双语试卷答案
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山东交通学院期末考试工程材料(含热处理)(双语)课程试卷答案 第 1 页 共 3 页 得分 阅卷人 一、选择题(每小题2分,共20分) 1. In the following alloys, B cannot form eutectic phase diagram. A. Al-Si B. Cu-Ni C. Pb-Sn D. Fe-C 2. In the following methods, A can eliminate (消除) dendritic segregation (枝晶偏析). A. homogenizing(均匀化退火) B. isothermal annealing(等温退火) C. recrystallization(再结晶退火) D. step hardening(分级淬火) 3. The most suitable steel for carburizing (渗碳) is D . A. 45 B. 40Cr C. 40MB D. 20CrMnTi 4. The structure obtained from 40Cr after quenching and high temperature tempering is called A . A. tempered sorbite B. tempered troostite C. tempered martensite D. perlite 5. In the following aluminum alloys, A cannot be strengthened by heat treatment. A. Rust-proof aluminum alloys B. Duralumin alloys (硬铝) C. Ultralumin alloys (超硬铝) D. Wrought aluminum alloys 6. In the following steel designations, C belongs to heat resistant steel. A. 20CrMnTi B. 9SiCr C. 15CrMo D. 60Si2Mn 7. The shape of graphite in the cast iron KTZ450-06 is D . A. nodular(球状) B. worm(蠕虫状) C. lamellar(片状) D. agglomerate(团絮状) 8. The main function of W in W18Cr4V is C . A. to improve hardenability B. to make grains fine C. to improve the hardness at high temperatures D. solid solution strengthening 9. In the following materials, B is the most suitable one to make a rod piece(杆类零件) of φ10mm. It’s known that the piece works under medium alternate pulling and pressing load (交变拉压载荷) and is requested to have uniform (均匀的) properties in the cross section(截面). A. T12 B. 40Cr C. 1Cr13 D. ZGMn13 10. In casting industry, cast irons of B composition are widely used. A. hypoeutectic B. eutectic C. hypereutectic D. hypoeutectoid 得分 阅卷人 二、判断题(每小题2分,共16分)
1. Interstitial solid solution (间隙固溶体) can only be limit solid solution. ( √ ) 2. When eutectoid steels are heated above the critical temperature, it is impossible to get pearlite(P) during the process of continuous cooling. ( × ) 3. Heat treatment processes of cast irons mainly change the shape and distribution (分布) of graphite(石墨). ( × ) 4. When eutectoid steels are heated to austenite,the structures obtained after cooling depend mainly on the heating temperature of steels. ( × ) 5. Austenitic stainless steels (奥氏体不锈钢) are the most widely used stainless steels.( √ ) 6. HSn62-1 is called “admiralty brass”(海军黄铜) because of its perfect corrosion resistance. ( √ ) 7. To improve the strength of stainless steels,the carbon content should be as high as possible. ( × ) 8. The solid solution of carbon in the γ-Fe is called Ferrite. ( × )
得分 阅卷人 三、简答题(本题共24分)
1. What are the influencing factors of graphitization (石墨化) of cast irons? (本题5分) 答:影响铸铁石墨化的因素主要是化学成分和冷却速度。 化学成分:C、Si是强烈促进石墨化的元素,含量越高,越易获得灰口组织;但含量过高时,易生成过多而粗大的石墨,降低铸铁的铸造性能。P可微弱促进石墨化。S、Mn是阻碍石墨化的元素。(3分) 冷却速度:冷却速度越快,原子扩散越不易进行,越易生成渗碳体,反之,冷却速度越慢,原子扩散越充分,越促进石墨化的进行。(2分)
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试卷适用班级
班级 姓名 学号 山东交通学院期末考试工程材料(含热处理)(双语)课程试卷答案 第 2 页 共 3 页 2. The melting point of Pb is 327℃, try to analyze (分析) what kind of processing it is when Pb is processed at 20℃ (cold work or hot work)?(本题5分) 答:金属的再结晶温度TR与熔点Tm的关系为:TR≈0.4Tm (T的单位为K) TR≈0.4×(327+273)=240 K ~-33℃ (3分) 即铅的再结晶温度为-33℃,其加工温度20℃高于再结晶温度,因此属于热加工。(2分)
3. Draw the crystal planar index (1 1 1) and the crystal orientation index [2 0 1] in the body centered cubic. (本题4分)
答:
(晶面指数和晶向指数各2分。) 4. Explain the following phenomenon (现象) on the basis of Fe-Fe3C phase diagram: It’s more difficult to cut steels such as T8, T10, T12 than 10, 20 and the saw (锯) tends to become worn (磨损的) easily. (本题5分)
答:T8、T10、T12钢属于过共析钢,其中Fe3C的含量比亚共析钢10、20钢高,(2分) 因而强度和硬度高,故锯切费力,锯条易磨损。(3分)
5. What's slip (滑移)?What’s the direction of slip? (本题5分) 答:滑移指在切应力作用下,晶体一部分相对于另一部分沿一定晶面发生相对滑动。(2分) 滑移总是沿晶体中原子排列最紧密的晶面和该晶面上原子排列最紧密的晶向进行。(3分)
得分 阅卷人 四、解释分析题(每题10分,共40分)
1. The Fe-Fe3C phase diagram is as follows. You are requested to analyze the cooling process of T12 steel, including important reactions and different structures obtained at different areas.
答:T12钢的冷却过程:如图所示,T12钢的成分线与各特征线分别交于1、2、3、4点。 1点对应温度之上,全部为液相;到达1点时合金开始结晶,生成奥氏体A 晶核并逐渐长大;1~2点之间,液相逐渐减少,奥氏体逐渐增多。到达2点时,结晶结束,液相全部转变为奥氏体。(3分) 2~3点之间,随温度下降,相的组成不变,为奥氏体单相区;到达3点时,从奥氏体中逐渐析出二次渗碳体Fe3CⅡ,并且随着温度下降,3~4点之间,Fe3CⅡ逐渐增多。(2分)
温度下降到4点时,奥氏体发生共析反应:。(3分) 4点温度之下,共析反应完成,最后的室温组织为P+ Fe3CⅡ。(2分)
试卷适用班级
班级
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