Project on Concrete Mix Design for various grades of Concrete

AIM

To calculate the concrete mix design for M35 grade concrete with fly ash and M50 grade concrete without Fly ash.

INTRODUCTION

Concrete is one of the most widely used and versatile building materials. It is made up of different ingredients which have various functions. Its properties as fresh and hardened concrete are a function of the right mix of constituents of concrete, which, if not done properly, will adversely affect the properties of concrete. Hence, the right proportion of all the constituents is more important. Concrete mix design is a scientific and systematic process of choosing an economical, relative proportion of various ingredients from available materials, which gives cohesive concrete of the desired workability in a fresh state, design strength, and durability in the hardened stage. The ingredients of a concrete cement mix are shown in fig. 1. Fly ash is a microscopically partitioned residue of thin coal combustion, it is carried by groove gases and collected by electrostatic precipitates. In fig. 2, fly ash powder is shown.

Fig. 1 Concrete Mix Ingredients

Fig. 2 Fly Ash Powder

EQUATIONS

1. f'ck = fck + 1.65 S

Where, f’ck = Target average compressive strength at 28 days

fck = Characteristic compressive strength at 28 days

S = Standard deviation from Table 1 of IS 10262: 2009

PROCEDURE

PART 1: Calculation of M35 grade concrete using Fly ash as part of the replacement of Ordinary Portland Cement (OPC). The procedure is shown in the following steps given below;

I. Stipulations for Proportioning

1. Grade designation: M35
2. Type of cement: OPC 43 grade conforming to IS 8112
3. Type of mineral admixtures: fly ash conforming to IS 3812-Part 1
4. Maximum nominal size of aggregate: 20mm
5. d Minimum cement content: 320 kg/`m^3` (From Table 5 of IS 456: 2000 - Exposure condition)
6. Maximum water-cement ratio: 0.45 (From Table 5 of I5 456: 2000-Exposure condition)
7. Workability: 100 mm (slump value for pumpable concrete)
8. Exposure condition: Severe (for reinforced concrete)
9. Method of concrete placing: Pumping
10. Degree of supervision: Good
11. Type of aggregate: Crushed angular aggregate
12. Maximum cement content: 450 kg/`m^3`
13. Chemical admixture type: Superplasticizer

II. Test Data for Materials

1. Cement used: OPC 43 grade conforming to IS 8112
2. Specific gravity of cement: 3.15
3. Fly ash: Conforming to IS 3812-Part 1
4. Specific gravity of fly ash: 2.2
5. Chemical admixture: Superplasticizer conforming to IS 9103
6. Specific gravity of:

• Coarse aggregate: 2.74
• Fine aggregate: 2.74

g. Water absorption:

• Coarse aggregate: 0.5%
• Fine aggregate: 1.0%

h. Free Surface moisture:

• Coarse aggregate: Nil
• Fine aggregate: Nil

i. Sieve Analysis:

• Coarse aggregate:

• Fine aggregate: Conforming to grading Zone I of Table 4 of IS 383

III. Target Strength for Mix proportioning

f'ck = fck + 1.65 S

Where, f’ck = Target average compressive strength at 28 days

fck = Characteristic compressive strength at 28 days

S = Standard deviation from Table 1 of IS 10262: 2009 given below

Hence, the grade of concrete is M35.

Standard deviation, S = 5 N/`mm^2`

Therefore, Target strength, fck = 35 + (1.65 x 5) = 43.25 N/`mm^2`

IV. Selection of Water-Cement Ratio

STEP 1: Calculate the water-cement ratio from the graph, from code IS 10261:2019 with respect to the calculated target mean strength (fck)

Here, target mean strength = 43.25 N/`mm^2`

Hence, we used curve 2

Hence, we assume w/c ratio = 0.38

STEP 2: Calculate the w/c ratio from another graph. A-F curve line indicates 7-day strength of cement, tested according to IS: 4031-1968 with respect to calculated target mean strength (fck)

Here, target mean strength = 43.25 N/`mm^2`

Hence, we used curve line E

Hence, assume w/c ratio = 0.41

STEP 3: Calculate the water-cement ratio from the maximum free water cement ratio according to the environment exposure condition as per table 5, IS 456:2000

Exposure condition = Severe

Hence, w/c ratio = 0.45

Now comparing all the values 0.38, 0.41, 0.45. The lower value of the above three values is adopted as the W/C ratio for mix design. Hence water-cement ratio is taken as 0.38.

