Project on Concrete Mix Design for various grades of Concrete
The Concrete Mix Design for M35 grade concrete with fly ash.
TARGET STRENGTH
Fck=fck+1.64s from table1-IS10262
Fck=43.25N/mm2
SELECTION OF WATER-CEMENT RATIO
Adopted W/C ratio=.45 from table5-IS10262
SELECTION OF WATER CONTENT
Maximum Water Content=186 L from clause 4.2 of IS1026 recommends increasing the water content by 3% to obtain an additional 25 mm slump. To obtain a slump of 50 mm in addition to our concrete, the water content must be increased by 6%.
Therefore, Water-content for 100mm Slump=186+(0.06*186) =197.16l
for use of super plasticizers we can reduce the water content to gain strength in the concrete.
Percentage of Reduction in water content =20%
Water content requirement after super plasticizers=198 – (0.20*198)=157.73
Therefore,Adopted Water Content=106l
CALCULATION OF CEMENT AND FLY ASH CONTENT
Adopted Water Cement Ratio=.45
Total Cementitious (Cement +Fly ash) requirement= Water Content / WaterCement ratio=160/0.45= 355.55kg
for use of flying ash the amount of cementitious content should be increased to get good performance and strength. Here we increase the cementitious content by 10%.
Final Cementitious Content (Cement+Fly ash)= 356*1.10 =391.6kg
From table 5 of IS 456 Minimum Cement Content
required for ‘severe’ exposure=340KG
SO Adopted Cementious content=391.6 kg
FOR 30% replacement
Fly ash content requirement=391.6*0.30=117.48=118KG
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
From table 3 of is 10262 forw/c =0.5 Coarse Aggregate ratio=0.62
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=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 | Based SuperPlasticizer | 2 L |
The Concrete Mix Design for M50 grade concrete without fly ash.
1.Target Strength for Mix Proportioning
ftarget = fck + 1.65 x s
from table 1 of IS:10262-2009 ,
standard deviation, s = 5.0 N/mm2
Therefore, target strength = 50 + 1.65 x 5.0 = 58.25 N/mm2
2.Selection of Water/Cement Ratio
Table 5 of IS:456:2000
Maximum water-cement ratio = 0.35
3.Selection of Water Content
From Table 2 of IS:456:2000, max water content =186 litres [for 25mm to 50mm slump range] for 20 mm aggregate.
Water content for 100 mm slump =186+186x[6/100] =197 ltrs.
Based on trials with Super plasticizer water content reduction of 27% has been achieved, Hence the arrived water content = 197 – [197 x (27/100)] = 143 ltrs.
4.Calculation of Cement Content
W/c ratio = 0.35
Cement content = 143/0.35 = 410 kg/m3
From Table 5 of IS:456:2000, min cement content for ‘Extreme exposure’ condition = 360 kg/m3
410 kg/m3 > 360 kg/m3, hence, O.K.
5. Proportion of Volume of Coarse Aggregate and Fine Aggregate Content
From Table 3 of IS:10262-2009 vol of coarse aggregate corresponding to 20 mm size aggregate and fine aggregate [Zone II] for a water-cement ratio of 0.50 =0.62.
Here, the water-cement ratio is 0.35. Therefore, the vol of CA is required to be increased to decrease the FA content. As the w/c ratio is lower by 0.10. the proportion of the vol of CA is increased by 0.02 [at the rate of -/+ 0.01 for every ± 0.05 change in the W/C ratio].
Therefore, the corrected proportion of the vol 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, vol of coarse aggregate = 0.65 x 0.9 = 0.585.
Vol of fine aggregate content =1–0.585 =0.415
6. Mix Calculations
a) Vol of concrete = 1m3
b) Vol of cement = (Mass of Cement/ Its Specific Gravity ) X (1/1000)
= (410/3.12) X (1/1000)= 0.131 m3
c) Vol of Water = (Mass of water/specific gravity of water) X (1/1000)
= (143/1) X (1/1000)= 0.143 m3
d) Vol of Superplasticizer = (Mass of Superplasticizer / Sp. gr. of admixture) X (1/1000)
= (8.2/1.145) X (1/1000)= 0.007 m3
e) Vol of all aggregate = (a – (b + c + d))
= 1 – (0.131 + 0.143+0.007)= 0.719 m3
e) Mass of CA = Vol of All Aggregate X Vol of CA X Sp.gr. of CA X 1000 = 0.719 x 0.585 x 2.74 x 1000= 1152.48 kg
f) Mass of FA = Vol of all aggregate X vol of FA X Sp. gr of FA X 1000= 0.719 x 0.415 x 2.56 x 1000= 763.86 kg
7. Concrete Mix Proportions
- 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