Evaluate Young modulus of elasticity - Ultimate stress:
Q: The following observations were made during tensile test on mild steel specimen 40 mm in diameter and 200 mm long. Elongation with 40 kN load (within the limit of proportionality), L = 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.
Sol.: (i) Young's modulus of elasticity E: Stress, σ= P/A
= 40/[ ? /4(0.04)2]
= 3.18 × 104 kN/m^{2}
Strain, e = δL/L = 0.0304/200 = 0.000152
E = stress/ strain = 3.18 × 104/0.000152
= 2.09 × 108 kN/m^{2} .......ANS
(ii) Yield point stress:
Yield point stress = yield point load/ Cross sectional area
= 161/[ ? /4(0.04)^{2}]
= 12.8 × 104 kN/m^{2} .......ANS
(iii) Ultimate stress:
Ultimate stress = maximum load/ Cross sectional area
= 242/[? /4(0.04)^{2}]
= 19.2 × 104 kN/m^{2} .......ANS
(iv) Percentage elongation:
Percentage elongation = (length of specimen at fracture - original length)/ Original length
= (249-200)/200
= 0.245 = 24.5% .......ANS