Challenge

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

INTRODUCTION:-

A) M35 grade concrete

Grade designation:- M35

Type of cement:- OPC 43 grade conforming to IS 8112

Type of mineral admixture:- Fly ash conforming to IS 3812 (Part-1)

Maximum nominal size of aggregate:- 20 mm

Minimum cement content:- 320 kg/m^3 (From table 5 of IS 456:2000-exposure condition

Maximum water cement ratio:- 0.45

Workability:- 100 mm (Slump value for pumpable concrete)

Exposure condition:- Severe

Method of concrete placing:- Pumping

Degree of supervision:- Good

Type of aggregate:- Crushed angular aggregate

Maximum cement content:- 450 kg/m^3

Chemical admixture type:- Superplasticiser

Specific gravity of cement:- 3.15

Specific gravity of fly ash:- 2.2

Specific gravity of coarse aggregate & fine aggregate:- 2.74

B) Grade designation:- M50

Type of cement:- OPC 43 grade conforming to IS 8112

Maximum nominal size of aggregate:- 20 mm

Minimum cement content:- 320 kg/m^3 (From table 5 of IS 456:2000-exposure condition

Maximum water cement ratio:- 0.45

Workability:- 100 mm (Slump value for pumpable concrete)

Exposure condition:- Severe

Method of concrete placing:- Pumping

Degree of supervision:- Good

Type of aggregate:- Crushed angular aggregate

Maximum cement content:- 450 kg/m^3

Chemical admixture type:- Superplasticiser

Specific gravity of cement:- 3.15

Specific gravity of fly ash:- 2.2

Specific gravity of coarse aggregate & fine aggregate:- 2.74

PROCEDURE:-

A) a) Target strength for mix proportioning:-

f'ck = fck + 1.65s = 35 + 1.65*5 = 43.25 N/mm^2 where f'ck denotes target average compressive strength in 28 days , fck denotes characteristic compressive strength at 28 days and s represents standard deviation

To get the s value from table 1 of IS 10262:2009

b) Selection of w/c ratio:- From table 5 of IS 456:2000 for severe exposure, maximum w/c ratio = 0.45

Adopt water cement ratio as 0.40

c) Selection of water content:- From table 2 of IS 10262:2009, maximum water content for 20 mm aggregate = 186 litre for 25-50 mm slump

Estimated water content for 100 mm slump = 186+(6*186/100) = 197 litres

As superplasiciser is used the water content can be reduced upto 30%

Based on trials, water content reduction of 29% is achieved with plasticiser.

Hence the arrived water content = 197*(100-29)/100 = 140 litre

d) Calculation of cement content:- water-cement ratio = 0.4

Cementitious material(Cement + fly ash) = 140/0.4 = 350 kg/m^3

From table 5 of IS 456:2000, minimum cement content for severe exposure condition = 320 kg/m^3

350 kg/m^3 > 320 kg/m^3. Hence OK

Let the increase of 10% cementitious material based on experience and trial method

Cementitious material content = 350*1.10 = 385 kg/m^3

Water content = 140 kg/m^3

So w/c ratio = 140/385 = 0.364

Fly ash @ 30% total cementitious material content = 385*30% = 115 kg/m^3

Cement (OPC) = 385-115 = 270 kg/m^3

e) Proportion of volume of coarse aggregate and fine aggregate content :- From table 3 of IS 10262:2009, volume of coarse aggregate corresponding to 20 mm size aggregate and fine aggregate (zone 1 ) for w/c ratio of 0.5 = 0.6.In the present case the water cement ratio is 0.40.

