Project on Concrete Mix Design for various grades of Concrete
Calculate the Concrete Mix Design for M35 grade concrete with fly ash
solution : data base
step 1 : stipulation for proportion
- M35 grade concrete with fly ash
- cement used - opc 33grade
- Type of fly ash admixture -flyash
- Chemical admixture -super plasticizers
- nominal size - 20mm aggregate
- slump =75 mm
- exposure condition - severe
- zone of fine aggregate - zone 2
IS 10262 -2019
1 . Target mean strength:
f'ck = fck + 1.65 S
f'ck = 35 +1.65*S
S- standard deviation (from table 2) = 5 N/MM^2
f"ck = 35 + 1.65*5
f'ck =43.25 N/mm^2
2.W/C ratio
w/c ratio = 0.38
Table 5 -page no.20
severe -0.45
between 0.38 and 0.45
lower value = 0.38
3.water content
slump value =75mm
for every 25mm 3%needs to be incresed
186 +3/100*186=192kg
Reducing w/c ratio 20-30%
(192*30/100)-192
=134 lit
4.Cement content
W/C =0.38 ( Got from 2nd step )
134/0.38 =c
C=352.63 =353KG
Cementatious content = 10%
353+ 30/100*353 = 388.3 (cement +flyash )
Flyash = 30% of cementatious concrete
30/100*388-388 = 116.4
388-116.4 =271.6 kg
cement =271.6 kg
cement save =353- 271
=1 .1/2-bag of cement
w/c= 134/388
= 0.35
5 ) calculation of Coarse aggregate and fine aggregates
0.20 nominal size - 0.62 (from table 5)
volume of coarse aggregate =0.62
(0.5-w/c/0.05 *0.01)+ volume of aggregates
(0.5-0.35/0.05)*0.01+0.62
volume of coarse aggregate = 0.65
pumpable concrete=0.65*0.9
= 0.585 = 0.59
Volume of fine aggregate = 1-0.59
=0.41
mix calaculation
1.volume of concrete=1m^3
2.volume of concrete = 0.086
1-(0.086+0.052+0.006+0.134)
volume of aggregate = 0.72
in 1m^3 = 0.72 is total aggregate
=0.72 m^3
mass of coarse aggregate =0.72*0.59*2.75*1000
=1171.44 kg/m^3
mass of fine aggregate
0.41*0.72*2.74 *1000
= 809 kg/m^3
Mix proportion of trail no.1
cement -271 kg
water - 134 lit
fine aggregate-809 kg/m^3
coarse aggregate - 1171.44 kg/m^3
water cement -0.38
2. M50 grade concrete mix design without Fly ash
solution : data base
step 1 : stipulation for proportion
- M50 GRADE CONCRETE
- cement used - opc 53 grade
- Chemical admixture -super plasticizers
- nominal size - 20mm aggregate
- slump =50 mm
- exposure condition - severe
- zone of fine aggregate - zone 1
IS 10262 -2019
1 . Target mean strength:
f'ck = fck + 1.65 S
f'ck = 50 +1.65*S
S- standard deviation (from table 2) = 5 N/MM^2
f"ck = 50 + 1.65*5
f'ck =58.25 N/mm^2
2.Selection of Watre/Cement ratio
From table 5 of IS 456:2000
severe -0.45
= 0.45
3.water content
slump value =50 mm
=186 lit
reducing water content 25%
(186*25/100)
=139.5 lit
4.Cement content
W/C =0.45 ( Got from 2nd step )
139/0.45 =c
C=308.88 =309 kg/m^3
4 ) calculation of Coarse aggregate and fine aggregates
Volume of coarse aggregate per unit volume of total aggregate for corresponding to 20mm maximum size and zone 1 of fine aggregate
0.60-table 3 of IS-10262:2009
Volume of coarese aggregatte =0.60
(0.5-w/c/0.05 *0.01)+ volume of aggregates
(0.5-0.45/0.05)*0.01+0.60
volume of coarse aggregate = 1.01
pumpable concrete=1.01*0.9
0.90
Volume of fine aggregate = 1-0.90
=0.1
mix calaculation
1.volume of concrete=1m^3
2.volume of aggregate (C.A&F.A) =a-(b+c+D)
1-(0.098+0.139+0.006)
volume of aggregate = 0.75 m^3
in 1m^3 = 0.75m^3 is total aggregate
e=0.75m^3
mass of coarse aggregate = 0.75*0.56*2.74*1000
=1155 kg/m^3
mass of fine aggregate
0.75*0.1*2.74 *1000
= 206.25 kg/m^3
Mix proportion of trail no.1
cement -309 kg
water - 139 lit
fine aggregate-206.25 kg/m^3
coarse aggregate - 1155 kg/m^3
chemical admixture -7kg
water cement -0.45
Trial mix ratio = 1:2.35:2.86