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Project on Concrete Mix Design for various grades of Concrete

MMeet Mehta15 Feb 2022

QUESTION 1: Calculate the Concrete Mix Design for M35 grade concrete with fly ash.


1. AIM:

  • To design Concrete Mix Design for M35 grade concrete with fly ash

2. OBJECTIVES OF THE PROJECT:

  • To find out the proportions of each element used to design M35 grade concrete with fly ash.

3. GIVEN DATA:

  • Grade of concrete: M35
  • Type of cement: OPC - 43 grade [assumed]
  • Max. nominal size of aggregates - 20mm [Table 5, IS 456:2000]
  • Minimum cement content - 340 kg/m^3 [Table 5, IS 456:2000]
  • Maximum w/c ratio - 0.45 [Table 5, IS 456:2000]
  • Workability - 100mm as pumpable concrete [assumed]
  • Exposure Condition - Very Severe [assumed]
  • Concrete placing method - Pumping [assumed]
  • Supervision Required - Good
  • Type of aggregates - Crushed Aggregates
  • Chemical Admixture - Super Plasticizer
  • Zone for aggregates - Zone I
  • Specific gravity:
    • Cement - 3.15
    • Coarse aggregates - 2.74
    • Fine aggregates - 2.74
    • Admixture - 1.145
    • Fly ash - 2.2
  • Water Absorption:
    • Coarse aggregates - 0.5%
    • Fine aggregates - 1%
  • Free surface moisture content for both Coarse and Fine aggregates = 0%

4. BODY OF THE CONTENT:

  • Step 1: To find out targetted mean strength.
    • The formula for targetted mean strength(f'ck) can be given as
      f'ck = fck + 1.65(S)
      here, fck = strength to be gained by concrete in 28 days = 35 N/mm^2
      S = Standard of deviation respective to the grade of concrete = 5 [ from, IS 10262:2019 table 1]

    • Therefore f'ck = 43.5 N/mm^2
  • Step 2: To find out the w/c ratio using the IS codes.

    1. To select the w/c ratio using Table-5 in IS 456:2000

      • From table 5 we can conclude that the w/c ratio for using M35 grade concrete in very severe conditions can be given as 0.45

    2. To find the w/c ratio using the IS 10262:2019

      • By using the graph given below we find the value for the w/c ratio as 0.38




    3. Final answer: The smaller of the values from " 1. & 2. " is considered

      • Thus, assuming the w/c ratio to be 0.38

  • Step 3: Selection of water content.

    • From Table-4 of IS 10262:2019, we can determine that for slump value of 25mm to 50mm, the water content is 186kg. If the slump value is more than 50mm then for every 25mm slump 3% of total water content is added.
    • The slump value for the mix we are designing is 100mm.
    • Therefore, adding 6% to the total water content.
    • Total water content = 186 + ( 0.06 * 186 )
      = 197kg
    • Thus the water content can be given as 197kg.
    • Now as there is an admixture, i.e., Superplasticizer being used in this mix, there will be a reduction in water content by 29%.
    • Therefore the final water content can be given as:
      = 197 - (0.29 * 197)
      = 139.87kg
      ~ 140kg

