ME16A: INTRODUCTION TO
STRENGTH OF MATERIALS
COURSE
INTRODUCTION
Details of Lecturer

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Course Lecturer: Dr. E.I. Ekwue
Room Number: 216 Main Block,
Faculty of Engineering
Email: ekwue@eng.uwi.tt ,
Tel. No. : 662 2002 Extension 3171
Office Hours: 9 a.m. to 12 Noon. (Tue,
Wed and Friday)
COURSE GOALS



This course has two specific goals:
(i) To introduce students to concepts of
stresses and strain; shearing force and
bending; as well as torsion and deflection of
different structural elements.
(ii) To develop theoretical and analytical skills
relevant to the areas mentioned in (i) above.
COURSE OUTLINE
COURSE CONTENTS
1.
General Concepts – Stresses and strain, two and three-dimensional systems.
Generalized Hooke’s Law – stress-strain relationships.
2.
Properties of Materials – Tension, Compression, Hardness and Impact tests.
3.
Statically Determinate Stress Systems. St. Venant’s Principle. Stress Analysis of axially
loaded bars. Strains and deformations in axially loaded bars. Statically Indeterminate stress
systems
4.
Shear Force and Bending Moment in Beams. Mathematical relationships between load
intensity, shearing force and bending moment. Bending stresses in beams. Beams of
two materials.
5.
Analysis of Stresses in Two-Dimensions. Principal Stresses, Mohr’s Circle
6.
Deflection of Beams – Simple cases. Direct integration and moment-area method.
7.
Torsion of Circular Cross-Sections.
Course Objectives
Upon successful completion of this course,
students should be able to:

(i) Understand and solve simple problems
involving stresses and strain in two and three
dimensions.

(ii) Understand the difference between
statically determinate and indeterminate
problems.

(iii) Understand and carry out simple experiments
illustrating properties of materials in tension,
compression as well as hardness and impact tests.
COURSE OBJECTIVES CONTD.

(iv) Analyze stresses in two dimensions and
understand the concepts of principal
stresses and the use of Mohr circles to solve
two-dimensional stress problems.

(v) Draw shear force and bending moment
diagrams of simple beams and understand
the relationships between loading intensity,
shearing force and bending moment.


(vi) Compute the bending stresses in beams
with one or two materials.
OBJECTIVES CONCLUDED

(vii) Calculate the deflection of beams
using the direct integration and momentarea method.


(viii) Apply sound analytical techniques
and logical procedures in the solution of
engineering problems.
Teaching Strategies


The course will be taught via
Lectures.
Lectures will also
involve the solution of tutorial
questions. Tutorial questions are
designed to complement and
enhance both the lectures and the
students appreciation of the
subject.
Course work assignments will be
reviewed with the students.
Lecture Times
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Wednesday: 2.00 to 2.50 p.m.
Thursday: 11.10 a.m. to 12.00 noon
Friday: 1.00 to 1.50 p.m.
Lab Sessions: Two Labs per student on
Mondays (Details to be Announced Later)
Attendance at the Lectures and Labs is
Compulsory.
Time-Table For Labs
MONDAY 1:00 - 4:00 P.M.
Week
Group
1,5,9
2,6,10
3,7,11,
4,8,12
K
-
ME13A
ME16A
(3,7)
ME13A
L
ME13A
-
ME13A
ME16A
(4,8)
M
ME16A ME13A
(5,9)
-
ME13A
N
ME13A ME16A
(6,10)
ME13A
-
More Course Details

BOOK – Hearn, E.J. (1997), Mechanics of
Materials 1, Third Edition, Butterworth,
Heinemann

COURSE WORK
1. One Mid-Semester Test (20%);
2. Practical report (15%) and
3. End of Semester 1 Examination (65%).
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ME16A: CHAPTER ONE
STRESS AND STRAIN
RELATIONS
1.1 DIRECT OR NORMAL
STRESS

When a force is transmitted through a body,
the body tends to change its shape or
deform. The body is said to be strained.

Direct Stress = Applied Force (F)
Cross Sectional Area (A)

Units:
Usually N/m2 (Pa), N/mm2,
GN/m2 or N/cm2
Note: 1 N/mm2 = 1 MN/m2 = 1 MPa

MN/m2,
Direct Stress Contd.

