Project on Concrete Mix Design for various grades of Concrete

AIM:

To calculate the Concrete Mix Design for M35 grade Concrete with fly ash

EQUATIONS:

• Target mean strength : f’ck= fck +1.65 s

                                                 f’ck = target mean strength

                                                  fck= compressive strength at 28 days of concrete

                                                  s  =      standard deviation ( IS 10262:2009 , table -1)

• Volume of cement : (mass of cement /specfic gravity of cement )x1 /1000
• Volume of water : ( mass of water /specific gravity of water ) x1 /1000
• Volume of chemical admixtures : ( mass of chemical admixture / specific gravity  of chemical admixture )x 1/1000
• Volume of fly ash : (mass of fly ash/specific gravity of fly ash )x 1/1000
• Volume of concrete: 1 m^3
• Volume of aggregates : ( volume of concrete -( volume of cement +volume of fly ash + volume of water + volume of admixture )
• Mass of coarse aggregates: volume of aggregate x volume of coarse aggregate x specific gravity of coarse aggregate x1000
• Mass of fine aggregate : volume of aggregate x volume of fine aggregate x specific gravity of fine aggregate x1000

CODES FOR CONCRETE MIX DESIGN:

• IS 10262: 2009
• IS 456: 2000
• IS 8112
• IS 9103
• IS 3812 ( part 1)

OBJECTIVES:

• The main objective is to produce concrete as economically as possible
• To achieve desired workability, durability and minimum strength of M35 grade concrete

PROCEDURE AND CALCULATION:

1. TARGET MEAN STRENGTH

               F’ck = fck + 1.65 5

                     fck= 35 N/mm^2  (M35 )

Table 1- IS 10262:2009

s = 5 N/mm^2

f ‘ck = 35 + ( 1.65 x 5)

          =   43.25 N/mm^2

2. SELECTION OF WATER CEMENT RATIO :

Given :

                M35

            Assume:

              Exposure condition : Extreme ( for reinforced concrete)

              Table 5 – IS 456 : 2000

   Therefore,

               Extreme= w/c ratio= 0.40

                 M35  =   w/c ratio= 0.45

   Lower the water cement ratio better the workability

         0.40 < 0.45

    Water cement ratio = 0.40

Minimum cement content = 360 kg/m^3

3. SELECTION OF WATER CONTENT

               Assume :

                   Nominal size of aggregates= 20 mm

               Table 2 -IS 10262: 2009

  Minimum water content = 186 litres or kg

  Slump ranges from 25 to 50 mm

 As superplasticizers (20 % reduced)

= 186 x 80/100 = 148.8    (100-20 = 80)

Water content = 149 litres

4. CALCULATION OF CEMETITIOUS CONTENT ( CEMENT + FLY ASH)

 Water cement ratio = w/c= 0.40

 Water content =   W = 149 kg

 Cementitious content = c=149/0.40 = 372.5 = 370 kg/m^3

 Minimum cement content = 360 kg/m^3

   360< 370 kg/m^3 ( hence okay )

 Increase by 10 %

   370 x 0.10 + 370 = 407  kg/m^3

       W= 149 kg

 Water cement ratio = 149/407 = 0.366

Fly ash at 30 % total cementitious material content

= 407 x 30/100 = 122 kg/m^3

Cement (ordinary Portland cement ) = 407-122 = 285 kg/m^3

Saving of cement while using fly ash = 370 – 285 = 85 kg/ m^3

Fly ash = 122 kg/m^3

5.VOLUME OF COARSE AGGREGATES AND FINE AGGREGATES

Fine aggregates are sieved according to zone 1 of table 4 IS 383

Coarse aggregates are sieved according to table 2 IS 383

According ,

 Table 3 IS 10262: 2009

Nominal size of aggregates = 20 mm

 Fine aggregates ( zone 1)

 Therefore ,

    Volume of coarse aggregate = 0.60

Water cement ratio = 0.4

( +/- 0.01 for every  +/- 0.005) 

  So, volume of coarse aggregate  = 0.62

Volume of fine aggregate = 1-0.62 = 0.38

 6.MIX CALCULATION :

1. Volume of concrete = 1m^3
2. Volume of cement = (285 / 3.15) x1/1000= 0.090 m^3
3. Volume of fly ash = ( 122/2.2) x1/1000=0.055 m^3
4. Volume of water = ( 149/1)x 1/1000=0.149 m^3
5. Volume of admixture = ( 7.6/ 1.145)x1/1000=0.006 m^3
6. Volume of coarse aggregate and fine aggregate = a – ( b + c + d+
   e) = 1-( 0.090 + 0.055 +0.149+ 0.006 )= 0.7 m^3
7. Mass of coarse aggregate = 0.7 x 0.62 x 2.74 x1000 = 1189.16 kg/m^3
8. Mass of fine aggregate = 0.7 x 0.38 x 2.74 x1000 = 728 kg/m^3

RESULTS  AND CONCLUSION :

      TRIAL MIX:

         Cement = 285 kg/m^3

          Fly ash = 122 kg/m^3

          Water =    149 kg

          Coarse aggregate = 1189 kg/m^3

          Fine aggregate = 728 kg/m^3

         Chemical admixture = 7.6 kg/m^3

          Water cement ratio = 0.366

           Trial mix ratio :   285 :728 : 1189    =  1: 2.55 : 4.17  ( C: FA: CA)

• This trial mix ratio should match the workability ,durability, minimum strength of M35 grade concrete .
• If it does not satisfy the above criteria two or more trials for increase of 10 % should be done .

