Project on Concrete Mix Design for various grades of Concrete
Using Fly-Ash
Data required:
Grade designation: M35
Type of cement: OPC 43 grade conforming to IS 8112
Maximum nominal size of aggregate: 20mm
Exposure type: Severe RCC
Minimum cement content: 320 kg/m^3
Workability: 100mm slump
Method of concrete placing: pumping
Type of aggregate: Crushed angular
Maximum cement content: 450 kg/m^3
Chemical admixture : Super plasticizer
Specific gravity of cement: 3.15
Specific gravity of fine aggregate: 2.74
Specific gravity of coarse aggregate: 2.74
Coarse and fine aggregate: Grade 1
Calculations:
Target Strength=Fck+1.65s
F'ck=35+1.65*f
F'ck=43.25
So based on experience adopt w/c ratio as 0.40
0.40<0.45
Water Content from table 2 Is10262 2009
20mm aggregate=186litre
Estimated for 100mm=186+6/100*186=197litres.....Increasing 3% for every 25 mm of slump
Due to use of super plasticizer water content has to be reduced by 29%
Water content=197*0.71=140litres
Cement content:
Cementious material=140/0.4=350kg/m^3.......Watercontent/(w/c ratio)
350>320 for severe condition based on table of IS456 minimum cement condition
Increasing 10% proportion for every due to use of fly ash
Cement ontent=350*1.10=385kg/m^3
water content=140kg/m^3
W/c ratio=140/385=0.364
Fly Ash @30%=385*30/100=115kg/m^3
but for present case w/c taio=0.40
So we have to increase course aggregate to decrease fine aggrgate in order to match 0.364 w/c ratio
w/c ratio lower than 0.10 proportion of course aggrgate is increased by 0.02
0.40=0.62
As we are using pumpable concrete we have to aggrgate proportion by 10%
0.62*0.9=0.56.....Course aggregate
Vol. of Fine aggregate=1-0.56=0.44
Mix design:
a. Vol. of concrete1m^3
b.Vol. of water=Mass*1/specific gravity*1000=140*1/1*1000=0.140m^3
c.Vol. of chemical admixture=Mass*1/specific gravity*1000=7.6*1/1.145*1000=0.006m^3
d.Vo. of cement=Mass*1/Specific gravity*1000=270*1/3.15*1000=0.086m^3
e.Vol. of fly ash=Mass*1/specific gravity*1000=115*1/2.2*1000=0.052m^3
f.Vol. of aggregate(fine & course aggregate)=a-(b+c+d+e)=1-(0.086+0.052+0.140+0.0060=0.716m^3
Mass of course aggregate=f*vol. of course aggregate*specific garvity course aggregate*1000=0.716*0.56*2.74*1000=1098kg/m^3
Mass of fine aggregate=f*vol. fine aggregate*specific gravity of fine aggregate*1000=863kg/m^3
Mix proportion for trial
Cement=270kg/m^3
Fly Ash=115kg/m^3
WATER=140KG/m^3
fine aggregate=863kg/m^3
course aggregate=1098kg/m^3
w/c=0.45
Trial mix ratio= 1:2.24:2.85
Withoout fly ash
Data required:
Grade designation: M50
Type of cement: OPC 43 grade conforming to IS 8112
Maximum nominal size of aggregate: 20mm
Exposure type: Severe
Minimum cement content: 320 kg/m^3
Workability: 100mm slump
Method of concrete placing: pumping
Type of aggregate: Crushed angular
Maximum cement content: 450 kg/m^3
Chemical admixture : Super plasticizer
Specific gravity of cement: 3.15
Specific gravity of fine aggregate: 2.74
Specific gravity of coarse aggregate: 2.74
Coarse and fine aggregate: Grade 1
Calculations:
Target Stregth=Fck+1.65s=5061.65*5=58.25
selection of w/c ratio:
Table 5 of Is456 2000
Max w/c ratio=0.45
adopted w/c ratio=0.40
Selection of water content:
Table 2 IS10262 2009
aggregate size 20mm=186litres of water
Estimated for 100 mm slump
For every 25 mm increase by 3%
=186+6/100*186=197litres
Calculation of cement content:
Cement=140/0.40=350k/m^3
350>320for severe condition from table of IS 456
Vol. of course aggregate and fine aggregate
Table 3 Is 10262 2009
Vol. of course aggregate corresponds to 20mm size aggregate and fine aggregate(zone 1) for w/c ratio=0.60
In present case w/c ratio=0.40
Vol. of course aggregate needs to be increased to decrease fine aggregate
As w/c ratio lower by 0.10 the proportion of vol. of course aggregate is increased by 0.02
Corrected vol. 0.40=062
For pumpable concrete reduce volume by 10%
Vol of course aggregate=0.62*0.9=0.56
fine aggregate=1-0.56=0.44
mix calculations=
a.Vol. of concrete=1m^3
b.Vol of cement=mass*1/specific gravity*1000=350*1/3.015*1000=0.111m^3
c.Vol.of water=mass*1/specific gravity*1000=0.140m^3
d.vol. of chemical admixture=mass*1/specific gravity*1000=7.6*1/1.145*1000=0.006m^3
e.vol. of aggregate=a-(b+c+d)=1-(0.111+0.140+0.006)=0.743m^3
f. mass of course aggregate=e*vol. of course aggregate*specific gravity*1000=0.743*0.56*2.74*1000=1140kg/m^3
g.mass of fine aggregate=e*vol. of fine aggregate*specific garvity*1000=0.743*0.44*2.74*1000=896kg/m^3
Trial No. 1:
cement=350kg/m^3
water=140kg/m^3
Fine aggregate=896kg/m^3
course aggregate=1140kg/m^3
chemical admixture=7 kg/m^3
wlc ratio=0.4
Trial mix= 1:2.56:3.26