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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°. 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°. 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

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

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. $$ \begin{array}{lcc} \hline \begin{array}{l} \text { Grain } \\ \text { diameter } \\ (\mathbf{m m}) \end{array} & \begin{array}{c} \text { Yield } \\ \text { Strength } \\ \text { (MPa) } \end{array} & \begin{array}{c} \text { Heat } \\ \text { Treating } \\ \text { Time (h) } \end{array} \\ \hline 0.028 & 300 & 10 \\ \hline 0.010 & 385 & 1 \\ \hline \end{array} $$

A cylindrical specimen of a nickel alloy having an elastic modulus of 207 GPa (30\times10^{6 psi) and an } original diameter of \(10.2 \mathrm{~mm}(0.40\) in.) experiences only elastic deformation when a tensile load of \(8900 \mathrm{~N}\left(2000 \mathrm{lb}_{i}\right)\) is applied. Compute the maximum length of the specimen before deformation if the maximum allowable elongation is \(0.25 \mathrm{~mm}\) \((0.010\) in.).

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.

(a) What is the approximate ductility (\%EL) of a brass that has a yield strength of 345 MPa \((50,000 \mathrm{psi}) ?\) (b) What is the approximate Brinell hardness of a 1040 steel having a yield strength of 620 MPa \((90,000 \mathrm{psi}) ?\)
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