Project on Concrete Mix Design for various grades of Concrete

Concrete Mix Design for M35 grade concrete with fly ash

Aim

To determine the mix proportion for a concrete of M35 grade using fly ash

Stipulation for proportioning

1. Grade Designation : M35
2. Type of cement : OPC 53 grade
3. Type of mineral admixture : Fly ash conforming to IS 3812 Part 1
4. Maximum nominal size of aggregate : 20mm
5. Minimum Cement Content : 340 kg/m³
6. Maximum Water Cement ratio : 0.45
7. Workability : 100mm
8. Exposure condition : Very Severe
9. Method of pumping : Pumpable
10. Degree of supervision : Good
11. Type of aggregate : Crushed angular
12. Maximum Cement Content : 450 kg/m³
13. Chemical admixture : Super plasticizer

Test Date for materials

1. Cement used : OPC 53 grade
2. Specific gravity of cement : 3.15
3. Fly ash : Conforming to IS 3812 Part 1
4. Chemical Admixture : Super Plasticizer
5. Specific gravity of fly ash : 2.2
6. Specific gravity of

• Coarse aggregates : 2.74
• Fine aggregates : 2.74

6.Water absorption of

• Coarse aggregates : 0.5%
• Fine aggregates : 1%

7. Free surface moisture

• Coarse aggregates : Nil
• Fine aggregates : Nil

8. Sieve Analysis

• Coarse Aggregate

• Fine aggregate : Conforming to grading zone II of table 4 of IS 383

Step:1 Determination of target mean strength

Determination of target mean strength, f’_ck = f_ck + 1.65 S

The assumed standard deviation as per IS 10262:2019 S = 5 N/mm2

Targeted mean strength, f’_ck = 35 + 1.65 * 5

= 43.25 N/mm2

Step:2 Determination of water cement ratio

• As per IS 10262:2019, the assumed w/c from the graph 28^th day compressive strength of concrete to the water cement ratio = 0.38
• As per IS 456:2000, table 5 let’s consider the exposure condition as “Very Severe” then the w/c for plain concrete is 0.50 & for RCC is 0.45

The water-cement ratio directly affects the strength of the concrete. Lower the w/c higher the strength of concrete.

Hence the water/cement is considered as 0.38 which is lower value

Step:3 Selection of water content

As per IS 10262: 2019 Table 4

Therefore, the water content for 100 mm slump = 186 + 6*186/100

= 197 liters

As superplasticizers are used, we can reduce the water content to gain strength in the concrete. Water content can be reduced up to 20% & above.

So, water content = 197 * 0.8 = 157.6 Liters = 158 liters

Step:4 Calculation of cement content:

From step 2, w/c = 0.38

From step 3, w = 158

Therefore, Cementitious material (Cement + fly ash) = 158/0.38

= 415.8 kg/m³ = 416 kg/m3

From IS 456 :2000 table 5 the minimum cement content for very severe exposure condition is 340 kg

Therefore 416 >340

Hence ok

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

• Decide the % fly ash to be used based on project requirement & quality of fly ash
• In certain situations, the increase in cementitious material content may be warranted. The decision on increase in cementitious material content and its % may be based on experience & trials. (Assuming with 10% increase)

Cementitious material content = 416 * 1.10 = 456 kg/m³

Water Content = 158 kg/m³

So, w/c = 158/456 = 0.34

Fly ash @30% total cementitious material content = 456 * 30% = 137 kg/m³

Cement (OPC) = 456-137 = 319 kg/m³

Saving of cement while using fly ash = 416 - 319 = 97 kg/m³

Fly ash being utilized = 137 kg/m³

Step:5 Volume of aggregates

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

Since w/c is 0.34, the volume of coarse aggregates needs to be increased to decrease the fine aggregate content. As the w/c is lowered by 0.16, the proportion of volume of coarse aggregate, aggregate is increased by 0.032. Hence corrected proportion of volume of coarse aggregate for w/c of 0.34 = 0.652

