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Chapter 6: Problem 39

Consider a hypothetical material that has a grain diameter of \(2.1 \times 10^{-2} \mathrm{~mm}\). After a heat treatment at \(600^{\circ} \mathrm{C}\) for \(3 \mathrm{~h}\), the grain diameter has increased to \(7.2 \times 10^{-2} \mathrm{~mm}\). Compute the grain diameter when a specimen of this same original material (i.e., \(d_{0}=2.1 \times 10^{-2} \mathrm{~mm}\) ) is heated for \(1.7 \mathrm{~h}\) at \(600^{\circ} \mathrm{C}\). Assume the \(n\) grain diameter exponent has a value of 2

Short Answer

Expert verified
Answer: The grain diameter after heating for 1.7 hours is 4.8脳10鈦宦� mm.

Step by step solution

01

We are given the following information: - Initial grain diameter: \(d_0 = 2.1\times10^{-2}\mathrm{~mm}\) - Grain diameter after \(3\mathrm{~h}\) of heat treatment: \(d_{3h} = 7.2\times10^{-2}\mathrm{~mm}\) - Temperature: \(600^{\circ}\mathrm{C}\) - Heat treatment time for the desired grain diameter: \(t = 1.7\mathrm{~h}\) - Grain diameter exponent: \(n = 2\) #Step 2: Calculate the constant K using d_{3h}#

Using the given values of \(d_0\), \(d_{3h}\), and \( n\), we can solve for the constant \(K\): $$ K = \frac{d_{3h}^n - d_0^n}{t_{3h}} $$ Plug in the given values: $$ K = \frac{(7.2\times10^{-2})^2 - (2.1\times10^{-2})^2}{3\mathrm{~h}} $$ Calculate the value of \(K\): $$ K = 3.12\times10^{-3}\mathrm{~mm^2/h} $$ #Step 3: Calculate the grain diameter for heating time t#
02

Now that we have the constant \(K\), we can use it to calculate the grain diameter after heating for \(1.7\mathrm{~h}\): $$ d^n - d_0^n = K \cdot t $$ Rearrange the equation to find \(d\): $$ d = \sqrt[n]{d_0^n + K \cdot t} $$ Plug in the known values: $$ d = \sqrt[2]{(2.1\times10^{-2})^2 + (3.12\times10^{-3})(1.7)} $$ Calculate the grain diameter \(d\): $$ d = 4.8\times10^{-2}\mathrm{~mm} $$ #Step 4: State the final answer#

The grain diameter of the hypothetical material after heating at \(600^{\circ}\mathrm{C}\) for \(1.7\mathrm{~h}\) is \(4.8\times10^{-2}\mathrm{~mm}\).

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Key Concepts

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

Heat Treatment
Heat treatment is a controlled process used to alter the physical and sometimes chemical properties of a material. It's a crucial part of materials science that aims to improve various attributes such as strength, hardness, ductility, and toughness of metals and their alloys. By heating materials to a specific temperature and maintaining that temperature for a set period, followed by cooling at a certain rate, the microstructure of the material can be changed. This results in altering the material's physical properties to meet specific requirements.

For instance, in the case of our hypothetical material, heat treatment at 600掳C leads to grain growth. These grains, which are crystallites in the metal, determine many of its mechanical properties. Smaller grains can make a material stronger, while larger grains can enhance ductility. Grain size is also influenced by the duration of the heat treatment. As seen in the exercise, a 3-hour exposure resulted in significant growth in grain size, contributing to changes in the material's properties.
Grain Diameter Calculation
Calculating the grain diameter is a way to quantify changes in the microstructure of materials after heat treatment. Grain diameter calculations are typically guided by empirical relationships, involving constants that characterize the specific behavior of a material under certain conditions. One common relationship applied for grain growth as a result of heat treatment is: \[ d^n - d_0^n = K \cdot t \] where:
  • \( d \) is the final grain diameter
  • \( d_0 \) is the initial grain diameter
  • \( t \) is the time of heat treatment
  • \( n \) is the grain diameter exponent which is characteristic of the material's response to heat
  • \( K \) is a material-specific constant
The grain diameter exponent, \( n \), typically varies with the material and the heat-treatment conditions. In the exercise, an exponent of 2 is used, which relates to the specific kinetics of grain growth for the material in question. Using this relationship ensures that the calculated grain diameter accounts for the initial size, thermal exposure time, and the nature of the material's response.
Materials Science
Materials science is an interdisciplinary field that focuses on understanding and manipulating the properties of materials to create new products or enhance existing ones. It integrates knowledge from chemistry, physics, engineering, and even biology to innovate in materials technology. In materials science, the study of heat treatment and subsequent grain growth is essential, as these processes significantly influence the performance and reliability of materials in application.

The hypothetical exercise provided illustrates how materials scientists can predict changes in microstructure, such as grain size, that occur in response to thermal processes. By using mathematical models and understanding the properties of the material, they can tailor the heat treatment process to achieve a desired outcome. It's these principles of materials science that enable us to develop materials with properties optimized for specific uses, from aerospace components to medical implants, and everyday consumer goods.

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Most popular questions from this chapter

In Section \(2.6\) it was noted that the net bonding energy \(E_{N}\) between two isolated positive and negative ions is a function of interionic distance \(r\) as follows: $$ E_{N}=-\frac{A}{r}+\frac{B}{r^{n}} $$ where \(A, B\), and \(n\) are constants for the particular ion pair. Equation \(6.25\) is also valid for the bonding energy between adjacent ions in solid materials. The modulus of elasticity \(E\) is proportional to the slope of the interionic force-separation curve at the equilibrium interionic separation; that is, $$ E \propto\left(\frac{d F}{d r}\right)_{r_{0}} $$ Derive an expression for the dependence of the modulus of elasticity on these \(A, B\), and \(n\) parameters (for the two-ion system) using the following procedure: 1\. Establish a relationship for the force \(F\) as a function of \(r\), realizing that $$ F=\frac{d E_{N}}{d r} $$ 2\. Now take the derivative \(d F / d r\). 3\. Develop an expression for \(r_{0}\), the equilibrium separation. Because \(r_{0}\) corresponds to the value of \(r\) at the minimum of the \(E_{N}\)-versus-r curve (Figure \(2.8 b\) ), take the derivative \(d E_{N} / d r\), set it equal to zero, and solve for \(r\), which corresponds to \(r_{0}\). 4\. Finally, substitute this expression for \(r_{0}\) into the relationship obtained by taking \(d F / d r\).

A cylindrical rod of steel \(\left(E=207 \mathrm{GPa}, 30 \times 10^{6}\right.\) psi) having a yield strength of \(310 \mathrm{MPa}(45,000\) psi) is to be subjected to a load of \(11,100 \mathrm{~N}\) (2500 lb_{f } ). If the length of the rod is \(500 \mathrm{~mm}\) (20.0 in.), what must be the diameter to allow an elongation of \(0.38 \mathrm{~mm}(0.015 \mathrm{in} .)\) ?

List four major differences between deformation by twinning and deformation by slip relative to mechanism, conditions of occurrence, and final result.

(a) From Figure \(7.25\), compute the length of time required for the average grain diameter to increase from \(0.03\) to \(0.3 \mathrm{~mm}\) at \(600^{\circ} \mathrm{C}\) for this brass material. (b) Repeat the calculation, this time using \(700^{\circ} \mathrm{C}\).

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).
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