Project on Concrete Mix Design for various grades of Concrete
ANS :-
For M35 Grade
- Type of Cement = OPC
- Type of Admixture = Fly ash
- Maximum Nominal Size of Aggregate = 20mm
- Minimum Cement Content = 340kg ......................IS 456
- Maximum Cement Content = 450kg ......................IS 456
- Partial replacement of cement by Fly ash= 30% ................As per requirement
- Maximum W/C ratio = 0.45
- Workability =100mm
- Water absorption
Coarse aggregate = 0.5%
Fine aggregate = 1% - Sieve Analysis
Coarse aggregate = -
Fine aggregate = Zone 2
- Target strength for mix proportion of M35 grade concrete
Fck = fck+1.65s
Fck = 35+1.65X5
Fck = 43.25 N/mm2 - Selection of water cement ratio
Maximum W/C ratio =0.45
Adopted W/C ratio =0.45 - Selection of water content =186 liters
The water content above is for a slump of 25 to 50 mm, but according to the specification our concrete requires a slump of 100 mm. Therefore, to get the desired slump, the water content must be increased. Clause 4.2 of IS1026 recommends increasing the water content by 3% to obtain an additional 25 mm slump. Therefore, to obtain a slump of 50 mm in addition to our concrete, the water content must be increased by 6%.
Water-content for 100mm Slump = 186+(0.06*186) =197.16 liters
As we use super plasticizers we can reduce the water content to gain strength in the concrete. The amount of water-content to be reduced depends on the type of super plasticizer.
Percentage of Reduction in water content = 20%
Water content requirement after Super Plasticizer = 198-(0.20X198) =157.73 liters
Adopted Water Content = 160 liters - Calculation of cement and fly ash content
Adopted W/C ratio =0.45
Total Cementitious (Cement +Fly ash) requirement = Water Content / WaterCement ratio
= 160/0.45
= 355.55 kg
Here we have increased the cementitious content by 10%.
Final Cementitious Content (Cement+Fly ash) = 356X1.10
= 391.6 kg 0
Minimum Cement Content required for ‘severe’ exposure= 340 kg .............................AS per IS 456
Adopted Cementious content =391.6 kg > 340 kg
Fly ash content requirement = 391.6X0.30 ..................30% replacement
= 117.48~118 kg
Cement Content requirement =Cementious content – Fly ash
=391.6-117.48
=274.12
Calculation of Super Plasticiers @ 0.6 percent to the mass of cementitious content = = 391.6*(0.6/100) = 2.35 kg - Coarse & Fine aggregate
Coarse Aggregate ratio = 0.62 .................For W/c ratio =0.5
Coarse Aggregate ratio for 0.45 w/c ratio = 0.63
Final Fine Aggregate ratio = 0.37 .........................(1-0.63)=0.37 - Calculation of M35 Grade concrete
- Volume of Concrete= 1 m3.
- Total Volume of Cement =Cement/(Specific gravityX1000)
= 275/(3.15X1000) =0.087 m3 - Total Volume of Fly ash =Flyash/(Specific gravityX1000)
=118/(2.5X1000) =0.041 m3 - Volume of Water= Water /(Specific gravityX1000)
=160/(1X1000) = 0.160 m3. - Volume of Superplasticizers = Superplasticizers /(Specific gravityX1000)
=2.35/(1.12X1000) =0.002m3 - Total Aggregates requirement = Volume of Concrete -(Total Volume of Cement+Total Volume of Fly ash+Volume of Water+Volume of Superplasticizers)
=1 -(0.087 +0.041+0.160+0.002) = 0.71 m3 - Coarse Aggregate (C.A)= Total Aggregates requirement X Coarse Aggregate ratio X Specific gravity X 1000
=0.71X0.63X2.74X1000 =1225.6 kg - Fine Aggregate (F.A)= Total Aggregates requirement XFine Aggregate ratio X Specific gravity X 1000
=0.71X0.37X2.74X1000 =719.8 kg
| S.No. | MATERIALS | QUANTITY |
| 1 | Cement | 275 kg |
| 2 | Flyash | 118 kg |
| 3 | Fine Aggregate | 720 kg |
| 4 | Coarse Aggregate (20mm) | 1226 kg |
| 5 | WATER | 160 L |
