Project on Concrete Mix Design for various grades of Concrete

Project on : Concrete Mix Design for various grades of Concrete

 

 

 

 

 

Q. Calculate the Concrete Mix Design for M35 grade concrete with fly ash 

Ans:-

 

Step -1. Stipulations for proportioning :

 

• Grade designation:  M₃₅
• Type of cement:  OPC 43 grade
• Maximum nominal size of aggregate:  20mm
• Minimum cement content:  320 Kg ……….…. (From Table 5 of IS 456:2000)
• Maximum water-cement ratio:  0.45 ………… (From Table 5 of IS 456:2000)
• Workability:  100mm slump
• Exposure condition:  Sever ( for Reinforced concrete )
• Method of concrete placing:  Pumping
• Degree of supervision:  Good
• Type of aggregate: Crushed angular aggregate
• Chemical admixture type : Superplasticizer
• Maximum cement content:  450 kg/m³

 

 Step - 2. Test Data For Materials:

• Cement used: OPC 43 grade conforming to IS8112
• Specific gravity of cement:  3.15
• Flyash : Conforming to IS 3812 ( Part 1 )
• Specific gravity of flyash : 2.2
• Chemical admixture : Superplasticizer conforming to IS 9103
• Secific gravity of superplasticizer : 1.145
• Specific gravity of :   CA =2.74 and FA = 2.74
• Water absorption: CA  = 0.50% and FA = 1%
• Sieve analysis:  Conforming to grading Zone I (Table 2 & 4, IS: 383–1970)
• Free (surface) moisture
• Coarse aggregate:  Nill
• Fine aggregate:  Nill

 

Step - 3. Calculation of Target strength:

• Target mean strength (f’ck) = fck+1.65S

Where,

• fck= Compressive strength of concrete = 35MPa
• S = Standard deviation ( IS 10262 : 2009, Table 1 ) = 5MPa

 

 

 

• F’ck = 35 + ( 1.65 X 5 ) = 43.25 MPa

 

 

Step - 4. Selection of W/C ratio

Based on experience, water-cement ratio required for the target mean strength of 43.25MPa is 0.33 ( as IS 10262 : 2009 , Compressive strength vs W/C ratio graph )

Maximum water-cement ratio for sever exposure condition = 0.45 (From Table 5 of IS 456 -2000)

• 0.33<0.45, hence OK.

 

 

Step - 5. Selection of water content:

From Table 2 of IS 10262:2009, Max. Water content for 20mm aggregate = 186Kg (for 25mm to 50mm slump)

 

We have 50mm extra slump,so as per IS 10262:2009,We have to increase the water content by 6%

Hence,

• Water Content = 186 + (6 X 186/100) =197.16lit = 197lit .

 

• As the superplasticizer is used, Water content can be reduced upto 20% .
• Based on trials, Water content reduction of 29% is achieved with plasticizers.

 

• Arrived Water content = 197 – ( 29 X 197) / 100 = 139.87 = 140lit

 

Step - 6. Calculation of cement content :

• W/C = 0.33

        140 / C = 0.33

        C = 424.24 = 425 kg/m³

From Table 5 of IS 456, minimum cement content for ‘severe’ exposure condition = 320 kg/m³

 

425 kg/m³ > 320 kg/m³,

So take, cementatious material ( Cement + Flyash ) =  425 kg/m³

 

Now, The proportion a mix containing flyash the following steps are sujjested :

• The % of flyash needs to be decided based on project requirement and quality of materials.
• In some cases increase in cementatious material can be warrented. Increase in cementatious material content and its percentage may be based on experience and trial.( This example is with increase of 10 % cementatious material content )

Therefore,

• Cementatious Material Content = 425 + (10 X 425) / 100 = 467.5 = 468 kg/m³

 

• Water Content = 140 lit

 

• W/C ratio = 140 / 468 = 0.299

 

• Flyash at 30% of total cementatious material = 468 X 30/100 = 140 kg/m³

 

