The following data was recorded during tensile test made on a standard tensile t

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.

Explanation / Answer

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 = P/4 (25)2 = 490.87mm2
Final Area Af = P/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

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