| 6 | PCE based SuperPlasticizer | 2 L |
For M50 Grade
- Grade designation : M 50
- Type of cement : OPC 53 grade confirming to IS 8112
- Maximum nominal size of aggregate : 20 mm
- Minimum cement content : 360 kg/m3
- Maximum water-cement ratio : 0.35
- Workability : 50-75 mm (slump)
- Exposure condition : Extreme (for reinforced concrete)
- Method of concrete placing : Pumping
- Degree of supervision : Good
- Type of aggregate : Crushed angular aggregate
- Maximum cement content : 450 kg/m3
- Chemical admixture type : Super Plasticizer
Test data for materials
- Specific gravity of cement : 3.12
- Specific gravity of Coarse Aggregate : 2.74
- Specific gravity of Fine Aggregate : 2.56
Procedure for concrete mix design of M50 grade concrete
- Target strength for mix proportioning=ftarget= fck+ 1.65 x Standard Deviation
According to the table 1 of IS: 10262-2009 ,
standard deviation, s = 5.0 N/mm2
Target strength for mix proportioning=ftarget=50 + 1.65 x 5.0 = 58.25 N/mm2
- Selection of water cement ratio
Maxiumum water cement ratio= 0.35 ...................From IS 456 - Selection of water content
Maxiumum water content= 186liter ...................From IS 456 Table no. 2
(for 25 to 50 mm slump range) for 20 mm aggregate.
Maxiumum water content=186+(6/100) x 186 =197 litres.
(for 25 to 50 mm slump range) for 20 mm aggregate.
Based on Super plasticizer water content reduction of 27% has been achieved, Hence the arrived water content = 197 – [197 x (27/100)] = 143 litres. - Calaculation of cement content
Water-cement ratio = 0.35
Cement content = 143/0.35 = 410 kg/m3
From Table 5 of IS 456, minimum cement content for ‘Extreme exposure’ condition = 360 kg/m3
410 kg/m3 > 360 kg/m3, hence, O.K.
- Coarse aggregate and fine aggregate content
From Table 3 of IS 10262. volume of coarse aggregate corresponding to 20 mm size aggregate and fine aggregate (Zone II) for a water-cement ratio of 0.50 =0.62.
In the present case, the water-cement ratio is 0.35. Therefore. the volume of coarse aggregate is required to be increased to decrease the fine aggregate content. As the water-cement ratio is lower by 0.10. the proportion of the volume of coarse aggregate is increased by 0.02 (at the rate of -/+ 0.01 for every ± 0.05 change in the water-cement ratio).
Therefore. the corrected proportion of the volume of coarse aggregate for the water-cement ratio of 0.35 = 0.65.
For pumpable concrete, these values should be reduced by 10 percent.
Therefore, volume of coarse aggregate = 0.65 x 0.9 = 0.585.
Volume of fine aggregate content =1 – 0.585 =0.415
- Mix calculation
-
Volume of concrete = 1m3
Volume of cement = (Mass of Cement/ Its Specific Gravity ) X (1/1000)
= (410/3.12) X (1/1000)
= 0.131 m3
Volume of Water = (Mass of water/specific gravity of water) X (1/1000)
= (143/1) X (1/1000)
= 0.143 m3
Volume of Superplasticizer = (Mass of Superplasticizer / Specific gravity of admixture) X (1/1000)
= (8.2/1.145) X (1/1000)
= 0.007 m3
Volume of all aggregate = ( Volume of concrete-(Volume of cement+Volume of Water+Volume of Superplasticizer)
= 1 – (0.131 + 0.143+0.007)
= 0.719 m3
Mass of coarse aggregate = Volume of All Aggregate X Volume of Coarse Aggregate X Specific Gravity of Coarse Aggregate X 1000
= 0.719 x 0.585 x 2.74 x 1000
= 1152.48 kg
Mass of fine aggregate = Volume of all aggregate X volume of fine aggregate X Specific gravity of fine aggregate X 1000
= 0.719 x 0.415 x 2.56 x 1000
= 763.86 kg
Concrete Mix Proportion
- Cement = 410 kg/m3
- Water = 143 kg/m3
- Super plasticizer = 7 kg/m3
- Fine aggregates = 763.86 kg/m3
- Coarse aggregate = 1152.48 kg/m3
- W/c = 0.35