Question

The following observations were made during a tensile test on a mild steel specimen 40 mm in diameter and 200 mm long. Elongation with 40 kN load (within limit of proportionality), δL = 0.0304 mm 

Yield load =161 KN 

Maximum load = 242 KN 

Length of specimen at fracture = 249 mm 

Determine: 

(i) Young's modulus of elasticity 

(ii) Yield point stress 

(iii) Ultimate stress 

(iv) Percentage elongation

Answer 1

 (i) Young's modulus of elasticity E :

 Stress, σ = P/A 

= 40/[Π/4(0.04)²] = 3.18 × 10⁴ kN/m²

Strain, e = δL/L = 0.0304/200 = 0.000152 

E = stress/ strain = 3.18 × 10⁴/0.000152

= 2.09 × 108 kN/m²

(ii) Yield point stress: 

Yield point stress = yield point load/ Cross sectional area 

= 161/[Π/4(0.04)²]

= 12.8 × 10⁴ kN/m²

(iii) Ultimate stress:

Ultimate stress = maximum load/ Cross sectional area

= 242/[Π/4(0.04)²]

= 19.2 × 10⁴ kN/m²

(iv) Percentage elongation: 

Percentage elongation = (length of specimen at fracture - original length)/ Original length

= (249–200)/200

= 0.245 = 24.5%.