Project on Concrete Mix Design for various grades of Concrete

Calculate the Concrete Mix Design for M35 grade concrete with fly ash & M50 grade concrete without Fly ash

AIM:

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

OBJECTIVES:

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

Introduction :

Concrete mix design is a procedure of selecting the suitable ingredients of concrete and their relative proportions with an objective to prepare concrete of certain minimum strength, desired workability and durability as economically as possible.

Step By Step Procedure :

Step 1

Target mean strengt:

F'ck = fck +1.65xS

F'ck = Target mean

fck = Compressive strenght at 28 days of concrete

s = standard deviation value for the M35 grade concrete [ given in Table 8 of IS 456]

 fck = 35 N/mm^2  ,  s = 5 N/mm^2 

F'ck = 35 + 1.65 x 5

        = 43.25 N/mm^2 

 Target mean strengt = 43.25 N/mm^2 

step 2

Water Cement ratio :

Grade M35 = w/c ratio= 0.45 [ given in Table 5 of IS 456:2000]

assume Extreme ,

  Extreme = w/c ratio= 0.40

Water cement ratio = 0.40

Minimum cement content = 360 kg/m^3

step 3

Water  Content :

Assume :

Nominal size of aggregates= 20 mm

Minimum water content = 186 litres or kg [ given in Table 2 of IS 10262: 2009]

As superplasticizers (20 % reduced) (100-20 = 80)

= 186 x 80/100 = 148.8

Water content = 149 litres

step 4

 Cementious 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

 Take Minimum cement content = 360 kg/m^3

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 cement content

= 407 x 30/100

Fly ash = 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

step 5

Volume of fine aggregate and coarse aggregate

[Table 3 IS 10262: 2009]

Take Nominal size of aggregates = 20 mm

Fine aggregates ( zone 1)

Volume of coarse aggregate = 0.60

Water cement ratio = 0.4

(Adding 0.01 for every decrease and 0.05 and vice versa)

 volume of coarse aggregate = 0.62

Volume of fine aggregate = 1-0.62

                                         = 0.38

step 5

Mix calculatio :

 a ) Volume of concrete = 1m^3

b ) Volume of cement : (mass of cement /specfic gravity of cement )x1 /1000

Volume of cement = (285 / 3.15) x1/1000

                              = 0.090 m^3

c ) Volume of fly ash : (mass of fly ash/specific gravity of fly ash )x 1/1000

Volume of fly ash = ( 122/2.2) x1/1000

                              =0.055 m^3

d ) Volume of water : ( mass of water /specific gravity of water ) x1 /1000

Volume of water = ( 149/1)x 1/1000

                           =0.149 m^3

e ) Volume of chemical admixtures : ( mass of chemical admixture / specific gravity  of chemical admixture )x 1/1000

Volume of admixture = ( 7.6/ 1.145)x1/1000

                                  =0.006 m^3

f ) Volume of  aggregate= a – ( b + c + d+ e )

                                      = 1-( 0.090 + 0.055 +0.149+ 0.006 )

                                      = 0.7 m^3

g ) Mass of coarse aggregate= volume of aggregate x volume of coarse aggregate x specific gravity of  coarse aggregate x1000

Mass of coarse aggregate = 0.7 x 0.62 x 2.74 x1000

                                          = 1189.16 kg/m^3

h ) Mass of fine aggregate = volume of aggregate x volume of fine aggregate x specific gravity of fine aggregate x1000

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 the Concrete Mix Design for M50 grade Concrete without fly ash

OBJECTIVES:

• The main objective is to produce concrete as economically as possible
• To achieve desired workability, durability and minimum strength of M50 grade concrete without fly ash

Introduction :

Concrete mix design is a procedure of selecting the suitable ingredients of concrete and their relative proportions with an objective to prepare concrete of certain minimum strength, desired workability and durability as economically as possible.

Step By Step Procedure :

Step 1

Target mean strengt:

F'ck = fck +1.65xS

F'ck = Target mean

fck = Compressive strenght at 28 days of concrete

s = standard deviation value for the M35 grade concrete [ given in Table 8 of IS 456]

 fck = 50 N/mm^2  ,  s = 5 N/mm^2 

F'ck = 50 + 1.65 x 5

        = 58.25 N/mm^2 

 Target mean strengt = 58.25 N/mm^2 

step 2

Water Cement ratio :

Grade M50 = w/c ratio= 0.45 [ given in Table 5 of IS 456:2000]

assume Extreme ,

  Extreme = w/c ratio= 0.40

Water cement ratio = 0.40

Minimum cement content = 360 kg/m^3

step 3

Water  Content :

Assume :

Nominal size of aggregates= 20 mm

Minimum water content = 186 litres or kg [ given in Table 2 of IS 10262: 2009]

As superplasticizers (20 % reduced) (100-20 = 80)

= 186 x 80/100 = 148.8

Water content = 149 litres

step 4

 Cementious Content ( Cement + Fly ash)

 Water cement ratio = w/c= 0.40

 Water content =   W = 149 kg

 Cementitious content ,

                                   c=149/0.40

                                  = 372.5

                                  = 373 kg/m^3

 Take Minimum cement content = 360 kg/m^3

step 5

Volume of fine aggregate and coarse aggregate

[Table 3 IS 10262: 2009]

Take Nominal size of aggregates = 20 mm

Fine aggregates ( zone 1)

Volume of coarse aggregate = 0.60

Water cement ratio = 0.4

(Adding 0.01 for every decrease and 0.05 and vice versa)

 volume of coarse aggregate = 0.62

Volume of fine aggregate = 1-0.62

                                         = 0.38

step 5

Mix calculatio :

 a ) Volume of concrete = 1m^3

b ) Volume of cement : (mass of cement /specfic gravity of cement )x1 /1000

Volume of cement = (373 / 3.15) x1/1000

                              = 0.118 m^3

c) Volume of water : ( mass of water /specific gravity of water ) x1 /1000

Volume of water = ( 149/1)x 1/1000

                           =0.149 m^3

d) Volume of chemical admixtures : ( mass of chemical admixture / specific gravity  of chemical admixture )x 1/1000

Volume of admixture = ( 7.6/ 1.145)x1/1000

                                  =0.006 m^3

e ) Volume of  aggregate= a – ( b + c + d )

                                      = 1-( 0.118 + 0.149 + 0.006 )

                                      = 0.727 m^3

f ) Mass of coarse aggregate= volume of aggregate x volume of coarse aggregate x specific gravity of  coarse aggregate x1000

Mass of coarse aggregate = 0.727 x 0.62 x 2.74 x1000

                                          = 1235.02 kg/m^3

g ) Mass of fine aggregate = volume of aggregate x volume of fine aggregate x specific gravity of fine aggregate x1000

Mass of fine aggregate = 0.7 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