Challenge

Project on Concrete Mix Design for various grades of Concrete

Design M35 Grade concrete with Fly ash

Type of aggregate= Crushed angular aggregate

Type of cement is 53 grade conforming to IS8112

Size of aggregate= 20mm

Specific gravity oc cement=3.15

Workability: 100 mm slump

Exposure= Severe

Specific Gravity of fly ash= 2.2 (IS 3812 part 1)

Specific Gravity pf CA and FA=2.74

Admixture= Super-plasticizer

Water absorption: CA=0.5% ; FA=1%

(a) Target mean strenght of concrete:

F'ck=Fck+1.65S (From IS 10262, Table 1) at 28 days span

F'ck=35+1.65*5

F'ck=43.25 N/mm2

(b) Selection of w/c ratio:

Assumed severe condition

Min Cement content=340 kg/m3

Max w/c ratio =0.45 ( Table 5, IS 456)

Based on experience water cement content is 0.40<0.45 hence,Ok.

(c) Selection of water content:

Table 2 of IS 10262

20 mm aggregate is considered

Therefore; max water content for 25mm to 50mm slump is 186 litre

(3% added for every extra 25mm slump)

Considering 100 mm slump.( pumpable concrete)

So, 186+6/100*186=197 litres.

Super plastizer is used water content can be reduced upto 30%. Lets consider 29%

Therefore water content=197*0.71=140 litre.

(d) Selection of cement content:

Calculated by w/c ratio

Cement content ( cement + fly-ash)=140/.40= 350 kg/m3

350 kg/m3 > 340 kg/m3 ,HenceOK.

Increasing 10% of cementious matrerial

Cementious material content=350*1.10=385 Kg/m3

Water content=140 litre

W/c ratio= 140/385 =0.364

Fly ash @ 30% of total cementious material =385*30%=115 kg/m3

OPC= 385-115 =270 kg/m3

From Table 3 of IS 10262, Volume of CA of 20 mm size (Zone 1) 0.50 = 0.60

Water cement ratio is 0.40 (-/+ change of 0.01 foe every -/+ 0.05 change in water cement ratio)

So 0.10 is needed to be increased

So corrected value of Ca for w/c ratio of 0.10 is = 0.62

For pumpable concrete values are decreased by 10%.

So volume of coarse aggregate= 0.62*0.9=0.56

Volume of fine aggregate = 1-0.56=0.44.

(e) Mix calculation:

Volume of concrete= 1 m3

Volume of cement= 270/3.15 * 1/1000 = 0.086 m3

Volume of Fly-ash= 115/2.2 *1/1000= 0.052 m3

Volume of water= 140/1 * 1/1000 =0.140 m3

Volume of chemical admixture = 7.6/1.145 * 1/1000 = 0.006 m3

Volume of aggregate ( FA & CA)= (1-(.086+.052+.140+.006)) = 0.716 m3

Mass of CA = .716*.56*2.74*1000 =1098.63 Kg/m3

Mass of FA = .716*.44*2.74*1000 =863.20 Kg/m3

Design M50 Grade concrete

Type of aggregate= Crushed angular aggregate

Type of cement is 53 grade conforming to IS8112

Workability: 100 mm slump

Exposure= Severe

Specific gravity oc cement=3.15

Specific Gravity pf CA and FA=2.74

Admixture= Super-plasticizer

Water absorption: CA=0.5% ; FA=1%

(a) Target mean strenght of concrete:

F'ck=Fck+1.65S (From IS 10262, Table 1) at 28 days span

F'ck=50+1.65*5

F'ck=58.25 N/mm2

(b) Selection of w/c ratio:

Assumed severe condition

Cement content=360 kg/m3

w/c ratio =0.40 (Table 5, IS 456)

Based on experience water cement content is 0.38<0.40, hence Ok.

(c) Selection of water content:

Table 2 of IS 10262

20 mm aggregate is considered

Therefore; max water content for 25mm to 50mm slump is 186 litre

(3% added for every extra 25mm slump)

Considering 100 mm slump.( pumpable concrete)

So, 186+6/100*186=197 litres.

Super plastizer is used water content can be reduced upto 30%. Lets consider 29%

Therefore water content=197*0.71=140 litre.

(d) Selection of cement content:

Calculated by w/c ratio

Cement content=140/.38= 368.42 kg/m3

368.42 kg/m3 > 360 kg/m3 ,Hence not Ok

From Table 3 of IS 10262, Volume of CA of 20 mm size (Zone 1) 0.50 = 0.60

Water cement ratio is 0.38 (-/+ change of 0.01 foe every -/+ 0.05 change in water cement ratio)

So 0.12 is needed to be increased ( SO, (.01*.12)/.05=0.024)

So corrected value of Ca for w/c ratio of 0.38 is = 0.624

For pumpable concrete values are decreased by 10%.

So volume of coarse aggregate= 0.624*0.9=0.56

Volume of fine aggregate = 1-0.56=0.44.

(e) Mix calculation:

Volume of concrete= 1 m3

Volume of cement= 368.42/3.15 * 1/1000 = .116 m3

Volume of water= 140/1 * 1/1000 =0.140 m3

Volume of chemical admixture = 7.6/1.145 * 1/1000 = 0.006 m3

Volume of aggregate ( FA & CA)= (1-(.116+.140+.006)) = 0.738 m3

Mass of CA = .738*.56*2.74*1000 =1132.38 Kg/m3

Mass of FA = .738*.44*2.74*1000 =899.73 Kg/m3

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