If the concrete is pumpable then the values are reduced by 10%

Hence volume of coarse aggregate = 0.652*0.9 = 0.586 = 0.59m³

Volume of fine aggregate = 1-0.59 = 0.41m³

Step:6 Mix Calculation

1. Volume of concrete = 1m³
2. Volume of cement = Mass of cement / Specific gravity of cement x 1/1000

= 319/3.15 x 1/1000

= 0.101 m³

3. Volume of Fly ash = Mass of Fly ash / Specific gravity of Fly ash x 1/1000

= 137/2.2 x 1/1000

= 0.062 m³

4. Volume of water = Mass of water / Specific gravity of water x 1/1000

= 158/1 * 1/1000

= 0.158 m³

5.Volume of chemical admixture = Mass of chemical admixture / Specific gravity of chemical admixture x 1/1000

Mass of chemical admixture is 2% of the cementitious material

= 9.12 /1.145 x 1/1000

= 0.008 m³

6. Volume of aggregates = volume of concrete - (vol. of cement + vol. of fly ash + vol. water + vol. of chemical admixture)

= 1-(0.101+0.062+0.158+0.008)

= 0.671 m³

• Mass of coarse aggregate = vol. of aggregates * vol. of course aggregate * specific gravity of coarse aggregate * 1000

= 0.671*0.59*2.74*1000

= 1085 kg/m³

• Mass of fine aggregate = vol. of aggregates * vol. of fine aggregate * specific gravity of fine aggregate * 1000

= 0.671*0.41*2.74*1000

= 754 kg/m³

Mix Proportion

Cement = 319 kg/m³

Fly ash = 137 kg/m³

Water = 158 kg/m³

Fine aggregate = 754 kg/m³

Coarse aggregate = 1085 kg/m³

Chemical admixture = 9.12 kg/m³

Water cement ratio = 0.34

Trial Mix for M35 using fly ash = Cement: Fine aggregate: Coarse Aggregates = 1: 1.65: 2.37

Points to be noted:

• The total mass of coarse aggregate shall be divided into two fractions of 20 – 10 mm and 10 - 4.75 mm, in a suitable ratio, to satisfy the overall grading requirements for 20 mm max size aggregate as per Table 7 of IS 383.

• The slump shall be measured and the water content and dosage of admixture shall be adjusted for achieving the required slump based on trial, if required. The mix proportions shall be reworked for the actual water content and checked for durability requirements.

• Two more trials having variation of ± 10 percent of water-cementitious materials ratio in mix proportions shall be carried out and a graph between these water cementitious materials ratios and their corresponding strengths shall be plotted to work out the mix proportions for the given target strength for field trials. However, durability requirement shall be met.

Concrete Mix Design for M50 grade concrete without Fly ash

Aim

To determine the mix proportion for a concrete of M50 grade

Stipulation for proportioning

1. Grade Designation : M50
2. Type of cement : OPC
3. Maximum nominal size of aggregate : 20mm
4. Minimum Cement Content : 320 kg/m³
5. Maximum Water Cement ratio : 0.45
6. Workability : 75 mm
7. Exposure condition : Severe
8. Method of pumping : Pumpable
9. Degree of supervision : Good
10. Type of aggregate : Crushed angular
11. Maximum Cement Content : 450 kg/m³
12. Chemical admixture : Super plasticizer

Test Date for materials

1. Cement used : OPC
2. Specific gravity of cement : 3.15
3. Chemical Admixture : Super Plasticizer
4. Specific gravity of

• Coarse aggregates : 2.74
• Fine aggregates : 2.74

5.Water absorption of

• Coarse aggregates : 0.5%
• Fine aggregates : 1%

7. Free surface moisture

• Coarse aggregates : Nil
• Fine aggregates : Nil

8. Sieve Analysis

• Coarse Aggregate

• Fine aggregate : Conforming to grading zone II of table 4 of IS 383

Step:1 Determination of target mean strength

Determination of target mean strength, f’_ck = f_ck + 1.65 S

The assumed standard deviation as per IS 10262:2019 S = 5 N/mm2

Targeted mean strength, f’_ck = 50 + 1.65 * 5

= 58.25 N/mm2

Step:2 Determination of water cement ratio

• As per IS 10262:2019, the assumed w/c from the graph 28^th day compressive strength of concrete to the water cement ratio = 0.33
• As per IS 456:2000, table 5 let’s consider the exposure condition as “Severe” then the w/c for plain concrete is 0.50 & for RCC is 0.45

The water-cement ratio directly affects the strength of the concrete. Lower the w/c higher the strength of concrete.

