## 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.

## 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