31 pius augustine elasticity (b.sc)

• https://www.youtube.com/watch?v=j_cXxZ9w7T8
Mechanics of solids
Dr. Pius Augustine, SH College, Kochi

Material property play a
vital role…..
a greater lift force means that
the wings will bend more…

Inter atomic forces
Electrical interaction
between atoms
Short range nearly 10-10m.
(radius)
If distance between atoms is much greater
than range, no interaction

As the atoms come towards each other force is
attractive (-ve) and very close repulsive (+ve)
Inter atomic forces
Interatomic potential energy
between two identical atoms
as a function of distance
between their nuclei

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Force vs distance between atoms

Discussion on graphs
i. F = 0, when R tends to infinity
ii. As R decreases F is attractive
iii. At R = r0, F reduces to zero
iv. If attraction continue at r0, atoms
would coalesce. To prevent this,
repulsive force come into play

Discussion on graphs
v. Potential energy is min at r0 (max –ve), is
stable eqbm (F = 0)
vi. r0 is called normal or eqbm distance

Solids
Definite arrangement of molecules and resultant force on
each molecule is zero.
When deformed – molecules are either pulled apart or
pushed close to each other – equilibrium is disturbed.
(Refer graph) if inter molecular separation increase –
attractive force increase and atoms will be brought back to
equilibrium.
If compressed, intermolecular force of repulsion brings back
to equilibrium.

Liquids
No definite arrangement of molecules inside.
However, forces will resist any change in
equilibrium separation - hence posses
definite volume
Liquids and gases donot resist changes in their
shapes.

Rigid body
Not deformed under the action of various forces.
Distance between molecules remains fixed even
during motion
Note:
Rigid body is an idealization: donot bend, stretch, or
squash when forces act on them.
All real materials are elastic and do deform to some
extent.
Elastic properties are tremendously important.

Elasticity
Deforming force: which changes shape and size
Restoring force: develop inside body,
tries to bring back to initial condition,
due to cohesive force.
Elasticity: property of a body by virtue of which the
body regains its original shape and size when the
deforming force is removed.

Elastic energy
When body regains its original state, force
of elasticity acts through a distance.
Work done by elastic force, stored as Pot.
Energy.
Eg. Compressed steam issuing out of a
boiler possesses elastic energy.

Perfectly elastic body
Regains immediately and perfectly on removal
of deforming force.
Ideal case
Nothing in practice
Eg. Quartz, phosphor bronze, steel, glass, ivory
etc.

Perfectly plastic body
Does not show even tendency to
regain, on removal of deforming
force.
Ideal
Nothing in practice
Eg. Putty , wax, mud , plasticine etc.

Stress???
For different kinds of deformation, stress is
a quantity characterizes the strength of
the forces causing the deformation.

Compressional stress

Torsional stress
Bottom - fixed

Shear stress

Stress
Internal restoring force acting per unit area
of a deformed body
Unit - dyne/ cm2 or N/m2
Neglecting plastic behavior of body
Deforming force = Restoring force

Tensile (longitudinal)stress: over unit
cross sectional area when length of body
increases in the direction of deforming
force
Compressional stress: length decreases
Normal stress: Deforming force acts
normal to area

Tangential (shearing) stress: force
tangentially over an area. Body
sheared.
If the body is subjected to a uniform
and equal force from all sides, then
the stress is called hydrostatic stress.

Strain
Deforming force changes length, volume or
shape of the body. (said to be strained)
Strain produced is measured as the ratio
of change in configuration to the original
configuration of the body.
No unit, no dimension.

Tensile or longitudinal or linear strain: within elastic
limit ratio of change in length to original length
Bulk or volumetric strain: ∆v/v
Shear strain: ratio of relative displmt.of one plane to
its distance from the fixed plane (tanθ)
Relative displacement of two parallel planes unit
apart

Relation between volume and linear strains
Consider a homogeneous cubic material – side unity
Volume V1 = 1 m3. (unit volume)
Cube is allowed to have uniform dialation by a small amount x
(ie. increase in length per unit original length = x)
ie. Linear strain = x
New Volume V2 = (1+x)3 = 13 + 3x (neglecting higher powers)
Increase in volume = 1 + 3x – 1 = 3x This is increase in
volume/unit volume = 3 times linear strain

Show that shear strain = extensional strain +
compressional strain.
Locked
F
45o
A A’ B B’
C D
θ
P
x
Face of the cube ABCD
Tangential force F
CD is fixed
AA’ = x (very small)
Without much error, Angle PAA’ = 45o = Angle AA’P
Angle APA’ = 90o. Similarly Angle BB’Q = 45o
L
Q
x

