.
Example 3 A circular pipe of length 1 m, internal diameter 40mm and external diameter 50mm, is
subjected to an axial load of 100kN and shortens by 1mm. Calculate the stress and
strain in the pipe. [Note: the weight of the pipe may be ignored.]
Example 4 A solid circular rod of diameter 10mm, length 1m and density 8  10
3
kg m
-3
, stands on
flat surface. Calculate the maximum stress in the rod.
2 Mechanical Properties of Materials
When a body is stretched, or compressed, internal stresses and strains are set up inside the material.
The relationship between stress and the strain depends on the magnitude of the stress and on the
material. Figure 2.1 shows a typical stress-strain relationship for a steel bar (not to scale).
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Figure 2.1 - Typical stress-strain relationship for a steel bar
The stress-strain relationship shows a number of distinct regions of behaviour as the stress increases.
OA: linear elasticity
Up to the ‘proportional limit’, 
A
, stress and strain are linearly related (Hooke’s law)
 = E , E = slope of line = ‘Young’s modulus of elasticity’
AB non-linear elasticity
Increasing stress produces greater strain than expected by the linear relationship (the bar
becomes a 'softer' spring).
BC: perfect plasticity (‘yielding’)
Beyond the ‘yield stress’, 
B
, the material yields with no increase in applied load. It behaves
like a spring of zero stiffness.
[Note: total elongation  10–15 greater than that in the linear elastic region.]
CD: strain hardening
From C to D, the material experiences changes in its crystalline structure (the stiffness of the
spring now increases again). The largest stress the material can withstand occurs at D:

D
= ultimate stress
DE: necking
The bar starts to constrict at some point along its length and the cross-sectional area decreases
significantly. At point E the bar breaks. [Note: the dotted line CE’ is called the ‘true stress-
strain curve’ because it is based on the actual cross-sectional area where the necking is
occurring. The solid line CE is based on the original cross-sectional area of the bar.]

Figure 2.2 - Necking of bar
Elastic behaviour is said to occur if the material returns to its original state after the load is removed.
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

unloading
loading

Figure 2.3 - Linear elastic behaviour


unloading
loading

Figure 2.4 - Non-linear elastic behaviour
Hysteresis is when a material does not return to its original state after the load is removed. When this
happens, a residual strain is left inside the material.
Creep is an effect that arises over long time-scales due to long-term loading of the material (e.g. piano
strings going out of tune).

time, t
elongation,



Figure 2.5 - Creep of bar under tension over long
timescale

time, t


Figure 2.6 - Stress relaxation of wire over long
timescale
3 Line of action of axial force in a prismatic bar
In many situations of practical interest it is not necessary to consider the detailed stress distribution
inside a body. A major simplification arises from the fact that the net effect of the stresses can be
represented by a single force P acting in the axial direction through the centre of mass of the cross-
section, given by
P = A
4 Poisson’s Ratio
When a prismatic bar is extended axially, it contracts in the lateral direction. Provided the material is
homogeneous, with uniform elastic properties in all directions, and the axial force is uniform
throughout the material, then
lateral strain  -(longitudinal strain) (Why is there a negative sign?)
We define Poisson’s ratio,  (Greek nu), as




strainallongitutin
strainlateral
so that






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Poisson’s Ratio for Common Materials
Cork 0
Metals 0.25 – 0.35
Concrete 0.1 – 0.2
Rubber 0.5
5 Dilatation
When a body is stretched it changes its volume. The fractional change in volume is called the
dilatation, e.

Figure 5.1 - Change in shape of rectangular bar under tension
Consider a prismatic bar of rectangular cross-section (Figure 5.1) stretched in the x-direction from its
original length, a, to length a+a. As a result of being stretched in the x-direction, the bar contracts in
the y-direction, from b to b - b, and in the z-direction, from c to c - c. The dilatation, e, is given
by






  
1111
abc
1c1b1a
1
V
V
V
VV
volumeoriginal
volumeinchange
e
2
0
1
0
01










If  << 1 (which is generally true for elastic deformations), we may approximate the expression on the
right hand side, using the binomial theorem, yielding








 211211211e
Finally, putting
E

  (from Hooke’s law), we obtain the following approximate expression for the
dilatation of the bar
 


