Project on Concrete Mix Design for various grades of Concrete

Aim:

To calculate the Mix design for M35 grade concrete with fly ash as well the M50 grade without fly ash as per IS 456 & IS 10262 code books.

Introduction:

Mix design refers to making the optimum proportions of each and every ingredients in the concrete mix to achieve the required or targeted properties of the fresh and hardened concrete mix say slump, compressive strength etc.

Procedures:

• Calculation of Target Strength
• Finding the Water cement ratio
• Calculation of cement content
• Calculation of quantities like chemical admixture, fly ash
• Calculation of Proportion of aggregate
• Calculation of aggregate quantity (Coarse and Fine)

For performing the calculation part, we ned to refer to IS 456 and IS 10262 of its latest versions.

Also, let us assume some of required data to perform the calculation of mix design:

1. Type of cement : OPC 43 grade
2. Maximum nominal aggregate size : 20mm
3. Exposure condition type : Severe
4. Workability : 75mm
5. Method of placing : Pumping
6. Type of aggregate : Crushed angular
7. Chemical Admixture : Superplasticiser
8. Specific gravity of cement : 3.15
9. Specific gravity of fly ash : 2.20
10. Specific gravity of coarse agregate : 2.74
11. Specific gravity of fine aggregate : 2.65
12. Specific gravity of admixture : 1.145 (@1% of usage with respect to the cement usage)
13. Let us consider that the fine aggregate falls under the sieve of Zone-II

Calculation Part:

1. Target mean strength, f’ck = fck + 1.65*S

From IS 10262-2019, Table.No:2 Standard deviation “S” is 5.

f’ck = 35+ 1.65*5

= 43.25 N/mm2

Or

f’ck = fck + X

From IS 10262-2019, Table.No:1 Value of “X” is 6.5.

f’ck = 35 + 6.5

= 41.5 N/mm2, whichever calculated f’ck value is greater.

Hence, the value of f’ck = 43.25 N/mm2

2. With respect to the Target mean strength and grade of cement, by referring to graph of IS 10262-2019, w/c is 0.4 As per IS 456 Table.No:5 with respect to the exposure condition and the grade of the concrete maximum water cement ratio is 0.45 > 0.40 of actual value. Hence it is okay.
3. With respect to the nominal maximum size of the aggregate 20mm, water content is 186 kg/m3 as per IS 10262-2019 Table.No:4. As per IS 10262-2019 Clause.No:5.3 water content has to be increased by 3% for increase in slump value of 25mm. so the water content = 186 + (3%*186)

W= 191.58 kg/m3. Since the superplasticiser is being used the water content will be reduced by 30%, so the final water content is w= 191.58*0.7 = 134 kgm3

4. Cement content, c= w/0.40

= 134/0.40 = 335 kg/m3 > 320kg/m3 which is the minimum cement content as per IS 456 Table. No:5

5. As per IS 10262, Clause.No:5.4 Volume of cement found out has to be increased by 10 %. Hence, the volume of cementitious material = 334*1.10 = 367.40 kg/m3

Fly ash will be used at the 20% of the cementitious material. So, the proportion of cementitious material is cement: Flyash-80:20

6. As per IS 10262-2019, volume of coarse aggregate to total unit of aggregate is 0.66 with respect to the nominal maximum size of aggregate as well the sieve analysis of fine aggregate that falls in Zone-II.

From the Table.No:5 of IS 10262-2019, volume of coarse aggregate to total unit of aggregate is given for w/c ratio of 0.50. Here the w/c is 0.40.For every -/+ of 0.05 value in w/c ratio, volume of coarse aggregate ahs to be +/- by 0.01 value. So, the ration will be of 0.66+(0.01*2) =0.68

Since the concrete is pumpable, the aggregate proportion will be reduced by 10%. Hence the volume of aggregate = 0.68*0.9 = 0.612

Volume of fine aggregate = 1-0.612 = 0.388

7. Total Volume of concrete = 1 m3
8. Volume of cement = (Mass of cement*1/Specific gravity of cement*1000)

= (0.8*367.4*1/3.15*1000)

= 0.093 m3 (a)

9. Volume of fly ash = (Mass of fly ash*1/Specific gravity of cement*1000)

= (20% *367.4*1/2.2*1000)

= 0.033 m3 (b)

10. Volume of water = (Mass of water*1/Specific gravity of water*1000)

= (134*1/1*1000)

= 0.134 m3 (c)

11. Volume of Chemical Admixture = (Mass of admixture*1/Specific gravity of admixture*1000)

= (3.67*1/1.145*1000)

= 0.003 m3 (d)

