Project on Concrete Mix Design for various grades of Concrete
1.Calculate the Concrete Mix Design for M35 grade concrete with fly ash & M50 grade concrete without Fly ash
Answer:
Data required for concrete mix design
- Grade of concrete
- Minimum size of aggregate
- Min cement content
- Min w/c ratio
- Workability in terms of slump
- Exposure condition
- Maximum temperature at the pouring point
- Early age strength
- Grading zone of fine aggregate
- Type of aggregate
- Max cement content
- Admixture used
- Specific gravity of all material used
- Moisture content (free)
Procedure for the concrete mix design
Step: 1
Target mean strength of concrete, ft = fck + 1.65 S
Where, fck = compressive strength of concrete at 28 days
S = standard deviation obtained from table
Step: 2 Selection of w/c ratio
From table-5 IS 456: 2000, it can be taken depending upon the exposure condition
Step:3 selection of water content
From table-2 IS 10262: 2009
The maximum water content for 25mm to 50mm slump = 186 kg = 186 lit
(for additional slump 3% may be added for every 25mm slump)
Step:4 Selection of cement content
It can be calculated from w/c ratio
Min cement content is taken from table-5 of IS 456: 2000
Step: 5 Volume of coarse aggregate and fine aggregate content
The proportion of aggregate determined from table-3 of IS 10262: 2009 (selected depending upon the fine aggregate zone)
Step:6 Mix calculation
The mix calculation per unit volume of concrete shall be as follows
- Volume of concrete =1m^3
- Volume of cement =
- Volume of water =
- Volume of chemical admixtures =
- Volume of total aggregate (FA& CA) = (a-(b + c + d))
- Mass of coarse aggregate = e × Vol of CA × Specific gravity of CA × 1000
- Mass of fine aggregate = e × Vol of FA × Specific gravity of FA × 1000
Step:7 Mix proportions
Cement = kg/m^3
Water = kg/m^3
Fine aggregate = kg/m^3
Coarse aggregate = kg/m^3
Chemical admixture = kg/m^3
w/c ratio =
Step:8 Mix Ratio
Cement: Fine aggregate: Coarse aggregate
Concrete mix design for M35 with fly ash
Step:1
ft = fck + 1.65 S
ft = 35 + (1.65*5)
= 35 + 8.25
= 43.25 N/mm^2
Step: 2 w/c ratio
From table-5 IS 456: 2000,
For severe exposure condition of reinforced concrete w/c = 0.45
So take w/c = 0.40
0.40 < 0.45 hence OK
Step:3 water content
From table-2 IS 10262: 2009
The maximum water content for 25mm to 50mm slump for 20 mm aggregate size = 186 kg = 186 lit
(for additional slump 3% may be added for every 25mm slump)
Water content = 186 + (6/100) *186
Water content = 197.16 kg
It can be reduced by 20% by admixture but in trials it has been found it reduced 29%. So the water content is 71% of the calculated water content
Water content required = 0.71*197.16 = 140kg
Step:4 cement content
w/c =0.4
cementitious material (cement + fly ash) = 140/0.4 = 350 kg/m^3
Min cement content from table-5 of IS 456: 2000 is 320 kg
350 kg > 320 kg, hence OK
Cementitious material content = 1.10*350 = 385kg/m^3
Water content = 140 kg/m^3
So, w/c ratio = 140/385 = 0.364
Fly ash @ 30% of cementitious material content = 0.30*385 = 115 kg/m^3
Cement (OPC) = 385-115 = 270 kg/m^3
Step: 5 Volume of coarse aggregate and fine aggregate content
The proportion of aggregate determined from table-3 of IS 10262: 2009 (selected depending upon the fine aggregate zone)
For zone I and nominal aggregate size 20mm
The proportion of coarse aggregate is 0.60 of total volume of aggregate
It is given for w/c =0.50
But in we have w/c = 0.40
So for every +/- 0.05 w/c ratio the proportion of coarse aggregate adjusted with -/+ 0.01
