Using the data in Problem 6.30 and Equations 6.15, 6.16, and 6.18a, generate a true stress–true strain plot for stainless steel. Equation 6.18a becomes invalid past the point at which necking begins; therefore, measured diameters are given in the following table for the last three data points, which should be used in true stress computations.
Load |
| Length |
| Diameter |
|
N | lbf | Mm | in. | mm | in. |
159,500 | 35,850 | 54.864 | 2.160 | 12.22 | 0.481 |
151,500 | 34,050 | 55.880 | 2.200 | 11.80 | 0.464 |
124,700 | 28,000 | 56.642 | 2.230 | 10.65 | 0.419 |
(6.15)
(6.16)
(6.18a)
Problem 6.30
A cylindrical specimen of stainless steel having a diameter of 12.8 mm (0.505 in.) and a gauge length of 50.800 mm (2.000 in.) is pulled in tension. Use the load–elongation characteristics shown in the following table to complete parts (a) through (f).
Load |
| Length |
|
N | lbf | mm | in. |
0 | 0 | 50.800 | 2.000 |
12,700 | 2,850 | 50.825 | 2.001 |
25,400 | 5,710 | 50.851 | 2.002 |
38,100 | 8,560 | 50.876 | 2.003 |
50,800 | 11,400 | 50.902 | 2.004 |
76,200 | 17,100 | 50.952 | 2.006 |
89,100 | 20,000 | 51.003 | 2.008 |
92,700 | 20,800 | 51.054 | 2.010 |
102,500 | 23,000 | 51.181 | 2.015 |
107,800 | 24,200 | 51.308 | 2.020 |
119,400 | 26,800 | 51.562 | 2.030 |
128,300 | 28,800 | 51.816 | 2.040 |
149,700 | 33,650 | 52.832 | 2.080 |
159,000 | 35,750 | 53.848 | 2.120 |
160,400 | 36,000 | 54.356 | 2.140 |
159,500 | 35,850 | 54.864 | 2.160 |
151,500 | 34,050 | 55.880 | 2.200 |
124,700 | 28,000 | 56.642 | 2.230 |
| Fracture |
|
|
(a) Plot the data as engineering stress versus engineering strain.
(b) Compute the modulus of elasticity.
(c) Determine the yield strength at a strain offset of 0.002.
(d) Determine the tensile strength of this alloy.
(e) What is the approximate ductility, in percent elongation?
(f) Compute the modulus of resilience.
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