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.
Where f'ck is the Target mean strength
fck is the compressive strength at 28 of concrete.
s is the Standard deviation ( IS 10262:2009, table-I)
- Volume of cement = (Mass of cement/ Specific gravity of cement) x 1/1000
- Volume of water = (Mass of water/ Specific gravity of water) x 1/1000
- Volume of chemical admixture = (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 M3
- Volume of aggregates = Volume of concrete - (Volume of cement + Volume of water + Volume of admixtures + Volume of Fly ash)
- Mass of coarse aggregate: Volume of aggregate x Volume of coarse aggregate x Specific gravity coarse aggregate x 1000
- Mass of fine aggregate: Volume of aggregate x Volume of fine aggregate x Specific gravity of fine aggregate x 1000
CODES FOR CONCRETE MIX DESIGN :
- IS 10262-2009
- IS 456-2000
- IS 8112
- IS 9103
- IS 3812 (Part - I)
To achieve desired workability, Durability, and minimum strength of M 35 grade of concrete.
As per Table -1 ( IS 10262-2009 )
fck is 35 N/mm2 for M=30 grade of concrete.
s is 5 N/mm2
(1) Target mean strength = f'ck = fck + 1.65 s
=> 35 + 1.65 x 5
=> 43.25 N/mm2
(2) Selection of water cement ratio = Table - 5 (IS : 456-2000) therefore,
M 35 = w/c = 4.0
Minimum cement content= 360kg/m3
(3) SELECTION OF WATER CONTENT:
The nominal size of the aggregate is 20mm
(Table-2 IS : 10262-2000).
Minimum water content = 186 lit
Slump ranges from 25-50mm. As superplasticizers (20% ) reduced.
=> 186 x 80/100 = 148.8
Water content = 149 lit
(4) CALCULATION OF CEMENT CONTENT
W/c = 0.4
Water content is 149 kg
Cement content C = 149/0.40
= 372.5 = 373 kg/M3
Increase by 10%
370 x 0.10 +370 = 407 kg/m3
W=149 kg
Water cement ratio = 149/407 = 0.366
Fly ash at 30% total cementitious material content
= 407 x 30/100 = 122 kg/m3
cement = 407-122=285 kg/m3
saving of cement while using fly ash =370-285=85 kg/m3
Fly ash =122 kg/m3
(5) VOLUME OF COARSE AGGREGATE AND FINE AGGREGATE
Fine aggregate is sieved according to zone 1 of Table IS-383
Coarse aggregate is sieved according to table 2 IS-383
According,
Table 3 IS: 10262-2009.
Nominal size of aggregate = 20mm
fine aggregate (zone 1)
Therefore,
Volume of coarse aggregate = 0.60
water cement ratio = 0.40
so, Volume of coarse aggregate = 0.62
Volume of fine aggregate = 1-0.62 = 0.38
6) MIX CACULATION:
- volume of concrete = 1m3
- Volume of cement = 285/3.15 x 1/1000 = 0.090m3
- Volume of fly ash = 122/2.2 x 1/1000 = 0.055m3
- Volume of water = 149/1 x 1/1000 = 0.149m3
- Volume of admixture = 7.6/1.145 x 1/1000 = 0.006m3
- Volume of coarse aggregate and fine aggregate = a- (b+c+d+e) = 1-(0.090+0.055+0.149+0.006)=0.7m3
- Mass of coarse aggregate = 0.7 x 0.62 x 2.74 x 1000 = 1189.16 kg/m3
- Mass of fine aggregate = 0.7 x 0.38 x 2.74 x 1000 = 728 kg/m3
AIM
To calculate the M 50 Grade without fly ash
EQUATIONS
1) Target mean strength F'ck = fck + 1.65 s
- fck is 50 N/mm2 for (M50) Table 1-IS 10262-2009.
- s is 5 N/mm2
=> 50+ (1.65 x 5)
=> 58.25 N/mm2
2) SELECTION OF WATER CEMENT RATIO
- water cement ratio = 0.40 for (M 50) Table 5-IS 456 : 2000
- Minimum cement content = 360 kg/m3
3) SELECTION OF WATER CONTENT
- Nominal size of aggregate = 20mm (Table 2-IS:10262-2009)
- Minimum water content = 186 lit
- Slump range from 25-50mm
- As super plasticizeer (20% reduced)= 186 x 80/100 = 148.8
water content : 149 lit
4) CALCULATION OF CEMENT CONTENT
- water cement ratio = 0.40
- water content = 149 kg
- cement content = 149/0.40 = 372.5
=> 373kg/m3
(5) VOLUME OF COARSE AGGREGATES AND FINE AGGREGATES:
Fine aggregate is sieved according to zone 1 of table 4 IS-383
Coarse aggregate is sieved according to table 2 IS-383
According, to Table 3 IS-10262: 2009
Nominal size aggregate = 20mm
Fine aggregates (zone - 1)
Therefore, the volume of coarse aggregate = 0.60
Water cement ratio = 0.4
so, the volume of coarse aggregate = 0.62
The volume of fine aggregate = 1-0.62 = 0.38
(6) MIX CALCULATION :
- Volume of concrete = 1m3
- Volume of cement = (373/3.15) x 1/1000 = 0.118 m3
- Volume of water = (149/1) x 1/1000 = 0.149 m3
- Volume of admixture = (7.6/1.145) x 1/1000 = 0.006
- Volume of coarse aggregate and fine aggregate = a-(b+c+d) = 1-(0.118+0.149+0.006) = 0.727 m3
- Mass of coarse aggregate = 0.727 x 0.62 x 2.74 x 1000 = 1235.02 kg/m3
- Mass of fine aggregate = 0.727 x 0.38 x 2.74 x 1000 = 756.9 kg/m3