LECTURE

1 INTRODUCTION

AND REVIEW

Preamble: Engineering
science is usually
subdivided
into number of topics
such
as 1.
Solid Mechanics
2.
Fluid Mechanics
3.
Heat Transfer
4.
Properties
of materials
and soon Although
there are close links
between them in terms of
the physical
principles
involved
and methods of analysis
employed.
The solid
mechanics as a subject
may be defined as a branch of applied
mechanics
that deals with behaviours
of solid
bodies subjected
to various
types of
loadings.
This is usually
subdivided
into further
two streams i.e Mechanics of
rigid
bodies or simply Mechanics and Mechanics of deformable
solids.
The mechanics of deformable
solids
which is branch of applied
mechanics is
known by several
names i.e.
strength
of materials,
mechanics of materials
etc.
Mechanics of rigid
bodies:
The mechanics of rigid
bodies is primarily
concerned with the static
and dynamic behaviour
under external
forces
of
engineering
components and systems which are treated
as infinitely
strong and
undeformable
Primarily
we deal here with the forces
and motions associated
with
particles
and rigid
bodies.
Mechanics

of

deformable

Mechanics
of
The mechanics

solids

solids:
of deformable

solids

is

more

concerned

with

the

internal

forces

and associated
changes in the geometry of the components involved.
of particular
importance
are the properties
of the materials
used, the strength
of which will
determine whether the components fail
by breaking
in service,
and the stiffness
of which will
determine whether the amount of deformation
they suffer
is
acceptable.
Therefore,
the subject
of mechanics of materials
or strength
of
materials
is central
to the whole activity
of engineering
design.
Usually
the
objectives
in analysis
here will
be the determination
of the stresses,
strains,
and deflections
produced by loads.
Theoretical
analyses and experimental
results
have an equal roles in this field.
Analysis
of stress
and strain
.
Concept of stress
: Let us introduce
the concept of stress
as we know that the
main problem of engineering
mechanics of material
is the investigation
of the
internal
resistance
of the body, i.e.
the nature of forces
set up within
a body
to balance the effect
of the externally
applied
forces.
The externally
applied
forces
are termed as loads.
These externally
applied
forces may be due to any
one

of

the

reason.

(i)
due to service
conditions
(ii)
due to environment
in which the
component works
(iii)
through contact
with other members
(iv)
due to fluid
pressures
(v)
due to gravity
or inertia
forces.
As we know that in mechanics of deformable
solids,
externally
applied
forces
acts on a body and body suffers
a deformation.
From equilibrium
point of view,
this action
should be opposed or reacted by internal
forces which are set up
within
the particles
of material
due to cohesion.
These internal
forces
give rise to a concept of stress.
Therefore,
let us define
a stress

Therefore,

Stress:

let

stress

where

A is

the

is

us define

defined

area

of

a term

stress

as the force

the

intensity

or force

per unit

Here we are using an assumption
that the total
force or total
any material
is uniformly
distributed
over its cross
section.
But the stress
distributions
may be for from uniform,
with local
stress

If

known

the force

as

stress

carried

sectional
area,
A,
load dP, of the total

As a particular
stress
defined mathematically
Units :
The basic
are N / m2 (or Pa)
MPa = 106 Pa ;

area.

section

load

carried

regions

of high

concentrations.

by a component

is

we must consider
force
P',
Then

generally
as
units

holds

of stress

GPa = 109 Pa;

not

uniformly

a small
definition

true
in S.I

area,
of

only
units

distributed
dA'
which
stress
is

at a point,
i.e.

over
carries

therefore

its

cross

a small

it

is

(International

system)

an equivalent

to MPa.

KPa = 103 Pa

Some times N / mm2 units are also used, because this is
while US customary unit is pound per square inch psi.

by

TYPES

OF STRESSES

.

only two basic

stresses

other

either

stresses

exists
are

: (1)

similar

to

normal stress
these

basic

and (2)
stresses

of these e.g. bending stress is a combination tensile,
stresses.
Torsional
stress,
as encountered in twisting

shear shear stress.
or

are

a combination

compressive and shear
of a shaft is a shearing

stress.

Let us define the normal stresses
Normal stresses
: we have defined
are

normal

to

the

normal stresses

areas

and shear stresses in the following
sections.
stress as force per unit area. If the stresses

concerned,

are generally

then

these

are

termed

denoted by a Greek letter

as normal

stresses.

The

( s )

This is also known as uniaxial
state of stress,
because the stresses acts only
in one direction
however, such a state rarely exists,
therefore
we have biaxial
and triaxial
state of stresses where either the two mutually perpendicular
normal stresses acts or three mutually perpendicular
normal stresses acts as
shown in the figures below :
Tensile or compressive stresses
: The normal stresses can be either
tensile
compressive whether the stresses acts out of the area or into the area
Bearing
bearing
Shear

Let

Stress
stress

stresses

us consider

: when one object presses against another, it is
( They are in fact the compressive stresses ).

referred

or

to a

.

now the

situation,

where

the

cross

sectional

area

of

a block

of material
is subject to a distribution
of forces which are parallel,
rather
than normal, to the area concerned. Such forces are associated with a shearing
of the material,
and are referred
to as shear forces.
The resulting
force
interistes

are

known

as

shear

The resulting
force intensities
stress being equal to
where

P is

the

total

force

and

the particular
stress generally
define shear stress at a point

stresses.

are known as shear stresses,

A the

area

over

which

it

acts.

holds good only at a point
as

the mean shear

As we

therefore

know

that

we can

The greek symbol t ( tau ) ( suggesting tangential
) is used to denote shear
stress.
However, it must be borne in mind that the stress ( resultant
stress )
at any point in a body is basically
resolved into two components s and t one
acts perpendicular
and other parallel
to the area concerned, as it is clearly
defined in the following
figure.
The single shear takes place on the single plane and the shear area is the cross
- sectional
of the rivett,
whereas the double shear takes place in the case of
Butt joints
of rivetts
and the shear area is the twice of the X - sectional
area
of the
rivett.
LECTURE 2
ANALYSIS
OF STERSSES

General State of stress at a point : Stress at a point in a material
body has
been defined as a force per unit area. But this definition
is some what
ambiguous since it depends upon what area we consider at that point.
Let us,
consider a point 'q'
in the interior
of the body
Let us pass a cutting
shown

plane through a pont 'q'
perpendicular

to the X - axis

as

below

The corresponding force components can be shown like this
dFx = sxx. dax dFy
= txy. dax
dFz = txz. dax
where dax is the area surrounding the point 'q'
when the cutting plane A r is to
x

-

axis.

In a similar
way it can be assummed that
point 'q'
perpendicular
to the y - axis.
shown

below

the cutting plane is passed through the
The corresponding force components are

The corresponding
force components may be written
as dFx = tyx.
day dFz = tyz. day
where day is the area surrounding
the point
'q'
when the cutting

day dFy = syy.
A r is

to

In the last it can be considered
that the cutting
plane is passed through the
point
'q'
perpendicular
to the 2 - axis.
The corresponding
force components may be written
as dFx = tzx. daz
dFy = tzy. daz
dFz = szz. daz
where daz is the area surrounding
the point
'q'
when the cutting
plane A r is

to

y

2

-

-

plane

axis.

axis.

