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Chapter 7: Problem 5

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

Short Answer

Expert verified
Why or why not? Answer: No, all metals do not have the same slip system. The slip system can vary between different metals, depending on their crystal structures and atomic bonding properties. The variability among slip systems in metals is caused by differences in their crystal structures (FCC, BCC, or HCP) and the type and strength of atomic bonding, which affect the way dislocations move through the crystal lattice. The number and orientation of slip planes and directions within a given metal also depend on its specific crystal structure and bonding properties.

Step by step solution

01

Define a Slip System

A slip system is a combination of a slip plane and a slip direction that allows deformation to occur in a crystalline material. In a slip system, a dislocation moves through the crystal lattice by causing the atoms to shift along the slip plane in a specific direction. This process is essential for understanding the plastic deformation of metals.
02

Discuss Slip Systems for Different Metals

No, all metals do not have the same slip system. The slip system can vary between different metals, depending on their crystal structures and atomic bonding properties.
03

Explain Why Slip Systems Vary Among Metals

The reason for variability among slip systems in metals lies in the differences in their crystal structures and bonding. Some metals have a face-centered cubic (FCC) structure, while others might have a body-centered cubic (BCC) or hexagonal close-packed (HCP) structure. The type and strength of atomic bonding among metals also vary, affecting the way dislocations move through the crystal lattice. Additionally, the number and orientation of slip planes and directions within a given metal depend on the specific crystal structure of that material. Therefore, the slip systems among metals can differ significantly based on their unique combination of crystal structure and bonding properties.

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Most popular questions from this chapter

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.

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.

Briefly cite the differences between recovery and recrystallization processes.

The lower yield point for an iron that has an average grain diameter of \(5 \times 10^{-2} \mathrm{~mm}\) is 135 MPa (19,500 psi). At a grain diameter of \(8 \times\) \(10^{-3} \mathrm{~mm}\), the yield point increases to \(260 \mathrm{MPa}\) \((37,500 \mathrm{psi})\). At what grain diameter will the lower yield point be \(205 \mathrm{MPa}(30,000 \mathrm{psi})\) ?

(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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