Project on Concrete Mix Design for various grades of Concrete
Calculation of concrete mix design for M35 grade concrete with fly ash.
Step1) Target mean strength f'ck=fuck+1.65s
fuck=35 N/mm^2(M35)
From table 1 of IS 10262:2009
s= 5N/mm^2
f'ck =35+(1.65*5)
= 43.25N/mm^2
Step2) selection of water cement ratio:-
Given:M35
Assume:
Exposure condition : Extreme (for reinforced concrete)
Table 5 - IS 456:2000
therefore , extreme= w/c ratio =0.40
M35 =w/c ratio= 0.45
Lower the water cement ratio better the workability
0.40<0.45
Water cement ratio= 0.40
Minimum cement content = 360kg/m^3
Step3) Selection of water content:-
Assume:
Nominal size of aggregate= 20mm
Table 2- IS 10262:2009
Minimum water content = 186 litres or kg
Slump ranges from 25 to 50 mm
As super plasticizers (20% reduced)
= 186*80/100= 148.8. (100-20=80)
Water content =149 litres
Step4) Calculation of cementitious content (cement+ fly ash)
Water cement ratio w/c = 0.40
Water content W= 149kg
Cementitious content c= 149/0.40 = 372.5=370kg/m^3
Minimum cement content = 360 kg/m^3
360<370 kg/m^3 Hence ok.
Increase by 10%
370*0.10+370= 407kg/m^3
W=149kg
Water cement ratio =149/407= 0.366
Fly ash at 30% total cementitious material content= 407*30/100= 12kg/m^3
Cement (OPC) = 407 - 122= 285kg/m^3
Saving of cement while using fly ash = 370- 285= 85 kg /m^3
Fly ash = 122kg/m^3
Step 5) volume of coarse Aggregate and fine Aggregate
Fine Aggregate are sieved according to zone 1 of table 4 IS 383
Coarse Aggregate are 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
(+/-0.01 for every +/- 0.005)
So, volume of coarse agg= 0.62
Volume of fine Aggregate = 1- 0.62 = 0.38
Step6) mix calculation:-
- Volume of concrete =1m^3
- Volume of cement= (285/3.15)*1/1000=0.090m^3
- Volume of fly ash =(122/2.2)*1/1000=0.055m^3
- Volume of water= (149/1)*1/1000=0.149m^3
- Volume of admixture= (7.6/1.145)*1/1000=0.006m^3
- Volume of coarse Aggregate and fine Aggregate = - (b+c+d+e) = 1-(0.090+0.055+0.149+0.006)= 0.7m^3
- Mass of coarse Aggregate= 0.7* 0.62*2.74*1000= 1189.16 kg/m^3
- Mass of fine Aggregate = 0.7*0.38*2.74*1000= 728kg/m^3
Step7) result and conclusion:- trial mix:
Cement= 285 kg/m^3
Fly ash = 122 kg/m^3
Water= 149 kg
Coarse Aggregate = 1189kg/m^3
Fine Aggregate = 728 kg /m^3
Chemical admixture = 7.6 kg/m^3
Water cement ratio= 0.366
Trial mix ratio = 285:728:1188 = 1:2.55:4.17
= cement: FA: CA
This trial mix ratio should match the workability, durability, minimum strength of M35 grade concrete.
If it does not satisfy the above criteria two or more trails for increase of 10% should be done .
Calculation of concrete mix design for M50 grade concrete without fly ash.
Step1) Target mean strength f'ck=fuck+1.65s
fuck=50N/mm^2(M50)
From table 1 of IS 10262:2009
s= 5N/mm^2
f'ck =50+(1.65*5)
= 58.25N/mm^2
Step2) selection of water cement ratio:-
Given:M50
Assume:
Exposure condition : Extreme (for reinforced concrete)
Table 5 - IS 456:2000
therefore , extreme= w/c ratio =0.40
M50 =w/c ratio= 0.40
Lower the water cement ratio better the workability
0.40=0.40
Water cement ratio= 0.40
Minimum cement content = 360kg/m^3
Step3) Selection of water content:-
Assume:
Nominal size of aggregate= 20mm
Table 2- IS 10262:2009
Minimum water content = 186 litres or kg
Slump ranges from 25 to 50 mm
As super plasticizers (20% reduced)
= 186*80/100= 148.8. (100-20=80)
Water content =149 litres
Step4) Calculation of cementitious content
water cement ratio w/c = 0.40
Water content W= 149kg
Cementitious content c= 149/0.40 = 372.5=373kg/m^3
Minimum cement content = 360 kg/m^3
360<373 kg/m^3 Hence ok.
Step 5) volume of coarse Aggregate and fine Aggregate
Fine Aggregate are sieved according to zone 1 of table 4 IS 383
Coarse Aggregate are 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
(+/-0.01 for every +/- 0.005)
So, volume of coarse agg= 0.62
Volume of fine Aggregate = 1- 0.62 = 0.38
Step6) mix calculation:-
- Volume of concrete =1m^3
- Volume of cement= (373/3.15)*1/1000=0.118m^3
- Volume of water= (149/1)*1/1000=0.149m^3
- Volume of admixture= (7.6/1.145)*1/1000=0.006m^3
- Volume of coarse Aggregate and fine Aggregate = - (b+c+d+e) = 1-(0.118+0.149+0.006)= 0.727m^3
- Mass of coarse Aggregate = 0.727*0.62*2.74*1000= 1235.02 kg/m^3
- Mass of fine Aggregate = 0.727*0.38*2.74*1000= 756.9kg/m^3
Step7) result and conclusion:- trial mix:
Cement= 373 kg/m^3
Water= 149 kg
Coarse Aggregate = 1235.02kg/m^3
Fine Aggregate = 756.9 kg /m^3
Chemical admixture = 7.6 kg/m^3
Water cement ratio= 0.40
Trial mix ratio = 373:756.9:1235.02 = 1:2.02:3.31
= cement: FA: CA
This trial mix ratio should match the workability, durability, minimum strength of M50 grade concrete.
If it does not satisfy the above criteria two or more trails for increase of 10% should be done .