Project on Concrete Mix Design for various grades of Concrete

MIX PROPORTIONING FOR A CONCRETE OF M35 GRADE USING FLY ASH

Grade Designation = M35

Type of cement = OPC 53 Grade

Type of mineral admixture = Fly ash confirming to IS 3812 (Part - 1)

Exposure condition = Severe

Workability = 125mm (Pumping)

Chemical admixture type = Superplasticizer

Specific gravity of cement = 3.14

Specific gravity of fly ash = 2.2

Specific gravity of coarse aggregate = 2.74

Specific gravity of fine aggregate = 2.65

Specific gravity of chemical admixture = 1.145

Step – 1: Target Mean Strength

F’_ck = f_ck + 1.65XS

= 43.25 N/mm²

Step – 2: Calculation of w/c ratio

From Fig – 1 of IS 10262:2019, for target mean strength of 43.25 N/mm² and cement confirming to OPC 53 grade, the w/c ratio is 0.46.

But the maximum free w/c ratio as per IS 456:2000 for severe exposure condition is 0.45. So, adopt w/c ratio of 0.45.

Step – 3: Calculation of water content

Assume required slump of 125 mm

As per table 4 of IS 10262:2019 water content for a slump of 50 mm and 20 mm aggregates is 186 lit.

So, as per CL 5.3 of IS 10262:2019, the corrected water content is 202.74 lit.

Since super plasticizer is used, as per CL 5.3 and Annex G of IS 10262:2019, water content can be reduced further by 23% at the rate of 1% by weight of cementitious material.

Hence corrected water content = 156 Lit.

Step – 4: Calculation of cement content

w/c = 0.45

hence cement content = 346.67 Kg/m³. As per table 5 of IS 456:2000 minimum cement content is 340 Kg/m³.

Assuming a fly ash replacement of 30% by weight of cement based on trials and several successful results on this percentage replacement. As per CL-5.4.1 of IS 10262:2019, if fly ash content is 20% or more, it is suggested to increase cementitious content by 10%.

Hence cement content = 381.34 Kg/m³.

Hence revised w/c = 0.41

Fly ash = 381*0.3 = 114.3 Kg/m³

Corrected cement content = 381-114 = 267 Kg/m³

According to Table 5 of IS 456:2000, Cement content = 381 Kg/m³ > 340 Kg/m³. Hence OK.

Step – 5: Calculation of volume of aggregates

As per table 5 of IS 10262:2019, volume of coarse aggregates is 0.62 for a w/c ratio of 0.5.

Hence, as per CL 5.5.1 of IS 10262:2019, the corrected volume of coarse aggregates is 0.638.

Since the concrete has to be placed using pump, As per CL – 5.5.2 of IS 10262:2019 suggests that it is desirable to reduce the coarse aggregate content determined by 10%.

Hence corrected Coarse Aggregate Volume = 0.574

The volume of fine aggregates is 1-0.574 = 0.426.

Step – 6: Mix Proportioning

1. Volume of Concrete = 1 m³
2. Volume of Cement =

=0.085 m³

1. Volume of Chemical Admixture =

= 0.0518 m³

1. Volume of Chemical Admixture =

= 0.00332 m³

1. Volume of water =

= 0.156 m³

1. Volume of aggregate = a-(b+c+d+e) = 704 m³
2. Mass of CA = f × volume of CA × specific gravity of CA × 1000 = 1107 Kg
3. Mass of FA = f × volume of FA × specific gravity of FA × 1000 = 795 Kg

MIX PROPORTIONS:

Cement = 267 Kg/m³ Free w/c ratio = 0.41

Fly ash = 114 Kg/m³

Water = 156 Kg/m³

Fine Aggregate = 795 Kg/m³

Coarse Aggregate = 1107 Kg/m³

Chemical Admixture = 3.81 Kg/m³

MIX PROPORTIONING FOR A CONCRETE OF M50 GRADE

Grade of concrete = M50

Type of cement = OPC 53 Grade

Exposure condition = Severe

Workability = 115mm (Pumping)

Chemical admixture type = Superplasticizer

Specific gravity of cement = 3.14

Specific gravity of coarse aggregate = 2.74

Specific gravity of fine aggregate = 2.65

Specific gravity of chemical admixture = 1.145

Step – 1: Target Mean Strength

F’_ck = f_ck + 1.65XS

= 58.25 N/mm²

Step – 2: Calculation of w/c ratio

From Fig – 1 of IS 10262:2019, for target mean strength of 58.25 N/mm², the w/c ratio is 0.35.

But the maximum free w/c ratio as per IS 456:2000 for severe exposure condition is 0.45. So, adopt w/c ratio of 0.35.

Step – 3: Calculation of water content

Assume required slump of 115 mm

As per table 4 of IS 10262:2019 water content for a slump of 50 mm and 20 mm aggregates is 186 lit.

So, as per CL 5.3 of IS 10262:2019, the corrected water content is 200.508 lit.

Since super plasticizer is used, as per CL 5.3 and Annex G of IS 10262:2019, water content can be reduced further by 23% at the rate of 1% by weight of cementitious material.

Hence corrected water content = 154.39 Lit.

Step – 4: Calculation of cement content

w/c = 0.35

hence cement content = 441.11 Kg/m³. As per table 5 of IS 456:2000 minimum cement content for severe exposure condition is 320 Kg/m³.

Hence cement content = 441.11 Kg/m³ > 320 Kg/m³, Hence OK.

Step – 5: Calculation of volume of aggregates

As per table 5 of IS 10262:2019, volume of coarse aggregates is 0.62 for a w/c ratio of 0.5.

Hence, as per CL 5.5.1 of IS 10262:2019, the corrected volume of coarse aggregates is 0.65.

The volume of fine aggregates is 1-0.65 = 0.35.

Step – 6: Mix Proportioning

1. Volume of Concrete = 1 m³
2. Volume of Cement =

=0.141 m³

1. Volume of Chemical Admixture =

= 0.00385 m³

1. Volume of water =

= 0.154 m³

1. Volume of aggregate = a-(b+c+d) = 701 m³
2. Mass of CA = e × volume of CA × specific gravity of CA × 1000 = 1248 Kg
3. Mass of FA = e × volume of FA × specific gravity of FA × 1000 = 650 Kg

MIX PROPORTIONS:

Cement = 441 Kg/m³

Free w/c ratio = 0.35

Water = 154 Kg/m³

Fine Aggregate = 650 Kg/m³

Coarse Aggregate = 1248 Kg/m³

Chemical Admixture = 4.41 Lit/m³