Project on Concrete Mix Design for various grades of Concrete
1.Calculate the Concrete Mix Design for M35 grade concrete with fly ash & M50 grade concrete without Fly ash?
Answer - Calculate the Concrete Mix Design for M35 grade concrete with fly ash :-The strength of concrete may vary depending on the quality of the material, the mixing method and the construction procedures. Therefore the target strength (Fck) must be greater than the required strength of the concrete (fck).
Table 1 of IS456 gives the standard deviation value (s) for different grade concretes.
| Fck | = fck+1.65s | “S” from Table 1- IS10262 |
| Fck | = 35+1.65×5 | |
| Fck | = 43.25 N/mm2 |
SELECTION OF WATER-CEMENT RATIO
The maximum water-cement ratios for different grades of concrete are given in Table 5-IS456. You can reduce the water-cement ratio based on your needs.
The water-cement ratio directly affects the strength of the concrete.
| Maximum W/C ratio | =0.45 | TABLE 5-IS456 |
| Adopted W/C ratio | =0.45 |
SELECTION OF WATER CONTENT
| Maximum Water Content | =186 L | TABLE 2- IS1026 |
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 L |
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 SP | =198 – (0.20*198) |
| =157.73 | |
| Adopted Water Content |
=160 L |
| Adopted Water Cement Ratio | =0.45 |
| Total Cementitious (Cement +Fly ash) requirement | = Water Content / WaterCement ratio |
| =160/0.45 | |
| =355.55 kg |
Since flying ash is used, the amount of cementitious content (cement + fly ash) should be increased to get good performance and strength. The percentage of increasing cementitious content depends on experience and research. Here we have increased the cementitious content by 10%.
| Final Cementitious Content (Cement+Fly ash) |
= 356*1.10 | |
| =391.6 kg | ||
| Minimum Cement Content required for ‘severe’ exposure |
=340kg | Table 5-IS456 |
| Adopted Cementious content | =391.6 kg | > 340kg |
| Fly ash content requirement | = 391.6 * 0.30 | 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 AND FINE AGGREGATE
| Coarse Aggregate ratio | = 0.62 | TABLE3-IS10262 (For W/C =0.5) |
| Coarse Aggregate ratio for 0.45 w/c ratio |
= 0.63 | (adding 0.01 for every decrease .05 and vice versa) |
| Final Fine Aggregate ratio | = 0.37 | (1-0.63) = 0.37 |
MIX CALCULATIONS OF M35 GRADE CONCRETE
| A | Volume of Concrete | = 1 cu.m |
| B | Total Volume of Cement | = Cement/(S.G*1000) |
| = 275/(3.15*1000) | ||
| = 0.087 cu.m | ||
| C | Total Volume of Fly ash | = Flyash/(S.G*1000) |
| = 118/(2.5*1000) | ||
| = 0.041 cu.m | ||
| D | Volume of Water | =Water /(S.G*1000) |
| = 160/(1*1000) | ||
| = 0.160 cu.m | ||
| E | Volume of Superplasticizers | = Superplasticizers /(S.G*1000) |
| = 2.35/(1.12*1000) | ||
| = 0.002 cu.m | ||
| F | Total Aggregates requirement | = A – (B+C+D+E) |
| =1 -(0.087 +0.041+0.160+0.002) | ||
| =0.71 cu.m | ||
| G | Coarse Aggregate (C.A) | = F * C.A ratio * S.G * 1000 |
| = 0.71*0.63 *2.74 *1000 | ||
| = 1225.6 kg | ||
| I | Fine Aggregate (F.A) | = F * F.A ratio * S.G * 1000 |
| = 0.71*0.37*2.74 *1000 | ||
| =719.8 kg |
Cement Flyash Sand Aggregate Water and Superplasticizer needed for M35 grade concrete
| 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 |
To find the design mix ratio, divide the calculated value of all materials by the weight of cement. Therefore Mix Ratio of M35 Grade concrete by weight is Cement : Flyash: F.A : C.A : Water = 1:0.4:2.6:4.46:0.58
*. Calculate the Concrete Mix Design for M50 grade concrete :-
The mix design M-50 grade (Using Admixture –Sikament) provided here is for reference purpose only. Actual site conditions vary and thus this should be adjusted as per the location and other factors.
Parameters for mix design M50
Grade Designation = M-50
Type of cement = O.P.C-43 grade
Brand of cement = Vikram ( Grasim )
Admixture = Sika [Sikament 170 ( H ) ]
Fine Aggregate = Zone-II
Sp. Gravity
Cement = 3.15
Fine Aggregate = 2.61
Coarse Aggregate (20mm) = 2.65
Coarse Aggregate (10mm) = 2.66
Minimum Cement (As per contract) =400 kg / m3
Maximum water cement ratio (As per contract) = 0.45
Mix Calculation: –
1. Target Mean Strength = 50 + ( 5 X 1.65 ) = 58.25 Mpa
2. Selection of water cement ratio:-
Assume water cement ratio = 0.35
3. Calculation of water: –
Approximate water content for 20mm max. Size of aggregate = 180 kg /m3 (As per Table No. 5 , IS : 10262 ). As plasticizer is proposed we can reduce water content by 20%.
Now water content = 180 X 0.8 = 144 kg /m3
4. Calculation of cement content:-
Water cement ratio = 0.35
Water content per cum of concrete = 144 kg
Cement content = 144/0.35 = 411.4 kg / m3
Say cement content = 412 kg / m3 (As per contract Minimum cement content 400 kg / m3 )
Hence O.K.
5. Calculation for C.A. & F.A.: [ Formula’s can be seen in earlier posts]-
Volume of concrete = 1 m3
Volume of cement = 412 / ( 3.15 X 1000 ) = 0.1308 m3
Volume of water = 144 / ( 1 X 1000 ) = 0.1440 m3
Volume of Admixture = 4.994 / (1.145 X 1000 ) = 0.0043 m3
Total weight of other materials except coarse aggregate = 0.1308 + 0.1440 +0.0043 = 0.2791 m3
Volume of coarse and fine aggregate = 1 – 0.2791 = 0.7209 m3
Volume of F.A. = 0.7209 X 0.33 = 0.2379 m3 (Assuming 33% by volume of total aggregate )
Volume of C.A. = 0.7209 – 0.2379 = 0.4830 m3
Therefore weight of F.A. = 0.2379 X 2.61 X 1000 = 620.919 kg/ m3
Say weight of F.A. = 621 kg/ m3
Therefore weight of C.A. = 0.4830 X 2.655 X 1000 = 1282.365 kg/ m3
Say weight of C.A. = 1284 kg/ m3
Considering 20 mm: 10mm = 0.55: 0.45
20mm = 706 kg .
10mm = 578 kg .
Hence Mix details per m3
Increasing cement, water, admixture by 2.5% for this trial
Cement = 412 X 1.025 = 422 kg
Water = 144 X 1.025 = 147.6 kg
Fine aggregate = 621 kg
Coarse aggregate 20 mm = 706 kg
Coarse aggregate 10 mm = 578 kg
Admixture = 1.2 % by weight of cement = 5.064 kg.
Water: cement: F.A.: C.A. = 0.35: 1: 1.472: 3.043