Question

A concrete column is to be designed at a w/c ratio of 0.40 in an environment where an aggressive F/T persists. A maximum size of aggregate of 19 mm was used. A slump of 30 mm 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 2.8, and an absorption of 2%, a BSG (SSD) of 2.5, whereas the coarse aggregate have a bulk density of 1450 kg/m3; an absorption of 3% and a BSG (SSD) of 2.50. The moisture content for FA and CA was 4 and 1%, respectively. 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 according to ACI 211.1.-91 , table 6.3.3 for 35 mm slump and nominal aggregate size of 50 mm , amount of water required per cubic meter of concrete is 168 kg

=> Amount of water = 168 kg per concrete

Now, since concrete is exposed to excessive freezing/thawing, class of exposure is severe

According to Table 6.3.3 approximate air content in concrete for 19 mm aggregate = 6% or 0.06

Given, water cement ratio is 0.40

Hence, amount of cement = 168 / 0.40 = 420 kg

Now, according to table 6.3.6 for nominal aggregate size of 19 mm and assuming fineness modulus of 2.8 volume of coarse aggregate is 0.62 m^3

=> Amount of coarse aggregate = Bulk Density x volume

Given,density of coarse aggregate as 1450 kg/m^3

=> Amount of coarse aggregate = 1450 x 0.62 = 899 kg

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

=> Fine aggregate volume = 1 - [(168/1000) + (420 / 3.15 x 1000) + (899/ 2.5 x 1000) + 0.06]

=> Fine aggregate volume = 1 - 0.720 = 0.28 m3

=> Amount of fine aggregate = Volume x Specific gravity x Water density = 0.28 x 2.5 x 1000 = 700 kg

Following is batch weight of concrete before moisture correction :

Component	Amount (kg)
Cement	420
Water	168
Coarse aggregate	899
Fine aggregate	700

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 provided by coarse aggregate = (0.01 - 0.03) x 899 kg = - 17.98 kg

Water provided by fine aggregate = (0.04- 0.02) x 700 kg = 14 kg

=> Actual amount of water to be added = 168 - (-17.98)  - 14 = 171.98 kg 172 kg

Due to moisture, actual amount of coarse and fine aggregate is needed to be increase

=> Corrected amount of coarse aggregate = 1.01 x 899 = 908 kg

=>  Corrected amount of fine aggregate = 1.04 x 700 = 728 kg

Hence, following is batch weight of concrete after moisture correction :

Component	Amount (kg)
Cement	420
Water	172
Coarse aggregate	908
Fine aggregate	728