Project on Concrete Mix Design for various grades of Concrete
To calculate the concrete mix design for M35 grade of concrete with fly ash
target the mean strength ; f'ck = fck +1.65
compressive strength at 28 days of concrete S = standard deviation (IS 10262:2009,table-1)
volume of cement : (massof cement/specific gravity of cement) X 1/1000
volume of water : (mass of water/specific gravity of water)X1/1000
volume of chemical admixtrures : (mass of chemical admixtrure/specific gravity of chemical admixtrure)X1/1000
volume of fly ash : (mass of fly ash/specific gravity of fly ash)X1/1000
volume of concrete : 1m^3
volume of aggregates : ( volume of cement+ volume of fly ash+ volume of water+ volume of admixtrure)
mass of coarse aggregates : volume of aggregate X volume of coarse agregate X specific gravity of coarse aggregate X 1000
massof fine aggregate : volume of aggregate x volume of fine agggregate X specific gravity of fine aggregate X 1000
1. target mean strength
table 1-IS 10262:2009
S = 5n/mm^2
f'ck=35+(1.65X5)
=43.25N/mm^2
2. selection of water cement ratio
M35, assume exposure condition: extreme
table 5- IS 456:2000
therefore, extreme=w/c ratio=0.40
M35=w/c ratio=0.45
lowe the water cement ratio better the workability 0.40<0.45
minimum cement cintent=360kg/m^3
3. selection of water content :
assuem; nominal size of aggregate=20mm
table 2-IS 10262:2009
minimum water content=186liter or kg
slump ranges from 25 to 50mm
super plasticizers 20% reduced
=186X80/100=148.8 (100-20=80)
water content=149liters
4. calculation of cementatious content (cement+flyash)
water cemnt ration=w/c=0.40
water content=w=149kg
cementatious material=c=149/0.40=375.5=0.370kg/m^3
minimum cement content=360kg/m^3
360<370 hence okay
increase by 10%
370X0.10+370=407kg/m^3
w=149kg
water cement ratio=149/407=0.366
fly ash at 30% total cementatiuos material content=407X30/100=122kg/m^3
cement(ordinary portland cement)=407-122=285kg/m^3
fly ash=122kg/m^3
5. volume of coarse aggreagtes and fine aggregates
fine aggregates are sieved according to zone 1 to table 4 IS 383
coarse aggregate are sieved according to table 2 IS 282
acording, table 3IS 10262:2009
nominal size of aggregate=0.60
water cement ratio=0.40
so,volume of corase aggregate = 0.62
volume of fine aggregate=1-0.62=0.38
6. mix calculations:
1. volume of concrete=1m^3
2. volume of cement=(285/3.15)X1/1000=0.090m^3
3. volume of fly ash= (122/2.20X1/1000=0.055m^3
4. volume of water=(149/1)X1/1000)=0.149m^3
5. volume of admixtrure=(7.6/1.145)X1/1000=0.006m^3
6. volume of corase aggregate and fine aggregate=a-(b+c+d+e)=1-(0.090+0.055+0.149+0.006)=0.7m^3
7. mass of corase aggregate=0.7X0.62X2.74X1000=1189.16kg/m^3
8. mass of fine aggregate=0.7X0.38X2.74X100728kg/m^3
TO CALCULATE M50 GRADE OF CONCRETE WITHOUT FLY ASH
target mean strength:f'ck=fck+1.65
copressive strength at 28days of concrete
S standard deviation IS 10262:2009 table-1)
1. target mean strength
f'ck=fck+1.65
fck=50N/mm^2
table 1-IS 10262;2009
s=5N/mm^2
f'ck=50+(1.65X5)=58.25N/mm^2
2.selection of water cement ratio
assume , exposure condition: extreme forreinforced 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
water cement ratio=0.40
minimum cement content=360kg/m^3
3. selection of water content
assume, nominal size if aggregate=20mm
table 2-IS10262:2009
minimum water content= 186liters or kg
slimp ranges from 25 to 50mm
=186X80/100=148.8
water content=149liters
4. calculation of cement content
water cement reatio=w/c 0.40
water content=w=149kg
cement content=c=149/0.40=375.5=373kg/m^3
minimum cement cintent=360kg/m^3
360<373 hence okay
5. volume of coarse aggregate and fine aggregate
fine aggregate are sieves according to zone 1 of table 4 IS 383
coarse aggregates are sieved according to zone 2 of table IS 383
according, table 3 IS 10262:2009
nominal size of aggegates=20mm
fine aggregate(zone1)
volume of coarse aggregate=0.60
water cement ratio=0.40
so,volume of coares aggregate=0.62
volume of fine aggregate=1-0.62=0.38
6. mix calculations:
1.volume of concrete=1m^3
2.volume of cement=(373/3.15)X1/1000=0.118m^3
3.volume of water=149/1X1/1000=0.149m^3
4.volume of admixture=7.6/1.145X1/1000=0.006
5.volume of coares aggreagate=1-0.118+0.149+0.006=0.727m^3
6.mass of coares aggregate=0.727X0.62X2.74X1000=1235.02m^3
7.mass of fine aggregate=0.727X0.38X2.74X1000=756.9m^3