Project on Concrete Mix Design for various grades of Concrete
Target Strength:
f’ck = fck + 1.65 S
From Table 2 of IS 10262: 2019, S = 5
f’ck = 40 + 1.65 x 5 = 48.25 N/mm^2
or
f’ck = fck + X
From Table 1 X = 6.5
= 40 + 6.5
Approximate Air Content:
From Table 3 of IS 10262: 2019, the approximate amount of entrapped air in normal concrete is 1.5% for 10mm aggregate
Selection of water-cement ratio:
From Table 5 of IS 456, maximum w/c = 0.45
From As per IS 10262: 2019w/c = 0.36
0.36 < 0>
Therefore w/c ratio = 0.36
Selection of water content:
From IS 10262: 2019 Table 4, Water content for 10mm size aggregate = 208 kg
For 110 mm slump, water content = 208 +(0.072 x 208) = 222.976 lit
Based on the trial water content reduction of 23% is considered while using a superplasticizer.
Now water content= 222.976*0.77 = 171.69 kg/m^3
Calculation of cement content:
Water cement ratio = 0.36
Water content as per concrete = 171.69 kg/m^3
Cementitious content = 171.69/0.36 = 476.92 kg/m^3
As per IS 456, the minimum cement content is 320 kg/m^3
Hence it is OK
Since fly ash is going to be replaced with cement, based on trials we need to add 10%
New Cementitious content = 476.92 x 1.10 = 524.61 kg/m^3
New w/c = 171.69/524.61 = 0.327
25% of cement is going to be replaced by fly ash
Fly ash = 524.61 x 25% = 131.16 kg/m^3
Cement = 393.48 kg/^3
Superplasticizer = 1.5% of weight of cementitious content = 0.015 x 524.6 = 7.86 kg/m^3
Proportioning of the volume of coarse aggregate and fine aggregate:
As per IS :10262: 2019 Table5
For Zone 2, Volume of coarse aggregate = 0.50
For w/c 0.5 = 0.50
For w/c 0.327 = ?
X = [((0.5-0.327)*0.01)/0.05]
X=0.0346
Proportioning Volume of coarse aggregate = 0.50 + 0.028 = 0.5346
Proportioning of fine aggregate = 1- 0.528 = 0.4654
Mix Calculation:
Volume of concrete = 1m^3
Volume of cement= weight of cement /(specific gravity *1000)
= 393.48/(3.17 * 1000) = 0.124 m^3
Volume of fly ash = 131.16/(2.2*1000) = 0.0596
Volume of water = 171.69/(1*1000) = 0.171 m^3
Volume of admixture = 7.86/(1.10 *1000) = 0.0071 m^3
The total volume of other material except for aggregates=0.015+ 0.124+0.0596+0.171 +0.0071 = 0.3767 m^3
Volume of coarse aggregate and fine aggregate = 1- 0.3767 = 0.6233 m^3
Mass of coarse aggregate = 0.6233 x 0.5346 x 2.75 x 1000 = 916.34 kg
Mass of Fine aggregate = 0.6233 x 0.4654 x 2.64 x 1000 = 765.82 kg
Mix proportions:
- Cement = 393.48 kg/m^3
- Fly ash = 131.16 kg/m^3
- Water = 171.69 kg/m^3
- Admixture = 7.86 kg/m^3
- Fine aggregate = 765.82 kg/m^3
- Coarse aggregate = 916.34 kg/m^3
- w/c = 0.327
Free (surface) moisture:
Coarse aggregate: Free moisture = Total moisture content – Water absorption
= 2.0 – 0.5 = 1.5 percent
Fine aggregate: Free moisture = Total moisture content – Water absorption
= 5.0 – 1.0 = 4.0 percent
Adjustment of water content:
- Coarse aggregate in wet condition = Weight of coarse aggregate x 1.5 moisture content
- Coarse aggregate in wet condition = 916.34 x 1.015 = 930.0851 kg
- Fine aggregate in wet condition = 765.82 x 1.04 = 796.75 kg
Water content in the aggregates:
- Coarse aggregate = 930.085 - 916.34 = 13.74
- Fine aggregate = 796.75 - 765.82 = 30.93
Required amount of water to be added = 171.69 - 13.74 - 30.93
The required amount of water to be added = 127.02 kg/m^3
Mix proportions:
Cement = 393.48 kg/m^3
Fly ash = 131.16 kg/m^3
Water = 127.23 kg/m^3
Admixture = 7.86 kg/m^3
Fine aggregate = 796.75 kg/m^3
Coarse aggregate = 930.08 kg/m^3
w/c = 0.327