Direct stress may be tensile,  t or
compressive, c and result from forces
acting perpendicular to the plane of the
cross-section
Tension
Compression
1.2 Direct or Normal Strain


When loads are applied to a body,
some deformation will occur resulting to
a change in dimension.
Consider a bar, subjected to axial
tensile loading force, F. If the bar
extension is dl and its original length
(before loading) is L, then tensile strain
is:
Direct or Normal Strain Contd.
F
F
L
dl
Direct Strain ( ) = Change in Length
Original Length
i.e.  = dl/L

Direct or Normal Strain Contd.



As strain is a ratio of lengths, it is
dimensionless.
Similarly, for compression by amount,
dl: Compressive strain = - dl/L
Note: Strain is positive for an increase
in dimension and negative for a
reduction in dimension.
1.3 Shear Stress and Shear Strain



Shear stresses are produced by
equal and opposite parallel forces
not in line.
The forces tend to make one part
of the material slide over the other
part.
Shear stress is tangential to the
area over which it acts.
Shear Stress and Shear Strain
Contd.
C
x
C’
D
D’
F
P
L

A
S
Q
R
B
Shear strain is the distortion produced by shear stress on
an element or rectangular block as above. The shear
strain, 
(gamma) is given as:

= x/L = tan 
Shear Stress and Shear Strain
Concluded
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
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For small  ,
 
Shear strain then becomes the change
in the right angle.
It is dimensionless and is measured in
radians.
1.3 Complementary Shear Stress
1
P
Q
2
a
2
S
1
R
Consider a small element, PQRS of the material in the
last diagram. Let the shear stress created on faces PQ
and RS be  1
Complimentary Shear Stress
Contd.



The element is therefore subjected to a
couple and for equilibrium, a balancing
couple must be brought into action.
This will only arise from the shear stress on
faces QR and PS.
Let the shear stresses on these faces be
. 2
Complimentary Shear Stress
Contd.

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
Let t be the thickness of the material at
right angles to the paper and lengths of
sides of element be a and b as shown.
For equilibrium, clockwise couple =
anticlockwise couple
i.e. Force on PQ (or RS) x a = Force
on QR (or PS) x b
1 x b t x a  2
x at x b
i. e.  1
 2
Complimentary Shear Stress
Concluded


Thus: Whenever a shear stress occurs on
a plane within a material, it is automatically
accompanied by an equal shear stress on
the perpendicular plane.
The direction of the complementary shear
stress is such that their couple opposes that
of the original shear stresses.
1.4 Volumetric Strain
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Hydrostatic stress refers to tensile or
compressive stress in all dimensions
within or external to a body.
Hydrostatic stress results in change in
volume of the material.
Consider a cube with sides x, y, z. Let
dx, dy, and dz represent increase in
length in all directions.
i.e. new volume = (x + dx) (y + dy) (z +
dz)
Volumetric Strain Contd.
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Neglecting products of small quantities:
New volume = x y z + z y dx + x z dy + x y dz
Original volume = x y z
V = z y dx + x z dy + x y dz
Volumetric strain,  v = z y dx + x z dy + x y dz
xyz
 v = dx/x + dy/y + dz/z
v

x  y  z
Strains Contd.
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Note: By similar reasoning, on area x y
a
 x
 y
Also: (i) The strain on the diameter of a
circle is equal to the strain on the
circumference.
(ii) The strain on the area of a circle, is
equal to twice the strain on its diameter.
(iii) Strain on volume of a sphere, is equal
to three times the strain on its diameter.
Strains Contd.
(iv ) Given  D
and
length of
and
L
as strains on the diameter
a cylinder ,
Strain on the volume is
v
 2D
These
 L
can be proved using the theorem
of small errors
Examples
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(i) Diameter, D = 2 x radius, r i.e. D = 2 r
Taking logs: log D = log 2 + log r
Taking differentials: dD/D = dr/r
Also: Circumference, C = 2 r
i.e. log C = Log2
+ log r
dC/C = dr/r = dD/D
c
i.e. the strain on the circumference,
D
= strain on the diameter,
Strains Contd.
(iv)

Volume of a cylinder, V =
r2 L where L is the length
Taking logs:

log V = log
Taking differentials:
dV/V = 2 dr/r + dL/L
i.e.
v  2 D   L
+ 2 log r + log L
v  2 D   L

