Project on Concrete Mix Design for various grades of Concrete
1. Aim:- To determine the Mix proportioning of concrete for M30 grade of concrete using flyash and without using Flyash as a part of replacement of OPC
Introduction:- Concrete mix design involves a process of preparation in which mix ingredients creates the required strength and durability for the concrete structure. Because every ingredient in the mix consists of different properties. In this mix proportioning of concrete using flyash for M30 grade of concrete we are using IS code is IS 10652: 2000.
Procedure:- Stipulation:-
A. Grade designation: M35
B. Type of Cement: OPC 43 grade confirming to IS 8112
C. Type of Mineral Admixture: Flyash confirming to IS 3812(part 1)
D. Maximum nominal size of aggregate: 20mm
E. Minimum Cement content : 340 kg/m^3 (from table-5 of IS 456-2000 exposure condition)
F. Maximum water cement ratio: 0.45(from table 5 of IS 456-2000 exposure condition)
G. Workability: 100mm (slump value for pumpable concrete)
H. Exposure condition: Very 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: 450Kg/m^3
A. Target Mean strength of concrete :-
Target Mean strength of concrete, f'ck = fck + 1.65 S
Where,
'fck' is the compressive strength of the concrete for 28 days
'S' is the standard deviation, S = 5 N/mm^2
Therefore, Target strength, f'ck = 35 + 1.65x5
= 43.25N/mm^2
B. Selection of Water-Cement ratio:-
From table-5 of IS 456:2000 for severe exposure,
Maximum water-cement ratio = 0.45
Based on experiment, adopt water-cement ratio as 0.40
0.40 < 0.45, Hence ok
C. Selection of Water Content:-
From table 2 of IS 10262:2009, Maximum water content for 20mm aggregate = 186 lit/ for 25mm to 50mm
Estimated water content for 100mm slump
= 186+6/100 x 186
= 197 lit.
As superplasticizers is used, water content can be reduced upto 30%.
Based on trails water content reduction of 29% is achieved with plasticizers.
Hence, the arrived water content = 197x0.71
= 140 lit
Calculation of Cement Content:-
water-cement ratio = 0.4
Cementious Material (Cement + Flyash) content = 140/0.4
= 350 kg/m^3
From table-5 of IS 456-2000 minimum cement content for very servere condition = 340kg/m^3
350 kg/m^3 > 340kg/m^3
Hence ok.
Now to proportion a mix containing fly ash, following steps are suggested:-
a. The percentage of flyash needs to be decided based on project requirement and quality of materials.
b. In some cases increase by cementeous material content be warranted. Increase in cementeous material content and its percentage may be based on experience and trials (This example is with increase of 10% cementeous material content).
cementeous material content = 350x1.10 = 385kg/m^3
Water content = 140kg/m^3
so, water cement ratio = 140/385
= 0.364
Fly ash @ 30% total cementeous material content = 385 x 30% = 115kg/m^3
Cement(OPC) = 385 -115 = 270 kg/m^3
Saving of cement while using flyash = 350-270 = 80kg/m^3
Flyash being utilized = 115kg/m^3
Proportion of Volume of Coarse aggregate and fine aggregate:-
from table-3 of IS 10262:2009, Volume of coarse aggregate corresponding to 20mm size aggregate and fine aggregate (zone1) for water cement ratio of 0.5 = 0.6
In the present case of W/c ratio is 0.40
Therefore, the volume of coarse aggregate needs to be increased to decrease the fine aggregate content.
As, w/c ratio is lower by 0.10, the proportion of volume of coarse aggregate is increased by 0.02
Therefore, Corrected volume of coarse aggregate for w/c ratio of 0.40 = 0.62
For pumpable concrete, these values are reduced by 10%.
Therefore, volume of coarse aggregate = 0.62 x 0.9 = 0.56
Volume of fine aggregate = 1-0.56
= 0.44
Mix calculation:-
a. Volume of concrete = 1m^3
b. volume of cement = mass of cement / specific gravity of cement x 1/1000
= 270/3.15 x 1/1000
= 0.086m^3
c. Volume of Flyash = 115/2.2 x 1/1000
= 0.052m^3
d. Volume of water = Mass of water/ specific gravity of water x 1/1000
= 140/1 x 1/1000
= 0.140 m^3
e. Volume of chemical admixture = 7.6/1.145 x 1/1000
= 0.006m^3
f. Volume of aggregate (FA & CA) = (a -(b+c+d+e))
= (1-(0.086 + 0.52 + 0.40 + 0.006))
g. Mass of CA = f x volume of CA x specific gravity of CA x 1000
= 0.716 x 0.56 x 2.74 x 1000
= 1098 kg/m^3
h. Mass of FA = f x volume of FA x specific gravity of FA x 1000
= 0.716 x 0.44 x 2.74 x 1000
= 863kg/m^3
Mix Proportions for trial Number :-
Cement = 270 kg/m^3
Fly ash = 115 kg/m^3
water = 140 kg/m^3
FA = 863 kg/m^3
CA = 1098 kg/m^3
Chemical admixture = 7kg/m^3
water/cement ratio = 0.364
Trail mix ratio = 1 : 2.4 :2.85
2. Aim:- To determine the Mix proportioning of concrete for M30 grade of concrete without using Flyash as a part of replacement of OPC
Introduction:- Here instead of flyash, we are using super plasticizer. Super plasticizer adiitives are used in making the high strength of concrete. Plsticizers are chemical compounds that enable the production of concrete with approximately 15% less water content.
