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Project on Concrete Mix Design for various grades of Concrete

PPriyanka Badiger15 Jan 2023

AIM:The mix design for M50 grade concrete without fly ash

INTRODUCTION:

1. Stipulation for proportioning

  1. Grade designation: M50
  2. Type of cement: OPC 43 grade conforming to IS 8112
  3. Maximum nominal size of aggregate: 20mm
  4. Manimum cement content: 320kg/m^3(From table 5 of IS 456:2019 Exposure condition)
  5. Maximum water-cement ratio: 0.45(from table 5 of 456:2019-exposure condition)
  6. Workability: 100mm (slump value for pumpable concrete)
  7. Exposure condition: Severe(for reinforced concrete)
  8. Method of concrete placing: Pumping
  9. Degree of supervision: Good
  10. Type of aggregate: Crushing angular aggregate
  11. Maximum cement content: 450kg/m^3
  12. Chemical admixture type: Superplasticizer

2.Test data for materials:

  1. cement used: OPC43 grade conforming to IS 8112
  2. Specific gravity of cement: 3.15
  3. chemical admixture: Superplasticizer conforming to IS9103
  4. Specific gravity of: 1)coarse aggregate: 2.74 2)Fine aggregate: 2.74
  5. Water absorption: 1)coarse aggregate: 0.5% 2)Fine aggregate: 1.0%
  6. Free Surface moisture: 1)coarse aggregate:Nill 2)Fine aggregate: Nill

3.Target strength for Mix proportioning:

ft = fck+1.65 S

fck= 50, S=5

Therefore, the target strength,

ft = 50+1.65*5

ft = 58.25N/MM^2

where, ft= Target average compressive strength at 28 days

fck= characteristics compressive strength at 28 days

S= standard deviation

4.Selection of Water-cement ratio

From IS 10262:2019, water-cement ratio as 0.28

5.Selection of water content

from Table 4 of IS 10262:2019, maximum water cement for 20mm aggregate=186lit(for 25-50mm slump range)

Estimated water content for 100mm slump=186+(6/100)*186

= 197litre

As superplasticizer is used, the water content can be reduced up to 30%

Hence the arrieved water content =197* 0.70= 137.9 = 138litre

6.Calculation of cement content:

Water cement ratio = 0.28

cement content = 138/0.28 = 493kg/m^3

from table 5 of IS 456 minmum cement content for 'sever' exposure condition = 320 kg/m^3

493kg/m^3> 320kg/m^3, Hence OK

7.Proportion of volume of coarse aggregate and fine aggregate content:

From table 5 of IS 10262:2019, volume of coarse aggregate corresponding to 20mm size aggregate and fine aggregate (Zone 1) for water-cement ratio is 0.5

in the present case the water cement ratio is =0.28

As water-cement ratio is lower by 0.10, the proportion of 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,corrected volume of coarse aggregate for the water-cement ratio is of 0.28= 0.644

For pumping concrete these value are reduction by 10%

Therefore, volume of coarse aggregate=0.644*0.9=0.58

Volume of fine aggregate content=1-0.58 = 0.42

8.Mix calculation: The mix calculation per unit volume of concrete shall be as follows:

  1. Volume of concrete = 1m^3
  2. volume of cement = ( mass of cement/specific gravity of cement)*(1/1000) =495/3.15*1/1000 = 0.157m^3
  3. volume of water= (mass of water/specific gravity of water)*(1/1000) = (138/1)*1/1000 = 0.138m^3
  4. volume of chemical admixture = (mass of chemical admixture/specific gravity of admixture)*1/1000 (7.6/1.145)*1/1000 = 0.006m^3
  5. volume of aggregate(FA AND CA) = (a-(b+c+d)) (1-(0.157+0.138+0.006)) = 0.699m^3
  6. Mass of coarse aggregate =e*volume of coarse aggregate aggregate * specific gravity of coarse aggregate * 1000 =0.699*0.58*2.74*1000 = 1111kg/m^3
  7. mass of fine aggregate= e*volume of fine aggregate*specific gravity of fine aggregate*1000 =0.699*0.42*2.74*1000 =807kg/m^3

9.Mix proportion for trial number 1:

  1. Cement = 493kg/m^3
  2. Water = 138kg/m^3
  3. Fine aggregate =807kg/m^3
  4. Coarse aggregate =1111kg/m^3
  5. Chemical admixture =0.7kg/m^3
  6. Water-cement ratio =0.28litre

RESULT:

Trial Mix Ratio =1:1.63:2.25

==============================================================================

Q: Formulate the mix design for M35 grade concrete with fly ash

AIM:The mix design for M35grade concrete with fly ash

INTRODUCTION:

