Project on Concrete Mix Design for various grades of Concrete
Q.1. Calculate the Concrete Mix Design for M35 grade concrete with fly ash.
Ans. Step 1:- stipulations for proportioning
- Garde designation : M35
- Type of cement : opc 43 grade conforming to IS 8112
- Type of mineral admixture : Fly ash conforming to IS3812(part 1)
- Maximum nominal size of aggregate : 20 mm
- Minimum cement content :320 kg/m3 (from table 5 of Is 456:2000 exposure condition)
- Maximum water cement ratio : 0.45 (from table 5 of IS 456:2000 exposure condition)
- Work ability :100 mm (slump value for pumpable concrete)
- Exposure condition : severe (for reinforced concrete)
- .method of concrete placing : pumping
- Degree of supervision : Good
- Type of aggregate : crushed angular aggregate
- Maximum cement content : 450 kg/m3
- Chemical admixture type : super plasticizer
Step2:- Test data for material
- cement used : opc 43 grade Is 8112
- Specific gravity of cement : 3.15
- Fly : conforming to Is 3812 (part 1)
- Specific gravity of fly ash : 2.2
- Chemical admixture : super plasticizer conforming to Us 9103
- Specific gravity of coarse aggregate :2.74 & fine aggregate : 2.74
- Water absorption coarse aggregate : 0.5% & fine aggregate1%
- Free surface moisture : nill
Step3:- Target strength for mix proportioning
F'ck = fck+1.65s
S= standard deviation ( from Table 1 of Is 10262:2009)
(S = 5N/mm2)
F'ck = 35+1.65*5
F'ck= 43.25N/mm2
Step4:- Selection of water - cement ratio
From table 5 of Is456:2000 for severe exposure, maximum water cement ratio =0.45
Based on experience , adopt water cement ratio as 0.40
O.4<0.45 Hence ok
Step5:- Selection of water content
From table 2 of Is 10262:2009 ,maximum water content for 20 mm aggregate = 186 litre (for 25 mm to 50 mm slump range)
Estimated water content for 100 mm slump = 186+6/100*186= 197 litre
As super plasticizer is used , the water content can be reduced up to 30%.
Based on trials water content reduction of 29% is achieved with plasticizer .
Water content = 197*0.71= 140 litre
Step6:- Calculation of cement contant:-
Water-cement ratio = 0.40
Cementitious material (cement + fly ash) content=140/0.40= 350 kg
From table 5 Is 456 minimum cement content for severe exposure condition = 320 kg/m3
350>320 kg/m3 Hence ok
This example is with increase of 10% cementitious material content.
Cementitious material content = 350*1.10=385 kg/m3
Water content = 140 kg/m3
So, water cement rato = 140/385 = 0.364
Fly ash @ 30% total cementitious material content
= 385*30%= 115 kg/m3
Cement (opc) = 385-115
= 270 kg/m3
Saving of cement while using fly ash = 350-270 = 80 Kg/m3
Fly ash being utilized = 115 kg /m3
Step7:- proportion of volume of concrete aggregate and fine aggregate content
Table 3 Is 10262:2009 , volume of coarse aggregate corresponding to 20mm sieve aggregate and fine aggregate for water cement ratio of 0.5= 0.6
In present case water cement ratio is 0.40
Therefore volume of coarse aggregate for the water cement ratio of 0.40=0.62
For pumpable concrete these value are reduced by 10%.
Therefore volume of coarse aggregate = 0.62*0.9= 0.56
Volume of fine aggregate = 1-0.56= 0.44
Step8:- mix calculation:-
A. Volume of concrete = 1m3
B. Volume of cement = mass of cement /specific gravity of cement *1000
= 270/3.15*1000= 0.086 m3
C. Volume of fly ash = mass of fly ash /spe. Gravity of fly ash*1000
= 115/2.2*1000 = 0.052
D. Volume of water = mass of water /sp.gravity of water*1000
= 140/1*1000= 0.140 m3
E.volume of chemical admixture = mass of chemical admixture / sp. Gravity of admixture*1000
= 7.6/1.145*1000 = 0.006m3
F. Volume of aggregate= A-(B+C+D+E)
= 1-(0.086+0.052+0.140+0.006)
=0.716m3
G. Mass of coarse aggregate = f* volume of coarse aggregate * sp.gravity of coarse aggregate*1000
= 0.716*0.56*2.74*1000
= 1098 kg/m3
H. Mass of fine aggregate =f* volume of fine aggregate*sp.gravity of fine aggregate *1000
= 0.716*0.44*2.74*1000=863 kg/m3
Step9:- mix proportions for trial number:-
Cement= 270 kg/m3
Fly ash = 115 kg/m3
Water = 140 kg/m3
Fine aggregate= 863 kg/m3
Coarse aggregate = 1098 Kg/m3
Chemical admixture = 7 kg/m3
Water cement ratio 0.364
Trial mix ratio =1:2.24:2.85
Q.2. Calculate the Concrete Mix Design for M50 grade concrete without fly ash.