V. Selection of Water content

From Table 2 of IS 10262: 2009,

Maximum water content for 20mm aggregate = 186 litre (for 25mm to 5Omm slump range)

Estimated water content for 100 mm slump = 186 + ((6/100) * 186) = 197 litre

As a superplasticizer is used, the water content can be reduced up to 30%.

Based on trials, water content reduction of 29% is achieved with plasticizer.

Hence, the arrived water content= 197 x 0.71 = 140 litre

VI. Calculation of Cement Content

Water-cement ratio = 0.4

Cementitious material (cement + fly ash) content = 140/0.4 = 350 kg/`m^3`

From Table 5 of IS 456,

Minimum cement content for 'severe' exposure condition = 320 kg/`m^3`

Here, 350 kg/`m^3` > 320 kg/`m^3`, Hence OK

Now, to proportion a mix containing fly ash the following steps are suggested:

1. The percentage of fly ash needs to be decided based on project requirements and the quality of materials
2. In some cases, an increase in cementitious material can be warranted. An increase in cementitious material content and its percentage may be based on experience and trial.

Here, an increase of 10% cementitious material content is taken.

Cementitious material content = 350 x 1.10 = 385 kg/`m^3`

Water content = 140 kg/`m^3`

So, Water-cement ratio = 140/385 = 0.364

Fly ash 30% total cementitious material content = 385 x 30% = 115 kg/`m^3`

Cement (OPC) = 385-115 = 270 kg/`m^3`

Saving of cement, while using fly ash = 350 - 270 = 80 kg/`m^3`

Fly ash being utilized = 115 kg/`m^3`

VII. Proportion of Volume of Coarse Aggregate and Fine Aggregate Content

From Table 3 of IS 10262: 2009,

The volume of coarse aggregate corresponding to 20mm size aggregate and fine aggregate (Zone 1) for the water-cement ratio of 0.5 = 0.6

In the present case, the water cement ratio is 0.38

• Therefore, the volume of coarse aggregate needs to be increased to decrease the fine aggregate content.
• As the water-cement ratio is lower by 0.12, the proportion of the volume of coarse aggregate should be increased by 0.024.

Therefore, the corrected volume of coarse aggregate for the water-cement ratio of 0.38 = 0.624

For pumpable concrete these values are reduced by 10%

Therefore, volume of coarse aggregate = 0.624 x 0.9 = 0.56

Volume of fine aggregate content = 1 -0.56 = 0.44

VIII. Mix Calculations

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

1. Volume of Concrete = 1 `m^3`
2. Volume of Cement = Mass of Cement/Specific gravity of Cement x 1/1000 = (270/3.15) x (1/1000) = 0.086 `m^3`
3. Volume of Fly ash = Mass of fly ash/Specific gravity of fly ash x 1/1000 = (115/2.2) x (1/1000) = 0.052 `m^3`
4. Volume of Water = Mass of Water/Specific gravity of Water × 1/1000 = (140/1) x (1/1000) = 0.140 `m^3`
5. Volume of Chemical Admixtures = Mass of Chemical Admixtures /Specific gravity of Chemical Admixtures × 1/1000 = (7.6/1.145) x (1/1000) = 0.006 `m^3`
6. Volume of Aggregate (CA and FA) = (a -(b + c + d + e)) = (1-(0.086+0.052+0.140+0.006)) = 0.716 `m^3`
7. Mass of Coarse Aggregate (CA) = f × Volume of CA × Specific gravity of CA × 1000 = 0.716 x 0.56 x 2.74 x 1000 = 1098 kg/`m^3`
8. Mass of Fine Aggregate (FA) = f × Volume of FA × Specific gravity of FA × 1000 = 0.716 x 0.44 x 2.74 x 1000 = 863 kg/`m^3`

