Project on Concrete Mix Design for various grades of Concrete
1. Concrete mix design for M35 grade concrete with fly ash.
Step 1- Stipulations for proportioning
Grade of designation = M35
Type of cement = OPC43 grade
Type of mineral admixture = Fly ash
maximum nominal size of aggregate = 20mm
minimum cement content = 320kg/m^3 (from table 5 IS 456 - exposure condition)
maximum water cement ratio = 0.45 from table 5 IS456
workability = 100mm slump value
exposure condition = severe
Type of aggregate = crushed angular
maximum cement content = 450 kg/m^3
The values of fine aggregate should conform to zone 1 of table 4 of IS383.
Step 2 - Target mean strength of concrete,
f'ck = fck + 1.65s
where fck is characteristics compressive strength at 28 days,
s = standard deviation (IS 10262:2009,table 1)
for M35, s = 5N/mm^2
f'ck = 43.25 N/mm^2
Step 3 - water cement ratio
From table 5 of IS 456 2000 it can be taken depending upon the exposure condition,
Assume the weather condition is severe, then water cement ratio is 0.45.
From figure 1 of IS10262 -2019 the water cement ratio is 0.38.
therefore, water cement ratio is taken as 0.38
Step 4 - selection of water content
From table 4 of IS10262-2009, the maximum water content for 25mm-50mm slump is 186Kg= 186 lit
Estimated water content for 100mm slump is 186 + 6/100*186 = 197lit
As superplasticizer is used, the water content can be reduced upto 30% and above.
Based on trials water content reduction of 29% is achieved with plasticizer.
Hence arrived water content is 197*0.71 = 140 lit
Step 5- selection of cement content
It is calculated from water cement ratio
w = 140
w/c = 0.38
Cementious material content = w/0.38 = 140/0.38 =368Kg/m^3
From table 5 of IS456 minimum cement content for sever exposure condition is 320Kg/m^3
368>320 hence ok.
To proportion a mix containing fly ash the following steps are adopted,
a) the percentage of fly ash needs to be decided based on the project requirement.
b) In some cases increase in cementious content can be warranted. Increase in cementious material content may be based on experience and trial. So lets increase only 10% cementious material content.
Cementious material content = 368*1.10 = 404.8kg/m^3
water content = 140 kg/m^3
w/c ratio = 140/404.8 = 0.345
fly ash @30% total cementious material content = 404.8*30% = 121kg/m^3
cement (OPC)= 404.8-121 = 283 Kg/m^3
saving of cement while using fly ash = 368-283 = 85Kg/m^3
Fly ash being utilised = 121kg/m^3
Step 6 - volume of coarse aggregate and fine aggregate
This can be determined from table 3 IS10262 - 2009. Volume of coarse aggregate corresponding to 20mm size aggregate and fine aggregate (zone 1) for water cement ratio = 0.6
In our case the w/c ratio is 0.38
therefore the volume of coarse aggregate needs to be increased to decrease the fine aggregate content.
As in this case the water cement ratio is lower by 0.12, the proportion of volume of coarse aggregate is increased by 0.024 (at the rate of -/+ 0.01 for every for every +/- 0.05 change in water cement ratio.
Therefore corrected volume of coarse aggregate for the water cement ratio of 0.38 = 0.62
For pumpable concrete these values are reduced by 10%
therefore volume of coarse aggregate = 0.624*0.9 = 0.56
volume of fine aggregate content = 1-0.56 = 0.44
Step 7- Mix calculations,
a)volume of concrete = 1 m^3
b)volume of cement = mass of cement/specific gravity of cement * 1/1000
= 283/3.15*1/1000 = 0.089 m^3
c) volume of fly ash = mass of fly ash/specific gravity of fly ash*1/1000
= 121/2.2*1/1000 = 0.055 m^3
d) volume of water = mass of water/specific gravity of water * 1/1000
= 140/1 * 1/1000 = 0.140 m^3
e) volume of chemical admixture = mass of chemical admixture/sp gravity * 1/1000
= 7.6/1.145*1/1000 = 0.006m^3
f) volume of aggregate = (a-(b+c+d+e))
