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The stress-strain diagram for a steel alloy having an original diameter of 0
The stress-strain diagram for a steel alloy having an original diameter of 0.5 in. and a gauge length of 2 in. is given in the figure. Calculate the modulus of elasticity, the yield strength, the ultimate stress, and the fracture stress. If the specimen is loaded until it is stressed to 50 ksi, determine the approximate amount of elastic recovery and the increase in the gauge length after it is unloaded.
Expert Solution
Answer:
1. Modulus of elasticity:
Cosider the elastic line in the Lower strain scale: The line crosses the point (30 ksi stress, 0.001 strain)
E = 30/0.001 = 30000 ksi = 30 msi
2. Yield strength is corresponding to the line passing through B in Lower strain scale ~= 35 ksi
3. Ultimate strength ~= 63 ksi
4. fracture stress = 50 ksi
5. Corresponding to stress of 50 ksi, elastic strain ~= 0.0012 and plastic strain ~= 0.05
So elastic recovery = 0.0012*2 = 0.0024 in
Increase in gauge length/permanent deformation = 0.05*2 = 0.1 in.
PFA
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