Chapter 6
Q. 6.1
Tensile Testing of Aluminum Alloy
Convert the load and change in length data in Table 6-1 to engineering stress and strain and plot a stress strain curve.
| 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 |
Step-by-Step
Verified Solution
For the 1000-lb load:
S=\frac{F}{A_{0}}=\frac{1000 \ lb}{(\pi /4)(0.505 \ in.)^2}=4,993 \ psi
e=\frac{\Delta l}{l_{0}}=\frac{0.001 \ in.}{2.000 \ in.}=0.0005 \ in./in.
The results of similar calculations for each of the remaining loads are given in Table 6-1 and are plotted in Figure 6-5.