Thus, from the foregoing
discussion
it is amply clear that there is nothing like
stress
at a point
'q'
rather we have a situation
where it is a combination
of
state of stress
at a point q. Thus, it becomes imperative
to understand
the term
state of stress
at a point
'q'.
Therefore,
it becomes easy to express astate of
stress
by the scheme as discussed
earlier,
where the stresses
on the three
mutually
perpendiclar
planes are labelled
in the manner as shown earlier.
the
state of stress
as depicted
earlier
is called
the general or a triaxial
state of
stress
that can exist
at any interior
point of a loaded body.
Before defining
the general state of stress
at a point.
Let us make overselves
conversant

with

the

notations

for

the

stresses.

we have already
chosen to distinguish
between normal and shear
stress with the help of symbols s and t
Cartesian
- co-ordinate
system
In the Cartesian
co-ordinates
system, we make use of the axes, X, Y and 2 Let
us

consider

the

stresses

acting

small

element

of

the

material

and

show

the

various

normal

the faces

Thus, in the Cartesian
co-ordinates
system the normal stresses
have been
represented
by sx, syand sz.
Cylindrical
- co-ordinate
system
In the Cylindrical
- co-ordinate
system we make use of co-ordinates
r, q and 2.
Thus, in the Cylindrical
co-ordinates
system, the normal stresses
i.e components
acting
over a element is being denoted by sr, sqand sz.
Sign convention
: The tensile
forces
are termed as ( +ve ) while the compressive
forces
are termed as negative
( -ve ).
First
sub
script
: it indicates
the direction
of the normal to the surface.
Second subscript
: it indicates
the direction
of the stress.
It may be noted that in the case of normal stresses
the double script
notation
may be dispensed with as the direction
of the normal stress
and the direction
of
normal

to

the

surface

of

the

element

on which

it

acts

is

the

same.

Therefore,

a

single
subscript
notation
as used is sufficient
to define the normal stresses.
Shear Stresses
: with shear stress
components, the single
subscript
notation
is
not practical,
because such stresses
are in direction
parallel
to the surfaces
on which they act. we therefore
have two directions
to specify,
that of normal
to

the

surface

and

the

stress

itself.

To do this,

we stress

we attach
two subscripts
to the symbol '
t'
, for shear
In cartesian
and polar co-ordinates,
we have the stress
the figures.
txy

,

tyx

,

tyz

,

tzy

,

tzx

,

txz

trq

,

tqr

,

tqz

,

itself.

To do this,

stresses.
components

tzq

,tzr

,

as shown in

trz

So as shown above, the normal stresses
and shear stress
components
indicated
on a small element of material
seperately
has been combined and
depicted
on a single
element.
Similarly
for a cylindrical
co-ordinate
system
us shown the normal and shear stresses
components separately.
Now let

us combine

the

normal

and shear

stress

components

as shown below

let

:

Now let us define
the state of stress
at a point formally.
State of stress
at a point
By state of stress
at a point,
we mean an information
which is required
at that
point such that it remains under equilibrium.
or simply a general state of
stress
at a point involves
all the normal stress
components, together
with all
the shear stress
components as shown in earlier
figures.

Therefore,
sx

txy

we need nine
txz

sy

tyx

tyz

components,
sz

tzx

to define

the state

of stress

at a point

tzy

If we apply the conditions
of equilibrium
which are as follows:
5 FX = 0 ; 5 M
X = 0 5 Fy = 0 ; 5 M y = 0 5 F2 = 0 ; 5 M 2 = 0
Then we get
txy = tyx
tyz = tzy
tzx = txy
Then we will
need only six components to specify
the state of stress
at a point
i.e

sx

,

sy,

sz

,

txy

,

tyz

,

tzx

Now let

us define
the concept of complementary
shear
Complementary shear stresses:
The existence
two sides of the element induces complementary
shear
sides of the element to maintain
equilibrium.
on
of
is
to

stresses.
of shear
stresses

stresses
on any
on the other two

planes AB and CD, the shear stress
t acts. To maintain
the static
equilibrium
this element,
on planes AD and BC, t'
should act, we shall
see that t'
which
known as the complementary
shear stress would come out to equal and opposite
the t . Let us prove this thing for a general case as discussed
below:

The figure
shows a small rectangular
element with sides of length
parallel
to X and y directions.
Its thickness
normal to the plane
in 2
direction.
All nine normal and shear stress
components may
element,
only those in X and y directions
are shown.
Sign convections
for shear stresses:
Direct
stresses
or normal
tensile

+ve

- compressive

Shear stresses:
the

element

c.c.w

- tending

Dx, Dy
of paper is
act on the
stresses

D2
-

ve

to turn

the element

c.w +ve.

- tending

to turn

ve.

The resulting
forces
applied
to the element are in equilibrium
in X and y
direction.
( Although
other normal and shear stress
components are not shown,
their
presence does not affect
the final
conclusion
).
Assumption
: The weight of the element is neglected.
Since the element is a static
piece of solid
body, the moments applied
to it
must also be in equilibrium.
Let '0'
be the centre of the element.
Let us
consider
the axis through the point
'0'.
the resultant
force associated
with
normal stresses
sx and sy acting
on the sides of the element each pass through
this

axis,

Now forces
the forces
txy

. D x

and

therefore,

have

on top and bottom
on left
and right

no moment.

surfaces
produce a couple which must be balanced
hand faces
Thus,
tyx . D x . D 2 . D y =

by

. D 2 . D y

In other word, the complementary
shear stresses
are equal in magnitude.
The same form of relationship
can be obtained
for the other two pair of shear
stress
components to arrive
at the relations
LECTURE 3 Analysis
of Stresses:
Consider a point
q'
in some sort of structural
member like
as shown in
figure
below. Assuming that at point exist.
q'
a plane state of stress
exist.
i.e.
the state of state stress
is to describe
by a parameters
sx, sy and txy
These stresses
could be indicate
a on the two dimensional
diagram as shown
below:

This is a commen way of representing
the stresses.
It must be realize
a that the
material
is unaware of what we have called
the x and y axes. i.e.
the material
has to resist
the loads irrespective
less of how we wish to name them or whether
they are horizontal,
vertical
or otherwise
further
more, the material
will
fail
when the stresses
exceed beyond a permissible
value.
Thus, a fundamental
problem
in engineering
design is to determine
the maximum normal stress
or maximum shear
stress
at any particular
point in a body. There is no reason to believe
apriori
that sx, sy and txy are the maximum value.
Rather the maximum stresses
may
associates
themselves with some other planes located
at q'.
Thus, it becomes
imperative
to determine
the values of sq and tq. In order tto achieve this let
us consider
the following.
Shear

stress:

