STRESS,
STRAIN AND DEFORMATION OF SOLIDS
TWO
MARKS QUESTIONS:
1.
A rod of
diameter 30mm and length 400mm was found to elongate 0.35mm when it was
subjected to a load of 65KN. compute the modulus of elasticity of the material
of this rod.
(APR/MAY 2011)
Sol: Elongation
∂L = PL/AE
E=PL/A∂L
= 65x10³x400
π/4
x(30)²x0.35
= 105.092 N/mm²
2. What
is strain energy and write its unit in S.I. System? (APR/MAY 2011, Nov/Dec 2013)
When an elastic material is deformed
due to application of external force, internal resistance is developed in the
material of the body. Due to deformation, some work is done by the internal
resistance developed in the body, which is stored in the form of energy. This
energy is known as strain energy. It is expressed in N-m.
3. State Hooke’s law. (APR/MAY 2010, May/June 2013)
It states that when a material is loaded,
within its elastic limit, the stress is directly proportional to the strain.
Stress
α strain
σ α
e
E = σ / e Unit is N/mm²
Where, E – Young’s
Modulus,
σ – Stress,
e - Strain.
4. Define
Bulk modulus. (APR/MAY 2010)
When a body is stressed, within its
elastic limit, the ratio of direct stress to the corresponding volumetric
strain is constant. This ratio is known as Bulk Modulus.
Bulk Modulus = Direct
stress/ Volumetric strain
5. Define
Poison’s Ratio. (May/June 2009)
When a body is stressed within its
elastic limit, the ratio of lateral strain to the longitudinal strain is
constant for a given material.
Poisson’s ratio, or 1/m = Lateral
strain/Longitudinal strain.
6. What
is Thermal stress? (May/June 2009)
When a material is free to expand or
contract due to change in temperature, no stress and strain will be developed
in the material. But when the material is rigidly fixed at both the ends, the
change in length is prevented. Due to change in temperature, stress will be
developed in the material. Such stress is known as thermal stress.
7. The strain induced in an MS bar of rectangular section having width
equal to twice the depth is 2.5x10^-5. The bar is subjected to a tensile load
of 4KN. Find the section dimensions of the bar Take E=0.2x10^6 N/mm². (Nov/Dec’
2008)
Given: e= 2.5x10^-5, P=4x10^3 N, E=0.2x10^6
N/mm²
To find: Breadth b,
Depth d
Sol: Stress, σ = E x e = 0.2x10^6 x 2.5x10^-5
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A = P/ σ = 4x10^3/5
= 800
mm²
Area, A = b x d [b = 2d]
= 2d x d
800 = 2d²
d
= 20mm; b=2d=2 x 20 = 40
mm.
Result: 1. Breadth b = 40mm 2. Depth d=
20mm
8. Define Proof Resilience and Modulus
of Resilience.
(Nov/Dec’2008, May/June 2013,
Nov/Dec’2013)
The maximum strain energy that can
be stored in a material within its elastic limit is known as proof resilience.
It is the Proof resilience of the
material per unit volume.
Modulus
of resilience = Proof resilience
Volume of the body
9. The Young’s modulus and the shear modulus of material are 120Gpa and 45Gpa
respectively. What is its Bulk modulus?
(APR/MAY 2008)
Given:
Young’s Modulus
E = 120 G Pa = 120x10^9 Pa =120x10^9 N/m² = 120x10³ N/mm²
Shear Modulus, G
= 45Gpa = 45 x 10³ N/mm²
Solution: Young’s Modulus, E = 9KG
3K + G
120x10³
= 9 x K x 45 x 10³
3
K + 45 x 10³
120x10³ [3 K + 45 x 10³] = 9 x K x 45 x
10³
3 K + 45 x
10³ = 3.375 K
45 x 10³ = 0.375 K
|
10. Calculate the instantaneous stress produced in a bar of cross sectional
area 1000 mm² and 3m Long by the sudden
application of a tensile load of unknown magnitude, if the instantaneous Extension
is 1.5 mm. Also find the corresponding load. Take E = 200 G pa. (APR/MAY 2008)
Given: A = 1000 mm²; L = 3 m = 3000 mm;
∂L = 1.5 mm; E = 200GPa = 200 x 10³ N/mm²
Solution: ∂L
= σ x L / E
1.5 = σ x
3000 / 200 x 10³
|
σ = 2P / A
P = σ x A / 2
= 100 x 1000
2
|
11. Give the relation between modulus of
elasticity and modulus of rigidity. (May/June
2007)
E = 2G [1 + 1/m]
Where,
E – Young’s Modulus, N/mm²,
G
– Modulus of rigidity, N/mm²,
1/m
– Poisson’s ratio.