Show that shear strain = extensional strain +
compressional strain.
Locked
F
45o
A A’ B B’
C D
θ
P
x
Extensional Strain = B’Q/BC = (x/L) (1/√2)2.
= x/2L = ½ (x/L) = θ/2
Similarly along DA compressional strain = θ/2
Shear strain θ = compressional strain + extensional strain
L
Q
∆BB’Q Cos45 = B’Q/x
∆DBC Cos45 = BD/BC = L/BC
B’Q = x Cos45 BC = L/ Cos45

Hooke’s law
“ within elastic limit, the stress is directly
proportional to strain”
Stress / strain = const.
Constant is known as modulus of elasticity.
3 types. i. Young’s modulus (Y)
ii. Bulk modulus (K)
iii. Shear modulus (η)

Stress – Strain graph
Note: Hooke’s law is not really a general law but an experimental
finding that is valid only over a limited range.
Horizontal scale is not uniform. Strain corresponds to OA is
very small.

Young’s modulus of elasticity
Y = longitudinal stress / longitudinal strain
Y = (F/A)/(e/L) = FL/Ae
A= 1, L =1, e = 1, Y = F
Young’s modulus of elasticity is defined as the force
required to extend a wire of unit length and unit
area of cross section, through unity
Or it is defined as the longitudinal stress
required to double the length of the wire.Dr. Pius Augustine, SH College, Kochi

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Young’s modulus and slope of F-r graph
For extremely small change in interatomic
separtaion, the F-r curve is just a straight
line on both sided of the equilibrium (r0)
The force ∆F that appears on distortion is proportional to stress and (r-r0).
ie. ∆r is proportional to the strain.
Hence slope (∆F/∆r) at r = r0 is proportional to the Y of the material.
ie. Larger the interatomic bonding, greater is the slope and
higher will be Y Dr. Pius Augustine, SH College, Kochi

If a stress of 1 kg/mm2 is applied to a
wire, what will be the percentage increase
in its length? E = 1.0 x 1011 Pa.
Hint: 1 kg/mm2 = 1 kgf/mm2
= 9.8 N/10-6 m2.

A heavy wire is suspended from a roof but
no weight is attached to its lower end. Is it
under stress ?
It is under stress, because the
weight of the heavy wire acts as
the deforming force

What are elastomers ?
Elastic substances which can be subjected
to a large value of strain, are called
elastomers.

What is elastic fatigue?
Loss in the strength of a material
caused due to repeated
alternating strains to which the
material is subjected.
Eg. In torsion pendulum, rate of
decay of torsional oscillations is
greater than fresh wire.

Two identical discs (A and B) suspended by
identical wires
A is given tortional vibrations daily
After a few days both of them are given
vibrations
Vibrations of A die out soon
Lord Kelvin suggested that wire of A is tired or
fatigued.

A hand wire is broken by bending it
repeatedly in opposite direction. Why?
Due to elastic fatigue
Loss in strength of the material due to
repeated alternating strains, to which the
wire is subjected.

Why a spring balance does not give correct
measurement, when it has been used for a
long time?
Due to elastic fatigue
Takes longer time to recover original
configuration
So does not show correct measurement

Why are bridges declared unsafe after
a long use ?
Loses its elastic strength due to
repeated alternating strains.

What do you mean by elastic after effect ?
Twisted quartz fibre regains immediately, where as
glass fibre will take hours to return to original state
after the removal of torque.
This delay in regaining original state after the removal
of the deforming force is called elastic after effect.
Eg: In galvanometers quartz or phosphor bronze is
used as the elastic after effect is negligible.

What is the basis of deciding thickness of metallic ropes
used in cranes to lift heavy weights?
Elastic limit and factor of safety
Let, load to be lifted is 104kg
Taking factor of safety = 10kg, total load = 105kg
Maximum stress = (105 x 9.8) / πr2
Maximum stress must not exceed elastic limit of steel (20 x
107 N/m2)
Equating and solving, r = 33.2 cm
For flexibility, number of wires are twisted

Maximum height of a mountain
At the base of mountain
P = hdg (d= 3*103kg/m3)
Pressure at the bottom must be less than elastic
limit of the rock supporting the mountain
(= 3 * 108 N/m2)
hdg< 3 * 108 Solving, h<10km
This h is nearly equal to height of mount everest.

What is meant by anisotropic nature of
crystalline solid?
The physical properties like thermal
conductivity, electrical conductivity,
compressibility etc have different values
in different directions.

