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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

(a) Equations \(6.4 a\) and \(6.4 \mathrm{~b}\) are expressions for normal \(\left(\sigma^{\prime}\right)\) and shear \(\left(\tau^{\prime}\right)\) stresses, respectively, as a function of the applied tensile stress \((\sigma)\) and the inclination angle of the plane on which these stresses are taken ( \(\theta\) of Figure 6.4). Make a plot showing the orientation parameters of these expressions (i.e., \(\cos ^{2} \theta\) and \(\sin \theta \cos \theta\) ) versus \(\theta\). (b) From this plot, at what angle of inclination is the normal stress a maximum? (c) At what inclination angle is the shear stress a maximum?

An undeformed specimen of some alloy has an average grain diameter of \(0.050 \mathrm{~mm}\). You are asked to reduce its average grain diameter to \(0.020 \mathrm{~mm}\). Is this possible? If so, explain the procedures you would use and name the processes involved. If it is not possible, explain why.

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

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).

A specimen of copper having a rectangular cross section \(15.2 \mathrm{~mm} \times 19.1 \mathrm{~mm}(0.60 \mathrm{in} . \times 0.75 \mathrm{in}\).) is pulled in tension with \(44,500 \mathrm{~N}\left(10,000 \mathrm{lb}_{\mathrm{f}}\right)\) force, producing only elastic deformation. Calculate the resulting strain.
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