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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

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.

To provide some perspective on the dimensions of atomic defects, consider a metal specimen that has a dislocation density of \(10^{4} \mathrm{mm}^{-2}\). Suppose that all the dislocations in \(1000 \mathrm{~mm}^{3}\left(1 \mathrm{~cm}^{3}\right)\) were somehow removed and linked end to end. How far (in miles) would this chain extend? Now suppose that the density is increased to \(10^{10} \mathrm{~mm}^{-2}\) by cold working. What would be the chain length of dislocations in \(1000 \mathrm{~mm}^{3}\) of material?

(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.

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

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 \(12.0 \mathrm{MPa} \cdot \mathrm{mm}^{1 / 2}\left(1740 \mathrm{psi} \cdot \mathrm{mm}^{1 / 2}\right)\)
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