What are tensor physical quantities ?
Physical quantities having different values
in different directions are called tensors
Eg. Stress, moment of inertia

What is elastic limit ?
It is the maximum stress on, whose
removal, the bodies regain their
original dimension

Do liquids possess rigidity ?
No
Because they have no shapes of
their own

A B
C
D
Slope of region OA gives Y

Stress – strain graph explanation
Region OA - elastic region. If stress is removed any
where in between, regains original condition. Obey
Hooke’s law. A is proportional or elastic limit.
Beyond elastic region is called plastic region.
A to B, proportionality does not hold.
AB is curved and takes a slight bend. Slight increase in
stress produces larger strain. Point B is yield point. If
stress is removed in the region, does not regain
original. Permanent set.

stress – strain graph explanation
A to B – doesnot obey Hooke’s law.
O to B – if the load is gradually removed a curve
is traced to regain original length.
ie. OB is reversible region and forces are
conservative.
Energy put to cause deformation is recovered
when stress is removed.

stress – strain graph explanation
BC region more stress than in the region AB is
required to produce same strain. Continue
upto C called ultimate stress.
Along CD strain increases rapidly and stress
decreases. Continue to breaking point.

Ductile : For some material a large amount of
plastic deformation takes place between the
elastic limit and the fracture point. They are
ductile materials. Hence can be drawn in to
wires .
Eg: Cu, Al, soft iron wire

Brittle : Break as soon as the stress is increased
beyond elastic limit. Cannot be drawn into
wires.
Eg: glass, ceramics, steel piano string.
Malleable: hammered into thin sheet .
Eg: Au, Ag,Pb etc

Vulcanized rubber can be stretched more than 7 times
original length by stress.
Stress is not proportional to strain.
But regains original length if stress is removed.
It follows different curve for increasing and decreasing
stress – behavior is called elastic hysteresis.
Forces is non conservative – friction involved.

Elastic hysteresis : when stress is applied on vulcanized rubber,
no portion of the curve obeys Hooke’s law. But it regains original
condition on removal of stress.
Non coincidence of curves for increasing and decreasing stress is
known as elastic hysterisis.
Area of hysterisis loop = energy dissipated Dr. Pius Augustine, SH College, Kochi

Stress required to cause actual fracture of a material is
called breaking stress or ultimate strength or tensile
strength.
Two materials – say two types of steel may have very
similar elastic constants but vastly different breaking
stresses

While parking your car on a crowded street, you accidentally back
into a steel post. You pull forward unitl the car no longer touches
the post and then get out to inspect the damage. What does your
rear bumper look like if the strain in the impact was i) less than at
the proportional limit, ii) greater than at the proportional limit but
less than yield point iii) greater than at the yield point, but less
than at the fracture point or iv) greater than at the fracture point?
Ans: i) and ii) bumper would regain original shape
iii) Permanent dent can be expected
iv) Bumper will be torn or broken. Dr. Pius Augustine, SH College, Kochi

Area within stress strain graph is energy density
Slope of the graph is elasticity of the material

Work done in stretching a wire
Wire of length ‘L’ and cross – section A stretched by
a force of F acting along its length.
Y = FL / Ax or F = YAx/L
To provide additional extension of dx
dW = F dx = (YAx /L)dx

Work done in stretching a wire
Work done in stretching an amount l
Intergrating over limit 0 to l
W = ½ stretching force x extension = ½ F e
This work is stored and called strain energy
W/ unit volume = energy density = (½ Fe)/Ae
= ½ stress x strain.

If an object is immersed in a fluid (liquid or gas) at
rest, the fluid exerts a force on any part of the
object’s surface, which is perpendicular to the
surface.
This perpendicular force per unit area is called
Pressure.
Pressure plays the role of stress in volume

Bulk modulus of elasticity
Ratio of volume stress to volume strain
B = -dP/ (dV /V)
-ve sign – as pressure increases, volume decreases
B is a positive quantity.
Bulk modulus of a gas can be isothermal (P)or adiabatic
(γP)
Compressibility is reciprocal of B - unit is Pa-1
Compressibility is the fractional decrease in volume
produced per unit increase in pressure

Work done/unit volume in volume strain
dl
Let V be the original volume of a gaseous
system, which is compressed by v
It is a gradual change and let dv is a small
fractional decrease in the process (say when r
decrease by dl) A – surface area of the body
dW = F dl P A dl = PdV
Total work done W = ∫ PdV = ∫ PdV = ∫ (K/V) v dV
= (K/V) v2/2 = ½ (Kv/V) x v = ½ P x v
Bulk modulus K = P/(v/V)
0
V
0
V
0
V

Work done/unit volume in volume strain
dl
Bulk modulus K = P/(v/V)
Total work done in compressing the volume V
by v under the application of pressure P
W = (K/V) v2/2 = ½ (Kv/V) x v = ½ P x v
= ½ x applied pressure x reduction in volume
Work done to cause unit reduction in volume
W/V = ½ (Kv/V) x v /V = ½ stress x strain

Compressibility of few substances
1. Mercury 3.7 x 10-11 Pa-1 = 3.8 x 10-6 atm-1
2. Glycerine 21 x 10-11 Pa-1 = 21 x 10-6 atm-1
3. Water 45.8 x 10-11 Pa-1 = 46.4 x 10-6 atm-1
With each atmosphere increase in pressure, the volume
of water decrease by 46.4 parts per million.
Materials with small bulk modulus and large
compressibility are easier to compress. Dr. Pius Augustine, SH College, Kochi

A hydraulic press contains 0.25m3 (250 L) of oil. Find
the decrease in volume of the oil when it is subjected
to a pressure increase ∆P = 1.6 x 107 Pa (about 160
mm or 2300 psi). The bulk modulus of the oil is B = 5 x
109Pa (aprox 5 x 104 atm), and its compressibility is 20
x 10-6 atm-1.
∆V = - K V ∆P = - 8 x 10-4 m3.

