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

                   

              

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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