Respuesta :
Complete Question:
Grain diameter 1 (mm) = 4.4E-02
Yield stress 1 (MPa) = 131
Grain diameter 2 (mm) = 7.7E-03
Yield Stress 2 (MPa) = 268
The yield strength for an alloy that has an average grain diameter, d1, is listed above as Yield Stress 1 . At a grain diameter of d2, the yield strength increases Yield Stress 2. At what grain diameter, in mm, will the yield strength be 217 MPa
Answer:
d = 1.3 * 10⁻² m
Explanation:
According to the Hall Petch equation:
[tex]\sigma_y = \sigma_0 + k/\sqrt{d} \\[/tex]
At [tex]d_{1} = 4.4 * 10^{-2} mm[/tex], [tex]\sigma_{y1} = 131 MPa = 131 N/ mm^2[/tex]
[tex]131 = \sigma_0 + k/\sqrt{4.4 * 10^{-2}} \\k = 27.45 - 0.2096 \sigma_0[/tex]
At [tex]d_{2} = 7.7 * 10^{-3} mm[/tex], [tex]\sigma_{y2} = 131 MPa = 268 N/ mm^2[/tex]
[tex]268 = \sigma_0 + (27.45 - 0.2096 \sigma_0)/\sqrt{7.7 * 10^{-3}} \\23.5036 = 27.47 - 0.1219 \sigma_0\\ \sigma_0 = 32.45 N/mm^2[/tex]
k = 27.45 - 0.2096(32.45)
k = 20.64
At [tex]\sigma_y = 217 MPa[/tex], reapplying Hall Petch law:
[tex]\sigma_y = \sigma_0 + k/\sqrt{d} \\[/tex]
[tex]217 =32.45 + 20.64/\sqrt{d} \\217 - 32.45 = 20.64/\sqrt{d}\\184.55 = 20.64/ \sqrt{d} \\\sqrt{d} = 20.64/184.55\\\sqrt{d} = 0.11184\\d = 0.013 mm[/tex]
d = 1.3 * 10⁻² m
