土木工程材料模拟题及答案

(2) set W/C and use cement properly. W/C is the decisive factor in concrete solidity. Strictly control the maximum W/C and guarantee the plentiful usage of cement （2分）

(3) select excellent crushed stones and crude aggregates. Improve the fine and crude aggregates gradation. Select the crude aggregates with larger diameter within the permitted diameter ranges. Reduce aggregates voidage and specific surface area. （2分）

(4) Use air entraining admixture and water reducing admixture to improve the anti-permeability and frost-resistance ability. （2分） (5) Enforce the construction quality control （1分）

3. (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) 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）

4. (1) Cohesion. Solid or half solid petrol asphalt can be expressed by penetration

degree index. Liquid petrol asphalt can be expressed by standard cohesion. （1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）

5. 计算题 （共18分）

1. 解：孔隙率 P ＝(1 － ρ0/ρ)× 100 ％＝(1-1.8/2.7)× 100 ％ =33 ％；

重量吸水率 mw ＝(m水/m)× 100 ％＝ [(1020-920)/920] × 100 ％ =11 ％； 开口孔隙率＝ V开 /V0 =[ (1020-920)/(920/1.8) ] × 100 ％＝ 19.6 ％ 闭口孔隙率＝ 33 ％－ 19.6 ％＝ 13.4 ％

所以，该材料的孔隙率、重量吸水率、开口孔隙率及闭口孔隙率分别为： 33 ％； 11 ％； 19.6 ％； 13.4 ％。 （5分）

2. 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 。（6分）

3. 解：设混凝土n天的抗压强度为fn，龄期为n ，则 fn＝Algn ＋ B ①

又有： f7 ＝ Alg7 ＋ B ② f14 ＝ Alg14＋ B ③

由式 ②③得 ： A=16.61 ； B ＝ 6.96 ，

代入式 ①得： f28 ＝ 16.61lg28＋ 6.96 ＝ 31MPa （7分）

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