Project on Concrete Mix Design for various grades of Concrete
MIX DESIGN FOR M35 GRADE CONCRETE WITH FLY ASH
AIM:
To calculate the Concrete Mix Design for M35 grade Concrete with fly ash
EQUATIONS:
• Target mean strength : fck= fck +1.65 s
f'ck = target mean strength
fck= compressive strength at 28 days of concrete
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 m13
• 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:
1. TARGET MEAN STRENGTH
F'ck=fck + 1.65 5
fck= 35 N/mm^2 (M35)
Table 1- IS 10262:2009
Table 1 Assumed Standard Deviation
S= 5 N/mm^2
f'Ck = 35 + (1.65 x 5)
43.25 N/mm^2
2. SELECTION OF WATER CEMENT RATIO:
Given :
M35
Assume:
Exposure condition : Extreme (for reinforced concrete)
Therefore,
Table 5 - IS 456 : 2000
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
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 x80/100 = 148.8 (100-20 = 80)
water content = 149 litres
4. 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 × 0.10 + 370 = 407 kg/m^3
W= 149 kg
Water cement ratio = 149/407 = 0.366
Flv ash at 30 % total cementitious material content
= 407 x 30/100 = 122 kg/m13
Cement (ordinary Portland cement ) = 407-122 = 285 kg/m^3
India (+
Saving of cement while using fly ash = 370 - 285 = 85 kg/ m13
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:
1. Volume of concrete = 1m^3
2. Volume of cement = (285 / 3.15) x1/1000= 0.090 m13
3. Volume of fly ash = (122/2.2) x1/1000=0.055 m13
4. Volume of water = (149/1)x1/1000=0.149 m^3
5. Volume of admixture = (7.6/ 1.145 x1/1000=0.006 m13
6. Volume of coarse aggregate and fine aggregate = a - (b + c + d+
e) = 1- (0.090 + 0.055 +0.149+ 0.006 ) = 0.7 mA3
7. Mass of coarse aggregate = 0.7 x 0.62 x 2.74 x1000 = 1189.16 kg/m13
8. Mass of fine aggregate = 0.7 x 0.38 x 2.74 x1000 = 728 kg/m13
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
MIX DESIGN FOR M50 GRADE WITHOUT FLY ASH
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 M50 grade concrete
PROCEDURE AND CALCULATION:
1. TARGET MEAN STRENGTH
F'ck=fck + 1.65 5
fck= 50 N/mm^2 (M50)
Table 1- IS 10262:2009
S= 5 N/mm12
f'ck = 50 + (1.65 x5)
= 58.25 N/mm12
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
Table 5 Minimum Cement Content, Maximum Water-Cement Ratio and Minimum Grade of Concrete
for Different Exposures with Normal Weight Aggregates of 20 mm Nominal Maximum Size
3. SELECTION OF WATER CONTENT
Assume:
Nominal size of aggregates= 20 mm
Table 2 -IS 10262: 2009
Minimum water content = 186 lites 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:
1. Volume of concrete =1m^3
2. Volume of cement = (373/3.15 x1/1000 = 0.118 m13
3. Volume of water = (149/1 x1/1000= 0.149 m^3
4. Volume of admixture = (7.6/1.145 x1/1000 = 0.006
5. Volume of coarse aggregate and fine aggregate = a - (b + c + d) = 1- ( 0.118 + 0.149 + 0.006 ) = 0.727 m^3
6. Mass of coarse aggregate = 0.727 x 0.62 x2.74 x1000 = 1235.02 kg/m^3
7. 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/m13
Water = 149 kg
Coarse aggregate = 1235.02 kg/m^3
Fine aggregate = 756.9 kg/m^3
Chemical admixture = 7.6 kg/m13
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