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Chapter 7: Problem 20

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

Expert verified
Answer: Small-angle grain boundaries are less effective in hindering the slip process because they have a smaller misorientation angle and a lesser degree of lattice mismatch between neighboring grains, which makes it easier for dislocations to move across the grain boundary and continue along the slip plane. In contrast, high-angle grain boundaries have a larger misorientation angle and a greater lattice mismatch, posing a more significant obstacle for dislocation movement and interfering more effectively with the slip process.

Step by step solution

01

Defining grain boundaries and the slip process

Grain boundaries are the interfaces between different grains, or crystallites, in a polycrystalline material. The slip process refers to the dislocation movement within a grain, leading to plastic deformation in the material.
02

Identifying small-angle and high-angle grain boundaries

Small-angle grain boundaries have a small misorientation angle (less than 15 degrees) between adjacent grains, while high-angle grain boundaries have a larger misorientation angle (greater than 15 degrees). The misorientation angle determines the extent of lattice mismatch at the grain boundary.
03

Understanding the role of grain boundaries in the slip process

Grain boundaries act as obstacles to the slip process. When a dislocation (a line defect in the crystal structure) encounters a grain boundary, it may be stopped, absorbed, or forced to change its direction. The effectiveness of grain boundaries in hindering the slip process is determined by the misorientation angle or the degree of lattice mismatch between the neighboring grains.
04

Explaining the effectiveness of small-angle grain boundaries

In small-angle grain boundaries, the lattice mismatch between adjacent grains is relatively small, and therefore, the structures of neighboring grains are more compatible. As a result, dislocations can more easily move across the grain boundary and continue along the slip plane, causing less interference in the slip process.
05

Comparing to high-angle grain boundaries

In contrast, high-angle grain boundaries have a larger misorientation angle, leading to a greater degree of lattice mismatch between neighboring grains. This makes it more challenging for dislocations to move across the grain boundary, causing a more significant hindrance in the slip process. Therefore, high-angle grain boundaries are more effective in interfering with the slip process compared to small-angle grain boundaries.

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Most popular questions from this chapter

Consider a single crystal of some hypothetical metal that has the FCC crystal structure and is oriented such that a tensile stress is applied along a [102] direction. If slip occurs on a (111) plane and in a [101] direction, compute the stress at which the crystal yields if its critical resolved shear stress is \(3.42 \mathrm{MPa}\).

(a) Show, for a tensile test, that $$ \% \mathrm{CW}=\left(\frac{\epsilon}{\epsilon+1}\right) \times 100 $$ if there is no change in specimen volume during the deformation process (i.e., \(A_{0} l_{0}=A_{d} l_{d}\) ). (b) Using the result of part (a), compute the percent cold work experienced by naval brass (the stress-strain behavior of which is shown in Figure 6.12) when a stress of 400 MPa \((58,000\) psi) is applied.

Two previously undeformed cylindrical specimens of an alloy are to be strain hardened by reducing their cross-sectional areas (while maintaining their circular cross sections). For one specimen, the initial and deformed radii are \(16 \mathrm{~mm}\) and \(11 \mathrm{~mm}\), respectively. The second specimen, with an initial radius of \(12 \mathrm{~mm}\), must have the same deformed hardness as the first specimen; compute the second specimen's radius after deformation.

An undeformed specimen of some alloy has an average grain diameter of \(0.040 \mathrm{~mm}\). You are asked to reduce its average grain diameter to \(0.010 \mathrm{~mm}\). Is this possible? If so, explain the procedures you would use and name the processes involved. If it is not possible, explain why.

(a) What is the approximate ductility (\%EL) of a brass that has a yield strength of \(275 \mathrm{MPa}\) \((40,000 \mathrm{psi}) ?\) (b) What is the approximate Brinell hardness of a 1040 steel having a yield strength of 690 \(\mathrm{MPa}(100,000 \mathrm{psi})\) ?
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