Project on Concrete Mix Design for various grades of Concrete
AIM: Calculate the concrete mix design for M50 grade concrete without Fly ash.
Test Data for Materials
The following data are collected by performing standard tests in a laboratory.
Specific gravity of cement : 3.12
Specific gravity of Coarse Aggregate : 2.74
Specific gravity of Fine Aggregate : 2.56
Minimum cement content : 360 kg/m3
STEPS AND CALCULATIONS :
1. 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)
F’ck = fck + 1.65 5
fck= 50 N/mm^2
Table 1- IS 10262:2009
s = 5 N/mm^2
f ‘ck = 50 + ( 1.65 x 5)
= 58.25 N/mm^2
2. CALCULATION OF WATER CEMENT RATIO :
Exposure condition : Extreme ( for reinforced concrete)
Table 5 – IS 456 : 2000
water-cement ratio = 0.35
3. CALCULATION OF WATER CONTENT
From Table 2 -IS 10262: 2009,
maximum water content =186 litres (for 25 to 50 mm slump range) for 20 mm aggregate.
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 = 0.35
Cement content = 149/0.35 = 426 kg/m3
From Table 5 of IS 456:2000,
minimum cement content for ‘Extreme exposure’ condition = 360 kg/m3
426 kg/m3 > 360 kg/m3, hence, O.K
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
( +/- 0.01 for every +/- 0.05 in the water-cement ratio )
So, 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 )x1 /1000
=(426 / 3.12) x1/1000
= 0.136 m^3
3. Volume of water = ( mass of water /specific gravity of water ) x1 /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)x1/1000
=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 :
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
AIM: To calculate the Concrete Mix Design for M35 grade Concrete with fly ash
Test Data for Materials
The following data are collected by performing standard tests in a laboratory.
Specific gravity of cement : 3.15
Specific gravity of Coarse Aggregate : 2.74
Specific gravity of Fine Aggregate : 2.56
Minimum cement content : 360 kg/m3
STEPS AND CALCULATION:
1. 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)
F’ck = fck + 1.65 5
fck= 35 N/mm^2
Table 1- IS 10262:2009
s = 5 N/mm^2
f ‘ck = 35 + ( 1.65 x 5)
= 43.25 N/mm^2
2. CALCULATION OF WATER CEMENT RATIO :
Exposure condition : Extreme ( for reinforced concrete)
Table 5 – IS 456 : 2000
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 superplasticizers (20 % reduced)
= 186 x 80/100 = 148.8 (100-20 = 80)
Water content = 149 litres
4. CALCULATION OF CEMENT CONTENT With FLY ASH
Water cement ratio = w/c= 0.40
Water content = W = 149 kg
Cement 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
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 +/- 0.05 in the water-cement ratio)
So, 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 )x1 /1000
=(285 / 3.15) x1/1000
= 0.090 m^3
3. Volume of fly ash = (mass of fly ash/specific gravity of fly ash )x 1/1000
= (122/2.2) x1/1000
=0.055 m^3
4. Volume of water = ( mass of water /specific gravity of water ) x1 /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)x1/1000
=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 :
Cement = 285 kg/m^3
Fly ash = 122 kg/m^3
Water = 149 kg
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