Project on Concrete Mix Design for various grades of Concrete
Calculate the Concrete Mix Design for M35 grade concrete with fly ash
Aim : To calculate the concrete mix ratio of ingredients for M35 grade of concrete with fly ash content as mineral admixture.
Introduction : Concrete mix design is done for concrete above M25 grade as below that grade of concrete, the ratio is prefixed. The concrete mix design is done in accordance with the Indian Standard Codes namely, IS 456:2000 and IS 10262:2009. In this mix design fly ash content is used for reducing the volume of the cement used. The stipulations considered for this concrete mix design are
Grade : M35
Cement: OPC 43 grade conforming to IS 8112
Maximum nominal size of aggregate : 20mm
Minimum cement content : 320 kg/m3
Maximum water-cement ratio : 0.45
Workability : 100mm slump(for pumpable concrete)
Exposure condition : Severe(for reinforced concrete)
Mineral admixture : Fly ash content
Chemical admixture : Superplasticizer
Procedure :
1. Calculation of Target Mean Strength
f'ck=fck+1.65*S
where, fck=compressive strength of concrete
S=Standard deviation
f'ck=35+(1.65*5)
=43.25 N/mm2
2.Selection of W/C ratio
From Table 5 of IS 456:2000 the w/c ratio for severe exposure condition is 0.45
From the graphs the w/c ratio is 0.4 which is less than 0.45 hence it is adopted.
3.Selection of water content
From table 2 of IS 10262:2009 the water content for 20mm size aggregate is 186 litres for 25mm to 50mm slump
Here the workability of the concrete is 100mm slump hence the value of water content is increased by 6%
=186+(6/100*186)
=197 litres
As superplasticizer is used the water content can be decreased by 29% based on trial so,
=197*0.71
=140 litres.
4.Selection of cement content
Cementitous material(Cement + Flyash content)=W/w/c
=140/0.4
=350 kg/m3
which is greater than 320 kg/m3 hence ok.
5.Proportion of Fly ash content
Due to the fly ash content the volume of cementitous material is increased. Based on trial 10% value is increased so,
=350*1.10
=385 kg/m3
water-cement ratio=140/385=0.36
Fly ash content @ 30% cementitous material=385*30%=115 kg/m3
Cement content=385-115=270 kg/m3
6.Calculation of volume of Coarse aggregate and Fine aggregate
From table 3 of IS 10262:2009 the value of coarse aggregate for 20mm size aggregate, fine aggregate (Zone 1) and w/c ratio of 0.5=0.6
In this case the w/c ratio is 0.4 hence the value is increase by 0.2 so,
0.4=0.62
For pumpable concrete the values are decreased by 10% so,
=0.62*0.9
=0.56
The volume of fine aggregate is
=1-0.56
=0.44
7.Mix Calculations
Volume of concrete=1m3
Volume of cement=(Mass of cement/specific gravity of cement)*(1/1000)
=(270/3.15)*(1/1000)
=0.086m3
Volume of fly ash =(Mass of fly ash/specific gravity of fly ash)*(1/1000)
=(115/2.2)*(1/1000)
=0.052m3
Volume of water=(Mass of water/specific gravity of water)*(1/1000)
=(140/1)*(1/1000)
=0.140m3
Volume of admixture=(Mass of admixture/specific gravity of water)*(1/1000)
Mass of admixture=2% of cementitous material
=2% of 385
=7.7
Volume of admixture=(7.7/1.145)*(1/1000)
=0.006m3
Volume of Coarse and Fine aggregate=(vol of concrete-(vol of cement+vol of fly ash+vol of water+vol of admixture))
=(1-(0.086+0.052+0.140+0.006))
=0.716m3
Mass of coarse aggregate=Vol of aggregate*vol of coarse aggregate*specific gravity of coarse aggregate*1000
=0.716*0.56*2.74*1000
=1098 kg/m3
Mass of fine aggregate=Vol of aggregate*vol of fine aggregate*specific gravity of fine aggregate*1000
=0.716*0.44*2.74*1000
=863 kg/m3
8.Mix proportions
Cement=270 kg/m3
Fly ash=115 kg/m3
Water=140 kg/m3
Coarse aggregate=1098 kg/m3
Fine aggregate=863 kg/m3
W/C ratio=0.36
Admixture=7 kg/m3
Trial Mix Ratio =1:2.24:2.85
The ratio arrived is based on 1st trial, usually 2 or more trial will be done to fix the final mix ratio.
