Project on Concrete Mix Design for various grades of Concrete

PROJECT 1

AIM

To calculate the Concrete Mix Design for M35 grade concrete with Fly ash.

OBJECTIVE

To achieve specified characteristics at specified age, workability of fresh concrete and durability requirements.

CODES FOR CONCRETE MIX DESIGN

• IS 10262: 2009
• IS 456: 2000

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 x 1000
• Mass of fine aggregate : Volume of aggregate x Volume of fine aggregate x Specific gravity of fine aggregate x 1000

PROCEDURE AND CALCULATION

1. TARGET MEAN STRENGTH

f’ck = fck + 1.65 5

fck= 35 N/mm^2 for M35

As per Table 1 of IS 10262:2009,

Standard deviation, s = 5 N/mm^2

f ‘ck = 35 + ( 1.65 x 5)

= 43.25 N/mm^2

2. SELECTION OF WATER CEMENT RATIO

Assume:

Exposure condition : Extreme ( for Reinforced concrete)

As per Table 5 of IS 456:2000,

Max. w/c ratio for Extreme condition = 0.4

Max. w/c ratio for M35 = 0.45

0.40 < 0.45

Take,

Water cement ratio = 0.40

Minimum cement content = 360 kg/m^3

3. SELECTION OF WATER CONTENT

Assume,

Nominal size of aggregates = 20 mm

As per Table 2 of IS 10262: 2009,

For 20mm aggregates, Max. water content = 186 L (Slump ranges from 25 to 50 mm)

As superplasticizers (20 % reduced)

Water content = 186 x 80/100 = 148.8

= 149 L

4. CALCULATION OF CEMETITIOUS CONTENT

We've ,

w/c ratio = 0.4

Water content = 149 L

Cementitious content (cement + fly ash) = 149/0.4

= 372.5 kg/m^3

Take,

Cementitious content = 370 kg/m^3

Minimum cement content = 360 kg/m^3

360< 370 kg/m^3, Hence okay.

Because of the use of fly ash, cement content is increased by 10%.

Cement content = 370 x 1.10 = 407 kg/m^3

Water cement ratio = 149/407 = 0.366

Fly ash at 30 % total cementitious material content = 407 x 30/100 = 122 kg/m^3

Cement = 407 - 122 = 285 kg/m^3

Saving of cement while using fly ash = 370 - 285 = 85 kg/ m^3

5. VOLUME OF COARSE AGGREGATES AND FINE AGGREGATES

Fine aggregates are sieved according to Zone 1.

According to Table 3 of IS 10262: 2009,

Volume of coarse aggregate corresponding to 20 mm size aggregate and fine aggregate (Zone I) for water-cement ratio of 0.50 = 0.60

We've, Water cement ratio = 0.4

As the w/c ratio is lower by 0.10, the proportion of volume of coarse aggregate is increased by 0.02.

So, Volume of coarse aggregate = 0.62

Volume of fine aggregate = 1 - 0.62

= 0.38

6. MIX CALCULATION

a) Volume of concrete = 1 m^3

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

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

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

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

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

= 1 - (0.09+0.055+0.149+0.006)

= 0.7 m^3

g) Mass of coarse aggregate = f x volume of coarse aggregate x Specific gravity of coarse aggregate x 1000

= 0.70 x 0.62 x 2.74 x 1000 = 1189.16 kg/m^3

h) Mass of fine aggregate = f x volume of fine aggregate x Specific gravity of fine aggregate x 1000

= 0.70 x 0.38 x 2.74 x 1000 = 728 kg/m^3

RESULTS

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

CONCLUSION

• This trial mix ratio should match the workability, durability and minimum strength of M35 grade concrete.
• If it does not satisfy the above criteria, two or more trials for variation of ± 10 percent of w/c ratio should be done.

PROJECT 2

AIM

To calculate the Concrete Mix Design for M50 grade concrete without Fly ash.

OBJECTIVE

To achieve specified characteristics at specified age, workability of fresh concrete and durability requirements.

CODES FOR CONCRETE MIX DESIGN

• IS 10262: 2009
• IS 456: 2000

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 x 1000
• Mass of fine aggregate : Volume of aggregate x Volume of fine aggregate x Specific gravity of fine aggregate x 1000

PROCEDURE AND CALCULATION

1. TARGET MEAN STRENGTH

f’ck = fck + 1.65 5

fck= 50 N/mm^2 for M50

As per Table 1 of IS 10262:2009,

Standard deviation, s = 5 N/mm^2

f ‘ck = 50 + ( 1.65 x 5)

= 58.25 N/mm^2

2. SELECTION OF WATER CEMENT RATIO

Assume:

Exposure condition : Extreme ( for Reinforced concrete)

As per Table 5 of IS 456:2000,

Max. w/c ratio for Extreme condition = 0.4

Max. w/c ratio for M50 = 0.4

Take,

Water cement ratio = 0.40

Minimum cement content = 360 kg/m^3

3. SELECTION OF WATER CONTENT

Assume,

Nominal size of aggregates = 20 mm

As per Table 2 of IS 10262: 2009,

For 20mm aggregates, Max. water content = 186 L (Slump ranges from 25 to 50 mm)

As superplasticizers (20 % reduced)

Water content = 186 x 80/100 = 148.8

= 149 L

4. CALCULATION OF CEMETITIOUS CONTENT

We've ,

w/c ratio = 0.4

Water content = 149 L

Cementitious content = 149/0.4

= 372.5 kg/m^3

Take,

Cementitious content = 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.

According to Table 3 of IS 10262: 2009,

Volume of coarse aggregate corresponding to 20 mm size aggregate and fine aggregate (Zone I) for water-cement ratio of 0.50= 0.60

We've, Water cement ratio = 0.4

As the w/c ratio is lower by 0.10, the proportion of volume of coarse aggregate is increased by 0.02.

So, Volume of coarse aggregate = 0.62

Volume of fine aggregate = 1 - 0.62

= 0.38

6. MIX CALCULATION

a) Volume of concrete = 1 m^3

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

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

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

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

= 1 - (0.118+0.149+0.006)

= 0.727 m^3

f) Mass of coarse aggregate = e x volume of coarse aggregate x Specific gravity of coarse aggregate x 1000

= 0.727 x 0.62 x 2.74 x 1000 = 1235.02 kg/m^3

g) Mass of fine aggregate = e x volume of fine aggregate x Specific gravity of fine aggregate x 1000

= 0.727 x 0.38 x 2.74 x 1000 = 756.9 kg/m^3

RESULTS

TRIAL MIX:

Cement = 373 kg/m^3

Water = 149 kg

Coarse aggregate = 1235 kg/m^3

Fine aggregate = 757 kg/m^3

Chemical admixture = 7.6 kg/m^3

Water cement ratio = 0.4

Trial mix ratio = 373 :756.9 : 1235.02

= 1: 2.02: 3.31

CONCLUSION

• This trial mix ratio should match the workability, durability and minimum strength of M50 grade concrete.
• If it does not satisfy the above criteria, two or more trials for variation of ± 10 percent of w/c ratio should be done.