Project on Concrete Mix Design for various grades of Concrete
Procedure:
Proportions
a. Grade designation : M35
b. Type of cement : OPC 43 grade conforming to IS 8112
c. Type of mineral admixture: Fly ash confirming to IS 3812 ( Part 1)
d. Maximum nominal size of aggregates : 20mm
e. Minimum cement content : 320 kg/m^3 (Table :5 IS 456-2000,based on exposure condition)
f. Maximum water-cement ratio : 0.45 (Table :5 IS 456-2000,based on exposure condition)
g. Workability :100mm slump value of pumpable concrete
h. Exposure condition : Severe ( for reinforced concrete)
i. Method of concrete placing : Pumping
j. Degree of supervision : Good
k. Type of aggregate: Crushed angular aggregate
l. Maximum cement content : 450 kg/m^3.
m. Chemical admixture type : Superplaticizer.
Data
a. Cement used : OPC 43 grade confirming to IS 8112
b. Specific gravity of cement : 3.15
c. Fly ash : Conforming to IS 3812 part 1
d. Specific gravity of flyash : 2.2
e. Chemical admixture : Superplaticizer conforming to IS 9103
f. Specific gravity of :
Coarse aggregate : 2.74
Fine aggregate : 2.74
g. Water absorption :
Coarse aggregate : 0.5%
Fine aggregate : 1%
Target mean strength
Target mean strength f'ck = fck +1.65s
= 35 +1.65*5 =43.25 Nmm2
fck- Characteristic compressive strength
s- Standard deviation
Selection of water-cement ratio
Maximum water-cement ratio =0.45 (for severe condition ,from table 5 of IS 456-2000)
On experience water- cement ratio =0.4
0.4 <0.45, Hence OK.
Calculation of water content
Maximum water content for 20mm aggregate =186 litres ( Table 2 of IS 1026:2009) for 25mm to 50mm slump range
For 100 mm slump ,= 186+(6/100)*186 =197 litres
As superplaticizer is used, the water content is reduced upto 30%
Based on trials water content reduction of 29% is achieved with plasticizer
Hence arrived water content = 197*0.71 = 140 litres
Calculation of cement content
Water - cement ratio = 0.4
Cement +flyash = 140/0.4 =350 kgm3
In table 5 of IS 456-2000, the cement content for severe exposure =320 kgm3
350 kgm3 > 320kgm3 Hence OK.
In this increase of 10% material content is considered.
Cementatious material content = 350*1.1 =385kgm3
Water content = 140kgm3
So, water- cement ratio = 140/385 =0.364
Fly ash at 30% total cementatious material content = 35-30% =115kg/m3
Cement (OPC) =385-115 =270kgm3
Saving of cement while using flyash = 350-270=80kgm3
Fly ash beign utilized = 115kgm3
volume of coarse aggregate and fine aggregate content
From table 3 of IS 1026:2009 ,volume of coarse aggregate corresponding to 20 mm size aggregate and fine aggregate (Zone 1 ) for water- cement ratio =0.6
In present case the water - cement ratio is 0.4
Therefore the volume of coarse aggregate needs to be increased to decrease the fine aggregate content .
As the water- cement ratio is lowered by 0.1 ,the proportion of volume of coarse aggregate is increased by 0.02( at the rate of +/- 0.01 for every +/- 0.05 change in water- cement ratio)
Hence the corrected value ofcoarse aggregate for the water-cemnt ratio = 0.64
( If coarse aggregate is not of angular type, then also the volume of coarse aggregate may be required to be increased suitably ,based on experience)
For pumpable concrete, these are reduced by 10 %
Volume of coarse aggregate = 0.64*0.9 = 0.576
Volume of fine aggregate = 1-0.576 =0.424
Mix calculations
Mix calculation per unit volume of concrete:
a. Volume of concrete = 1 m^3.
