Project on Concrete Mix Design for various grades of Concrete
Concrete Mix Design for M35 grade concrete with fly ash
Step 1 calculation of target mean strength
target mean strength Ft = Fck + 1.655*S
where
Ft = target average compressive strength at 28 days
Fck = charecteristic compressive strength at 28 days
S = standarsd deviation
Fck = 35 N/mm2 (for m35 grade of concrete )
S = 5 ( table 8 of IS 456 -2000)
Ft = 35 + 1.65*5
= 43.25 N/mm2
Step 2 Calculation of water cement ratio
take water cement ratio from IS 10262-2019 and IS 456-2000 by Using target mean strength value.
from graph the water cement ratio is .38 (code book of IS 10262-2019 )
from table 5 the water cement ratio is .45 ( code book of IS 456-2000 )
The lower value of the above two values is adopted as water cement ratio for mix design
Hence our water cement ratio is 0.38.
Step 3 Selection of water content
From Table 2 of IS 456, maximum water content =186 litres (for 25 to 50 mm slump range) for 20 mm aggregate.
Estimated water content for 100 mm slump =186+(6/100) x 186 =197 liter.
(Note: If Super plasticizer is used, the water content can be reduced upto 20% and above.)
Based on trials with Super plasticizer water content reduction of 25% has been achieved,
Hence the arrived water content = 197-[197 x (25/100)] = 147 litres.
step 4 Calculation of cement content
Water-cement ratio =0.45
Cement content = 147/0.38 = 387 kg/m3
From Table 5 of IS 456, minimum cement content for ‘Very Severe‘ exposure condition = 340 kg/m3
387 kg/m3 > 340 kg/m3, hence, O.K.
Increase by 10 %
387 x 0.10 + 387 = 426 kg/m^3
W= 147 kg
Water cement ratio = 147/426 = 0.345
Fly ash at 30 % total cementitious material content= 426 x 30/100 = 127.8 kg/m^3
Cement (ordinary Portland cement ) = 426-127.8 = 298 kg/m^3
Saving of cement while using fly ash = 387 – 298 = 89 kg/ m^3
Fly ash = 127.8 kg/m^3
Step5 VOLUME OF COARSE AGGREGATES AND FINE AGGREGATES
Fine aggregates are sieved according to zone 1 of table 4 IS 383
Coarse aggregates are sieved according to table 2 IS 383
According ,
Table 3 IS 10262: 2009
Nominal size of aggregates = 20 mm
Fine aggregates ( zone 1)
Therefore ,
Volume of coarse aggregate = 0.60
Water cement ratio = 0.38
( +/- 0.01 for every +/- 0.005)
= (.5-water cement ratio/.05)*.01 + W
= (.5-.38/.05)+.01 + .6
So, volume of coarse aggregate = 0.624
Volume of fine aggregate = 1-0.624 = 0.376
Step 6 MIX CALCULATION :
- Volume of concrete = 1m^3
- Volume of cement = (298 / 3.15) x1/1000= 0.0946 m^3
- Volume of fly ash = ( 127.8 /2.2) x1/1000=0.058 m^3
- Volume of water = ( 147/1)x 1/1000=0.147 m^3
- Volume of admixture = ( 7.6/ 1.145)x1/1000=0.006 m^3
- Volume of coarse aggregate and fine aggregate = a – ( b + c + d+
e) = 1-( 0.0946 + 0.058 +0.147+ 0.006 )= .6944 = 0.7 m^3 - Mass of coarse aggregate = 0.7 x 0.624 x 2.74 x1000 = 1196.83 kg/m^3
- Mass of fine aggregate = 0.7 x 0.376 x 2.74 x1000 = 721 kg/m^3
RESULTS AND CONCLUSION :
TRIAL MIX:
Cement = 298 kg/m^3
Fly ash = 127.8 kg/m^3
Water = 147 kg
Coarse aggregate = 1196.83 kg/m^3
Fine aggregate = 721 kg/m^3
Chemical admixture = 7.6 kg/m^3
Water cement ratio = 0.345
Trial mix ratio : 298 :721 : 1196 = 1: 2.419 : 4.013 ( C: FA: CA)
- This trial mix ratio should match the workability ,durability, minimum strength of M35 grade concrete .
- If it does not satisfy the above criteria two or more trials for increase of 10 % should be done .
