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2011年全国材料员考试试题一(2)

    11-14 22:49:11    浏览次数: 995次    栏目:材料员考试试题

标签:材料员试题,历年材料员考试真题,http://www.deyou8.com 2011年全国材料员考试试题一(2),
  1. 名词解释 (每题2分,共14分)
  2. 表观密度是指材料在自然状态下,单位体积的质量。
  3. 抗渗性是指材料抵抗压力水渗透的能力。
  4. 初凝时间是指水泥加水拌合起至标准稠度静浆开始失去可塑性的所需的时间。
  5. 混凝土拌合物和易性是指混凝土拌和物易于施工操作并能获致质量均匀、成型密实的性质。
  6. 混凝土的强度等级是指按混凝土立方体抗压强度来划分的强度等级。
  7. Q235-AZ是指碳素结构钢的屈服点大于等于235MPa,质量等级为 A,脱氧程度为镇静钢。
  8. 石油沥青的粘滞性是反应沥青材料内部其相对流动的一种特性,以绝对粘度表示,是沥青的重要指标之一。
  9. 判断题 (对的划√,不对的划×, 每题1分,共16分)

    1. × 2. × 3. × 4. × 5. × 6. × 7. × 8. √ 9. √ 10. × 11. × 12. √ 13. × 14. × 15.× 16. √

  10. 填空题( 每题1分,共20分)
  11. 软化系数,大
  12. 热量,膨胀,游离水,收缩
  13. 45min, 390min
  14. W/C 、砂率、单位用水量
  15. 流动性、保水性,沉入度、分层度
  16. 屈服、σ02
  17. 增大,差,小
  18. 问答题(共30分)
    1. (1)Water resistance refers to the performance that the material shows in resisting the water destruction. It is remarked with softening index.(2分)

    (2)Anti-permeability is defined as the ability resisting water permeation by pressure. It is expressed by permeability index and permeability gradation.(2分)

    (3)Anti-freezing refers to the ability that the saturated materials still keep its strength and integrality after repeated thawing and freezing circles. It is expressed by anti-freezing gradation. (2分)

  19. (1)When mixed with fixed amount of cement, there will be small W/C and dry concrete with small slump, so it is hard to form dense shapes. If W/C is lower, the concrete is easy to be broken and has poor performance in viscidity so that the hardened concrete strength and durability will reduce. If W/C is higher, the concrete not only has thin concentration and larger slump but it is easy to segregate, laminate and bleed as well, therefore the hardened concrete strength and durability will reduce. With proper W/C the concrete can be formed into dense and uniform shapes. (4分)

    (2)Concrete strength depends on W/C on the condition of the same cement type and strength degree. Within the certain range of W/C (the concrete can be formed into dense shapes.), the lower W/C, the higher the compressive strength becomes. As W/C becomes lower (the concrete can not be formed into dense shapes.), the porosity will become larger and the strength will reduce. As W/C becomes higher, the strength will reduce. (5分)

  20. (1)Concrete mixture mobility sets slump and Vebe Consistometer as its index. Slump is used for larger mobility concrete mixture while Vebe Consistometer is used for dry and hard concrete mixture.(4分)

    (2)It is recommended to select smaller slump for cement efficiency and high quality concrete under the principle of construction feasibility and dense vibration. In construction the concrete mixture slump is decided upon the component cross section, supporting steel spacing and construction method. It is recommended to select larger slump when the cross section is small or the supporting steel spacing is wider or the concrete mixture is mixed by manual vibration, otherwise smaller slump is recommended when the cross section is larger or the supporting steel spacing is narrower or the concrete mixture is mixed by machine vibration. (6分)

  21. (1)Cohesion. Solid or half solid petrol asphalt can be expressed by penetration degree index. Liquid petrol asphalt can be expressed by standard cohesion.(2分)

    (2)Plasticity. It can be expressed by elongation index.(1分)

    (3)Temperature stability. It can be expressed by softening point index.(1分)

    (4)Atmosphere stability. It can be expressed by evaporation loss and penetration degree after evaporation index.(1分)

  22. 计算题(共20分)
  23. 解:设该材料干燥状态得质量为 m ,则有
    ρ = m/V (1)
    ρ0 = m/V0 (2)
    mw = m × 100 %/m (3)
    由式(1)和(2)得:V = V0 -V=m(ρ - ρ0)/ρ0 ρ
    由式(3)得: V = V = m mw
    所以, V/V= (1 %× 2.56 × 2.65)/(5.65-2.56)
    = 0.75 V/V = 1-0.75 = 0.25
    该岩石中开口孔隙与闭口孔隙所占的比例分别为 0.75 和 0.25 。 (5分)
  24. 解:按试验室配合比, 1m3 混凝土中各种材料得用量为:
    C= 300Kg ; W = 300 × 0.56 = 168Kg ; S = 300 × 1.92 = 576Kg ; G = 300 × 3.97 = 1191Kg ;
    所以,施工配合比为:
    C= 300Kg ; W = 168-576 × 5 % -1191 × 1 %= 127Kg ;
    S = 576 ×(1 + 5 %)= 605Kg ; G = 1191 × (1+1 % ) = 1203Kg ;
    施工配合比为: C=300Kg; W127Kg;S=605Kg;G=1203Kg。(7分)

  25. 解:设水泥的质量为 CKg ,则 W = 0.62CKg ; S = 2.43CKg ; G = 4.71CKg ;
    按假定表观密度法有: C+S+G+W= ρ0h
    所以,C + 0.62C + 2.43C + 4.71C = 2400
    由上式可得: C= 274Kg ; W = 170Kg ; S = 666Kg ; G = 1290Kg 。
    所以,各种材料的用量为: C= 274Kg ; W = 170Kg ; S = 666Kg ; G = 1290Kg 。 (8分)