Required:
Prove
statements.
the
other
two
1.5 Elasticity and Hooke’s Law
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All solid materials deform when they are
stressed, and as stress is increased,
deformation also increases.
If a material returns to its original size and
shape on removal of load causing
deformation, it is said to be elastic.
If the stress is steadily increased, a point is
reached when, after the removal of load, not
all the induced strain is removed.
This is called the elastic limit.
Hooke’s Law

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
States that providing
the limit of
proportionality of a material is not exceeded,
the stress is directly proportional to the strain
produced.
If a graph of stress and strain is plotted as
load is gradually applied, the first portion of
the graph will be a straight line.
The slope of this line is the constant of
proportionality called modulus of Elasticity, E
or Young’s Modulus.
It is a measure of the stiffness of a material.
Hooke’s Law
Modulus of Elasticity, E =
Direct stress 

Direct strain 
Also: For Shear stress: Modulus of rigidity or shear modulus, G =
Shear stress 

Shear strain 
Also: Volumetric strain,  v
is proportional to hydrostatic
stress, 
within the elastic range
i.e. :
 / v
 K
called bulk modulus.
Stress-Strain Relations of Mild
Steel
Equation For Extension
From the above equations:

F/A F L
E 



dl / L A dl
dl 
F L
AE
This equation for extension is
very important
Extension For Bar of Varying Cross
Section
For a bar of varying cross section:
P
A1
A2
L1
A3
L2
dl 
L
M
N
F L1 L2 L3


E A1 A2 A3
L3
O
P
Q
P
Factor of Safety
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
The load which any member of a machine
carries is called working load, and stress
produced by this load is the working stress.
Obviously, the working stress must be less
than the yield stress, tensile strength or the
ultimate stress.
This working stress is also called the
permissible stress or the allowable stress or
the design stress.
Factor of Safety Contd.

Some reasons for factor of safety
include the inexactness or inaccuracies
in the estimation of stresses and the
non-uniformity of some materials.
Factor of safety =
Ultimate or yield stress
Design or working stress
Note:
Ultimate stress is used for materials e.g.
concrete which do not have a well-defined yield point,
or brittle materials which behave in a linear manner
up to failure. Yield stress is used for other materials
e.g. steel with well defined yield stress.
1.7 Practical Class Details

Each Student will have two practical
classes: one on :



Stress/strain characteristics and
Hardness and impact tests.
(i) The stress/strain characteristics
practical will involve the measurement
of the characteristics for four metals,
copper, aluminium, steel and brass
using a tensometer.
Practical Class Details Contd.




The test will be done up to fracture of the
metals.
This test will also involve the accurate
measurement of the modulus of elasticity for
one metal.
There is the incorporation of an
extensometer for accurate measurement of
very small extensions to produce an
accurate stress-strain graphs.
The test will be done up to elastic limit.
Practical Class Details Contd.

(ii)
The hardness test will be
done using the same four metals
and the Rockwell Hardness test.

The impact test with the four
metals will be carried out using the
Izod test.
1.8 MATERIALS TESTING

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

1.8.1. Tensile Test: This is the most
common test carried out on a material.
It is performed on a machine capable of
applying a true axial load to the test
specimen. The machine must have:
(i)
A means of measuring the applied
load and
(ii) An extensometer is attached to the test
specimen to determine its extension.
Tensile Test Contd.



Notes: 1. For iron or steel, the limit of
proportionality and the elastic limit are
virtually same but for other materials like
non-ferrous materials, they are different.
2. Up to maximum or ultimate stress, there
is no visible reduction in diameter of
specimen but after this stress, a local
reduction in diameter called necking occurs
and this is more well defined as the load falls
off up to fracture point.
Original area of specimen is used for
analysis.
Results From a Tensile Test
E
Stress up to lim it of proportionality
Strain
(a)
Modulus of Elasticity,
(b)
Yield Stress or Proof Stress (See below)
(c)
Percentage elongation =
Increase in gauge length
x 100
Original gauge length
(d)
Percentage reduction in area =
Original area  area at fracture
x 100
Original area
(e)
Tensile Strength =
Maximum load
Original cross sec tional area
The percentage of elongation and percentage reduction in area give an indication of th e
ductility of the material i.e. its ability to withstand strain without fracture occurring.
Proof Stress