If the applied
load P consists
of two equal and opposite
parallel
forces
not in
the same line,
than there is a tendency for one part of the body to slide
over
or shear from the other part across any section
LM. If the cross section
at LM
measured parallel
to the load is A, then the average value of shear stress
t =

P/A . The shear stress
is tangential
to the area over which it acts.
If the shear stress varies
then at a point then t may be defined
as
Complementary shear stress:
Let ABCD be a small rectangular
element of sides X, y and 2 perpendicular
to the
plane of paper let there be shear stress
acting
on planes AB and CD . It is
obvious that these stresses
will
from a couple ( t . xz )y which can only be
balanced by tangential
forces
on planes AD and BC. These are known as
complementary
shear stresses.
i.e.
the existence
of shear stresses
on sides AB
and CD of the element implies
that there must also be complementary
shear
stresses
on to maintain
equilibrium.
Let t'
be the complementary
shear stress
induced on planes
AD and BC. Then for
the equilibrium
( t . xz )y = t'
( yz )x
t = t'
Thus, every shear stress
is accompanied by an equal complementary
shear stress.
Stresses
on oblique
plane: Till
now we have dealt with either
pure normal direct
stress
or pure shear stress.
In many instances,
however both direct
and shear
stresses
acts and the resultant
stress
across any section
will
be neither
normal
nor tangential
to the plane.
A plane stse of stress
is a 2 dimensional
stae of stress
in a sense that the
stress
components in one direction
are all zero i.e
s2 = tyz = tzx = 0
examples of plane state of stress
includes
plates and shells.
Consider the general case of a bar under direct
load F giving
rise to a stress

syvertically
The
stress
acting
atapointis represented
bythestresses
acting
on
the faces of the element enclosing
the point.
The stresses
change with the inclination
of the planes passing through that
point i.e.
the stress
on the faces of the element vary as the angular position
of the element changes.
Let the block be of unit depth now considering
the equilibrium
of forces
on the triangle
portion
ABC Resolving
forces
perpendicular
to BC, gives
sq.BC.1 = sysinq
. AB . 1 but AB/BC = sinq or AB = BCsinq
Substituting
this value in the above equation,
we get
sq.BC.1 = sysinq
. BCsinq
. 1 or

(1)

Now resolving

the forces

parallel

again AB = BCcosq
sysinqcosq

If

q = 900 the BC will

direct

stress

or

to BC

tq.BC.1
(2)

be parallel

normal

tq.BC.1

= sycosq

. BCsinq

= sy cosq

. ABsinq

. 1 or

=

to AB and tq = 0, i.e.

tq

there

will

. 1

be only

stress.

By examining the equations
(1) and (2), the following
conclusions
may be drawn
(i)
The value of direct
stress
sq is maximum and is equal to sy when q = 900.
(ii)
The shear stress
tq has a maximum value of 0.5 sy when q = 450
(iii)
The stresses
sq and sq are not simply the resolution
of sy
Material
subjected
to pure shear:
Consider the element shown to which shear stresses
have been applied
to the
sides

AB

and

DC

Complementary shear stresses
of equal value but of opposite
effect
are then set
up on the sides AD and BC in order to prevent the rotation
of the element.
Since
the applied
and complementary
shear stresses
are of equal value on the x and y
planes.
Therefore,
they are both represented
by the symbol txy.
Now consider
the equilibrium
of portion
of PBC
Assuming unit depth and resolving
normal to PC or in the direction
of sq
sq.PC.1

= txy.PB.cosq.1+

txy.BC.sinq.1

Now writing
PB and BC in terms
PB/PC = sinq BC/PC = cosq
sq = 2txysinqcosq
(1)

Now resolving
txy

. PBsinq

-ve sign

= txy.PB.cosq

+ txy.BC.sinq

of PC so that it cancels out from the two sides
sq.PC.1 = txy.cosqsinqPC+
txy.cosq.sinqPC

sq = txy.2.sinqcosq

forces
- txy

parallel

to PC or in

the direction

tq.then

txyPC

. 1 =

. BCcosq

has been put because

this

component

is

in

the same direction

as that

of

txyPC

. 1

tq.

again
= txy

converting
. PB.sin2q

= -txycos2q

the various
quantities
- txy . PCcos2q
or

in terms of PC we have
= -[ txy (cos2q - sin2q)

]

(2)

the negative
sign means that
Let us examine the equations

the sense of tq is opposite
(1) and (2) respectively

to that

of assumed one.

From equation

(1)

i.e,

The equation
(1)
Let us take into

sq = txy

sin2q

represents
that the maximum value of sq is txy when q = 450.
consideration
the equation
(2) which states
that
tq = - txy

cos2q
It indicates
thatthemaximum
value
of tq is txywhen
q =00or900.it hasa
value zero when q = 450.
From equation
(1) it may be noticed
that the normal component sq has maximum and
minimum values of +txy (tension)
and -txy (compression)
on plane at 1 450 to the
applied
shear and on these planes the tangential
component tq is zero.
Hence the system of pure shear stresses
produces and equivalent
direct
stress
system, one set compressive
and one tensile
each located
at 450 to the
original
shear directions
as depicted
in the figure
below:
Material
subjected
to two mutually
perpendicular
direct
stresses:
Now consider
a rectangular
element of unit depth, subjected
to a system of two
direct
stresses
both tensile,
sx and syacting
right
angles to each other.
for

equilibrium

= sy

sin

converting
sy

sin2q

of the

q . AB.1

cos

AB and BC in

ABC, resolving

perpendicular

(3)
Now resolving

terms

of AC so that

AC cancels

sq . AC.1

out from

the sides

that cos2q - sin2q = cos2q or (1 - cos2q)/2
= sin2q
cos2q)/2
= cos2q
transformations
the expression
for sq reduces to
= 1/2sy
(1 + cos2q)
the various
terms we get

parallal

to

AC

sq.AC.1=

-txy..cosq.AB.1+

ve sign appears because this component is
Again converting
the various
quantities

cancels

to AC

q . BC.1

sq =

+ sxcos2q

Futher,
recalling
Similarly
(1 +
Hence by these
cos2q) + 1/2sx
On rearranging
The
AC.

portion

+ sx

out

from

the

two

(1 -

txy.BC.sinq.1

in the same direction
as that of
in terms of AC so that the AC

sides.

