Project on Concrete Mix Design for various grades of Concrete

Aim :- To Calculate the Concrete Mix Design for M35 grade concrete with fly ash

Introduction:-

Data collection

Grade of concrete = M35

Extreme weather condition = severe Exposure condition

Grade of cement = 43 (OPC)

Type of mineral admixture = fly ash conforming to IS 3812 (Part 1)

Maximum size of aggregate 20mm

Minimum cement content = 320 kg/m³ from table 5 IS456 2000 Exposure condition

Maximum w/c ratio = 0.45 from table 5 IS456 2000 Exposure condition

Workability 100mm (slump value for pump-able concrete)

Method of concrete placing – pumping

Degree of supervision – good

Type of aggregate- cursed angular aggregate

Maximum cement content 450 kg/m³

Chemical admixture type – super plasticizer confirming to IS 9103

Test data for materials

Cement used = opc 43 grade conforming to IS 8112

Specific gravity of cement 3.15

Fly ash = conforming to IS3812(Part 1)

Specific gravity of fly ash = 2.2

Specific gravity of coarse aggregate 2.74

Specific gravity of fine aggregate 2.74

Water absorption coarse aggregate 0.5 %

Water absorption fine aggregate 1.0 %

Free surface moisture coarse aggregate Nil

Free surface moisture fine aggregate Nil

Fck = 35 N/mm²

Standard deviation for M35 grade concrete is 5N/mm²

ft = fck + 1.65S

ft= 35 + 1.65*5

Target mean strength of concrete ft= 43.25 N/mm²

Step 1

From grape 1 w/c ratio is 0.37

Step 2

From grape 2 w/c ratio is 0.42

Step 3

From exposure condition w/c ratio is 0.45

Lowest w/c ratio is 0.37

So water cement ratio for mix design M35 grade of concrete and for severe exposure condition is 0.37

Selection of water content

From table 2 of IS 10262: 2009 maximum water content for 20mm aggregate = 186litre (for 25mm to 50mm slump range)

Estimated water content for 100 mm slump = 186 + 6*186/100

= 197 litre

As super plasticizer is used the water content can be reduced up to 30% and above

Based on trial water content reduction of 29% is achieved with plasticizer

Hence arrived water content = 197*0.71 = 140 litre

Calculation of cement content

W/C ratio 0.37

Cementitions material (cement+fly ash) content = 140/0.37 = 380 kg/m³

From table 5 of IS 456 minimum cement content for severe exposure condition = 320 kg/m³

380 kg/m³ > 320 kg/m³ hence ok

Now to proportion a mix containing fly ash the following steps are suggested

1. The percentage of fly ash needs to be decided on project requirement. And quality of material.
2. In some case increasing cementation material can be warranted. Increase in cementitions material content and its percentage may be based on experience and trial

Cementitions material content = 380 x 1.1 = 418 kg/m³

Water content = 140 kg/m³

So w/c ration = 140/418 =0.334

Fly ash @30% total cementitious material content = 418 x 30% = 126 kg/m³

Cement OPC = 418-126 = 292 kg/m³

Saving of cement while using fly ash = 380-292 = 88 kg/m³

Fly ash being utilized = 126 kg/m³

Proportion of volume of coarse aggregate and fine aggregate content

From table 3 of IS 10262:2009, volume of coarse aggregate corresponding to 20mm size Aggregate and fine aggregate (zone1) for water cement ratio 0.6

In present case the water cement ratio is 0.37

Therefore the volume of coarse aggregate needs to be increased to decreased the fine aggregate content.

As water-cement ratio is lower by 0.1 the proportion of volume of coarse aggregate is increased by 0.02 (at the rate of -/+ 0.01 for every +/- 0.05 change in w/c ratio)

There for corrected volume of coarse aggregate for the water cement ratio of 0.37=0.62

For pump able concrete these values are reduced by 10%

Therefore volume of coarse aggregate = 0.62 x 0.9 = 0.56

Volume of fine aggregate content = 1-0.56 =0.44

Mix calculation

1. Volume of concrete =1 m³
2. Volume of cement = mass of cement / specific gravity of cement x1/1000

Volume of cement = 292/3.15 x 1/1000

Volume of cement = 0.0926 m³

1. Volume of fly ash = mass of cement / specific gravity of cement x1/1000

Volume of fly ash = 126/2.2 x1/1000

Volume of fly ash =0.0572 m³

1. Volume of water = mass of water / specific gravity of water x1/1000

Volume of water = 140/1 x 1/1000

Volume of water =0.140 m³

1. Volume of chemical admixture =

= mass of chemical admixture / specific gravity of admixture x1/1000

= 7.6/1.145x1/1000

=0.006 m³

1. Volume of aggregate (FA &CA) = (a-(b+c+d+e))

Volume of aggregate (FA &CA) =(1-(0.0926+0.0572+0.140+0.006))

