Project on Concrete Mix Design for various grades of Concrete

                                      Project on Concrete Mix Design for various grades of Concrete

 

   AIM : 

                      To determine the concrete mix design for various grade of concrete. Now I determine the two types grade of            concrete one is M35 and another one is M50 grade. Concrete is a heterogeneous and hardened mass obtained from a            mixture of cement, sand, coarse aggregate and water, in a certain proportion. 

                     

   INTRODUCTION : 

                      Concrete is a material made of cement/cementitious materials, water, and aggregates. It has been used for            many years in Earth's history as a building material for structures such as bridges, houses, dams, highways, sidewalks,
        skyscrapers. Concrete has several properties such as strength and durability and can be cast into any shape, and form          and texture, and be colored for aesthetics.

                      The grade of concrete varies greatly in accordance with the changing proportion of its constituent materials.            The proportion and the ratio, in which the materials should be mixed together to obtain a certain grade of the concrete,
        already been specified by IS 456:2000

                      Has Unfortunately, there are some challenges still faced today in delivering uniform and high quality concrete
        and therefore, concrete technologies continue to be developed to modify and improve both properties of products,
        fresh and hardened concrete.

 

   DATA OBTAINED,

•            

  Data Required	Mix Design = M35	Mix Design = M50
  Grade of concrete	M35 N/mm^2	M50 N/mm^2
  Maximum size of aggregate	20 mm	20 mm
  Type of cement	OPC 43 Grade conforming (IS 8112)	OPC 53 Grade conforming (IS 8112)
  Maximum w/c ratio	0.50 (Table 5 of IS 456:2000)	0.45 (Table 5 of IS 456:2000)
  Workability in terms of slump	100 - 120 mm	100 - 120 mm
  Exposure condition	Moderate (For reinforced concrete)	Severe (For reinforced concrete)
  Method of concrete placing	pumping	pumping
  Type of mineral admixture	Fly ash ( specific gravity 2.2)	-
  Type of aggregate	Crushed angular aggregate	Crushed angular aggregate
  Minimum cement content	300 Kg/m^3 (Table 5 of IS                                          456:2000)	450 Kg/m^3 (Table 5 of IS                                          456:2000)
  Degree of supervision	Good	Good
  Chemical admixture type	Superplasticizer	Superplasticizer

                         

 

  PROCEDURE : 

  Specific gravity :

  1. Specific gravity of cement  3.15

• Coarse aggregate   : 2.74
• Fine aggregate       : 2.74            

  2. Water absorption

• Coarse aggregate  : 0.5%
• Fine aggregate      : 1.0%

  3. Free surface moisture

• Coarse aggregate  : NIL
• Fine aggregate      : NIL

 4. Sieve analysis

• Coarse aggregate  : Sieve analysis as per table 2 of IS 383
• Fine aggregate      : Sieve analysis as per grading zone I of table 4 of IS 383

 5. Specific gravity of fly ash = 2.2 

 

  EQUATIONS USE :

    1. Target mean strength

• fck =fck+1.65s

   Where,

• fck target average compressive strength at 28 days
• fck characteristics compressive strength at 28 days, and
• S standard deviation

    2. Cement content

• Cement content = water content/ water-cement ratio

    3. Volume Calculations

• (a) Volume of concrete
• (b) Volume of cement = [Mass of cement]/ [Specific Gravity of Cement] x 1/1000
• (c) Volume of water = [Mass of water]/ [Specific Gravity of water] x 1/1000
• (d)Volume of chemical admixture = [Mass of admixture]/[Specific Gravity of admixture] x 1/1000
• (e) Volume of all in aggregate =[a-(b+c+d)] 
• (f) Mass of coarse aggregate= e x Volume of Coarse Aggregate x Specific Gravity of Fine Aggregate x 1000
• (g) Mass of fine aggregate= e x Volume of Fine Aggregate x Specific Gravity of Fine Aggregate x 1000

 

  #  PART - 1 :   MIX DESIGN M35

 

  STEP : 1

• Target strength for mix proportion is obtained from

                       f'ck = fck + 1.65 S

                          S = Assumed Standard deviation (Table I : IS 10262 : 2009)

                       f'ck = Target average compressive strength at 28 days

                        fck = Characteristic compressive strength at 28 days

                          S = 5.0 N/mm^2

             Thus,  f'ck = 35 + 1.65 (5) = 43.25 N/mm^2

  STEP : 2

• Selection of water-cement ratio:

             From table 5 of IS 456:2000 for moderate exposure, maximum water-cement ratio = 0.50 

  STEP : 3            

• Selection of water content 

            From table 2 of IS 10262:2009, Maximum water content for 20mm aggregate = 186 liter ( for 25mm to 50mm                                                                                                                                                                            slump )  

