Elasticity (Lecture-1)
[Properties of matter, Elasticity, Rigid body, Perfect elastic,
Plastic body, Elastic Limit, Breaking stress,
Hook’s Law]
General Properties of matter
Solid
The particles (ions, atoms or molecules) are packed closely
together. The forces between particles are strong enough so
that the particles cannot move freely but can only vibrate.
In crystalline solids, the particles (atoms, molecules, or ions) are
packed in a regularly ordered, repeating pattern.
Glasses and other non-crystalline, amorphous solids without longrange order are not thermal equilibrium ground states
Solid
Liquid
gas
plasma
Liquid
Structure of a classical monatomic liquid. Atoms
have many nearest neighbors in contact, yet no longrange order is present.
Gas
The spaces between gas molecules are very big. Gas
molecules have very weak or no bonds at all. The
molecules in "gas" can move freely and fast.
Plasma
In a plasma, electrons are ripped away from their
nuclei, forming an electron "sea". This gives it the
ability to conduct electricity.
Phase Transitions
Elasticity
If an external force is applied to a material, it
causes deformation in molecular structure of that
material. By removing this force, material turns its
original shapes; this process is known as
elasticity of material.
It is the ability of returning its original shape after
removing the applied stress.
Perfectly elastic body: If the deformation of a body produced by a given
deforming force at a given temperature remains unchanged, the body is said
to be perfectly elastic.
Perfectly plastic body: If any body remains deformed and shows no tendency
to regain its original condition on the removal of the deforming force the
body is said to be perfectly plastic.
Load: Any combination of external forces acting on a body, its own weight
along with forces connected with it, whose net effect is to deform the body,
is referred to as a load.
Stress
The ratio of force to area, , is defined as stress measured in N/m2. Thus
if F be the deforming force applied uniformly over an area A, then
stress = F/A
if F be the deforming force be inclined to the surface, F makes angle α with
the normal of the surface, then
shearing stress = Fcosα/ A
And normal stress = Fsinα/ A
Strain
The ratio of the change in length to length, , is defined as strain (a
unitless quantity). In other words,
stress = Y × strain
i) Longitudinal or tensile strain:(The changing length per unit length).
ii) Volume strain: (The changing volume per unit volume)
iii) Shearing strain: (The changing shape per unit shape)
Rigid body
In physics, a rigid body is an idealization of a solid body in
which deformation is neglected. In other words, the distance
between any two given points of a rigid body remains
constant in time regardless of external forces exerted on it.
Even though such an object cannot physically exist due to
relativity, objects can normally be assumed to be perfectly
rigid if they are not moving near the speed of light.
Elastic Limit:
Elastic limit is defined by the load or the stress at which the body just
cases to return to its original form, when the load is removed.
In Solid, if the stress is gradually increased, the strain too increases with
Hooke’s law until a point is reached at which the linear relationship
breaks down. The value of the stress for which Hooke’s law just causes
to be obeyed is called the elastic limit.
Breaking stress:
The forces per unit area of a body for which the body breaks is called
breaking stress
Breaking stress = Breaking force/Area
Hooke's law:
— first stated formally by Robert Hooke (1678) of England
In contemporary language, extension is directly proportional to force,
a simple object that's essentially one-dimensional.
Hooke's law can be generalized to …
“stress is proportional to strain”
when strain refers to a change in some spatial dimension (length, angle,
or volume) compared to its original value and stress refers to the cause
of the change (a force applied to a surface).
stress  strain
Or stress = constant  strain
The coefficient that relates a particular type of stress to the
strain that results is called a modulus (plural, moduli).
Modulus = Stress/Strain
Stress-strain diagram of a material:
If a wire or a bar is subjected to gradually increasing stress and a graph
