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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

The critical resolved shear stress for iron is 27 MPa (4000 psi). Determine the maximum possible yield strength for a single crystal of Fe pulled in tension.

(a) Show, for a tensile test, that $$ \% \mathrm{CW}=\left(\frac{\epsilon}{\epsilon+1}\right) \times 100 $$ if there is no change in specimen volume during the deformation process (i.e., \(A_{0} l_{0}=A_{d} l_{d}\) ). (b) Using the result of part (a), compute the percent cold work experienced by naval brass (the stress-strain behavior of which is shown in Figure 6.12) when a stress of 400 MPa \((58,000\) psi) is applied.

(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.

Is it possible for two screw dislocations of opposite sign to annihilate each other? Explain your answer.

Consider a single crystal of silver oriented such that a tensile stress is applied along a \([001]\) direction. If slip occurs on a (111) plane and in a [ \(\overline{101}\) ] direction, and is initiated at an applied tensile stress of \(1.1 \mathrm{MPa}\) (160 psi), compute the critical resolved shear stress.
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