(4)
Conclusions

:

The following
conclusions
may be drawn from equation
(3) and (4)
(i)
The maximum direct
stress would be equal to sx or sy which ever is the
greater,
when q = 00 or 900
(ii)
The maximum shear stress
in the plane of
the applied
stresses
occurs when q = 450
LECTURE 4 Material
subjected
to combined direct
and shear stresses:
Now
consider
a complex stress
system shown below, acting
on an element of material.
The stresses
sx and sy may be compressive
or tensile
and may be the result
of
direct
forces
or as a result
of bending.The
shear stresses
may be as shown or
completely
reversed and occur as a result
of either
shear force or torsion
as
shown in the figure
below:
As per the double subscript
notation
the shear stress
on the face BC
should be notified
as tyx , however, we have already
seen that for a pair of
shear stresses
there is a set of complementary
shear stresses
generated such
that tyx = txy
By looking
at this state of stress,
it may be observed that this state of stress
is

(i)

combination

Material

of

formulas

deserved

sq = tyx

sin2

(ii)

Material

case

the

two

different

subjected

various

are

as

q

tq

subjected

of stress

shear.

In this

case the various

follows

= - tyx

cos

2 q

to two mutually

formula's

To get the required
respective
equations

cases:

to pure stae

derived

are

perpendicular
as

direct

stresses.

equations
for the case under consideration,let
for the above two cases such that

These are the equilibrium
on material
proportions

equations
for
and are equally

In this

follows.

us add the

stresses
at a point.
They do not depend
valid
for elastic
and inelastic

behaviour

This eqn gives two values of 2q that differ
by 1800 .Hence the
maximum and minimum normal stresses
occurate
900 apart.

planes

on which

From the

triangle

it

may be determined
Substituting

values

of cos2 q and sin2

q in

equation

the

(5) we get

This shows that the values oshear stress
is zero on the
Hence the maximum and minimum values of normal stresses
shearing
stress.
The maximum and minimum normal stresses
principal
stresses,
and the planes on which they act are
the solution
of equation

principal
planes.
occur on planes of zero
are called
the
called
principal
plane

will
yield
two values of 2q separated
by 1800 i.e.
two values of q separated
by
900 .Thus the two principal
stresses
occur on mutually
perpendicular
planes
termed principal
planes.
Therefore
the two
dimensional
complex stress
system can now be reduced to the
equivalent
system of principal
stresses.
Let us recall
value

of

that

maximum

for

the case of a material

shear

subjected

to direct

stresses

the

stresses

Therefore,it
can be concluded that the equation
(2) is a negative
reciprocal
of
equation
(1) hence the roots for the double angle of equation
(2) are 900 away
from the corresponding
angle of equation
(1).
This means that the angles that angles that locate
the plane of maximum or
minimum shearing
stresses
form angles of 450 with the planes of principal
stresses.

Futher,

by making

the

triangle

we get

Because of root the difference
in sign convention
arises
from the point of view
of locating
the planes on which shear stress
act. From physical
point of view
these sign have no meaning.
The largest
stress
regard less of sign is always know as maximum shear stress.
Principal
plane inclination
in terms of associated
principal
stress:
we know that the equation
yields
two values of q i.e.
the
inclination
of the two principal
planes on which the principal
stresses
s1 and
s2 act. It is uncertain,however,
which stress
acts on which plane unless
equation.
is used and observing
which one of the two principal
stresses
is
obtained.
Alternatively
we can also find the answer to this problem in the
following
manner
Consider once again the equilibrium
depth, Assuming AC to be a principal
and

the

shear

stress

is

of a triangular
plane on which

5

GRAPHICAL

of unit
sp acts,

zero.

Resolving
the forces
horizontally
we get:
sx .BC . 1 + txy .AB . 1 = sp . cosq . AC
by BC we get
LECTURE

block of material
principal
stresses

SOLUTION

dividing

MOHR'S
STRESS

the above equation

through

CIRCLE

The transformation
equations
for plane stress
can be represented
in a graphical
form known as Mohr's
circle.
This grapical
representation
is very useful
in
depending the relationships
between normal and shear stresses
acting
on any
inclined
plane at a point in a stresses
body.
To draw a Mohr's
stress
circle
consider
a complex stress
system as shown in the figure
The above system represents
a complete stress
system for any condition
of
applied
load in two dimensions
The Mohr's
stress
circle
is used to find out graphically
the direct
stress
s and
sheer stress
t on any plane inclined
at q to the plane on which sx acts.The
direction
of q here is taken in anticlockwise
direction
from the BC.
STEPS:

In order to do achieve the desired
objective
we proceed in the following
manner
(i)
Label the Block ABCD.
(ii)
Set up axes for the direct
stress
(as
abscissa)
and shear stress
(as ordinate)
(iii)
Plot the stresses
on two
adjacent
faces e.g. AB and BC, using the following
sign convention.

Direct stresses - tensile
positive;
compressive, negative
Shear stresses
tending to turn block clockwise,
positive
tending to turn block
counter clockwise,
negative
[ i.e shearing stresses are +ve when its movement about the centre of the
element is clockwise ]
This gives two points on the graph which may than be labeled as respectively
to
denote stresses on these planes.
(iv)
Join .
(v)
The point P where this line cuts the s axis is than the
centre of Mohr's
stress circle
and the line joining
is diameter.
Therefore the
circle
can now be drawn.
Now every point on the circle
then represents a
state of stress on some plane through C.
Proof:
Consider any point Q on the circumference
of the circle,
such that PQ
makes an angle 2q with BC, and drop a perpendicular
from Q to meet the s axis at
N.Then OQ represents the resultant
stress on the plane an angle q to BC. Here we
have assumed that sx > sy
Now let us find out the coordinates
of point Q. These are ON and QN.
From the figure drawn earlier
ON = OP + PN
OP = OK + KP
OP = sy + 1/2 ( sxsy)
= sy / 2 + sy / 2 + sx / 2 + sy / 2
= ( sx + sy ) / 2
PN = Rcos( 2q - b )
hence ON = OP + PN
= ( sx + sy ) / 2 + Rcos( 2q - b )
= ( sx + sy ) / 2 + Rcos2q cosb + Rsin2qsinb
now make

the

Thus,

ON = 1/2

substitutions

sy)sin2q

- txycos2q

for

Rcosb

and

Rsinb.

( sx + sy ) + 1/2 ( sx - sy )cos2q + txysin2q
(1)
Similarly
QM = Rsin( 2q - b )
= Rsin2qcosb - Rcos2qsinb
Thus, substituting
the values of R cosb and Rsinb, we get
QM = 1/2

( sx -

(2)

If we examine the equation (1) and (2), we see that this is the same equation
which we have already derived analytically
Thus the co-ordinates
of Q are the normal and shear stresses on the plane
inclined
at q to BC in the original
stress system.
N.B: Since angle PQ is 2q on Mohr's
circle
and not q it becomes obvious that
angles are doubled on Mohr's
circle.
This is the only difference,
however, as
They are measured in the same direction
and from the same plane in both figures.
Further points to be noted are :
(1) The direct stress is maximum when Q is at M and at this point obviously the
sheer stress is zero, hence by definition
OM is the length representing
the
maximum principal
stresses s1 and 2q1 gives the angle of the plane q1 from BC.
Similar OL is the other principal
stress and is represented by s2
(2) The maximum shear stress is given by the highest point on the circle
and is
represented by the radius of the circle.
This follows that since shear stresses and complimentary sheer stresses have the
same value; therefore
the centre of the circle will always lie on the s axis
midway between sx and sy . [ since +txy & -txy are shear stress & complimentary
shear stress so they are same in magnitude but different
in sign. ]
(3) From the above point the maximum sheer stress i.e.
the Radius of the Mohr's
stress

circle

would

while the direct
sx and sy i.e

be

stress

(4) As already defined
components are zero.
sheer

stress

is

on the plane of maximum shear must be mid

may between

the principal
planes are the planes on which the shear
Therefore are conclude that on principal
plane the

zero.

