Project on Concrete Mix Design for various grades of Concrete
AIM:Calculate the Concrete Mix Design for M35 grade concrete with fly ash
Ans Garde of Concrete = M35
PROCEDURE AND CALCULATION:
- TARGET MEAN STRENGTH:
F’ck = fck + 1.65 5
fck= 35 N/mm^2
s = 5 N/mm^2
f ‘ck = 35 + ( 1.65 x 5)
= 43.25 N/mm^2
2. SELECTION OF WATER CEMENT RATIO : 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 ( Table : 5 ( Is 456:2000))
3. SELECTION OF WATER CONTENT:
From Table 2 of Is 10262 : 2009 , Maximum water content for 20mm aggregate =186 lts ( For Slump 25 to 50mm Slump Range )
Water content (For Slump 25 to 50 mm)= 186 lts
As Superplasticizer is used , the water content can be reduced up to 20 %
Hence the arrived water content = 186 * 0.80 =148.8 = 149 lts
4. CALULATION OF CEMENT CONTENT :
Water cement ratio = w/c= 0.40
Cementitious material ( Cement + fly ash ) Content =149/0.40 = 372.5 = 370 kg/m^3
Hence Table 5 of Is 456 minium cement content for Extreme =360 kg/m^3
Minimum cement content = 360 kg/m^3
370 kg/m^3 > 360 Kg/ m^3 ( hence okay )
Now , to proportion a mix containing fly ash the following steps are suggested
a) The percentage of fly ash needs to be decided based on project requirement and quality of materials
b) In Some Cases increase in cementitious materials can be warranted . Increase in cementious material content and it percentage may be based on experience and tiral(Increase 10% cementitious material content )
cementitious material = 370 *0.1+370 =407 kg/m^3
water content = 149 kg/m^3
so, water cement ratio =149/407 = 0.36
Fly ash at 30 % total cementitious material content = 407 x 30/100 = 122 kg/m^3
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
From Table 3 of Is 10262 : 2009 , Volume of coarse aggregate corresponding to 20mm size aggregate and fine aggregate ( Zone - l ) for water - cement ratio of =0.4
There fore , the volume of coarse aggregate needs to be incresed to decrease the fine aggregate content ,
As w/c ratio is lower by 0.10 , the proportion of volume of coarse aggregate is increased by 0.02
( +/- 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 = ( mass of cement /specfic gravity of cement )x (1/1000) =(285 / 3.15) x1/1000= 0.090 m^3
- Volume of fly ash =( mass of Fly ash /specfic gravity of fly ash )x (1/1000) =( 122/2.2) x1/1000=0.055 m^3
- Volume of water = ( mass of water /specfic gravity of water )x (1/1000) =( 149/1)x 1/1000=0.149 m^3
- Volume of admixture = ( mass of chemical admixture /specfic gravity of admixture )x (1/1000) =( 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 :
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)
AIM:Calculate the Concrete Mix Design for M50 grade concrete without Fly ash
Garde of Concrete = M50
PROCEDURE AND CALCULATION:
- TARGET MEAN STRENGTH:
F’ck = fck + 1.65 5
fck= 50 N/mm^2
s = 5 N/mm^2
f ‘ck = 50 + ( 1.65 x 5)
= 58.25 N/mm^2
2. SELECTION OF WATER CEMENT RATIO : 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
Water cement ratio = 0.40
Minimum cement content = 360 kg/m^3 ( Table : 5 ( Is 456:2000))
3. SELECTION OF WATER CONTENT:
From Table 2 of Is 10262 : 2009 , Maximum water content for 20mm aggregate =186 lts ( For Slump 25 to 50mm Slump Range )
Water content (For Slump 25 to 50 mm)= 186 lts
As Superplasticizer is used , the water content can be reduced up to 20 %
Hence the arrived water content = 186 * 0.80 =148.8 = 149 lts
4. CALULATION 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
373 >360 kg/m^3 ( hence okay )
5.VOLUME OF COARSE AGGREGATES AND FINE AGGREGATES
From Table 3 of Is 10262 : 2009 , Volume of coarse aggregate corresponding to 20mm size aggregate and fine aggregate ( Zone - l ) for water - cement ratio of =0.4
There fore , the volume of coarse aggregate needs to be incresed to decrease the fine aggregate content ,
As w/c ratio is lower by 0.10 , the proportion of volume of coarse aggregate is increased by 0.02
( +/- 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 = ( mass of cement /specfic gravity of cement )x (1/1000) =(373/3.15)x1/1000 = 0.118 m^3
- Volume of water = ( mass of water /specfic gravity of water )x (1/1000) =( 149/1)x1/1000= 0.149 m^3
- Volume of admixture =( mass of chemical admixture /specfic gravity of admixture )x (1/1000) = ( 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 :
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)