Properties of matter
Subhasish Chandra
Assistant Professor
Institute of Forensic Science, Nagpur
Elasticity
2Concepts of Physics Part 1 by H. C. Verma
Rigid Body: A body is said to be rigid if a force applied to it produces
no or negligible change in its shape and size.
Elastic Body: If a body completely regains its original shape and size
on the removal of the deforming forces, the body is said
to be perfectly elastic. This property of the body is called
elasticity.
Plastic Body: If a body completely retains its altered shape and size on
the removal of the deforming forces, the body is said to
be perfectly plastic.
Isotropic Body: If the material of which the body is composed has same
physical properties in every direction, the body is said to
be isotropic.
Definition
3
When external forces acting on a body produce some deformation of thebody, internal restoring forces between adjacent molecules on either side ofa section through the body are set up in it.
These forces tend to restore the body to its original shape and size. Whenevery portion of the body is in equilibrium, the internal restoring force perunit area is called the stress.
Stress is defined as
Stress = Component of Force normal to Area
Area on which the force acts
Stress = F / A
Dimension: [ML-1T-2] Unit: N/m2
Stress is not a scalar quantity. It is neither a vector quantity as no specificdirection can be assigned to it.
For an isotropic body, stress can be resolved into six components. Hence, itis a symmetric tensor. There are three types of stress
▪ Tensile or Longitudinal Stress
▪ Volume or Bulk Stress
▪ Tangential or Shear Stress StressFundamentals of Physics by Halliday, Resnick, Walker
Elasticity
4
External forces acting on a body produce a change in its shape and
size. The fractional change is called strain.
There are three types of strain
▪ Tensile or Longitudinal Strain
It is produced by applying equal and opposite forces along the
length of deformation. It is given as δL/L.
▪ Bulk or Volume Strain
It is produced by uniform increase of pressure on the surface of
the body. It is given as -δV/V.
▪ Shear or Tangential Strain
It is produced by tangential force acting on the surface of a
body, the opposite end being fixed. It is the angular deformation
θ.Strain
Elasticity
Concepts of Physics Part 1 by H. C. Verma
5
For stresses within the elastic limit of a material, the strain
produced is directly proportional to the stress applied,
This constant has a definite value for any given material and has
the units and dimensions of stress.
Hooke’s Law
elasticity of modulusstrain
stress
constantstrain
stress
strainstress
Elasticity
Fundamentals of Physics by Halliday, Resnick, Walker
6
Young’s Modulus
Bulk Modulus
Modulus of Rigidity
Moduli of Elasticity
LA
FL
LL
AF
Strain alLongitudin
StressalLongitudinY
V
PV
VV
P
Strain Volume
StressVolumeK
A
FAF
Strain angentialT
StressTangential
Concepts of Physics Part 1 by H. C. Verma
Elasticity
7
It is a commonly observed fact that when we stretch a string or a
wire, it becomes longer but thinner.
A longitudinal strain produced in the wire is accompanied by a
transverse or lateral strain of an opposite kind in a direction at
right angles to the direction of the applied force.
Within elastic limit, the lateral strain is proportional to the
longitudinal strain.
Poisson’s Ratio
strain allongitudin
strain lateral
stressallongitudin
strainlateral
stressallongitudin
strainallongitudin
1Y
Fundamentals of Physics by Halliday, Resnick, Walker
Elasticity
8
Stretching of Wire
Elasticity
Concepts of Physics Part 1 by H. C. Verma
9
Suppose a wire of length L and
cross-sectional area A is
stretched by an external force F.
If the extension produced in the
wire is x,
Work done is given by,
Elastic Potential
EnergyPhysics for Degree Students by C. L. Arora, P. S. Hemne
xL
AYF
xA
FLY
2
0
0
2l
L
AYW
xdxL
AYW
FdxW
FdxdW
l
l
This work is stored into the wire as
its elastic potential energy.
