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Chapter 7: Problem 41

An uncold-worked brass specimen of average grain size \(0.01 \mathrm{mm}\) has a yield strength of \(150 \mathrm{MPa}(21,750 \mathrm{psi}) .\) Estimate the yield strength of this alloy after it has been heated to \(500^{\circ} \mathrm{C}\) for \(1000 \mathrm{s}\), if it is known that the value of \(\sigma_{0}\) is \(25 \mathrm{MPa}(3625 \mathrm{psi})\).

Short Answer

Expert verified
Question: Estimate the yield strength after the brass has been heated to 500°C for 1000 seconds, given the initial data and Hall-Petch equation. Answer: Due to the lack of information on activation energy (Q) and pre-exponential constant (Dₒ), we cannot find the new grain size nor estimate the new yield strength. More information is required to solve this problem.

Step by step solution

01

Write the Hall-Petch equation for initial yield strength

First, we will write the Hall-Petch equation relating the initial yield strength (σ_yield) with the initial grain size (d) and the value of σ₀: \(σ_{\text{yield}} = σ_0 + k \cdot d^{-\frac{1}{2}}\) In this equation, k is the material's constant, and d is the average grain diameter.
02

Substitute given values for initial yield strength

Now we will apply the given values of σ₀ (25 MPa), σ_yield (150 MPa) and average grain size (0.01 mm or \(1\times 10^{-5}\) m) into the equation: \(150\,\text{MPa} = 25\,\text{MPa}+k\cdot(1\times 10^{-5}\,\text{m})^{-\frac{1}{2}}\)
03

Solve for the material constant, k

In this step, we will solve the equation for k: k = (150 MPa - 25 MPa) \(\cdot\) \((1\times 10^{-5}\,\text{m})^{\frac{1}{2}}\) k ≈ 125 MPa · sqrt(\(1\times 10^{-5}\,\text{m}\)) k ≈ 1.25 MPa·m\(^{1/2}\)
04

Determine the new grain size at 500°C and 1000 seconds

Now, we need to find the new grain size at 500°C for 1000 seconds. We will use the Arrhenius equation for grain growth: \(d_{\text{new}} ^2 = d_{\text{old}} ^2 + 2\cdot D_0\cdot t\cdot e^{-\frac{Q}{RT}}\) In this equation: - \(d_{\text{new}}\) is the new average grain diameter. - \(d_{\text{old}}\) is the old average grain diameter (0.01 mm) - \(D_0\) is a pre-exponential constant - t is the time, 1000 seconds - Q is the activation energy for grain growth - R is the gas constant, 8.314 J/mol·K - T is the absolute temperature in K (500°C + 273.15 K = 773.15 K) It's important to notice that the values of D₀ and Q are not given. Therefore, in this particular exercise, we cannot find the new grain size d_new.
05

Estimating the new yield strength

As we cannot find the value of d_new using the provided information, it's impossible to calculate the new yield strength of the alloy using the Hall-Petch equation. To estimate the yield strength of this alloy after it has been heated to 500°C for 1000 seconds, we would need more information about the activation energy (Q) and/or the pre-exponential constant (Dₒ) specific to the uncold-worked brass.

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Most popular questions from this chapter

A single crystal of zinc is oriented for a tensile test such that its slip plane normal makes an angle of \(65^{\circ}\) with the tensile axis. Three possible slip directions make angles of \(30^{\circ}, 48^{\circ}\) and \(78^{\circ}\) with the same tensile axis. (a) Which of these three slip directions is most favored? (b) If plastic deformation begins at a tensile stress of \(2.5 \mathrm{MPa}(355 \mathrm{psi})\), determine the critical resolved shear stress for zinc.

The lower yield point for an iron that has an average grain diameter of \(1 \times 10^{-2} \mathrm{mm}\) is \(230 \mathrm{MPa}(33,000 \mathrm{psi}) .\) At a grain diameter of \(6 \times 10^{-3} \mathrm{mm},\) the yield point increases to 275 MPa \((40,000\) psi ) . At what grain diameter will the lower yield point be \(310 \mathrm{MPa}\) \((45,000 \mathrm{psi}) ?\)

Briefly explain why HCP metals are typically more brittle than \(\mathrm{FCC}\) and \(\mathrm{BCC}\) metals

Consider a metal single crystal oriented such that the normal to the slip plane and the slip direction are at angles of \(60^{\circ}\) and \(35^{\circ},\) respectively, with the tensile axis. If the critical resolved shear stress is \(6.2 \mathrm{MPa}(900 \mathrm{psi}),\) will an applied stress of \(12 \mathrm{MPa}(1750 \mathrm{psi})\) cause the single crystal to yield? If not, what stress will be necessary?

Briefly explain why small-angle grain boundaries are not as effective in interfering with the slip process as are high-angle grain boundaries.
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