• Cement ( OPC ) = 468 – 140 = 328 kg/m³

 

• Saving of cement while using flyash = 425 - 328 = 97 kg/m³

 

• Flyash being utilized = 140 kg/m³

 

Step - 7. Proportion of volume of coarse aggregate and Fine aggregate content :

From Table 3 of IS 10262- 2009,

 

For Nominal maximum size of aggregate = 20 mm,

Volume of coarse aggregate for water-cement ratio of 0.50 =0.60 (Corresponding Zone 1)

In this case w/c = 0.4 lower by 0.10. So, the proportion of volume of coarse aggregate is increased by 0.02 (at the rate of ± 0.01 for every ± 0.05 change in water-cement ratio). 

Therefore,corrected volume of CA for W/C ratio 0.33 = 0.634

• For pumpable concrete,this value is reduced by 10%

Volume of CA = 0.634– (10 X 0.634) / 100 = 0.57

Volume of FA = 1 – 0.57 = 0.43

 

Step - 8. Mix Calculation:

The mix calculation per unit volume of concrete shall be as follows:-

• Volume of concrete = 1m³.

 

• Volume of cement= (Mass of Cement )/ (Specific gravity of cement) X 1/1000 =  328 / 3.15 X 1 / 1000 =  104m³.

 

• Volume of flyash = (Mass of flyash )/ (Specific gravity of flyash) X 1/1000 = (140 / 2.2) X (1 / 1000) = 0.064 m³

 

• Volume of Water = (Mass of Water )/ (Specific gravity of Water) X 1/1000 = 140 / 1 X 1 / 1000 = 0.140 m³

 

• Volume of chemical admixture = (Mass of chemical admixture )/ (Specific gravity of chemical admixture) X 1/1000 = 7.6 / 1.145 X 1 / 1000 = 0.006 m³

 

• Volume of aggregate ( CA+FA) = [ Volume of concrete - (Volume of cement + Volume of flyash + Volume of Water + Volume of chemical admixture ) = [ 1 – (0.104+0.064+0.140+0.006)] =    686 m³.

 

• Mass of coarse aggregate = ( Volume of aggregate X volume of CA X specifi gravity of CA X 1000 ) = ( 0.686 X 0.57 X 2.74 X 1000) = 1071.40 Kg/ m³.

 

• Mass of Fine aggregate = ( Volume of aggregate X volume of FA X specifi gravity of FA X 1000 ) = ( 0.686 X 0.43 X 2.74 X 1000) = 808.25 Kg/ m³.

 

Step - 9. Mix proportions :

• Cement = 328 Kg/ m³.
• Water = 140 Kg/ m³.
• Fine aggregate = 808.25 Kg/ m³.
• Coarse aggregate = 1071.40 Kg/ m³.
• W/C ratio = 0.299

 

Mix Ratio:

Mix ratio = 1 : 1.73 : 2.29

 

 

 

Q. Calculate the Concrete Mix Design for M50 grade concrete without Fly ash

Ans:- 

 

Step - 1.Stipulations for proportioning

• Grade designation:  M₅₀
• Type of cement:  OPC 43 grade
• Maximum nominal size of aggregate:  20mm
• Minimum cement content:  320 Kg ……….…. (From Table 5 of IS 456:2000)
• Maximum water-cement ratio:  0.45 ………… (From Table 5 of IS 456:2000)
• Workability:  75mm slump
• Exposure condition:  Sever
• Method of concrete placing:  Pumping
• Degree of supervision:  Good
• Type of aggregate: Crushed angular aggregate
• Maximum cement content:  450 kg/m³
• Chemical Admixture : Not used

 

 Step - 2.Test Data For Materials:

• Cement used: OPC 43 grade conforming to IS8112
• Specific gravity of cement:  3.15
• Specific gravity of :   CA =2.74 and FA = 2.74
• Water absorption:Coarse aggregate = 0.50% and FA = 1%
• Sieve analysis:  Conforming to grading Zone 1(Table 2 & 4, IS: 383–1970)
• Free (surface) moisture
• Coarse aggregate:  Nill
• Fine aggregate:  Nill