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

                                                                          =197 liter 

           Where as, 

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

                  Based on trials with super plasticizer water content reduction of 20% is achieved,

                  Hence the arrived water content = 197 X 0.8 = 157 liters

 STEP : 4

• Calculation of cement content:

                  Water-cement ratio = 0.50

                  Cementitious material (Cement + Fly ash) content = 157/0.50 = 314 kg/m^3

                  From table 5 of IS 456 Minimum cement content for moderate exposure = 314 Kg/m^3

                                            314 kg/m^3 > 300 kg/m^3, Hence ok

  STEP : 5

• Calculation of fly ash content:

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

               a) The percentage of fly ash needs to be decided based on project requirement and quality of materials

               b) In some cases increase in cementitious material can be warranted. Increase in cementitious material content                         and its percentage may be based on experience and trial (This example is with increase of 10% cementitious                         material content)

                                                    

                     Cementitious material content                               = 314 x 1.10 = 345 Kg/mA3

                     Water content                                                      = 157 Kg/m^3

                     So, Water-cement ratio                                         = 157/345 = 0.455

                     Fly ash @ 30% total cementitious material content  = 345 x 30% = 103 Kg/m^3

                     Cement (OPC)                                                      = 345-103 = 242 Kg/m^3   

                     Saving of cement while using fly ash                      = 314-242 = 72 Kg/m^3 and

                     Fly ash being utilized                                            = 103 Kg/m^3

 

                     Proportion of volume of coarse aggregate and fine aggregate:

                     Volume of coarse aggregate and fine aggregate for w/c ratio 0.5 0.6 (Table 3:IS 10262:2009)

                     In this case, w/c ratio is 0.4. Therefore, volume off coarse aggregate is required to be increased to decrease                         the fine aggregate content. As the w/c ratio is reduced by O.1, the proportion of volume of coarse aggregate                         is increased by 0.02 ( at the rate of -/+ 0.01 for every t 0.05 change in w/c ratio).

                     Thus, the corrected proportion of volume of coarse aggregate for w/c ratio of 0.4 0.62.

                     For pumpable concrete, it can be reduced upto 10%.

                     Hence, volume of coarse aggregate = 0.62x0.9 = 0.56.

                     Volume of fine aggregate = 1-0.56 0.44.

  STEP : 6

• Mix calculation  

            The mix calculation per unit volume of concrete shall be as follow:

               a) Volume of concrete = 1 m^3

               b) Volume of cement  = (Mass of cement/Specific gravity of cement) X (1/1000) 

                                               = (242/3.15) X (1/1000)

                                               = 0.0768 m^3 

               c) Volume of fly ash   = (Mass of fly ash/Specific gravity of fly ash) X (1/1000)

                                                                                  Specific gravity of fly ash = 2.2 

                                              = (103/2.2) X (1/1000)

                                              = 0.0468 m^3 

               d) Volume of water   = (Mass of water/Specific gravity of water) X (1/1000)

                                              = (157/1) X (1/1000)

                                              = 0.157 m^3 

               e) Volume of chemical

                   admixture            = (Mass of chemical admixture/Specific gravity of admixture) X (1/1000)

                                              = (7.6/1.145) X (1/1000)

                                              = 0.006 m^3 

              f) Volume of aggregate

                               (FA & CA) = (a-(b+c+d+e))

                                              = (1-(0.076+0.046+0.157+0.006))

                                              = 0.715 m^3   

             g) Mass of coarse aggregate 

                                              = f X Volume of coarse aggregate X Specific gravity of coarse aggregate X 1000

                                              = 0.715 X  0.56 X 2.74 X 1000

                                              = 1097 Kg/m^3  

             h) Mass of fine aggregate

                                              = f X Volume of fine aggregate X  Specific gravity of fine aggregate X 1000

                                              = 0.715 X 0.44 X 2.74 X 1000

                                              = 862 Kg/m^3

     Result :

•  

  Cement	242   Kg/m^3
  Fly ash	103   Kg/m^3
  Water	157   Kg/m^3
  Fine aggregate	862   Kg/m^3
  Coarse aggregate	1097 Kg/m^3
  Chemical admixture	7      Kg/m^3
  Water-cement ratio	0.455
  Trial Mix Ratio	1 : 2.49 : 3.18

 

 Conclusion : 

• The slump shall be measured and the water content and dosage of admixture shall be adjusted for achieving the required slump on trial. The mix proportion shall be reworked for the actual water content and checked for durability requirements
• Two more trials having variation of 10% of water-cement ratio in Clause X shall be carried out and a graph between three water-cement ratios and corresponding strengths shall be plotted to work out the mix proportions for the given target strength for field trials. However, durability shall be met.

