Project on Concrete Mix Design for various grades of Concrete
AIM : To Calculate the Concrete Mix Design for M35 grade concrete with fly ash.
Procedure :
Mix proportion of concrete (M35) using fly ash as a part of replacement of OPC
Step 1: Stipulation for proportioning
Grade- M35
Type of cement- OPC 43
Type of mineral admixture- Fly ash
Maximum nominal size of aggregate- 20mm (crushed angular aggregates )
Exposure condition- moderate
Minimum cement content- 300kg/m^3 (From table 5 IS 456: 2000)
Maximum water cement ratio- 0.5 (From table 5 IS 456: 2000)
Workability- 100mm (pumpable concrete)
Maximum cement content- 450kg/m^3
Chemical admixture- Superplasticizers (IS 9103)
Step 2: Test data of material
Type of cement- OPC 43 grade conforming to IS 8112
Specific gravity of cement- 3.15
Fly ash- Conforming to IS 3812 (part 1)
Specific gravity of fly ash- 2.2
Specific gravity of coarse aggregate- 2.74
Specific gravity of fine aggregate- 2.74
Water absorption coarse aggregate- 0.5%
Water absorption fine aggregate- 1%
Moisture content- 0
Step 3 : Target strength of M35
fck' =fck +1.65*S
fck' =35 +1.65*5 = 43.25 N/mm^2
Step 4 : Selection of W/C ratio
Maximum water cement ratio- 0.5 (From table 5 IS 456: 2000)
As per experience water cement ratio- 0.45
Hence 0.45<0.5 , so 0.45 is Ok
Step 5 : Calculation of water content
water content = 186 kg from table 2 of IS 10262: 2009 for 20mm aggregate (For slump of 50mm)
For slump of 100mm = 1.06*186= 197 lit. (for every 25mm slump value water content increased by 3% )
Superplasticers are used so 30% reduction in water content = 197*0.7 =137.9 litres
Step 6 : Calculation of cement content and fly ash
W/C = 0.45
137.9/C = 0.45
Cementitious material(cement +fly ash) C = 306.44 kg/m^3
Minimum cement content- 300kg/m^3 (From table 5 IS 456: 2000) for moderate exposure condtion.
Based experiances and trial 10% increase in cementitious material content
cementitious material content = 1.10*306.44 = 337.084 Kg/m^3
Water content = 137.9 kg/m^3
W/C ratio = 137.9/337.084 = 0.4
Fly ash occupy 34% of cementitious material content = 337.084*0.34 =114.86 kg/m^3
Cement content = 337.084-114.86 = 222.22 kg/m^3
Fly ash content =114.86 kg/m^3
Step 7 : Proportion volumn of coarse aggregate and fine aggregate content
From table 3 of IS 10262 :2009 volumn of coarse aggregate corressponding to 20mm size aggregate & fine aagregate of zone 1 for W/C ratio 0.5 = 0.6
W/C ratio in this mix is 0.45 so volumn of coarse aggregate corressponding to 20mm size aggregate & fine aagregate of zone 1 is = 0.61 (for every +/-0.05 change in W/C ratio the aggregate ratio changed by -/+0.01)
For pumpable concrete 10% reduction = 0.61*0.9 = 0.549
Therefore volume coarse aggregate = 0.549
volume fine aggregate = 1- 0.549 =0.451
Step 8 : Mix calculation
The mix calculation per unit volume of concrete as follows :
- Volumn of concrete : 1m^3
- Volume of cement : Mass of cement/(specific gravity of cement*1000)
- Volumn of Water : Mass of water/ (specific gravity of water*1000)
- Volumn of chemical admixture : Mass of chemical admixture/ (specific gravity of admixture*1000)
- Volume of Fly ash : Mass of FA/(specific gravity of FA*1000)
- Volume of fine and coarse aggregate : [a-(b+c+d+e)]
- Mass of coarse aggregate : f *volume of CA *specific gravity of CA *1000
- Mass of fine aggregate : f *volume of FA *specific gravity of FA *1000
b. Volume of cement= 222.22/(3.15*1000) =0.0705 m^3
c. Volumn of Water = 137.9/(1*1000) = 0.137 m^3
d. Volumn of chemical admixture = 6.12/(1.145*1000) = 0.00534 m^3
e. Volume of Fly ash = 114.86/(2.2*1000) = 0.0522 m^3
f. Volume of fine and coarse aggregate = 0.734 m^3
g. Mass of coarse aggregate : 0.734*0.549*2.74*1000 =1104.12 kg/m^3
h. Mass of fine aggregate : 0.734*0.451*2.74*1000 =907.03 kg/m^3
Result : Trial mix number 1
Cement= 222.22 kg/m^3
Fly ash = 114.86 kg/m^3
Water =137.9 kg/m^3
chemical admixture = 6.12 kg/m^3
fine aggregate = 907.03 kg/m^3
coarse aggregate = 1104.12 kg/m^3
W/C ratio = 0.45
Trial Mix Ratio = 1 : 4.08 : 4.96
AIM : To Calculate the Concrete Mix Design for M50 grade concrete without Fly ash.
