Construction Technology_Concrete Mix Design_Project 1
1. Calculate the concrete Mix Design for M35 grade concrete with fly ash
a. Grade Designation = M35
b. Type of cement = OPC 43 grade conforming to IS 8112
c. Type of mineral admixtures = Fly ash conforming to IS 3812 (Part 1)
d. Max nominal size of aggregates = 20MM
e. Minimum cement content = 320kg/m^3 (from table 5 IS 456:2000 exposure condition)
f. Max water-cement ratio = 0.45 (from table 5 IS 456:2000 exposure condition)
g. workability = 100mm (Slump value for pumpable concrete)
h. Exposure condition = severe (for reinforced concrete)
i. Method of concrete placing = Pumping
j. Degree of supervision = good
k. Type of aggregates = crushed angular aggregates
l. Maximum cement content = 450kg/m^3
m. Chemical admixture = super plasticizer
Test data
a. cement used = OPC 43 conforming to IS 8112
b. specific gravity of cement = 3.15
c. Fly ash = conforming to IS 3812 (part 1)
d. specific gravity of fly ash = 2.2
e. chemical admixture = super plasticizer conforming to IS 9103
f. specific gravity of course aggregate = 2.74 specific gravity of fine aggregate= 2.74
g. water absorption of course aggregates = 0.5% , water absorption of fine aggregate= 1.0%
h. Free surface moisture for course aggregate = NIL , free surface moisture for fine aggregare = NIL
i. Sieve analysis- course aggregate conforming to table 2 of IS 383
fine aggregate conforming to grading zone 1 of table 4 of IS 383
* Target mean strength
f'ck = fck +1.65S (from table 1 IS 10262:2009) S=5N/mm2
= 35+1.65*5 = 43.25 N/mm^2
from table 5 IS 456:2000 max w/c ratio = 0.45
Based on experience adopt w/c ratio = 0.40
0.40 < 0.45, hence OK
Selection of water cement ratio
from table 2 of IS 10262:2009, max water content for 20 mm aggregate = 186 litre , for 25 to 50 mm slump range
Estimated water content for 100 mm slump = 186+6/100 *186 = 197 litres
As super plasticizer is used, the water content can be reduced by 30%
Based on water content reduction of 29% is achived with plasticizer
Hence the arrived water content = 197* 0.71= 140 litres
* calculation of cement content
w/c ratio = 0.40
cementitious material (cement+flyash)= 140/0.40= 350kg/m^3
from table 5 of IS 456:2000 min cement content for severe exposure condition = 320kg/m^3
350kg/m^3 >320kg/m^3, Hence OK
Proportion of mix containing fly ash
a. percentage of fly ash needs to be decided based on project requirment & quality of materials
b. In some cases increases in cementitious material can be warranted. increase in cementitious material content and percentage may be used on experience and trials. (this is with increase of 10% of cementitious material content)
cementitious material content = 350*1.10 =385kg/m^3
water content = 140kg/m^3
water-cement ratio= 140/385=0.364
fly ash @ 30% total cementitious material= 385*30%=115kg/m^3
cement (OPC)= 385-115=270kg/m^3
saving of cement while using flyash= 350-270=80kg/m^3
flyash being utilisez=115kg/m^3
Proportion of volume of course aggregate and fine aggregate
from table 3 of IS 10262:2009, volume of course aggregate corresponding to 20mm size aggregate and fine aggregate (Zone 1) w/c ratio = 0.6
In present case w/c ratio = 0.40
therefore volume of course aggregate needs to be increased to decreased fine aggregate content
as w/c ratio is lower by 0.10, the proportion of volume of course aggregate is increased by 0.02 at the water of -/+0.01 for every -/+0.05 change in w/c ratio
corrected volume of course aggregate for w/c ratio of 0.4 =0.62
for pumpable concrete these values are reduced by 10%
volume of course aggregate=0.62*0.9=0.56
volume of aggregate content=1-0.56=0.44
Mix calculation
the mix calculation per unit volume of concrete
a. volume of concrete= 1m^3
b. volume of cement= mass of cement/specific gravity of cement*1/1000
= 270/3.15*1/1000=0.086m^3
c. volume of flyash= mass of flyash/specific gravity of flyash*1/1000
= 115/2.2*1/1000=0.052m^3
d. volume of water= mass of water/specific gravity of water*1/1000
= 140/1*1/1000=0.140m^3
e. volume of chemical admixture= mass of chemical admixture/specific gravity of chemical admixture*1/1000
= 7.6/1.145*1/1000=0.006m^3
f. volume of aggregate(Fine aggregate & Course aggregate)=(a-(b+c+d+e))
