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'* modulus of elasticity

(ii) Yield point stress

(iii) Ultimate stress

(iv) Percentage elongation

Answer 1

(i) Young'* 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%.