Procedure :
Mix proportion of concrete (M50)
Step 1: Stipulation for proportioning
Grade- M50
Type of cement- OPC 43
Maximum nominal size of aggregate- 20mm (crushed angular aggregates )
Exposure condition- moderate
Minimum cement content- 300kg/m^3 (From table 5 IS 456: 2000)
Maximum water cement ratio- 0.5 (From table 5 IS 456: 2000)
Workability- 100mm (pumpable concrete)
Maximum cement content- 450kg/m^3
Chemical admixture- Superplasticizers
Step 2: Test data of material
Type of cement- OPC 43 grade conforming to IS 8112
Specific gravity of cement- 3.15
Specific gravity of coarse aggregate- 2.74
Specific gravity of fine aggregate- 2.74
Water absorption coarse aggregate- 0.5%
Water absorption fine aggregate- 1%
Moisture content- 0
Step 3 : Target strength of M50
fck' =fck +1.65*S
fck' =50 +1.65*5 = 58.25 N/mm^2
Step 4 : Selection of W/C ratio
Maximum water cement ratio- 0.5 (From table 5 IS 456: 2000)
As per experience water cement ratio- 0.45
Hence 0.45<0.5 , so 0.45 is Ok
Step 5 : Calculation of water content
water content = 186 kg from table 2 of IS 10262: 2009 for 20mm aggregate (For slump of 50mm)
For slump of 100mm = 1.06*186= 197 lit. (for every 25mm slump value water content increased by 3% )
Superplasticers are used so 30% reduction in water content = 197*0.7 =137.9 litres
Step 6 : Calculation of cement content
W/C = 0.45
137.9/C = 0.45
C = 306.44 kg/m^3
Minimum cement content- 300kg/m^3 (From table 5 IS 456: 2000) for moderate exposure condtion.
Step 7 : Proportion volumn of coarse aggregate and fine aggregate content
From table 3 of IS 10262 :2009 volumn of coarse aggregate corressponding to 20mm size aggregate & fine aagregate of zone 1 for W/C ratio 0.5 = 0.6
W/C ratio in this mix is 0.45 so volumn of coarse aggregate corressponding to 20mm size aggregate & fine aagregate of zone 1 is = 0.61 (for every +/-0.05 change in W/C ratio the aggregate ratio changed by -/+0.01)
For pumpable concrete 10% reduction = 0.61*0.9 = 0.549
Therefore volume coarse aggregate = 0.549
volume fine aggregate = 1- 0.549 =0.451
Step 8 : Mix calculation
The mix calculation per unit volume of concrete as follows :
- Volumn of concrete : 1m^3
- Volume of cement : Mass of cement/(specific gravity of cement*1000)
- Volumn of Water : Mass of water/ (specific gravity of water*1000)
- Volumn of chemical admixture : Mass of chemical admixture/ (specific gravity of admixture*1000)
- Volume of fine and coarse aggregate : [a-(b+c+d)]
- Mass of coarse aggregate : f *volume of CA *specific gravity of CA *1000
- Mass of fine aggregate : f *volume of FA *specific gravity of FA *1000
b. Volume of cement= 306.44/(3.15*1000) =0.0972 m^3
c. Volumn of Water = 137.9/(1*1000) = 0.137 m^3
d. Volumn of chemical admixture = 6.12/(1.145*1000) = 0.00534 m^3
f. Volume of fine and coarse aggregate = 0.760 m^3
g. Mass of coarse aggregate : 0.760*0.549*2.74*1000 =1143.23 kg/m^3
h. Mass of fine aggregate : 0.760*0.451*2.74*1000 =995.82 kg/m^3
Result : Trial mix number 1
Cement= 306.44 kg/m^3
Water =137.9 kg/m^3
chemical admixture = 6.12 kg/m^3
fine aggregate = 995.82 kg/m^3
coarse aggregate = 1143.23 kg/m^3
W/C ratio = 0.45
Trial Mix Ratio = 1 : 3.24 : 3.73