Project on Concrete Mix Design for various grades of Concrete

1. Aim: To compute the mix design for M35 grade of concrete with fly ash.

Intoduction: It is mix design done to know the propoertins of the ingredients exactly needed to achieve the target strength at 28 days. Here fly ash is used as partial replacement for the cement. This reduces the shrinkage in the concrete components after hardening and it is economical too.

Procedure

Stipulations for proportioning

1. Grade designation :M30
2. Type of cement : OPC 43 grade conforming to IS 269
3. Type of mineral admixture: Fly ash conforming o IS 3812 (Part 1)
4. Maximum nominal size of the aggregate :20 mm
5. Workability :100 mm (slump)
6. Method of placing : Chute(non-pumpable)
7. Degree of site control : Good
8. Type of aggregate : Crushed angular aggregate
9. Maximum cement content : 450 kg/m3

Test data for materials

1. Cement used : OPC 43 grade conforming to IS 269
2. Fly ash: conforming o IS 3812 (Part 1)
3. Specific gravity of fly ash: 2.2
4. Specific gravity of cement : 3.15
5. Specific gravity of coarse aggregate : 2.74
6. Specific gravity of fine aggregate : 2.65

Procedure

Target strength for mix proportioning

f'ck = fck + 1.65S

= 35 + (1.65*5) From table 2, of IS 10262: 2019 Standard deviation, S = 5 N/mm2

= 43.25 N/mm2

Selection of water-cement ratio

From fig 1, of IS 10262: 2019 the free water-cement ratio required for the target strength 43.25 N/mm2 is 0.39 for OPC 43 grade curve I.e., from curve 2

Selection of water content

From Tabel 4, of IS 10262: 2019 water content = 186 kg (for 50 mm slump)

Estimated water content for 75 mm slump = 186 + (6*186)/100 = 197.16 kg

We could say 197 kg of water

Calculation of cement content

Water-cement ratio = 0.39

Cement content = 197/0.39 = 505.12 kg

Say cement content = 505 kg/ cubic meter

This mix design is with increase in 10% of fly ash.

Cementitious material content = 505 * 1.10 = 555.5

Water content = 197 kg/ cubic-meter

So, water-cementitious ratio = 197/555.5 = 0.354

Fly ash @ 30% of total cementitious material content = 555.5 * 30% = 166.65 = 166 kg/cubic-meter

cement content (OPC) = 555.5 -166 = 389.5 = 389 kg/cubic-meter.

Proportioning of volume of coarse aggregate and fine aggregate

From Table 5, of IS 10262: 2019 the proportionate volume of coarse aggregate corresponding to 20 mm size aggregate and fine aggregate (zone 2) for water-cement ratio 0.50 = 0.62.

In the present case the water-cement ratio is 0.39. Therefore, volume of coarse aggregate is to be increased to decrease fine aggregate content. As the water –cement ratio is lower by 0.11, the volume of coarse aggregate is increased by 0.022. therefore, corrected proportion of volume of coarse aggregate for water –cement ratio 0.39 = 0.62 + 0.022 = 0.642.

Volume of fine aggregate = 1-0.642 = 0.358.

Mix calculation

The mix calculations per unit volume of concrete shall be as follows:

a.Total volume = 1 m3

b.Volume of cement = (mass of cement/specific gravity of cement) *(1/1000)

= (389/3.15) * (1/1000) = 0.123 m3

c. Volume of fly ash = (mass of fly ash/specific gravity of fly ash) *(1/1000)

= (166/2.2) * (1/1000) = 0.0754 m3

d.Volume of water = (mass of water/specific gravity of water) *(1/1000)

= (197/1) * (1/1000) = 0.197 m3

e.Volume of all in aggregate = [a - (b + c + d)] = [1 - (0.123 + 0.0754 +0.197 )] = 0.6046

f.Mass of coarse aggregate = g x Volume of coarse aggregate x specific gravity of coarse aggregate x 1000

