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Chapter 7: 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

Expert verified
Answer: Small-angle grain boundaries are less effective in interfering with the slip process because they have a low dislocation density and smaller misorientation. This results in less distorted atomic arrangements, allowing dislocations to glide or climb through the grain boundaries more easily. In contrast, high-angle grain boundaries have a higher misorientation and more significant lattice distortion, creating a more significant barrier for dislocation movement, effectively restricting the slip process and impacting the material's mechanical properties.

Step by step solution

01

Understanding Grain Boundaries

Grain boundaries are interfaces between different crystallites (grains) in a polycrystalline material. Grain boundaries are formed due to the misalignment of atomic planes when the crystal structure of a material is formed. There are two types of grain boundaries: small-angle grain boundaries and high-angle grain boundaries.
02

Small-Angle Grain Boundaries

Small-angle grain boundaries have a relatively small misorientation between the two adjoining grains. They are formed when the misorientation angle is less than 15掳. These boundaries consist of an array of dislocations, and since the misorientation is small, the dislocation density in these boundaries is low.
03

High-Angle Grain Boundaries

High-angle grain boundaries occur when the misorientation angle between the two grains is greater than 15掳. In these boundaries, there is a higher lattice distortion, and the atoms' arrangement is more random compared to small-angle grain boundaries. This leads to more significant defects and irregularities in the material structure.
04

The Slip Process

The slip process is the primary mechanism of plastic deformation in materials. It occurs when dislocations in the crystal structure move under an applied stress, causing one plane of atoms to slide over another plane. The ease of the slip process is essential in determining a material's mechanical properties like ductility and strength.
05

Effect of Small-Angle Grain Boundaries on the Slip Process

Since small-angle grain boundaries have a low dislocation density and smaller misorientation, the atoms in these boundaries are not as distorted as in high-angle grain boundaries. The slip process can occur relatively easily across these boundaries as the dislocations can glide or climb through the low-density dislocation arrays. Therefore, small-angle grain boundaries are less effective in interfering with the slip process.
06

Effect of High-Angle Grain Boundaries on the Slip Process

High-angle grain boundaries have a higher misorientation and more significant lattice distortion, with a random arrangement of atoms. This causes a more significant barrier to dislocation movement, as they face more significant resistance and distortion when trying to move across these boundaries. As a result, high-angle grain boundaries effectively interfere with the slip process, making them more efficient in restricting plastic deformation and impacting the material's mechanical properties.

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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 [112] direction. If slip occurs on a (111) plane and in a [011] direction, and the crystal yields at a stress of \(5.12 \mathrm{MPa}\), compute the critical resolved shear stress.

Briefly explain why HCP metals are typically more brittle than FCC and BCC metals.

Is it possible for two screw dislocations of opposite sign to annihilate each other? Explain your answer.

As noted in Section \(3.15\), for single crystals of some substances, the physical properties are anisotropic that is, they depend on crystallographic direction. One such property is the modulus of elasticity. For cubic single crystals, the modulus of elasticity in a general \([u v w]\) direction, \(E_{\text {uww }}\) is described by the relationship $$ \begin{aligned} \frac{1}{E_{u v w}}=& \frac{1}{E_{(100)}}-3\left(\frac{1}{E_{(100)}}-\frac{1}{E_{\\{111)}}\right) \\\ &\left(\alpha^{2} \beta^{2}+\beta^{2} \gamma^{2}+\gamma^{2} \alpha^{2}\right) \end{aligned} $$ where \(E_{\\{100)}\) and \(E_{\langle 111\rangle}\) are the moduli of elasticity in the \([100]\) and \([111]\) directions, respectively; \(\alpha, \beta\), and \(\gamma\) are the cosines of the angles between \([u v w]\) and the respective [100], [010], and [001] directions. Verify that the \(E_{\\{110\rangle}\) values for aluminum, copper, and iron in Table \(3.4\) are correct.

(a) Define a slip system. (b) Do all metals have the same slip system? Why or why not?
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