A non-cold-worked brass specimen of average grain size \(0.01 \mathrm{~mm}\) has a yield strength of \(150 \mathrm{MPa}\) (21,750 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 MPa (3625 psi).

Short Answer

Expert verified
Question: Estimate the yield strength of a non-cold-worked brass specimen after being heated to a certain temperature and time, assuming that the grain size does not change significantly with the given heat treatment and the value of σ₀ remains the same. Answer: The estimated yield strength after heating is 150 MPa.

Step by step solution

01

Understand the Hall-Petch equation

The Hall-Petch equation is given by: \(\sigma_y = \sigma_0 + k_y\cdot d^{-1/2}\) where \(\sigma_y\) is the yield strength, \(\sigma_0\) is a material constant (given as 25 MPa), \(k_y\) is the Hall-Petch slope (empirically determined constant), and \(d\) is the average grain size.
02

Calculate the Hall-Petch slope for the initial condition

For the initial condition, we have the yield strength (\(\sigma_{y1}\)), grain size (\(d_1\)), and \(\sigma_0\), so we can rearrange the Hall-Petch equation to find the Hall-Petch slope. From the equation: \(\sigma_y = \sigma_0 + k_y\cdot d^{-1/2}\), We have: \(k_y = (\sigma_y - \sigma_0) \cdot d^{1/2}\) Given: \(d_1 = 0.01\:mm = 0.01\cdot10^{-3}m\) \(\sigma_{y1} = 150\:MPa = 150\cdot10^6\:Pa\) \(\sigma_{0} = 25\:MPa = 25\cdot10^6\:Pa\) Plug in these values to find the Hall-Petch slope: \(k_y = (150\cdot10^6 - 25\cdot10^6) \cdot (0.01\cdot10^{-3})^{1/2}\) \(k_y = 125\cdot10^6 \cdot 10^{-1.5}\) \(k_y = 1.25\cdot10^7\:Pa\cdot m^{1/2}\)
03

Determine the grain size after the heating process

To do this, we need to use the grain growth equation with the given temperature and time: \(d^3 - d_1^3 = K(t)\) where \(K(t)\) is a material property, and \(d\) is the grain size after heating. However, in this exercise, we are not provided with the grain growth rate constant. Therefore, we cannot determine the grain size after the heating process. Instead, we will assume that the grain size does not change significantly with the given heat treatment and use the initial grain size for our calculation.
04

Calculate the yield strength after the heating process

Using the Hall-Petch equation and assuming the same grain size, we can estimate the yield strength after the heating process. Given: \(\sigma_0 = 25\:MPa = 25\cdot10^6\:Pa\) Using the initial grain size: \(d = d_1 = 0.01\cdot10^{-3}m\) \(k_y = 1.25\cdot10^7\:Pa\cdot m^{1/2}\) (from step 2) Because we don't have information about the changes in the value of \(\sigma_0\) due to the heating process, we will assume that \(\sigma_0\) remains the same. So, we can calculate the estimated yield strength after heating using the Hall-Petch equation: \(\sigma_{y2} = \sigma_0 + k_y\cdot d^{-1/2}\) \(\sigma_{y2} = 25\cdot10^6 + 1.25\cdot10^7\cdot(0.01\cdot10^{-3})^{-1/2}\) \(\sigma_{y2} = 25\cdot10^6 + 1.25\cdot10^7\cdot10^{1.5}\) \(\sigma_{y2} = 25\cdot10^6 + 125\cdot10^6\) \(\sigma_{y2} = 150\cdot10^6\:Pa\) As we assumed that the heating process did not cause any significant changes in the grain size or the value of \(\sigma_0\), we get the same yield strength as the initial yield strength (\(\sigma_{y2} = 150\:MPa\)).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Yield Strength
Understanding yield strength is pivotal when studying materials science, especially in the context of mechanical properties of metals and alloys. Yield strength represents the stress at which a material begins to deform plastically. This is the point on the stress-strain curve that marks the end of elastic behavior and the start of permanent deformation. Knowing the yield strength of a material is critical because it defines the allowable limit to which the material can be loaded without causing permanent deformation.

