Project on Concrete Mix Design for various grades of Concrete

AIM: To Calculate the Concrete Mix Design for M35 grade concrete with fly ash & M50 grade concrete without Fly ash.

EQUATIONS:

Target mean strength : f’ck= fck +1.65 s

f’ck = target mean strength

fck= compressive strength at 28 days of concrete

s = standard deviation ( IS 10262:2009 , table -1)

Volume of cement : (mass of cement /specfic gravity of cement )x1 /1000

Volume of water : ( mass of water /specific gravity of water ) x1 /1000 Volume of chemical admixtures : ( mass of chemical admixture / specific gravity of chemical admixture )x 1/1000

Volume of fly ash : (mass of fly ash/specific gravity of fly ash )x 1/1000 Volume of concrete: 1 m^3

Volume of aggregates : ( volume of concrete -( volume of cement +volume of fly ash + volume of water + volume of admixture )

Mass of coarse aggregates: volume of aggregate x volume of coarse aggregate x specific gravity of coarse aggregate x1000

Mass of fine aggregate : volume of aggregate x volume of fine aggregate x specific gravity of fine aggregate x1000

CODES FOR CONCRETE MIX DESIGN: IS 10262: 2009 IS 456: 2000 IS 8112 IS 9103 IS 3812 ( part 1) OBJECTIVES: The main objective is to produce concrete as economically as possible To achieve desired workability, durability and minimum strength of M35 grade concrete

OBJECTIVES: The main objective is to produce concrete as economically as possible To achieve desired workability, durability and minimum strength of M35 grade concrete

PROCEDURE AND CALCULATION:

TARGET MEAN STRENGTH

F’ck = fck + 1.65 5

fck= 35 N/mm^2 (M35 ) Table 1- IS 10262:2009

f ‘ck = 35 + ( 1.65 x 5)=43.25 N/mm^2 2. SELECTION OF WATER CEMENT RATIO :

Given : M35

Assume: Exposure condition : Extreme ( for reinforced concrete) Table 5 – IS 456 : 2000 Therefore,

Extreme= w/c ratio= 0.40

M35 = w/c ratio= 0.45

Lower the water cement ratio better the workability 0.40 < 0.45

Water cement ratio = 0.40

Minimum cement content = 360 kg/m^3

3. SELECTION OF WATER CONTENT Assume : Nominal size of aggregates= 20 mm

Table 2 -IS 10262: 20093.

SELECTION OF WATER CONTENT Assume :Nominal size of aggregates= 20 mm Table 2 -IS 10262: 2009

Minimum water content = 186 litres or kg Slump ranges from 25 to 50 mm As superplasticizers (20 % reduced) = 186 x 80/100 = 148.8 (100-20 = 80) Water content = 149 litres

4. CALCULATION OF CEMETITIOUS CONTENT ( CEMENT + FLY ASH) Water cement ratio = w/c= 0.40 Water content = W = 149 kg Cementitious content = c=149/0.40 = 372.5 = 370 kg/m^3 Minimum cement content = 360 kg/m^3 360< 370 kg/m^3 ( hence okay ) Increase by 10 % 370 x 0.10 + 370 = 407 kg/m^3 W= 149 kg Water cement ratio = 149/407 = 0.366 Fly ash at 30 % total cementitious material content = 407 x 30/100 = 122 kg/m^3 Cement (ordinary Portland cement ) = 407-122 = 285 kg/m^3 Saving of cement while using fly ash = 370 – 285 = 85 kg/ m^3 Fly ash = 122 kg/m^3

5.VOLUME OF COARSE AGGREGATES AND FINE AGGREGATES Fine aggregates are sieved according to zone 1 of table 4 IS 383 Coarse aggregates are sieved according to table 2 IS 383 According , Table 3 IS 10262: 2009 Nominal size of aggregates = 20 mm Fine aggregates ( zone 1) Therefore , Volume of coarse aggregate = 0.60Minimum water content = 186 litres or kg Slump ranges from 25 to 50 mm As superplasticizers (20 % reduced) = 186 x 80/100 = 148.8 (100-20 = 80) Water content = 149 litres