12. Write the concept used for finding stresses
in compound bars. (May/June 2007)
(i)
Elongation or contraction in each bar is equal. So, the strains induced in
those bars are also equal.
Change
in length of bar (1) = Change in length of bar (2)
P1L1 = P2L2
A1E1 A2E2
(ii) The sum of loads carried by
individual materials of a composite member is equal to the
Total load applied on the
member.
Total
load, P = Load carried by bar (1) + Load carried by bar (2)
|
13. Estimate the load carried by a bar if the
axial stress is 10 N/mm² and the diameter of bar is 10 mm. (Nov/Dec’ 2006)
Given: Stress, σ = 10 N/mm²; Diameter D
= 10 mm
Solution: Stress = Load /
Area
σ = Load / π/4 x (D) ²
10 = Load / π/4 x (10) ²
|
14.
What is the strain energy stored when a bar of 6 mm diameter 1 m length is
subjected to an axial Load of 4 KN, E = 200 KN/mm². (Nov/Dec’ 2006)
Given: Diameter D = 6 mm; Area, A = π/4
x (D) ² = π/4 x (6) ² = 28.274 mm²;
Load,
P = 4 KN = 4 x 10³ N; Length, L = 1 m = 1000 mm;
Young’s
Modulus E = 200 KN/mm² = 200 x 10³ N/mm²
Solution:
Volume
of the bar, V = A x L
= 28.274 x 1000
|
Stress, σ = Load / Area = P/A = 4 x 10³ / 28.274
|
Strain energy stored, U = σ²/2E x V =
(141.472)²/2x200x10³ x 28274
|
15. A circular rod 2 m long and 15 mm diameter
is subjected to an axial tensile load of 30 KN. Find the elongation of the rod
if the modulus of elasticity of the material of the rod is 120 KN/mm².(May/June 2006)
Given:
L = 2 m = 2000
mm; D = 15 mm; Load P = 30 KN = 30 x 10³ N;
Modulus of Elasticity E
= 120 KN/mm² = 120 x 10³ N/mm²
To find: (a) Stress (b) Strain (c)
Elongation.
Solution:
Stress = Load /
Area = P/A = 30 x 10³/ π/4 x (D) ² = 30 x 10³/ π/4 x (15) ²
|
|
Young’s modulus,
E = stress/strain = σ/e;
e = σ/E =
169.85/120 x 10³ = 0.00141
Strain e = Change in length / Original length
e = ∂L/L; ∂L = e x L = 0.001415 x 2000 = 2.83 mm
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16.
Define Strain energy and write its unit. (May/June
2006)
When an elastic material is deformed
due to application of external force, internal resistance is developed in the
material of the body. Due to deformation, some work is done by the internal
resistance developed in the body, which is stored in the form of energy. This
energy is known as strain energy. It is expressed in N-m.
17.
Define Factor of safety. (Oct’98)
It is defined as the ratio of
ultimate tensile stress to the permissible stress (working Stress).
Factor of Safety =
Ultimate stress / Permissible stress
18.
Define Modulus of Rigidity. (May 2003)
When a body is stressed, within
its elastic limit, the ratio of shearing stress to the corresponding shearing
strain is constant. This ratio is known as Modulus of rigidity.
Modulus of rigidity Or
Shear modulus, G = Shearing stress / Shearing strain.
19. Define Bulk modulus. (Oct’2000)
When a body is subjected to a
uniform direct stress in all the three mutually Perpendicular directions, the
ratio of the direct stress to the corresponding volumetric strain is found to
be a constant is called as the bulk modulus of the material and is denoted by K.
20. What do you understand by a compound bar? (Dec’2004)
A composite member is composed of
two or more different materials which are joined together so that the system is
elongated or compressed as a single unit.
21. What are the types of elastic constants? (APRIL 2000)
There are three types of elastic
constant.
1. Modulus of Elasticity or Young’s
Modulus, E
2. Bulk Modulus, K
3. Shear Modulus or Modulus of
Rigidity, G.
22. Define strain energy density. (May 2004, 2003)
Strain energy density is defined as
the maximum strain energy that can be stored in a material within the elastic
limit per unit volume. It is also known modulus of resilience.
23. What is stability? (Dec’ 2003)
The stability may be defined as an
ability of a material to withstand high load without major deformation.
