Project on Concrete Mix Design for various grades of Concrete
A) CONCRETE MIX DESIGN FOR M35 GRADE WITH FLY ASH
I} for M35 grade; fck = 35 MPa
from IS 456:2000, Table 8 ;
Standard deviation = 5.0
Therfore, Targeted mean strength = 35+(1.65*5.0) = 43.25 MPa.
II} w/c ratio selection;
From IS 456:2000, Table 5;
Maximum water content = 0.45 and Minimum cement content = 320 Kg/m3
Now adopt w/c ratio = 0.4
III} For nominal size of maxm aggregate of 20 mm the water content is 186 Kg/m3 for slump of 25 to 50 mm.
There is addition of 3% of water content over each 25 mm slump above 50 mm slump.
therefore here we increase the 6% of water content.
modified water content = 1.06*186 = 197.16 Kg/m3.
IV} calculation of cement content;
cement content = (Water content)/(w/c ratio) = 197.16/0.4 = 492.9 Kg/m3. > 320 Kg/m3 ... HENCE OK
now here we use the Fly Ash in concrete,
Therefore consider there is 10% increase in cementetious material,
Therefore cementetious material content = 492.9*1.1= 541.2 Kg/m3
Modified w/c ratio = 197.16/541.2 = 0.364 < 00.4
Therefore consider min w/c ratio = 0.364
Let Fly ash of 30% total cementitious material content
Fly ash content = 541.2*0.3 = 162.36 Kg/m3
cement content = 541.2-162.36 = 378.84 Kg/m3
V} Now volume of C.A. and F.A., From IS 10262:2009,Table 3:
For 20 mm nominal size aggregate,
Consider zone I: For w/c ratio of 0.5 there we consider volume of C.A. of 0.6
so for our w/c ratio of 0.364 == apporx 0.35
there is increase in proportion of CA by 0.03
Corrected proportion of C.A. = 0.6+0.03 = 0.63
for pumpable concrete these values are reduced by 10%.
volume of C.A. = 0.63*0.9 = 0.567
Volume of F.A. = 1-0.567 = 0.433
VI} mix calculations:
a) Volume of concrete = 1m3
b) volm of cement = 378.84/(3.15*1000)= 0.1203 m3
c) volm of water = 197.16/1000 = 0.197 m3
d) volume of chemical admixtures = (378.98*2)/(1000*100*1.145) = 0.0066 m3
e) volm of C.A. and F.A. = 1-(0.1203+0.197+0.0066) = 0.676 m3
f) Mass of C.A. = 0.676*0.567*2.74*1000 = 1050.34 Kg/m3
g) Mass of F.A. = 0.676 * 0.433*2.74*1000 = 802.02 kg/m3
Trial ratio = C:FA:CA = 1 : 2.117 : 2.773
B) Mix design for M50 grade without fly ash.
I} M50 : fck = 50 MPa :- S.D. = 5.0 [From IS 456:2000]
targated mean strength = 50 + (1.65*5) = 58.25 MPa
II} from IS 456:2000, Table 5;
maximum w/c ratio = 0.45 and Minimum cement content = 320 Kg/m3.
III} For nominal size of maxm aggregate of 20 mm the water content is 186 Kg/m3 for slump of 25 to 50 mm.
There is addition of 3% of water content over each 25 mm slump above 50 mm slump.
therefore here we increase the 6% of water content.
modified water content = 1.06*186 = 197.16 Kg/m3.
IV} calculation of cement content;
cement content = (Water content)/(w/c ratio) = 197.16/0.4 = 492.9 Kg/m3. > 320 Kg/m3 ... HENCE OK
V} Now volume of C.A. and F.A., From IS 10262:2009,Table 3:
For 20 mm nominal size aggregate,
Consider zone I: For w/c ratio of 0.5 there we consider volume of C.A. of 0.6
so for our w/c ratio of 0.4
there is increase in proportion of CA by 0.04
Corrected proportion of C.A. = 0.6+0.04 = 0.64
for pumpable concrete these values are reduced by 10%.
volm of CA = 0.64 *0.9 = 0.576
volm of FA = 01-0.576 = 0.424
VI} mix calculations:
a) Volume of concrete = 1m3
b) volm of cement = 492.9/(3.15*1000)= 0.1565 m3
c) volm of water = 197.16/1000 = 0.197 m3
d) volume of chemical admixtures = (492.9*2)/(1000*100*1.145) = 0.0086 m3
e) volm of C.A. and F.A. = 1-(0.1565+0.197+0.0086) = 0.638 m3
f) Mass of C.A. = 0.638*0.576*2.74*1000 = 1006.92 Kg/m3
g) Mass of F.A. = 0.638 * 0.424*2.74*1000 = 741.202 kg/m3
Trial ratio = C:FA:CA = 1 : 1.5 : 2.05