Please Explain and Show step by step Concrete is needed for a sidewalk that is 2
ID: 1713929 • Letter: P
Question
Please Explain and Show step by step Concrete is needed for a sidewalk that is 20 ft long, 5 ft wide, and 6 inches thick. A mixture has been designed as 0.50:1:2.50:3.00 based on the ratios for water cement Fine Aggregate: Coarse Aggregate with 10% air voids since it would be exposed. The ratios are based on the aggregate in the SSD condition. Material densities along with the ratios are given in the following table. SSD Density Absorption Ratio 1 2.50 3.00 lb/ft3 197 150 165 62.4 nla Cement FA CA nla 5 4 nla nla water 0.50 Air (10.0%) Dry density Based on the information given 1. Determine the amount of cement, in pounds, that is needed for this sidewalk. Show all calculations. (10 points)Explanation / Answer
Volume of concrete required in sidewalk ( wet volume) = 20*5*6/12= 50 cft
Dry volume of concrete = 1.54 * 50 = 77 cft
Mix proportion = 0.5 : 1: 2.5 : 3
As nothing has been mentioned in question thus assuming that the above proportioning is volumetric
Total volume after deducting air voids = 0.9* 77= 69.3 cft
Volume of cement = 69.3/ ( 0.5+ 1+2.5+3) =9.9 cft
Weight of cement in lb = 9.9 * 197 =1950.3 lb
For question b it is given that weight of cement= 940 lbs
Volume of cement = 940/ 197 = 4.77 cft
Volume of coarse aggregate required as per proportioning = 3* 4.77 = 14.31 cft
Thus weight of coarse aggregate required = 165 * 14.31 = 2361.15 lb
For question 3 it has been given that in reality 8 percent moisture is contained in coarse aggregates in field however at ssd only 4 % absorption therefore volume of additional water = 0.04 * 14.31 = 0.5724 cft
Thus volume of coarse aggregates required to compensate = 14.31/ 0.96 = 14.91 cft
Thus weight of coarse aggregates required to be added at actual site conditions for 10 bags of cement = 14.91* 165 = 2460.15 lb
For question in part 4
Fine aggregate deficient water volume = 1% = 1*(4.77* 2.5 )/100 = 0.12 cft
Coarse aggregate excess water volume = 4% = 4*(4.77* 3) /100= 0.5724 cft
Thus amount of water required to be added to mix = 4.77* 0.5 +0.12 - 0.5724 = 1.9326 cft