PROBLEM 1.1

          A rod 300 cm long and of diameter 4.0 cm is subjected to an axial pull of 20 KN. If the modulus of Elasticity of the material of the rod is 4.0 x 10⁵ N/mm²; Determine:

(i) The Stress, 

(ii) The Strain, and 

(iii) The Elongation of the rod.

Sol:

Given,  Length of the rod, L = 300 cm,

              Diameter,               D = 4.0 cm = 40 mm, 

              Area,                       A = (π/4) D² = (π/4) 40²  = 400 π mm²

              ^{Axial pull,              P = 20 KN = 20,000 N}

             Modulus of Elasticity, E  = 4.0 x 10⁵ N/mm²

(i) The Stress (σ) is given by equation as

                         σ = P/A = 20,000/400 π  = 15.92 N/mm² 

                 

(ii) Using the equation, the strain is obtained as 

                         E = σ/e

Strain              e = σ/E = 15.92 / 4.0 x 10⁵ = 0.0000398

(iii) Elongation is obtained by using equation as

                          e = dL/L

Elongation,     dL = e x L  = 0.0000398 x 300 = 0.01194 cm

                   

              

Credits: Books.google.co.in

                Scribd.com

MODULUS OF ELASTICITY (OR) YOUNG’S MODULUS

          The ratio of Tensile stress to the corresponding strain (or) Compressive stress to the corresponding strain is a constant.

          This ratio is known as Modulus of Elasticity (or) Young's Modulus and is denoted by ‘E’.

HOOK'S LAW

WHAT IS HOOK’S LAW?

          Hook’s law (or) Law of Elasticity was discovered in 1660 by the English scientist Robert Hook. Hook's law states that "when a material is loaded within elastic limit, the stress is proportional to the strain produced". This means that the ratio of the stress to the corresponding strain is constant within the elastic limit. This constant is called Modulus of Elasticity (or) Modulus of Rigidity (or) Elastic Modulii. 

Hook’s law (or) Law of Elasticity

Credits: Wkipedia

PROBLEM 1.2

Find the minimum diameter of a steel wire, which is used to raise a load of 6000 N, if the stress in the rod is not to exceed 105 MN/m² .

Sol:

As per given data:

            Load           P = 6000 N

            Stress        σ = 105 MN/m²    = 105 x 10 ⁶ N/m²            (1 MN = 10⁶ N)

                                  = 105 N/mm²                                        ( 10⁶ N/m² = 1 N/mm²)

 Let       D = The diameter of the rod in mm

Area      A = (π/4) D²

We know that, Stress  σ = Load/Area = P/A

                                    105 = 6000/(π/4) D²

                                     D² = 6000 x 4 / π x 105

                                     D   = 72.75