LECTURE 1 INTRODUCTION
The solid mechanics may be defined as a branch of applied
mechanics( includes fluid mechanics, solid mechanics etc) 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 and
Mechanics of deformable solids.
The mechanics of deformable solids is also
known by strength of materials.
Mechanics of rigid bodies: The mechanics
of rigid bodies treats materials as infinitely strong and undeformable.
Mechanics of deformable solids : The mechanics 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 : 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
Stress:
stress is defined as the internal force per unit area.
Where A is the area of the section
Here we are using an assumption
that the total force or total load carried by any material is uniformly
distributed over its cross – section. But the stress distributions may be for
from uniform, with local regions of high stress known as stress concentrations.
If the force carried by a component is not uniformly distributed over its
cross – sectional area, A
Units :
The basic units of stress in S.I units i.e. (International
system) are N/m^2 (or Pa).
MPa = 10^6 Pa ; GPa = 10^9 Pa; KPa =
10^3 Pa
Some times N/mm^2 units are also used, because this is an
equivalent to MPa.
TYPES OF STRESSES :
only two basic stresses exists : (1) normal stress and (2) shear
shear stress. Other stresses either are similar to these basic stresses or are a
combination of these e.g. bending stress is a combination tensile, compressive
and shear stresses. Torsional stress, as encountered in twisting of a shaft is a
shearing stress.
Normal stresses : We have defined stress as force per unit area. If the
stresses are normal to the areas concerned, then these are termed as normal
stresses. The normal stresses are generally denoted by a Greek letter ( 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 or
compressive whether the stresses acts out of the area or into the area
Bearing Stress : When one object presses against another, it is referred

to a bearing stress ( They are in fact the compressive stresses ).
Shear stresses :
Let us consider 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 stresses. The resulting force
intensities are known as shear stresses, the mean shear stress being equal to
Where P is the total force and A the area over which it acts.
As we
know that the particular stress generally holds good only at a point therefore
we can define shear stress at a point as
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.
LECTURE 2 ANALYSIS OF STERSSES
General State of stress at a point : Stress astate of stress at a point
is 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.
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 itself. To do this,
we attach two subscripts to the symbol ' t' , for shear stresses.
Now let us combine the normal and shear stress components as shown below : 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, we need
nine components, to define
Complementary shear stresses:
The existence
of shear stresses on any two sides of the element induces complementary shear
stresses on the other two sides of the element to maintain equilibrium.
Sign convections for shear stresses:
Direct stresses or normal
stresses
- tensile +ve
- compressive –ve
Shear stresses:
tending to turn the element C.W +ve.
- tending to turn the element C.C.W –
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 ‘O' be the centre of the element. Let us consider
..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:
...
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 sorces 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.
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 sz = 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
sy vertically
The stress acting at a point is represented by the stresses 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 to BC
tq.BC.1 = sy
cosq . ABsinq . 1
again AB = BCcosq
tq.BC.1 = sycosq .
BCsinq . 1 or tq = sysinqcosq
(2)
If q = 900 the BC will be parallel to AB and tq = 0, i.e. there will be only
direct stress or normal 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.
The equation (1) represents that the maximum value of sq is txy when q = 450.
Let us take into consideration the equation (2) which states that
tq = - txy
cos2q.
It indicates that the maximum value of tq is txy when q = 00 or 900. it
has a 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 ± 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.
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 combination of two different cases:
(i) Material subjected to pure stae of stress shear. In this case the various
formulas deserved are as follows
sq = tyx sin2 q
tq = - tyx cos 2 q
(ii) Material subjected to two mutually perpendicular direct stresses. In
this case the various formula's derived are as follows.
To get the required equations for the case under consideration,let us add the
respective equations for the above two cases such that
These are the equilibrium equations for stresses at a point. They do not depend
on material proportions and are equally valid for elastic and inelastic
behaviour
This eqn gives two values of 2q that differ by 1800 .Hence the planes on which
maximum and minimum normal stresses occurate 900 apart.
From the triangle it may be determined

Substituting the values
of cos2 q and sin2 q in equation (5) we get
This shows that the values oshear stress is zero on the principal planes.
Hence the maximum and minimum values of normal stresses occur on planes of zero
shearing stress. The maximum and minimum normal stresses are called the
principal stresses, and the planes on which they act are called principal plane
the solution of equation
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 that for the case of a material subjected to direct stresses the
value of maximum shear 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 of a triangular block of material of unit
depth, Assuming AC to be a principal plane on which principal stresses sp acts,
and the shear stress is zero.
Resolving the forces horizontally we get:
sx .BC . 1 + txy .AB . 1 = sp . cosq . AC
dividing the above equation through
by BC we get
LECTURE 5 GRAPHICAL SOLUTION – MOHR'S STRESS 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 ( sx- sy)
= 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 substitutions for Rcosb and Rsinb. Thus,
ON = 1/2 ( 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 sy)sin2q - txycos2q
(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 be
While the direct stress on the plane of maximum shear must be mid – may between
sx and sy i.e
(4) As already defined the principal planes are the planes on which the shear
components are zero.
Therefore are conclude that on principal plane the
sheer stress is 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 for a complex stress problems using
Mohr's circle is a very powerful technique, 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.
ILLUSRATIVE
Salient points of Mohr's stress circle:
1. complementary shear stresses
(on planes 900 apart on the circle) are equal in magnitude
2. The
principal planes are orthogonal: points L and M are 1800 apart on the circle
(900 apart in 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 =
OL
s2 = OM
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
co-ordinates 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 AB and BC are
diametrically opposite thus, what ever may be their orientation, they will
always lie on the diametre or we can say that their sum won't change, it can
also be seen from analytical relations
We know
on plane BC; q = 0
sn1 = 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
CONCEPT OF 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 of the material and is a nondimensional
Quantity i.e. it has no units.
Since in practice, the extensions of materials
under load are very very small, it is often convenient to measure the strain in
the form of strain x 10-6 i.e. micro strain, when the symbol used becomes m Î.
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 is non – dimensional i.e. it has no unit.So we have
two types of strain i.e. normal stress & shear stresses.
Hook's Law : A material is said to be elastic if it returns to its original
unloaded dimensions when load is removed.
Hook's law therefore states that
Stress (
) 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.
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. We will have the following relation.