= (1-(0.086+0.052+0.140+0.006))
= 0.716m^3
g. Mass of course aggregate= f*volume of course aggregate * specific gravity of course aggregate*1000
= 0.716*0.56*2.74*1000
= 1098kg/m^3
h. Mass of fine aggregate= f*volume of fine aggregate * specific gravity of fine aggregate*1000
= 0.716*0.44*2.74*1000
= 863kg/m^3
Mix Proportion
a. cement= 270kg/m^3
b. fly ash = 115kg/m^3
c. water= 140kg/m^3
d. fine aggregate= 863kg/m^3
e. course aggregate= 1098kg/m^3
f. chemical admixture= 7kg/m^3
g. w/c ratio= 0.364
trial mix ratio= 1:2.24:2.85
M50 grade concrete without flyash
a. Grade Designation = M50
b. Type of cement = OPC 43 grade conforming to IS 8112
c. Max nominal size of aggregates = 20MM
d. Minimum cement content = 320kg/m^3 (from table 5 IS 456:2000 exposure condition)
e. Max water-cement ratio = 0.45 (from table 5 IS 456:2000 exposure condition)
f. workability = 100mm (Slump value for pumpable concrete)
g. Exposure condition = severe (for reinforced concrete)
h. Method pf concrete placing = Pumping
i. Degree of supervision = good
j. Type of aggregates = crushed angular aggregates
k. Maximum cement content = 450kg/m^3
l. Chemical admixture = super plasticizer
Test data
a. cement used = OPC 43 conforming to IS 8112
b. specific gravity of cement = 3.15
c. chemical admixture = super plasticizer conforming to IS 9103
d. specific gravity of course aggregate = 2.74 specific gravity of fine aggregate= 2.74
e. water absorption of course aggregates = 0.5% , water absorption of fine aggregate= 1.0%
f. Free surface moisture for course aggregate = NIL , free surface moisture for fine aggregare = NIL
g. Sieve analysis- course aggregate conforming to table 2 of IS 383
fine aggregate conforming to grading zone 1 of table 4 of IS 383
* Target mean strength
f'ck = fck +1.65S (from table 1 IS 10262:2009) S=5N/mm2
= 50+1.65*5 = 58.25 N/mm^2
from table 5 IS 456:2000 max w/c ratio = 0.45
Based on experience adopt w/c ratio = 0.40
0.40 < 0.45, hence OK
Selection of water cement ratio
from table 2 of IS 10262:2009, max water content for 20 mm aggregate = 186 litre , for 25 to 50 mm slump range
Estimated water content for 100 mm slump = 186+6/100 *186 = 197 litres
As super plasticizer is used, the water content can be reduced by 30%
Based on water content reduction of 29% is achived with plasticizer
Hence the arrived water content = 197* 0.71= 140 litres
* calculation of cement content
w/c ratio = 0.40
cementitious material (cement)= 140/0.40= 350kg/m^3
from table 5 of IS 456:2000 min cement content for severe exposure condition = 320kg/m^3
350kg/m^3 >320kg/m^3, Hence OK
Proportion of volume of course aggregate and fine aggregate
from table 3 of IS 10262:2009, volume of course aggregate corresponding to 20mm size aggregate and fine aggregate (Zone 1) w/c ratio = 0.6
In present case w/c ratio = 0.40
therefore volume of course aggregate needs to be increased to decreased fine aggregate content
as w/c ratio is lower by 0.10, the proportion of volume of course aggregate is increased by 0.02 at the water of -/+0.01 for every -/+0.05 change in w/c ratio
corrected volume of course aggregate for w/c ratio of 0.4 =0.62
for pumpable concrete these values are reduced by 10%
volume of course aggregate=0.62*0.9=0.56
volume of aggregate content=1-0.56=0.44
Mix calculation
the mix calculation per unit volume of concrete
a. volume of concrete= 1m^3
b. volume of cement= mass of cement/specific gravity of cement*1/1000
= 350/3.15*1/1000=0.111m^3
c. volume of water= mass of water/specific gravity of water*1/1000
= 140/1*1/1000=0.140m^3
d. volume of chemical admixture= mass of chemical admixture/specific gravity of chemical admixture*1/1000
= 7.6/1.145*1/1000=0.006m^3
e. volume of aggregate(Fine aggregate & Course aggregate)=(a-(b+c+d))
= (1-(0.111+0.140+0.006))
= 0.743m^3
f. Mass of course aggregate= e*volume of course aggregate * specific gravity of course aggregate*1000
= 0.743*0.56*2.74*1000
= 1140kg/m^3
g. Mass of fine aggregate= e*volume of fine aggregate * specific gravity of fine aggregate*1000
= 0.743*0.44*2.74*1000
= 896kg/m^3
Mix Proportion
a. cement= 350kg/m^3
b. water= 140kg/m^3
c. fine aggregate= 896kg/m^3
d. course aggregate= 1140kg/m^3
e. chemical admixture= 7kg/m^3
f. w/c ratio= 0.4
trial mix ratio= 1:2.56:3.26