Challenge

Project on Concrete Mix Design for various grades of Concrete

AIM:- To Calculate the Concrete Mix Design for M35 grade concrete with fly ash & M50 grade concrete without Fly ash.

INTRODUCTION:- The process of selecting suitable ingredients of concrete and determining their relativeamounts with the objective of producing a concrete of the required, strength, durability, and workability as economically as possible, is termed the concrete mix design. In this project we are designing two conrete mixes , M35 grade concrete with fly ash and M50 without Fly ash.

Part 1:- To calculate M35 concrete with flyash.

Step 1:-

Stipulations for proportioning of M35 grade concrete.

  • Garde of concrete                                        = M35                                                                                  
  • Type of cement                                            = OPC 43 
  • Type of admixture                                        = fly ash
  • Maximum nominal size of coarse aggregate    = 20 mm
  • Minimum cement content                              = 340 kg/m^3 (table 5 of IS 456 exposure condition)
  • Maximum cement content                             = 450 kg/m^3 (table 5 of IS 456 exposure condition)
  • Type of aggregate                                         = Crushed angular aggregates 
  • Workabilty                                                    = 100mm
  • Method of concrete placing                            = Pumping
  • Exposure condition                                        = Very severe 
  • Maximum water cement content                     = 0.45 (table 5 of IS 456 exposure condition)
  • Chemical admixture                                       = Super plasticizer

1)  Target strength for mix proportion 

f'ck=fck+1.65S 

where,

f'ck = target strength of concrete after 28 days 

fck = characteristics strength of concrete or grade

S = standard deviation

from table 1 of IS 10262:2009, standard deviation S=5 N/mm^2.

therefore,

f'ck= 35+1.65 (5)

f'ck= 43.25 n/mm^2

2) Selection of water cement ratio 

According to table 5 of Is 456:2000 exposure condition is assumed very severe 

therefore,

   vere severe W/C=0.45

   Minimum cement content = 340 kg/m^3

3) Selection of water content 

 Nominal size of aggregates assunmed are of 20 mm

therefore, 

According to Table 2 of IS 10262:2009 

Minimum water content for 20 mm aggregates for slump range from 25mm to 50mm is = 186 litres

But our slump ranges till 100 mm therefore according to clause 4.2 3% water is to incresed for every 25 mm 

186+(6/100)*186=197 litres 

But, since we are using super plasticizer water content can be reduced by 20%

therefore, 197-(20/100*197)= 157 litres 

Water content= 157 litres 

4) Calculations of Cementious content 

 W/c ratio= 0.45

 therefore, total cementious content = 157/0.45=348 kg/m^3

The cemnetious content is incresed by 10% to attain good strength and workabilty. this increment is done based on trial and previous experiences.

therefore total cemetious content = 348 * 1.10= 382 Kg/m^3

Water cement ratio = 157/382= 0.4 

fly ash content is assumed 30%= 382*0.3 = 114 kg/m^3

cement required = 382-114= 268 Kg/m^3

quantity of cement saved by adding fly ash = 348-268 = 80 Kg/m^3

5) Volume of coarse aggregate and fine aggregate

From table 3 of IS 10262:2009 

Volume of coarse aggregate for 20 mm size in zone 1 of water cement ratio is 0.60

the proportion of volume of coarse aggregate is changed at a rate of +/- 0.01 for every +/- 0.05 change in water cement ratio.

Water cement ratio= 0.45

As the method of concrete placing is pumping 10% is reduced = 0.60 * 0.9= 0.54

Volume of coarse aggregate =0.54

Volume of fine aggregate = 1 - 0.54= 0.46 

6) Estimation of concrete mix calculation 

  1. Volume of concrete = 1 m3
  2. Volume of cement = (Mass of cement / Specific gravity of cement) x (1/1000) = (268/3.15) x (1/1000) = 0.085 m3
  3. volume of fly ash = (mass of fly ash / specific gravity of fly ash ) x (1/1000) = (114/2.2)x (1/1000) = 0.051 m^3 
  4. Volume of water = (Mass of water / Specific gravity of water) x (1/1000) = (157/1) x (1/1000) = 0.157 m3
  5. volume of chemical admixture = (mass of admix/ specific gravity of admix) x (1/1000) = (7.76/1.14) x (1/1000) = 0.006 m^3
  6. Total Volume of Aggregates = (a-(b+c+d+e)) =1- (0.085+0.51+0.157+0.006) = 0.701 m3
  7. Mass of coarse aggregates = f X Volume of Coarse Aggregate X Specific Gravity of Coarse Aggregate X 1000 = 0.701 X 0.54 X 2.70 X 1000 = 1022 kgs/m3
  8. Mass of fine aggregates = f X Volume of Fine Aggregate X Specific Gravity of fine Aggregate X 1000 = 0.701 X 0.46 X 2.70 X 1000 = 870 kgs/m3

7) Final mix proportions 

Cement 268 Kg/m^3
Fly ash  114 kg/m^3
Water content  157 litres 
Fine aggregate  870 kgs/m3
Coarse aggregate  1022 kgs/m3
Chemical admixture  7.76 kg/m3

Trial Mix Ratio = 268:870:1022 = 1:3.24:3.81

  • This trial mix ratio should be durable, have workability and minum strength for M35 concrete .
  • Two more trial having a variation of 10% in water cement ratio and based on that appropriate trial mix is selected. 

