Project on Concrete Mix Design for various grades of Concrete

Aim:

To prepare the Mix-Design of the concrete for the grade M35 with fly ash and M50 without fly ash as per Indian Standards.

Introduction:

Mix design of concrete is the proportioning of concrete ingredients which are used to provide strength and durability for the structures using the following standards

IS Method: Concrete mix proportioning guidelines (IS 10262-2019) and recommended guidelines for concrete mix design IS 10262 -1982.

Step by step procedure:

1)Concrete mix design for M35 grade with fly ash

1.Data required

1. Grade designation: M35 with 10% of flyash
2. Type of cement: OPC 43 Grade conforming IS 8112
3. Maximum nominal size of aggregate: 20mm
4. Minimum cement content: 340 kg/m³
5. Workability: 100mm slump
6. Exposure condition: Very Severe
7. Method of concrete placing: Pumping
8. Type of aggregate: Crushed Angular Aggregates
9. Maximum cement content: 450 kg/m^3
10. Chemical admixture type: Super Plasticizer
11. Specific gravity of cement: 3.15
12. Specific gravity of fly ash: 2.2
13. Specific gravity of fine aggregate: 2.61
14. Specific gravity of coarse aggregate: 2.73
15. Coarse and fine aggregate: Grade 1

2.TARGET STRENGTH FOR MIX PROPORTIONING

f’ck =fck + 1.65 s

where

f’ck = target average compressive strength at 28 days,

fck = characteristics compressive strength at 28 days

s = standard deviation.

Standard Deviation, s = 5 N/mm^2 (From Table I of IS 10262:2009)

Therefore, target strength = 35 + 1.65 x 5 = 43.25 N/mm^2

3.SELECTION OF WATER CEMENT RATIO

From the Table 5 of IS 456:2000, for severe exposure condition maximum Water Cement Ratio = 0.45

Adopt w/c = 0.41

4.SELECTION OF WATER CONTENT

Maximum water content for 20 mm aggregate = 186 litre (From Table 2 of IS 10262:2009)

For 75mm slump,water content = 186+ (0.03x186) = 191.58 litres.

Super plasticizer reduces 29% of water content.

Hence new water content = [197 x (71/100)] = 139.87 litre.

5.CALCULATION OF CEMENT CONTENT

w/c Ratio = 0.41

Cementitious content = 139.87/0.41 = 341.14 > 340 kg/m^3

Therefore Cementitious content = 341.14 kg/m^3

Here cement is replaced bt 20% fly ash, therefore increse the total cementitious content by 10%

Cementitious content = 1.10X341.14 = 375.25 kg/m^3

New water cement ratio = 139.87/375.25 = 0.37

20% fly ash of cementitious content = 75.05 kg/m^3

Cement = 375.25 - 75.05 = 300.2 kg/m^3

2% of superplaticizer of Cementitious content = (2/100)X375.25 = 7.505 kg/m^3

6.PROPORTION OF VOLUME OF COARSE AGGREGATE AND FINE AGGREGATE

From Table 3 of (IS 10262:2009) for 0.5 w/c,Volume of coarse aggregate =0.6

As per A7 IS 10262:2009, for w/c = 0.37 the volume of coarse aggregate = 0.626

Since the concrete is pumpable, volume of coarse aggregate is reduced by 10%

volume of coarse aggregate =0.626 x 0.9 = 0.56.

Volume of fine aggregate content = 1 – 0.56= 0.44.

7.MIX CALCULATIONS

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

a) Volume of concrete = 1 m^3

b) Volume of cement = Mass of cement / (Specific Gravity of Cement x 1000)

= 300.2/(3.15 x 1000)

= 0.095 m^3

c) Volume of fly ash = Mass of fly ash / (Specific Gravity of fly ash x 1000)

= 75.05/(2.2x1000) = 0.034 m^3

d) Volume of water = Mass of water / (Specific Gravity of water x 1000)

= 139.87/(1 x 1000)

= 0.139 m^3

e)Super plastiziser is upto 2% of mass of total cementitious content

= (7.505/1.15)X(1/1000) = 0.006 m^3

f) Volume of all in aggregate = a-(b+c+d+e)

= 1-(0.095+0.034+0.139+0.006)

= 0.726 m^3

g) Mass of coarse aggregate= f x Volume of Coarse Aggregate x Specific Gravity of coarse aggregate x 1000

