a round steel alloy bar with a diameter of 0.75 in and a gauge length of 3.0 in
ID: 1821700 • Letter: A
Question
a round steel alloy bar with a diameter of 0.75 in and a gauge length of 3.0 in was subjected to tension, with the results shown in the accompanying table. using a computer spread sheet program, plot the stress-strain relationship. from the graph, determine the young's modulus of the steel alloy and the deformation corrsponding to a 8225 lb load.load(lb) Deformation 10^-4 in.
2000 11.28
4000 22.54
6000 33.80
8000 45.08
10000 56.36
12000 67.66
Is the deformation equal to the strain? Or do I have to do: (3-0.001128)/3...etc?
Explanation / Answer
The deformation is solved for by using the young's modulus.
Young's modulus = E = (Stress1 - Stress2)/(Strain1-Strain2)
Using points at 2000 and 4000 lbs
Stress1 = (load/area) = 2000/1.767 = 1131.86
Stress2 = 4000/1.767 = 2263.72
Strain1 = (11.28*10-4)/(3) = 0.000376
Strain2 = (22.54*10-4)/(3) = 0.000751
E = (1131.86 - 2263.72) / (0.000376 - 0.000751) = -1131.86 / -0.000375 = 3,018,293.3
For the next step you need the equation
= /E L = *L = P/A = Stress L = Change in Length L = Length
= Load/Area = 8225/1.767 = 4654.78
= /E = 4654.75 / 3,018,293.3 = 0.001542
L = * L = 0.001542 * 3 = 0.0138 in.
Hope This helps :)
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