Using the data in Problem 7.29 and Equations 7.15, 7.16, and 7.18a, generate a true stress–true strain plot for aluminum. Equation 7.18a becomes invalid past the point at which necking begins; therefore, measured diameters are given in the following table for the last four data points, which should be used in true stress computations.
Load | Length | Diameter | |||
N | lbf | mm | in. | mm | in. |
46,100 | 10,400 | 56.896 | 2.240 | 11.71 | 0.461 |
44,800 | 10,100 | 57.658 | 2.270 | 11.26 | 0.443 |
42,600 | 9,600 | 58.420 | 2.300 | 10.62 | 0.418 |
36,400 | 8,200 | 59.182 | 2.330 | 9.40 | 0.370 |
(7.15)
(7.16)
(7.18a)
Problem 7.29
A cylindrical specimen of aluminum having a diameter of 0.505 in. (12.8 mm) and a gauge length of 2.000 in. (50.800 mm) 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 |
7,330 | 1,650 | 50.851 | 2.002 |
15,100 | 3,400 | 50.902 | 2.004 |
23,100 | 5,200 | 50.952 | 2.006 |
30,400 | 6,850 | 51.003 | 2.008 |
34,400 | 7,750 | 51.054 | 2.010 |
38,400 | 8,650 | 51.308 | 2.020 |
41,300 | 9,300 | 51.816 | 2.040 |
44,800 | 10,100 | 52.832 | 2.080 |
46,200 | 10,400 | 53.848 | 2.120 |
47,300 | 10,650 | 54.864 | 2.160 |
47,500 | 10,700 | 55.880 | 2.200 |
46,100 | 10,400 | 56.896 | 2.240 |
44,800 | 10,100 | 57.658 | 2.270 |
42,600 | 9,600 | 58.420 | 2.300 |
36,400 | 8,200 | 59.182 | 2.330 |
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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