IX. Mix Proportions for M35 Grade of Concrete

• Cement = 270 kg/`m^3`
• Fly ash = 115 kg/`m^3`
• Water = 140 Kg/`m^3`
• Fine aggregate = 863 kg/`m^3`
• Coarse aggregate = 1098 kg/`m^3`
• Chemical admixture = 7 kg/`m^3`
• Water-cement ratio = 0.364

Therefore, The mix Ratio for M35 grade of concrete = 1: 2.24: 2.85

PART 2: Calculation of M50 grade concrete using Ordinary Portland Cement (OPC). The procedure is shown in the following steps given below;

I. Stipulations for Proportioning

1. Grade designation: M50
2. Type of cement: OPC 43 grade conforming to IS 8112
3. Type of mineral admixtures: fly ash conforming to IS 3812-Part 1
4. Maximum nominal size of aggregate: 20mm
5. Minimum cement content: 320 kg/`m^3` (From Table 5 of IS 456: 2000 - Exposure condition)
6. Maximum water-cement ratio: 0.45 (From Table 5 of I5 456: 2000-Exposure condition)
7. Workability: 100 mm (slump value for pumpable concrete)
8. Exposure condition: Severe (for reinforced concrete)
9. Method of concrete placing: Pumping
10. Degree of supervision: Good
11. Type of aggregate: Crushed angular aggregate
12. Maximum cement content: 450 kg/`m^3`
13. Chemical admixture type: Superplasticizer

II. Test Data for Materials

1. Cement used: OPC 43 grade conforming to IS 8112
2. Specific gravity of cement: 3.15
3. Chemical admixture: Superplasticizer conforming to IS 9103
4. Specific gravity of:

• Coarse aggregate: 2.74
• Fine aggregate: 2.74

e. Water absorption:

• Coarse aggregate: 0.5%
• Fine aggregate: 1.0%

f. Free Surface moisture:

• Coarse aggregate: Nil
• Fine aggregate: Nil

g. Sieve Analysis:

• Coarse aggregate:

• Fine aggregate: Conforming to grading Zone I of Table 4 of IS 383

III. Target Strength for Mix proportioning

f'ck = fck + 1.65 S

Where, f’ck = Target average compressive strength at 28 days

fck = Characteristic compressive strength at 28 days

S = Standard deviation from Table 1 of IS 10262: 2009 given below

Hence, the grade of concrete is M35.

Standard deviation, S = 5 N/`mm^2`

Therefore, Target strength, fck = 50 + (1.65 x 5) = 58.25 N/`mm^2`

IV. Selection of Water-Cement Ratio

STEP 1: Calculate the water-cement ratio from the graph, from code IS 10261:2019 with respect to the calculated target mean strength (fck)

Here, target mean strength = 58.25 N/`mm^2`

Hence, we used curve 3

Hence, we assume w/c ratio = 0.34

STEP 2: Calculate the w/c ratio from another graph. A-F curve line indicates 7-day strength of cement, tested according to IS: 4031-1968 with respect to calculated target mean strength (fck)

Here, target mean strength = 58.25 N/`mm^2`

Hence, we used curve line F

Hence, assume w/c ratio = 0.35

STEP 3: Calculate the water-cement ratio from the maximum free water cement ratio according to the environmental exposure condition as per table 5, IS 456:2000

Exposure condition = Severe

Hence, w/c ratio = 0.45

Now, comparing all the values 0.34, 0.35, 0.45. The lower value of the above three values is adopted as the W/C ratio for mix design. Hence water-cement ratio is taken as 0.34.