 

AIM:

To calculate M50 grade  without fly ash

EQUATIONS:

• Target mean strength : f’ck= fck +1.65 s

                                                 f’ck = target mean strength

                                                  fck= compressive strength at 28 days of concrete

                                                  s  =      standard deviation ( IS 10262:2009 , table -1)

• Volume of cement : (mass of cement /specific gravity of cement )x1/1000
• Volume of water : ( mass of water/specific gravity of water ) x1 /1000
• Volume of chemical admixtures : ( mass of chemical admixture/ specific gravity of chemical admixture )x 1/1000
• Volume of concrete: 1 m^3
• Volume of aggregates : ( volume of concrete -( volume of cement + volume of water+volume of chemical admixture) )
• Mass of coarse aggregates: volume of aggregate x volume of coarse aggregate x specific gravity of coarse aggregate x1000
• Mass of fine aggregate : volume of aggregate x volume of fine aggregate x specific gravity of fine aggregate x1000

CODES FOR CONCRETE MIX DESIGN:

• IS 10262: 2009
• IS 456: 2000
• IS 8112
• IS 9103
• IS 3812 ( part 1)

OBJECTIVES:

• The main objective is to produce concrete as economically as possible
• To achieve desired workability, durability and minimum strength of M 50 grade concrete

PROCEDURE AND CALCULATION:

1. TARGET MEAN STRENGTH

               F’ck = fck + 1.65 5

                     fck= 50 N/mm^2  (M50 ) 

Table 1- IS 10262:2009

s = 5 N/mm^2

         f ‘ck = 50 + ( 1.65 x 5)

         =   58.25 N/mm^2

2.  SELECTION OF WATER CEMENT RATIO :

             Given :

                M50

            Assume:

              Exposure condition : Extreme ( for reinforced concrete)

              Table 5 – IS 456 : 2000

             Therefore,

               Extreme= w/c ratio= 0.40

                 M50  =   w/c ratio= 0.40

             Lower the water cement ratio better the workability

               0.40

              Water cement ratio = 0.40

               Minimum cement content = 360 kg/m^3

3. SELECTION OF WATER CONTENT

               Assume :

                   Nominal size of aggregates= 20 mm

               Table 2 -IS 10262: 2009

Minimum water content = 186 litres or kg

  Slump ranges from 25 to 50 mm

 As superplasticizers (20 % reduced)

= 186 x 80/100 = 148.8    (100-20 = 80)

Water content = 149 litres

 4.CALCULATION OF CEMENT CONTENT 

 Water cement ratio = w/c= 0.40

 Water content =   W = 149 kg

 Cement  content = c=149/0.40 = 372.5 = 373 kg/m^3

 Minimum cement content = 360 kg/m^3

   360< 373 kg/m^3 ( hence okay )

 5.VOLUME OF COARSE AGGREGATES AND FINE AGGREGATES

Fine aggregates are sieved according to zone 1 of table 4 IS 383

Coarse aggregates are sieved according to table 2 IS 383

 

According ,

 Table 3 IS 10262: 2009

Nominal size of aggregates = 20 mm

 Fine aggregates ( zone 1)

 Therefore ,

    Volume of coarse aggregate = 0.60

Water cement ratio = 0.4

(+/- 0.01 for every  +/- 0.005)

So, volume of coarse aggregate  = 0.62

Volume of fine aggregate = 1-0.62 = 0.38

6.MIX CALCULATION :

1. Volume of concrete = 1m^3
2. Volume of cement = (373/3.15)x1/1000 = 0.118 m^3
3. Volume of water = ( 149/1)x1/1000= 0.149 m^3
4. Volume of admixture = ( 7.6/1.145)x1/1000 = 0.006
5. Volume of coarse aggregate and fine aggregate = a – ( b + c + d) = 1-( 0.118 + 0.149 +  0.006 )= 0.727 m^3
6. Mass of coarse aggregate = 0.727 x 0.62 x 2.74 x1000 = 1235.02 kg/m^3
7. Mass of fine aggregate = 0.727 x 0.38 x 2.74 x1000 = 756.9 kg/m^3

 

RESULTS  AND CONCLUSION :

      TRIAL MIX:

         Cement = 373  kg/m^3

          Water =    149 kg

          Coarse aggregate = 1235.02 kg/m^3

          Fine aggregate = 756.9 kg/m^3

         Chemical admixture = 7.6 kg/m^3

          Water cement ratio = 0.40

           Trial mix ratio :   373 :756.9 : 1235.02    =  1:  2.02:  3.31  ( C: FA: CA)

• This trial mix ratio should match the workability ,durability, minimum strength of M50 grade concrete .
• If it does not satisfy the above criteria two or more trials for increase of 10 % should be done .