Question

Using the absolute volume method, a colleague started a mix design for a simple concrete foundation in a mild climate and got through Step 5 and has asked you to finish the work. They have already determined the mix needs:
• a w/c ratio of 0.45
• a 2” slump
• 1830 lb of gravel in the oven dry conditions,
• and 3.5% air provided by air entrainer.
The available gravel (coarse aggregate) is subangular with a NMSA = 3⁄4”. It has a specific gravity, Gs = 2.7, with a moisture content of 1.8% and an Absorption of 2.2%. The available sand (fine aggregate) has a specific gravity, Gs =2.45, with a moisture content of 3.8% and an absorption of 2.4%. Gs of cement = 3.15.

For each cubic yard of concrete for this mix, determine:
• the total weight of water, cement, moist sand, and moist gravel required. Don’t worry about the amount of
air entrainer product in fluid ounces.
• Summarize the total mix design (neglecting the required fluid ounces of air entrainer)

Course aggregate:

Fine aggregate:

Cement:

Water:

Answer 1

Ans) Let the total volume of trial mix be 1 cubic yard (27 )  then according to ACI 211.1-91 , table 6.3.3 , for 3/4 in aggregate and 2 in slump , amount of water required per cubic yard of concrete is 280 lb/cy

=> Water requied = 280 lb/cy

Also, Amount of cement = Amount of water / w-c ratio

=> Amount of cement required = 280 lb / 0.45 = 622.22 lb

=> Volume of cement required = Amount / (Specific gravity x 62.4)

=> Volume of cement required = 622.22 / (3.15 x 62.4) = 3.165

Given, amount of coarse aggregate required = 1830 lb

=> Volume of coarse aggregate = Amount / (Specific gravity x 62.4)

=> Volume of coarse aggregate = 1800 / (2.7 x 62.4) = 10.68

Now, according to absolute volume method,

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

=> Fine aggregate volume = 27 - [(280/62.4) + 3.165 + 10.68 + 0.035(27)]

=> Fine aggregate volume = 27 - 19.28 = 7.72 ft3

=> Amount of fine aggregate = Volume x Specific gravity x Water unit weight = 7.72 x 2.45 x 62.4 = 1180.23 lb

Correction in mixing water :

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

Water absorbed by coarse aggregate = (0.022 - 0.018) x 1830 lb = 7.32 lb

Water absorbed by fine aggregate = (0.024 - 0.038) x 1180.23 = -16.52 lb

  => Actual amount of water to be added = 280 lb + 7.32 - 16.52 = 270.80 lb

Hence, summary of mix design is as follows :

Component	Amount (lb)
Cement	622.22
Water	270.80
Coarse aggregate	1830.00
Fine aggregate	1180.23