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.

**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.

**2****Hooke’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

**3****Young’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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Civil Engineering : Building Components and Structures : Elasticity |

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