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Chapter 7: Problem 21

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

Short Answer

Expert verified
HCP metals are typically more brittle than FCC and BCC metals because they have fewer slip systems. The limited number of slip systems in HCP metals makes it difficult for them to deform plastically under stress, resulting in a higher tendency to fracture. Conversely, FCC and BCC metals have many more slip systems, allowing them to accommodate higher stress levels by deforming plastically, making them less brittle.

Step by step solution

01

Understand the crystal structures of HCP, FCC, and BCC metals

Hexagonal close-packed (HCP), face-centered cubic (FCC), and body-centered cubic (BCC) are common crystal structures of metals. These structures define the arrangement of atoms in their unit cells and affect their mechanical properties such as ductility and brittleness. In an HCP structure, the atoms are closely packed in layers, with the top and bottom layers being hexagonal and the middle layer slightly shifted. An FCC structure has atoms located at the corners and faces of the cube. The unit cell is densely packed, which provides good slip systems. A BCC structure has atoms positioned at the corners of the cube, and one atom is present in the cube's center.
02

Understand slip systems

A slip system refers to the combination of a slip plane (a plane along which atoms can move) and a slip direction (the direction of atomic movement). More slip systems in a crystal structure usually result in higher ductility because it allows the material to deform plastically under stress.
03

Discuss the differences in slip systems of HCP, FCC, and BCC metals

FCC and BCC structures have more slip systems compared to HCP structures. In an FCC arrangement, there are 12 slip systems in total, while a BCC structure has 48 slip systems. In comparison, HCP structures usually have only three or six slip systems.
04

Relate the slip systems to brittleness in HCP, FCC, and BCC metals

Since HCP metals have fewer slip systems, they tend to be more brittle. Fewer slip systems make it difficult for the material to deform plastically under stress, resulting in a higher tendency to fracture. FCC and BCC metals, on the other hand, have many more slip systems, allowing them to accommodate higher stress levels by deforming plastically. This makes them less brittle than HCP metals. In conclusion, the brittleness of metals mainly depends on their slip systems. Due to the limited number of slip systems in HCP metals compared to FCC and BCC metals, HCP metals are typically more brittle.

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Most popular questions from this chapter

(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 \(\mathrm{MPa}(100,000 \mathrm{psi})\) ?

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 \(3.42 \mathrm{MPa}\).

One slip system for the BCC crystal structure is \(\\{110\\}(111\rangle\). In a manner similar to Figure \(7.6 b\), sketch a \\{110\\}-type plane for the BCC structure, representing atom positions with circles. Now, using arrows, indicate two different \(\langle 111\rangle\) slip directions within this plane.

An uncold-worked brass specimen of average grain size \(0.008 \mathrm{~mm}\) has a yield strength of 160 MPa \((23,500\) psi). Estimate the yield strength of this alloy after it has been heated to \(600^{\circ} \mathrm{C}\) for \(1000 \mathrm{~s}\), if it is known that the value of \(k_{y}\) is \(12.0 \mathrm{MPa} \cdot \mathrm{mm}^{1 / 2}\left(1740 \mathrm{psi} \cdot \mathrm{mm}^{1 / 2}\right)\)

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: $$ \begin{array}{lcr} \hline & \begin{array}{c} \text { Circular } \\ \text { (diameter, } \mathbf{m m}) \end{array} & \begin{array}{c} \text { Rectangular } \\ (\mathbf{m m}) \end{array} \\ \hline \text { Original dimensions } & 15.2 & 125 \times 175 \\ \text { Deformed dimensions } & 11.4 & 75 \times 200 \\ \hline \end{array} $$ Which of these specimens will be the hardest after plastic deformation, and why?
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