Chapter 6
Q. 6.4
Ductility of an Aluminum Alloy
The aluminum alloy in Example 6-1 has a final length after failure of 2.195 in. and a final diameter of 0.398 in. at the fracture surface. Calculate the ductility of this alloy.
Step-by-Step
Verified Solution
% Elongation = \frac{l_{f} \ – \ l_{0}}{l_{0}} × 100=\frac{2.195 \ – \ 2.000}{2.000} × 100= 9.75%
% Reduction in area= \frac{A_{0} \ – \ A{f}}{A_{0}} × 100
=\frac{(\pi /4)(0.505)^2 \ – \ (\pi /4)(0.398)^2}{(\pi /4)(0.505)^2} × 100
= 37.9%
The final length is less than 2.205 in. (see Table 6-1) because, after fracture, the elastic strain is recovered.
| Table 6-1 The results of a tensile test of a 0.505-in. diameter aluminum alloy test bar, initial length (l_{0}) = 2 in. | |||
|
Load (lb) |
Change in Length (in.) |
Calculated | |
| Stress (psi) | Strain (in./in.) | ||
| 0 | 0.000 | 0 | 0 |
| 1000 | 0.001 | 4,993 | 0.0005 |
| 3000 | 0.003 | 14,978 | 0.0015 |
| 5000 | 0.005 | 24,963 | 0.0025 |
| 7000 | 0.007 | 34,948 | 0.0035 |
| 7500 | 0.030 | 37,445 | 0.0150 |
| 7900 | 0.080 | 39,442 | 0.0400 |
| 8000 (maximum load) | 0.120 | 39,941 | 0.0600 |
| 7950 | 0.160 | 39,691 | 0.0800 |
| 7600 (fracture) | 0.205 | 37,944 | 0.1025 |