High carbon steels, cast iron and most of the
non-ferrous alloys do not exhibit a well
defined yield as is the case with mild steel.
For these materials, a limiting stress called
proof stress is specified, corresponding to a
non-proportional extension.
The non-proportional extension is a specified
percentage of the original length e.g. 0.05,
0.10, 0.20 or 0.50%.
Determination of Proof Stress
Stress
Proof Stress
P
A
Strain
The proof stress is obtained by drawing AP parallel to the initial
slope of the stress/strain graph, the distance, OA being the strain
corresponding to the required non-proportional extension e.g. for
0.05% proof stress, the strain is 0.0005.
1.8.2




Hardness Test
The hardness of a material is determined by
its ability to withstand indentation. There are
four major hardness tests.
(i) Rockwell Hardness Test: This uses an
indentor with a 120o conical diamond with a
rounded apex for hard materials, or steel ball
for softer materials.
A minor load, F is applied to cause a small
indentation as indicated in Fig. (a) below.
The major load, Fm is then applied and
removed after a specified time to leave load
F still acting. The two stages are shown as
(b) and (c).
Rockwell Hardness Test
Hardness Test Contd.



Thus the permanent increase in the
depth of penetration caused by the
major load is d mm. The Rockwell
hardness number, HR is:
HR = K - 500 d
Where: K is a constant with value of
100 for the diamond indentor and 130
for the steel indentor.
1.8.3




Impact Testing
The toughness of a material is defined
as its ability to withstand a shock
loading without fracture. Two principal
impact tests are the:
Izod and the
Charpy tests.
A test specimen is rigidly supported
and is impacted by a striker attached to
a pendulum.
Impact Test Concluded

The difference in height from which a
pendulum is released and the height to
which it rises after impact gives a
measure of the energy absorbed by the
specimen and this is recorded on a dial
mounted on a tester.
Example on Elongation

A flat plate of steel, 1 cm thick,
and of trapezoidal form tapers
from 5 cm width to 10 cm width
in a length of 40 cm.
Determine the elongation under
an axial force of 50 kN. E = 2 x
107 N/cm2.
Diagram of a Trapezoidal Steel Plate
t
P
dx
B1
P
B2
x
L
Solution
Consider a length, dx at a distance, x from width, B 1,
Width at that section
B2  B1
x  B1  Kx
L
B2  B1
where K 
L
 B1 
Area (Ax) of chosen c/section = ( B1 + K x ) t. If the length ‘dx’
elongates an amount du under load, its strain is:
du P 1
 .
dx A E
Solution Contd.
Total extension of bar, u
u
z
u
P
tE
u
L
0
P
Ax E
z
dx 
z
dx

0 B  kx
1
B K L
P
ln 1
KtE
B1
L
L
0
P
dx
( B1  Kx ) t E
P
L
ln B1  Kx
KtE
0
Solution Contd.
Substituting back for K,
B  B2  B1
P
ln 1
B  B1
B1
( 2
)t E
L
B
P
u
ln 2
B  B1
B1
( 2
)t E
L
u
In problem, t = 1 cm, B1 = 5 cm, B2 = 10 cm, L = 40 cm, P = 50,000 N, E = 2 x 107 N/cm 2
50,000 N
u
(
10  5
) x 1 cm x 2 x10 7
40
ln
10
 0.01386 cm
5
Solution Concluded
Substituting back for K,
u
u
B1  B2  B1
P
ln
B2  B1
B1
(
)t E
L
B
P
ln 2
B  B1
B1
( 2
)t E
L
In problem, t = 1 cm, B1 = 5 cm, B2 = 10 cm, L = 40 cm,
P = 50,000 N, E = 2 x 107 N/cm2
u
50,000 N
10  5
(
) x 1 cm x 2 x10 7
40
10
ln
5
 0.01386 cm
1.9 Lateral Strain and Poisson’s Ratio



Under the action of a longitudinal
stress, a body will extend in the
direction of the stress and contract in
the transverse or lateral direction
(see Fig. below).
The
reverse
occurs
under
a
compressive load.
Stress Effects
P
P
Longitudinal Tensile Stress Effect
P
P
Longitudinal Compressive Stress Effect
Poisson’s Ratio
Lateral strain is proportional to the longitudinal strain,
with the constant of proportionality called ‘Poisson’s ratio’ with symbol,
Mathematically,