When heat treatment is applied, the yield strength can be altered. This change is due to the microstructural transformations that occur at elevated temperatures. For instance, heating can cause the dislocation density within the grains to decrease, potentially reducing yield strength. However, if heating leads to precipitation hardening, the yield strength may increase due to the impediments to dislocation motion created by the precipitates. Understanding and predicting these changes is essential in material design and processing.
Grain Size
The grain size of a metallic or polycrystalline material significantly influences its mechanical properties. Grains are individual crystals within a metal that have misorientations at their boundaries. Generally, materials with smaller grains have higher yield strengths and better toughness due to the grain boundary area serving as a barrier to dislocation motion.

This concept is quantified in the Hall-Petch equation, where yield strength is inversely proportional to the square root of the grain size. However, it's important to understand that grain size can change under certain conditions, such as during heat treatment. Heat treatment can cause the grains in a metal to grow larger, especially when held at high temperatures for an extended period. The growth rate often depends on temperature and time. Despite this potential change, the assumption of constant grain size is made in some calculations for simplicity, as was the case in the given exercise where the Hall-Petch slope was calculated with initial grain size data.
Heat Treatment
The process of heat treatment involves heating and cooling a material in a specific way to alter its microstructure and, as a result, its mechanical properties. Heat treatments such as annealing, quenching, and tempering are designed to achieve desired outcomes like increased hardness, enhanced ductility, improved toughness, or specific residual stress states.

During heat treatment, factors such as temperature, time, and cooling rate will determine the final grain size and phase distribution within the metal. An increase in temperature can lead to grain growth, which may reduce yield strength due to the Hall-Petch relationship, whereas other heat treatments can lead to the formation of new, harder phases or solution strengthening. In the exemplified problem, the yield strength estimation after heat treatment was derived under the assumption that the heat treatment did not significantly modify either the grain size or the initial value of \(\sigma_0\), though in real-world applications, such modifications should be anticipated and accounted for.

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Most popular questions from this chapter

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

Experimentally, it has been observed for single crystals of a number of metals that the critical resolved shear stress \(\tau_{\mathrm{crs}}\) is a function of the dislocation density \(\rho_{D}\) as $$ \tau_{\text {crss }}=\tau_{0}+A \sqrt{\rho_{D}} $$ where \(\tau_{0}\) and \(A\) are constants. For copper, the critical resolved shear stress is \(0.69 \mathrm{MPa}\) (100 psi) at a dislocation density of \(10^{4} \mathrm{~mm}^{-2}\). If it is known that the value of \(\tau_{0}\) for copper is \(0.069 \mathrm{MPa}\) (10 psi), compute \(\tau_{\mathrm{crss}}\) at a dislocation density of \(10^{6} \mathrm{~mm}^{-2}\).

A cylindrical specimen of a nickel alloy having an elastic modulus of 207 GPa \(\left(30 \times 10^{6}\right.\) psi) and an original diameter of \(10.2 \mathrm{~mm}(0.40 \mathrm{in}\).) experiences only elastic deformation when a tensile load of \(8900 \mathrm{~N}\left(2000 \mathrm{lb}_{\mathrm{f}}\right)\) is applied. Compute the maximum length of the specimen before deformation if the maximum allowable elongation is \(0.25 \mathrm{~mm}\) \((0.010 \mathrm{in} .)\)

An aluminum bar \(125 \mathrm{~mm}\) (5.0 in.) long and having a square cross section \(16.5 \mathrm{~mm}(0.65 \mathrm{in}\).) on an edge is pulled in tension with a load of 66,700 \(\mathrm{N}\left(15,000 \mathrm{lb}_{\mathrm{f}}\right)\) and experiences an elongation of \(0.43 \mathrm{~mm}\left(1.7 \times 10^{-2}\right.\) in.). Assuming that the deformation is entirely elastic, calculate the modulus of elasticity of the aluminum.

Explain the differences in grain structure for a metal that has been cold worked and one that has been cold worked and then recrystallized.

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