Isothermal and Adiabatic Elasticity of Ideal gas
Isothermal PV = RT, PdV + VdP = 0
P = - V(dP/dV) = -dP * (dV/V) = Bulk Stress(dP)/Volume strain
Isothermal elasticity = Pressure.
Adiabatic PVγ = constant
Vγ dP + PγVγ-1 dV = 0
dP/dV = -γ P/V
-dP * (dV/V) = γ P
Adiabatic elasticity = γ times isothermal elasticity

What will be the density of lead under a pressure of 2
x 108 Pa? Density of lead = 11.4 x 103 kg/m3. K = 8 GPa
Hint: V2 < V1
(V1-V2)/V1 = ∆P/K = 1/40
So, V2/V1 = 39/40
Mass const = ρ1V1 = ρ2V2
ρ2 - ? Ans: 11.69 x 103 kg/m3.

Rigidity modulus or shear modulus η
Ratio of shearing stress to shearing strain within
elastic limit
G or η = F/Aθ
Strain
= Tanθ ≈ θ = x/h
Note: shear modulus apply only to solid materials.
Gases and liquids donot have definite shape
dx
x
L

Poisson’s ratio (σ)
It is the ratio of lateral strain to longitudinal strain within
elastic limit.
Wire of length l and diameter x
Longitudinal strain α = dl/l
Lateral strain β = dx/x
Poisson’s ratio σ = β/α
Theoretical value of σ lies b/w 1 and 0.5
For most of materials it is b/w 0.2 and 0.4

Show that maximum value of Poisson’s
ratio = 0.5
Differentiating equation for volume v = πD2L/4
dV = π/4 (D2 x dL + L x 2D x dD)
For max: dV = 0
Solving –(dD L / dLD) = -1/2
-σ = -1/2 = - 0.5

Work done/unit volume in shearing strain
Shear modulus η = F/Aθ
= FL/Ax = FL/L2x = F/Lx
F = η Lx
Work done for small displacement
dW = F dx
Total work done
W = ∫ F dx = η L ∫ x dx = F L x2 = ½ F x
Lx 2
Work done/unit volume = ½ F x = ½ (F/L2) (x/L)
L3
= ½ stress x strain
0
x
0
x

Work done/unit volume in shearing strain
Shear modulus η = F/Aθ
= FL/Ax = FL/L2x = F/Lx
F = η Lx
Work done for small displacement
dW = F dx

Elastic After Effect
When deforming force is removed –there will be a delay in
regaining original shape or size. This delay is called elastic
after effect
It depends on the nature of the material.
Quartz, phosphor bronze, silver, gold etc show less elastic
after effect.
ie. regain original condition immediately after the removal of
deforming force.
Hence used in galvanometers, electrometers etc.Dr. Pius Augustine, SH College, Kochi

Thermal stress
Rod of length lo at room temperature clamped at both ends .
Temp increases by ∆T, final length is l
l = lo(1+α ∆T)
Linear strain = (l-lo)/l0 = α∆T
Stress = Y * strain = Y * α∆T
Force excerted by the rod due to heating
= thermal stress x area
= Y * α∆T * A

Which is more elastic – steel or rubber? Why?
Steel
Modulus of elasticity = stress / strain
If same force is applied on steel and rubber
wires of equal length and area of cross
section, strain produced in rubber is more.
Hence elasticity of steel is more.

What are the factors on which modulus of
elasticity of a material depends ?
Nature of material and the manner
in which it is deformed.

What is the origin of stress ?
Deforming force displaces atoms from their
actual positions. These displaced atoms
exert an opposing force, which appears
as stress.

Name the factors which affect the
property of elasticity of a solid
i. Presence of impurities
ii. Change of temperature.
iii. Effect of hammering, annealing and rolling.

Breaking stress
Ratio of maximum load to which
the wire is subjected to original
cross section.

If only diameter of a wire is doubled, how will the
following parameters be affected?
i. extension for same load
ii. Load for same extension
Y = mgL/Ae = 4mgL / πD2e
i. For same load (mg), - D2e is a constant.
Doubling of D, make D2 four times, hence e
become 1/4th.
ii. For same ‘e’ - Mg/D2 is a constant.
Four time D2 will make mg four times.