is plotted between the stress applied and the corresponding strain
produced. This curve follows the general characteristics:
•Up to the point A, the wire is perfectly elastic. Hooke’s law governing
the linear relationship between stress and strain is fully obeyed, straight
portion OA of the curve.
From A to B, stress and strain are not proportional, but if the load is
removed at any point, the material will return its original length. It is
elastic limit and the stress is elastic strength of the wire.
Beyond the point B, strain increase more rapidly, but not obey Hooke’s
law. The increases in length is now partly elastic and partly plastic.
Beyond the point D, this is represented by the wavy portion ED of the
curve. The stress corresponing to the point D is referred to as the
yielding stress.
Beyond the point E, the yielding comes to stop. Stress must be
increased for producing strain, which is now plastic in nature. The
cross-section of the wire decreases faster at some section of it and a
neck is developed there. This results in an automatic increase in
stress until a point G is reached at which fracture takes place, is
known as fracture point.
If the large plastic deformation takes place betn the elastic limit and
fracture point , the metal is said to be ductile or short deformation,
is said to be brittle.
The stress corresponding to the point G represents the maximum
stress to which the wire can be subjected and is measured by the
maximum load, is called the breaking stress. The maximum stress
within elastic limit is called working stress.
Factor of Safety: The ratio between the breaking stress and working
stress is called the factor of safety or factor of ignorance.
Different Types of Elasticity:
Specifying how stress and strain are to be measured, including directions, allows for
many types of elastic moduli to be defined. The three primary ones are:
Young's modulus (E) describes linear elasticity, or elasticity of
length, the tendency of an object to deform along an axis when
opposing forces are applied along that axis; it is defined as the ratio
of tensile stress to tensile strain within elastic limit. It is often
referred to simply as the elastic modulus.
If a force F, applied normally to a cross-sectional area A, produced a change
in length l in the original length L, then
Tensile stress = F/A and
Tensile strain = l/L
Young’s modulus, Y=tensile stress/tensile strain
= (F/A)/(l/L) = F.L/lA
The bulk modulus (K) describes volumetric elasticity, or the
tendency of an object to deform in all directions when
uniformly loaded in all directions; it is defined as
volumetric stress over volumetric strain, and is known as
incompressibility of the material (K).
If F be the force applied normally and uniformly on a surface area A, then
the
stress = force applied per unit area, pressure = P = F/A
Let v is the change in volume in an original volume V.
Then strain = -v/V, Minus indicates, if pressure increases volume
decreases. Hence Bulk modulus,
K= (F/A)/(-v/V) = -F.V/v.A = -PV/v
Compressibility of the material of the body is the reciprocal of bulk
modulus (1/K).
The shear modulus or modulus of rigidity (G or ) describes an
object's tendency to shear (the deformation of shape at constant
volume) when acted upon by opposing forces; it is defined as shear
stress over shear strain.
Modulus of rigidity: The modulus of rigidity of a material may
be defined as the tangential or shearing stress per shearing strain.
In this case, while there is no change in the volume of the body,
there is a change in the shape of the body.
Let the lower face of the rectangular solid cube be fixed. A tangential force
F be applied to its upper face The point shifts and the lines joining the two
faces turn through an angle θ.
The coefficient of rigidity or the modulus of rigidity, η
η = Tangential stress/tangential strain = (F/A)/ θ
Since θ is very small, tan θ = θ Thus tan θ =l/L, θ = l/L,
Hence η = (F/A)/(l/L) = F.L/l.A
Problems
1. A wire 8 m long and 1 mm in diameter, supports a
mass of 8 kgm. It is stretched by 1 mm. Calculate
the Young’s modulus of the wire.
2. A pressure of 84 Kpa decreases the volume of 2004 L
of water by 0.004 percent. Compute the
compressibility of water.
Poisson's ratio
On being stretch, a wire becomes longer but thinner, although
