Question

A concrete for a weave breaker is to be designed at a w/c ratio of 0.45. A maximum size of aggregate of 25 mm was used hence a water content of 200 kg/m3 (with 2% air) was required to achieve satisfactory workability. Both the coarse and fine aggregate conform to the grading requirements of ASTM C33. The fine aggregate having a fineness modulus of 2.4, and an absorption of 2%, a BSG (SSD) of 2.6, whereas the coarse aggregate have a bulk density of 1500 kg/m3; an absorption of 3% and a BSG (SSD) of 2.50. The moisture content for both aggregate is 2%. Compute mixture proportions (in kg/m^3) before and after moisture corrections following ACI procedure.

Answer 1

Ans) Let the total volume of trial mix be 1 cubic meter then,

Amount of water required per cubic meter concrete = 200 kg / concrete (GIven)

Water cement ratio = 0.45 (Given)

Amount of air = 2% or 0.02 (Given)

Hence, amount of cement = amount of water / w-c ratio

=> Cement required = 200  / 0.45 = 444.44 kg

Now, according to ACI 211.1-91 , table 6.3.6 for manximuml aggregate size of 25 mm and fineness modulus of 2.4 volume of coarse aggregate is 0.71 m^3

=> Amount of coarse aggregate = Density x volume

Given,density of coarse aggregate as 1500 kg/m^3

=> Amount of coarse aggregate = 1500 x 0.71 = 1065 kg

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

=> Fine aggregate volume = 1 - [(200/1000) + (444.44 / 3.15 x 1000) + (1065/ 2.5 x 1000) + 0.02]

=> Fine aggregate volume = 1 - 0.787 = 0.213 m3

=> Amount of fine aggregate = Volume x Specific gravity x Water density = 0.213 x 2.6 x 1000 = 553.80 kg

Mixture proportion before moisture corrections are as follows :

Component	Amount (kg / concrete)
Cement	444.44
Water	200
Coarse aggregate	1065
Fine aggregate	553.80

Now,

Since both aggregates has moisture and absorption capacity, so they will absorb provide some water to concrete as well as absorb some water from concrete so in order to maintain constant water cement ratio, amount of mixing water needs to be corrected

Water absorbed by coarse aggregate = (0.03 - 0.02) x 1065 kg = 10.65 kg

Water absorbed by fine aggregate = (0.02 - 0.02) x 553.8 kg = 0

=> Corrected amount of water to be added = 200 + 10.65 + 0 = 210.65 kg

Due to moisture, actual amount of both aggregates needed to be increased so,

  Correct amount of coarse aggregate = 1.02 x 1065 = 1086.3 kg

  Correct amount of fine aggregate = 1.02 x 553.8 = 564.88 kg

Mixture proportion after moisture corrections are as follows :

Component	Amount (kg / concrete)
Cement	444.44
Water	210.65
Coarse aggregate	1086.3
Fine aggregate	564.88