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110年 - 110 國立臺灣大學_碩士班招生考試_部分系所:材料科學(A)#100703
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3.Fundamentals of dislocation slip (11%):
(e) Some engineering alloys are extremely ductile at room temperature although they do not have closely-packed slip planes. Why? (2%)
相關申論題
(a) Martensitic transformation always generates hard and brittle product phase. (2%)
#421497
(b) It is very common to find a twinning system in metal that both the twinning plane and twinning shear direction are rational. (2%)
#421498
(c) Compared to homogeneous nucleation, heterogeneous nucleation has a smaller critical free energy barrier (△G') and critical radius (r*) both. (2%)
#421499
(d) We only need solid solution treatment and aging treatment for effective precipitation hardening. (2%)
#421500
(e) When a crystal is deformed in compression, the slip plane rotates so that it tends to become perpendicular to the stress axis. (2%)
#421501
(a) Write down all seven crystal systems. Which two systems do not have any perpendicular interaxial angles (that is,α≠ 90°, β≠ 90°, γ≠ 90°?(3%)
#421502
(b) The interplanar spacing is a function of the Miller indices and lattice parameters. For example, for crystal structures having cubic symmetry:d Derive the interplanar spacing for crystal structures having tetragonal symmetry. (3%)
#421503
(c) A BCT unit cell in martensitic steel is simply a body-centered cube that has been elongated along the c-axis. Due to the reduced symmetry from BCC, the martensite (002) and (200) peaks spit in X-ray diffraction pattern may be observed. Today a martensitic steel sampie with unknown carbon concentration is detected to have 2θ = 5.7° difference between (002) and (200) planes while using X-ray diffraction with Cr-Ka radiation (wavelength = 0.22897 nm). From a linear relationship between carbon content and cla ratio: , and lattice parameter a = 0.2868 nm, evaluate the carbon concentration of this martensitic steel. (4%)
#421504
(a) What is the definition of the Burgers vector? Also, write down the Burgers vector b for FCC, BCC, and HCP structures. (2%)
#421505
(b) Zinc has an HCP crystal structure, a cla ratio of 1.856. In addition to slip on the basal plane, zinc has been observed to slip on a unique slip system. Draw one schematic HCP unit cell (such as the figure shown below) to illustrate basal slip AND another HCP unit cell to illustrate slip system. (2%)
#421506
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