its length increases, its diameter or cross-section decreases.
Within the elastic limit, the ratio between the contraction or
lateral strain to the elongation or linear strain for a given
tensile stress is constant for a body of a given material, is
called the Poisson’s ratio for the material. Thus
Poisson’s ratio (σ)
=lateral strain/linear strain
= secondary strain/primary strain
where lateral strain means a strain in a direction at right angle
to the applied force, linear and lateral strains per unit stress are
denoted by α and β respectively. Thus
σ = β/α
Shear is equivalent to compression and extension
When a cube is sheared, its shape is altered but its volume as
well as its thickness remain unchanged.
ABCD in Fig. represents a section of a cube, the length of
whose side is l, when a force is applied to the upper side AD is
sheared through a small angle θ into new position A'D'.
Since shear is extremely small, then the extension and
compression may be expressed angle θ (= ˂ABA' = ˂DCD').
Let A'M and DL be perpendiculars on to AC and BD'
respectively.
As shear is very small, triangles are isosceles triangles so that
LD' = DD' cos45 = DD' / √2
Since AD= l then BD= BL= √(l²+l²) =l. √2
Extension along the diagonal BD
=LD'/BD = (DD'/√2). (1/l.√2) = DD'/2l
But DD'/l = DD'/CD = θ
But DD'/l = DD'/CD = θ
So extension along BD= θ/2
Similarly the compression along AC
= AM/AC = (AA' /√2).(1/AC) =(AA' /√2 .)(1/l.√2) = AA'/2l =
θ/2
Thus shear is equivalent to compression and extension, each of
value θ /2.
# Equivalence of a shearing stress to an equal tensile
and an equal compressive stress at right angles to one
another
Let a tangential force F be applied to the upper face AD and
the lower face BC is fixed. If l is the length of each side of the
cube, then the tangential or shearing stress to which the face
AD is subjected is F/l²= F/A.
If the cube were free to move, it try to moved along the
direction of the force F.
But the lower face is fixed, an equal and opposite force comes
into play on this face to form a couple F.l which tends to rotate
the cube clockwise.
But since the cube does not rotate, it means that there must be an equal and
opposite couple acting on it due to forces F acting along AB and CD. This
couple is also equal to F.l and tends to rotate the cube in an anticlockwise
direction to keep the cube in equilibrium.
Now the resultant of the forces F and F along AD and CD is F.√2 along
OD while the resultant of the forces F and F along AB and CB is F. √2
along OB where O is the point of intersection of the diagonals BD and AC
of the face ABCD.
Similarly, there is an equal but inward pull on diagonal AC, tending to
shorten its length.
Thus, a tangential force (F) applied to one face of the cube brings into play
a tensile force F. √2 along one diagonal and an equal compressive force F.
√2 along the other AC, perpendicular to it.
If the cube is now cut into two halves by a plane through diagonal
BD and perpendicular to the plane of the paper, then
the face of each half parallel to BD will have an area l.√2.l=
l².√2, with an outward force F.√2 acting on it.
The tensile stress acting along BD is, therefore, equal to F.
√2/l². √2 = F/I²= F/A.
Similarly the cube is cut into two halves along the diagonal
AC, a compressive force F.√2 will be found.
The compressive stress along AC will be F. √2/l². √2 = F/l²=
F/A.
Thus a tangential or a shearing stress is equivalent to an equal
tensile and an equal compressive stress at right angles to each
other.
# Relations among the elastic constants
Let a cube of unit edge, acted upon by a unit tension along one
edge. If α be the increase per unit length per unit tension along
the direction of the force. Then
Stress = force per unit area = 1
strain = α/1 = α, therefore,
Young’s modulus Y = stress/strain
= 1/α ----------------- (i)
Bulk modulus, K = stress/volume strain
= 1/3(α-2β) or ------------- (ii)
Compressibility, 1/K = 3(α-2β),
Hence, the modulus of rigidity,
η = 1/2(α+β) ------------------ (iii)
# Relations among the elastic constants
Let a cube of unit edge, acted upon by a unit tension along one
edge. If α be the increase per unit length per unit tension along
the direction of the force. Then
Stress = force per unit area = 1
strain = α/1 = α, therefore,
Young’s modulus Y = stress/strain
