Project on Concrete Mix Design for various grades of Concrete
- Calculate the concrete mix design for M35 grade concrete with fly ash
Given,
M35 grade concrete with fly ash
Mix Design:
- 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
- b) Obtain the w/c ratio:
for f’ck = 43.25 N/mm2, w/c = 0.38
Severe exposure condition , from table 5 of IS 456-2000, w/c = 0.45
The lower value is taken. w/c = 0.38
- c) Selection of water content:
From table 2 of IS 10262,
Max water content = 186 litres
Considering slump as 75 mm = 186 + (3/100)186= 191.58 litres
Superplasticizers are added and thereby reducing the water demand by say 30%
Water content = 191.58 x 0.70 = 134.10 litres
- d) Calculation of cement content:
w/c = 0.38
Cement content = 134.10/0.38 = 352.89 kg/m3
To proportion a mix containing fly ash, the following steps are to be followed.
The % of flyash needs to be decided based on project requirement and quality of materials
Increase in cementitious material content is by 10%
Cementitious material content = 352.89 x 1.10 = 388.17 kg/m3
Water content = 134.10 litres
w/c ratio = 134.10 / 388.17 = 0.34
Flyash at 30% of cement = 388.17 x (30/100) = 116.45 kg/m3
Cement = 388.17 – 116.45. = 272 kg/m3
Saving of cement using flyash =352.89 – 272 = 80.89 kg/m3
- 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 = 0.62
(0.5-W/C /0.05x0.01)+0.62=(0.5-0.34/0.05x0.01)+0.62=0.652
Vol of CA = 0.652
Assuming pumpable concrete, the volume of CA has to be reduced by 10%
Vol of CA = 0.652x0.9 =0.586 m3
Vol of FA = 1- 0.586=0.414 m3
- f) Mix Calculations:
1) Volume of concrete = 1m3
2) Volume of cement = (Mass of cement/ Sp. Gravity of cement) x (1/1000)
= (272/3.15)x0.001 = 0.086 m3
3) Volume of fly ash= (Mass of flyash/ Sp. Gravity of flyash) * (1/1000)
= (116.45/2.2)x0.001 = 0.052 m3
4) Volume of water = (Mass of water/ Sp. Gravity of water) * (1/1000)
= (134.10/1)x(1/1000) = 0.134 m3
5) Volume of chemical admixture = (Mass of chemical admix/ Sp. Gravity ) * (1/1000)
= (7.76/1.145)*(1/1000) = 0.0067 m3
6) Volume of aggregates = (1) – ((2)+(3)+(4)+(5))
= 1 – (0.086+0.052+0.134+0.0067) = 0.721m3
7) Mass of CA = Volume of aggregates * Volume of CA * Sp. Gravity of CA *1000
= 0.721x 0.586 x 2.74x1000= 1157.66 kg/m3
8) Mass of FA =0.721x0.414x2.74x1000 = 783.84 kg/m3
- g) Mix proportions:
Cement = 272 kg/m3
Fly ash = 116.45 kg/m3
Water = 134.10 kg/m3
FA = 783.84 kg/m3
CA = 1157.66 kg/m3
Mix ratio = 1:2.87:4.25
-
Calculate the concrete mix design for M35 grade concrete without fly ash
Given,
M35 grade concrete without fly ash
Mix Design:
- 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
- b) Obtain the w/c ratio:
for f’ck = 43.25 N/mm2, w/c = 0.38
Severe exposure condition , from table 5 of IS 456-2000, w/c = 0.45
The lower value is taken. w/c = 0.38
- c) Selection of water content:
From table 2 of IS 10262,
Max water content = 186 litres
Considering slump as 75 mm = 186 + (3/100)186= 191.58 litres
Superplasticizers are added and thereby reducing the water demand by say 30%
Water content = 191.58 x 0.70 = 134.10 litres
- d) Calculation of cement content:
w/c = 0.38
Cement content = 134.10/0.38 = 352.89 kg/m3
- 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 = 0.62
(0.5-W/C /0.05x0.01)+0.62=(0.5-0.38/0.05x0.01)+0.62=0.652
Vol of CA = 0.644
Assuming pumpable concrete, the volume of CA has to be reduced by 10%
Vol of CA = 0.652x0.9 =0.579 m3
Vol of FA = 1- 0.586=0.421 m3
- f) Mix Calculations:
1) Volume of concrete = 1m3
2) Volume of cement = (Mass of cement/ Sp. Gravity of cement) x (1/1000)
= (352.89/3.15)x0.001 = 0.112m3
3) Volume of water = (Mass of water/ Sp. Gravity of water) x (1/1000)
= (134.10/1)x(1/1000) = 0.134 m3
4) Volume of chemical admixture = (Mass of chemical admix/ Sp. Gravity ) x (1/1000)
= (7.05/1.145)x(1/1000) = 0.0061 m3
5) Volume of aggregates = (1) – ((2)+(3)+(4)+(5))
= 1 – (0.112+0.134+0.0061) = 0.74m3
6) Mass of CA = Volume of aggregates x Volume of CA x Sp. Gravity of CA x 1000
= 0.74x 0.579 x 2.74x1000= 1173.98 kg/m3
7) Mass of FA =0.74x0.421x2.74x1000 = 853.61 kg/m3
- g) Mix proportions:
Cement = 352.89 kg/m3
Water = 134.10 kg/m3
FA = 853.61 kg/m3
CA = 1173.98 kg/m3
Mix ratio = 1:2.41:3.32