Elasticity
When an external force acts on a
body, it undergoes some deformation. The property by which a body returns to
its original shape after the removal of external load is elasticity.
1Elastic limit
The limiting value of stress up
to and within which the entire deformation disappears on removal of the
external forces is called the elastic limit of the material. The body returns
to its original shape after the removal of the loading when the intensity of stress
is within a certain limit. The limit is called elastic limit.
2Hooke’s law
Hooke’s law states that
when an elastic material is stressed within elastic limit, the stress is
proportional to the strain.
K = stress/strain
3Young’s Modulus
The ratio of the axial stress to
the corresponding axial strain, with in elastic limit is called the
young’s modulus
E= Axial stress/Axial strain
4 Shear modulus
The ratio of the shear stress to
the corresponding shear strain is a constant, which is called shear modulus
G = Shear stress/shear strain
5 Bulk Modulus
When a body is subjected to
uniform direct stress in all the three mutually perpendicular direction, the
ratio of the direct stress to the corresponding volumetric strain is found to
be a constant which is called the bulk modulus
K = direct Stress/ Volumetric Strain
6 Poisson’s ratio
Poisson’s ratio = Lateral strain/ Longitudinal
Strain
7 Relation between elastic constants
E =2G ( 1+r)
E=3K (1-2r)
E = 9KG/ 3 K +G
E= Young’s modulus
G= Shear modulus
K=Bulk modulus
R= poisson’s ratio
8 Factor of safety
Factor safety = Ultimate stress/ Allowable stress
Problem
1. A mild steel rod of 12mm
diameter and 200 mm length elongates 0.085 mm under an axial
pull of 10kN. Determine the young’s modulus of the material.
Load P= 10KN
Cross sectional Area = (3.14 x 12 2)/4
Stress = 10000/113.1 = 88.4 N/mm2
Strain = 0.085/ 200 =
4.25 x10 -4
Young’s modulus = 88.4 / 4.25x 10-4 = 2.08
x 10-5 N/mm
2. A bar of 30 mm diameter is subjected to a pull of 60 kN.
The measured extension on
gauge length of 200 mm is 0.09 mm
and the change in diameter is 0.0039 mm. Calculate the
Poisson’s ratio and the values of the three moduli.
Dia of bar d = 30mm
Area A= 3.14 x 302 = 706.86 mm2
Pull P = 60kN
Length l = 200 mm
Change in length = 0.09 mm
Change in diameter = 0.0039mm
Stress = P/A = (60 x103)/ 706.86 = 84.9 N/mm2
Strain = P/E =84.9/ E
e = 0.09/200 = 0.00045
84.9/ E = 0.00045
= 1.886 x
105 N/mm2
Poisson’s ratio
1/m = 0.00013/0.00045 =
0.289
Modulus of rigidity
G = ( 1.886 x 10 5)/2(0.289 +1) = 7.3157 x 10 4 N/mm2
Bulk modulus
K= (1.886 x10 5)/3 ( 1-(2 x 0.289) = 1.489 x 10 5 N/mm2
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