(5) Since the resultant
of two stress at 900 can be found from the parallogram
of vectors as shown in the diagram.Thus,
the resultant
stress on the plane at q
to BC is given by OQ on Mohr's
Circle.
(6) The graphical
method of solution
circle
is a very powerful technique,

for a complex stress problems using Mohr's
since all the information
relating
to any

plane within
the stressed
element is contained
in the single
construction.
It
thus, provides
a convenient
and rapid means of solution.
which is less prone to
arithmetical
errors
and is highly
recommended.
LECTURE 6 ILLUSRATIVE PROBLEMS: Let us discuss few representative
problems
dealing
with complex state of stress
to be solved either
analytically
or
graphically.
PROB 1:

A circular

what is

the Value

value

of

50

MN/m2

bar

40

mm diameter

of shear

stress

carries

on the

an

planes

axial

tensile

on which

load

the

normal

of

105

stress

kN.

has a

tensile.

PROB 2:
For a given loading
conditions
the state of stress
in the wall of a
cylinder
is expressed as follows:
(a)
85 MN/m2 tensile
(b)
25 MN/m2
tensile
at right
angles to (a)
(c)
Shear stresses
of 60 MN/m2 on the planes
on which the stresses
(a) and (b) act; the sheer couple acting
on planes
carrying
the 25 MN/m2 stress
is clockwise
in effect.
Calculate
the principal
stresses
and the planes on which they act. what would be the effect
on these
results
if owing to a change of loading
(a) becomes compressive
while stresses
(b) and (c) remain unchanged
Salient
points
of Mohr's
stress
circle:
1.
complementary
(on planes 900 apart on the circle)
are equal in magnitude
principal
planes are orthogonal:
points
L and M are 1800 apart
(900

apart

in

shear stresses
2.
The
on the circle

material)

3.
There are no shear stresses
on principal
planes:
point L and M lie on normal
stress
axis.
4.
The planes of maximum shear are 450 from the principal
points
D and E are 900 , measured round the circle
from points
L and M.
5.
The maximum shear stresses
are equal in magnitude and given by points
D and
E
6.
The normal stresses
on the planes of maximum shear stress
are equal
i.e.
points
D and E both have normal stress
co-ordinate
which
is equal to
the two principal
stresses.
As we know that the circle
represents
all possible
states
of normal and
shear stress
on any plane through a stresses
point in a material.
Further we
have seen that the co-ordinates
of the point
Q are seen to be the same as
those derived from equilibrium
of the element.
i.e.
the normal and shear stress
components on any plane passing through the point can be found using Mohr's
circle.
worthy of note:
1. The sides AB and BC of the element ABCD, which are 900 apart,
are represented
on the circle
by and they are 1800 apart.
2. It has been shown that Mohr's
circle
represents
all possible
states
at a
point.
Thus, it can be seen at a point.
Thus, it,
can be seen that two planes LP
and PM, 1800 apart on the diagram and therefore
900 apart in the material,
on
which shear stress
tq is zero. These planes are termed as principal
planes and
normal stresses
acting
on them are known as principal
stresses.
Thus ,
s1 =
0L

S2 = 0M

3. The maximum shear stress
in an element is given by the top and bottom points
of the circle
i.e by points
J1 and J2 ,Thus the maximum shear stress would be
equal to the radius of i.e.
tmax= 1/2( s1- s2 ),the
corresponding
normal stress
is obviously
the distance
OP = 1/2 ( sx+ sy ) , Further
it can also be seen that
the planes on which the shear stress
is maximum are situated
900 from the
principal
planes ( on circle
), and 450 in the material.
4.The minimum normal stress
is just as important
as the maximum. The algebraic
minimum stress
could have a magnitude greater
than that of the maximum principal
stress

if

the

state

of

stress

were

such

that

the

centre

of

the

circle

is

to

the

left
of orgin.
i.e.
if
s1 = 20 MN/m2 (say)
s2 = -80 MN/m2 (say)
Then
tmaxm = ( s1 - s2 / 2 ) = 50 MN/m2
If should be noted that the principal
stresses
are considered
a maximum or
minimum mathematically
e.g. a compressive
or negative
stress
is less than a
positive

stress,

irrespective

or

numerical

value.

5. Since the stresses
on perpendular
faces of any element are given by the coordinates
of two diametrically
opposite
points on the circle,
thus, the sum of
the two normal stresses
for any and all orientations
of the element is constant,
i.e.
Thus sum is an invariant
for any particular
state of stress.
Sum of the two normal stress
components acting
on mutually
perpendicular
planes
at a point in a state of plane stress
is not affected
by the orientation
of
these planes.

This

can

be

also

understand

from

the

circle

Since

diametrically
opposite
thus, what ever may be their
always lie on the diametre
or we can say that their
also be seen from analytical
relations
we know
sn1

AB

and

BC are

orientation,
they will
sum won't
change, it can
on plane BC; q = 0

= sx

on plane AB; q = 2700
sn2 = sy
Thus sn1 + sn2= sx+ sy
6. If s1 = s2, the Mohr's
stress
circle
degenerates
into a point and no shearing
stresses
are developed on xy plane.
7. If sX+ sy= 0, then the center of Mohr's
circle
coincides
with the origin
of s
-

t

co-ordinates.

LECTURE

7-ANALYSIS

OF STRAINS

CONCEPTOF STRAIN: if a bar is subjected
to a direct
load, and hence a stress
the bar will
change in length.
If the bar has an original
length
L and changes
by an amount dL, the strain
produce is defined as follows:
Strain

is

a measure

of

the

deformation

Quantity
i.e.
it has no units.
under load are very very small,

the form of strain

of

the

material

and

is

a nondimensional

Since in practice,
the extensions
of materials
it is often convenient
to measure the strain
in

X 10-6 i.e.

micro strain,

when the symbol used becomes m I.

Sign convention
for strain:
Tensile
strains
are positive
whereas compressive
strains
are negative.
The strain
defined
earlier
was known as linear
strain
or
normal

strain.