If only length of a wire is doubled , how will the
following parameters be affected?
i. extension for same load
ii. Load for same extension
Y = mgL/Ae
i. For same mg, L/ e is a constant. If L is doubled, e
also becomes doubled.
ii. For same e, mgL is a constant. When L is
doubled, mg becomes half.

Mention one situation where the restoring force
is not equal and opposite to the applied force.
Beyond elastic limit
Which elasticity is possessed by all three
states of matter ?
Bulk modulus

What is the value of Y or K for a perfectly rigid
body ?
infinite
What is the value of shear modulus of a liquid?
Zero

Why do springs become slack after a
prolonged use ?
Elasticity of the material decreases due to
repeated deformation in the long run.
Elastic limit also decreases and springs become
deformed permanently

Why are electric poles given hollow structure?
Hollow shaft is stronger than a solid shaft
made from the same and equal amounts
of material.

Why is spring made of steel and not of copper ?
A better spring will be one, in which a large
restoring force is developed on being
deformed
This in turn depends on elasticity of material.
Y for steel > Y for copper.

A wire of length 10m and diameter 2mm
elongates 0.2mm when stretched by a weight of
0.55kg. Calculate the Y of the material of the
wire.
Ans. 8.58 x 1010 N/m2

Four identical hollow cylindrical columns of steel
support a big structure of mass 50,000 kg. The
inner and outer radii of each column are 30 cm
and 40 cm respectively. Assuming load
distribution to be uniform, calculate the
compressional strain of each column. Y = 200 Gpa.
F = (50,000 * 9.8)/4 A = π(0.42 – 0.32)

A structural steel rod has a radius of 10 mm
and length of 1 m. A 100 kN force
stretches it along its length.
Calculate i) stress ii) strain ii) elongation.
(given Y = 2 x 1011 N/m2)

A steel wire of length 4.7 m and cross section
3 x 10-5 m2 stretches by the same amount as
copper wire 3.5 m and cross section 4 x 10-5 m2
under a given load. What is the ratio of the Y of
steel to that of Cu?
Ans. 1.8

The length of a wire increases 8 mm when a
weight of 5 kg is hung. If all conditions are
same, but the radius of the wire is doubled,
what will be the increase in its length ?
Ans . 2 mm

Relation connecting Y, K, η and σ
P
P
A B
CD
FE
G H
Cube of unit side length
P – Normal outward stress on upper and
lower surfaces – which produce strain
that direction.
Other 4 faces (front, back, left and right)
experience lateral contraction.
Note the terms below
λ - longitudinal strain per unit applied stress
μ - lateral strain per unit applied stress

Continue…
Applied stress = P
λ - longitudinal strain per unit applied stress
Total longitudinal strain = λP
Young’s modulus = longitudinal stress
longitudinal strain
= P/λP
Y = 1/λ Dr. Pius Augustine, SH College, Kochi

Case 1 Shear modulus η
Keeping the stress P on upper and lower
surfaces, apply a normal compression
stress ‘P’ on left and right faces.
It produce a compression in horizontal
direction inward and an extension upward
and perpendicular to the screen.Dr. Pius Augustine, SH College, Kochi

Case 1 Shear continue…..
Along the upward and downward direction,
the lateral strain developed = μP
Total extensional strain experienced by upper
and lower surfaces
= λP + μP = P (λ + μ)
Similarly total compresional strain
experienced by horizontal surface = P (λ + μ)

Case 1 Shear continue….
Shear strain = compresional strain + extensional strain
= 2 [P (λ + μ)]
Rigidity modulus η = stress/shearing strain
= P/ Sh. Strain
= P/ 2 [P (λ + μ)]
= 1/ 2 (λ + μ)
(λ + μ) = 1/2η

Case 2 Bulk modulus
Pull all the three faces (both directions)
Normal outward stress acting on 6 faces
Along each direction – there will be one
extension and two compresions

Case 1 bulk modulus continue ……
Resultant extensional strain = λP – μP – μP
= P (λ - 2μ)
Volume strain produced = 3 x linear strain
= 3 P (λ - 2μ)
Bulk modulus K = P
3 P (λ - 2μ)
(λ - 2μ) = 1/3K
6 P on 6 areas
So stress = P

Case 1 Case 1 Interconnections Y, K, η
Y = 1/λ (λ + μ) = 1/(2η) ……..…….. (A)
(λ - 2μ) = 1/3K …………… (B)
Lateral strain μ = (3K - 2η)
18 Kη
……. (C)
………(E)
………(D)

Poisson ratio
σ = lateral strain/ longitudinal strain
= μ/λ
Poisson ration σ = (3K - 2η)
2(3K + η)
………(F)

Poisson ratio
From (F) 6kσ + 2ησ = 3K – 2η
3K (1-2σ) = 2η (1+σ)

