Project on Concrete Mix Design for various grades of Concrete

1. Mix design for M35 concrete with fly ash-

Grade of concrete- M35

Exposure condition- severe

Type of mineral admixture- flyash

Type of cement- opc 43

Aggregate size – 20 mm

Workability – 100mm slump

Concrete placing – pumping

Admixture – superplasticizer

Min cement content – 320kg/m3 (IS 456:2000)

• Target mean strength = fck+1.65s

=35+1.65*(5) s= standard deviation for M35

= 43.25 N/mm2

• Selection of w/c ratio

Moderate exposure max w/c ratio is 0.4.

• Selection of water content

From IS 10262:2009, max water content for 20 mm aggregate is 186 litres.

Estimated water content for 100 mm slump

= 186+6/100*186.

= 197

• As superplasticizer is used water content is reduced up to 30% and above

Based on trial water content reduction of 30% achieved with plasticizer.

= 197*0.70= 140 litres

• Calculation of cement content

w/c ratio = 0.4

= 140/0.40

=350 kg/m3

350 > 320 hence ok

• Percentage of fly ash needs to be decided-

For this increase of 10% cementatious material content –

Cementatious material content= 350*101=385kg/m3

Water content- 140 kg/m3

w/c- 140/385 = 0.364

fly ash @ 30% of total cementatious material content =385*30%

=115 kg/m3

Cement opc = 385-115= 270 kg/m3.

• Proportion of volume of coarse aggregate and fine aggregate table 3 IS 10262:2009

Vol of coarse aggregate corresponding to 20 mm size aggregate and fine aggregate for zone 1 for w/c ratio = 0.60

Present w/c ratio = 0.4

Hence vol coarse aggregate need to be increased to decreased fine aggregate content

So corrected vol of coarse aggregate for the w/c ratio of 0.4 = 0.62

For pumpable concrete value are reduced by 10%

Vol of coarse aggregate 0.62*0.9= 0.56

Vol of fine aggregate 1-0.56= 0.44

• Mix calculation

1. Vol of concrete 1 m3
2. Vol of cement = mass of cement/ specific gravity *1/100

= 270/3.15*1/100

= 0.086 m3

1. Vol of fly ash = mass of fly ash/ specific gravity *1/100

= 115 / 2.2 *1/1000

= 0.052 m3

1. Vol of water = mass of water/ specific gravity of water*1/1000

= 140/1*1/1000

= 0.140 m3

1. Vol of chemical admixture = mass of chemical admixture/specific gravity admixture

= 7/1.145*1/1000

= 0.006 m3

1. Vol of (CA and FA) = (a-(b+c+d+e))

= 1-(0.086+0.052+0.140+0.006)

= 0.716 m3

1. Mass of CA = e*vol of CA *SG of CA *1000

= 0.716*0.56*2.74*1000

= 1098 kg/m3

1. Mass of FA = e*vol of FA *SG of FA *1000

= 0.716*0.44*2.74*1000

= 863 kg/m3

Finally

Cement = 270 kg/m3

Fly ash = 115 kg/m3

Water = 140 litres

FA = 863 kg/m3

CA = 1098 kg/m3

So, ratio 1:2.24:2.85

2. Mix design for M50 concrete -

Grade of concrete- M50

Exposure condition- severe

Type of cement- opc 43

Aggregate size – 20 mm

Workability – 100mm slump

Concrete placing – pumping

Admixture – superplasticizer

Min cement content – 320kg/m3 (IS 456:2000)

• Target mean strength = fck+1.65s

=50+1.65*(5) s= standard deviation for M50

= 58.25 N/mm2

• Selection of w/c ratio

Moderate exposure max w/c ratio is 0.4.

• Selection of water content

From IS 10262:2009, max water content for 20 mm aggregate is 186 litres.

Estimated water content for 100 mm slump

= 186+6/100*186.

= 197

• As superplasticizer is used water content is reduced up to 30% and above

Based on trial water content reduction of 30% achieved with plasticizer.

= 197*0.70= 140 litres

• Calculation of cement content

w/c ratio = 0.4

= 140/0.40

=350 kg/m3

350 > 320 hence ok

• Proportion of volume of coarse aggregate and fine aggregate table 3 IS 10262:2009

Vol of coarse aggregate corresponding to 20 mm size aggregate and fine aggregate for zone 1 for w/c ratio = 0.60

Present w/c ratio = 0.4

Hence vol coarse aggregate need to be increased to decreased fine aggregate content

So corrected vol of coarse aggregate for the w/c ratio of 0.4 = 0.62

For pumpable concrete value are reduced by 10%

Vol of coarse aggregate 0.62*0.9= 0.56

Vol of fine aggregate 1-0.56= 0.44

• Mix calculation

1. Vol of concrete 1 m3
2. Vol of cement = mass of cement/ specific gravity *1/100

= 350/3.15*1/100

= 0.111 m3

1. Vol of water = mass of water/ specific gravity of water*1/1000

= 140/1*1/1000

= 0.140 m3

1. Vol of chemical admixture = mass of chemical admixture/specific gravity admixture

= 7/1.145*1/1000

= 0.006 m3

e.Vol of (CA and FA) = (a-(b+c+d))

= 1-(0.111+0.140+0.006)

= 0.743 m3

1. Mass of CA = e*vol of CA *SG of CA *1000

= 0.743*0.56*2.74*1000

= 1140 kg/m3

1. Mass of FA = e*vol of FA *SG of FA *1000

= 0.743*0.44*2.74*1000

= 896 kg/m3

Finally

Cement = 350 kg/m3

Water = 140 litres

FA = 896 kg/m3

CA = 1140 kg/m3

So, ratio 1:2.56:3.26