For Two dimensional strain: system, the stress in the third direction becomes
zero i.e sz = 0 or s3 = 0
Although we will have a strain in this direction owing to stresses s1& s2 .
Hence the set of equation as described earlier reduces to
Hence a strain can exist without a stress in that 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 denoted by Îv.
So let us determine the expression for the volumetric strain.
Volumetric Strain:
C
Strains on an oblique plane
(a) Linear strain
C the shear stress is replaced by half the shear
strain
Shear strain: To determine the shear stain in the direction 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 ½ 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 the plane strain condition
Plane Strain :
In xy plane three strain components may exist as can be seen from the following
figures:
Therefore, a strain at any point in body can be characterized by two axial
strains i.e Îx in x direction, Îy in y - direction and gxy the shear strain.
In the case of normal strains subscripts have been used to indicate the
direction of the strain, and Îx , Îy are defined as the relative 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 as shown
in Fig. 1, 2, and 3 respectively is termed as the plane strain state.
Thus, the plane strain condition is defined only by the components Îx , Îy , gxy
: Îz = 0; gxz= 0; gyz= 0
It should be noted that the plane stress is not the stress system associated
with plane strain. The plane strain condition is associated with three
dimensional stress system and plane stress is associated with three dimensional
strain system.
LECTURE 8
PRINCIPAL STRAIN
For the strains on an oblique plane we have an oblique we have two equations
which are identical in form with the equation defining the direct stress on any
inclined plane q .
Since the equations for stress and strains on oblique planes are identical in
form, so it is evident that Mohr's stress circle construction can be used
equally well to represent strain conditions using the horizontal axis for linear
strains and the vertical axis for half the shear 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 stresses
themselves.
The direction of the stresses and therefore associated strains are therefore
normal (i.e. at 900) to the directions of the planes. Since angles are doubled
in Mohr's stress circle construction it follows therefore that for a true
similarity of working a relative rotation of axes of 2 x 900 = 1800 must be
introduced. This is achieved by plotting positive sheer strains vertically
downwards on the strain circle construction.
The sign convention adopted for the strains is as follows: Linear Strains :
extension - positive
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
to x - direction the convention for the shear strains are bit difficult. The
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 strain is always to be drown on the top of Îx .If the
shear stain gxy is given ]
Moh's strain circle
For the 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 strain circle :
Since the transformation equations for plane strain are similar to those for
plane stress, we can employ a similar form of pictorial representation. This is
known as Mohr's strain circle.
The main difference between Mohr's stress circle and stress circle is that a
factor of half is attached to the shear strains.
Points X' and Y' represents the strains associated with x and y
directions with Î and gxy /2 as co-ordiantes
Co-ordinates of X' and Y' points are located as follows :
CONSTRUCTION : In this we would like to locate the points x' & y' instead of AB
and BC as we have done in the case of Mohr's stress circle.
steps 1. Take normal or linear strains 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 Mohr's 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 a more accurate and convenient
method of measuring strains. A typical strain gage is shown below.
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 Îx , Îy 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 a rosette, together with the
in – plane normal strains they record, the state of strain at the region of the
surface concerned can be found. Let us consider the general case shown in the
figure below, where three strain gages numbered 1, 2, 3, where three strain
gages numbered 1, 2, 3 are arranged at an angles of q1 , q2 , q3 measured c.c.w
from reference direction, which we take as x – 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 components Îx , Îy , Îz and gxy referred to x and y coordiantes as
This is a set of three simultaneous linear algebraic equations for the three
unknows Îx, Îy , gxy to solve these equation is a laborious one as far 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 Îa , Îb , Îc in the direction of
lines OA, OB and OC.
Thus
Thus, Îx , Îy and gxy can easily be determined from the strain 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 - STRAIN RELATIONS
Stress – Strain 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 material:
The elastic perfectly plastic

material is having the characteristics as shown below:
(vi) Elastic – Plastic material:
The elastic plastic material
exhibits a stress Vs strain diagram as depicted in the figure below:
Elastic Stress – strain 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 for stresses not greater than the
proportional 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. It will cause an increase in
dimension in X-direction whereas the dimensions in y and z direction will be
decreased.
Therefore the resulting strains in three directions are
Similarly
let us consider that normal stress sy alone is acting and the resulting 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's a shear stress will produce only its
corresponding shear strain and will not influence the axial strain. Thus, we can
write Hook's law 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 the xy plane then sx , sy , txy will be the only stress components
acting on the parrallepiped. This combination of stress components is called the
plane stress 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.
Îz= gzx = gzy = 0 and the non – zero terms would be Îx, Îy & gxy
i.e. if strain components Îx, Îy and gxy and angle q are specified, the strain
components Îx', Îy' 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)
E = Young's Modulus of Rigidity
= Stress / strain
(ii) G =
Shear Modulus or Modulus of rigidity
= Shear stress / Shear strain
(iii) g = Possion's ratio
= - lateral strain / longitudinal strain
(iv) K = Bulk Modulus of elasticity
= Volumetric stress /
Volumetric strain
Where
Volumetric strain = sum of linear stress in x, y and z direction.
Volumetric stress = stress which cause the change in volume.