Ans. Step 1:- stipulations for proportioning
- Garde designation : M50
- Type of cement : opc 43 grade conforming to IS 8112
- Type of mineral admixture : Fly ash conforming to IS3812(part 1)
- Maximum nominal size of aggregate : 20 mm
- Minimum cement content :320 kg/m3 (from table 5 of Is 456:2000 exposure condition)
- Maximum water cement ratio : 0.45 (from table 5 of IS 456:2000 exposure condition)
- Work ability :100 mm (slump value for pumpable concrete)
- Exposure condition : severe (for reinforced concrete)
- .method of concrete placing : pumping
- Degree of supervision : Good
- Type of aggregate : crushed angular aggregate
- Maximum cement content : 450 kg/m3
- Chemical admixture type : super plasticizer
Step2:- Test data for material
- cement used : opc 43 grade Is 8112
- Specific gravity of cement : 3.15
- Chemical admixture : super plasticizer conforming to Us 9103
- Specific gravity of coarse aggregate :2.74 & fine aggregate : 2.74
- Water absorption coarse aggregate : 0.5% & fine aggregate1%
- Free surface moisture : nill
Step3:- Target strength for mix proportioning
F'ck = fck+1.65s
S= standard deviation ( from Table 1 of Is 10262:2009)
(S = 5N/mm2)
F'ck = 50+1.65*5
F'ck= 58.25N/mm2
Step4:- Selection of water - cement ratio
From table 5 of Is456:2000 for severe exposure, maximum water cement ratio =0.45
Based on experience , adopt water cement ratio as 0.30
O.3<0.45 Hence ok
Step5:- Selection of water content
From table 2 of Is 10262:2009 ,maximum water content for 20 mm aggregate = 186 litre (for 25 mm to 50 mm slump range)
Estimated water content for 100 mm slump = 186+6/100*186= 197 litre
As super plasticizer is used , the water content can be reduced up to 30%.
Based on trials water content reduction of 29% is achieved with plasticizer .
Water content = 197*0.71= 140 litre
Step6:- Calculation of cement contant:-
Water-cement ratio = 0.30
Cement content=140/0.30= 467 kg
From table 5 Is 456 minimum cement content for severe exposure condition = 320 kg/m3
467>320 kg/m3 Hence ok.
Step7:- proportion of volume of concrete aggregate and fine aggregate content
Table 3 Is 10262:2009 , volume of coarse aggregate corresponding to 20mm sieve aggregate and fine aggregate for water cement ratio of 0.5= 0.6
In present case water cement ratio is 0.30
Therefore volume of coarse aggregate for the water cement ratio of 0.30=0.62
For pumpable concrete these value are reduced by 10%.
Therefore volume of coarse aggregate = 0.62*0.9= 0.56
Volume of fine aggregate = 1-0.56= 0.44
Step8:- mix calculation:-
A. Volume of concrete = 1m3
B. Volume of cement = mass of cement /specific gravity of cement *1000
= 467/3.15*1000= 0.148 m3
C. Volume of water = mass of water /sp.gravity of water*1000
= 140/1*1000= 0.140 m3
D.volume of chemical admixture = mass of chemical admixture / sp. Gravity of admixture*1000
= 7/1.145*1000 = 0.006m3
E. Volume of aggregate= A-(B+C+D)
= 1-(0.148+0.140+0.006)
=0.706m3
F. Mass of coarse aggregate = E* volume of coarse aggregate * sp.gravity of coarse aggregate*1000
= 0.706*0.56*2.74*1000
= 1083 kg/m3
G. Mass of fine aggregate =E* volume of fine aggregate*sp.gravity of fine aggregate *1000
= 0.706*0.44*2.74*1000=851 kg/m3
Step9:- mix proportions for trial number:-
Cement= 467 kg/m3
Water = 140 kg/m3
Fine aggregate= 851 kg/m3
Coarse aggregate = 1083 Kg/m3
Chemical admixture = 7 kg/m3
Water cement ratio 0.3
Trial mix ratio =1:2.56:3.26