Lateral strain
Direct or longitudinal strain

For most metals, the range of
is 0.28 to 0.33.
.
1.10 Thermal Strain
Most structural materials expand when heated,
in accordance to the law:
  T
where  is linear strain and
 is the coefficient of linear expansion;
T is the rise in temperature.
That is for a rod of Length, L;
if its temperature increased by t, the extension,
dl =

L T.
Thermal Strain Contd.
As in the case of lateral strains, thermal strains
do not induce stresses unless they are constrained.
The total strain in a body experiencing thermal stress
may be divided into two components:
Strain due to stress,   and
That due to temperature,  T .
Thus:


=  + T

 T
=
E
1.11. Principle of Superposition





It states that the effects of several actions
taking place simultaneously can be
reproduced exactly by adding the effect of
each action separately.
The principle is general and has wide
applications and holds true if:
(i) The structure is elastic
(ii) The stress-strain relationship is linear
(iii) The deformations are small.
1.12 General Stress-Strain
Relationships
1.12 General Stress-Strain
Relationships
For the element of material as in Figure above
subjected to uniaxial stress,  x , the ensuing strain
is as shown in (b).
Strain in x direction,
x 
x
E
Strains in y and z directions as a
result of strain in x –direction
=
  x and    x  
x
E
each
Note: The negative sign indicates contraction.
General Stress-Strain Relationships
Contd.
For an element subjected to triaxial stresses,
 x,
y
 z , the total strain in x direction will be
and
due to  x and lateral strains due to  y
and  z .
Using the principle of superposition, the resultant strain in x-direction is:
x 
x
E

i. e.  x 
 y
E

 z
E
1
{ x   ( y   z )}
E
y 
1
{ y   ( x   z )}
E
z 
1
{ z   ( x   y )}
E
Generalised Hooke’s Law in three dimensions
General Stress-Strain Relationships
Contd.

In the case of shear
strain, there is no lateral strain,
hence the shear stress/shear
strain relationship is the same
for both uniaxial and complex
strain systems.
Note:
Plain Stress and Plain Strain




A plain stress condition is said to exist when
stress in the z direction is zero.
The above equations may be applied for
but strain in the z direction is not zero.
Also plain strain condition exists when the
strain in z direction is zero.
Using strain in Z direction as zero in this
case does not mean that stress in the z
direction is zero.
Strain Caused by Stress and
Temperature
In addition to strain caused by stress, there may also be thermal strain
due to change in temperature. The general form of the stress/strain
relations is:
x 
1
{ x   ( y   z )}   t
E
y 
1
{ y   ( x   z )}   t
E
z 
1
{ z   ( x   y )}   t
E
 xy 
 xy
G
; 
yz

 yz
G
;  zx 
 zx
G
Try On Your Own
1  2
( x   y   z )
Show that :  v 
E
Example
Example: A plate of uniform thickness 1 cm and dimension 3 x 2 cm is acted upon by
the loads shown. Taking E = 2 x 107 N/cm2, determine  x
0.3.
and y . Poisson’s ratio is
42 kN
y
18 kN
2 cm 18 kN
x
42 kN
3cm
Solution
x 
18000 N
 9000 N / cm2
2cm x 1cm
y 
42000 N
 14000 N / cm2
3cm x 1cm
Hooke’s law in two dimensions states that:
and
x 
1
1
6
[ x    y ] 
[
9000

0
.
3
(
14000
]

240
x
10
E
2 x 107
y 
1
1
6
[ y    x ] 
[
14000

0
.
3
(
9000
]

565
x
10
E
2 x 107
1.13 Relationship between Elastic
Modulus (E) and Bulk Modulus, K
It has been shown that :  v   x   y   z
1
 x   ( y   z )
E
For hydrostatic stress,  x   y   z  
x 
i. e.
x 
1

  2    1 2 
E
E
Similarly ,  y and  z are each
v   x   y  z

1 2 
E
 Volumetric strain
3
1 2 
E
3
E
1 2 
v 
v
Volumetric or hydrostatic stress 

