Project on Concrete Mix Design for various grades of Concrete
MIX DEDSIGN FOR M35 CONCRETE WITH FLY ASH AS PARTIAL REPLACMENET FOR CEMENT:
1.
target mean strength:
ft=fck+1.65s
=35+1.65(5.0)
=43.25 N
2.SELECTION OF WATER CEMENT RATIO:
(considering weather condition as severe)
w/c ratio is 0.45, we take w/c ratio as 0.40 with experience since the water is less the strength is more
w/c ratio is o.40
3. SELECTION OF WATER CONTENT:
assuming slump value as 50 mm the water content is 185 litre.
4. SELECTION OF CEMENT CONTENT:
water content=186 kg
w/c=0.40
=186/0.40
=465 kg/m^3
FOR replacing flyash with cement with 30 %
cementitious material=465 x 1.10
=511kg/m^3
water content =186
w/c ratio =186/511=0.36
flyash @ 30% on total cementitious material= 511 x (30/100)=153 kg/m^3
cement used = 511-153
= 358 kg/m^3
cement saved by using flyash= 465-358=107 kg/m^3
flyash used = 153kg/m^3
MIX DESIGN:
volume of concrete=1 m^3
volume of cement= (358/3.15) x 0.001
= 0.113 m^3
volume of fly ash =(153/2.2) x 0.001
=0.069 m^3
volume of water =(186/1) x 0.001
=0.186 m^3
volume of F.A and C.A=
volume of concrete - volume of all other materials used
f= 1-0.368
f= 0.632 m^3
mass of C.A=F x volume of C.A x S.G of C.A x 1000
=1039kg/m^3
mass of F.A =F x volume of F.A x S.G of F.A x 1000
=692 KG/m^3
weight of the ingredients of the concrete:
cement:368 kg/m^3
coarse aggregate:1039 kg/m^3
fine aggregate:692 kg/m^3
MIX DESIGN WITHOUT PARTIAL REPLACEMENT OF FLY ASH FOR CEMENT:
1.
target mean strength:
ft=fck+1.65s
=35+1.65(5.0)
=43.25 N
2.SELECTION OF WATER CEMENT RATIO:
(considering weather condition as severe)
w/c ratio is 0.45, we take w/c ratio as 0.40 with experience since the water is less the strength is more
w/c ratio is o.40
3. SELECTION OF WATER CONTENT:
assuming slump value as 50 mm the water content is 185 litre.
4. SELECTION OF CEMENT CONTENT:
water content=186 kg
w/c=0.40
=186/0.40
=465 kg/m^3
5. MIX CALCULATION
volume of the concrete:1m^3
volume of the cement= (465/3.15) x 0.001
= 0.147 m^3
volume of the water = (186/1) x 0.001
= 0.186 m^3
volume of the aggregate (C.A and F.A)
f=volume of concrete-volume of other ingredient of the concrete
=1-(0.147+0.186)
=0.667m^3
mass of the C.A:
=F x volume of C.A x S.G of C.A x 1000
=0.667 X 0.60 X 2.74 X 1000
=1096 kg/m^3
mass of the F.A:
=F x volume of C.A x S.G of C.A x 1000
=0.667 X 0.40 X 2.74 X 1000
=731 KG/m^3
mix design:
cement:465 kg/m^3
coarse aggregate:1096 kg/m^3
fine aggregate:731kg/m^3
reference are fromthe IS CODE 456:2000
IS CODE 10262:2009