Project on Concrete Mix Design for various grades of Concrete
Calculate the Concrete Mix Design for M35 grade concrete with fly ash & M50 grade concrete without Fly ash
Mix Design for M35 grade concrete with fly ash :-
Assumptions:-
Type of cement = OPC
Aggregate normal size = 20 mm
Exposure condition seviour
slump value =75 mm
Superplastisizer
Fine aggregate Zone-II
Methode of concrete placing by pumping
Grade of concrete design = M35
Step-1
Targate mean Strength = Fck' = Fck + 1.65*S
= 35 + 1.65 * 5 = 43.25 N/mm^2
Step-2
from table no 5 IS 10262:2009 for seviere exposure W/C = 0.45
so we take 0.4 < 0.45 Hence OK.
Step-3
Selection of water content
from table 2 of IS 10262:2009
Max water content for 20 mm aggregate = 186lits for 25mm to 50mm slump value range
We know that every Increase 25mm slump 3% water added
for 75 mm slump 3% added
Step-4
Established water content = 186+3/100*186 = 191.6 litse
As per superplastisizer is used the water content can be reduce upto 30%
Arrived water content = 191.6*0.7 = 134 lits
Step-5
Calculate cement content:-
Water cement ratio take lower value =W/C =0.4
Cementitious material (cement + fly ash) CEMENTITIOUS =134/0.4 = 335 kg/M^3
from table 5 IS 456 minumum cement for seveier condition =320 kg/M^3
335 > 320 kg/M^3 Hence OK
Now to preparation a mix containing fly ash following steps suggested
a) Percentage of fly ash needed to be design based on project requirement and quality of material
b) In some cases increase in cementitious material content and its percentage based on experience and trail
(with increase of 10% cementitious material content)
Cementitious material content = 335*1.10 = 369 kg/m^3
Water content = 134 kg/m^3
water cement ratio =134/369 = 0.363
fly ash 30% total cementitious material content = 369* 0.3 = 111 kg/m^3
Cement OPC = 369-111 = 258 kg/m^3
sacving of cement while using fly ash =335-258 = 77 kg/m^3
step-6
Proportion of volume of cource aggregate and fine aggregate from table 3 IS 10262:2009
Volume of C.A 20mm aggregate ZONE-II
In the precent case of w/c ratio = 0.4
As water cement ratio lower by 0.10 the proporrtion of volume of C.A Increased by 0.62
(at the rate of -/+ 0.01 for every +- 0.05 change of water cement ratio)
Corrected volume of C.A for water-cement ratio 0.40 = 0.62
plumb concrete the volume of cource aggregate = 0.62 * 0.9 = 0.56
volume of fine aggregate = 1-0.56 = 0.44
Srep-7
a) Volume of concrete = 1m^3
b) Volume of cement = 258/3.15*1/1000 = 0.08 m^3
c) volume of fly ash = 111/2.2 *1/1000 = 0.050 m^3
d) Volume of water = 134/1*1/1000 = 0.134 m^3
e) Volume of chemical admixtures 2% by mass of cemetition material = 7/1.145*1/1000 = 0.006 m^3
f) Volume of Aggregate Fine aggregate and cource aggregate = (a-(b+c+d+e)) = (1-(0.08+0.050+0.134+0.006)) =0.73 m^3
g) Mass of Cource aggregate = f * volume of C.A * sp gravity of C.A * 1000 = 0.73*0.56*2.74*1000 = 1120 Kg/m^3
h) Mass of fine aggregate = f * volume of F.A * sp gravity of F.A * 1000 = 0.73*0.44*2.74*1000 = 880 Kg/m^3
step-8
Mix proportion
cement = 258 kg/m^3
fly ash = 111 Kg/m^3
water 134 Kg/m^3
Fine aggregate = 880 Kg/m^3
Cource aggregate = 1120 Kg/m^3
Chemical Admixture 7 Kg/m^3
water cement ratio = 0.363
Trial mix = 1:3:4
Mix Design for M50 grade concrete :-
Assumptions:-
Type of cement = OPC
Aggregate normal size = 20 mm
Exposure condition seviour
slump value =75 mm
Superplastisizer
Fine aggregate Zone-II
Methode of concrete placing by pumping
Grade of concrete design = M50
Step-1
Targate mean Strength = Fck' = Fck + 1.65*S
= 50 + 1.65 * 5 = 58.25 N/mm^2
Step-2
from table no 5 IS 10262:2009 for seviere exposure W/C = 0.45
so we take 0.4 < 0.45 Hence OK.
Step-3
Selection of water content
from table 2 of IS 10262:2009
Max water content for 20 mm aggregate = 186lits for 25mm to 50mm slump value range
We know that every Increase 25mm slump 3% water added
for 75 mm slump 3% added
Step-4
Established water content = 186+3/100*186 = 191.6 litse
As per superplastisizer is used the water content can be reduce upto 30%
Arrived water content = 191.6*0.7 = 134 lits
Step-5
Calculate cement content:-
Water cement ratio take lower value =W/C =0.4
Cementitious material (cement + fly ash) CEMENTITIOUS =134/0.4 = 335 kg/M^3
from table 5 IS 456 minumum cement for seveier condition =320 kg/M^3
335 > 320 kg/M^3 Hence OK
step-6
Proportion of volume of cource aggregate and fine aggregate from table 3 IS 10262:2009
Volume of C.A 20mm aggregate ZONE-II
In the precent case of w/c ratio = 0.4
As water cement ratio lower by 0.10 the proporrtion of volume of C.A Increased by 0.62
(at the rate of -/+ 0.01 for every +- 0.05 change of water cement ratio)
Corrected volume of C.A for water-cement ratio 0.40 = 0.62
plumb concrete the volume of cource aggregate = 0.62 * 0.9 = 0.56
volume of fine aggregate = 1-0.56 = 0.44
Srep-7
a) Volume of concrete = 1m^3
b) Volume of cement = 335/3.15*1/1000 = 0.106 m^3
c) Volume of water = 134/1*1/1000 = 0.134 m^3
d) Volume of chemical admixtures 2% by mass of cemetition material = 7/1.145*1/1000 = 0.006 m^3
e) Volume of Aggregate Fine aggregate and cource aggregate = (a-(b+c+d+e)) = (1-(0.08+0.050+0.134+0.006)) =0.754 m^3
f) Mass of Cource aggregate = f * volume of C.A * sp gravity of C.A * 1000 = 0.73*0.56*2.74*1000 = 1157 Kg/m^3
g) Mass of fine aggregate = f * volume of F.A * sp gravity of F.A * 1000 = 0.73*0.44*2.74*1000 = 909 Kg/m^3
step-8
Mix proportion
cement = 335 kg/m^3
fly ash = 111 Kg/m^3
water 134 Kg/m^3
Fine aggregate = 909 Kg/m^3
Cource aggregate = 1157 Kg/m^3
Chemical Admixture 7 Kg/m^3
water cement ratio = 0.4
Trile mix = 1:2.71:3.45