2η (1+σ) = 3K (1-2σ)
If σ is +ve, LHS will be a +ve number
ie. RHS should be +ve
3K (1-2σ) is +ve
(1-2σ) is +ve 2σ < 1
σ < 0.5
ie. +ve limiting value of σ is 0.5

2η (1+σ) = 3K (1-2σ)
If σ is -ve, RHS will be a +ve number ie. LHS should be +ve
2η (1+σ) is +ve (1+σ) is +ve σ > -1
(-1 and above – towards zero) ie. -ve limiting value of σ is -1
Combining two results
-1 < σ < 0.5
total spread 1.5
Note: -ve value of σ is not permissible. Because lateral elongation
is always associated with longitudinal elongation.
Value lies between 0 and 0.5 and in most materials it
is between 0.2 to 0.4 Dr. Pius Augustine, SH College, Kochi

Beam
A bar or a rod having length very large its diameter or
cross section or thickness
It may be visualized as thin layers or co axial cylinders
(filaments)

Neutral Surface
Layers above the neutral surface will be compressed
and layers below will be elongated.
ie. as we move from top to bottom on these layers,
(compressed to elongated), there is a layer in between,
which is neither compressed, nor elongated – It is called
neutral surface

Plane of bending and neutral axis
F
r
Torque τ = r X F
It is the moment of bending
The plane in which this couple acts is called plane of bending.
(diagram – it is the plane of the screen)
Plane of bending is perpendicular to the neutral surface
Line of intersection of plane of bending and neutral
surface is called neutral axis.

Bending Moment – Derive expression
Moment due to elastic reaction which balance the
bending torque.

Assumptions
1. Only bending is considered – not shearing or compress
(only Y not K or η)
2. Y is same for both extension and compression
3. Length of the beam is very large compared to thickness
(l>>t)
4.Weight of the beam is very small compared to the applied
force (W<<<F)
5.Cross section of the beam remains unchanged during
bending. ie. geometric moment of inertia constant.
6. During bending curvature is very small (1/R)

Bending moment continue….
NS – neutral surface
Extension in GH layer = G1H1 – GH = (R+Z)θ – Rθ = zθ
R – radius of curvature of NS
GH layer at z below NS (EF = GH = Rθ)

Along the layers or filament – there is force
Below NS – extensional force
Above NS – compression force
Small area ∆a on GH, experiences the above stress

This is true for all sections of the filament
Sum of the moments of the extensional and compresional force
across the areas of all filaments about NS (which forms restoring
couple, to balance)

∑∆a z2 (= AK2) represents a quantity quite similar to the
moment of inertia and is called geometrical moment of
inertia or second moment of inertia of the bar about
an axis on the neutral surface at right angles to the
neutral axis.
Geometrical moment of inertia of a beam is defined as its moment
of inertia if it has a unit mass per unit area

Flexural rigidity = YI – defined as the external bending
moment required to bend the beam into an arc of unit
radius. (in the above equation R =1)
Regular beam I = bd3/12 b – breadth, d – thickness
Rod I = πr4/4

∑az2 is MI of the beam about NS = AK2. A – area of
cross section, K- radius of gyration about NS
For rectangular cross-section
A = b x d and K2 = d2/12. b – breadth and d- width
I = AK2 = (b x d) d2/12 = bd3/12
For circular cross-section A = πr2 and K2 = r2/4
r – radius
I = AK2 = πr2 x r2/4 = πr4/4

Depression at the free end of a cantilever
When a cantilever is loaded, it bends
The filaments in the interior of a cantilever under the
action of a bending couple
Cantilever is a beam fixed horizontally at one end

Depression at the free end of a cantilever
A cantilever is loaded (W1=Mg) at
the free end is shown
AB is neutral axis
Maximum depression at the free
end (δ). Weight of the beam is
assumed negligible
Radius of curvature is not a constant. ie. bent cantilever takes
the shape of a parabola.

Complete expression for radius of curvature from
differential calculus is
For small bending dy/dx <<< 1 and hence denominator drops to 1

Depression - continue…..
At the point P, moment of
bending couple = Mg (l-x)
Internal bending moment = YI/R
At balanced state.
d2y = Mg (l-x)
dx2 YI
On integrating
dy = Mg [lx-(x2/2)] + C1
dx YI
Y – Young’s modulus
y - depression at P
R – radius of curvature
of NS at P
Boundary condition at x = 0 (fixed end), no variation in y,
ie. dy/dx = 0 So, C1 = 0 Dr. Pius Augustine, SH College, Kochi

Depression - continue….
Boundary condition at x = 0 (fixed end), no variation in y,
ie. y = 0 So, C2 = 0
Integrate again
y = Mg [(lx2/2)-(x3/6)] + C2
YI
Depression at x from fixed end
At free end x = L and y = δ

Y from cantilever
For rectangular beam I = bd3/12

Double cantilever – Non uniform bending
If a bar is supported at two knife edges A and B, l meter
apart in a horizontal plane so that equal lengths of the bar
project beyond the knife edges and a weight W is
suspended at the middle point, it act as a double lever.