Hence the water/cement is considered as 0.33 which is lower value

Step:3 Selection of water content

As per IS 10262: 2019 Table 4

Therefore, the water content for 75 mm slump = 186 + 3*186/100

= 191.6 liters

As superplasticizers are used, we can reduce the water content to gain strength in the concrete. Water content can be reduced up to 23% based on trial data

So, water content = 191.6 * 0.77 = 157.6 Liters = 148 liters

Step:4 Calculation of cement content:

From step 2, w/c = 0.33

From step 3, w = 149.4

Therefore, Cementitious material = 148/0.33

= 448.4 kg/m³ = 449 kg/m3

From IS 456 :2000 table 5 the minimum cement content for very severe exposure condition is 320 kg

Therefore 449 >320

Hence ok

Step:5 Volume of aggregates

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

Since w/c is 0.33, the volume of coarse aggregates needs to be increased to decrease the fine aggregate content. As the w/c is lowered by 0.17, the proportion of volume of coarse aggregate, aggregate is increased by 0.034. Hence corrected proportion of volume of coarse aggregate for w/c of 0.33 = 0.654

If the concrete is pumpable then the values are reduced by 10%

Hence volume of coarse aggregate = 0.654*0.9 = 0.588 m³

Volume of fine aggregate = 1-0.588 = 0.412m³

Step:6 Mix Calculation

1. Volume of concrete = 1m³
2. Volume of cement = Mass of cement / Specific gravity of cement x 1/1000

= 449/3.15 x 1/1000

= 0.142 m³

3. Volume of water = Mass of water / Specific gravity of water x 1/1000

= 148/1 * 1/1000

= 0.148 m³

4.Volume of chemical admixture = Mass of chemical admixture / Specific gravity of chemical admixture x 1/1000

Mass of chemical admixture is 2% of the cementitious material

= 8.98 /1.145 x 1/1000

= 0.0078 m³

6. Volume of aggregates = volume of concrete - (vol. of cement + vol. of fly ash + vol. water + vol. of chemical admixture)

= 1-(0.142+0.148+0.0078)

= 0.632 m³

• Mass of coarse aggregate = vol. of aggregates * vol. of course aggregate * specific gravity of coarse aggregate * 1000

= 0.632*0.59*2.74*1000

= 1022 kg/m³

• Mass of fine aggregate = vol. of aggregates * vol. of fine aggregate * specific gravity of fine aggregate * 1000

= 0.632*0.41*2.74*1000

= 710 kg/m³

Mix Proportion

Cement = 449 kg/m³

Water = 148 kg/m³

Fine aggregate = 710 kg/m³

Coarse aggregate = 1022 kg/m³

Chemical admixture = 8.98 kg/m³

Water cement ratio = 0.33

Trial Mix for M50 grade of concrete = Cement: Fine aggregate: Coarse Aggregates = 1: 1.58 : 2.27

Points to be noted:

• The total mass of coarse aggregate shall be divided into two fractions of 20 – 10 mm and 10 - 4.75 mm, in a suitable ratio, to satisfy the overall grading requirements for 20 mm max size aggregate as per Table 7 of IS 383.

• The slump shall be measured and the water content and dosage of admixture shall be adjusted for achieving the required slump based on trial, if required. The mix proportions shall be reworked for the actual water content and checked for durability requirements.

• Two more trials having variation of ± 10 percent of water-cementitious materials ratio in mix proportions shall be carried out and a graph between these water cementitious materials ratios and their corresponding strengths shall be plotted to work out the mix proportions for the given target strength for field trials. However, durability requirement shall be met.