Calculate the Concrete Mix Design for M50 grade concrete
Aim : To calculate the concrete mix ratio of ingredients for M50 grade of concrete.
Introduction : Concrete mix design is done for concrete above M25 grade as below that grade of concrete, the ratio is prefixed. The concrete mix design is done in accordance with the Indian Standard Codes namely, IS 456:2000 and IS 10262:2009. The stipulations considered for this concrete mix design are
Grade : M50
Cement: OPC 43 grade conforming to IS 8112
Maximum nominal size of aggregate : 20mm
Minimum cement content : 320 kg/m3
Maximum water-cement ratio : 0.45
Workability : 100mm slump(for pumpable concrete)
Exposure condition : Severe(for reinforced concrete)
Chemical admixture : Superplasticizer
Procedure :
1. Calculation of Target Mean Strength
f'ck=fck+1.65*S
where, fck=compressive strength of concrete
S=Standard deviation
f'ck=50+(1.65*5)
=58.25 N/mm2
2.Selection of W/C ratio
From Table 5 of IS 456:2000 the w/c ratio for severe exposure condition is 0.45
From the graphs the w/c ratio is 0.35 which is less than 0.45 hence it is adopted.
3.Selection of water content
From table 2 of IS 10262:2009 the water content for 20mm size aggregate is 186 litres for 25mm to 50mm slump
Here the workability of the concrete is 100mm slump hence the value of water content is increased by 6%
=186+(6/100*186)
=197 litres
As superplasticizer is used the water content can be decreased by 29% based on trial so,
=197*0.71
=140 litres.
4.Selection of cement content
Cementitous material=W/w/c
=140/0.35
=400 kg/m3
which is greater than 320 kg/m3 hence ok.
5.Calculation of volume of Coarse aggregate and Fine aggregate
From table 3 of IS 10262:2009 the value of coarse aggregate for 20mm size aggregate, fine aggregate (Zone 1) and w/c ratio of 0.5=0.6
In this case the w/c ratio is 0.35 hence the value is increase by 0.3 so,
0.35=0.63
For pumpable concrete the values are decreased by 10% so,
=0.63*0.9
=0.57
The volume of fine aggregate is
=1-0.57
=0.43
6.Mix Calculations
Volume of concrete=1m3
Volume of cement=(Mass of cement/specific gravity of cement)*(1/1000)
=(400/3.15)*(1/1000)
=0.126m3
Volume of water=(Mass of water/specific gravity of water)*(1/1000)
=(140/1)*(1/1000)
=0.140m3
Volume of admixture=(Mass of admixture/specific gravity of water)*(1/1000)
Mass of admixture=2% of cementitous material
=2% of 400
=8
Volume of admixture=(8/1.145)*(1/1000)
=0.007m3
Volume of Coarse and Fine aggregate=(vol of concrete-(vol of cement+vol of water+vol of admixture))
=(1-(0.126+0.140+0.007))
=0.727m3
Mass of coarse aggregate=Vol of aggregate*vol of coarse aggregate*specific gravity of coarse aggregate*1000
=0.727*0.57*2.74*1000
=1135 kg/m3
Mass of fine aggregate=Vol of aggregate*vol of fine aggregate*specific gravity of fine aggregate*1000
=0.727*0.43*2.74*1000
=856 kg/m3
7.Mix proportions
Cement=400 kg/m3
Water=140 kg/m3
Coarse aggregate=1135 kg/m3
Fine aggregate=856 kg/m3
W/C ratio=0.35
Admixture=8 kg/m3
Trial Mix Ratio =1:2.14:2.83
The ratio arrived is based on 1st trial, usually 2 or more trial will be done to fix the final mix ratio.