= 0.6046 x 0.642 x 2.74 x 1000 = 1063 kg

g.Mass of fine aggregate = g x Volume of fine aggregate x specific gravity of fine aggregate x 1000

= 0.6046 x 0.358 x 2.65 x 1000 = 573 kg

Mix proportion for trial number 1

Cement = 389 kg/m3

Fly ash = 166 kg/m3

Water = 197 kg/m3

Coarse aggregate = 1063 kg/m3

Fine aggregate = 573 kg/m3

Free water-cement ratio =0.354

Therefore, Mix proportion of M30 grade concrete = 1: 1.03: 1.91

2. Aim: To compute the mix design for M50 grade of concrete without fly ash.

Intoduction: The exact proportions of various materials of the M50 grade concrete is computed using code IS 10262: 2019.

Stipulations for proportioning

1. Grade designation :M50
2. Type of cement : OPC 53 grade conforming to IS 269
3. Maximum nominal size of the aggregate :20 mm
4. Workability :75 mm (slump)
5. Method of placing : Chute(non-pumpable)
6. Degree of site control : Good
7. Type of aggregate : Crushed angular aggregate
8. Maximum cement content : 450 kg/m3

Test data for materials

1. Cement used : OPC 53 grade conforming to IS 269
2. Specific gravity of cement : 3.15
3. Specific gravity of coarse aggregate : 2.74
4. Specific gravity of fine aggregate : 2.65

Procedure

Target strength for mix proportioning

f'ck = fck + 1.65S

= 50 + (1.65*5) From table 2, of IS 10262: 2019 Standard deviation, S = 5 N/mm2

= 58.25 N/mm2

Selection of water-cement ratio

From fig 1, of IS 10262: 2019 the free water-cement ratio required for the target strength 58.25 N/mm2 is 0.34 for OPC 43 grade curve I.e., from curve 3

Selection of water content

From Tabel 4, of IS 10262: 2019 water content = 186 kg (for 50 mm slump)

Estimated water content for 75 mm slump = 186 + (3*186)/100 = 191.58 kg

We could say 192 kg of water

Calculation of cement content

Water-cement ratio = 0.29

Cement content = 192/0.34 = 564.7 kg

Proportioning of volume of coarse aggregate and fine aggregate

From Table 5, of IS 10262: 2019 the proportionate volume of coarse aggregate corresponding to 20 mm size aggregate and fine aggregate (zone 2) for water-cement ratio 0.50 = 0.62.

In the present case the water-cement ratio is 0.34. Therefore, volume of coarse aggregate is to be increased to decrease fine aggregate content. As the water –cement ratio is lower by 0.16, the volume of coarse aggregate is increased by 0.032. therefore, corrected proportion of volume of coarse aggregate for water –cement ratio 0.29 = 0.62 + 0.032 = 0.652.

Volume of fine aggregate = 1-0.652 = 0.348.

Mix calculation

The mix calculations per unit volume of concrete shall be as follows:

a.Total volume = 1 m3

b.Volume of cement = (mass of cement/specific gravity of cement) *(1/1000)

= (564.7/3.15) * (1/1000) = 0.179 m3

c.Volume of water = (mass of water/specific gravity of water) *(1/1000)

= (192/1) * (1/1000) = 0.192 m3

d.Volume of all in aggregate = [a - (b + c)] = [1 - (0.179 + 0.192)] = 0.629

e.Mass of coarse aggregate = g x Volume of coarse aggregate x specific gravity of coarse aggregate x 1000

= 0.629 x 0.652 x 2.74 x 1000 = 1123 kg

f.Mass of fine aggregate = g x Volume of fine aggregate x specific gravity of fine aggregate x 1000

= 0.629 x 0.348 x 2.65 x 1000 = 580 kg

Mix proportion for trial number 1

Cement = 564.7 kg/m3

Water = 192 kg/m3

Coarse aggregate = 1123 kg/m3

Fine aggregate = 580 kg/m3

Free water-cement ratio =0.34

Therefore, Mix proportion of M30 grade concrete = 1: 1.02: 1.98