Shear strain:
The tangent of the angle through which two adjacent
sides of an
element rotates
relative
to their
initial
position
is termed shear strain.
In
many cases the angle is very small and the angle it self is used, ( in
radians
).
Shear strain:
shear stresses
acts along the surface.
This

Shear

strain

is

measured

in

radians

& hence

has no unit.So
we have two types
Hook's
Law : A material
is said

of strain
i.e.
to be elastic

unloaded

removed.

dimensions

when

Hook's
law therefore

load

states

is

that

Stress

is

non

dimensional

i.e.

it

normal stress & shear stresses.
if it returns
to its original

(

) a strain(

)

Modulus of elasticity
: within
the elastic
limits
of materials
i.e.
within
the
limits
in which Hook's
law applies,
it has been shown that
Stress / strain
= constant.
This constant
is given by the symbol E and is
termed as the modulus of elasticity
or Young's
modulus of elasticity
Thus The value of Young's
modulus E is generally
assumed to be the same in
tension
or compression
and for most engineering
material
has high, numerical
value

of

the

order

of

200

GPa

Poisson's
ratio:
If a bar is subjected
to a longitudinal
stress
there will
be a
strain
in this direction
equal to s / E . There will
also be a strain
in all
directions
at right
angles to s . The final
shape being shown by the dotted
lines.

It

has

been

observed

that

for

an elastic

materials,

the

lateral

strain

is

proportional
to the longitudinal
strain.
Poison's
ratio
( m ) = - lateral
strain
/ longitudinal
strain
For most engineering
materials
the value of m his between 0.25 and 0.33.
Three
dimensional
state of strain
: Consider an element subjected
to three
mutually
perpendicular
tensile
stresses
sx , syand sz as shown in the figure
below.

If

sy and s2 were not

definition

present

the strain

of Young's
modulus of Elasticity

in

the x direction

from

E would be equal to

the basic

Ix:

sX/ E

The effects
of sy and s2 in x direction
are given by the definition
of Poisson's
ratio
m '
to be equal as -m sy/ E and -m sz/ E
The negative
sign indicating
that if syand sz are positive
i.e.
tensile,
these
they tend to reduce the strain
in x direction
thus the total
linear
strain
is x
direction
is given by
Principal
strains
in terms of stress:
In the absence of shear stresses
on the
faces of the elements let us say that sx , sy , sz are in fact the principal
stress.
The resulting
strain
in the three directions
would be the principal

strains.
i.e.wewillhave
thefollowing
relation.
For Two dimensional
zero

i.e

s2

strain:

= O or

s3

system,

the stress

Although we will have a strain in this
Hence the set of equation as described
Hence

a strain

in the third

direction

becomes

= 0

can

exist

without

direction
owing to stresses
earlier
reduces to
a stress

in

that

s1& s2

direction

Hydrostatic
stress : The term Hydrostatic
stress is used to describe a state of
tensile
or compressive stress equal in all directions
within or external
to a
body. Hydrostatic
stress causes a change in volume of a material,
which if

expressed per unit of original

volume gives a volumetric strain

So let us determine the expression for the volumetric
strain.
Volumetric Strain:
Consider a rectangle solid of sides
the action of principal
stresses s1 , s2 , s3 respectively.

Then ii

, i2 , and T3 are the corresponding linear

of theArectangle becomes
( x + I1
hence

. x );

( y + I2

strains,

denoted by TV.

x, y and 2 under

than the dimensions

A

. y );

( 2 + I3

. 2 )

the

ALITER : Let a cuboid of material
having initial
sides of Length x, y and 2. If
under some load system, the sides changes in length by dx, dy, and dz then the
new volume ( x + dx ) ( y + dy ) ( 2 +dz )
New volume = xyz + yzdx + xzdy + xydz
Original
volume = xyz
Change in volume = yzdx +xzdy + xydz

Volumetric strain

= ( yzdx +xzdy + xydz ) / xyz = ix+ iy+ iz

Neglecting
Volumetric

the products of epsilon's
since the strains
strains
in terms of principal
stresses:

As

that

we

know

Strains
(a)

on an oblique

Linear

are sufficiently

small.

plane

strain

Consider

plane. The strains

a rectangular

block of material

OLMN as shown in the xy

along ox and oy are TX and iy , and gxy is the shearing

strain.

Then it is required to find an expression for iq, i.e the linear strain in a
direction inclined at q to OX, in terms of ix ,iy , gxy and q. Let the diagonal
OM be of length

'a'
then ON = a cos q and OL = a sin q , and the increase

length of those under strains
original

length

are ixacos q and iya sin q ( i.e.

) respectively.

If

M moves to M,

then

strain

in

x

the movement of M

parallel to x axis is ixacos q + gxy sin q and the movementparallel
axis is iyasin q

to the y

Thus the movement of M parallel
to OM , which since the strains
are small is
practically
coincident
with MM. and this would be the summation of portions
and (2) respectively
and is equal to
This

expression

is identical

stress on any inclined
replacing
Shear

txy i.e.

strain:

To

in form with

the equation

defining

(1)

the direct

plane q with ix and iy replacing sx and sy and % gxy

the shear stress
determine

the

shear

is

replaced

stain

in

by half
the

direction

the shear strain
OM consider

the

displacement of point P at the foot of the perpendicular
from N to OM and the
following
expression can be derived as
In the above expression % is there so as to keep the consistency with the stress
relations.

Futher -ve sign in the expression occurs so as to keep the consistency of sign
convention,
because OM moves clockwise with respect to OM it is considered to
be negative strain.
The other relevant
expressions are the following
Let us now define
Plane

Strain

the

plane strain

condition

.

In xy plane three
figures:

strain

components may exist

as can be seen from the following

Therefore,

strains

a strain

i.e

at any point

Ix in X direction,

In the case of normal

direction

strains

of the strain,

in

body can be characterized

Iy in y - direction
subscripts

by two axial

and gxy the shear strain.

have been used to indicate

and Ix , Iy are defined

as the relative

the

changes in

length in the co-ordinate
directions.
with shear strains,
the single
subscript
notation
is not practical,
because such
strains
involves
displacements
and length which are not in same direction.The
symbol and subscript
gxy used for the shear strain
referred
to the x and y
planes.
The order of the subscript
is unimportant.
gxy and gyx refer
to the same
physical
quantity.
However, the sign convention
is important.The
shear strain
gxy is considered
to be positive
if it represents
a decrease the angle between
the sides of an element of material
lying
parallel
the positive
X and y axes.
Alternatively
we can think of positive
shear strains
produced by the positive
shear

stresses

and

viceversa.

Plane strain
:
An element of material
subjected
only to the strains
in Fig. 1, 2, and 3 respectively
is termed as the plane strain
state.

Thus, the plane strain condition
: I2 = 0; gxz= 0; gyz= 0
It should be noted
with plane strain.
dimensional
stress
strain

is defined

as shown

only by the components Ix , Iy , gxy

that the plane stress
is not the stress
system associated
The plane strain
condition
is associated
with three
system and plane stress
is associated
with three dimensional

system.