= 1/α ----------------- (i)
Bulk modulus, K = stress/volume strain
= 1/3(α-2β) or ------------- (ii)
Compressibility, 1/K = 3(α-2β),
Hence, the modulus of rigidity,
η = 1/2(α+β) ------------------ (iii)
# Prove that 1/K= 3(α-2β)
Let ABCDEFGH is a unit cube and let Tx, Ty and Tz be the
forces per unit area acting perpendicular to the faces BEHC
and AFGD, ABCD and EFGH, and ABEF and CHGD
respectively as Fig. Let α be the increase in length per unit
tension along the direction of the force and β is the contraction
produced per unit length per unit tension in the direction
perpendicular to the force.
The elongation produced in the edge AB due to the force Tx is
α.Tx
contractions produced by Ty Tz are β.Ty and β.Tz, Similarly for
BE and BC.
Thus the lengths of the edges become
AB=1+α.Tx-β.Ty-β.Tz
BE=1+α.Ty-β.Tz-β.Tx
BC=1+α.Tz-β.Tx-β.Ty
Hence the volume of the cube will now become
(1+α.Tx-β.Ty-β.Tz)(1+α.Ty-β.Tz-β.Tx)(1+α.Tz-β.Tx-β.Ty)
=1+(α-2β)(Tx+Ty+Tz)
If Tx=Ty=Tz=T
neglecting squares and products of α and β are very small
compared to other
The volume of the cube becomes
1+(α-2β)3T
The increase in volume is
1+(α-2β)3T-1= (α-2β)3T
Thus Bulk modulus, K= stress/volume strain
K=(T/1)/3T(α-2β)
K=1/3(α-2β)
1/K=3(α-2β)
# Prove that η = 1/2(α+β)
Let the top face ABHG, of a cube the length of whose side is
L, be sheared by a shearing force F, the lower face of the cube
being fixed. As a result of the shear A takes up the position A'
and B, the position B', the angle ADA' being equal angle to the
angle BCB' =θ As a result the diagonal DB is increased to DB'
and the diagonal AC shortened to A'C.
Then the stress=F/area of the face ABHG = F/L2 = T
BE=let the displacement AA' =BB' =l
then
shear strain =l/L=θ Thus the modulus of rigidity
=stress/strain= T/θ
The extension of the diagonal DB, due to tensile stress along DB
is α.T, and that due to compressive strain along AC is β.T where
α and β are the linear and lateral strain per unit stress
respectively. Therefore, total extension in length of the diagonal
DB is
DB×(α.T+β.T) =DB× T(α+β)
=L 2T(α+β) [DB=(L2+L2 ]
Let BE be the perpendicular dropped from B on DB' Then DB=
EB'.
Now B'BC
EB=EB'=BB'cos45=l(1/2)
Hence
L2T(α+β)= l/2,
LT/l =1/2(α+β),
T/(l/L) =1/2(α+β), T/θ = 1/2(α+β)
But the modulus of rigidity,  = stress/strain
=1/2(α+β)
# Relation among Y, K, and σ
Let a cube of unit edge, acted upon by a unit tension along one edge. If α be
the increase per unit length per unit tension along the direction of the
force and β is the laternal strain. Then
Young’s modulus,
Y = 1/α
----------------- (i)
Bulk modulus,
K = 1/3(α-2β) or ------------- (ii)
And, the modulus of rigidity, η = 1/2(α+β) ------------------ (iii)
We know,
Bulk modulus, K
= stress/volume strain
= 1/3(α-2β)
= 1/3α(1-2β/α)
= (1/α)/3(1-2.(β/α))
But σ = β/α and Y = 1/ α
So
K = Y/3(1-2σ)
or
Y = 3K (1-2σ)
Let a cube of unit edge, acted upon by a unit tension along one edge. If α be the
increase per unit length per unit tension along the direction of the force.
Then
Young’s modulus,
Y = 1/α
----------------- (i)
Bulk modulus,
K = 1/3(α-2β) or ------------- (ii)
And, the modulus of rigidity, η = 1/2(α+β) ------------------ (iii)
Multiplying (iii) by 2
2 η = 1/(α+β)
1/η = 2α+2β -------------(iv)
From rearranging relation (ii)
1/3K = α-2β --------------(v)
Adding eqns. (iv) and (v),
3α =(1/η) + (1/3K) = 3K+η/3Kη
Or α = 3K+η/9Kη But α = 1/Y
Hence, 1/Y = 3K+η/9Kη
9/Y = 3K+η/Kη
= 3K/Kη +η/Kη
= 3/η +1/K, This is the relation connecting the three elastic constant.
H.W
1. Relation among Y, η, and σ ( Y=2(1+σ))
2. Relation among K, η and σ ( σ=3K-2/(6K+2))
# Work done in a strain
The word done is stored up in the body in the form of potential
energy which may be called elastic potential energy, energy of
strain or strain energy.
(i) Longitudinal strain:
Let F be the stretching force applied to a vertical wire, fixed at
the upper end by supporting a load at its lower end. If l is the
resulting stretch produced in the wire whose original length L
and area of cross-section is A then
Y = (F/A)/(l/L)= F.L/l.A
or F = YlA/L
where Y – Young’s modulus
If there is an additional small increase dl in the length of the
wire, then the word done
F.dl = YAl.dl /L
Hence, the total work done, W =
l
l
YAl
1 YAl 2
0 F .dl  0 L .dl  2 L
1  YAl 
1
 