= 1-(0.089+0.055+0.140+0.006) = 0.71
g) mass of coarse aggregate = 0.71 * 0.56*2.74*1000 = 1089 Kg/m^3
h) mass of fine aggregate = 0.71 * 0.44 * 2.74 * 1000 = 856 Kg/m^3
Step 8- mix proportion
cement = 283Kg/m^3
fly ash = 121Kg/m^3
fine aggregate = 856 Kg/m^3
coaarse aggregate = 1089 kg/m^3
total mix ratio = 1: 3.02 : 3.84
2. Concrete mix design for M50 grade of concrete.
Step 1 - Stipulations for proportioning
Grade of designation = M50
Type of cement = OPC43 grade
maximum nominal size of aggregate = 20mm
minimum cement content = 320kg/m^3 (from table 5 IS 456 - exposure condition)
maximum water cement ratio = 0.45 from table 5 IS456
workability = 100mm slump value
exposure condition = severe
Type of aggregate = crushed angular
maximum cement content = 450 kg/m^3
The values of fine aggregate should conform to zone 1 of table 4 of IS383.
Step 2 - Target mean strength of concrete,
f'ck = fck + 1.65s
where fck is characteristics compressive strength at 28 days,
s = standard deviation (IS 10262:2009,table 1)
for M50, s = 5N/mm^2
f'ck = 58.25 N/mm^2
Step 3 - water cement ratio
From table 5 of IS 456 2000 it can be taken depending upon the exposure condition,
Assume the weather condition is severe, then water cement ratio is 0.45.
From figure 1 of IS10262 -2019 the water cement ratio is 0.33.
therefore, water cement ratio is taken as 0.33
Step 4 - selection of water content
From table 4 of IS10262-2009, the maximum water content for 25mm-50mm slump is 186Kg= 186 lit
Estimated water content for 100mm slump is 186 + 6/100*186 = 197lit
As superplasticizer is used, the water content can be reduced upto 30% and above.
Based on trials water content reduction of 29% is achieved with plasticizer.
Hence arrived water content is 197*0.71 = 140 lit
Step 5- selection of cement content
It is calculated from water cement ratio
w = 140
w/c = 0.33
Cement content = w/0.33 = 140/0.33 = 424 Kg/m^3
From table 5 of IS456 minimum cement content for sever exposure condition is 320Kg/m^3
424>320 hence ok.
Step 6 - volume of coarse aggregate and fine aggregate
This can be determined from table 3 IS10262 - 2009. Volume of coarse aggregate corresponding to 20mm size aggregate and fine aggregate (zone 1) for water cement ratio = 0.6
In our case the w/c ratio is 0.33
therefore the volume of coarse aggregate needs to be increased to decrease the fine aggregate content.
As in this case the water cement ratio is lower by 0.17, the proportion of volume of coarse aggregate is increased by 0.034 (at the rate of -/+ 0.01 for every for every +/- 0.05 change in water cement ratio.
Therefore corrected volume of coarse aggregate for the water cement ratio of 0.33 = 0.63
For pumpable concrete these values are reduced by 10%
therefore volume of coarse aggregate = 0.63*0.9 = 0.57
volume of fine aggregate content = 1-0.56 = 0.43
Step 7- Mix calculations,
a)volume of concrete = 1 m^3
b)volume of cement = mass of cement/specific gravity of cement * 1/1000
= 424/3.15*1/1000 = 0.134 m^3
c) volume of water = mass of water/specific gravity of water * 1/1000
= 140/1 * 1/1000 = 0.140 m^3
d) volume of chemical admixture = mass of chemical admixture/sp gravity * 1/1000
= 7.6/1.145*1/1000 = 0.006m^3
e) volume of aggregate = (a-(b+c+d))
= 1-(0.134+0.140+0.006) = 0.72
g) mass of coarse aggregate = 0.72 * 0.57*2.74*1000 = 1124.5 Kg/m^3
h) mass of fine aggregate = 0.72 * 0.43* 2.74 * 1000 = 848.3 Kg/m^3
Step 8- mix proportion
cement = 424 Kg/m^3
fine aggregate = 848.3 Kg/m^3
coaarse aggregate = 1124.5 kg/m^3
total mix ratio = 1:2:2.65