12. Volume of entrapped air = 0.01 m3 (e)
13. Volume of all aggregate = 1-(0.093+0.033+0.134+0.003+0.01) = 0.727 (e)
14. Mass of coarse aggregate = e*specific gravity*Volume of aggregate*1000

= 0.727*2.74* 0.612*1000 = 1219 kg/m3

15. Mass of fine aggregate = e*specific gravity*Volume of aggregate*1000

= 0.727*2.65* 0.388*1000 = 747.5 kg/m3

Result:

Proportion of cementitious material: Fine aggregate: Coarse aggregate for Mix design of M35 grade with fly ash is 367.4:747.5:1219 @ w/c of 0.40

Mix ratio for M35 grade is 1:2.03:3.32 @ w/c of 0.40

For Mix design of M50:

1. Type of cement : OPC 43 grade
2. Maximum nominal aggregate size : 20mm
3. Exposure condition type : Severe
4. Workability : 75mm
5. Method of placing : Pumping
6. Type of aggregate : Crushed angular
7. Chemical Admixture : Superplasticiser
8. Specific gravity of cement : 3.15
9. Specific gravity of coarse aggregate : 2.74
10. Specific gravity of fine aggregate : 2.65
11. Specific gravity of admixture : 1.145 (@1% of usage with respect to the cement usage)
12. Let us consider that the fine aggregate falls under the sieve of Zone-II

Calculation Part:

1. Target mean strength, f’ck = fck + 1.65*S

From IS 10262-2019, Table.No:2 Standard deviation “S” is 5.

f’ck = 50+ 1.65*5

= 58.25 N/mm2

Or

f’ck = fck + X

From IS 10262-2019, Table.No:1 Value of “X” is 6.5.

f’ck = 50 + 6.5

= 56.5 N/mm2, whichever calculated f’ck value is greater.

Hence, the value of f’ck = 48.25 N/mm2

2. With respect to the Target mean strength and grade of cement, by referring to graph of IS 10262-2019, w/c is 0.30. As per IS 456 Table.No:5 with respect to the exposure condition and the grade of the concrete maximum water cement ratio is 0.45 > 0.30 of actual value. Hence it is okay.
3. With respect to the nominal maximum size of the aggregate 20mm, water content is 186 kg/m3 as per IS 10262-2019 Table.No:4. As per IS 10262-2019 Clause.No:5.3 water content has to increase by 3% for increase in slump value of 25mm. so the water content = 186 + (3%*186)

W= 191.58 kg/m3. Since the superplasticiser is being used the water content will be reduced by 30%, so the final water content is w= 191.58*0.7 = 134 kgm3

4. Cement content, c= w/0.30

= 134/0.30 = 446.67 kg/m3 > 320kg/m3 which is the minimum cement content as per IS 456 Table. No:5

5. As per IS 10262-2019, volume of coarse aggregate to total unit of aggregate is 0.66 with respect to the nominal maximum size of aggregate as well the sieve analysis of fine aggregate that falls in Zone-II.

From the Table.No:5 of IS 10262-2019, volume of coarse aggregate to total unit of aggregate is given for w/c ratio of 0.5. Here the w/c is 0.30. For every -/+ of 0.05 value in w/c ratio, volume of coarse aggregate has to be +/- by 0.01 value. So, the ratio will be of 0.66+(0.04) =0.70

Since the concrete is pumpable, the aggregate proportion will be reduced by 10%. Hence the volume of aggregate = 0.70*0.9 = 0.63

Volume of fine aggregate = 1-0.63 = 0.37

6. Total Volume of concrete = 1 m3
7. Volume of cement = (Mass of cement*1/Specific gravity of cement*1000)

= (446.67*1/3.15*1000)

= 0.14 m3 (a)

8. Volume of water = (Mass of water*1/Specific gravity of water*1000)

= (134*1/1*1000)

= 0.134 m3 (b)

9. Volume of Chemical Admixture = (Mass of admixture*1/Specific gravity of admixture*1000)

= (4.47*1/1.145*1000)

= 0.0039 m3 (c)

10. Volume of entrapped air = 0.01 m3 (d)
11. Volume of all aggregate = 1-(0.14+0.134+0.0039+0.01) = 0.712 (e)
12. Mass of coarse aggregate = e*specific gravity*Volume of aggregate*1000

= 0.712*2.74* 0.63*1000 = 1229.05 kg/m3

13. Mass of fine aggregate = e*specific gravity*Volume of aggregate*1000

= 0.712*2.65* 0.37*1000 = 698.12 kg/m3

Result:

Proportion of Cement: Fine aggregate: Coarse aggregate for Mix design for M50 grade without fly ash is 446.67:698.12:1229.05 @ w/c of 0.30

Mix ratio for M50 grade is 1:1.56:2.75 @ w/c of 0.30