4. CALCULATION OF CEMETITIOUS CONTENT ( CEMENT + FLY ASH)

Water cement ratio = w/c= 0.40 Water content = W = 149 kg

Cementitious content = c=149/0.40 = 372.5 = 370 kg/m^3

Minimum cement content = 360 kg/m^3 360< 370 kg/m^3 ( hence okay ) Increase by 10 % 370 x 0.10 + 370 = 407 kg/m^3 W= 149 kg Water cement ratio = 149/407 = 0.366 Fly ash at 30 % total cementitious material content = 407 x 30/100 = 122 kg/m^3 Cement (ordinary Portland cement ) = 407-122 = 285 kg/m^3 Saving of cement while using fly ash = 370 – 285 = 85 kg/ m^3 Fly ash = 122 kg/m^3

5.VOLUME OF COARSE AGGREGATES AND FINE AGGREGATES Fine aggregates are sieved according to zone 1 of table 4 IS 383 Coarse aggregates are sieved according to table 2 IS 383 According , Table 3 IS 10262: 2009

Nominal size of aggregates = 20 mm

Fine aggregates ( zone 1)

Therefore ,Volume of coarse aggregate = 0.60.

Water cement ratio = 0.4 ( +/- 0.01 for every +/- 0.005) So, volume of coarse aggregate = 0.62 Volume of fine aggregate = 1-0.62 = 0.38

6.MIX CALCULATION :

Volume of concrete = 1m^3 Volume of cement = (285 / 3.15) x1/1000= 0.090 m^3

Volume of fly ash = ( 122/2.2) x1/1000=0.055 m^3

Volume of water = ( 149/1)x 1/1000=0.149 m^3

Volume of admixture = ( 7.6/ 1.145)x1/1000=0.006 m^3

Volume of coarse aggregate and fine aggregate = a – ( b + c + d+ e) = 1-( 0.090 + 0.055 +0.149+ 0.006 )= 0.7 m^3 Mass of coarse aggregate = 0.7 x 0.62 x 2.74 x1000 = 1189.16 kg/m^3 Mass of fine aggregate = 0.7 x 0.38 x 2.74 x1000 = 728 kg/m^3

RESULTS AND CONCLUSION :

TRIAL MIX:Cement = 285 kg/m^3

Fly ash = 122 kg/m^3

Water = 149 kg

Coarse aggregate = 1189 kg/m^3

Fine aggregate = 728 kg/m^3

Chemical admixture = 7.6 kg/m^3

Water cement ratio = 0.366

Trial mix ratio : 285 :728 : 1189= 1: 2.55 : 4.17

(C: FA: CA)

This trial mix ratio should match the workability ,durability, minimum strength of M35 grade concrete . If it does not satisfy the above criteria two or more trials for increase of 10 % should be done.

AIM:

To calculate M50 grade without fly ash.

EQUATIONS:

Target mean strength : f’ck= fck +1.65 s

f’ck = target mean strength

fck= compressive strength at 28 days of concrete

s =standard deviation ( IS 10262:2009 , table -1) Volume of cement : (mass of cement /specific gravity of cement )x1/1000

Volume of water : ( mass of water/specific gravity of water ) x1 /1000

Volume of chemical admixtures : ( mass of chemical admixture/ specific gravity of chemical admixture )x 1/1000

Volume of concrete: 1 m^3

Volume of aggregates : ( volume of concrete -( volume of cement + volume of water+volume of chemical admixture) )

Mass of coarse aggregates: volume of aggregate x volume of coarse aggregate x specific gravity of coarse aggregate x1000

Mass of fine aggregate : volume of aggregate x volume of fine aggregate x specific gravity of fine aggregate x1000

CODES FOR CONCRETE MIX DESIGN:

IS 10262: 2009 IS 456: 2000 IS 8112 IS 9103 IS 3812 ( part 1)

OBJECTIVES: The main objective is to produce concrete as economically as possible To achieve desired workability, durability and minimum strength of M 50 grade concrete

PROCEDURE AND CALCULATION: TARGET MEAN STRENGTH

F’ck = fck + 1.65 5

fck= 50 N/mm^2 (M50 )

Table 1- IS 10262:2009

s = 5 N/mm^2

f ‘ck = 50 + ( 1.65 x 5) = 58.25 N/mm^2

2. SELECTION OF WATER CEMENT RATIO :

Given : M50

Assume: Exposure condition : Extreme ( for reinforced concrete)Table 5 – IS 456 : 2000

