2) The following data is obtained from a tensile test of circular mild steel specimen in a UTM.
Diameter of the specimen = 3 cm
Length of the specimen = 20 cm
Load at the elastic limit = 250 KN
AL at load of 150 kN = 0.21 mm
Maximum load = 380 KN
Total extension = 60 mm
Final diameter of the specimen at fracture = 2.25 cm .
Determine the following:
a) Elastic Modulus
b) Stress at elastic limit
c) Percentage of elongatio d) Percentage of decrease in area
e) Ultimate stress. (10 marks)
Answers
Answer:
1. A tensile test was conducted on a mild steel bar. The following data was obtained from the test: i) Diameter of the steel bar = 3cm ii) Gauge length of the bar =20cm iii) Load at elastic limit =250kN iv) Extension at a load of 150kN = 0.21mm v) Maximum load =380kN vi) Total extension =60mm vii) Diameter of the rod at the failure = 2.25cm Determine: a) The Young’s modulus b) The stress at elastic limit c) The percentage elongation d) The percentage decrease in area
Answer:
Hope it helps you.
Explanation:
(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