Volumetric strain
v
E
i. e. E  3 K 1  2  and K 
3 1 2 
Bulk Modulus, K 
Maximum Value For Poisson’s
Ratio
From the equation, if v = 0.5, the value of K becomes infinitely large.
Hence the body is incompressible. If v > 0.5, K becomes negative
i.e. the body will expand under hydrostatic pressure which is
inconceivable. It may be concluded that the upper limit of Poisson’s ratio
is 0.5.
Note:
2 G 1 
K
3 1 2 
Where: G is Shear Modulus
and
E  2 G 1 
1.14 Compound Bars
A compound bar is one comprising two or more parallel elements, of different materials,
which are fixed together at their end. The compound bar may be loaded in tension or
compression.
1
F
2
F
2
Section through a typical compound bar consisting of a circular bar (1) surrounded by a
tube (2)
1.14.1 Stresses Due to Applied Loads
in Compound Bars
If a compound bar is loaded in compression by a force, F,
Since the rod and tube are of the same length and must remain
together, the two materials must have the same strain i.e.
1  2
S tra in 
S tre ss
E
i. e
1 2

, 
E1
E2
2

 1E 2
..... ( 1 )
E1
Where E1 and E2 are the elastic moduli of materials 1 and 2 respectively.
Also: The total load, F must be shared by the two materials, i.e. F = F 1 + F2
Where: F1 and F2 are the loads in the individual elements.
Compound Bars Contd.
Now: as force = stress x area: Then: F =  1 A1   2 A2
...............(2)
Where A1 and A2 are the areas of materials 1 and 2 respectively.
Substituting for
F   1 A1 
1 
 2 from Eqn. 1 into Eqn 2:
 1 E 2 A2
E1
F E1
E1 A1  E 2 A2
L
M
N
  1 A1 
and  2 
E 2 A2
E1
O
P
Q
F E2
E1 A1  E 2 A2
1.14.2 Temperature stresses in
compound bars
1
1
2
2
L
L 1 T
(a)
1
L 2 T
2
F
F
FL
A1 E 1
{b}
1
F
2
(c)
F
FL
A2 E 2
Temperature stresses in compound
bar Contd.
Consider a compound bar, see (a) above of length, L consisting of 2
different materials (1) and (2) having coefficients of expansion
1
and  2 respectively with  1 > 2 . If the bar is subjected to a
1
uniform temperature rise, T and the right hand fixing released,
 1 the bar (1) will expand more than (2) as shown in diagram (b).
However, because of the end fixing, free expansion cannot occur.
Diagram (c) shows that the end fixing must supply a force which
decreases the length of bar (1) and increases the length of bar (2)
until equilibrium is achieved at a common length.
As no external forces are involved, a self equilibrating
(balancing force system is created).
Temperature Stresses Contd.
Free expansions in bars (1) and (2) are L 1T
and
L 2 T respectively.
Due to end fixing force, F: the decrease in length of bar (1) is
FL
FL
and the increase in length of (2) is
.
A1 E1
A2 E2
L
FL
FL
L 1 T 
 L 2 T 
A1 E1
A2 E2
1
1
i. e. F[

]  T ( 1   2 )
A1 E1 A2 E2
L
A E AE O
 T (
M
P
E
E
A
A
N
Q
2
2
1
1 
2
2
At Equilibrium:
i. e.  1 A1
1
1
1
2
1
1
2
T ( 1   2 ) A2 E1 E2
A1 E1  A2 E2
1
L 1 T
(a)
1
L 2 T
2
2)
F
F
1
F
2
(c)
2 
FL
A1 E1
{b}
T ( 1   2 ) A1 E1 E2
A1 E1  A2 E2
Note: As a result of Force, F, bar (1) will be in compression while (2) will be in tension.
F
FL
A2 E 2
Example

A steel tube having an external diameter of
36 mm and an internal diameter of 30 mm
has a brass rod of 20 mm diameter inside it,
the two materials being joined rigidly at their
ends when the ambient temperature is 18
0C.
Determine the stresses in the two
materials: (a) when the temperature is
raised to 68 0C (b) when a compressive
load of 20 kN is applied at the increased
temperature.
Example Contd.


