Project on Concrete Mix Design for various grades of Concrete

PROJECT 1

1) Calculate the concrete mix design for M35 grade concrete with fly ash

Given,

M35 grade concrete with fly ash

Mix Design:

1. a) Target Mean strength:

Grade of concrete = M35

fck = 35 N/mm2

From Table 1 of IS 10262-2009,

S = 5 N/mm2

f’ck = fck + 1.65 S

= 35 + (1.65 * 5)

= 43.25 N/mm2

1. b) Obtain the w/c ratio:

From the graph 1, for the corresponding 28 days compressive strength which is the target mean strength ,obtain w/c ratio

w/c = 0.34

From graph 2, for f’ck = 43.25 N/mm2, extend line till curve E (41.5-48N/mm2) horizontally and then dropping a vertical line to determine w/c ratio

w/c = 0.42  

Assuming ‘Severe’ exposure condition , from table 5 of IS 456-2000,

w/c = 0.45

The lower value is taken. w/c = 0.34

1. c) Selection of water content:

From table 2 of IS 10262,

Max water content = 186 litres

Considering slump as 100 mm = 186 + (6/100)186

= 197.16 litres

Assuming Superplasticizers are added and thereby reducing the water demand by say 29%

Water content = 197.16 * 0.71 = 140 litres

1. d) Calculation of cement content:

w/c = 0.34

Cement content = 140/0.34 = 411.76 kg/m3

To proportion  a mix containing fly ash, the following steps are to be followed.

1. The % of flyash needs to be decided based on project requirement and quality of materials
2. Increase in cementitious material content is by 10%

Cementitious material content = 411.76 * 1.10 = 452.94 kg/m3

Water content = 140 litres

w/c ratio = 140 / 452.94 = 0.31

Flyash at 30% of cement = 452 .94 * (30/100) = 135.88 kg/m3

Cement = 452.94 – 135.88 = 317.058 kg/m3

Saving of cement using flyash = 411.76 – 317.058 = 94.70 kg/m3

1. e) Proportion of volume of CA and FA:

From table 3 of IS 10262, Volume of CA corresponding to 20mm size aggregate and FA( zone 1 ) for w/c ratio of 0.5 = 0.60

In our case, the w/c ratio is 0.31, which is lower by 0.19 (0.5-0.31=0.19à0.19/0.05=3.8à3.8*0.01=0.038)

the volume of CA is increased by 0.038,

Vol of CA = 0.638

Assuming pumpable concrete, the volume of CA has to be reduced by 10%

Vol of CA = 0.638 * 0.9 = 0.574 m3

Vol of FA = 1- 0.574=0.426 m3

1. f) Mix Calculations:

1) Volume of concrete = 1m3

2)  Volume of cement = (Mass of cement/ Sp. Gravity of cement) * (1/1000)

= (317.058/3.15)*0.001 = 0.101 m3

3) Volume of flyash= (Mass of flyash/ Sp. Gravity of flyash) * (1/1000)

= (135.88/2.2)*0.001 = 0.062 m3

4) Volume of water = (Mass of water/ Sp. Gravity of water) * (1/1000)

= (140/1)*(1/1000) = 0.14 m3

5) Volume of chemical admixture = (Mass of chemical admix/ Sp. Gravity ) * (1/1000)

= (7.6/1.145)*(1/1000) = 0.006 m3

6) Volume of aggregates = (1) – ((2)+(3)+(4)+(5))

= 1 – (0.101+0.062+0.14+0.006) = 0.691m3

7) Mass of CA = Volume of aggregates * Volume of CA * Sp. Gravity of CA *1000

0.691 * 0.574 * 2.74*1000= 1086.78 kg/m3

8) Mass of FA =0.691 * 0.426*2.74*1000 = 806.57 kg/m3

1. g) Mix proportions:

Cement = 317.058 kg/m3

Flyash = 135.88 kg/m3

Water = 140 kg/m3

FA = 806.57 kg/m3

CA = 1086.78 kg/m3

w/c = 0.31

Mix ratio = 1:2.54:3.43

2) Calculate the concrete mix design for M50 grade concrete without fly ash

Given,

M50 grade concrete without fly ash

Mix Design:

1. a) Target Mean strength:

Grade of concrete = M50

fck = 50 N/mm2

From Table 1 of IS 10262-2009,

S = 5 N/mm2

f’ck = fck + 1.65 S

= 50 + (1.65 * 5)

= 58.25 N/mm2

1. b) Obtain the w/c ratio:

Assuming ‘Severe’ exposure condition , from table 5 of IS 456-2000,

w/c = 0.45

Based on experience, w/c = 0.40

1. c) Selection of water content:

From table 2 of IS 10262,

Max water content = 186 litres

Considering slump as 100 mm = 186 + (6/100)186

= 197.16 litres

Assuming Superplasticizers are added and thereby reducing the water demand by say 29%

Water content = 197.16 * 0.71 = 140 litres

1. d) Calculation of cement content:

w/c = 0.40

Cement content = 140/0.40 = 350 kg/m3

1. e) Proportion of volume of CA and FA:

From table 3 of IS 10262, Volume of CA corresponding to 20mm size aggregate and FA( zone 1 ) for w/c ratio of 0.5 = 0.60

In our case, the w/c ratio is 0.4, which is lower by 0.1

the volume of CA is increased by 0.02,

Vol of CA = 0.62

Assuming pumpable concrete, the volume of CA has to be reduced by 10%

Vol of CA = 0.62 * 0.9 = 0.56 m3

Vol of FA = 1- 0.56=0.44 m3

1. f) Mix Calculations:

1) Volume of concrete = 1m3

2)  Volume of cement = (Mass of cement/ Sp. Gravity of cement) * (1/1000)

= (350/3.15)*0.001 = 0.111 m3

3) Volume of water = (Mass of water/ Sp. Gravity of water) * (1/1000)

= (140/1)*(1/1000) = 0.14 m3

4) Volume of chemical admixture = (Mass of chemical admix/ Sp. Gravity ) * (1/1000)

= (7.6/1.145)*(1/1000) = 0.006 m3

5) Volume of aggregates = (1) – ((2)+(3)+(4))

= 1 – (0.111+0.14+0.006) = 0.743m3

6) Mass of CA = Volume of aggregates * Volume of CA * Sp. Gravity of CA *1000

0.743 * 0.56 * 2.74*1000= 1140 kg/m3

7) Mass of FA =0.743 * 0.44*2.74*1000 = 896 kg/m3

1. g) Mix proportions:

Cement = 350 kg/m3

Water = 140 kg/m3

FA = 896 kg/m3

CA = 1140 kg/m3

Chemical admix = 7.6 kg/m3

w/c = 0.40

Mix ratio = 1:2.56:3.26