A double cantilever (non uniform bending) can be
visualized as two cantilevers – each clamped at
centre ‘O’ and Mg/2 force is acting upwards at L/2
(free end).
Depression at ‘O’ is equivalent to elevation at A or B
from the lowest point ‘O’.
Elevation varies along the length of the cantilever.
So, aproach is – find elevation at any point in
between and integrate. Dr. Pius Augustine, SH College, Kochi

l/2l/2
P
P O O
Mg/2Mg/2
x (l/2 – x)
Moment of deflecting couple = (Mg/2) PB
= (Mg/2) (l/2 – x)
The point P is equilibrium
Bending moment will balance deflecting couple
(Mg/2) (l/2 – x) = YI/R = YI (d2y/dx2)
Integrate twice and arrive at (as before C1 and C2 are zero)
y = Mg lx2 - x3
2YI 4 6

l/2l/2
P
P O O
Mg/2Mg/2
x (l/2 – x)
Deflection at P,
At far end A or B, x = l/2 and y = δ
Substitute and solve δ = Mg (3 l3 - l3)
2YI 48
For rectangular bar
I = bd3/12.
Y = Mgl3
4bd3δ

Precautions to be taken while doing experiment
1. Knife edges should be rigid and fixed on a rigid support.
2. The knife-edges should be at equal distances from the
centre of the bar.
3. The weights should be placed or removed from the
hanger gently and should be increased or decreased
gradually in equal steps.
4. The maximum load used should be such that it keeps the
bar within elastic limits.
5. Thickness should be measured very accurately at a
number of points along the bar.
Note that the thickness is a small quantity and its cube is used
in calculations. So a small error in its measurement will
cause three times large error in the result.

Girder
For a beam loaded at middle
δ proportional to l3 1/b 1/d3 and 1/Y
When beam is used as girder, δ – should be minimum
To minimize δ - decrease l, or increase b and/or d.

Girder continue…
When a beam is used as girder, bending should be
minimum, so the side having greater magnitude of
cross section should be used as the depth (vertical).
Also Y- of the material used should be high (steel)
When loaded upper and lower surface would
experience maximum stress and middle is neutral
surface.
Upper and lower regions should be much stronger.Dr. Pius Augustine, SH College, Kochi

Girder continue… I shaped girders??
In the middle region, stress is minimum
Even if material is minimum in the middle, will
not be an issue.
So can save money without
affecting the strength.

Elevation at the midpoint of a beam supported
symmetrically and loaded equally at the ends.
UNIFORM BENDING
Reaction = W = Mg
Projection beyond the peg ‘a’ on either side
EF = δ, is the elevation of the midpoint above the knife edges
Bending forms a circular shape

Uniform bending continue…
δ
R
R-δ
2R-δ
L/2L/2C D
o
P
Q
PQ = δ
CD = Length of the bar
From ∆ OPD
(L/2)2 + (R-δ)2 = R2
L2 = 2Rδ - δ2
4
δ2 - very small and can be neglected

UNIFORM BENDING - continue……
Consider two half sections
Two equal and opposite
forces (Mg) acting at ‘a’
apart constitute a
couple – which is
balanced by the bending
moment of the bar.
Same with other half section.
W = Mg
Mg
W = Mg
Mg
Mg a = YI = YI 8δ
R L2
δ = Mg a l2
8YI
Y = Mg a l2
8δI

UNIFORM BENDING - continue……
W = Mg
Mg
W = Mg
Mg
δ = Mg a l2
8YI
Y = Mg a l2
8δI
I = bd3/12
Y = 12Mg a l2 = 3Mg a l2
8δbd3 2bd3 δ

Cause shear in the material
So rigidity modulus.
Upper end fixed
Lower end is twisted
φ - angle of shear
(shear strain)
θ – angle of twist
Rigidity modulus and Torsional Couple
B C
φ

Consider a cylinder of radius x
And thickness dx.
BC/x = θ BC /L = φ
xθ = Lφ
Shearing strain φ = xθ/L
Rigidity modulus η
Rigidity modulus – continue…..

