Project on Concrete Mix Design for various grades of Concrete
1)Ans :- concrete mix design for M35 grade concrete with fly ash
geade designation M35
type of cement opc 43 grade
Target strengh for mix proportioning
fck=fck+1.65 S
target strengh = 35+(1.65*5)= 43.25N/mm2
water cement ratio is 0.45
maximum water content for 20mm aggregate =186kg
186+(6/100)*186=197litre
water content 197 litres.
calculation of cement content
cement content =water content /w-c ratio =(197/0.45)= 438kgs
cementition material= 438*1.10=482kgs
water content =145 liters
fly ash @30%= 482*30%= 144
cement opc = 482-144=338
saving cement = 438-338= 100
a) volume of cement = 1m3
b) volume of cement = mass of cement /specific gravity of cement*1/1000
338/3.15*1/1000= 0.1073
c) volume of fly ash = mass of fly ash /specific gravity of fly ash *1/1000
144/2.2*1/1000 = 0.0654
d) volume of water = mass of water / specific gravity of water *1/1000
145/1*1/1000 =0.145
e) volume of chemical admixture = mass of chemical admixture / specific gravity of chemical admixture*1/1000
7.6/1.145*1/1000 = 0.0066
f) volume of aggregate (FA +CA) = (a-(b+C+D+E)
(1-(0.1073+0.0654+0.145+0.0066)= 1109 m3
mass of coares aggregate = f* volume of coares aggregate * specific gravity of coares aggregate *1000
1109*0.56*2.74*1000 = 1701 m3
mass of fine agregate = f*volume of fine aggregate * specific gravity of fine aggregate *1000
1109*0.44*2.74*1000 = 1337
cement = 338 kg/m3
fly ash = 144 kg /m3
water = 145
fine aggregate = 1337kg /m3
coares aggregate = 1701 kg/m3
chemical admixture= 7 kg/m3
water cement ratio = 0.45
Trial mix ratio = 1:3.95: 5.03
concrete mix design for M50 grade concrete without fly ash concrete
grade of concrete = M50
type of cement = 53 grade opc
Target strenght of proportinig
fck =50+(1.65*5) = 58.25 N/mm2
water cement ratio is = 0.35
maximum water content for 20mm aggregate = 186kg
186+6/100*186 = 197 liters
water content 197 liters
cement content = water content / w-c ratio = 197/0.35 = 562 kgs
cementition material = 562*1.10= 618
water content = 140 liters
cement opc = 618 kg
a) volume of concrete = 1m3
b)volume of cement = mass of cement /specific gravity of cement *1/1000
618/3.15*1/1000 = 0.1961 kgs
c) volume of water = mass of water / specific gravity of water *1/1000
140/1*1/1000 =0.14 kgs
d) volume of chemical admixture = mass of chemical admixture/specific gravity of chemical admixture*1/1000
7.6/1.145*1/1000=0.0066kgs
e) volume of aggregate = FA+CA =a- (b+C+D)
1-(0.1961+0.14+0.066+)= 0.9505 kgs
mass of coares aggregate = f*volume of coares aggregate * specific gravity of coares aggregate *1000
0.9505*0.56*2.74*1000 =1458 kgs
mass of fineaggregate = f* volume of fine aggregate * specific gravity of fine aggregate *1000
0.9505*0.44*2.74*1000 =1145 kgs
cement = 618 kgs
water = 140 kgs
fine aggregate = 1145 kgs
coares aggregate = 1458 kgs
chemical admixture = 7 kgs
water cement ratio = 0.35
Trial mix ratio = 1 : 1.85:2.35