LECTURE

8

PRINCIPAL

STRAIN

For the strains
on an oblique
plane we have an oblique
which are identical
in form with the equation
defining
inclined
plane q
Since

the equations

form,

so it

is

equally

well

strains

and

for

evident

stress
that

to represent
the

and strains

strain

vertical

on oblique

Mohr's
stress

axis

circle

conditions

for

half

the

planes

are identical

construction

using
shear

we have two equations
the direct
stress
on any

the

can

horizontal

in

be used

axis

for

linear

strain.

It should be noted, however that the angles given by Mohr's
stress
circle
refer
to the directions
of the planes on which the stress
act and not the direction
of
the
The

stresses
direction

normal
in

themselves.
of the
stresses

(i.e.

at 900)

Mohr'sstress

similarity
introduced.
Linear

therefore

circle

construction

associated

on

the

strain

convention

Strains

circle

adopted

: extension

strains

of the planes.
it

follows

of working a relative
rotation
This is achieved by plotting

downwards

The sign

and

to the directions

Since

therefore

are

angles

that

for

therefore

are doubled
a true

of axes of 2 x 900 = 1800 must be
positive
sheer strains
vertically

construction.

for

the strains

- positive

is

as follows:

compression

- negative

{ Shear of strains
are taken positive,
when they increase
the original
right
angle of an unstrained
element.
}
Shear strains
: for Mohr's
strains
circle
sheer strain
gxy - is +ve referred
x

-

direction

the

convention

for

the

shear

strains

are

bit

difficult.

The

to

first

subscript
in the symbol gxy usually
denotes the shear strains
associated
with
direction.
e.g. in gxy represents
the shear strain
in x - direction
and for
gyx represents
the shear strain
in y - direction.
If under strain
the line
associated
with first
subscript
moves counter clockwise
with respect to the
other line,
the shearing
strain
is said to be positive,
and if it moves
clockwise
it is said to be negative.

N.B: The positive
shear

stain

Moh's
strain

For the

shear strain

gxy is

given

is always to be drown on the top of Ix .If

the

]

circle

plane

strain

conditions

can we derivate

the following

relations

A typical
point P on the circle
given the normal strain
and half the sheer
strain
1/2gxy associated
with a particular
plane. we note again that an angle
subtended at the centre of Mohr's
circle
by an arc connecting
two points
on the
circle
is twice the physical
angle in the material.
Mohr

Since
plane

strain

circle

.

the transformation
equations
for plane strain
are similar
to those for
stress,
we can employ a similar
form of pictorial
representation.
This

is

known as Mohrs
strain
circle.
The main difference
between
Mohrs
factor
of half
is attached
to the

Points

directions

X

and Y

stress
circle
shear
strains.

represents

the strains

and

stress

circle

associated

with

is

that

a

x and y

with I and gxy /2 as co-ordiantes

Co-ordinates
In x

of X and Y points are located as follows
direction,
the strains
produced, the strains

xy are Ix and - gxy /2
where as in the Y - direction,

the strains

produced by sx,and

- t

are produced by Iy and + gxy are

produced by sy and + txy
These co-ordinated
are consistent
with our sign notation
( i.e.
stresses produces produce +ve shear strain & vice versa )

+ ve shear

on the face AB is txy+ve i.e strains are ( Iy, +gxy /2 ) where as on the face
BC, txy is negative hence the strains are ( Ix, - gxy /2 )
A typical
point P on the circle
gives the normal strains
and half the shear
strain,
associated with a particular
plane we must measure the angle from X

axis (taken as reference)
derived

with

reference

CONSTRUCTION

In this
have

steps
1. Take

in

with

angle measuring in

the c.c.w

direction

.

we would like

done

as the required formulas for Iq , -1/2 gq have been

to x-axis

the

normal

case

or linear

to locate

the points

of

stress

Mohrs

strains

x

& y

instead

of AB and BC as we

circle.

on x-axis,

whereas

half

of

shear

strains

are

plotted on y-axis.
2. Locate the points X and y
3. Join X and y and draw the Mohrs strain
circle
4. Measure the required parameter from this construction.
Note: positive
shear strains
are associated with planes carrying
positive
shear stresses and negative strains with planes carrying negative shear
stresses.
ILLUSTRATIVE

EXAMPLES

Use of strain Gauges
Although we can not measure stresses within a structural
member, we can measure
strains,
and from them the stresses can be computed, Even so, we can only
measure strains
on the surface.
For example, we can mark points and lines
on the
surface and measure changes in their spacing angles. In doing this we are of
course only measuring average strains
over the region concerned. Also in view of
the very small changes in dimensions, it is difficult
to archive accuracy in the
measurements

In practice,
electrical
strain gage provide
method of measuring strains.
A typical
strain gage is shown below.

a more accurate

and convenient

The gage shown above can measure normal strain in the local plane of the surface
in the direction
of line PQ, which is parallel
to the folds of paper. This
strain is an average value of for the region covered by the gage, rather than a
value

at

any particular

point.

The strain gage is not sensitive
to normal strain in the direction
perpendicular
to PQ, nor does it respond to shear strain.
therefore,
in order to determine the
state of strain at a particular
small region of the surface, we usually need
more than one strain gage.
To define a general two dimensional state of strain,
we need to have three

pieces of information,

such as Ix , Iy and gxy referred

to any convenient

orthogonal co-ordinates
x and y in the plane of the surface. we therefore
need
to obtain measurements from three strain gages. These three gages must be
arranged at different
orientations
on the surface to from a strain
rossett.
Typical examples have been shown, where the gages are arranged at either
450 or
600

to

each

other

as

shown

below

:

A group of three gages arranged in a particular
fashion is called a strain
rosette.
Because the rosette is mounted on the surface of the body, where the
material
is in plane stress,
therefore,
the transformation
equations for plane

strain

to

calculate

the

strains

in

various

directions.

Knowing the orientation
of the three gages forming
in
plane normal strains
they record,
the state of
surface
concerned can be found. Let us consider
the
figure
below, where three strain
gages numbered 1,
gages numbered 1, 2, 3 are arranged at an angles of
from

reference

direction,

which

we take

as X

a rosette,
together
with the
strain
at the region of the
general case shown in the
2, 3, where three strain
q1 , q2 , q3 measured c.c.w

axis.

Now, although
the conditions
at a surface,
on which there are no shear or normal
stress
components. Are these of plane stress
rather
than the plane strain,
we
can still
use strain
transformation
equations
to express the three measured

normal strains

in terms of strain

and y co-ordiantes
This

is

a set

unknows ix,

components ix , iy , iz and gxy referred

to X

as

of three

simultaneous

linear

algebraic

iy , gxy to solve these equation

equations

is a laborious

for

the

one as far

three

as

manually is concerned,
but with computer it can be readily
done.Using
these
later
on, the state of strain
can be determined
at any point.
Let us consider
a 450 degree stain
rosette
consisting
of three electrical
resistance
strain
gages arranged as shown in the figure
below :

The gages A, B,C measure the normal strains
lines

0A,

ia , ib , ic in the direction

of

0B and 0C.

