Project on Concrete Mix Design for various grades of Concrete
Aim: to determine the mix proportion of M30 Grade concrete using fly ash.
1] Stipulation for proportioning
a. Grade of desigination= M35
b. Type of cement= OPC43
c.Type of mineral admixture= Fly ash
d.Maximum nominal size of aggregate= 20mm
e.Minimum cement content= 320kg/m^2
f.Maximum W/C ratio=0.45
g.Workability=100mm Slump
h.Exposure condition= severe
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
m.Chemical admixture type= Superplasticizer
2] TEST DATA FOR MATERIAL:
a.Cement used= OPC 43 Grade
b.The specific gravity of cement=3.15
c.Fly ash= Conformation to IS 3812 part 1
d. The specific gravity of fly ash= 2.2
e.Chemical Admixture= Superplasticizer
f. Specific gravity of
CA= 2.74
FA= 2.74
g.Water absorption
CA=0.5%
FA=1.0%
h. Free surface moisture
CA=NIL
FA=NIL
3] TARGET MEAN STRENGTH FOR MIX PROPORTION
F'ck= fck+1.65xS
From Table 2 of IS 10262:2019
Standard deviation, S=5 N/mm^2
Therefore, Target mean strength = 35+1.65x5
=43.25 N/mm^2
4] SELECTION OF WATER-CEMENT RATIO:
From Table of IS 456:2000 for severe exposure, maximum water-cement ratio = 0.45
Based on the experience, Adopt the Water-Cement ratio as 0.40
0.40<0.45, Hence ok
5] SELECTION OF WATER CONTENT:
From table 4 of IS 10262:2019
Maximum water content for 20mm aggregate =186 liters ( For 25 to 50mm slump range )
Estimated water content for 100mm slump= 186+6/100x186
= 197 lites
As Superplasticizer is used, the water content can be reduced by up to 30%. Based on trials water content reduction of 29% is achieved with plasticizer
Hence the arrived water content was 197x0.71=140 liter
6] CALCULATION OF CEMENT CONTENT
W/C ratio= 0.4
Cementitious material ( cement+ fly ash) content = 140/0.4= 350kg/m^3
From Table 5 of IS 456 minimum cement content for severe Exposure conditions = 320kg/m^3
350 kg/m^3> 320kg/m^3
Now to proportion, a mix containing fly ash the following steps are suggested
a. The percentage of fly ash needs to be decided based on the project requirement and the quality of material
b. In some cases increases in cementitious material can be warranted. Increases in cementitious material content and percentage may be based on experience and trial ( This example is with the increase of 10% cementitious material )
Cementitious material content =350x1.10
= 385kg/m^3
Water content= 140 kg/m^3
So, W/C ratio= 140/385= 0.364
Fly ash @ 30%= 385x30%= 115 kg/m^3
Cement(OPC)= 385-115= 270kg/m^3
Saving for cement while using fly ash = 350-270
= 80 kg/m^3
fly ash = 115 kg/m^3
7] PROPORTION OF VOLUME OF COARSE AGGREGATE AND FINE AGGREGATE CONTENT
From Table of IS 10262: 2009, Volume of coarse aggregate corresponding to 20mm size coarse aggregate & fine aggregate
(zone1) for water cement ratio of 0.5=0.6
In the present case, the water-cement ratio is 0.4
-Therefore, the volume of coarse aggregate needs to be increased in order to decrease the volume of fine aggregate
- As the water-cement ratio is lower by 0.10, the proportion volume of coarse aggregate is increased by 0.02 (at the rate of -/+0.01 for every +/-0.05 change in water-cement ratio)
Therefore, the corrected volume of coarse aggregate for the water-cement ratio is 0.4=0.62
For pumpable concrete, these values are reduced by 10%
Therefore, The volume of coarse aggregate = 0.62x0.9=0.56
The volume of Fine aggregate= 1-0.56=0.44
8] MIX CALCULATION:
This mix calculation per unit volume of concrete shall be as follow
a. volume of concrete=1m^3
b. volume od cement = Massof cement /specific Gravity of cement x 1/1000
= 270/3.15x1/1000
= 0.086m^3
c. volume of fly ash = Mass of fly ash/S.G of fly ash x1/1000
= 115/2.2x1/1000
=0.052m^3
d.volume of water = Mass of water/S.G of water x1/1000
= 140/1 x1/1000
=0.140m^3
e. volume of chemical admixture= Mass of chemical admixture/S.G of Chemical admixture x1/1000
= 7.6/1.145x1/1000
=0.006m^3
f.Volume of aggregate( FA& CA)= (a-(b+c+d+e))
=(1-(0.086+0.052+0.140+0.006)
=0.716m^3
g. Mass of coarse aggregate=fx volume of coarse aggregate x S.G of C.Ax 1000
=0.716x0.56x2.74x1000
=1098kg/m^3
h. Mass of fine aggregate=Fx volume of fine aggregate xS.G of fine aggregates 1000
= 0.716x0.44x2.74x1000
=863kg/m^3
9] MIX PROPORTION FOR TRIAL NUMBER
cement =270 kg/m^3
fly ash= 115kg/m^3
water=140 kg/m^3
Fine aggregate=863kg/m^3
coarse aggregate=1098kg/m^3
Chemical admixture=7 kg/m^3
W/c ratio=0.364
trial mix ratio= 1:2.24:2.85
10]The slump shall be measured and the water content and dosage of admixture shall be adjusted for achieving
the required slump based on trial if required. The mix proportions shall be reworked for the actual water
content and checked for durability requirements.