V. Selection of Water content

From Table 2 of IS 10262:2009,

Maximum water content for 20mm aggregate = 186 litre (for 25mm to 5Omm slump range)

Estimated water content for 100 mm slump = 186 + ((6/100) x 186) = 197 litre

As a superplasticizer is used, the water content can be reduced up to 20% and above

Based on trials, water content reduction of 29% is achieved with plasticizer.

Hence, the arrived water content= 197 x 0.71 = 140 litre

VI. Calculation of Cement Content

Water-cement ratio = 0.4

Cement content = 140/0.4 = 350 kg/`m^3`

From Table 5 of IS 456,

Minimum cement content for 'severe' exposure condition = 320 kg/`m^3`

Here, 350 kg/`m^3` > 320 kg/`m^3`, Hence OK

VII. Proportion of Volume of Coarse Aggregate and Fine Aggregate Content

From Table 3 of IS 10262:2009,

The volume of coarse aggregate corresponding to 20mm size aggregate and fine aggregate (Zone 1) for the water-cement ratio of 0.5 = 0.6

In the present case, the water cement ratio is 0.34

• Therefore, the volume of coarse aggregate needs to be increased to decrease the fine aggregate content.
• As the water-cement ratio is lower by 0.16, the proportion of the volume of coarse aggregate should be increased by 0.032.

Therefore, the corrected volume of coarse aggregate for the water-cement ratio of 0.34 = 0.632

For pumpable concrete these values are reduced by 10%

Therefore, volume of coarse aggregate = 0.632 x 0.9 = 0.57

Volume of fine aggregate content = 1 -0.57 = 0.43

VIII. Mix Calculations

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

1. Volume of Concrete = 1 `m^3`
2. Volume of Cement = Mass of Cement/Specific gravity of Cement x 1/1000 = (350/3.15) x (1/1000) = 0.111 `m^3`
3. Volume of Water = Mass of Water/Specific gravity of Water × 1/1000 = (140/1) x (1/1000) = 0.140 `m^3`
4. Volume of Chemical Admixtures = Mass of Chemical Admixtures /Specific gravity of Chemical Admixtures × 1/1000 = (7/1.145) x (1/1000) = 0.006 `m^3`
5. Volume of Aggregate (CA and FA) = (a -(b + c + d)) = (1-(0.111+0.140+0.006)) = 0.743 `m^3`
6. Mass of Coarse Aggregate (CA) = e × Volume of CA × Specific gravity of CA × 1000 = 0.743 x 0.57 x 2.74 x 1000 = 1160 kg/`m^3`
7. Mass of Fine Aggregate (FA) = f × Volume of FA × Specific gravity of FA × 1000 = 0.743 x 0.43 x 2.74 x 1000 = 875 kg/`m^3`

IX. Mix Proportions for M50 Grade of Concrete

• Cement = 350 kg/`m^3`
• Water = 140 Kg/`m^3`
• Fine aggregate = 875 kg/`m^3`
• Coarse aggregate = 1160 kg/`m^3`
• Chemical admixture = 7 kg/`m^3`
• Water-cement ratio = 0.34

Therefore, The mix Ratio for M50 grade of concrete = 1: 2.5: 3.31

RESULTS

1. The mix ratio for M35 grade of concrete = 1: 2.24: 2.85
2. The mix ratio for M50 grade of concrete = 1: 2.5: 3.31

CONCLUSION

The mix ratio for the M35 grade of concrete is obtained as 1: 2.24: 2.85 and the mix ratio for the M50 grade of concrete is obtained as 1: 2.5: 3.31. For both the cases, the slump shall be measured and the water content and dosage of admixture shall be adjusted for achieving the required slump on trial. The mix proportion shall be reworked for the actual water content and checked for durability requirements. Two more trials having a variation of 10% of the water-cement ratio in Clause X shall be carried out.