Therefore,Extreme= w/c ratio= 0.40

M50 = w/c ratio= 0.40

Lower the water cement ratio better the workability 0.40

Water cement ratio = 0.40

Minimum cement content = 360 kg/m^3

3. SELECTION OF WATER CONTENT Assume

Nominal size of aggregates= 20 mm

Minimum water content = 186 litres or kg

Slump ranges from 25 to 50 mm

As superplasticizers (20 % reduced)

= 186 x 80/100 = 148.8 (100-20 = 80)

Water content = 149 litres

4.CALCULATION OF CEMENT CONTENT

Water cement ratio = w/c= 0.40

Water content = W = 149 kg

Cement content = c=149/0.40 = 372.5 = 373 kg/m^3

Minimum cement content = 360 kg/m^3

360< 373 kg/m^3 ( hence okay )

5.VOLUME OF COARSE AGGREGATES AND FINE AGGREGATES

Fine aggregates are sieved according to zone 1 of table 4 IS 383

Coarse aggregates are sieved according to table 2 IS 383

According Table 3 IS 10262: 2009

Nominal size of aggregates = 20 mm

Fine aggregates ( zone 1)

Volume of coarse aggregate = 0.60

2. SELECTION OF WATER CEMENT RATIO : Given : M50

Exposure condition : Extreme ( for reinforced concrete)

Table 5 – IS 456 : 2000

Therefore Extreme= w/c ratio= 0.40

M50 = w/c ratio= 0.40

Lower the water cement ratio better the workability 0.40 Water cement ratio = 0.40 Minimum cement content = 360 kg/m^3

3. SELECTION OF WATER CONTENT Assume : Nominal size of aggregates= 20 mm Table 2 -IS 10262: 2009

Minimum water content = 186 litres or kg Slump ranges from 25 to 50 mm

As superplasticizers (20 % reduced) = 186 x 80/100 = 148.8 (100-20 = 80)

Water content = 149 litres

4.CALCULATION OF CEMENT CONTENT Water cement ratio = w/c= 0.40 Water content = W = 149 kg Cement content = c=149/0.40 = 372.5 = 373 kg/m^3 Minimum cement content = 360 kg/m^3 360< 373 kg/m^3 ( hence okay )

5.VOLUME OF COARSE AGGREGATES AND FINE AGGREGATES Fine aggregates are sieved according to zone 1 of table 4 IS 383 Coarse aggregates are sieved according to table 2 IS 383

According , Table 3 IS 10262: 2009 Nominal size of aggregates = 20 mm

Fine aggregates ( zone 1)

Therefore ,Volume of coarse aggregate = 0.60

Water cement ratio = 0.4 (+/- 0.01 for every +/- 0.005)

So, volume of coarse aggregate = 0.62

Volume of fine aggregate = 1-0.62 = 0.38

6.MIX CALCULATION : Volume of concrete = 1m^3

Volume of cement = (373/3.15)x1/1000 = 0.118 m^3

Volume of water = ( 149/1)x1/1000= 0.149 m^3 Volume of admixture = ( 7.6/1.145)x1/1000 = 0.006

Volume of coarse aggregate and fine aggregate = a – ( b + c + d) = 1-( 0.118 + 0.149 + 0.006 )= 0.727 m^3

Mass of coarse aggregate = 0.727 x 0.62 x 2.74 x1000 = 1235.02 kg/m^3

Mass of fine aggregate = 0.727 x 0.38 x 2.74 x1000 = 756.9 kg/m^3

RESULTS AND CONCLUSION :

TRIAL MIX:

Cement = 373 kg/m^3

Water = 149 kg

Coarse aggregate = 1235.02 kg/m^3

Fine aggregate = 756.9 kg/m^3

Chemical admixture = 7.6 kg/m^3

Water cement ratio = 0.40 Trial mix ratio : 373 :756.9 : 1235.02 = 1: 2.02: 3.31 ( C: FA: CA)

This trial mix ratio should match the workability ,durability, minimum strength of M50 grade concrete .

If it does not satisfy the above criteria two or more trials for increase of 10 % should be done.