Project on Concrete Mix Design for various grades of Concrete
1Ans.The Concrete Mix Design for M35 grade concrete with fly ash
a) Data Collection:
Grade of concrete = M35
fck = 35 N/mm2
S = 5 N/mm2
f’ck = fck + 1.65 S
= 35 + (1.65 * 5)
= 43.25 N/mm2
- b) Obtain the w/c ratio:
From the graph, for the corresponding 28 days compressive strength which is the target mean strength ,obtain w/c ratio
w/c = 0.36
Based on the given exposure condition , from table 5 of IS 456, w/c = 0.45
The lower value is taken. w/c = 0.36
- c) Selection of water content:
From table 2 of IS 10262,
Max water content = 186 litres
Considering slump as 100 mm = 186 + (6/100)186
= 197.18 litres
as flyash is used water content can be reduced upto 30%
basded on trails water content reduction is 29%
water content = 197.18*0.71=140 liters
- d) Calculation of cement content:
w/c = 0.36
Cement content = 140/0.36 = 388.8 kg/m3
388.8 > 320
cementitious material content = 388.8*1.10=427.68kg/m3
water content =140 liters
so w/c ratio = 140/427.68=0.32
fly ash 30% = 427.68 *30%=128 kg/m3
cement (opc)=427.68-128.304=300kg/m3
saving of cement while using fly ash = 388.8-299.376=89kg/m3
flyash being utilized = 128kg/m3
- e) Proportion of volume of CA and FA:
From table 3 of IS 10262, Volume of CA corresponding to 20mm size aggregate and FA( zone 1 ) for w/c ratio of 0.5 = 0.60
In our case, the w/c ratio is 0.36, which is lower by 0.11, the volume of CA is increased by 0.024,
Vol of CA = 0.624
Vol of FA = 1- 0.624 = 0.376
- f) Mix Calculations:
1) Volume of concrete = 1m3
2) Volume of cement = (Mass of cement/ Sp. Gravity of cement) * (1/1000)
= (299/3.15)*0.001 = 0.094m3
3)volume of flyash = (mass of fly ash/sp.gravity of flyash)*(1/10000)
=(128/2.2)*(1/1000)=0.058
4) Volume of water = (Mass of water/ Sp. Gravity of water) * (1/1000)
= (140/1)*(1/1000) = 0.14m3
5)volume of chemical admixture = (mass of chemical admixture/sp. gravity of admixture)*(1/1000)
=7/1.145*1/1000=0.007
6) Volume of aggregates = (1-(2+3+4+5))
= 1 – (0.094+0.058+0.14+0.007) = 0.701 m3
5) Mass of CA = Volume of aggregates * Volume of CA * Sp. Gravity of CA *1000
0.701 * 0.624 * 2.74*1000= 1190 kg/m3
6) Mass of FA =0.701 * 0.376*2.74*1000 = 710 kg/m3
- g) Mix proportions:
Cement = 300 kg/m3
fly ash = 128 kg/m3
Water = 140 kg/m3
FA = 710 kg/m3
CA = 1190 kg/m3
w/c = 0.36
Mix ratio = 1:1.65:2.78
2Ans.M50 grade concrete without Fly ash
a) Data Collection:
Grade of concrete = M50
fck = 50 N/mm2
S = 5 N/mm2
f’ck = fck + 1.65 S
= 30 + (1.65 * 5)
= 58.25 N/mm2
- b) Obtain the w/c ratio:
From the graph, for the corresponding 28 days compressive strength which is the target mean strength ,obtain w/c ratio
w/c = 0.29
Based on the given exposure condition , from table 5 of IS 456, w/c = 0.45
The lower value is taken. w/c = 0.29
- c) Selection of water content:
From table 2 of IS 10262,
Max water content = 186 litres
Considering slump as 75 mm = 186 + (6/100)186
= 197.18 litres
- d) Calculation of cement content:
w/c = 0.29
Cement content = 197.18/0.29 = 679.93 kg/m3
- e) Proportion of volume of CA and FA:
From table 3 of IS 10262, Volume of CA corresponding to 20mm size aggregate and FA( zone 1 ) for w/c ratio of 0.5 = 0.60
In our case, the w/c ratio is 0.39, which is lower by 0.11, the volume of CA is increased by 0.022,
Vol of CA = 0.642
Vol of FA = 1- 0.642 = 0.358
- f) Mix Calculations:
1) Volume of concrete = 1m3
2) Volume of cement = (Mass of cement/ Sp. Gravity of cement) * (1/1000)
= (679.93/3.15)*0.001 = 0.215 m3
3) Volume of water = (Mass of water/ Sp. Gravity of water) * (1/1000)
= (197.18/1)*(1/1000) = 0.197 m3
4) Volume of aggregates = (1) – ((3)+(2))
= 1 – (0.215+0.197) = 0.982 m3
5) Mass of CA = Volume of aggregates * Volume of CA * Sp. Gravity of CA *1000
0.982* 0.642 * 2.74*1000= 1727.41kg/m3
6) Mass of FA =0.982 * 0.358*2.74*1000 = 963.26 kg/m3
- g) Mix proportions:
Cement = 679.92 kg/m3
Water = 197.18 kg/m3
FA = 1727.41 kg/m3
CA = 963.26kg/m3
w/c = 0.29
Mix ratio = 1:1.41:2.54