Project on Concrete Mix Design for various grades of Concrete
AIM: To calculate the Concrete Mix Design for M35 grade Concrete with Fly ash.
INTRODUCTION: Mix design is a process of proportioning the materials required for preparation of concrete to attain the desired strength and durability of the concrete.
Test Data of materials:
The following data are collected from standard lab tests:
Specific gravity of cement = 3.15
Concrete = M35 = fck = 35N/mm^2
Specific gravity of Coarse Aggregate = 2.74
Specific gravity of Fine Aggregate = 2.56
Slump value = 50mm
Minimum cement content = 360 kg/m^3
PROCEDURE:
STEPS AND CALCULATION:
1. TARGET MEAN STRENGTH:
ft = fck +1.65 S
ft = target mean strength
fck = compressive strength at 28 days of concrete
S = standard deviation ( IS 10262:2009 , table -1)
ft = fck + 1.65 5
fck= 35 N/mm^2
Table 1- IS 10262:2009
S = 5 N/mm^2
ft = 35 + ( 1.65 x 5)
Targeted mean strength = 43.25 N/mm^2
2. CALCULATION OF WATER CEMENT RATIO:
Exposure condition : Extreme ( for reinforced concrete)
Table 5 – IS 456 : 2000
Taking water-cement ratio = 0.40
3. CALCULATION OF WATER CONTENT:
Nominal size of aggregates= 20 mm
Table 2 -IS 10262: 2009
Minimum water content = 186 litres (Slump ranges from 25 to 50 mm)
As superplasticizer used water content reduced by 20%
Based on trails with super plasticizer water reduced 20%. Hence the arrived water content = 186-(186*(20/100)) = 149 litre.
4. CALCULATION OF CEMENT CONTENT With FLY ASH:
Water cement ratio = w/c= 0.40
Water content = W = 149 litre
Cement content = c =149/0.40 = 372.5 = 370 kg/m^3
Minimum cement content = 360 kg/m^3
From Table 5 of IS 456 minimum cement content is taken
360< 370 kg/m^3 ( Hence satisfied )
The increase by 10 % is done by trial and error method.
Cement material content = 370 x 0.10 + 370 = 407 kg/m^3
Water content = 149 kg/m^3
Now, the 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 which is used in the mix = 122 kg/m^3
5. VOLUME OF COARSE AGGREGATES AND FINE AGGREGATES:
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
(at the rate of +/- 0.01 for every +/- 0.05 change in the water-cement ratio)
So, corrected volume of coarse aggregate for the water-cement ratio 0.40 = 0.62
Therefore volume of coarse aggregate = 0.62
Volume of fine aggregate = 1-0.62 = 0.38
6. MIX CALCULATIONS:
1. Volume of concrete = 1m^3
2. Volume of cement = (mass of cement/specfic gravity of cement) x (1/1000)
=(285/3.15) x (1/1000)
= 0.090 m^3
3. Volume of fly ash = (mass of fly ash/specific gravity of fly ash) x (1/1000)
= (122/2.2) x (1/1000)
=0.055 m^3
4. Volume of water = (mass of water/specific gravity of water) x (1/1000)
= (149/1) x (1/1000)
=0.149 m^3
5. Volume of admixture = (mass of chemical admixture/specific gravity of chemical admixture)x (1/1000)
=(7.6/1.145) x (1/1000)
=0.006 m^3
6. Volume of coarse aggregate and fine aggregate =(volume of concrete)-(volume of cement+volume of fly ash+volume of water+volume of admixture)
= a - ( b + c + d+ e)
= 1- (0.090 + 0.055 + 0.149 + 0.006)
= 0.7 m^3
7. Mass of coarse aggregate = volume of aggregate x volume of coarse aggregate x specific gravity of coarse aggregate x1000
= 0.7 x 0.62 x 2.74 x1000
= 1189.16 kg/m^3
8. Mass of fine aggregate = volume of aggregate x volume of fine aggregate x specific gravity of fine aggregate x1000
= 0.7 x 0.38 x 2.56 x1000
= 680.96 kg/m^3
RESULTS: Mix design of M35 grade concrete with fly-ash is designed.
