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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
Step-by-Step
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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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