# Question 6.1: Tensile Testing of Aluminum Alloy Convert the load and chang......

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
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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.

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