Project on Concrete Mix Design for various grades of Concrete

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

AIM:

To calculate the concrete mix design for M35 grade concrete with fly ash

EQUATIONS:

• Target mean strength : f‘ ck= fck+1.65S

f’ ck =target mean strength

fck= compressive strength at 28 day of concrete

S= Standard deviation (IS 10262:2009, Table -1)

• volume of cement : ( Mass of cement /Specific gravity of cement ) x 1/1000
• Volume of water : ( Mass of water / Specific gravity of water ) x 1/1000
• 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 m³
• 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 coarse aggregate x 1000
• Mass of fine aggregate : Volume aggregate x Volume of fine aggregate x Specific gravity of fine aggregate x 1000

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 M35 grade concrete

PROCEDURE AND CALCULATION :

1. TARGET MEAN STRENGTH

f’ ck = fck + 1.65 S

fck= 35 N/mm² ( M35)

Table 1- IS10262 :2009

S= 5 N/mm²

f’ ck= 35+ (1.65x5)

= 43.25 N/mm²

1. 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 = 360kg /m³

3. SELECT THE 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 Super plasticizers ( 20 % reduced )

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

Water content = 149 litres

4 . CALCULATION OF CEMENTITIOUS 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 /m³

Minimum cement content = 360kg / m³

360 < 370 kg / m³ ( hence okay )

Increase by 10%

370x 0.10 +370 = 407 kg /m³

W= 149 kg

Water cement ratio =149/407 =0.366

Fly ash at 30% total cementitious material content

= 407 x 30/100= 122kg /m³

Cement ( ordinary Portland cement ) = 407 -122= 285 kg/m³

Saving of cement while using fly ash = 370-285 = 85 kg/ m³

Fly ash = 122 kg /m³

5. VOLUME OF COARESE AGGREGATE AND FINE AGGREGATE

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 sizes of aggregate = 20 mm

Fine aggregate (Zone I )

Therefore ,

Volume of coarse aggregate = 0.60

Water cement ratio =0.40

(+/ -0.01 for every +/- 0.005)

So, volume of coarse aggregate =0.62

Volume of fine aggregate = 1-0.62 =0.38

6.MIX CALCUALTION

1. Volume of concrete = 1m³
2. Volume of cement = (285/3.15)x 1000=0.090 m³
3. Volume of fly ash =(122/2.2)x1/1000= 0.055m³
4. Volume of water =(149 /1)x 1/1000=0.149m³
5. Volume of admixture =(7.6/1.145)x1/1000=0.006m³
6. Volume of coarse aggregate and fine aggregate = a-(b+c+d)= 1+(0.090+0.055+0.006) =0.7 m³
7. Mass of coarse aggregate = 0.7x0.62x 2.74x1000= 1189.16 kg/m³
8. Mass of fine aggregate = 0.7x 0.38x 2.74 x 1000= 728kg /m³

RESULTS AND CONCLUSION :

TRIAL MIX :

Cement = 285 kg /m³

Fly Ash= 122kg /m³

Water = 149kg

Coarse aggregate= 1189kg/m³

Fine aggregate = 728kg/m³

Chemical admixture = 7.6kg /m³

Water cement ratio= 0.366

Trail mix ratio: 285: 728: 1189= 1:2.55:4.17 ( C: FA: CA)

• This trail mix ratio should match 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+1.65S

f’ ck =target mean strength

fck= compressive strength at 28 day of concrete

S= Standard deviation (IS 10262:2009, Table -1)

• volume of cement : ( Mass of cement /Specific gravity of cement ) x 1/1000
• Volume of water : ( Mass of water / Specific gravity of water ) x 1/1000
• Volume of chemical admixtures : ( Mass of chemical admixture / specific gravity of chemical admixture) x 1/1000
• Volume of concrete: 1 m³
• Volume of aggregates : ( Volume of concrete – (volume of cement + Volume of water + Volume of admixture )
• Mass of coarse aggregates : Volume of aggregate x volume of coarse aggregate x Specific gravity coarse aggregate x 1000
• Mass of fine aggregate : Volume aggregate x Volume of fine aggregate x Specific gravity of fine aggregate x 1000

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 M50 grade concrete

PROCEDURE AND CALCULATION :

1. TARGET MEAN STRENGTH

f’ ck = fck + 1.65 S

fck= 50 N/mm² ( M50)

Table 1- IS10262 :2009

S= 5 N/mm²

f’ ck= 50+ (1.65x5)

= 58.25 N/mm²

1. 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 = 360kg /m³

1. SELECT THE 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 Super plasticizers ( 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

Cementitious content = C= 149/ 0.40 = 372.5 =373 kg /m³

Minimum cement content = 360kg / m³

360< 373 kg / m³ ( hence okay )

5.VOLUME OF COARESE AGGREGATE AND FINE AGGREGATE

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 sizes of aggregate = 20 mm

Fine aggregate (Zone I )

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

MIX CALCUALTION

1. Volume of concrete = 1m³
2. Volume of cement = (285/3.15)x 1000=0.118 m³
3. Volume of water =(149 /1)x 1/1000=0.149m³
4. Volume of admixture =(7.6/1.145)x1/1000=0.006m³
5. Volume of coarse aggregate and fine aggregate = a-(b+c+d)= 1+(0.090+0.055+0.006) =0.727 m³
6. Mass of coarse aggregate = 0.7x0.62x 2.74x1000= 1235.02 kg/m³
7. Mass of fine aggregate = 0.7x 0.38x 2.74 x 1000= 756.9kg /m³

RESULTS AND CONCLUSION :

TRIAL MIX :

Cement = 373 kg /m³

Water = 149kg

Coarse aggregate= 1235.02kg/m³

Fine aggregate = 756kg/m³

Chemical admixture = 7.6kg /m³

Water cement ratio= 0.40

Trail mix ratio: 373: 756.9: 1235.02 = 1:2.02:3.31 ( C: FA: CA)

• This trail mix ratio should match 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.