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Chapter 7: Problem 6

(a) Compare planar densities (Section \(3.11\) and Problem 3.60) for the \((100),(110)\), and (111) planes for FCC. (b) Compare planar densities (Problem 3.61) for the (100), (110), and (111) planes for BCC.

Short Answer

Expert verified
Question: Compare the planar densities of the (100), (110), and (111) planes for FCC and BCC crystal structures. Answer: For the (100) and (110) planes, the planar density is higher in the FCC structure, and for the (111) plane, the planar density is the same in both structures.

Step by step solution

01

Calculate number of atoms per plane for each plane

For the FCC structure, the lattice points for each plane are at the corners and face centers. Now let's count the number of atoms on each plane: - For the (100) plane, there are 4 atoms at the corners and 4 atoms at the face centers, making 4 complete atoms. - For the (110) plane, there are 4 atoms at the corners and 2 at the face centers, making 3 complete atoms. - For the (111) plane, there are 3 atoms at the corners and 3 at the face centers, making 3 complete atoms.
02

Calculate the area of the planes

Let a be the length of the unit cell edge. Then the area of each plane can be calculated as follows: - For the (100) plane, area = \(a^2\) - For the (110) plane, area = \(a^2\sqrt{2} / 2\) - For the (111) plane, area = \(a^2\sqrt{3} / 2\)
03

Calculate the planar density

Now we will calculate the planar density for each plane (number of atoms per area): - For the (100) plane, planar density = \(4 / a^2\) - For the (110) plane, planar density = \(3 / (a^2\sqrt{2} / 2)\) - For the (111) plane, planar density = \(3 / (a^2\sqrt{3} / 2)\) (b) BCC structure
04

Calculate number of atoms per plane for each plane

For the BCC structure, the lattice points for each plane are at the corners and body center. Now let's count the number of atoms on each plane: - For the (100) plane, there are 4 atoms at the corners, making 1 complete atom. - For the (110) plane, there are 4 atoms at the corners and 1 at the body center, making 2 complete atoms. - For the (111) plane, there are 3 atoms at the corners and 1 at the body center, making 2 complete atoms.
05

Calculate the area of the planes

Let a be the length of the unit cell edge. Then the area of each plane can be calculated as follows: - For the (100) plane, area = \(a^2\) - For the (110) plane, area = \(a^2\sqrt{2} / 2\) - For the (111) plane, area = \(a^2\sqrt{3} / 2\)
06

Calculate the planar density

Now we will calculate the planar density for each plane (number of atoms per area): - For the (100) plane, planar density = \(1 / a^2\) - For the (110) plane, planar density = \(2 / (a^2\sqrt{2} / 2)\) - For the (111) plane, planar density = \(2 / (a^2\sqrt{3} / 2)\) Upon comparing the planar densities for the FCC and BCC structures, we can observe that for the (100) and (110) planes, the planar density is higher in the FCC structure, and for the (111) plane, the planar density is the same in both structures.

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Most popular questions from this chapter

A cylindrical specimen of a nickel alloy having an elastic modulus of 207 GPa (30\times10^{6 psi) and an } original diameter of \(10.2 \mathrm{~mm}(0.40\) in.) experiences only elastic deformation when a tensile load of \(8900 \mathrm{~N}\left(2000 \mathrm{lb}_{i}\right)\) is applied. Compute the maximum length of the specimen before deformation if the maximum allowable elongation is \(0.25 \mathrm{~mm}\) \((0.010\) in.).

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

The lower yield point for an iron that has an († average grain diameter of \(1 \times 10^{-2} \mathrm{~mm}\) is 230 MPa (33,000 psi). At a grain diameter of \(6 \times 10^{-3}\) \(\mathrm{mm}\), the yield point increases to \(275 \mathrm{MPa}(40,000\) psi). At what grain diameter will the lower yield point be \(310 \mathrm{MPa}(45,000 \mathrm{psi}) ?\)

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 \(10^{6} \mathrm{~mm}^{-2}\).

Consider a cylindrical specimen of some hypothetical metal alloy that has a diameter of \(10.0\) \(\mathrm{mm}(0.39 \mathrm{in} .) .\) A tensile force of \(1500 \mathrm{~N}\left(340 \mathrm{lb}_{\mathrm{f}}\right)\) produces an elastic reduction in diameter of \(6.7 \times 10^{-4} \mathrm{~mm}\left(2.64 \times 10^{-5}\right.\) in.). Compute the elastic modulus of this alloy, given that Poisson's ratio is \(0.35\).
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