PE = (1/2)(force) (extension)
PE = (1/2)(stress) (strain) (volume)
l L
lAYU
lL
AYU
2
1
2
2
ALL
l
L
lYU
2
1
Elasticity
10
Relationship Corner
Elasticity Y, K and σ Y, η and σ
213
1213
1)2(3
)2(3
)2(3
KY
K
K
P
PK
strainVolume
stressVolumeK
P strainVolume
12
112
1)(2
)(2
)(2
Y
P
P
strainhearS
stresshearS
P strainearSh
Elements of Properties of Matter by D. S. Mathur
11
Relationship Corner
Elasticity Y, K and η
On Adding
K, η and σ
22
213
12
213
Y
K
Y
Y
KY
26
23
2
2
6
3
12
)21(3
K
K
Y
K
YK
Y
KY
319
33
KY
Y
K
Y
Physics for Degree Students by C. L. Arora, P. S. Hemne
12
Relationship Corner
Elasticity Maximum value of σ
This shows that σ will be
maximum, when Y/6K is
minimum.
Since Y and K are both positive,
the minimum value of Y/6Kcannot be less than zero.
maximum value of σ = 0.5
Minimum value of σ
This shows that σ will be
minimum, when Y/2η is
maximum.
Since Y and η are both positive,
the minimum value of Y/2η
cannot be less than zero.minimum value of σ = – 1
K
Y
KY
62
1
213
12
12
Y
Y
Elements of Properties of Matter by D. S. Mathur
13
Bending of Beams
Elasticity
A rod of uniform rectangular or
circular cross-section whose length
is much greater as compared to its
thickness is called a beam.
A beam can be considered to be
made up of thin plane layers
parallel to each other. Each layer is
considered to be made up of a
number of longitudinal filaments.
During the bending of the beam,
the layers of the material in the
upper half are extended and those
in the lower half are compressed.▪ There is one layer in the middle,
whose length remains unchanged.
This is called Neutral Surface.
▪ The filament in the neutral surface,
which passes through the centres of
cross-sections, is called the Neutral
Axis or neutral filament.
▪ The plane containing the neutral
axis and the centre of curvature is
called the Plane of Bending.
Physics for Degree Students by C. L. Arora, P. S. Hemne
14
Elasticity
The bending of a beam ofuniform cross-section clampedhorizontally at P and loaded atthe other end Q with a mass M isshown in the plane of bending.
The external torque producesinternal longitudinal forcesacting along the filaments at thecross-section.
.
Let R be the radius of curvature,
SO, of the neutral axis at S
KL be the length of the filament
in extension
y be the distance of KL above
the neutral axis
dθ be the angle subtended at the
centre of curvature O by KL
dA be an element of area of
cross-section of the filament at
K
dF be the longitudinal force on
the filament at K
Elements of Properties of Matter by D. S. Mathur
Bending of Beams
15
Elasticity
Longitudinal strain along KL is,
Longitudinal stress is, dF/dA.
Young’s modulus, Y, is
Longitudinal extensional forces above the
neutral surface and compressional forces
below the neutral surface form a system of
anti-clockwise couples.
The resultant anti-clockwise moment of the
couples about the line of intersection of the
neutral surface with cross section is called
the Internal Bending Moment (IBM).
The moment of the force is,
The resultant moment of all such forces
acting over the entire cross-section is the
internal bending moment. Hence,
But, ∫y2dA=I, where 𝐼 is the geometric
moment of inertia.
R
y
Rd
RddyR
ST
STKL
ydAR
YdF
Ry
dAdFY
dAyR
YydA
R
YyydF 2
dAyR
YydFIBM 2
IR
YIBM
Bending of Beams
Physics for Degree Students by C. L. Arora, P. S. Hemne
16
Elasticity
External forces acting on SQ part of therod are
▪ Weight Mg acting verticallydownwards at Q.
▪ Weight mg of the part SQ actingdownwards through the centre ofgravity G of the part SQ.
The resultant gives rise to an equal andopposite reaction at the cross-section Sforming a clockwise couple. The momentof the couple is called the ExternalBending Moment (EBM).
Taking moment of Mg and mg about
the line of intersection of the neutral
surface with the cross-section at S, we
get external bending moment
l is the length PQ of the rod and x is
the distance of the cross-section from
P.
