Project on Concrete Mix Design for various grades of Concrete
1)
a)data collection
grade of concrete =M35
f'ck=35
S=5N/mm^2
f'ck =fck+1.65S
f'ck=35+(1.65*5)=43.25N/mm^2
b)obtain the w/c ratio
from the graph, the corresponding 28days compressive strenth
w/c=0.43
Based on the given expossed condition from table 5 of IS 456 ,
w/c=0.45
The lower value is taken w/c=0.43
c)selection of water content from table 2 of IS 10262
max water content =186 liters
consider slump as 75mm=186+(3/100)186 =191.58lit
d)as superplastisizers added the water content can be redused upto 20%
Hence the arrived water content =197*0.80=156.6
2)calculate the cement concrete
w/c ratio=0.43
cementanious material (cement+fly ash)content =(156/0.43) =364.7kg/m^3
from table 5 of 456 ,minimum very seviere exposure is 340kg/m^3
364.7kg/m^3>340kg/m^3 hence okay
b)cementanious material can be warrented based on experience and trail increse of 10% cementanious material content
cementanious material content =364*1.10=400kg/m^3
water content =156kg/m^3
so, water content ratio =156/400=0.39
fly ash @30% total cementanious material content =400*30%=120kg/m^3
cement (opc) =400-120=280kg/m^3
saving og cement whilw using flyash =364-280=84
flyash being utilized =120kg/m^3
3)proportion of volume of coarse aggregate and fine aggregate content
From the table 3 of IS 10262:2009 for 20mm size (zone 1) for w/c ratio 0.5=0.6
in present case the water cement ratio is 0.43
as water cement ratio is lower than 0.05, the proportion of volume of coarse aggregate is incresed by 0.01
therefore corrected volume of coarse aggregate for the water cement ratio is 0.43=0.61
for pumpable concrete the value are reduced by 10%
Therefore volume of coarse aggregate =0.61*0.9=0.549
volume of fine aggregate content= 1-0.549=0.46
4)Mix ratio
The mix calculation per unit volume of concrete shall be as follows
a) volume of concrete =1m^3
b)volume of cement=(280/3.15)*(1/1000)=0.088
c)volume of fly ash=(120/2.2)*(1/1000)=0.0545
d)volume of water =(156/1)*(1/1000)=0.156
e)volume of aggregate (FA&CA)=(a-(b+c+d))=(1-(0.088+0.0545+0.156))=0.701kg/m^3
f)mass of coarse aggregate=e*volume of coarse aggregate*specific aggregate of CA*1000
=0.706*0.56*2.74*1000=1083.28kg/m^3
g)mass of fine aggregate =e*volume of fine aggregate *specific gravity of FA*1000
=0.706*0.44*2.74*1000=851kg/m^3
mix proportions
| Cement |
280kg/m^3 |
| fly ash | 120kg/m^3 |
| Water | 156kg/m^3 |
| Fine aggregate | 863kg/m^3 |
| coarse aggregate | 1083.28kg/m^3 |
| water cement ratio | 0.43 |
| Trail mix ratio |
1:2.15:2.70 |
M50 grade without flyash
grade of concrete =M50
f'ck=50N/mm^2
S=5N/mm^2
f'ck=fck+1.65S=50+(1.65*5)=58.25
b)obtain the water cement ratio
from the graph w/c =0.35
c) selection of water content
from table 2 of IS 10262
Max water content =186lit
consider slump value as 75mm =186(3/100)*186=191.58 liters
d)calculate cement content
w/c =0.35
cement content =191.58/0.35=547.14kg/m^3
i.e)proportion of volume of CA and FAFrom the table 3 of IS 10262, volume of CA corresponding to 20mm size aggredate FA (Z0ne 1) for w/c ratio of 0.50=0.60
In our case the water cement ratio is 0.35 which is lower by 0.15 the volume of CA is incresed by 0.03
volume of CA=0.633
volume of FA=1-0.633=0.367
f)Mix calculations
1. volume of concrete =1m^3
2.volume of cement =(547.14/3.15)*(1/1000)=0.173m^3
3.volume of warer =(191.58/1)*(1/1000)=0.19158m^3
4.volume of aggregates=1-(0.173+0.19158)=0.63542
5.Mass of CA=volume of aggregate *volume of CA* Sp.G of CA * 1000
=0.635*0.662*2.74*1000=1151.81kg/m^3
6.mass of FA =0.635*0.378*2.74*1000=657.68kg/m^3
g)Mix proportions
| Cement | 547.14kg/m^3 |
| Water | 191.58kg/m^3 |
| FA | 657.68kg/m^3 |
| CA | 1151.81kg/m^3 |
| W/C | 0.35kg/m^3 |
| Mix ratio | 1:1.2:2.10 |