= 0.726x 0.56 x 2.73 x 1000

= 1110 kg

h) Mass of fine aggregate= f x Volume of Fine Aggregate x Specific Gravity of Fine Aggregate x 1000

= 0.724x 0.44x 2.61 x 1000

= 834 kg

8.MIX PROPORTIONS

Cement = 300.2 kg/m^3

Fly ash = 75.05 kg/m^3

Fine aggregate = 834 kg/m^3

Coarse aggregate = 1110 kg/m^3

w/c = 0.37

Admixture = 7.505 kg/m^3

Result:

The mix proportions arrived: 1:2.22:2.96

2)Concrete mix design for M50 grade without fly ash

1.Data required

1. Grade designation: M50
2. Type of cement: OPC 43 Grade conforming IS 8112
3. Maximum nominal size of aggregate: 20mm
4. Minimum cement content: 360 kg/m³
5. Workability: 100mm slump
6. Exposure type: Extreme (reinforced)
7. Method of concrete placing : Pumping
8. Type of aggregate : Crushed Angular Aggregates
9. Maximum cement content : 450 kg/m^3
10. Chemical admixture type : Super Plasticizer
11. Specific gravity of cement: 3.15
12. Specific gravity of fine aggregate: 2.61
13. Specific gravity of coarse aggregate: 2.73
14. Coarse and fine aggregate: Grade 1

2.TARGET STRENGTH FOR MIX PROPORTIONING

f’ck =fck + 1.65 s

where

f’ck = target average compressive strength at 28 days,

fck = characteristics compressive strength at 28 days

s = standard deviation.

Standard Deviation, s = 5 N/mm^2 (From Table I of IS 10262:2009)

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

3.SELECTION OF WATER CEMENT RATIO

From the Table 5 of IS 456:2000, for severe exposure condition maximum Water Cement Ratio = 0.40

Adopt w/c = 0.33

4.SELECTION OF WATER CONTENT

Maximum water content for 20 mm aggregate = 186 litre (From Table 2 of IS 10262:2009)

For 100mm slump,water content = 186+ (0.06x186) = 197 litres.

Super plasticizer reduces 29% of water content.

Hence new water content = [197 x (71/100)] = 139.87 litre.

5.CALCULATION OF CEMENT CONTENT

w/c Ratio = 0.33

Cementitious content = 139.87/0.33 = 424 > 340 kg/m^3

Therefore Cementitious content = 424 kg/m^3

2.5% of superplaticizer of Cementitious content = (2.5/100)X424 = 10.6 kg/m^3

6.PROPORTION OF VOLUME OF COARSE AGGREGATE AND FINE AGGREGATE

From Table 3 of (IS 10262:2009) for 0.5 w/c, Volume of coarse aggregate =0.6

As per A7 IS 10262:2009, for w/c = 0.33 the volume of coarse aggregate = 0.636

Since the concrete is pumpable, volume of coarse aggregate is reduced by 10%

volume of coarse aggregate =0.636 x 0.91 = 0.578

Volume of fine aggregate content = 1 – 0.578= 0.421

7.MIX CALCULATIONS

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

a) Volume of concrete = 1 m^3

b) Volume of cement = Mass of cement / (Specific Gravity of Cement x 1000)

= 424/(3.15 x 1000)

= 0.134 m^3

c) Volume of water = Mass of water / (Specific Gravity of water x 1000)

= 139.87/(1 x 1000)

= 0.139 m^3

d)Super plastiziser is upto 2% of mass of total cementitious content

= (10.6/1.15)X(1/1000) = 0.0092 m^3

e) Volume of all in aggregate = a-(b+c+d+e)

= 1-(0.134+0.139+0.0092)

= 0.717 m^3

f) Mass of coarse aggregate= f x Volume of Coarse Aggregate x Specific Gravity of coarse aggregate x 1000

= 0.717x 0.578 x 2.73 x 1000

= 1131 kg

g) Mass of fine aggregate= f x Volume of Fine Aggregate x Specific Gravity of Fine Aggregate x 1000

= 0.717x 0.421x 2.61 x 1000

= 787 kg

8.MIX PROPORTIONS

Cement = 424 kg/m^3

Fine aggregate = 787 kg/m^3

Coarse aggregate = 1131 kg/m^3

w/c = 0.33

Admixture = 10.6 kg/m^3

Result:

The mix proportions arrived: 1:1.85:2.66