.l  F .l
2 L 
2
= ½ (stretching force X stretch or elongation)
Now the volume of the wire of length L
and cross-sectional area, A = L x A
Strain energy per unit volume
= ½ (F.l)/ (L.A)
= ½ (F/A) (l/L)
= ½ stress x strain

1 mg
l
x
2 r 2
L
where
So
the
F  mg
strain
and
A  r 2
energy
per unit volume
1 YAl l
.
2 LA L
1
l 2

Y( )
2
L

= ½ Young’s modulus X (longitudinal strain)2
# Volume strain:
Let p be the stress which is force per unit area or pressure
which is applied normally over an area A of a body of
volume V such that its volume decreases by v. Then the
bulk modulus, K ,
v
K  p /(  )
V
K .v
p
V
Omitting the minus sing.
If dx is a small movement in the direction of p, then the
work done
= p.A.dx
=p.dv
Hence W, the total work done,
v
W 

0
v
p.dv 

0
Kv
K 1 2
.dv 
. v
V
V 2
1 Kv
.v
2 V
1
1
 p.v  stress
2
2

x change in volume
Hence, work done per unit volume
1 pv 1 v
 p.
2 V
2 V
1
 stress x strain
2

Again strain per unit volume
1 Kv 2
1
v
1

 V  K .( ) 2  Bulk mod ulus X ( volume strain) 2
2 V
2
V
2
Limiting values of Poisson’s ratio (σ)
We know
Y  3K (1  2 ) and Y  2 (1   )
Hence 2 (1   )  3K (1  2 )
3K
1 

2 1  2
Both
3K
is positive therefore
2
1 
is also positive.
1  2
if  is a positive quantity, as K and  are always positive,
1
and
2
if  is a negative quantity, for  and K to be positive,
1  2  0 or  
1    0 or   1
3K
will be positive only when  lies
2
1
between 1 and
2
Hence
2.
Show that a volume strain is equivalent to three mutually
perpendicular linear strains, each equal in magnitude to one
third of the volume strain.
Soln: Let a cube of unit edge and compressed uniformly and equally in all
directions.
Then, the volume strain = decrease in volume/original volume = V/1 and
the linear strain along each edge = decrease in length/original length =
x/1
Since the length of each edge decreases by x,
the reduced length of each edge = 1-x,
hence the reduced volume = (1-x)3 therefore,
the decrease in volume, V = 1- (1-x)3= 1-(1+3x2-3x+x3) =3x
(neglecting higher power of x)
thus x = V/3
Thus a volume strain is equivalent to three linear strains, each equal to
V/3
Limiting values of Poisson’s ratio (σ)
We know
Y  3K (1  2 ) and Y  2 (1   )
Hence 2 (1   )  3K (1  2 )
3K
1 