To calculate M50 grade without fly ash
Step 1 calculation of target mean strength
- Target mean strength : f’ck= fck +1.65 s
f’ck = target mean strength
fck= compressive strength at 28 days of concrete
s = standard deviation ( IS 10262:2009 , table -1)
- Volume of cement : (mass of cement /specific gravity of cement )x1/1000
- Volume of water : ( mass of water/specific gravity of water ) x1 /1000
- Volume of chemical admixtures : ( mass of chemical admixture/ specific gravity of chemical admixture )x 1/1000
- Volume of concrete: 1 m^3
- Volume of aggregates : ( volume of concrete -( volume of cement + volume of water+volume of chemical admixture) )
- Mass of coarse aggregates: volume of aggregate x volume of coarse aggregate x specific gravity of coarse aggregate x1000
- Mass of fine aggregate : volume of aggregate x volume of fine aggregate x specific gravity of fine aggregate x1000
CODES FOR CONCRETE MIX DESIGN:
- IS 10262: 2009
- IS 456: 2000
- IS 8112
- IS 9103
- IS 3812 ( part 1)
OBJECTIVES:
- The main objective is to produce concrete as economically as possible
- To achieve desired workability, durability and minimum strength of M 50 grade concrete
PROCEDURE AND CALCULATION:
- TARGET MEAN STRENGTH
F’ck = fck + 1.65 5
fck= 50 N/mm^2 (M50 )
Table 1- IS 10262:2009
s = 5 N/mm^2
f ‘ck = 50 + ( 1.65 x 5)
= 58.25 N/mm^2
Step 2 Calculation of water cement ratio
Given :
M50
Assume:
Exposure condition : Extreme ( for reinforced concrete)
Table 5 – IS 456 : 2000
Therefore,
Extreme= w/c ratio= 0.40
M50 = w/c ratio= 0.40
Lower the water cement ratio better the workability
0.40
Water cement ratio = 0.40
Minimum cement content = 360 kg/m^3
Step 3 Selection of water content
Assume :
Nominal size of aggregates= 20 mm
Table 2 -IS 10262: 2009
Minimum water content = 186 litres or kg
Slump ranges from 25 to 50 mm
As superplasticizers (20 % reduced)
= 186 x 80/100 = 148.8 (100-20 = 80)
Water content = 149 litres
step 4 Calculation of cement content
Water cement ratio = w/c= 0.40
Water content = W = 149 kg
Cement content = c=149/0.40 = 372.5 = 373 kg/m^3
Minimum cement content = 360 kg/m^3
360< 373 kg/m^3 ( hence okay )
Step 5 VOLUME OF COARSE AGGREGATES AND FINE AGGREGATES
Fine aggregates are sieved according to zone 1 of table 4 IS 383
Coarse aggregates are sieved according to table 2 IS 383
According ,
Table 3 IS 10262: 2009
Nominal size of aggregates = 20 mm
Fine aggregates ( zone 1)
Therefore ,
Volume of coarse aggregate = 0.60
Water cement ratio = 0.4
(+/- 0.01 for every +/- 0.005)
So, volume of coarse aggregate = 0.62
Volume of fine aggregate = 1-0.62 = 0.38
Step 6 MIX CALCULATION :
- Volume of concrete = 1m^3
- Volume of cement = (373/3.15)x1/1000 = 0.118 m^3
- Volume of water = ( 149/1)x1/1000= 0.149 m^3
- Volume of admixture = ( 7.6/1.145)x1/1000 = 0.006
- Volume of coarse aggregate and fine aggregate = a – ( b + c + d) = 1-( 0.118 + 0.149 + 0.006 )= 0.727 m^3
- Mass of coarse aggregate = 0.727 x 0.62 x 2.74 x1000 = 1235.02 kg/m^3
- Mass of fine aggregate = 0.727 x 0.38 x 2.74 x1000 = 756.9 kg/m^3
RESULTS AND CONCLUSION :
TRIAL MIX:
Cement = 373 kg/m^3
Water = 149 kg
Coarse aggregate = 1235.02 kg/m^3
Fine aggregate = 756.9 kg/m^3
Chemical admixture = 7.6 kg/m^3
Water cement ratio = 0.40
Trial mix ratio : 373 :756.9 : 1235.02 = 1: 2.02: 3.31 ( C: FA: CA)
- This trial mix ratio should match the workability ,durability, minimum strength of M50 grade concrete .
- If it does not satisfy the above criteria two or more trials for increase of 10 % should be done .