Concrete Mix Design M25 SCC [IS CODE]

In order to determine the mix design quantities of different ingredients for a particular grade of concrete, few steps are required which are as follows:

1.       Determine the target compressive strength

The target compressive strength can be calculated by the following formula

                                T_fck = T_cs + 1.65 SD ……… [IS CODE: 10262-2009]

                Where, T_fck is the target compressive strength of concrete.

                             T_cs is the compressive strength of concrete.

                             SD is the Standard deviation.

            The grade of concrete to be designed is M25, then          

                                T_fck = 25 + 1.65 x 4 = 31.6 N/mm²

                                      

2.       Determine the water-cement ratio of the mix & free water content.

Determining the water cement ratio is critical for any mix design, no IS Code states any particular w/c ratio for any given mix. IS code only provides with the maximum w/c ratio that is allowed in a mix. Practically based on several trials w/c ratio is determined.

Said that, for M25 grade of concrete let’s consider w/c ratio 0.35.

                         & Free water content as 158 Lts.

                                        w/c = 0.35

From the ratio we getcementitious content = 158/0.35 = 451 Kg > 300 Kg, Hence OK.

Here IS code 456-2000 specifies the minimum cementitious content for M25 grade of concrete is 300 Kgs. (With nominal maximum size of aggregate being 20 mm)

3.       Calculation for weight of admixture

Considering 0.7% admixture by weight of cementitious material which gives us

                                0.70 x 451 / 100 = 3.157 Kg.

4.       Calculation of fly ash content.

Considering 26 % fly ash content by weight of cementitious material which gives us

                                26 x 451 / 100 = 117 Kg.

Therefore, cement content becomes = 451 – 117 = 334 Kg.

5.       Calculation of absolute volume of Cement, fly ash, admixture & water.

We can get the absolute volume of ingredients by

                                Volume = Mass / (Specific gravity x 1000)

For Cement:         C_Vol = 334 / (3.15 x 1000) = 0.106 M³.

For Fly ash:           F_Vol = 117 / (2.23 x 1000) =  0.0525 M³.

For Admixture:    A_Vol = 3.157 / (1.11 x 1000) = 0.0028 M³.

For Water:           W_Vol = 158 / (1 x 1000) = 0.158 M³.

6.       Calculation for volume of Fine & Coarse aggregates.

Total volume of concrete is 1 M³,

Therefore volume of aggregates comes to = 1 – ( C_Vol + F_Vol + A_Vol + W_Vol ) = 0.681 M³.

Considering volume of fine aggregates to be 49 %,

Which gives us = (49 x 0.681) / 100 = 0.337 M³.

Therefore, volume of coarse aggregate becomes = 0.681 – 0.337 M³ = 0.344 M³.

Now considering ratio of 10 mm CA to 20 mm CA to be 60:40,

Then volume of 10 mm CA becomes = (60 x 0.334) / 100 = 0.206 M³.

Therefore, volume of 20 mm CA comes to = (0.344 – 0.206) = 0.138 M³.

7.       Calculation for mass of Fine & Coarse aggregates.

Mass of aggregate = Volume of aggregate x Specific gravity of aggregate x 1000.

Mass of fine aggregate = 0.337 x 2.760 x 1000 = 930 Kg.

Mass of Coarse aggregate 10 mm = 0.206 x 2.840 x 1000 = 585 Kg.

Mass of Coarse aggregate 20 mm = 0.138 x 2.870 x 1000 = 396 Kg.