Project on Concrete Mix Design for various grades of Concrete
AIM:
To calculate the Concrete Mix Design for M35 grade Concrete with fly ash
EQUATIONS:
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 /specfic 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 fly ash : (mass of fly ash/specific gravity of fly ash )x 1/1000
Volume of concrete: 1 m^3
Volume of aggregates : ( volume of concrete -( volume of cement +volume of fly ash + volume of water + volume of 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 M35 grade concrete
PROCEDURE AND CALCULATION:
TARGET MEAN STRENGTH
F’ck = fck + 1.65 5
fck= 35 N/mm^2 (M35 )
Table 1- IS 10262:2009
s = 5 N/mm^2
f ‘ck = 35 + ( 1.65 x 5)
= 43.25 N/mm^2
- SELECTION OF WATER CEMENT RATIO :
Given :
M35
Assume:
Exposure condition : Extreme ( for reinforced concrete)
Table 5 – IS 456 : 2000
Therefore,
Extreme= w/c ratio= 0.40
M35 = w/c ratio= 0.45
Lower the water cement ratio better the workability
0.40 < 0.45
Water cement ratio = 0.40
Minimum cement content = 360 kg/m^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
- CALCULATION OF CEMETITIOUS CONTENT ( CEMENT + FLY ASH)
Water cement ratio = w/c= 0.40
Water content = W = 149 kg
Cementitious content = c=149/0.40 = 372.5 = 370 kg/m^3
Minimum cement content = 360 kg/m^3
360< 370 kg/m^3 ( hence okay )
Increase by 10 %
370 x 0.10 + 370 = 407 kg/m^3
W= 149 kg
Water cement ratio = 149/407 = 0.366
Fly ash at 30 % total cementitious material content
= 407 x 30/100 = 122 kg/m^3
Cement (ordinary Portland cement ) = 407-122 = 285 kg/m^3
Saving of cement while using fly ash = 370 – 285 = 85 kg/ m^3
Fly ash = 122 kg/m^3
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
6.MIX CALCULATION :
Volume of concrete = 1m^3
Volume of cement = (285 / 3.15) x1/1000= 0.090 m^3
Volume of fly ash = ( 122/2.2) x1/1000=0.055 m^3
Volume of water = ( 149/1)x 1/1000=0.149 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-( 0.090 + 0.055 +0.149+ 0.006 )= 0.7 m^3
Mass of coarse aggregate = 0.7 x 0.62 x 2.74 x1000 = 1189.16 kg/m^3
Mass of fine aggregate = 0.7 x 0.38 x 2.74 x1000 = 728 kg/m^3
RESULTS AND CONCLUSION :
TRIAL MIX:
Cement = 285 kg/m^3
Fly ash = 122 kg/m^3
Water = 149 kg
Coarse aggregate = 1189 kg/m^3
Fine aggregate = 728 kg/m^3
Chemical admixture = 7.6 kg/m^3
Water cement ratio = 0.366
Trial mix ratio : 285 :728 : 1189 = 1: 2.55 : 4.17 ( 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 .
AIM:
To calculate M50 grade without fly ash
EQUATIONS:
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
- SELECTION 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
- 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
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 )
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
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 .