 

 Step - 3.Calculation of Target strength:

• Target mean strength (f’ck) = fck+1.65S

Where,

• fck= Compressive strength of concrete = 50MPa
• S = Standard deviation ( IS 10262 : 2009, Table 1 ) = 5MPa

 

 

• F’ck = 50 + ( 1.65 X 5 ) = 58.25 MPa

 

 

 Step - 4.Selection of W/C ratio :

Based on experience, water-cement ratio required for the target mean strength of 58.25 MPa is 0.40

Maximum water-cement ratio for sever exposure condition = 0.45 (From Table 5 of IS 456 -2000)

• 40<0.45, hence OK.

 

 

Step - 5.Selection of water content:

From Table 2 of IS 10262:2009, Max. Water content for 20mm aggregate = 186Kg (for 25mm to 50mm slump)

 

We have 25mm extra slump,so as per IS 10262:2009,We have to increase the water content by 3%

Hence,

• Water Content = 186 + (3 X 186/100) =191.20lit= 191lit .

( As the chemical admixture is not used so, we need not to reduce the Water content )

 

 

 Step - 6.Calculation of cement content :

• W/C = 0.4

        191 / C = 0.4

        C = 477.5 Kg = 478kg

From Table 5 of IS 456, minimum cement content for ‘severe’ exposure condition = 320 kg/m3

478 kg/m3 > 320 kg/m³,

The maximum cement content in concrete for M50 limited to 450 kg. per cum of concrete as per Para 8.2.4.2 of IS 456

So take, cement content = 450Kg

 

Step - 7.Proportion of volume of coarse aggregate and Fine aggregate content :

From Table 3 of IS 10262- 2009,

 

For Nominal maximum size of aggregate = 20 mm,

Volume of coarse aggregate for water-cement ratio of 0.50 =0.60 (Corresponding Zone 1)

In this case w/c = 0.4 lower by 0.10. So, the proportion of volume of coarse aggregate is increased by 0.02 (at the rate of ± 0.01 for every ± 0.05 change in water-cement ratio). 

Therefore,corrected volume of CA for W/C ratio 0.40 = 0.62

• For pumpable concrete,this value is reduced by 10%

Volume of CA = 0.62 – (10 X 0.62) / 100 = 0.56

Volume of FA = 1 – 0.56 = 0.44

 

 

Step - 8.Mix Calculation:

The mix calculation per unit volume of concrete shall be as follows:-

• Volume of concrete = 1m³.

 

• Volume of cement= (Mass of Cement )/ (Specific gravity of cement) X 1/1000 =  450 / 3.15 X 1 / 1000 =  143m³.

 

• Volume of Water = (Mass of Water )/ (Specific gravity of Water) X 1/1000 = 191 / 1 X 1 / 1000 = 0.191 m³.

 

• Volume of aggregate ( CA+FA) = [Volume of concrete – (Volume of cement + Volume of Water ] = [ 1 – (0.143+0.191)] = 666 m³.

 

• Mass of coarse aggregate = ( Volume of aggregate X volume of CA X specifi gravity of CA X 1000 ) = ( 0.666 X 0.56 X 2.74 X 1000) = 1021.91 Kg/ m³.

 

• Mass of Fine aggregate = ( Volume of aggregate X volume of FA X specifi gravity of FA X 1000 ) = ( 0.666 X 0.44 X 2.74 X 1000) = 802.92 Kg/ m³.

 

 

Step - 9. Mix proportions :

• Cement = 450 Kg/ m³.
• Water = 191 Kg/ m³.
• Fine aggregate = 802.92 Kg/ m³.
• Coarse aggregate = 1021.91 Kg/ m³.
• W/C ratio = 0.40

 

• Mix Ratio:

Mix ratio = 1 : 1.78 : 2.27