Project on Concrete Mix Design for various grades of Concrete

 

 AIM: Calculate the concrete mix design for M50 grade concrete without Fly ash.

Test Data for Materials

The following data are collected by performing standard tests in a laboratory.

Specific gravity of cement : 3.12

Specific gravity of Coarse Aggregate : 2.74

Specific gravity of Fine Aggregate : 2.56

Minimum cement content : 360 kg/m³

 

STEPS AND CALCULATIONS :

 1. 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)

                                                F’ck = fck + 1.65 5

                                               fck= 50 N/mm^2  

Table 1- IS 10262:2009

s = 5 N/mm^2

f ‘ck = 50 + ( 1.65 x 5)

        = 58.25 N/mm^2

 

2. CALCULATION OF WATER CEMENT RATIO :

Exposure condition : Extreme ( for reinforced concrete)

              Table 5 – IS 456 : 2000

              water-cement ratio = 0.35 

 

3. CALCULATION OF WATER CONTENT

From Table 2 -IS 10262: 2009,

maximum water content =186 litres (for 25 to 50 mm slump range) for 20 mm aggregate.

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 = 0.35

Cement content = 149/0.35 = 426 kg/m³

From Table 5 of IS 456:2000,

minimum cement content for ‘Extreme exposure’ condition = 360 kg/m³ 

426 kg/m³ > 360 kg/m³, hence, O.K

 

5. VOLUME OF COARSE AGGREGATES AND FINE AGGREGATES

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 = 0.35

( +/- 0.01 for every  +/- 0.05 in the water-cement ratio ) 

  So, volume of coarse aggregate  = 0.65

Volume of fine aggregate = 1-0.65 = 0.35

 

6. MIX CALCULATIONS :

1. Volume of concrete = 1m^3

 

2. Volume of cement = (mass of cement /specfic gravity of cement )x1 /1000

                                   =(426 / 3.12) x1/1000

                                   = 0.136 m^3

 

 

3. Volume of water =  ( mass of water /specific gravity of water ) x1 /1000

                                = ( 149/1)x 1/1000

                                =0.149 m^3

 

4. Volume of admixture = ( mass of chemical admixture / specific gravity  of chemical admixture )x 1/1000

                                   =( 8.2 / 1.145)x1/1000

                                   =0.007 m^3

 

5. Volume of coarse aggregate and fine aggregate = ( volume of concrete -( volume of cement + volume of water + volume of admixture )

                                                                         = a – ( b + c + d)

                                                                         = 1-( 0.136 + 0.149 + 0.007 )

                                                                         = 0.708 m^3

 

6. Mass of coarse aggregate =  volume of aggregate x volume of coarse aggregate x specific gravity of coarse aggregate x1000

                                     = 0.708 x 0.60 x 2.74 x1000

                                     = 1163.95 kg/m^3

 

7. Mass of fine aggregate = volume of aggregate x volume of fine aggregate x specific gravity of fine aggregate x1000

                                     = 0.708 x 0.35 x 2.56 x1000

                                     = 634.36 kg/m^3

 

RESULTS :

         Cement = 426 kg/m^3

         Water = 149 kg

         Coarse aggregate = 1163.95 kg/m^3

         Fine aggregate = 634.36 kg/m^3

         Chemical admixture = 8.2 kg/m^3

         Water cement ratio = 0.35

 

 

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

Test Data for Materials

The following data are collected by performing standard tests in a laboratory.

Specific gravity of cement : 3.15

Specific gravity of Coarse Aggregate : 2.74

Specific gravity of Fine Aggregate : 2.56

Minimum cement content : 360 kg/m³

 

 STEPS AND CALCULATION:

 

 1. 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)

                                                F’ck = fck + 1.65 5

                                                fck= 35 N/mm^2  

Table 1- IS 10262:2009

s = 5 N/mm^2

f ‘ck = 35 + ( 1.65 x 5)

        = 43.25 N/mm^2

 

2. CALCULATION OF WATER CEMENT RATIO :

Exposure condition : Extreme ( for reinforced concrete)

              Table 5 – IS 456 : 2000

              Water cement ratio = 0.40

 

3. CALCULATION OF WATER CONTENT

Nominal size of aggregates= 20 mm

Table 2 -IS 10262: 2009

Minimum water content = 186 litres (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 With FLY ASH

 Water cement ratio = w/c= 0.40

 Water content =   W = 149 kg

 Cement 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

 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  +/- 0.05 in the water-cement ratio) 

  So, volume of coarse aggregate  = 0.62

Volume of fine aggregate = 1-0.62 = 0.38

 

6. MIX CALCULATIONS :

1. Volume of concrete = 1m^3

 

2. Volume of cement = (mass of cement /specfic gravity of cement )x1 /1000

                                   =(285 / 3.15) x1/1000

                                   = 0.090 m^3

 

3. Volume of fly ash = (mass of fly ash/specific gravity of fly ash )x 1/1000

                                  = (122/2.2) x1/1000

                                  =0.055 m^3

 

4. Volume of water =  ( mass of water /specific gravity of water ) x1 /1000

                                = ( 149/1)x 1/1000

                                =0.149 m^3

 

5. Volume of admixture = ( mass of chemical admixture / specific gravity  of chemical admixture )x 1/1000

                                   =( 7.6/ 1.145)x1/1000

                                   =0.006 m^3

 

6. Volume of coarse aggregate and fine aggregate = ( volume of concrete -( volume of cement +volume of fly ash + volume of water + volume of admixture )

                                                                         = a – ( b + c + d+ e)

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

                                                                         = 0.7 m^3

 

7. Mass of coarse aggregate =  volume of aggregate x volume of coarse aggregate x specific gravity of coarse aggregate x1000

                                     = 0.7 x 0.62 x 2.74 x1000

                                     = 1189.16 kg/m^3

 

8. Mass of fine aggregate = volume of aggregate x volume of fine aggregate x specific gravity of fine aggregate x1000

                                     = 0.7 x 0.38 x 2.56 x1000

                                     = 680.96 kg/m^3

 

RESULTS :

         Cement = 285 kg/m^3

         Fly ash = 122 kg/m^3

         Water =    149 kg

         Coarse aggregate = 1189 kg/m^3

         Fine aggregate = 680.96 kg/m^3

         Chemical admixture = 7.6 kg/m^3

         Water cement ratio = 0.36