Elements of Properties of Matter by D. S. Mathur
2
SQmgSQMgEBM
GSmgSQMgEBM
2
xlmgxlMgEBM
Bending of Beams
17
ElasticityThe external bending moment sets up
the internal forces. Therefore for
equilibrium of the part SQ of the beam,
If the beam is light, mg can be
neglected as compared to Mg. Then
xlMgIR
Y
2
xlmgxlMgI
R
Y
EBMIBM
Bending of Beams
Physics for Degree Students by C. L. Arora, P. S. Hemne
18
Cantilever
Elasticity
Let PQ be a long beam of uniform cross-
section clamped horizontally at one end P
and loaded with a mass M at the other end
Q. Let l be the length of the beam exposed
from the support. PQ be the original
position of the neutral axis and PQ’ be the
position when the beam is loaded. Let C be
the centre of curvature and R be the radius
of curvature of the neutral surface at C. The
beam is assumed to be light and hence
weight of the part CQ is neglected as
compared to the load. The depression
produced is δ.
For equilibrium of the cross-section,
The total depression is assumed to be
small; hence radius of curvature is given
as
On integrating with respect to 𝑥,
K1is a constant.
At x = 0, the tangent to the neutral axis is
horizontal. Hence,𝑑𝑦
𝑑𝑥= 0 and K1 = 0.
Elements of Properties of Matter by D. S. Mathur
xlMgIR
Y
xlYI
Mg
Rdx
yd
12
2
1
2
2K
xlx
YI
Mg
dx
dy
2
2x
lxYI
Mg
dx
dy
19
Cantilever
Elasticity
To obtain depression, y, of the neutral
axis, we integrate
K2 is a constant.
At x = 0, y = 0, K2 = 0
2
32
62K
xlx
YI
Mgy
62
32 xlx
YI
Mgy
The depression of the neutral axis at
any point (x,y)
Let δ be the depression at the loaded
end Q. Substituting x = l and y = δ,
This is true for long thin beams.
xlYI
Mgxy 3
6
2
YI
Mgl
3
3
Physics for Degree Students by C. L. Arora, P. S. Hemne
20
Cantilever
Elasticity Rectangular Cross Section
For a beam of breadth b and
thickness d, the geometric moment
of inertia is given as 𝐼 =𝑏𝑑3
12
Circular Cross Section
For a beam radius r , the geometric
moment of inertia is given as 𝐼 =𝜋𝑟4
12
Elements of Properties of Matter by D. S. Mathur
Beam Supported at Both
Ends
The reaction at each end is Mg/2 acting
vertically upwards.
3
34
Ybd
Mgl
4
3
3
4
rY
Mgl
YI
Mgl
YI
lMg
483
223
3
21
Elasticity
It consists of a regular solid body suchas a circular disc suspended at thecentre of its cross-section by means ofa metal wire. The upper end of the wireis fixed to a rigid support.
The disc is rotated through a smallangle about the axis OO’. These arecalled torsional oscillations.
If the angular displacement is withinthe elastic limit of the material of thewire, the oscillations will be simpleharmonic.
When the pendulum is rotated through an
angle by an external torque, a restoring
torque is produced in the wire due to twist
of the lower end.
When the external torque is removed, the
restoring torque produces angular
acceleration tending to bring back the body
to the initial equilibrium position.
If θ radian is the instantaneous angular
displacement, the restoring torque is given
by
C is the torsional constant
r: Radius of Wire
η: Modulus of Rigidity
l: Length of Wire
C
Torsional Pendulum
l
rC
2
4
Physics for Degree Students by C. L. Arora, P. S. Hemne
22
Elasticity
For the body of moment of inertia 𝐼, the torque is given by
The motion is angular simple harmonic
as the angular acceleration is directly
proportional to the angular
displacement from the equilibrium
position and is directed towards that
position.
Elements of Properties of Matter by D. S. Mathur
Torsional Pendulum
ICT
2
I
C
dt
d
Cdt
dI
dt
dI
2
2
2
2
2
24
22
r
IlT
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