  • Step 4: To calculate the cement & fly ash content

    • w/c ratio = 0.33
    • water content = W = 140kg
    • Therefore, cement and fly ash content (C) = W/(w/c)
      = 140/0.38
      = 368.42kg > 340kg [OK]
    • As fly ash is there, there is a 10% increment in the total weight of cement and fly ash.
    • Therefore, total cementitious content =1.1 * 368.42 = 405.26 kg
    • Now w/c = 140/405.26 = 0.34 < 0.45 [OK]
    • Now fly ash is 30% of the total cementitious material.
    • Therefore, fly ash content = 0.30 * 405.26
      = 121.57kg
    • Thus the cement content for this concrete mix = 405.26 - 121.57
      = 283.69kg
  • Step 5: Volume of Coarse aggregates(C.A.) and Fine aggregates(F.A)
    • In Zone I, the content of C.A. = 0.6 m^3
    • But in the present scenario, the w/c ratio is 0.33.
    • Therefore, the volume of C.A. needs to be increased to decrease the F.A. content in comparison with C.A.
    • There will be an increment of 0.02 for every 0.10 less content.
    • Therefore, the volume for C.A. for w/c ratio of 0.33 = 0.65 m^3
    • Now as mentioned in the data above, the concrete to be used is pumpable. Therefore there will be a reduction of 10% in the C.A.
    • The new content of C.A. = 0.65 * 0.9 = 0.585 m^3
    • Therefore volume content of F.A. = 1 - 0.585 = 0.415 m^3
  • Step 6: Mix calculations
    • Volume of Concrete = 1 m^3
    • Volume of Cement = Mass of cement / ( 1000 * specific gravity of cement ) = 283.69 / ( 1000 * 3.15 ) = 0.09 m^3
    • Volume of Water = Mass of water / ( 1000 * specific gravity of water ) = 140 / ( 1000 * 1 ) = 0.14 m^3
    • Volume of admixtures = Mass of admixture / ( 1000 * specific gravity of admixtute ) = 7.4 / ( 1000 * 1.145 ) = 0.006 m^3
    • Volume of fly ash = Mass of fly ash / ( 1000 * specific gravity of fly ash ) = 121.57 / ( 1000 * 2.2 ) = 0.055 m^3
    • Volume of C.A. & F.A. = 1 - ( 0.09 + 0.14 + 0.006 + 0.055 ) = 0.709 m^3
    • Mass of C.A. = Specific gravity of C.A. * Volume of C.A. * volume of C.A. & F.A. * 1000 = 2.74 * 0.585 * 0.709 * 1000 = 1136.45 kg
    • Mass of F.A. = Specific gravity of F.A. * Volume of F.A. * volume of C.A. & F.A. * 1000 = 2.74 * 0.415 * 0.709 * 1000 = 806.20 kg
  • Step 7: Mix proportions
    • The weights of the materials can be given as
      • Cement = 283.69 kg/m^3
      • Water = 140 kg/m^3
      • F.A. = 1136.45 kg/m^3
      • C.A. = 806.20 kg/m^3
      • Fly ash = 121.57 kg/m^3
      • w/c = 0.38

  • Step 8: Final trial mix ratio
    • Cement : F.A. : C.A. = 1 : 2.84 : 4

5. CONCLUSION:

  • The final mix ratio for the M35 grade concrete can be given as 1 : 2.84 : 4



______________________________________________________________________________________________________________________________________________________

QUESTION 2: Calculate the Concrete Mix Design for M50 grade concrete without fly ash.


1. AIM:

  • To design Concrete Mix Design for M50 grade concrete without fly ash.

2. OBJECTIVES OF THE PROJECT:

  • To find out the proportions of each element used to design M35 grade concrete with fly ash.

3. GIVEN DATA:

  • Grade of concrete: M50
  • Type of cement: OPC - 43 grade [assumed]
  • Max. nominal size of aggregates - 20mm [Table 5, IS 456:2000]
  • Minimum cement content - 340 kg/m^3 [Table 5, IS 456:2000]
  • Maximum w/c ratio - 0.45 [Table 5, IS 456:2000]
  • Workability - 100mm as pumpable concrete [assumed]
  • Exposure Condition - Very Severe [assumed]
  • Concrete placing method - Pumping [assumed]
  • Supervision Required - Good
  • Type of aggregates - Crushed Aggregates
  • Chemical Admixture - Super Plasticizer
  • Zone for aggregates - Zone I
  • Specific gravity:
    • Cement - 3.15
    • Coarse aggregates - 2.74
    • Fine aggregates - 2.74
  • Water Absorption:
    • Coarse aggregates - 0.5%
    • Fine aggregates - 1%
  • Free surface moisture content for both Coarse and Fine aggregates = 0%

4. BODY OF THE CONTENT:

  • Step 1: To find out targetted mean strength.
    • The formula for targetted mean strength(f'ck) can be given as
      f'ck = fck + 1.65(S)
      here, fck = strength to be gained by concrete in 28 days = 50 N/mm^2
      S = Standard of deviation respective to the grade of concrete = 5 [ from, IS 10262:2019 table 1]

    • Therefore f'ck = 58.5 N/mm^2
  • Step 2: To find out the w/c ratio using the IS codes.

