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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
Explain your answer. Answer: No, not all metals have the same slip system. The type and number of slip systems in a metal depend on its crystal structure and specific atomic arrangement. Different metals have different crystal structures, leading to variations in their atomic arrangements and lattice geometry. This results in different planes and directions with the highest atomic density and packing for various crystal structures. These differences in slip systems among various metals affect their mechanical behaviors under stress and their overall properties.

Step by step solution

01

Define a slip system

A slip system is a combination of a slip plane and a slip direction within that plane, which controls the crystallographic motion of atoms within a material, particularly metals, under stress. It is essential for understanding the plastic deformation and the mechanical properties of materials. In a slip system, the slip plane is the plane with the highest density of atoms, and the slip direction is the direction with the highest atomic packing.
02

Discuss whether all metals have the same slip system

No, not all metals have the same slip system. The type and number of slip systems in a metal depend on its crystal structure and its specific atomic arrangement.
03

Explain the reasons behind the differences in slip systems

Different metals have different crystal structures, which leads to variations in their atomic arrangements and lattice geometry. Consequently, the planes with the highest atomic density and the directions with the highest atomic packing will be different for various crystal structures. For example, FCC (face-centered cubic) metals, such as aluminum and copper, have {111} planes as their slip planes and <110> directions as their slip directions, while BCC (body-centered cubic) metals, such as iron and chromium, have {110} planes as their slip planes and <111> directions as their slip directions. The differences in slip systems among various metals affect their mechanical behaviors under stress and their overall properties.

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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 345 MPa \((50,000 \mathrm{psi}) ?\) (b) What is the approximate Brinell hardness of a 1040 steel having a yield strength of 620 MPa \((90,000 \mathrm{psi}) ?\)

Experimentally, it has been observed for single crystals of a number of metals that the critical resolved shear stress \(\tau_{\text {crss }}\) is a function of the dislocation density \(\rho_{D}\) as $$ \tau_{\mathrm{crss}}=\tau_{0}+A \sqrt{\rho_{D}} $$ where \(\tau_{0}\) and \(A\) are constants. For copper, the critical resolved shear stress is \(0.69 \mathrm{MPa}\) (100 psi) at a dislocation density of \(10^{4} \mathrm{~mm}^{-2}\). If it is known that the value of \(\tau_{0}\) for copper is \(0.069 \mathrm{MPa}\) (10 psi), compute \(\tau_{\text {crss }}\) at a dislocation density of \(10^{6} \mathrm{~mm}^{-2}\).

The following yield strength, grain diameter, and heat treatment time (for grain growth) data were gathered for an iron specimen that was heat treated at \(800^{\circ} \mathrm{C}\). Using these data, compute the yield strength of a specimen that was heated at \(800^{\circ} \mathrm{C}\) for \(3 \mathrm{~h}\). Assume a value of 2 for \(n\), the grain diameter exponent. $$ \begin{array}{lcc} \hline \begin{array}{l} \text { Grain } \\ \text { diameter } \\ (\mathbf{m m}) \end{array} & \begin{array}{c} \text { Yield } \\ \text { Strength } \\ \text { (MPa) } \end{array} & \begin{array}{c} \text { Heat } \\ \text { Treating } \\ \text { Time (h) } \end{array} \\ \hline 0.028 & 300 & 10 \\ \hline 0.010 & 385 & 1 \\ \hline \end{array} $$

A single crystal of zinc is oriented for a tensile test such that its slip plane normal makes an angle of \(65^{\circ}\) with the tensile axis. Three possible slip directions make angles of \(30^{\circ}, 48^{\circ}\), and \(78^{\circ}\) with the same tensile axis. (a) Which of these three slip directions is most favored? (b) If plastic deformation begins at a tensile stress of \(2.5 \mathrm{MPa}\) (355 psi), determine the critical resolved shear stress for zinc.

A cylindrical rod of steel \(\left(E=207 \mathrm{GPa}, 30 \times 10^{\circ}\right.\) psi) having a yield strength of \(310 \mathrm{MPa}(45,000\) psi) is to be subjected to a load of \(11,100 \mathrm{~N}\) (2500 \(\left.\mathrm{Ib}_{i}\right)\). If the length of the rod is \(500 \mathrm{~mm}(20.0 \mathrm{in}\).), what must be the diameter to allow an elongation of \(0.38 \mathrm{~mm}(0.015\) in.)?
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