Torsional couple or Couple per unit twist
Area (ring having thickness dx) = 2πx dx
Force = stress * area = η x θ * 2πx dx
L
Moment of the force = F * Perpendicular distance
(on small area)
= η x2 θ 2π dx * x
L
= 2πηθ x3 dx
L Dr. Pius Augustine, SH College, Kochi

Torsional couple or Couple per unit twist
Moment of the force on entire cylinder of radius ‘r’ or
couple (τ) = 2πηθ ∫ x3 dx
L
= 2πηθ x4
L 4
= πηr4 θ = C θ
2L
C - couple per unit twist
0
r
0
r

Couple per unit twist or Torsional rigidity
C - couple per unit twist = πηr4
2L
Torsional rigidity depends on r (radius), η (shear
modulus of elasticity), L (length)
C is not a constant for a material
For a particular wire - η, r and L are constants
C- is also called restoring couple per unit twist

Note : For Cylindrical Pipe
Torque (τ) = πη θ [R4-r4]
2L
= = πη (R2-r2) θ (R2+ r2)
L 2
[R4-r4] = (R2)2-(r2)2
= (R2+ r2) (R2-r2)
Qn: Show that hollow shafts are better than solid ones of
the same mass and area of cross section.

Area of cross section
of the pipe (shaded
region) – A2
Area of solid
rod = A1
A1 = A2
Material used in both
are equal –
ie. equal mass
Torque (τ1) = πn θ R2*R2
2L
= A1 n θ * R2 (τ1) α R2
2L
1 2

Torque on pipe (τ2) = πn (R2-r2) θ (R2+ r2)
L 2
= A2 n θ (R2+ r2)
2 L
(τ2) α (R2+ r2)
If r = 0, eqn will be for a solid rod/wire having radius R
ie. couple is much greater due to added term r2.
Hollow cylinder is stronger than solid cylinder (r=0) with
same external radius.

Static Torsion – Rigidity modulus of a rod
AB – 50 cm long (L) and 0.25 cm (2r)
in diameter, rod – material of which η
is to be found out.
Pointer is adjusted to coincide central graduation
After making the rod in elastic mood by repeated load and unloading
both clockwise and anticlockwise directions, a weight is added.
Mg – weight added.
Reading of pointer is noted. Increase weight in steps and both clockwise
and anticlockwise in the same manner and find average twist for M kg
from the pointer readings noted.
A
B
Pointer
R

Static Torsion – continue………
360o = 2π rad
1o = 2π/360 rad
αo = α (2π/360 ) rad
R – radius of the pulley
External couple = Mg * R
Torsional couple = C θ = πηr4 θ
2L
πηr4 θ = Mg * R
2L
Rigidity modulus η = 2 Mg R L
π r4 θ
A
B
Pointer
R

PE stored when a wire is twisted
A wire is twisted from angle 0 to Θ by applying a
gradually increasing torque.
At some angle ‘θ’ in between, dW = ½ Cθ dθ
Total work done = ∫ ½ Cθ dθ = ½ C Θ2.
Stored as Pot. energy
0
Θ

Torsional Oscillations and Torsion Pendulum
Disc of the pendulum is turned a little and released.
Extreme end PE = ½ CΘ2.
On releasing PE due to restoring force acts within
the string, disc rotates, gain KE (which is maximum
at θ = 0) and then over shoots and go to other
extreme.
KE = ½ I ω2.
I – moment of inertia of the body about the axis of
suspension and ω – angular velocity
Angular oscillations are called torsional oscillations
and arrangement is called torsion pendulum Dr. Pius Augustine, SH College, Kochi

Total energy is conserved
½ Iω2 + ½ Cθ2 = constant
½ I(dθ/dt)2 + ½ Cθ2 = constant
Differentiate w.r. to θ
d2θ + C θ = 0
dt2 I
Compare with differential
equation for SHM
d2y + ω2 y = 0
dx2

T = 2π I 2L
πη r4

Torsion pendulum experiment
L – length of the pendulum
t – Mean time for 20 oscillations
T-time period r – radius of wire (screw gauge)
For regular shapes, moment of inertia depends
on mass and dimensions.
For disc I = MR2/2
R
Rigidity modulus of the material of the wire can
be determined

A light rod of length 2.00 m is suspended from the ceiling horizontally
by means of two vertical wires of equal length tied to its ends. One
of the wires is made of steel and is of cross-section 10-3 m2 and
other is of brass of cross section 2 x 10-3 m2. Find out the position
along the rod at which a weight may be hung to produce,
i) Equal stresses in both wires
ii) Equal strains on both wires
Y- for steel - 2 x 1011 N/m2 and for brass – 1011 N/m2.

Solution: i)
Stress in steel = stress in brass
Ts/As = TB/AB
Ts/TB = As/AB = ½
Taking moments about D, Ts x = TB (2-x) x = 1.33m
Solution: ii)
Strain in steel = strain in brass
Ts/YSAs = TB/YBAB solving Ts/TB = 1 and x = 1.0 m
Tsteel Tbrass
x 2-x
D

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Dr. Pius Augustine, Dept of Physics, Sacred Heart College, Thevara
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Dr. Pius Augustine, Asst. Professor, Sacred Heart College, Thevara, Kochi.

31 pius augustine elasticity (b.sc)

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