The following yield strength, grain diameter, and heat treatment time (for grain growth) data were gathered for an iron specimen that was heat treated at \(800^{\circ} \mathrm{C}\). Using these data, compute the yield strength of a specimen that was heated at \(800^{\circ} \mathrm{C}\) for \(3 \mathrm{~h}\). Assume a value of 2 for \(n\), the grain diameter exponent. \begin{tabular}{lcc} \hline Grain diameter ( mm) & Yield Strength (MPa) & Heat Treating Time (h) \\\ \hline \(0.028\) & 300 & 10 \\ \hline \(0.010\) & 385 & 1 \\ \hline \end{tabular}

The Hall-Petch equation relates the yield strength to grain size in crystalline materials. It is given by: \[\sigma_y = \sigma_0 + k_yd^{-n}\] Where: \(\sigma_y\) = Yield strength \(\sigma_0\) = yield strength for an infinitely large grain size \(k_y\) = the strengthening (Hall-Petch) constant \(d\) = grain diameter \(n\) = grain diameter exponent

We have 2 data points: (d=0.028 mm, σ_y=300 MPa), and (d=0.010 mm, σ_y=385 MPa). We are also given the value of n = 2. We will use these data points to derive a system of linear equations to find the values of \(k_y\) and \(\sigma_0\). Using the first data point (0.028 mm, 300 MPa), we get: \[300 = \sigma_0 + k_y(0.028)^{-2}\] Using the second data point (0.010 mm, 385 MPa), we get: \[385 = \sigma_0 + k_y(0.010)^{-2}\] We now have a system of equations: \[300 = \sigma_0 + k_y(784.00)\] \[385 = \sigma_0 + k_y(10000.00)\] This system of linear equations can be solved using various methods such as elimination, substitution, or matrix operations.

We can solve the system of linear equations using the elimination method. First, we will subtract the first equation from the second equation to eliminate the \(\sigma_0\) term: \[385 - 300 = k_y(10000.00) - k_y(784.00)\] \[85 = k_y(9216.00)\] Now we can solve for \(k_y\): \[k_y = \frac{85}{9216.00} \approx 0.009218 \mathrm{MPa\cdot mm^2}\] With the value of \(k_y\), we can now use either of the equations to solve for \(\sigma_0\). We will use the first equation: \[\sigma_0 = 300 - k_y(784.00) \approx 107.313 \mathrm{MPa}\]

We need to find the grain diameter associated with 3 hours of heat treatment time. Since the data points are not evenly spaced, we will use linear interpolation to estimate the grain diameter for a 3 hour heat treatment time: \[d_3 = d_1 + \frac{t - t_1}{t_2 - t_1}(d_2 - d_1)\] Where: \(d_3\) = grain diameter at 3 hours of heat treatment time \(d_1\) = grain diameter at 1 hour of heat treatment time \(d_2\) = grain diameter at 10 hours of heat treatment time \(t\) = target heat treatment time = 3 hours \(t_1\) = heat treatment time at first data point = 1 hour \(t_2\) = heat treatment time at second data point = 10 hours Substituting the given values, we get: \[d_3 = 0.010 + \frac{3 - 1}{10 - 1}(0.028 - 0.010) \approx 0.012 \: \mathrm{mm}\] Now we will use the Hall-Petch equation to find the yield strength for a specimen with a grain diameter of 0.012 mm: \[\sigma_y = \sigma_0 + k_yd^{-2}\] \[\sigma_y \approx 107.313 + 0.009218(0.012)^{-2} \approx 355.35 \: \mathrm{MPa}\] The yield strength of the iron specimen that was heated at 800°C for 3 h is approximately 355.35 MPa.