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Chapter 7: Problem 36

(a) What is the driving force for recrystallization? (b) For grain growth?

Short Answer

Expert verified
Answer: The driving force for recrystallization is the reduction in stored energy. Recrystallization occurs when new defect-free grains nucleate and grow, replacing the deformed grains, and releasing the stored energy. Question: What is the driving force for grain growth? Answer: The driving force for grain growth is the reduction in the total grain boundary energy. This reduction in energy is achieved by the elimination of smaller grains and growth of larger grains, leading to a state with lower surface energy.

Step by step solution

01

(Definitions)

Recrystallization is a process in which deformed grains of a crystalline material are replaced by a new set of defect-free grains that nucleate and grow until the original deformed grains are entirely consumed. This process occurs mainly in materials subjected to mechanical deformation, causing an increase in stored energy due to dislocations and other defects. Grain growth is the process in which grains increase in size by eliminating other grains with higher surface energy. It often takes place after recrystallization and occurs in most materials, including metals, ceramics, and polymers.
02

(Driving force for recrystallization)

The driving force for recrystallization is the reduction in the stored energy. When a material is mechanically deformed, the internal strain energy is increased, mainly due to dislocations and other defects in the crystal lattice. This high stored energy state is thermodynamically unfavorable. Recrystallization occurs when the new defect-free grains nucleate and grow, replacing the deformed grains, and releasing the stored energy. In summary, the driving force for recrystallization is the reduction in stored energy.
03

(Driving force for grain growth)

The driving force for grain growth is the reduction in the total grain boundary energy. In a polycrystalline material, the grain boundaries are the regions of higher surface energy. When grain growth occurs, the total surface area of grain boundaries decreases, which reduces the overall grain boundary energy. This reduction in energy is achieved by the elimination of smaller grains and growth of larger grains, leading to a state with lower surface energy. Therefore, the driving force for grain growth lies in minimizing the total grain boundary energy.

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Most popular questions from this chapter

The lower yield point for an iron that has an average grain diameter of \(5 \times 10^{-2} \mathrm{~mm}\) is 135 MPa (19,500 psi). At a grain diameter of \(8 \times\) \(10^{-3} \mathrm{~mm}\), the yield point increases to \(260 \mathrm{MPa}\) \((37,500 \mathrm{psi})\). At what grain diameter will the lower yield point be \(205 \mathrm{MPa}(30,000 \mathrm{psi})\) ?

Experimentally, it has been observed for single crystals of a number of metals that the critical resolved shear stress \(\tau_{\text {crsss }}\) 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 \(2.10 \mathrm{MPa}\) (305 psi) at a dislocation density of \(10^{5} \mathrm{~mm}^{-2}\). If it is known that the value of \(A\) for copper is \(6.35\) \(\times 10^{-3} \mathrm{MPa} \cdot \mathrm{mm}(0.92 \mathrm{psi} \cdot \mathrm{mm})\), compute \(\tau_{\text {crss }}\) at a dislocation density of \(10^{7} \mathrm{~mm}^{-2}\).

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

One slip system for the BCC crystal structure is \(\\{110\\}(111\rangle\). In a manner similar to Figure \(7.6 b\), sketch a \\{110\\}-type plane for the BCC structure, representing atom positions with circles. Now, using arrows, indicate two different \(\langle 111\rangle\) slip directions within this plane.

(a) A single crystal of a metal that has the BCC crystal structure is oriented such that a tensile stress is applied in the \([100]\) direction. If the magnitude of this stress is \(4.0 \mathrm{MPa}\), compute the resolved shear stress in the \([1 \overline{11}]\) direction on each of the \((110),(011)\), and \((10 \overline{1})\) planes. (b) On the basis of these resolved shear stress values, which slip system(s) is (are) most favorably oriented?
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