24. Give the relation for change in length of a bar
hanging freely under its own weight. (May 2005)
Change
in length, ∂L = PL/AE
Where, P – Axial load.
L – Length of the bar.
E – Young’s Modulus of the bar.
A – Area of the bar.
25. A brass rod 2 m long is fixed at both its ends.
If the thermal stress is not to exceed 76.5 N/mm², calculate the temperature
through which the rod should be heated. Take the values of α and E as 17 x 10^-6/K and 90Gpa respectively. (May
2005)
Solution: Thermal stress, σ = α T E
76.5 = 17 x 10^-6 x T x 90 x 10³
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26. Determine the Poisson’s ratio and bulk modulus
of a material for which young’s modulus is 1.2 x 10^5 N/mm² and modulus of rigidity is 4.8 x 10^4
N/mm².
(Dec’ 2004)
Solution:
Young’s modulus, E = 2G (1+1/m)
1.2 x 10^5 = 2 x 4.8 x 10^4 (1+1/m)
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27.
Give the relationship between Bulk modulus and Young’s modulus. (APR’96,
Oct’97)
E
= 3K (1+2/m)
Where,
E – Young’s modulus,
K – Bulk modulus,
1/m – Poisson’s ratio.
28.
Define shear stress and shear strain (Nov/Dec’2010,
Nov/Dec’2013)
The two equal and opposite forces
act tangentially on any cross sectional plane of a body tending to slide one
part of the body over the other part. The stress induced in that section is
called shear stress and the corresponding strain is known as shear strain.
29.
State the principle of superposition. (Nov/Dec’2010)
The total deformation is equal to
the algebraic sum of the deformation of the individual sections. This principle
of finding out the resultant deformation is known as principle of superposition
The
change in length of such member is given by,
∂L
= P1L1 + P2L2 + P3L3 + …………
AE
30.
State Volumetric strain. (Apr’ 2002, Nov/Dec’2013)
Volumetric
strain is defined as the ratio of change in volume to the original volume of
the body.
Volumetric
strain ev = Change in volume
Original volume
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31.
Define tensile stress and tensile strain.
The
stress induced in a body, when subjected to two equal and opposite pulls, as a
result of which there is an increase in length, is known as tensile stress. The
ratio of increase in length to the original length is known as tensile strain.
32.
Define longitudinal strain and lateral strain. (Nov/Dec-2012)
Longitudinal strain:-is defined as the deformation of the body per unit length in the
direction of the applied load.
Lateral
strain:-is defined as the deformation of the body per unit
width in the direction of the applied load.
33.
Define stress and strain. (Oct’97)
Stress:
The force of
resistance per unit area, offered by a body against deformation is known as
stress.
Strain:
The ratio of
change in dimension to the original dimension when subjected to an external
load is termed as strain and is denoted by e. It has no unit.
34.
Define modulus of rigidity (May’2003)
The
ratio of shear stress to the corresponding shear strain when the stress is within
the elastic limit is known as modulus of rigidity or shear modulus and is denoted
by C or G or N.
35.
Give example for gradually applied load and suddenly applied load.
Example for gradually applied load:
When
we lower a body with the help of a crane, the body first touches the platform
on which it is to be placed. On further releasing the chain, the platform goes
on loading till it is fully loaded by the body. This is the case of gradually
applied load.
Example for suddenly applied load:
When
we lower a body with the help of a crane, the body is first of all, just above
the platform on which it is to be placed. If the chain breaks at once at this
moment the whole load of the body begins to act on the platform. This is the
case of suddenly applied load.
36.
What is resilience? (May’2010)
The
strain energy stored by the body within elastic limit, when loaded externally
is called resilience.
37.
Distinguish between suddenly applied and impact load.
When
the load is applied all of a sudden and not step wise is called is suddenly
applied load. The load which falls from a height or strike and body with
certain momentum is called falling or impact load.
38.
Define proof resilience? ( Nov/Dec’2013)
The
maximum strain energy stored in a body up to elastic limit is known as proof
resilience.
39.
Define modulus of elasticity. (Apr’98)
The
ratio of tensile stress or compressive stress to the corresponding strain is known
as modulus of elasticity or young‘s
modulus and is denoted by E.
40.
State principal plane. (Apr’98)
The
planes which have no shear stress are known as principal planes. These planes
carry only normal stresses.
41.
Define Elasticity. (May 2012)
Elasticity is the tendency
of solid materials to return to their original shape after being deformed.
Solid objects will deform when forces are applied on them. If the material is
elastic, the object will return to its initial shape and size when these forces
are removed.
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