Project on Concrete Mix Design for various grades of Concrete
Ans 1.
AIM- To Calculate the Concrete Mix Design for M35 grade concrete with fly ash
CODES FOLLOWED FOR CONCRETE MIX DESIGN-
IS 456:2000
IS 3812 (PART 1)
IS 8112
IS 9103
IS 10262:2009
INTRODUCTION-
The equations/formulae to be followed here is:
- Target mean Strength : f'ck= fck+1.65s
where, f'ck= target mean strength, fck= compressive strength at 28 days of concrete, s= standard deviation [Referred from Table-1 IS 10262:2009]
- Volume of Cement= (Mass of Cement/ Specific gravity of cement)x 1/1000
- Volume of Water= (Mass of Water/ Specific gravity of Water)x 1/1000
- Volume of Chemical Admixtures= (Mass of Chemical Admixtures/ Specific gravity of Chemical admixtures)x 1/1000
- Volume of fly ash= (Mass of fly ash/ Specific gravity of fly ash)x 1/1000
- Volume of Concrete= 1m3
- Volume of Aggregates= [Volume of Concrete -(Volume of cement + Volume of fly ash + Volume of admixture + Volume of water)]
- Mass of Coarse Aggregates= Volume of aggregate x volume of coarse aggregate x specific gravity of coarse aggregate x 1000
- Mass of Fine Aggregates= Volume of aggregate x volume of fine aggregate x specific gravity of fine aggregate x 1000
PROCEDURE-
- Target mean Strength : f'ck= fck+1.65s, f'ck= 35+1.65x5.0, f'ck= 43.25 N/mm2
- Water Cement Ratio : From [Table-5 IS 456:2000] In very severe exposure, the maximum free water cement ratio for M35 grade of concrete is 0.45. In Extreme exposure, the maximum free water cement ratio is 0.4. As we know lesser the water cement ratio better is the workability. So, 0.40<0.45. Hence we take 0.4 as the water cement ratio. So, the minimum cement content is 360 kg/m3.
- Water Content : Assume Nominal Size of Aggregate = 20mm [from Table-2 IS 10262:2009]; minimum water content= 186 kg Water
Content= 186x Superplasticizer (20%) = 186x (100-20)/100 = 148.8 litres. Take Water Content as 149 litres.
- CALCULATION OF CEMENT + FLY ASH :
Data calculated:
Water cement ratio (w/c)= 0.4
Water Content (W)= 149 litres
Cement content= W/(w/c) = 149/0.4 = 372.5 kg/m3
Minimum cement content = 360 kg/m3
Now 360<372.5. Hence Ok.
Increase by 10%= 372.5 x 0.1 + 372.5 = 409.75 kg/m3
New Water cement ratio = 149/409.75= 0.36
Fly ash 30% = 409.75x 0.3= 122.925 kg/m3
Cement (OPC) = 409.75-122.925 = 286.825 kg/m3
Cement - cement with fly ash = 372.5-286.825 = 85.675 kg/m3
- Volume of Coarse Aggregates and Fine Aggregates:
Fine aggregates are sieved on the basis of zone 1 of table 4 of IS 383:1970
Coarse Aggregates are sieved on the basis of table 2 of IS 383:1970
From table 3 IS10262:2009, for Nominal Maximum size of aggregate 20mm.
Volume of Coarse Aggregate in Zone 1 for 20mm size of aggregate is 0.60
Volume of Fine Aggregate is 1-0.6 = 0.4
- Mix Calculations:
Volume of Concrete = 1m3
Volume of Cement = (Mass of Cement/ Specific gravity of cement)x 1/1000 = (286.825/3.15)x 1/1000 = 0.09m3
Volume of Fly ash = (Mass of fly ash/ Specific gravity of fly ash)x 1/1000 = (122.925/2.2)x 1/1000 = 0.0558m3
Volume of Water = (Mass of Water/ Specific gravity of Water)x 1/1000 = (149/1)x 1/1000 = 0.149m3
Volume of Chemical Admixtures= (Mass of Chemical Admixtures/ Specific gravity of Chemical admixtures)x 1/1000 = (7.6/1.145)x 1/1000 = 0.006
Volume of Aggregates= [Volume of Concrete -(Volume of cement + Volume of fly ash + Volume of admixture + Volume of water)] = [1-(0.09+0.0558+0.006+0.149)] = 0.6992m3
Mass of Coarse Aggregates= Volume of aggregate x volume of coarse aggregate x specific gravity of coarse aggregate x 1000 = 0.6992x0.6x2.74x1000 = 1149.4848 kg/m3
Mass of Fine Aggregates= Volume of aggregate x volume of fine aggregate x specific gravity of fine aggregate x 1000 = 0.6992x0.4x2.74x1000 = 766.32 kg/m3
RESULT-
Cement (OPC) = 286.825 kg/m3
Fly Ash = 122.925 kg/m3
Water Content = 149 litres
Coarse Aggregate = 1149.48 kg/m3
Fine Aggregate = 762.32 kg/m3
Chemical Admixture = 7.6 kg/m3
W/C ratio = 0.36
Mix Ratio = Cement: Sand (Fine Aggregate) : Coarse Aggregate = 286.825: 762.32: 1149.48 = 1: 2.65: 4
Ans 2.
