Project on Concrete Mix Design for various grades of Concrete
Calculate the M50 grade concrete without Fly ash
Concrete Mix Design as Per IS 10262:2019:
Steps to be followed:
1. Target Strength (Refer table 1 and table 2)
2. Approximate Air Content ( refer table3)
3. Selection of w/c ratio ( refer fig.1)
4. Selection of water content (refer table4)
5. Calculation of cement content
6. Proportion of volume of coarse aggregate (refer table5)
7. Mix calculation
Aim:
To Design the M50 grade of Concrete without flyash
Mix proportion of concrete
Grade Designation - M50
Type of cement - OPC 43
Maximum nomial size of aggregates - 10mm
Workability - 100mm
Exposure - severe
Minimum cement Content - 320kg/m^3
Test data:
1. Specific gravity of Cement: 3.17
2. Specific gravity of Fine aggregate: 2.64
3. Specific gravity of Coarse aggregate: 2.75
4. Specific gravity of Chemical admixture: 1.10
5. Water absorption of coarse aggregate: 0.5%
6. Water absorption of fine aggregate: 1%
7. Fine aggregate: Conforming to Zone 2 of Table 9 of IS 383
Target strength:
f^ck = fck + 1.65 S
From Table 2 of IS 10262:2019, S=5
f^ck= 50 + (1.65*5) = 58.25N/mm^2
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 10262:2019 w/c = 0.34
0.34<0.45
Therefore w/c ratio=0.38
Selection of water content:
From IS 10262:2019 Table 4, Water content for 10mm size aggregate = 208kg
For 100mm slump, water content = 208(0.06*206) = 220.48lit
Based on trial water content reduction of 23% is considered while using super plasticizer.
Now water content = 220.48*0.77 = 169.77kg/m^3
Calculation of Cement content:
Water cement ratio = 0.34
Water content as per concrete = 169.77kg/m^3
Cement content = 169.77/0.34 = 499.32kg/m^3
As per IS 456, minimum cement content is 320kg/m^3
Hence it is ok.
Superplasticizer = 1.5% of weight of cement content = 0.015*499.32 = 7.48kg/m^3
Proportioning of colume of coarse aggregate and fine aggregate:
As per IS 10262:2019 Table 5
For zone 2, Volume of coarse aggregate = 0.50
For w/c 0.5=0.50
For w/c 0.34= ?
X =[((0.5-0.34)*0.01)/0.05]
X = 0.032
Proportioning volume of coarse aggregate = 0.50 + 0.032 = 0.532
Proportioning volume of fine aggregate = 1 - 0.532 = 0.468
Mix calculation:
a) Volume of concrete = 1m^3
b) Volume of cement = weight of cement/(specific gravity * 1000)
= 499.32/(3.17*1000) = 0.157m^3
c) Volume of water = [ mass of water] / {[specific gravity of water] * 1000}
= 169.77/(1*1000) = 0.169m^3
d) Volume of Admixture = 7.48/(1.10*1000) = 0.0068
Total volume of other material except aggregates = 0.015+0.157+0.169+0.0068 = 0.3478m^3
Volume of Coarse aggregate and Fine aggregate = 1-0.3478 = 0.6522m^3
Mass of coarse aggregate = 0.6522*0.532*2.75*1000 = 954.16kg/m^3
Mass of Fine aggregate = 0.6522*0.468*2.64*1000 = 805.81kg/m^3
Result:
Mix proportions:
Cement = 499.32kg/m^3
Water = 169.77kg/m^3
Admixture = 7.48kg/m^3
Fine aggregate = 805.81kg/m^3
Coarse aggregate = 954.16kg/m^3
w/c = 0.34
2. Calculate the Concrete Mix Design for M35 grade concrete with fly ash
Concrete Mix Design as Per IS 10262:2019:
Steps to be followed:
1. Target Strength (Refer table 1 and table 2)
2. Approximate Air Content ( refer table3)
3. Selection of w/c ratio ( refer fig.1)
4. Selection of water content (refer table4)