2 1  2
Both
3K
is positive therefore
2
1 
is also positive.
1  2
if  is a positive quantity, as K and  are always positive,
1
and
2
if  is a negative quantity, for  and K to be positive,
1  2  0 or  
1    0 or   1
3K
will be positive only when  lies
2
1
between 1 and
2
Hence
2.
Show that a volume strain is equivalent to three mutually
perpendicular linear strains, each equal in magnitude to one
third of the volume strain.
Soln: Let a cube of unit edge and compressed uniformly and equally in all
directions.
Then, the volume strain = decrease in volume/original volume = V/1 and
the linear strain along each edge = decrease in length/original length =
x/1
Since the length of each edge decreases by x,
the reduced length of each edge = 1-x,
hence the reduced volume = (1-x)3 therefore,
the decrease in volume, V = 1- (1-x)3= 1-(1+3x2-3x+x3) =3x
(neglecting higher power of x)
thus x = V/3
Thus a volume strain is equivalent to three linear strains, each equal to
V/3
# Deformation by bending
A rod or a bar of uniform cross-section, circular or rectangular, the length
of which is very much greater than its thickness is called a beam. The
shearing stresses set up over any section of such a structure are negligibly
small.
When a beam is fixed at one end and a load is applied at the other end, it
bends due to the moment of the applied force. The plane of bending is the
same in which the couple acts. The beam, so bent from its normal shape,
will of course return to its original shape once the deforming forces are
withdrawn provided it has not been strained beyond the elastic limit.
When the bar AB is supported at two points equidistant from its middle
point C. A downward force p=mg is applied at C, the bar bends down at the
middle, the amount of depression is proportional to P in accordance with
Hooke’s law.
Let a small portion of the bend beam along the neutral axis
(the line of intersection of the neutral layer and the plane of
the applied couple is known as the neutral axis) form the arc
mn of a circle of radius R and let it subtend an angle θ at the
centre of curvature O. Let m'n' be the length of an element at a
distance x from the neutral axis.
In the unstrained condition, mn=m'n'.
Now
mn= Rθ and m'n'=(R+x)θ. So increase in length,
m'n'-mn=(R+x)θ-Rθ= xθ
strain= xθ/Rθ = x/R
Thus the strain in a layer is directly proportional to its distance
from the neutral axis.
Bending moment:
A bending moment is a measure of the average internal
stress induced in a structural element when an external
force or moment is applied to the element causing the
element to bend.
The internal stresses in a cross-section of a structural
element can be resolved into a resultant force and a
resultant couple, the resultant internal couple is called the
bending moment.
Let us consider a small element ABCD of the beam bent in the form
of an arc of a circle of radius R. The neutral surface EF subtends an
angle θ at the centre of curvature O. where EF=Rθ.
Let EJ be drawn parallel to FB cutting GH at K. Now GH is the length
of the stretched filament HK lying at a distance x above the neutral
surface so that KH=EF=Z is the outstretched length. Then
tensile strain= increase in length/original length= dz/Z.
Let α be the area of cross-section of the filament, then the
tensile stress = P/α where P is the magnitude of the
applied force. Now the Young’s modulus
Y=tensile stress/tensile strain = (P/α)/(dz/Z),
Or tensile stress, P/α=Y.(dz/Z)
= young’s modulus X strain
=Y.(xθ/Rθ) = Yx/R
Or
P = Yxα/R= stress(Yx/R) X area(x)
Then the moment of this force about EF
= P.x =(Yxα/R).x= Yx2α/R
The total moment of the forces,
= ∑ Yx2α/R
=Y/R ∑ x2α
M = YIg/R
The total moment of the forces,
M = YIg/R
The bending moment obviously balances the moment of the
external force acting at the section.
Flexural rigidity: It is defined as the external bending
moment required to produce a unit radius of curvature. Or
the quantity YIg measures the resistance of the beam to
bending and is called the flexural rigidity of the beam.
Thus the bending moment of the beam
= flexural rigidity/ R