Question

The following data was recorded during tensile test made on a standard tensile test specimen: Original diameter and gauge length =25 mm and 80 mm; 

Minimum diameter at fracture =15 mm; 

Distance between gauge points at fracture = 95 mm; 

Load at yield point and at fracture = 50 kN and 65 kN; 

Maximum load that specimen could take = 86 kN. 

Make calculations for 

(a) Yield strength, ultimate tensile strength and breaking strength 

(b) Percentage elongation and percentage reduction in area after fracture 

(c) Nominal and true stress and fracture. 

Answer 1

Given data: 

Original diameter =25 mm 

gauge length = 80 mm; 

minimum diameter at fracture =15 mm 

distance between gauge points at fracture = 95 mm 

load at yield point and at fracture = 50 kN 

load at fracture = 65 kN; 

maximum load that specimen could take = 86 kN. 

Original Area Ao = Π/4 (25)² = 490.87mm² 

Final Area Af = Π/4 (15)² = 176.72mm² 

(a) Yield Strength = Yield Load / Original Cross sectional Area 

= (50 × 103)/490.87 = 101.86 N/mm² 

Ultimate tensile Strength Maximum Load / Original Cross sectional Area 

= (86 × 103)/490.87 = 175.2 N/mm²

Breaking Strength = fracture Load / Original Cross sectional Area 

= (65 × 103)/ 490.87 = 132.42 N/mm²

(b) Percentage elongation = (distance between gauge points at fracture - gauge length)/ gauge length

= [(95 – 80)/80] × 100 = 18.75%

percentage reduction in area after fracture = [(Original Area – Final Area)/ Original Area] × 100 

= [(490.87 – 176.72)/ 490.87] × 100 = 64%

(c) Nominal Stress = Load at fracture / Original Area = (65 × 1000)/ 490.87 

= 132.42 N/mm²

True Stress = Load at fracture / Final Area = (65 × 1000)/ 176.72 

= 367.8 N/mm².