Now the corrected value of coarse aggregate for the water cement ratio of 0.40 = 0.62

For pumpable concrete, these values are reduced by 10%

Volume of coarse aggregate = 0.62*0.9 = 0.56

Volume of fine aggregate = 1-0.56 = 0.44

f) Mix calculations:-

Volume of concrete = 1m^3

Volume of cement = Mass of cement/Specific gravity of cement * 1/1000 = 270/3.15*1/1000 = 0.086 m^3

Volume of fly ash = Mass of fly ash/Specific gravity of fly ash *1/1000 = 115*1/(2.2*1000) = 0.052 m^3

Volume of water = Mass of water/Specific gravity of water * 1/1000 = 140/1*1/1000 = 0.140 m^3

Volume of chemical admixture = Mass of chemical admixture/Specific gravity of admixture * 1/1000 = 7.5/1.145*1/1000 = 0.006 m^3

Volume of aggregate = 1-(0.086+0.052+0.140+0.006) = 0.716 m^3

Mass of coarse aggregate = Volume of aggregate*Volume of coarse aggregate * Specific gravity of coarse aggregate * 1000 = 0.716*0.56*2.74*1000 = 1098 kg/m^3

Mass of fine aggregate = Volume of aggregate*Volume of fine aggregate*Specific gravity of fine aggregate*1000 = 0.716*0.44*2.74*1000 = 863 kg/m^3

B) a)

a) Target strength for mix proportioning:-

f'ck = 50 + 1.65s = 35 + 1.65*5 = 58.25 N/mm^2 where f'ck denotes target average compressive strength in 28 days , fck denotes characteristic compressive strength at 28 days and s represents standard deviation

To get the s value from table 1 of IS 10262:2009

b)

Selection of w/c ratio:- From table 5 of IS 456:2000 for severe exposure, maximum w/c ratio = 0.45

Adopt water cement ratio as 0.40

c) Selection of water content:- From table 2 of IS 10262:2009, maximum water content for 20 mm aggregate = 186 litre for 25-50 mm slump

Estimated water content for 100 mm slump = 186+(6*186/100) = 197 litres

As superplasiciser is used the water content can be reduced upto 30%

Based on trials, water content reduction of 29% is achieved with plasticiser.

Hence the arrived water content = 197*(100-29)/100 = 140 litre

d) Calculation of cement content:- w/c ratio = 0.4

Cement content = 140/0.4 = 350 kg/m^3

From table 5 of IS 456 minimum water cement content for severe exposure condition = 320 kg/m^3

350 kg/m^3 > 320 kg/m^3. Hence OK

e) Proportion of volume of coarse aggregate and fine aggregate:-

From table 3 of IS 10262:2009, volume of coarse aggregate corresponding to 20 mm size aggregate and fine aggregate (zone 1 ) for w/c ratio of 0.5 = 0.6.In the present case the water cement ratio is 0.40.

Now the corrected value of coarse aggregate for the water cement ratio of 0.40 = 0.62

For pumpable concrete, these values are reduced by 10%

Volume of coarse aggregate = 0.62*0.9 = 0.56

Volume of fine aggregate = 1-0.56 = 0.44

f) Mix calculations;-

Volume of concrete = 1 m^3

Volume of cement = Mass of cement/Specific gravity of cement*1/1000 = 350/3.15*1/1000 = 0.111 m^3

Volume of water = Mass of water/Specific gravity of water*1/1000 = 140/1*1/1000 = 0.140 m^3

Volume of chemical admixture = Mass of chemical admixture/Specific gravity of admixture*1/1000 = 7.6/1.145*1/1000 = 0.006 m^3

Volume of aggregates = 1-(0.111+0.140+0.006) = 0.743 m^3

Mass of coarse aggregate = Volume of aggregates*Volume of coarse aggregate*Specific gravity of coarse aggregate*1000 = 0.743*0.56*2.74*1000 = 1140 kg/m^3

Mass of fine aggregates = 0.743*0.44*2.74*1000 = 896 kg/m^3

RESULT:A) Mix proportion for M35 grade concrete

Cement = 270 kg/m^3

Fly ash = 115 kg/m^3

Water = 140 kg/m^3

Fine aggregate = 863 kg/m^3

Coarse aggregate = 1098 kg/m^3

Chemical admixture = 7 kg/m^3

w-c ratio = 0.364

Trial mix ratio = 1:2.24:2.85

B) Mix proportion for M50 grade concrete

Cement = 350 kg/m^3

Water = 140 kg/m^3

Fine aggregate = 896 kg/m^3

Coarse aggregate = 1140 kg/m^3

Chemical admixture = 7 kg/m^3

w/c ratio = 0.4

Trial mix ratio = 1:2.56:3.26

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