判断题 (对的划√,不对的划×, 每题1分,共15分)
1. × 2. √ 3. × 4. × 5. √ 6. √ 7. √ 8. × 9. × 10. × 11. × 12. √ 13. × 14. × 15.×
Durability is proposed for the description of concrete ability in resisting environment and medium influence and maintaining good utility properties. The following steps can be taken to enhance its durability:
(1) select cement type according to the specific construction environment and properties. (2分)
(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 (2分)
解:因为: f饱 = P饱 max /A ; f干 = P干 max /A ;
所以,软化系数= f饱 / f干 = P饱 max /P干 max
= 180/240=0.75 < 0.85
所以,该材料不适合用作经常与水接触的工程部位的结构材料。(5分)
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 ; W = 127Kg ; S = 605Kg ; G = 1203Kg 。 (6分)
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 。 (7分)
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