Project on Concrete Mix Design for various grades of Concrete
PROJECT ON CONCRETE MIX DESIGN :
AIM : To calculate the concrete mix design for M35 grade concrete with fly ash and M50 grade without fly ash.
PROCEDURE :
1) Assume data required -
Type of cement = OPC 43 Grade.
Maximum nominal size of aggregate = 20 mm.
Exposure condition = Severe
Fine aggreagate = Zone 1.
Workability = Slump value 75 mm.
Chemical admixture = Superplasticizer-normal.
Mineral admixture = Fly ash confirming to IS 3812 (part 1).
Type of aggregate = Crushed angular aggregate.
Method of concrete placing = Pumping.
The specific gravity of a) Cement = 3.15
b) Fly ash = 2.2
c) Coarse aggregate = 2.74
d) Fine aggregate = 2.74
e) Superplasticizer = 1.145
2) Procedure for concrete mix design of M35 grade concrete with fly ash -
Step-1: Calculation of target mean strength of concrete (f'ck) -
f'ck = fck + 1.65 S
OR
f'ck = fck + X
Whichever is heigher is taken.
Here,
f'ck = 35 N/mm2 ........... (Characteristics compressive strength at 28 days).
S = Standard deviation
S = 5 ............ (From IS 10262:2019, Table 2).
X = Factor based on grade of concrete
X = 6.5 ............ (From IS 10262:2019, Table 1).
Therefore, by putting these values in the above equations, we get -
f'ck = 35 + 1.65 X 5 = 43.25 N/mm2.
And
f'ck = 35 + 6.5 = 41.5 N/mm2.
Hence, 43.25 N/mm2 > 41.5 N/mm2
Adopt, f'ck = 43.25 N/mm2.
Step-2: Selection of Water-Cementitious material ratio -
From Fig.1 of IS 10262:2019 the free water-cementitious ratio required for the target mean strength of 43.25 N/mm2 is 0.387 for OPC 43 grade curve.
This is lower than the maximum value of 0.45 prescribed for 'severe' exposure condition for reinforced concrete as per Table 5 of IS 456:2000.
Therefore, 0.387 < 0.45 ............ Hence O.K.
Therefore, W/C = 0.387
Step-3: Selection of water content -
From Table 4 of IS 10262:2019, water content for 20 mm aggregate and 50 mm slump value = 186 kg/m3.
Estimated water content for 75 mm slump
(increasing at rate of 3% for every 25 mm slump) = 186 + (3/100) X 186 = 191.58 kg/m3.
As superplasticizer is used, the water content may be reduced.
Based on trial data, the water content reduction of 23% is considered while using superplasticizer at the rate of 1% by weight of cement.
Hence, the arrived water content (W) = 191.58 X 0.77 = 147.52 kg
W = 148 kg/m3
Step-4: Calculation of cement and fly ash content -
Water to cementitious material ratio = 0.387
Therefore, cementitious Material = W/0.387 = 148/0.387 = 382.43 kg/m3.
Here, we are using fly ash, so we have to increase cementitious material by 10%.
Therefore, cementitious material (C) = 382.43 X 1.10
C = 420.67 kg/m3
From Table 5 of IS 456:2000 minimum cementitious content for 'severe' exposure condition = 320 kg/m3.
Here, 420.67 kg/m3 > 320 kg/m3 ............ Hence O.K.
now, recalculate W/C ratio,
therefore, W/C = 148/420.67
W/C = 0.35
Fly ash at 30% of total cementitious material content,
therefore, Fly ash = 420.67 X 30% = 420.67 x (3/100)
Fly ash = 126.2 kg/m3
Now, cement content (c) = cementitious material - fly ash
c = 420.67 - 126.2
c = 294.47 kg/m3
Step-5: Proportion of volume of coarse aggregate and fine aggregate content -
From Table 5 of IS 10262:2019, volume of coarse aggregate corresponding to 20 mm size and fine aggregate zone 1 for the water-cement ratio of 0.5 = 0.6
Here, W/C ratio is 0.35, therefore, the volume of coarse aggregate needs to be increased to decrease the fine aggregate content.
As water-cement ratio is lowered by 0.15 the proportion of the volume of coarse aggregate is increased by 0.03 (at the rate of -/+0.01 for every +/-0.05 change in W/C ratio).
Therefore, corrected volume of coarse aggregate for W/C of 0.35 is = 0.6 + 0.03 = 0.63
For pumpable concrete these values may be reduced by up to 10%. Here 10% reduction is considered.
Therefore, volume of coarse aggregate = 0.63 X 0.9 = 0.567, say 0.57
Now, volume of fine aggregate content = 1 - 0.57 = 0.43
Step-6: Mix calculations -
The mix calculations per unit volume of concrete shall be as follows :
a) Total volume of concrete = 1 m3.
b) Volume of cement = (mass of cement/sp. gravity of cement)X(1/1000)
= (294.47/3.15)X(1/1000)
= 0.093 m3
c) Volume of fly ash = (mass of fly ash/sp. gravity of fly ash)X(1/1000)
= (126.2/2.2)X(1/1000)
= 0.057 m3
d) Volume of water = (mass of water/sp.gravity of water)X(1/1000)
= (148/1)X(1/1000)
= 0.148 m3
e) Volume of chemical admixture(superplasticizer) = (mass of superplasticizer/sp. gravity of superplasticizer)X(1/1000)
= (4.21/1.145)X(1/1000) ............. (mass of superplasticizer = 1% of cementitious mtrl)
= 0.0037 m3
f) Volume of all in aggregate = a - (b+c+d+e)
= 1 - (0.093+0.057+0.148+0.0037)
= 0.698 m3
g) Mass of coarse aggregate = f X volume of coarse aggregate X sp. gravity of coarse aggregate X 1000
= 0.698 X 0.57 X 2.74 X 1000 = 1090.13 kg/m3
= 1090 kg/m3
h) Mass of fine aggregate = f X volume of fine aggregate X sp. gravity of fine aggregate X 1000
= 0.698 X 0.43 X 2.74 X 1000 = 822.4 kg/m3
= 822 kg/m3
Step-7: Mix proportions for trial -
Cement = 294.5 kg/m3
Fly ash = 126.2 kg/m3
Water = 148 kg/m3
Fine aggregate = 822 kg/m3
Coarse aggregate = 1090 kg/m3
Superplasticizer = 4.21 kg/m3
W/C ratio = 0.35
3) Procedure for concrete mix design of M50 grade concrete without fly ash -
Step-1: Calculation of target mean strength of concrete (f'ck) -
f'ck = fck + 1.65 S
OR
f'ck = fck + X
Whichever is heigher is taken.
