properties of matter - subhasish chandra

Properties of matter

Subhasish Chandra

Assistant Professor

Institute of Forensic Science, Nagpur

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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


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



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



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.



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


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



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.


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



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.


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



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


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



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.


Torsional Pendulum

ICT

2

I

C

dt

d

Cdt

dI

dt

dI

2

2

2

2

2

24

22

r

IlT


properties of matter - subhasish chandra

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