Question

Compute the SSD mix proportions for the following mix design using the ACI method: Compressive Strength: 3,000 psi Slump: 4 inches Non-Air Entrained Concrete The materials have the following properties: | Cement: Type I, specific gravity = 3.15 | Fine Aggregate: Bulk specific gravity = 2.68; fineness modulus = 2.40 Coarse Aggregate: Maximum size aggregate = 1 inch; bulk specific gravity = 2.68; Dry-rodded unit weight = 100 lbs/ft3

Answer 1

Ans) Let the total volume of trial mix be 1 yd³ (27 ft³) . According to ACI 211.1.8 Table 6.3.3 , for a slump of 4 in, nominal aggregate size of 1 in and non-air entrained concrete, amount of water required per cubic yard of concrete is 325 lb

=> Amount of water = 325 lb per yd³ concrete

From table 6.3.3, for 4 in slump and 1 in aggregate, amount of entrapped air for non entrained concrete = 1.5 %

According to Table 6.3.4(a), for F'c = 3000 psi and non air entrained concrete, w/c ratio = 0.59

Hence, amount of cement = 325 / 0.59 = 550.85 lb

Now, according to table 6.3.6 for nominal aggregate size of 1 in and fineness modulus of 2.4 , volume of coarse aggregate is 0.71 yd³ or 19.17 ft³

=> Amount of coarse aggregate = dry rodded density x volume = 100 lb/ft³ x 19.27 ft³ = 1927 lb

Volume of fine aggregate = Total volume of concrete - Volume of water , cement , coarse aggregate and air

=> Fine aggregate volume = 27 - [(325/62.4) + (550.85 / 3.15 x 62.4) + (1927/ 2.68 x 62.4) + 0.015(27)]

=> Fine aggregate volume = 27 - 19.93 = 7.07 ft³

=> Amount of fine aggregate = volume x specific gravity x unit weight of water = 7.07 ft³ x 2.68 x 62.4 pcf = 1182.33 lb

Hence, mix proportions required for 1 cubic yard concrete is as follows:

Material	Amount (lb)
Cement	550.85
Water	325
Coarse aggregate	1927
Fine aggregate	1182.33