    1. To select the w/c ratio using Table-5 in IS 456:2000

      • From table 5 we can conclude that the w/c ratio for using M50 grade concrete in very severe conditions can be given as 0.45

    2. To find the w/c ratio using the IS 10262:2019

      • By using the graph given below we find the value for the w/c ratio as 0.33




    3. Final answer: The smaller of the values from " 1. & 2. " is considered

      • Thus, assuming the w/c ratio to be 0.33

  • Step 3: Selection of water content.

    • From Table-4 of IS 10262:2019, we can determine that for slump value of 25mm to 50mm, the water content is 186kg. If the slump value is more than 50mm then for every 25mm slump 3% of total water content is added.
    • The slump value for the mix we are designing is 100mm.
    • Therefore, adding 6% to the total water content.
    • Total water content = 186 + ( 0.06 * 186 )
      = 197kg
    • Thus the water content can be given as 197kg.
    • Now as there is an admixture, i.e., Superplasticizer being used in this mix, there will be a reduction in water content by 29%.
    • Therefore the final water content can be given as:
      = 197 - (0.29 * 197)
      = 139.87kg
      ~ 140kg

  • Step 4: To calculate the cement content

    • w/c ratio = 0.33
    • water content = W = 140kg
    • Therefore, cement content (C) = W/(w/c)
      = 140/0.37
      = 378.37kg > 340kg [OK]
    • Thus the cement content for this concrete mix = 378.37kg

  • Step 5: Volume of Coarse aggregates(C.A.) and Fine aggregates(F.A)
    • In Zone I, the content of C.A. = 0.6 m^3
    • But in the present scenario, the w/c ratio is 0.33.
    • Therefore, the volume of C.A. needs to be increased to decrease the F.A. content in comparison with C.A.
    • There will be an increment of 0.02 for every 0.10 less content.
    • Therefore, the volume for C.A. for w/c ratio of 0.33 = 0.63 m^3
    • Now as mentioned in the data above, the concrete to be used is pumpable. Therefore there will be a reduction of 10% in the C.A.
    • The new content of C.A. = 0.63 * 0.9 = 0.567 m^3
    • Therefore volume content of F.A. = 1 - 0.567 = 0.433 m^3

  • Step 6: Mix calculations
    • Volume of Concrete = 1 m^3
    • Volume of Cement = Mass of cement / ( 1000 * specific gravity of cement ) = 378.37 / ( 1000 * 3.15 ) = 0.12 m^3
    • Volume of Water = Mass of water / ( 1000 * specific gravity of water ) = 140 / ( 1000 * 1 ) = 0.14 m^3
    • Volume of admixtures = Mass of admixture / ( 1000 * specific gravity of admixtute ) = 7.4 / ( 1000 * 1.145 ) = 0.006 m^3
    • Volume of C.A. & F.A. = 1 - ( 0.12 + 0.14 + 0.006 ) = 0.734 m^3
    • Mass of C.A. = Specific gravity of C.A. * Volume of C.A. * volume of C.A. & F.A. * 1000 = 2.74 * 0.567 * 0.734 * 1000 = 1140.32 kg
    • Mass of F.A. = Specific gravity of F.A. * Volume of F.A. * volume of C.A. & F.A. * 1000 = 2.74 * 0.433 * 0.734 * 1000 = 870.83 kg

  • Step 7: Mix proportions
    • The weights of the materials can be given as
      • Cement = 378.37 kg/m^3
      • Water = 140 kg/m^3
      • F.A. = 1140.32 kg/m^3
      • C.A. = 870.83 kg/m^3
      • w/c = 0.37

  • Step 8: Final trial mix ratio
    • Cement : F.A. : C.A. = 1 : 2.3 : 3

5. CONCLUSION:

  • The final designed mix for the M35 grade of concrete can be given as 1 : 2.3 : 3
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