b. Volume of cement = (Mass of cement / Specific gravity of cement ) *1/1000
= (270/3.15 )*1/1000 =0.086 m3
c. Volume of flyash = (Mass of flyash / Specific gravity of flyash )*1/1000
= (115/2.2) *1/1000 = 0.052m3
d. Volume of water = ( Mass of water / Specific gravity of water)*1/1000
= (140/1)*1/1000 =0.140 m3
e. Volume of chemical admixture = ( Mass of admixture / specific gravity of admixture)*1/1000
=( 7.6/1.14)*1/1000 =0.006 m3
f. Volume of aggregate ( Fine and coarse) = a- (b+c+d+e)
= 1-(0.086+0.052+0.140+0.006) =0.716 m3
g. Mass of coarse aggregate = f* Volume of coarse aggregate * Specific gravity of coarse aggregate *1000
= 0.716*0.576*2.74*1000 =1130kgm3
h. Mass of fine aggregate = f* Volume of fine aggregate* Specific gravity of fine aggreate*1000
= 0.716*0.424*2.74*1000= 831kgm3
Mix proportion
Cement = 270 kgm
Fly ash = 115 kgm3
Water = 140kgm3
Fine aggregate = 831 kgm3
Coarse aggregate = 1130 kgm3
Chemical admixture = 7kgm3
Water- cement ratio = 0.364
Mix ratio 1: 2.15: 2.94
Mix design of M 50 grade concrete
Proportions:
a. Grade designation : M50
b. Type of cement : OPC grade 43 conforming to IS 8112
c. Maximum nominal sizeof aggregate: 20mm
d. Minimum cement content : 320 kgm3
e. Maximum water-cement ratio : 0.45 (Table :5 IS 456-2000,based on exposure condition)
f. Workability :100mm ,slump value of pumpable concrete
g. Exposure condition : Severe ( for reinforced concrete)
h. Method of concrete placing : Pumping
i. Degree of supervision : Good
j. Type of aggregate: Crushed angular aggregate
k. Maximum cement content : 450 kg/m^3.
l. Chemical admixture type : Superplaticizer.
Data
a. Cement used : OPC 43 grade confirming to IS 8112
b. Specific gravity of cement : 3.15
c. Chemical admixture : Superplaticizer conforming to IS 9103
d. Specific gravity of :
Coarse aggregate : 2.74
Fine aggregate : 2.74
Target mean strength
Target mean strength , f'ck =fck +1.65 s
= 50+1.65*5 =58.25 Nmm2
fck = Characteristic strength
s=standard deviation
water cement ratio
From Table :5 of IS 456-2000, by exposure condition
Here severe condition is considered , Maximum water-cement ratio=0.45
Based on experience water- cement ratio =0.4
0.4 <0.45 ,Hence OK.
water content
Maximum water content for 20mm aggregate =186 litres ( Table 2 of IS 1026:2009) for 25mm to 50mm slump range
For 100 mm slump ,= 186+(6/100)*186 =197 litres
As superplaticizer is used, the water content is reduced upto 30%
Based on trials water content reduction of 29% is achieved with plasticizer
Hence arrived water content = 197*0.71 = 140 litres
Calculation of cement content
Water - cement ratio = 0.4
Cement content = 140/0.4 =350 kgm3
In table 5 of IS 456-2000, the cement content for svere exposure =320 kgm3
350 kgm3 > 320kgm3 Hence OK.
volume of coarse aggregate and fine aggregate content
From table 3 of IS 1026:2009 ,volume of coarse aggregate corresponding to 20 mm size aggregate and fine aggregate (Zone 1 ) for water- cement ratio =0.6
In present case the water - cement ratio is 0.4
Therefore the volume of coarse aggregate needs to be increased to decrease the fine aggregate content .
As the water- cement ratio is lowered by 0.1 ,the proportion of volume of coarse aggregate is increased by 0.02( at the rate of +/- 0.01 for every +/- 0.05 change in water- cement ratio)
Hence the corrected value of coarse aggregate for the water-cemnt ratio = 0.64
( If coarse aggregate is not of angular type, then also the volume of coarse aggregate may be required to be increased suitably ,based on experience)
For pumpable concrete, these are reduced by 10 %
Volume of coarse aggregate = 0.64*0.9 = 0.576
Volume of fine aggregate = 1-0.576 =0.424
Mix calculation
For unit volume of concrete
a. Volume of concrete = 1 m3
b. Volume of cement = ( Mass of ceent / Specific gravity of cement ) *1/1000
=(350/3.15)*1/1000 =0.111 m3
c. Volume of water = (Mass of water/ Specific gravity of water )*1/1000
= (140/1)*1/1000 =0.140 m3
d. Volume of chemical admixture = (Mass of admixture/ecific gravity of admixture)*1/1000
= (7.6/1.145)*1/1000 =0.006m3
e. Volume of agggregate (fine adn coarse) = a-(b+c+d)
=1-( 0.111+0.140+0.006) =0.743m3
f. Mass of coarse aggregate = ex Volume of coarse aggregate x Specific gravity of coarse aggregate x 1000
=0.743*0.576*2.74*1000= 1173kgm3
g. Mass of fine aggregate =e* Volume of fine aggregate* Specific gravity of fine aggregate *1000
= 0.743*0.424*2.74*1000 =863 kgm3
Mix proportion
Cement = 350 kgm3
Water = 140 kg/m
Fine aggregate= 863 kgm3
Coarse aggregate = 1173 kgm3
Chemical admixture = 7 kgm
Water cement ratio =0.4
Mix ratio =1: 2.46: 3.35