Project on Concrete Mix Design for various grades of Concrete
Answer-: concrete mix design of M35 grade concrete with fly ash :
1. Data required:
A. Grade designation:- M35
B. Type of cement:-OPC 53
C. Type of admixture:- fly ash
D. Maximum nominal size of aggregate:-20mm
E. Minimum cement content -340 kg
F. Maximum Cement content:-450 kg
G. Partial replacement of cement by fly ash:-30%
H. Maximum water cement ratio:- .45
I. Slump value:- 75 mm
J. Chemical Admixture:- superplasticizer
2.Test data collected from laboratory:
A. Specific gravity of cement:-3.15
B. Specific gravity of fly ash:- 2.90
C. Mineral admixture:-fly ash
D. Chemical Admixture:-0.6%
E. Specific gravity of C.A:-2.74
F. Specific gravity of F.A:- 2.74
G. Specific gravity of PCE based :-1.12
H. Water absorption C.A:- 0.5%
I. Water absorption F.A:- 1%
J. Sieve analysis F.A:- zone 2
3.Target strength for mix Proportions of M35 Grade concrete:- Fçķ=Fçķ +1.65s
Where Fçķ= target strength and s= standard deviation
So, Fçk= 35+(1.65×5)= 43.25 N/mm²
4. Selection of Water cement ratio:- 0.45(Table 5- IS456)
5. Selection of Water content:- 186 kg (Table 2- IS10262) for slump 25mm-50mm.
But we need to find out water for 75 mm slump,which we can obtain by 3% increase in water content. Hence water content will be 191.58 litres.
Since we are using superplasticizer we can reduce water content by 20 % to gain desired strength from the mix. So,
Adopted water content=191.58-(191.58*20/100)litres
=153.26 (let's say 155 litres)
6. Calculation of cement and fly ash content
Here, we have water cement ratio=0.45
Water content =155 litres
So, cement +fly ash content=155/0.45
=344.4 kg
Since we have used fly ash as cement replacement, here we are increasing (cement +fly ash ) content by 10% for higher strength and workability .
So, 344.4 + (344.4*.1)
=378.84 kg (Adopted )
Since minimum cement content required for sever exposure is 340 kg. Hence ok.
Fly ash content
=(cement + fly ash content)*(0.3)
=113.65 kg
cement content
= 378.84-113.65
=265.188 kg
Now calculation of superplasticizer at 0.6%=2.27 kg
Coarse and fine aggregate:
Coarse aggregate ratio for w/c ratio.5 is 0.62
But final Coarse aggregate ratio for 0.45 w/c ratio will be 0.63.
And final fine aggregate will be 0.37.
Mix calculation for M35 grade of concrete.
A)Volume of concrete =1 cum
B) total volume of cement = cement/( specific gravity*1000)= 265.188/(3.15*1000)= 0.084 cum
C) total volume of fly ash = fly ash content/( specific gravity *1000)= 113.65/(2.90*1000)=0.039 cum
D) total volume of water= water content/(specific gravity *1000)= 0.155 cum
E) Volume of superplasticizer =2.27/(1.12*1000)
=0.002 cum
F)Total aggregate required = A-(B+C+D+E)
=1-(0.084+0.039+.155+0.002)=0.72 cum
G) Total Coarse aggregate = Volume of aggregate required*s.g*C.A ratio*1000= .72*2.74*.63*1000=1242.8 kg
H) total fine aggregate =.72*2.74*.37*1000=729.9 kg
Now we have the date
Cement=265 kg
Fly ash=114kg
Coarse aggregate =1243 kg
Fine aggregate =730 kg
Superplasticizer =2.27 kg
Water = 155 kg
Hence the design mix will be 1:.45:4.7:2.75:.6 for
Cement:fly ash:C.A:F.A:water
Concrete mix design for M50 grade of concrete:
Given data:
A) Grade designation:-M50
B) Maximum size of aggregate-20mm
C) Cement :-OPC 53
D) minimum cement content:-360kg /m³
E) Maximum water cement ratio:-0.35
F) Workability:-50 mm-70mm
G) Supervision :- good
H) Maximum Cement content:- 450kg /m³
I) chemical Admixture:- superplasticizer
Test data of materials:
Specific gravity of cement:- 3.15
Specific gravity of F.A:-2.56
Specific gravity of C.A:- 2.60
1.Target strength of mix Proportioning:
Ftarget =Fck+1.65S
= 50+1.65×5
=58.25 N/mm²
2. Selection of Water cement ratio
Maximum water cement ratio =0.35
3. Selection of Water content
Water content for 20 mm aggregate is 186 litre for 25mm-50mm slump. But we are mixing concrete for 50mm-75mm slump. So water content will increase by 3%( IS 456 table 5)
Hence , water content will be 186+(186×.03)=191.58 for 50mm-75mm slump.
But since we have used superplasticizer as an admixture, it reduced the water content by 25% experimentally. Hence Adopted water content is=143.685( let's say 144 litres)
4. Proportion of volume of fine aggregate and Coarse aggregate content
Since for 0.5 w/c ratio volume of coarse aggregate is .62. But here Adopted w/c ratio is .35 ,hence volume of coarse aggregate is 0.65 and volume of fine aggregate is 0.35.
5. Cement content
cement content= water content/w/c ratio
=144/0.35=411.42 kg( lets say 412kg)
Which is greater than minimum cement content I.e 360kg,. Hence okay.
Superplasticizer content at 1.2%=4.944kg
Mix calculation of M50 grade of concrete
A)Volume of concrete-1 cum
B)Volume of cement= total cement /specific gravity *1000= 412/(3.15*1000) =.130 cum
C) Volume of water= .144litres
D) Volume of superplasticizer =4.944/(1.2*1000)=0.004
E) Volume of all Aggregate =1-(0.130+0.144+0.004)=0.722 cum
F) weight of coarse aggregate = total volume of aggregate *specific gravity of C.A*ratio of C.A*1000=.722×2.6×.65×1000=1220.18 kg (1220 kg adopted)
G) weight to total fine aggregate =.722×.35×1000×2.56=646.912 kg ( 647 kg adopted)
Now we have calculated data
a) cement=412 kg
b) water = 144 litres
c) fine aggregate =647 kg
D) coarse aggregate =1220 kg
E) superplasticizer = 4.944 kg
So ratio of cement:water:C.A:F.A: admixture will be
1:.35:1.6:3:.01