Force (N) | time (s) | Elongation (t*0.05 mm/s) | Engineering stress(Mpa) | Engineering Strain |
0 | 0 | 0 | 0 | 0 |
1500 | 5 | 0.25 | 53.05978069 | 0.00625 |
5000 | 10 | 0.5 | 176.8659356 | 0.0125 |
9250 | 20 | 1 | 327.2019809 | 0.025 |
11250 | 25 | 1.25 | 397.9483551 | 0.03125 |
11750 | 30 | 1.5 | 415.6349487 | 0.0375 |
11500 | 32.5 | 1.625 | 406.7916519 | 0.040625 |
12000 | 35 | 1.75 | 424.4782455 | 0.04375 |
14250 | 50 | 2.5 | 504.0679165 | 0.0625 |
15250 | 60 | 3 | 539.4411036 | 0.075 |
16000 | 70 | 3.5 | 565.970994 | 0.0875 |
16750 | 85 | 4.25 | 592.5008843 | 0.10625 |
17250 | 100 | 5 | 610.1874779 | 0.125 |
17500 | 132.5 | 6.625 | 619.0307747 | 0.165625 |
17250 | 170 | 8.5 | 610.1874779 | 0.2125 |
17000 | 180 | 9 | 601.3441811 | 0.225 |
16500 | 190 | 9.5 | 583.6575875 | 0.2375 |
16250 | 195 | 9.75 | 574.8142908 | 0.24375 |
16000 | 200 | 10 | 565.970994 | 0.25 |
15000 | 210 | 10.5 | 530.5978069 | 0.2625 |
13250 | 220 | 11 | 468.6947294 | 0.275 |
12697.1 | 222.5 | 11.125 | 449.1368942 | 0.278125 |
Initial Length (mm) | Initial Diameter(mm) | Area = pi*r^2 | |
40 | 6 | 28.27 | |
| | | | | | | |
Maximum load capacity: 50,000 N Test rate: 3 mm/min - 0.05 mm/sSteel A Steel B Initial length, Lo 40 mm 40 mm Initial diameter, do6 mm 6 mm Final length, Lf 48 mm 52 mm Final diameter, df 4.2 mm3.1 mm Â
From the data given in excel file, calculate:
(b) Tensile strength, modulus of elasticity, modulus ofresilience, % elongation and % reduction in area.
(c) Plot both engineering stress strain curve and true stressstrain curve.
(d) Compare the tensile behaviour of steel A and B in oneparagraph