Therefore, Corrected volume of coarse aggregate for w/c 0.40 = 0.62
For pumpable concrete these value reduced by 10%
Volume of coarse aggregate = 0.62*0.9 = 0.56
Volume of fine aggregate = 1- 0.56 = 0.44
Step:6 Mix calculation
The mix calculation per unit volume of concrete shall be as follows
- Volume of concrete =1m^3
- Volume of cement =
= (270/3.15) *(1/1000)
= 0.086 m^3
- Volume of fly ash = (115/2.2) *(1/1000)
= 0.052 m^3
- Volume of water =
= (140/1) *(1/1000) = 0.140 m^3
- Volume of chemical admixtures =
= (7/1.145) *(1/1000) = 0.006 m^3
- Volume of total aggregate (FA& CA) = (a- (b + c + d + e))
= (1-(0.086+0.052+0.140+0.006)) = 0.716 m^3
- Mass of coarse aggregate = f × Vol of CA × Specific gravity of CA × 1000
= 0.716 *0.56*2.74*1000
= 1098 kg/m^3
- Mass of fine aggregate = f× Vol of FA × Specific gravity of FA × 1000
= 0.716*0.44*2.74*1000
= 863 kg/m^3
Step:7 Mix proportions
Cement = 270 kg/m^3
Fly ash = 115 kg/m^3
Water = 140 kg/m^3
Fine aggregate = 863 kg/m^3
Coarse aggregate = 1098 kg/m^3
Chemical admixture = 7 kg/m^3
w/c ratio = 0.364
Step:8 Mix Ratio
Cement: Fine aggregate: Coarse aggregate = 1 : 2.24 : 2.85
Concrete mix design for M50 without fly ash
Step:1
ft = fck + 1.65 S
ft = 50 + (1.65*5)
= 50 + 8.25
= 58.25 N/mm^2
Step: 2 w/c ratio
From table-5 IS 456: 2000,
For severe exposure condition of reinforced concrete w/c = 0.45
So take w/c = 0.40
0.40 < 0.45 hence OK
Step:3 water content
From table-2 IS 10262: 2009
The maximum water content for 25mm to 50mm slump for 20 mm aggregate size = 186 kg = 186 lit
(for additional slump 3% may be added for every 25mm slump)
Water content = 186 + (6/100) *186
Water content = 197.16 kg
It can be reduced by 20% by admixture but in trials it has been found it reduced 29%. So the water content is 71% of the calculated water content
Water content required = 0.71*197.16 = 140kg
Step:4 cement content
w/c =0.4
c = 140/0.4 = 350 kg
Min cement content from table-5 of IS 456: 2000 is 320 kg
350 kg > 320 kg, hence OK
Step: 5 Volume of coarse aggregate and fine aggregate content
The proportion of aggregate determined from table-3 of IS 10262: 2009 (selected depending upon the fine aggregate zone)
For zone I and nominal aggregate size 20mm
The proportion of coarse aggregate is 0.60 of total volume of aggregate
It is given for w/c =0.50
But in we have w/c = 0.40
So for every +/- 0.05 w/c ratio the proportion of coarse aggregate adjusted with -/+ 0.01
Therefore, Corrected volume of coarse aggregate for w/c 0.40 = 0.62
For pumpable concrete these value reduced by 10%
Volume of coarse aggregate = 0.62*0.9 = 0.56
Volume of fine aggregate = 1- 0.56 = 0.44
Step:6 Mix calculation
The mix calculation per unit volume of concrete shall be as follows
- Volume of concrete =1m^3
- Volume of cement =
= (350/3.15) *(1/1000)
= 0.111m^3
- Volume of water =
= (140/1) *(1/1000) = 0.140 m^3
- Volume of chemical admixtures =
= (7/1.145) *(1/1000) = 0.006 m^3
- Volume of total aggregate (FA& CA) = (a- (b + c + d))
= (1-(0.111+0.140+0.006)) = 0.743 m^3
- Mass of coarse aggregate = e × Vol of CA × Specific gravity of CA × 1000
= 0.743 *0.56*2.74*1000
= 1140 kg/m^3
- Mass of fine aggregate = e × Vol of FA × Specific gravity of FA × 1000
= 0.743*0.44*2.74*1000
= 896 kg/m^3
Step:7 Mix proportions
Cement = 350 kg/m^3
Water = 140 kg/m^3
Fine aggregate = 896 kg/m^3
Coarse aggregate = 1140 kg/m^3
Chemical admixture = 7 kg/m^3
w/c ratio = 0.4
Step:8 Mix Ratio
Cement: Fine aggregate: Coarse aggregate = 1 : 2.56 : 3.26