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Chapter 7: Problem 21

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

Short Answer

Expert verified
Answer: HCP metals are generally more brittle than FCC and BCC metals because of the limited slip systems available in the HCP structure. With fewer slip systems, HCP metals show limited dislocation movement and plastic deformation, making them more susceptible to fracture under stress. As a result, HCP metals are generally brittle compared to FCC and BCC metals, which have more slip systems and, consequently, higher ductility.

Step by step solution

01

Understanding different crystal structures

In order to analyze the given exercise, we need to have a clear understanding of HCP, FCC, and BCC structures. HCP, FCC, and BCC are different types of crystal structures formed by metallic elements. In HCP structure, the atoms are arranged in a hexagonal pattern with two layers. The second layer is placed in the hollows of the first layer, and the third layer is positioned in a way that it is identical to the first layer. FCC and BCC structures are both cubic in nature, with the atoms arranged in a three-dimensional pattern. In the FCC structure, additional atoms are present at the center of each face of the unit cell, while in BCC, a single atom is positioned in the center of the unit cell.
02

Relationship between crystal structures and mechanical properties

The mechanical properties of a metal are influenced by its crystal structure. In general, metals that can easily change their shape without breaking upon an applied force have a higher ductility, while metals that tend to fracture under stress are more brittle. One of the factors that influence ductility is the number of slip systems, which allow dislocations to move through the material. A greater number of slip systems results in a more ductile material, as it can accommodate more dislocation movement and plastic deformation.
03

Analyzing the slip systems in different crystal structures

The number of available slip systems in a crystal structure directly affects the ductility and brittleness of the material. FCC and BCC structures have a higher number of slip systems compared to HCP. FCC has 12 slip systems, which makes it relatively more ductile than the other two structures. BCC has six effective slip systems, making it moderately ductile. In contrast, HCP metals generally only have three slip systems available, which makes them less ductile and more brittle than FCC and BCC metals.
04

Conclusion

HCP metals are typically more brittle than FCC and BCC metals because of the limited slip systems available in the HCP structure. With fewer slip systems, HCP metals show limited dislocation movement and plastic deformation, making them more susceptible to fracture under stress. As a result, HCP metals are generally brittle compared to FCC and BCC metals, which have more slip systems and, consequently, higher ductility.

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Most popular questions from this chapter

Consider a single crystal of some hypothetical metal that has the BCC crystal structure and is oriented such that a tensile stress is applied along a [121] direction. If slip occurs on a (101) plane and in a \([\overline{1} 11]\) direction, compute the stress at which the crystal yields if its critical resolved shear stress is \(2.4 \mathrm{MPa}\)

Consider a metal single crystal oriented such that the normal to the slip plane and the slip direction are at angles of \(60^{\circ}\) and \(35^{\circ},\) respectively, with the tensile axis. If the critical resolved shear stress is \(6.2 \mathrm{MPa}(900 \mathrm{psi}),\) will an applied stress of \(12 \mathrm{MPa}(1750 \mathrm{psi})\) cause the single crystal to yield? If not, what stress will be necessary?

A single crystal of zinc is oriented for a tensile test such that its slip plane normal makes an angle of \(65^{\circ}\) with the tensile axis. Three possible slip directions make angles of \(30^{\circ}, 48^{\circ}\) and \(78^{\circ}\) with the same tensile axis. (a) Which of these three slip directions is most favored? (b) If plastic deformation begins at a tensile stress of \(2.5 \mathrm{MPa}(355 \mathrm{psi})\), determine the critical resolved shear stress for zinc.

Two previously undeformed specimens of the same metal are to be plastically deformed by reducing their cross-sectional areas. One has a circular cross section, and the other is rectangular; during deformation the circular cross section is to remain circular, and the rectangular is to remain as such. Their original and deformed dimensions are as follows: $$\begin{array}{lcc} \hline & \begin{array}{c} \text {Circular} \\ \text {(diameter, } \mathbf{m m} \text { ) } \end{array} & \begin{array}{c} \text {Rectangular} \\ (\mathbf{m m}) \end{array} \\ \hline \text { Original dimensions } & 18.0 & 20 \times 50 \\ \text { Deformed dimensions } & 15.9 & 13.7 \times 55.1 \\ \hline \end{array}$$ Which of these specimens will be the hardest after plastic deformation, and why?

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