The elastic properties of the brass specimen enables it to return to its
original length when stressed below the yield strength.
The correct responses are;
4. a. The final length is 90.0 m.
b. The final length is 97.2 mm.
Reasons:
a. Diameter of the brass alloy = 7.5 mm
Length of the specimen = 90.0 mm
Force applied = 6000 N
The equation for the applied stress, σ, is presented as follows;
[tex]\sigma = \dfrac{Force \ applied}{Area \ of \ specimen} = \dfrac{6000 \, N}{\pi \cdot \left(\dfrac{7.5 \times 10^{-3}}{2} \, m} \right)^2 } \approx 135.81 \ \mathrm{MPA}[/tex]
Depending on the cold working condition, 135.81 MPa is below the yield
strength, and the brass will return to its original condition when the force is
removed. The final length is remains as 90.0 m.
b. When the applied force is F = 16,500 N, we have;
[tex]\sigma = \dfrac{16,500\, N}{\pi \cdot \left(\dfrac{7.5 \times 10^{-3}}{2} \, m} \right)^2 } \approx 373.48\ \mathrm{MPA}[/tex]
The stress found for the force of 16,500 N is above the yield stress of
brass, and it is therefore, in the plastic region.
From the stress strain curve, the strain can be estimated by drawing a line
from the point of the 373.48 MP on the stress strain curve, parallel to the
elastic region to intersect the strain axis, which gives a value of strain
approximately, ε = 0.08.
The length of the specimen is given by the formula; [tex]l_i = l_0 \cdot (1 + \epsilon)[/tex]
Therefore;
[tex]l_i[/tex] = 90 × (1 + 0.08) = 97.2
The final length of the specimen, [tex]l_i[/tex] = 97.2 mm
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