Here,
f'ck = 50 N/mm2 ........... (Characteristics compressive strength at 28 days).
S = Standard deviation
S = 5 ............ (From IS 10262:2019, Table 2).
X = Factor based on grade of concrete
X = 6.5 ............ (From IS 10262:2019, Table 1).
Therefore, by putting these values in the above equations, we get -
f'ck = 50 + 1.65 X 5 = 58.25 N/mm2.
And
f'ck = 50 + 6.5 = 56.5 N/mm2.
Hence, 58.25 N/mm2 > 56.5 N/mm2
Adopt, f'ck = 58.25 N/mm2.
Step-2: Selection of Water-Cement ratio -
From Fig.1 of IS 10262:2019 the free water-cement ratio required for the target mean strength of 58.25 N/mm2 is 0.29 for OPC 43 grade curve.
This is lower than the maximum value of 0.45 prescribed for 'severe' exposure condition for reinforced concrete as per Table 5 of IS 456:2000.
Therefore, 0.29 < 0.45 ............ Hence O.K.
Therefore, W/C = 0.29
Step-3: Selection of water content -
From Table 4 of IS 10262:2019, water content for 20 mm aggregate and 50 mm slump value = 186 kg/m3.
Estimated water content for 75 mm slump
(increasing at rate of 3% for every 25 mm slump) = 186 + (3/100) X 186 = 191.58 kg/m3.
As superplasticizer is used, the water content may be reduced.
Based on trial data, the water content reduction of 23% is considered while using superplasticizer at the rate of 1% by weight of cement.
Hence, the arrived water content (W) = 191.58 X 0.77 = 147.52 kg
W = 148 kg/m3
Step-4: Calculation of cement content -
Water to cement ratio = 0.29
Therefore, cement = W/0.29 = 148/0.29 = 510.3 kg/m3.
C = 510.3 kg/m3
From Table 5 of IS 456:2000 minimum cement content for 'severe' exposure condition = 320 kg/m3.
Here, 510.3 kg/m3 > 320 kg/m3 ............ Hence O.K.
Step-5: Proportion of volume of coarse aggregate and fine aggregate content -
From Table 5 of IS 10262:2019, volume of coarse aggregate corresponding to 20 mm size and fine aggregate zone 1 for the water-cement ratio of 0.5 = 0.6
Here, W/C ratio is 0.29, therefore, the volume of coarse aggregate needs to be increased to decrease the fine aggregate content.
As water-cement ratio is lowered by 0.21 the proportion of the volume of coarse aggregate is increased by 0.042 (at the rate of -/+0.01 for every +/-0.05 change in W/C ratio).
Therefore, corrected volume of coarse aggregate for W/C of 0.29 is = 0.6 + 0.042 = 0.642
For pumpable concrete these values may be reduced by up to 10%. Here 10% reduction is considered.
Therefore, volume of coarse aggregate = 0.642 X 0.9 = 0.578, say 0.58
Now, volume of fine aggregate content = 1 - 0.58 = 0.42
Step-6: Mix calculations -
The mix calculations per unit volume of concrete shall be as follows :
a) Total volume of concrete = 1 m3.
b) Volume of cement = (mass of cement/sp. gravity of cement)X(1/1000)
= (510.3/3.15)X(1/1000)
= 0.162 m3
c) Volume of water = (mass of water/sp.gravity of water)X(1/1000)
= (148/1)X(1/1000)
= 0.148 m3
d) Volume of chemical admixture(superplasticizer) = (mass of superplasticizer/sp. gravity of superplasticizer)X(1/1000)
= (5.1/1.145)X(1/1000) ............. (mass of superplasticizer = 1% of cement)
= 0.0045 m3
e) Volume of all in aggregate = a - (b+c+d)
= 1 - (0.162+0.148+0.0.0045)
= 0.6855 m3
f) Mass of coarse aggregate = f X volume of coarse aggregate X sp. gravity of coarse aggregate X 1000
= 0.6855 X 0.58 X 2.74 X 1000 = 1089.4 kg/m3
= 1089.4 kg/m3
g) Mass of fine aggregate = f X volume of fine aggregate X sp. gravity of fine aggregate X 1000
= 0.6855 X 0.42 X 2.74 X 1000 = 822.4 kg/m3
= 788.9 kg/m3
Step-7: Mix proportions for trial -
Cement = 510 kg/m3
Water = 148 kg/m3
Fine aggregate = 788.9 kg/m3
Coarse aggregate = 1089.4 kg/m3
Superplasticizer = 5.1 kg/m3
W/C ratio = 0.29
RESULT :
1) The concrete mix design for M35 with fly ash gives us trial mix ratio of cement : fine aggregate : coarse aggregate as 1:2.79:3.7.
2) The concrete mix design for M50 without fly ash gives us trial mix ratio of cement : fine aggregate : coarse aggregate as 1:1.55:2.14.