 

  #  PART - 2 :   MIX DESIGN M50

 

   STEP : 1

• Target strength for mix proportion is obtained from

                       f'ck = fck + 1.65 S

                          S = Assumed Standard deviation (Table I : IS 10262 : 2009)

                       f'ck = Target average compressive strength at 28 days

                        fck = Characteristic compressive strength at 28 days

                          S = 5.0 N/mm^2

             Thus,  f'ck = 50 + 1.65 (5) = 58.25 N/mm^2

  STEP : 2

• Selection of water-cement ratio:

             From table 5 of IS 456:2000 for moderate exposure, maximum water-cement ratio = 0.45 

             Based on experience adopted water cement ratio = 0.44

             0.44 < 0.45, hence ok

             Adopted w/c ratio = 0.44

  STEP : 3            

• Selection of water content 

            From table 2 of IS 10262:2009, Maximum water content for 20mm aggregate = 186 liter ( for 25mm to 50mm                                                                                                                                                                            slump )  

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

                                                                          =197 liter 

           Where as, 

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

                  Based on trials with super plasticizer water content reduction of 20% is achieved,

                  Hence the arrived water content = 197 X 0.8 = 157 liters

 STEP : 4

• Calculation of cement content:

                  Water-cement ratio = 0.44

                  Cementitious material (Cement + Fly ash) content = 157/0.44 = 359 kg/m^3

                  From table 5 of IS 456 Minimum cement content for severe exposure = 320 Kg/m^3

                                            359 kg/m^3 > 320 kg/m^3, Hence ok

  STEP : 5

                     Proportion of volume of coarse aggregate and fine aggregate:

                     Volume of coarse aggregate and fine aggregate for w/c ratio 0.44 (Table 3:IS 10262:2009)

                     In this case, w/c ratio is 0.4. Therefore, volume off coarse aggregate is required to be increased to decrease                         the fine aggregate content. As the w/c ratio is reduced by O.1, the proportion of volume of coarse aggregate                         is increased by 0.02 ( at the rate of -/+ 0.01 for every t 0.05 change in w/c ratio).

                     Thus, the corrected proportion of volume of coarse aggregate for w/c ratio of 0.44 = 0.632

                     For pumpable concrete, it can be reduced upto 10%.

                     Hence, volume of coarse aggregate = 0.632x0.9 = 0.568

                     Volume of fine aggregate = 1-0.56 = 0.44.

  STEP : 6

• Mix calculation  

            The mix calculation per unit volume of concrete shall be as follow:

               a) Volume of concrete = 1 m^3

               b) Volume of cement  = (Mass of cement/Specific gravity of cement) X (1/1000) 

                                               = (359/3.15) X (1/1000)

                                               = 0.114 m^3 

               c) Volume of water   = (Mass of water/Specific gravity of water) X (1/1000)

                                              = (157/1) X (1/1000)

                                              = 0.157 m^3 

               d) Volume of chemical

                   admixture            = (Mass of chemical admixture/Specific gravity of admixture) X (1/1000)

                                              = (7.6/1.145) X (1/1000)

                                              = 0.006 m^3 

              e) Volume of aggregate

                               (FA & CA) = (a-(b+c+d+e))

                                              = (1-(0.114+0.157+0.006))

                                              = 0.722 m^3   

             f) Mass of coarse aggregate 

                                              = f X Volume of coarse aggregate X Specific gravity of coarse aggregate X 1000

                                              = 0.722 X  0.568 X 2.74 X 1000

                                              = 1123 Kg/m^3  

             g) Mass of fine aggregate

                                              = f X Volume of fine aggregate X  Specific gravity of fine aggregate X 1000

                                              = 0.722 X 0.44 X 2.74 X 1000

                                              = 821 Kg/m^3

     Result :

•  

  Cement	359   Kg/m^3
  Water	157   Kg/m^3
  Fine aggregate	821   Kg/m^3
  Coarse aggregate	1123 Kg/m^3
  Chemical admixture	7.18     Kg/m^3
  Water-cement ratio	0.44
  Trial Mix Ratio	1 : 2.28 : 3.12

 

 Conclusion : 

• The slump shall be measured and the water content and dosage of admixture shall be adjusted for achieving the required slump on trial. The mix proportion shall be reworked for the actual water content and checked for durability requirements
• Two more trials having variation of 10% of water-cement ratio in Clause X shall be carried out and a graph between three water-cement ratios and corresponding strengths shall be plotted to work out the mix proportions for the given target strength for field trials. However, durability shall be met

 RESULT OF M35, M50

• DESIGN MIX	TRIAL MIX RATIO
  M35	1 : 2.49 : 3.18
  M50	1 : 2.28 : 3.12