Cement = 285 kg/m^3
Fly ash = 122 kg/m^3
Water = 149 kg/m^3
Coarse aggregate = 1189 kg/m^3
Fine aggregate = 680.96 kg/m^3
Chemical admixture = 7.6 kg/m^3
Water cement ratio = 0.36
Trail Mix Ratio = 1:1.67:2.92
AIM: To calculate the Concrete Mix Design for M50 grade Concrete without Fly ash.
INTRODUCTION: Mix design is a process of proportioning the materials required for preparation of concrete to attain the desired strength and durability of the concrete.
Test Data of materials:
The following data are collected from standard lab tests:
Specific gravity of cement = 3.12
Concrete = M50 = fck = 50N/mm^2
Specific gravity of Coarse Aggregate = 2.74
Specific gravity of Fine Aggregate = 2.56
Slump value = 50mm
Minimum cement content = 360 kg/m^3
PROCEDURE:
STEPS AND CALCULATIONS :
1. TARGET MEAN STRENGTH:
ft = fck +1.65 S
ft = target mean strength
fck = compressive strength at 28 days of concrete
S = standard deviation ( IS 10262:2009 , table -1)
ft = fck + 1.65 S
fck= 50 N/mm^2
Table 1- IS 10262:2009
S = 5 N/mm^2
ft = 50 + ( 1.65 x 5)
Targeted mean strength = 58.25 N/mm^2
2. CALCULATION OF WATER CEMENT RATIO:
Exposure condition : Extreme ( for reinforced concrete)
Table 5 – IS 456 : 2000
Taking water-cement ratio = 0.35
3. CALCULATION OF WATER CONTENT:
Nominal size of aggregates= 20 mm
Table 2 -IS 10262: 2009
Minimum water content = 186 litres (Slump ranges from 25 to 50 mm)
As superplasticizer used water content reduced by 20%
Based on trails with super plasticizer water reduced 20%. Hence the arrived water content = 186-(186*(20/100)) = 149 litre.
4. CALCULATION OF CEMENT CONTENT:
Water-cement ratio = 0.35
Cement content = 149/0.35 = 426 kg/m^3
From Table 5 of IS 456:2000,
minimum cement content for ‘Extreme exposure’ condition = 360 kg/m^3
426 kg/m^3 > 360 kg/m^3 Hence satisfied.
5. VOLUME OF COARSE AGGREGATES AND FINE AGGREGATES:
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 = 0.35
(at the rate +/- 0.01 for every +/- 0.05 change in the water-cement ratio )
So, corrected volume of coarse aggregate for water-cement ratio = 0.35 = 0.65
Therefore volume of coarse aggregate = 0.65
Volume of fine aggregate = 1-0.65 = 0.35
6. MIX CALCULATIONS:
1. Volume of concrete = 1m^3
2. Volume of cement = (mass of cement/specfic gravity of cement) x (1/1000)
=(426/3.12) x (1/1000)
= 0.136 m^3
3. Volume of water = (mass of water/specific gravity of water) x (1/1000)
= (149/1) x (1/1000)
=0.149 m^3
4. Volume of admixture = (mass of chemical admixture/specific gravity of chemical admixture) x (1/1000)
=(8.2/1.145) x (1/1000)
=0.007 m^3
5. Volume of coarse aggregate and fine aggregate = (volume of concrete)-(volume of cement + volume of water + volume of admixture)
= a – ( b + c + d)
= 1-(0.136 + 0.149 + 0.007)
= 0.708 m^3
6. Mass of coarse aggregate = volume of aggregate x volume of coarse aggregate x specific gravity of coarse aggregate x1000
= 0.708 x 0.60 x 2.74 x1000
= 1163.95 kg/m^3
7. Mass of fine aggregate = volume of aggregate x volume of fine aggregate x specific gravity of fine aggregate x1000
= 0.708 x 0.35 x 2.56 x1000
= 634.36 kg/m^3
RESULTS: Mix design of M50 grade of concrete without fly-ash is designed.
Cement = 426 kg/m^3
Water = 149 kg
Coarse aggregate = 1163.95 kg/m^3
Fine aggregate = 634.36 kg/m^3
Chemical admixture = 8.2 kg/m^3
Water cement ratio = 0.35
Trial Mix Ratio = 1:1.48:2.73