Part 2:- To calculate M50 concrete without flyash.

Step 1:-

Stipulations for proportioning of M50 grade concrete.

  • Garde of concrete                                        = M50                                                                                
  • Type of cement                                            = OPC 53
  • Type of admixture                                        = Nil
  • Maximum nominal size of coarse aggregate    = 20 mm
  • Minimum cement content                              = 340 kg/m^3 (table 5 of IS 456 exposure condition)
  • Maximum cement content                             = 450 kg/m^3 (table 5 of IS 456 exposure condition)
  • Type of aggregate                                         = crushed angular aggregates 
  • Workabilty                                                    = 100mm
  • Method of concrete placing                            = Pumping
  • Exposure condition                                        = very severe 
  • Maximum water cement content                     = 0.45 (table 5 of IS 456 exposure condition)
  • chemical admixture                                       = Super plasticizer

1)  Target strength for mix proportion 

f'ck=fck+1.65S 

where,

f'ck = target strength of concrete after 28 days 

fck = characteristics strength of concrete or grade

S = standard deviation

from table 1 of IS 10262:2009, standard deviation S=5 N/mm^2.

therefore,

f'ck= 50+1.65 (5)

f'ck= 58.25 n/mm^2

2) Selection of water cement ratio 

according to table 5 of Is 456:2000 exposure condition is assumed very severe 

therefore,

   vere severe W/C=0.45

   Minimum cement content = 340 kg/m^3

3) Selection of water content 

 Nominal size of aggregates assunmed are of 20 mm

therefore, 

according to Table 2 of IS 10262:2009 

Minimum water content for 20 mm aggregates for slump range from 25mm to 50mm is = 186 litres

but our slump ranges till 100 mm therefore according to clause 4.2 3% water is to incresed for every 25 mm 

186+(6/100)*186=197 litres 

But, since we are using super plasticizer water content can be reduced by 20%

therefore, 197-(20/100*197)= 157 litres 

Water content= 157 litres 

4) Calculations of Cement content 

 W/c ratio= 0.45

 therefore, total cement content = 157/0.45=348 kg/m^3

Cement content = 348 kg/m^3

5) Volume of coarse aggregate and fine aggregate

from table 3 of IS 10262:2009 

volume of coarse aggregate for 20 mm size in zone 1 of water cement ratio is 0.60

the proportion of volume of coarse aggregate is changed at a rate of +/- 0.01 for every +/- 0.05 change in water cement ratio.

water cement ratio= 0.45

as the method of concrete placing is pumping 10% is reduced = 0.60 * 0.9= 0.54

volume of coarse aggregate =0.54

volume of fine aggregate = 1 - 0.54= 0.46 

6) Estimation of concrete mix calculation 

  1. Volume of concrete = 1 m3
  2. Volume of cement = (Mass of cement / Specific gravity of cement) x (1/1000) = (348/3.15) x (1/1000) = 0.110 m3
  3. Volume of water = (Mass of water / Specific gravity of water) x (1/1000) = (157/1) x (1/1000) = 0.157 m3
  4. volume of chemical admixture = (mass of admix/ specific gravity of admix) x (1/1000) = (7.76/1.14) x (1/1000) = 0.006 m^3
  5. Total Volume of Aggregates = (a-(b+c+d)) =1- (0.110+0.157+0.006) = 0.727 m3
  6. Mass of coarse aggregates = e X Volume of Coarse Aggregate X Specific Gravity of Coarse Aggregate X 1000 = 0.727 X 0.54 X 2.70 X 1000 = 1060 kgs/m3
  7. Mass of fine aggregates = e X Volume of Fine Aggregate X Specific Gravity of fine Aggregate X 1000 = 0.727 X 0.46 X 2.70 X 1000 = 903 kgs/m3

7) Final mix proportions 

Cement 348 Kg/m^3
Water content  157 litres 
Fine aggregate  903 kgs/m3
Coarse aggregate  1060 kgs/m3
Chemical admixture  7.76 kg/m3

Trial Mix Ratio = 348:903:1060 = 1:2.59:3.04

  • This trial mix ratio should be durable, have workability and minum strength for M35 concrete .
  • Two more trial having a variation of 10% in water cement ratio and based on that appropriate trial mix is selected. 

RESULTS:-

For M35 grade of concrete with flyash A Mix proportion of 1:3.24:3.81  is to be used with a water cement ratio of 0.4.  

For M50 grade of concrete without flyash A Mix proportion of 1:2.59:3.04 is to be used with a water cement ratio of 0.45.  

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