5. Calculation of cement content
6. Proportion of volume of coarse aggregate (refer table5)
7. Mix calculation
Aim:
To Design the M35 grade of Concrete with flyash
Mix proportion of concrete
Grade Designation - M35 with fly ash
Type of cement - OPC 53
Maximum nomial size of aggregates - 20mm
Workability - 110mm
Exposure - Very severe
Minimum cement Content - 340kg/m^3
Maximum w/c ratio: 0.45
Minimum grade: M35
Method of placing: Pumping
Maximum cement content: 450kg/m^3(As per IS 456:2000, CI 8.2.4.2)
Chemical admixture: Superplasticizer
Test data:
1. Specific gravity of Cement: 3.17
2. Specific gravity of Fly ash: 2.2
3. Specific gravity of Fine aggregate: 2.64
4. Specific gravity of Coarse aggregate: 2.75
5. Specific gravity of Chemical admixture: 1.10
6. Water absorption of coarse aggregate: 0.5%
7. Water absorption of fine aggregate: 1%
8. Fine aggregate: Conforming to Zone 2 of Table 9 of IS 383
Target strength:
f^ck = fck + 1.65 S
From Table 2 of IS 10262:2019, S=5
f^ck= 35 + (1.65*5) = 43.25N/mm^2
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 10262:2019 w/c = 0.46
0.45<0.46
Therefore w/c ratio=0.45
Selection of water content:
From IS 10262:2019 Table 4, Water content for 20mm size aggregate = 186kg
For 110mm slump, water content = 186+(0.072*186) =199.39 lit
Based on trial water content reduction of 23% is considered while using super plasticizer.
Now water content = 199.39*0.77 = 153.53kg/m^3
Calculation of Cement content:
Water cement ratio = 0.45
Water content as per concrete = 153.53kg/m^3
Cementitious content = 153.53/0.45 = 341.17kg/m^3
As per IS 456, minimum cement content is 340kg/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 cement= 341.17 * 1.10 = 375.287kg/m^3
New w/c = 153.53/375.287 = 0.41
25% of cementitious is going to be replaced by fly ash
Fly ash = 375.287 * 25% = 93.82kg/m^3
Cement = 281.46kg/m^3
Superplasticizer = 1.5% of weight of cement content = 0.015*375.287 = 5.62kg/m^3
Proportioning of colume of coarse aggregate and fine aggregate:
As per IS 10262:2019 Table 5
For zone 2, Volume of coarse aggregate = 0.62
For w/c 0.5=0.62
For w/c 0.41= ?
X =[((0.5-0.41)*0.01)/0.05]
X = 0.018
Proportioning volume of coarse aggregate = 0.62 + 0.018 = 0.638
Since the method of placing is pumping, volume of coarse aggregate should be reduced by 10%
New volume of coarse aggregate = 0.638 * 0.9 = 0.574
Proportioning volume of fine aggregate = 1 - 0.574 = 0.426
Mix calculation:
a) Volume of concrete = 1m^3
b) Volume of cement = weight of cement/(specific gravity * 1000)
= 481.46/(3.17*1000) = 0.088m^3
c) Volume of fly ash = 93.82/(2.2*1000) = 0.0426m^3
d) Volume of water = [ mass of water] / {[specific gravity of water] * 1000}
= 153.53/(1*1000) = 0.153m^3
e) Volume of Admixture = 5.63/(1.10*1000) = 0.0051m^3
Total volume of other material except aggregates = 0.088+0.0426+0.153+0.0051+0.01 = 0.291m^3
Volume of Coarse aggregate and Fine aggregate = 1-0.291 = 0.709m^3
Mass of coarse aggregate = 0.709*0.574*2.75*1000 = 1,119.15kg/m^3
Mass of Fine aggregate = 0.709*0.426*2.64*1000 = 797.36kg/m^3
Result:
Mix proportions:
Cement = 281.47kg/m^3
Fly ash = 93.82kg/m^3
Water = 153.53kg/m^3
Admixture = 5.63kg/m^3
Fine aggregate = 797.36kg/m^3
Coarse aggregate = 1,119.15 kg/m^3
w/c = 0.45