4. Material property play a
vital role…..
a greater lift force means that
the wings will bend more…
Dr. Pius Augustine, SH College, Kochi
5. Inter atomic forces
Electrical interaction
between atoms
Short range nearly 10-10m.
(radius)
If distance between atoms is much greater
than range, no interaction
Dr. Pius Augustine, SH College, Kochi
6. 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
Dr. Pius Augustine, SH College, Kochi
8. 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
Dr. Pius Augustine, SH College, Kochi
9. 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
Dr. Pius Augustine, SH College, Kochi
10. 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.
Dr. Pius Augustine, SH College, Kochi
11. 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.
Dr. Pius Augustine, SH College, Kochi
12. 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.
Dr. Pius Augustine, SH College, Kochi
13. 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.
Dr. Pius Augustine, SH College, Kochi
14. 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.
Dr. Pius Augustine, SH College, Kochi
15. 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.
Dr. Pius Augustine, SH College, Kochi
16. 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.
Dr. Pius Augustine, SH College, Kochi
17. Stress???
For different kinds of deformation, stress is
a quantity characterizes the strength of
the forces causing the deformation.
Dr. Pius Augustine, SH College, Kochi
21. 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
Dr. Pius Augustine, SH College, Kochi
22. 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
Dr. Pius Augustine, SH College, Kochi
23. 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.
Dr. Pius Augustine, SH College, Kochi
24. 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.
Dr. Pius Augustine, SH College, Kochi
25. 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
Dr. Pius Augustine, SH College, Kochi
26. 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
Dr. Pius Augustine, SH College, Kochi
27. 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
Dr. Pius Augustine, SH College, Kochi
28. 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
Dr. Pius Augustine, SH College, Kochi
29. 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)
Dr. Pius Augustine, SH College, Kochi
30. 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.
Dr. Pius Augustine, SH College, Kochi
31. 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
32. ro
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
33. 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.
Dr. Pius Augustine, SH College, Kochi
34. 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
Dr. Pius Augustine, SH College, Kochi
35. What are elastomers ?
Elastic substances which can be subjected
to a large value of strain, are called
elastomers.
Dr. Pius Augustine, SH College, Kochi
36. 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.
Dr. Pius Augustine, SH College, Kochi
37. 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.
Dr. Pius Augustine, SH College, Kochi
38. 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.
Dr. Pius Augustine, SH College, Kochi
39. 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
Dr. Pius Augustine, SH College, Kochi
40. Why are bridges declared unsafe after
a long use ?
Loses its elastic strength due to
repeated alternating strains.
Dr. Pius Augustine, SH College, Kochi
41. 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.
Dr. Pius Augustine, SH College, Kochi
42. 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
Dr. Pius Augustine, SH College, Kochi
43. 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.
Dr. Pius Augustine, SH College, Kochi
44. 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.
Dr. Pius Augustine, SH College, Kochi
45. What are tensor physical quantities ?
Physical quantities having different values
in different directions are called tensors
Eg. Stress, moment of inertia
Dr. Pius Augustine, SH College, Kochi
46. What is elastic limit ?
It is the maximum stress on, whose
removal, the bodies regain their
original dimension
Dr. Pius Augustine, SH College, Kochi
47. Do liquids possess rigidity ?
No
Because they have no shapes of
their own
Dr. Pius Augustine, SH College, Kochi
48. A B
C
D
Slope of region OA gives Y
Dr. Pius Augustine, SH College, Kochi
49. 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, doesnot regain original.
Permanent set.
Dr. Pius Augustine, SH College, Kochi
50. 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.
Dr. Pius Augustine, SH College, Kochi
51. 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.
Dr. Pius Augustine, SH College, Kochi
52. 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
Dr. Pius Augustine, SH College, Kochi
53. 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
Dr. Pius Augustine, SH College, Kochi
54. 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.
Dr. Pius Augustine, SH College, Kochi
55. 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
56. 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
Dr. Pius Augustine, SH College, Kochi
57. 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
58. Area within stress strain graph is energy density
Slope of the graph is elasticity of the material
Dr. Pius Augustine, SH College, Kochi
60. 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
Dr. Pius Augustine, SH College, Kochi
61. 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.
Dr. Pius Augustine, SH College, Kochi
62. 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
63. 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
Dr. Pius Augustine, SH College, Kochi
64. 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
Dr. Pius Augustine, SH College, Kochi
65. 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
Dr. Pius Augustine, SH College, Kochi
66. 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
67. 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.
Dr. Pius Augustine, SH College, Kochi
68. 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
Dr. Pius Augustine, SH College, Kochi
69. 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.
Dr. Pius Augustine, SH College, Kochi
70. 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
Dr. Pius Augustine, SH College, Kochi
71. 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
Dr. Pius Augustine, SH College, Kochi
72. 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
Dr. Pius Augustine, SH College, Kochi
73. 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
Dr. Pius Augustine, SH College, Kochi
74. 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
Dr. Pius Augustine, SH College, Kochi
75. 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
76. 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
Dr. Pius Augustine, SH College, Kochi
77. 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.
Dr. Pius Augustine, SH College, Kochi
78. What are the factors on which modulus of
elasticity of a material depends ?
Nature of material and the manner
in which it is deformed.
Dr. Pius Augustine, SH College, Kochi
79. 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.
Dr. Pius Augustine, SH College, Kochi
80. 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.
Dr. Pius Augustine, SH College, Kochi
81. Breaking stress
Ratio of maximum load to which
the wire is subjected to original
cross section.
Dr. Pius Augustine, SH College, Kochi
82. 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.
Dr. Pius Augustine, SH College, Kochi
83. 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.
Dr. Pius Augustine, SH College, Kochi
84. 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
Dr. Pius Augustine, SH College, Kochi
85. 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
Dr. Pius Augustine, SH College, Kochi
86. 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
Dr. Pius Augustine, SH College, Kochi
87. Why are electric poles given hollow structure?
Hollow shaft is stronger than a solid shaft
made from the same and equal amounts
of material.
Dr. Pius Augustine, SH College, Kochi
88. 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.
Dr. Pius Augustine, SH College, Kochi
89. 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
Dr. Pius Augustine, SH College, Kochi
90. 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)
Dr. Pius Augustine, SH College, Kochi
91. 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)
Dr. Pius Augustine, SH College, Kochi
92. 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
Dr. Pius Augustine, SH College, Kochi
93. 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
Dr. Pius Augustine, SH College, Kochi
94. 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.
Dr. Pius Augustine, SH College, Kochi
95. 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
Dr. Pius Augustine, SH College, Kochi
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Dr. Pius Augustine, SH College, Kochi
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Dr. Pius Augustine, SH College, Kochi
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Dr. Pius Augustine, Dept of Physics, Sacred Heart College, Thevara
we will
overcome
Thank You
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Dr. Pius Augustine, Asst. Professor, Sacred Heart College, Thevara, Kochi.