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Chapter 6: Problem 21

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

Short Answer

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Answer: Small-angle grain boundaries are less effective in interfering with the slip process compared to high-angle grain boundaries because they have a smaller misorientation angle between adjacent grains. This leads to a less significant change in the atomic arrangement across the boundary, allowing dislocations to move more freely across the boundary. In contrast, high-angle grain boundaries serve as stronger barriers to dislocation movement and the resulting plastic deformation due to their larger misorientation angle and more significant change in atomic arrangement.

Step by step solution

01

Definition of Grain Boundaries

Grain boundaries are the interfaces between crystallites or grains in a polycrystalline material. They have a significant impact on the mechanical and other properties of materials because they interfere with the slip process. The effectiveness of this interference depends on the angle between the adjacent grains.
02

Small-Angle Grain Boundaries

Small-angle grain boundaries are those with a relatively small misorientation angle (usually less than 15 degrees) between the adjacent grains. In this case, the atomic arrangement across the boundary is slightly changed, and it can be considered as an array of edge dislocations. Due to this small change in the atomic arrangement, the slip process is not significantly hindered, and dislocations can move relatively freely across the boundary.
03

High-Angle Grain Boundaries

High-angle grain boundaries have a larger misorientation angle (greater than 15 degrees) between the adjacent grains. In these cases, there is a more significant change in the atomic arrangement across the boundary, and there is no regular spacing of dislocations. This causes a higher degree of interference with the slip process. In fact, when a dislocation reaches a high-angle grain boundary, it may either be blocked, absorbed, or initiate the creation of a new dislocation in the neighboring grain, depending on the applied stress, temperature, and other factors.
04

Effect on Slip Process

The slip process, or dislocation movement, is crucial for the plastic deformation of materials. Small-angle grain boundaries do not significantly interfere with the slip process, so dislocations can move relatively freely across them. In contrast, high-angle grain boundaries act as more effective barriers for dislocation movement, which makes it harder for dislocations to propagate through the material and generate plastic deformation.
05

Conclusion

In summary, small-angle grain boundaries are not as effective in interfering with the slip process as high-angle grain boundaries because they have a smaller misorientation angle between adjacent grains, leading to a less significant change in the atomic arrangement across the boundary. This allows dislocations to move more freely across the boundary, while high-angle grain boundaries serve as stronger barriers to dislocation movement and the resulting plastic deformation.

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Most popular questions from this chapter

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 \(10^{6} \mathrm{~mm}^{-2}\).

Briefly cite the differences between the recovery and recrystallization processes.

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 \(d F / d r\).

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}

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 \(0.25 \mathrm{~mm}\) \((0.010 \mathrm{in} .)\)
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