AIM- To Calculate the Concrete Mix Design for M50 grade concrete without fly ash
CODES FOLLOWED FOR CONCRETE MIX DESIGN-
IS 456:2000
IS 3812 (PART 1)
IS 8112
IS 9103
IS 10262:2009
INTRODUCTION-
The equations/formulae to be followed here is:
- Target mean Strength : f'ck= fck+1.65s
where, f'ck= target mean strength, fck= compressive strength at 28 days of concrete, s= standard deviation [Referred from Table-1 IS 10262:2009]
- Volume of Cement= (Mass of Cement/ Specific gravity of cement)x 1/1000
- Volume of Water= (Mass of Water/ Specific gravity of Water)x 1/1000
- Volume of Chemical Admixtures= (Mass of Chemical Admixtures/ Specific gravity of Chemical admixtures)x 1/1000
- Volume of Concrete= 1m3
- Volume of Aggregates= [Volume of Concrete -(Volume of cement + Volume of fly ash + Volume of admixture + Volume of water)]
- Mass of Coarse Aggregates= Volume of aggregate x volume of coarse aggregate x specific gravity of coarse aggregate x 1000
- Mass of Fine Aggregates= Volume of aggregate x volume of fine aggregate x specific gravity of fine aggregate x 1000
PROCEDURE-
- Target mean Strength : f'ck= fck+1.65s, f'ck= 50+1.65x5.0, f'ck= 58.25 N/mm2
- Water Cement Ratio : From [Table-5 IS 456:2000] In Extreme exposure, the maximum free water cement ratio for M50 grade of concrete is 0.4. As we know lesser the water cement ratio better is the workability. Hence we take 0.4 as the water cement ratio. So, the minimum cement content is 360 kg/m3.
- Water Content : Assume Nominal Size of Aggregate = 20mm [from Table-2 IS 10262:2009]; minimum water content= 186 kg Water
Content= 186x Superplasticizer (20%) = 186x (100-20)/100 = 148.8 litres. Take Water Content as 149 litres.
- CALCULATION OF CEMENT :
Data calculated:
Water cement ratio (w/c)= 0.4
Water Content (W)= 149 litres
Cement content= W/(w/c) = 149/0.4 = 372.5 kg/m3
Minimum cement content = 360 kg/m3
Now 360<372.5. Hence Ok.
- Volume of Coarse Aggregates and Fine Aggregates:
Fine aggregates are sieved on the basis of zone 1 of table 4 of IS 383:1970
Coarse Aggregates are sieved on the basis of table 2 of IS 383:1970
From table 3 IS10262:2009, for Nominal Maximum size of aggregate 20mm.
Volume of Coarse Aggregate in Zone 1 for 20mm size of aggregate is 0.60
Volume of Fine Aggregate is 1-0.6 = 0.4
- Mix Calculations:
Volume of Concrete = 1m3
Volume of Cement = (Mass of Cement/ Specific gravity of cement)x 1/1000 = (372.5/3.15)x 1/1000 = 0.118m3
Volume of Water = (Mass of Water/ Specific gravity of Water)x 1/1000 = (149/1)x 1/1000 = 0.149m3
Volume of Chemical Admixtures= (Mass of Chemical Admixtures/ Specific gravity of Chemical admixtures)x 1/1000 = (7.6/1.145)x 1/1000 = 0.006
Volume of Aggregates= [Volume of Concrete -(Volume of cement + Volume of admixture + Volume of water)] = [1-(0.118+0.006+0.149)] = 0.727m3
Mass of Coarse Aggregates= Volume of aggregate x volume of coarse aggregate x specific gravity of coarse aggregate x 1000 = 0.727x0.6x2.74x1000 = 1195.188 kg/m3
Mass of Fine Aggregates= Volume of aggregate x volume of fine aggregate x specific gravity of fine aggregate x 1000 = 0.727x0.4x2.74x1000 = 796.792 kg/m3
RESULT-
Cement (OPC) = 372.5 kg/m3
Water Content = 149 litres
Coarse Aggregate = 1195.188 kg/m3
Fine Aggregate = 796.792 kg/m3
Chemical Admixture = 7.6 kg/m3
W/C ratio = 0.4
Mix Ratio = Cement: Sand (Fine Aggregate) : Coarse Aggregate = 372.5: 796.792: 1195.188 = 1: 2.13: 3.20