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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

(a) Compare planar densities (Section \(3.11\) and Problem 3.54) for the (100), (110), and (111) planes for FCC. (b) Compare planar densities (Problem 3.55) for the (100), (110), and (111) planes for BCC.

(a) From the plot of yield strength versus (grain diameter) \(^{-1 / 2}\) for a \(70 \mathrm{Cu}-30 \mathrm{Zn}\) cartridge brass, Figure \(7.15\), determine values for the constants \(\sigma_{0}\) and \(k_{y}\) in Equation \(7.7\). (b) Now predict the yield strength of this alloy when the average grain diameter is \(1.0 \times 10^{-3} \mathrm{~mm}\)

Briefly explain why small-angle grain boundaries are not as effective in interfering with the slip process as are high-angle grain boundaries.

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 \(205 \mathrm{MPa}(30,000 \mathrm{psi})\) ?

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.
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