2018-2019学年第一学期期末试卷
课程名称：材料科学导论 闭卷 A 卷 120 分钟
一、 名词解释（中文作答，每小题3分，共15分）
1. Crystalline imperfection
2. Diffusion flux
3. Elastic deformation
4. Dislocation density
5. Case hardening
二、简答题（共45分）
Why is this so?
(b) How to differentiate the edge, screw, and mixed dislocations? (10分)
2.(a) Why is interstitial diffusion normally more rapid than vacancy diffusion?
(b) What is this simple principle on which virtually all strengthening techniques rely?
List four strengthening mechanisms or techniques in metals. (10分)
3.For metal material, decreasing in the average grain size can result in increasing in strength and toughness at low temperatures, but it can weaken the resistance to creep at high temperatures. Why is this so? (10分)
4. Briefly explain why the hardness of tempered martensite diminishes with tempering time (at constant temperature) and with increasing temperature (at constant tempering time).(10分)
5. The following figure is the schematic representation of grains of austenite γ phase. Briefly draw up its microstructures of hypereutectoid stee l containing 0.90wt%C. (5分)
三、计算题（每题10分，共40分）
评阅人得分
1.(a) Compute and compare the linear densities of the [110], and [111] directions for FCC.
(b) Calculate and compare the planar densities of the (110) and (111) planes for FCC.
2. The preexponential (D0 ) and activation energy (Q) for the diffusion of carbon in
γ-iron are 2.3 × 10 - 5 m2/s and 148.0 kJ/mol, respectively. At approximately what temperature would a specimen of γ-iron have to be carburized for 2 h to produce the same diffusion result as at 900℃ for 15 h?
3. A cylindrical metal specimen having an original diameter of 12.6 mm and gauge length of 50.60 mm is pulled in tension until fracture occurs. The diameter at the point of fracture is 6.80 mm, and the fractured gauge length is 72.20 mm. Calculate the ductility in terms of percent reduction in area and percent elongation.
4. A structural component in the form of a wide plate is to be fabricated from a
MPa and 7075-T651 aluminum alloy that has a plane strain fracture toughness of 24m
a yield strength of 495 MPa. Calculate the maximum internal crack length allowable for the aluminum alloy component that is loaded to a stress one half of its yield strength. Assume that the value of Y is 1.30.