Procedure:-
I. Stipulations for proportioning:-
a. Grade designation: M50
b. Type of cement: OPC 43 grade confirming to IS 8112
c. Maximum nominal size of aggregate: 20mm
d. Minimum Cement content: 360kg/m^3
e. Maximum water-cement ratio: 0.45
f. Workability: 100 slump(Slump value for 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: 450kg/m^3
l. Chemical admixture type: Superplasticizer
II. Test data for materials:
a. Cement used: OPC 43 grade ocnfirming to IS 8112
b. Specific gravity of cement: 3.15
c. Chemical admixture: Super plasticizer confirming to IS 9103
d. Specific gravity of
- Coarse aggregate: 2.74
- Fine aggregate : 2.74
e. Water absorption:
- Coarse aggregate: 0.5%
- Fine aggregate : 1.0%
f. Free surface moisture:
- Coarse aggregate: Nil
- Fine aggregate : Nil
1. Target Mean strength of Concrete(f'ck):-
Here, target mean strength of concrete, f'ck = fck + 1.65S
where,
fck is the characteristics compressive strength ath 28days
'S' is the standard deviation
From table 1 of IS 10262:2009,
Standard deviation, S = 5N/mm^2
Therefore, Target strength = 50+ 1.65 X 5
= 58.25N/mm^2
Selection of Water Cement ratio:
From table 5 of IS 456:2000 for severe exposure,
Maximum water-cement ratio = 0.45
Based on experience, adopted water-cement ratio is 0.40
0.40 < 0.45, Hence ok.
Selection of water content:
From table 2 of IS 10262: 2009, Maximum water content for 20mm aggregate = 186 lit(for 25mm to 50mm slump range)
Estimated water content for 100 mm slump = 186 + 6/100 X 186
= 197 lit
As superplasticizer is used, the water content can be reduced up to 20% and above
Based on trials water content reduction of 29% is achieved with plasticizer.
Hence the arrived water content = 197 X 0.71 = 140lit
Calculation of cement content:
Water cement ratio = 0.4
Cement content = 140/0.4
= 350kg/m^3
from table 5 of IS 456 minimum cement content for "Severe" condition = 320kg/m^3
350 kg/m^3 > 320 kg/m^3, Hence Ok
Proportion of Volume of Coarse aggregate and Fine aggregate content:
From Table 3 of IS 10262:2009, volume of coarse aggregate corresponding to 20mm size aggregate and fine aggregate(Zone 1) for water cement ratio = 0.6
In the present case, the water-cement ratio is 0.40
Therefore, the volume of coarse aggregate need to be increased to decrease the fine aggregate content.
As water-cement ration is lower by 0.10, the proportion of volume of coarse aggregate is increased by 0.02 (at the rate of -/+0.01 every +_ 0.05 change in water-cement ratio)
Therefore, Corrected volume of Coarse aggregate for water cement ratio of 0.40 = 0.62
For pumpable concrete these values are reduced by 10%
Therefore, volume of coarse aggregate = 0.62 X 0.9
= 0.56
Volume of fine aggregate content = 1-0.56 = 0.44
Mix Calculation:
The mix calculation per unit volume of concrete shall be as follows:
a. volume of concrete = 1.^3
b. Volume of cement = Mass of cement / specific gravity of cement X 1/1000
= 350/3.15 X 1/1000
= 0.111 m^3
c. Volume of water = Mass of water / Specific gravity of water X 1/1000
= 140/1 X 1/1000
= 0.140 m^3
d. Volume of Chemical admixture = Mass of chemical admixture / specific gravity of admixture X 1/1000
= 7/1.145 X 1/1000
= 0.006m^3
e. Volume of aggregate(FA & CA) = (a - (b+c+d))
= (1-(0.111+0.140+0.0006))
= 0.0006m^3
f. mass of coarse aggregate = e x volume of Coarse aggregate x specific gravity of coarse aggregate X 1000
= 0.743 X 0.56 X 2.74 X 1000
= 1140 KG/M^3
g. Mass of fine aggregate1 = e X volume of fine aggregate x specific gravity of fine aggregate x 1000
= 0.743 x 0.44 x 2.74 x 1000
= 896 kg/m^3
Mix Proportion:
Cement = 350 kg/m^3
Water = 140 kg/m^3
Fine aggregate = 896 kg/m^3
Coarse aggregate = 1140 kg/m^3
Chemical admixture = 7kg/m^3
water-cement ratio = 0.4
Mix ratio = 1:2.56:3.26