stress strain

Question

6.2 An elastic rod is 5 m long and has a cross-sectional area of 1.5 cm 2 . The rod hangs vertically and stretches with 0,075 cm when a mass of 350 kg is Nor attached to its free end. Calculate the following: 6.2.1 The stress 6.2.2 The strain 6.2.3 Young's Modulus of elasticity for the material

Answer 1

6.2.1)

Find the stress in rod by using the formula:

σ = load / cross-sectional area

  = Weight / cross-sectional area

  = mg/cross-sectional area

  =(350)(9.81)/ (1.5 cm²  x (10^–2)²  m²/cm² )

  = 22.89x 10⁶ N/m²

   = 22.89 MPa

Answer 2

6.2.2)

Find the strain in the rod by using the formula:

ε =  Change in length/ original length

  =  [0,075 cm x 10^–2 m/cm]/ 5

 = 0.00015 m/m 

Answer 3

6.2.3)

Find the stress in rod by using the formula:

σ = load / cross-sectional area

  = Weight / cross-sectional area

  = mg/cross-sectional area

  =(350)(9.81)/ (1.5 cm²  x (10^–2)²  m²/cm² )

  = 22.89x 10⁶ N/m²

   = 22.89 MPa

Find the strain in the rod by using the formula:

ε =  Change in length/ original length

  =  [0,075 cm x 10^–2 m/cm]/ 5

 = 0.00015 m/m 

Find Young's Modulus of elasticity for the material by using stress - strain relation:

E = σ / ε

  = 22.89 / 0.00015

 =   152600 MPa