For brass: Modulus of elasticity = 80
GN/m2; Coefficient of expansion = 17 x
10 -6 /0C
For steel: Modulus of elasticity = 210
GN/m2; Coefficient of expansion = 11 x
10 -6 /0C
Solution
30
Brass rod
20
Steel tube
Area of brass rod (Ab) =
Area of steel tube (A s) =
 x 202
 314.16 mm2
4
 x (362  302 )
 31102
. mm2
4
As E s  311.02 x 10 m x 210 x 10 N / m 2  0.653142 x 108 N
6
1
 153106
.
x 10 8
As E s
2
9
36
Solution Contd.
Ab Eb  314.16 x 106 m2 x 80 x 109 N / m2  0.251327 x 108 N
1
 3.9788736 x 108
Ab Eb
T ( b   s )  50(17  11) x 106  3 x 104
With increase in temperature, brass will be in compression while
steel will be in tension. This is because expands more than steel.
1
1
i. e. F[

]  T ( b   s )
As E s Ab Eb
i.e. F[1.53106 + 3.9788736] x 10 -8 = 3 x 10 -4
F = 5444.71 N
Solution Concluded
Stress in steel tube =
5444.71N
 17.51N / mm2  17.51 MN / m2 (Tension)
2
31102
. mm
Stress in brass rod =
5444.71N
2
2

17
.
33
N
/
mm

17
.
33
MN
/
m
(Compression)
2
314.16 mm
(b) Stresses due to compression force, F’ of 20 kN
F ' Es
20 x 103 N x 210 x 109 N / m2
s 

 46.44 MN / m2 (Compression)
8
E s As  Eb Ab
0.653142  0.251327 x 10
F ' Eb
20 x 103 N x 80 x 109 N / m2
2
b 


17
.
69
MN
/
m
(Compression)
8
E s As  Eb Ab 0.653142  0.251327 x 10
Resultant stress in steel tube = - 46.44 + 17.51 = 28.93 MN/m2 (Compression)
Resultant stress in brass rod = -17.69 - 17.33 = 35.02 MN/m2 (Compression)
Example



A composite bar, 0.6 m long comprises a
steel bar 0.2 m long and 40 mm diameter
which is fixed at one end to a copper bar
having a length of 0.4 m.
Determine the necessary diameter of the
copper bar in order that the extension of
each material shall be the same when the
composite bar is subjected to an axial load.
What will be the stresses in the steel and
copper when the bar is subjected to an axial
tensile loading of 30 kN? (For steel, E = 210
GN/m2; for copper, E = 110 GN/m2)
Solution
0.2 mm
0.4 mm
F
40 mm dia
d
Let the diameter of the copper bar be d mm
Specified condition: Extensions in the two bars are equal
dlc  dls
dl   L 
Thus:

E
Fc Lc
FL
 s s
Ac Ec As E s
L
FL
AE
F
Solution Concluded
Also: Total force, F is transmitted by both copper and steel
i.e. Fc = Fs = F
i. e.
Lc
L
 s
Ac Ec As E s
Substitute values given in problem:
0.4 m
0.2 m

 d 2 / 4 m2 110 x 109 N / m2  / 4 x 0.0402 x 210 x 109 N / m2
2 x 210 x 0.0402 2
d 
m ; d  0.07816 m  7816
. mm.
110
2
Thus for a loading of 30 kN
30 x 103 N
 2387
. MN / m2
Stress in steel,  s 
2
6
 / 4 x 0.040 x 10
30 x 103 N
 9 MN / m2
Stress in copper,  c 
2
6
 / 4 x 0.07816 x 10
1.15 Elastic Strain Energy



If a material is strained by a gradually
applied load, then work is done on the
material by the applied load.
The work is stored in the material in the form
of strain energy.
If the strain is within the elastic range of the
material, this energy is not retained by the
material upon the removal of load.
Elastic Strain Energy Contd.
Figure below shows the load-extension graph of a uniform bar.
The extension dl is associated with a gradually applied load, P
which is within the elastic range. The shaded area represents
the work done in increasing the load from zero to its value
Load
P
Extension
dl
Work done = strain energy of bar = shaded area
Elastic Strain Energy Concluded
W = U = 1/2 P dl
(1)
Stress,  = P/A i.e P =  A
Strain = Stress/E
i.e dl/L =  /E ,
dl = (  L)/E
L= original length
Substituting for P and dl in Eqn (1) gives:
W = U = 1/2  A . (  L)/E =  2/2E x A L
A L is the volume of the bar.
i.e
U=

2
/2E x Volume
The units of strain energy are same as those of work i.e. Joules. Strain energy

per unit volume,
2
/2E is known as resilience. The greatest amount of energy that can
stored in a material without permanent set occurring will be when
elastic limit stress.

is equal to the