Question

Table 1: TENSILE TEST RESULTS OF A METAL SAMPLE with d= 7.42, lo = 40mm (4 marks) Load, KN Extension, mm Stress, MPa Strain 0 0 0 0 # 10 0.05 0.0002312 0.00125 17 0.08 0.0003931 0.002 4 25 0.11 0.0005781 0.00275 5 0.14 0.0006937 0.00350 30 34 0.20 0.005 37.5 0.40 0.01 0.0007862 8 0.0008672 2 0.0008903 5 0.0008325 3 38.5 0.60 0.015 36 0.90 0.0225 Cross-sectional Area (A) Modulus of Elasticity (E) Tensile Strength (ST) Percent Elongation (%EL) Sy PU E = S 7 %El. = Ал 4 Extension at fracture Gauge Length Ey

Answer 1

Data's given L. = 40 mm ; D. = 7.42m Modal calculation: Load 17 kN; Elongation =0.08 mm Stress is given by, F O= = 4F #D2 A 4x18x103 ax(7.42) o=0.0003931 MPa O= 2 Strain is given as, AL E L. 0.08 E = 40 E=0.002

Stress (Mpa) Vs strain 0.001 B 0.0009 с A 0.0008 0.0007 0.0006 Stress (MPa) 0.0005 0.0004 0.0003 0.0002 0.0001 0 0 0.005 0.01 0.015 0.02 0.025 Strain

a) Yield strength by 0.2% offset method is shown in the graph. Steps to obtain yield strength and yield strain: 1. Draw a line over the elastic limit. 2. Then, intersect the line with the 0.2% yield (i.e. 0.002 strain) 3. The point on the curve, where the line intersects is the yield point (point A) and corresponding stress is yield strength and strain is yield strain. Yield strength is 0.0008 MPa and Yield strain is 0.006 b) Modulus of Elasticity is given by, E= 0.0008 E = 0.006 E = 0.1333MPa c) Point C is failure strength. Ultimate strength (point B) is 0.00089 MPa Ultimate tensile load is 38.5 kN