 21
E
e
Question: What are the signs of  and e for a bar under axial compression?
Example 5 A steel pipe of length 1 m, of internal diameter 10 mm and outer diameter 15 mm, is
compressed by an axial load of 600 kN. Given the Young's modulus is 200 GPa and
Poisson’s ratio is 0.3, calculate (i) the cross-sectional area of the pipe, (ii) the axial
stress, (iii) the axial strain, (iv) the change in length of the pipe, (v) the lateral strain,
(vi) the increase in wall thickness, (vii) the change in volume, (viii) the dilatation.
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Figure 5.2 - Pipe under axial compression
6 Shear Stress and Strain
6.1 Shear stress and shear strain
Shear stress

A
P
gentialtan
, shear strain



h
,  = G, G is called the shear modulus.
6.1 Shear force on a bolt
Consider a bolt connection (a ‘bearing’) between two plates (Figure 1.4). Suppose the plates are of the
same thickness, t, the diameter of the bolt is d, and the forces applied to each plate are equal and
opposite. The cross-sectional area of the bearing, ie normal to the stress) is
A
b
= td
and the average bearing stress is
td
P
b
 .
The bearing stress acts normally to the axis of the bolt, and in the extreme, would cause the bolt to
shear laterally.
Example 6 A bearing pad (e.g. to support a bridge girder) is made of a linearly elastic material of
thickness, h, and lateral dimensions a, b. A steel plate on top of the pad exerts a
horizontal shear force, V. Derive an expression for the average shear stress in the
material and the horizontal displacement, d, of the plate.

Figure 6.1


Figure 6.2
Example 7 Hole Punch: Determine the shear stress in the plate and the average compressive stress
in the puncher shown in figure 6.3.
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Figure 6.3 - Hole punch (d = 10 mm, t = 5 mm, P = 100,000 N)
7 Design of Safe Structures
7.1 Allowable Stresses
In practice, most structures are designed with a considerable margin of safety. We define the ‘factor
of safety’, n, as
strengthrequired
strengthactual
n 
where the ‘actual strength’ is defined as the maximum loading the structure can withstand without
failing. Another useful measure is
allowable stress
safetyoffactor
strengthyield

from which we can derive the ‘allowable load’ as
allowable load = allowable stress


7.2 Analysis of Safe Structures
There are 2 basic approaches to the analysis of safe structures. The first approach applies when the
shape of the structure and the material properties are already specified. The problem then reduces to
determining the response of the structure to given loads. The second approach is more interesting
from a design point-of-view. It is to design a structure capable of supporting a given loading, i.e. to
choose the material, shape and size of the members, etc. For example, the cross-sectional area of a
member could be obtained from the formula
area
stress allowable
load

Example 8 Calculate the cross-sectional areas of bars AB and CD and the diameter of the pin
support at C, given the allowable stresses in AB and CD as 125 MPa (tensile stress) and
45 MPa (shear stress). The weights of the bars can be ignored. Pin C is in double

shear
.
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Figure 7.1
Example 9 The clevis pins in Figure 7.2 are 18 mm diameter and 2m apart. Find the average shear
stress in the pins, 
ave
, and the bearing stress between the steel plate and the clevis pins,

b
(weight density of steel = 77kN/m
3
). [Note: A clevis pin is in double shear.]

Figure 7.2
8 Change in Length of Axially-Loaded Members
8.1 Prismatic Bar

Figure 8.1
For a linearly elastic material, we have Hooke’s Law
P = k
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where k = stiffness constant (or spring constant),



 = E
where
,stressaxial
A
P
 E = Young’s modulus (or bulk modulus) and
.strain
L


 Hence,
we can write
L
E
A
P


or
E
L
EA
PL

 .
8.2 Stepped Bar
In many cases the cross-sectional area of a bar may change along its length. Consider the stepped bar
shown below. A load is applied at one end and the weight of the bar is ignored. The tension at any
point in the bar must be equal and opposite to the applied load. [What would happen if this were not
true?]
A
B
P

A

C

P

B

L

1


L

2


E

1

E

2


Figure 8.2
Total elongation,
2
2
1
1
21
EA
PL
EA
PL

In general, for a bar with n steps, this generalises to the expression




n
1i
i
i
n
1i
i
i
A
L
E
P
EA
PL
.