11]Two more trials having variation of ±1Opercent of water-cement ratio in I0 shall be carried out and a
graph between three water-cement ratios and their corresponding strengths shall be plotted to work out the mix
proportions for the given target strength for field trials. However, durability requirements shall be met.
Ami= To determine the mix proportion of M50 Grade
1]
1. Grade designation = M50
2. Type of cement = OPC43 Grade
3. Maximum nominal size of aggregate= 20mm
4. Minimum cement content= 320kg/m^3
5. Maximum W/C ratio= 0.45
6. Workability= 100mm
6. Exposure condition= Sever for reinforcement concrete
7. Method of concrete placing= pumping
8. Degree of supervision= Good
9. Type of aggregate= crushed angular aggregate
10. Maximum cement content=450kg/m^3
11. Chemical Admixture= Superplasticizer
2] TEST DATA FOR MATERIAL:
a.Cement used= OPC 43 Grade
b.The specific gravity of cement=3.15
c.Fly ash= Conformation to IS 3812 part 1
d. The specific gravity of fly ash= 2.2
e.Chemical Admixture= Superplasticizer
f. Specific gravity of
CA= 2.74
FA= 2.74
g.Water absorption
CA=0.5%
FA=1.0%
h. Free surface moisture
CA=NIL
FA=NIL
3] TARGET MEAN STRENGTH FOR MIX PROPORTION
F'ck= fck+1.65xS
From Table 2 of IS 10262:2019
Standard deviation, S=5 N/mm^2
Therefore, Target mean strength = 50+1.65x5
=58.25 N/mm^2
4] SELECTION OF WATER-CEMENT RATIO:
From Table of IS 456:2000 for severe exposure, maximum water-cement ratio = 0.45
Based on the experience, Adopt the Water-Cement ratio as 0.40
0.40<0.45, Hence ok
5] SELECTION OF WATER CONTENT:
From table 4 of IS 10262:2019
Maximum water content for 20mm aggregate =186 liters ( For 25 to 50mm slump range )
Estimated water content for 100mm slump= 186+6/100x186
= 197 lites
As Superplasticizer is used, the water content can be reduced by up to 30%. Based on trials water content reduction of 29% is achieved with plasticizer
Hence the arrived water content was 197x0.71=140 liter
6] CALCULATION OF CEMENT CONTENT
W/C ratio= 0.4
Cement content = 140/0.4= 350kg/m^3
From Table 5 of IS 456 minimum cement content for severe Exposure conditions = 320kg/m^3
350 kg/m^3> 320kg/m^3
7] PROPORTION OF VOLUME OF COARSE AGGREGATE AND FINE AGGREGATE CONTENT
From Table of IS 10262: 2009, Volume of coarse aggregate corresponding to 20mm size coarse aggregate & fine aggregate
(zone1) for water cement ratio of 0.5=0.6
In the present case, the water-cement ratio is 0.4
-Therefore, the volume of coarse aggregate needs to be increased in order to decrease the volume of fine aggregate
- As the water-cement ratio is lower by 0.10, the proportion volume of coarse aggregate is increased by 0.02 (at the rate of -/+0.01 for every +/-0.05 change in water-cement ratio)
Therefore, the corrected volume of coarse aggregate for the water-cement ratio is 0.4=0.62
For pumpable concrete, these values are reduced by 10%
Therefore, The volume of coarse aggregate = 0.62x0.9=0.56
The volume of Fine aggregate= 1-0.56=0.44
8] MIX CALCULATION:
This mix calculation per unit volume of concrete shall be as follow
a. volume of concrete=1m^3
b. volume od cement = Massof cement /specific Gravity of cement x 1/1000
= 350/3.15x1/1000
= 0.111m^3
c.volume of water = Mass of water/S.G of water x1/1000
= 140/1 x1/1000
=0.140m^3
d. volume of chemical admixture= Mass of chemical admixture/S.G of Chemical admixture x1/1000
= 7.6/1.145x1/1000
=0.006m^3
e.Volume of aggregate( FA& CA)= (a-(b+c+d))
=(1-(0.111+0.140+0.006)
=0.743m^3
f. Mass of coarse aggregate=ex volume of coarse aggregate x S.G of C.Ax 1000
=0.743x0.56x2.74x1000
=1140kg/m^3
g. Mass of fine aggregate=ex volume of fine aggregate xS.G of fine aggregates 1000
= 0.743x0.44x2.74x1000
=896kg/m^3
9] MIX PROPORTION FOR TRIAL NUMBER
cement =350 kg/m^3
water=140 kg/m^3
Fine aggregate=896kg/m^3
coarse aggregate=1140kg/m^3
Chemical admixture=7 kg/m^3
W/c ratio=0.0.4
trial mix ratio= 1:2.56:3.26
10]The slump shall be measured and the water content and dosage of admixture shall be adjusted for achieving
the required slump based on trial if required. The mix proportions shall be reworked for the actual water
content and checked for durability requirements.
11]Two more trials having variation of ±1Opercent of water-cement ratio in I0 shall be carried out and a
graph between three water-cement ratios and their corresponding strengths shall be plotted to work out the mix
proportions for the given target strength for field trials. However, durability requirements shall be met