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.

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