1. Stipulation for proportioning

  1. Grade designation: M35
  2. Type of cement: OPC 43 grade conforming to IS 8112
  3. type of mineral admixture: Fly ash
  4. Maximum nominal size of aggregate: 20mm
  5. Manimum cement content: 320kg/m^3(From table 5 of IS 456:2019 Exposure condition)
  6. Maximum water-cement ratio: 0.45(from table 5 of 456:2019-exposure condition)
  7. Workability: 100mm (slump value for pumpable concrete)
  8. Exposure condition: Severe(for reinforced concrete)
  9. Method of concrete placing: Pumping
  10. Degree of supervision: Good
  11. Type of aggregate: Crushing angular aggregate
  12. Maximum cement content: 450kg/m^3
  13. Chemical admixture type: Superplasticizer

2.Test data for materials:

  1. cement used: OPC43 grade conforming to IS 8112
  2. Specific gravity of cement: 3.15
  3. Fly ash: Conforming to IS 3812(Part 1)
  4. Specific gravity of fly ash : 2.2
  5. chemical admixture: Superplasticizer conforming to IS9103
  6. Specific gravity of: 1)coarse aggregate: 2.74 2)Fine aggregate: 2.74
  7. Water absorption: 1)coarse aggregate: 0.5% 2)Fine aggregate: 1.0%
  8. Free Surface moisture: 1)coarse aggregate:Nill 2)Fine aggregate: Nill

3.Target strength for Mix proportioning:

ft = fck+1.65 S

fck= 35, S=5

Therefore, the target strength,

ft = 35+1.65*5

ft = 43.25N/MM^2

where, ft= Target average compressive strength at 28 days

fck= characteristics compressive strength at 28 days

S= standard deviation

4.Selection of Water-cement ratio

From IS 10262:2019, water-cement ratio as 0.39

5.Selection of water content

from Table 4 of IS 10262:2019, maximum water cement for 20mm aggregate=186lit(for 25-50mm slump range)

Estimated water content for 100mm slump=186+(6/100)*186

= 197litre

As superplasticizer is used, the water content can be reduced up to 30%

Hence the arrieved water content =197* 0.70= 137.9 = 138litre

6.Calculation of cement content:

Water cement ratio = 0.39

cementitious material(cement+fly ash) content= 138/0.39 = 345kg/m^3

from table 5 of IS 456 minmum cement content for 'sever' exposure condition = 320 kg/m^3

345kg/m^3> 320kg/m^3, Hence OK

Now to proportion a mix containing fly ash the following steps are suggested:

  1. The percentage of fly ash needs to be decided based on project requirment and quality of materials
  2. In some cases increse in cementitious material can be warranted. increse in cementitious material content and its percentage may be based on experience and trial(this ex is with increse of 10% cementitious material content) cementitious material = 345*1.10=379.5kg/m^3 Fly ash @30% total cementitious content. =345*30%=103.5kg/m^3 Cement(OPC)= 241.5kg/m^3

7.Proportion of volume of coarse aggregate and fine aggregate content:

From table 5 of IS 10262:2019, volume of coarse aggregate corresponding to 20mm size aggregate and fine aggregate (Zone 1) for water-cement ratio is 0.5

in the present case the water cement ratio is =0.39

As water-cement ratio is lower by 0.10, the proportion of 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,corrected volume of coarse aggregate for the water-cement ratio is of 0.39= 0.652

For pumping concrete these value are reduction by 10%

Therefore, volume of coarse aggregate=0.652*0.9=0.59

Volume of fine aggregate content=1-0.58 = 0.41

8.Mix calculation: The mix calculation per unit volume of concrete shall be as follows:

  1. Volume of concrete = 1m^3
  2. volume of cement = ( mass of cement/specific gravity of cement)*(1/1000) =241.5/3.15*1/1000 = 0.077m^3
  3. volume of fly ash= (mass of fly ash/specific gravity of fly ash)*(1/1000) = (103.5/2.2)*1/1000 = 0.047m^3
  4. volume of water= (mass of water/specific gravity of water)*(1/1000) = (138/1)*1/1000 = 0.138m^3
  5. volume of chemical admixture = (mass of chemical admixture/specific gravity of admixture)*1/1000 (7.6/1.145)*1/1000 = 0.006m^3
  6. volume of aggregate(FA AND CA) = (a-(b+c+d+e)) (1-(0.077+0.047+0.138+0.006)) = 0.73m^3
  7. Mass of coarse aggregate =f*volume of coarse aggregate * specific gravity of coarse aggregate * 1000 =0.73*0.59*2.74*1000 = 1180kg/m^3
  8. mass of fine aggregate= f*volume of fine aggregate*specific gravity of fine aggregate*1000 =0.73*0.41*2.74*1000 =820kg/m^3

9.Mix proportion for trial number 1:

  1. Cement = 241.5kg/m^3
  2. Fly ash = 103.5kg/m^3
  3. Water = 138kg/m^3
  4. Fine aggregate = 820kg/m^3
  5. Coarse aggregate = 1180kg/m^3
  6. Chemical admixture = 0.7kg/m^3
  7. Water-cement ratio = 0.39litre

RESULT:

Trial Mix Ratio = 1:2.3:3.5

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