Volume of aggregate (FA &CA) =0.704 m³

1. Mass of coarse aggregate

= f x volume of coarse aggregate x specific gravity of fine aggregate x 1000

=0.704x0.56x2.74x1000

=1080 kg/m³

1. Mass of fine aggregate

= f x volume of fine aggregate x specific gravity of coarse aggregate x 1000

=0.704x0.44x2.74x1000

= 849 kg/m³

Mix proportion for trial number 1

Cement =292 kg/m³

Fly ash =126 kg/m³

Water =140 kg/m³

Fine aggregate =849 kg/m³

Coarse aggregate =1080 kg/m³

Chemical admixture =7 kg/m³

w/c ratio = 0.37

Result:-

Trial mix ratio = 1:2.03:2.583

Aim :- To Calculate the Concrete Mix Design for M50 grade concrete without fly ash

Introduction:-

Data collection

Grade of concrete = M50

Extreme weather condition = severe Exposure condition

Grade of cement = 43 (OPC)

Maximum size of aggregate 20mm

Minimum cement content = 320 kg/m³ from table 5 IS456 2000 Exposure condition

Maximum w/c ratio = 0.45 from table 5 IS456 2000 Exposure condition

Workability 100mm (slump value for pump-able concrete)

Method of concrete placing – pumping

Degree of supervision – good

Type of aggregate- cursed angular aggregate

Maximum cement content 450 kg/m³

Chemical admixture type – super plasticizer

Test data for materials

Specific gravity of cement 3.15

Specific gravity of coarse aggregate 2.74

Specific gravity of fine aggregate 2.74

Water absorption coarse aggregate 0.5 %

Water absorption fine aggregate 1.0 %

Free surface moisture coarse aggregate Nil

Free surface moisture fine aggregate Nil

Fck = 50 N/mm²

Standard deviation for M50 grade concrete is 5N/mm²

ft = fck + 1.65S

ft= 50 + 1.65*5

Target mean strength of concrete ft= 58.25 N/mm²

From exposure condition w/c ratio is 0.45

Lowest w/c ratio is 0.4

So water cement ratio for mix design M50 grade of concrete and for severe exposure condition is 0.4

Selection of water content

From table 2 of IS 10262: 2009 maximum water content for 20mm aggregate = 186litre (for 25mm to 50mm slump range)

Estimated water content for 100 mm slump = 186 + 6*186/100

= 197litre

As super plasticizer is used the water content can be reduced up to 30% and above

Based on trial water content reduction of 29% is achieved with plasticizer

Hence arrived water content = 197*0.71 = 140 litre

Calculation of cement content

W/C ratio 0.4

Cement content = 140/0.4 = 350 kg/m³

From table 5 of IS 456 minimum cement content for severe exposure condition = 320 kg/m³

350 kg/m³ > 320 kg/m³ hence ok

Proportion of volume of coarse aggregate and fine aggregate content

From table 3 of IS 10262:2009, volume of coarse aggregate corresponding to 20mm size Aggregate and fine aggregate (zone1) for water cement ratio 0.6

In present case the water cement ratio is 0.4

Therefore the volume of coarse aggregate needs to be increased to decreased the fine aggregate content.

As water-cement ratio is lower by 0.1 the proportion of volume of coarse aggregate is increased by 0.02 (at the rate of -/+ 0.01 for every +/- 0.05 change in w/c ratio)

There for corrected volume of coarse aggregate for the water cement ratio of 0.4=0.62

For pump able concrete these values are reduced by 10%

Therefore volume of coarse aggregate = 0.62 x 0.9 = 0.56

Volume of fine aggregate content = 1-0.56 =0.44

Mix calculation

1. Volume of concrete =1 m³
2. Volume of cement = mass of cement / specific gravity of cement x1/1000

Volume of cement = 350/3.15 x 1/1000

Volume of cement = 0.110 m³

1. Volume of water = mass of water / specific gravity of water x1/1000

Volume of water = 140/1 x 1/1000

Volume of water =0.140 m³

1. Volume of chemical admixture =

= mass of chemical admixture / specific gravity of admixture x1/1000

= 7.6/1.145x1/1000

=0.006 m³

1. Volume of aggregate (FA &CA) = (a-(b+c+d))

Volume of aggregate (FA &CA) =(1-(0.110+0.14+0.006))

Volume of aggregate (FA &CA) =0.743 m³

1. Mass of coarse aggregate

= e x volume of coarse aggregate x specific gravity of fine aggregate x 1000

=0.743x0.56x2.74x1000

=1140 kg/m³

1. Mass of fine aggregate

= e x volume of fine aggregate x specific gravity of coarse aggregate x 1000

=0.743x0.44x2.74x1000

= 896 kg/m³

Mix proportion for trial number 1

Cement =380 kg/m³

Water =140 kg/m³

Fine aggregate =896 kg/m³

Coarse aggregate =1140 kg/m³

Chemical admixture =7 kg/m³

w/c ratio = 0.4

Result :-

Trial mix ratio = 1:2.56:3.26