Thus

Thus, substituting
the relation
(3) in the equation
+ ic ) and other equation becomes ix = ia ; iy= ic
Since

the gages A and C are aligned

ix and iy directly
Thus, ix , iy and gxy can easily

with

(2) we get

the x and y axes,

gxy = 2ib-

they

be determined from the strain

give

( ia

the strains

gage readings.

Knowing these strains,
we can calculate
the strains
in any other directions
by
means of Mohr's
circle
or from the transformation
equations.
The 600 Rossett:
For the 600 strain
rosette,
using the same procedure we can
obtain following
relation.
LECTURE

9

Stress

STRESS

Strain

-

STRAIN

Relations:

RELATIONS

The Hook's
law,

states

that

within

the

elastic

limits

the stress
is proportional
to the strain
since for most materials
it is
impossible
to describe
the entire
stress
strain
curve with simple mathematical
expression,
in any given problem the behavior
of the materials
is represented
by
an idealized
stress
strain
curve, which emphasizes those aspects of the
behaviors
which are most important
is that particular
problem.
(i)

Linear

elastic

material:

A linear

elastic

material

is

one in

which the strain
is proportional
to stress
as shown below:
There are also other types of idealized
models of material
behavior.
(ii)
Rigid Materials:
It is the one which donot experience
any strain
regardless
of the applied
stress.
(iii)
Perfectly
plastic(non-strain
hardening):
A perfectly
plastic
i.e
non-strain
hardening
material
is shown below:
(iv)
Rigid Plastic
material(strain
hardening):
A rigid
plastic
material
i.e strain
hardening
is depicted
in the figure
below:
(v) Elastic
Perfectly
Plastic
is having the characteristics

material:
The elastic
as shown below:

perfectly

(vi)
Elastic
Plastic
material:
The elastic
plastic
material
stress Vs strain
diagram as depicted
in the figure
below:
Elastic

Stress

strain

plastic
exhibits

material
a

Relations

Previously
stress
strain
relations
were considered
for the
special
case of a uniaxial
loading
i.e.
only one component of stress
i.e.
the
axial
or normal component of stress was coming into picture.
In this section
we
shall
generalize
the elastic
behavior,
so as to arrive
at the relations
which
connect all the six components of stress with the six components of elastic
stress.
Futher,
we would restrict
overselves
to linearly
elastic
material.

Before writing
down the relations
let us introduce
a term ISOTROPY
ISOTROPIC: If the response of the material
is independent
of the orientation
of
the load axis of the sample, then we say that the material
is isotropic
or in
other words we can say that isotropy
of a material
in a characteristics,
which
gives us the information
that the properties
are the same in the three
orthogonal
directions
x y z, on the other hand if the response is dependent on
orientation
it is known as anisotropic.
Examples of anisotropic
materials,
whose properties
are different
in different
directions

are

(i) wood (ii)
Fibre reinforced
plastic
(iii)
Reinforced
concrete
HOMOGENIUS:A material
is homogenous if it has the same composition
through our
body. Hence the elastic
properties
are the same at every point in the body.
However, the properties
need not to be the same in all the direction
for the
material
to be homogenous. Isotropic
materials
have the same elastic
properties
in all the directions.
Therefore,
the material
must be both homogenous and
isotropic
in order to have the lateral
strains
to be same at every point in a
particular

component.

Generalized

Hook's
Law: we know that

proportional

for

stresses

not greater

than

the

limit.

These equation
expresses the relationship
between stress
and strain
(Hook's
law)
for uniaxial
state of stress
only when the stress
is not greater
than the
proportional
limit.
In order to analyze the deformational
effects
produced by
all

the

stresses,

we shall

consider

the

effects

of

one axial

stress

at

a time.

Since we presumably are dealing
with strains
of the order of one percent or
less.
These effects
can be superimposed
arbitrarily.
The figure
below shows the
general triaxial
state of stress.
Let us consider
a case when sx alone is acting.
dimension in X-direction
whereas the dimensions

It will
cause an increase
in y and z direction
will

in
be

decreased.

Therefore
the resulting
strains
in
let us consider
that normal stress

three directions
are
sy alone is acting
and the

resulting

Similarly
strains

are

Now let

us consider

the stress

sz acting

alone,

thus

the strains

produced

are

In the following
analysis
shear stresses
were not considered.
It can be shown
that for an isotropic
material'sa shear stress will
produce only its
corresponding
shear strain
and will
not influence
the axial
strain.
Thus, we can
write

Hook'slaw

for

the

individual

shear

strains

and

shear

stresses

in

the

following
manner.
The Equations
(1) through (6) are known as Generalized
Hook's
law and are the
constitutive
equations
for the linear
elastic
isotropic
materials.
when these
equations
isotropic
materials.
when these equations
are used as written,
the
strains
can be completely
determined
from known values of the stresses.
To
engineers
the plane stress
situation
is of much relevance
( i.e.
sz = txz = tyz
= 0 ), Thus then the above set of equations
reduces to
Hook's
law is probably
the most well known and widely
used constitutive
equations
for an engineering
materials.
However, we can not say that all the
engineering
materials
are linear
elastic
isotropic
ones. Because now in the
present times,
the new materials
are being developed every day. Many useful
materials
exhibit
nonlinear
response and are not elastic
too.
Plane Stress:
In many instances
the stress
situation
is less complicated
for
example if we pull one long thin wire of uniform section
and examine
small
parallepiped
where x
axis coincides
with the axis of the wire
So if we take
acting
on the
plane

stress

the xy plane
parrallepiped.

then sx , sy , txy
This combination

will
be the only stress
of stress
components is

components
called
the

situation

A plane stress
may be defined
as a stress
condition
in which all components
associated
with a given direction
( i.e the z direction
in this example ) are

zero

Plane strain:
If we focus our attention
on a body whose particles
all lie in the
same plane and which deforms only in this plane. This deforms only in this
plane. This type of deformation is called as the plane strain,
so for such a
situation.

i2: gzx = gzy = 0 and the non zero terms would be ix, iy & gxy
i.e. if strain components ix, Ty and gxy and angle q are specified, the strain
components ix,
Ty and gxy with respect to some other axes can be determined.
ELASTIC

CONSTANTS

In considering
the elastic
behavior of an isotropic
materials
under, normal,
shear and hydrostatic
loading,
we introduce a total of four elastic
constants
namely E, G, K, and g .
It turns out that not all of these are independent to the others. In fact,
given
any two of them, the other two can be foundout . Let us define these elastic
constants

(i)
(ii)

E = Young's
Modulus of Rigidity
G = Shear Modulus or Modulus of rigidity

(iii)
(iv)

g = Possion's
ratio
K = Bulk Modulus of elasticity

= Shear

Volumetric

stress

us

Shear

/ strain

strain

= - lateral

strain / longitudinal
= Volumetric
stress /

strain

where
Volumetric
strain
Volumetric
stress = stress
Let

/

= Stress

find

the

relations

= sum of linear
stress in X, y and 2 direction.
which cause the change in volume.
between

them

strain