Project on Concrete Mix Design for various grades of Concrete

Concrete mix design for M35 grade concrete with flyash:

• Target mean strength f’ck = fck+1.65s

= 35+1.65(5)

= 43.25N/mm²

• Selection of water-cement ratio

From table 5 of IS 456:2000, is 0.45

By experience we consider water cement ratio as 0.40

• Selection of water content

assume max nominal size of aggregate as 20mm, with 100mm slump,

acc to is 10262:2009, table 2, maximum water content = 186 liters

186+(186*6/100) = 197 liter

• Calculation of cement content:

We know water cement ratio as 0.40 and water content as 197liter

Cement content = 197/0.4 = 492.5 kg

Cementatious material content = 492.5*1.10 = 541.75kg

Water content = 197 L

So, water cement ratio = 197/541.75 = 0.364

Flyash @ 30% total = 541.75 *30% = 162.525kg/ m³

Cement (OPC) = 541.75-162.525 = 379.23kg/ m³

Amount of cement saved is = 492.5-379.23 = 113.27kg/ m³

• Volume of coarse aggregate and fine aggregate:

For 20mm maximum nominal size of aggregate, assuming zone1 for water cement ratio = 0.6

But in present case water cement ratio is 0.4, hence corrected volume of CA 0.40 = 0.62

Volume of CA = 0.62

Volume of FA = 0.38

• Mix calculations:

3. Volume of concrete = 1m³.
4. Volume of concrete = mass of cement/sp. gr. of cement =

(mass of cement/sp. gr. of cement)*1/1000

(379.23kg /3.15)*1/1000 = 0.120 m³.

1. Volume of water = (197/1)*1/1000 = 0.197
2. volume of flyash = (162.525/2.2)*(1/1000) = 0.0738kg/m³
3. Volume of all aggregate = 1-(0.120+0.197+0.0738) = 0.609
4. mass of CA = d*vol. of CA *sp. gr. of CA*1000

= 0.609*0.62*2.67*1000 = 1008.13kg/m³

1. mass of FA = d*vol. of FA*sp. gr. of FA*1000

= 0.609*0.38*2.67*1000 = 617.89kg/ m³.

• Mix proportion trial 1:

Cement:379.23 kg/ m³

Water:197 L

FA: 617.89 kg/ m³

CA:1008.13 kg/ m³

Flyash:162.525 kg/ m³

Water-cement ratio:0.4

Trial mix ratio:1:1.1:1.8

Concrete mix design for M50 grade concrete without flyash:

• Target mean strength f’ck = fck+1.65s

= 50+1.65(5)

= 58.25N/mm².

• Selection of water-cement ratio

From table 5 of IS 456:2000, is 0.45

By experience we consider water cement ratio as 0.40.

• Selection of water content

Assume max nominal size of aggregate as 20mm, with 100mm slump,

Acc to is 10262:2009, table 2, maximum water content = 186 liters

186+ (186*6/100) = 197 liter.

• Calculation of cement content:

We know water cement ratio as 0.40 and water content as 197liter

Cement content = 197/0.4 = 492.5 kg

• Volume of coarse aggregate and fine aggregate:

For 20mm maximum nominal size of aggregate, assuming zone1 for water cement ratio = 0.6

But in present case water cement ratio is 0.4, hence corrected volume of CA 0.40 = 0.62

Volume of CA = 0.62

Volume of FA = 0.38

• Mix calculations:

3. Volume of concrete = 1m³.
4. Volume of concrete = mass of cement/sp. gr. of cement =

(mass of cement/sp. gr. of cement)*1/1000

(492.5 /3.15)*1/1000 = 0.156 m³.

1. Volume of water = (197/1)*1/1000 = 0.197
2. Volume of all aggregate = 1-(0.156+0.197) = 0.647m³
3. mass of CA = d*vol. of CA *sp. gr. of CA*1000

= 0.0647*0.62*2.67*1000 = 1071.04kg/m³

1. mass of FA = d*vol. of FA*sp. gr. of FA*1000

= 0.647*0.38*2.67*1000 = 656.44kg/ m³.

• Mix proportion trial 1:

Cement: 492.5 kg/ m³

Water: 197 L

FA: 656.44kg/ m³

CA: 1071.04kg / m³

Water-cement ratio: 0.4

Trial mix ratio: 1:1.3:2.17