Project on Concrete Mix Design for various grades of Concrete

1 Answer )

Aim of the Project :

 To calculate the Concrete Mix Design for M35 grade concrete with fly ash.

The mix design calculation of M35 Grade Concrete with fly ash and Poly Carboxylite type superplasticizer as per Indian Standard Codebook IS10262(2009) – Guidelines for Concrete Mix Proportioning.

The following equations need to be considered :

                                                                            f’ck= fck+1.65s

here f’ck is the mean target strength for concrete mix design

fck is characteristic compressive strength at 28 days

s  is the standard deviation

The volume of cement =1m^3

The volume of cement = (mass of cement /specific gravity of cement ) *1/1000

The volume of water = (mass of water/specific gravity of water )*1/1000

The volume of chemical admixture =   (mass of chemical admixture/specific gravity of admixture) *     *1/1000

The volume of aggregates = (volume of concrete –(volume of cement +volume of fly ash+volume of water +volume of admixture

Mass of coarse aggregate = volume of aggregate *volume of coarse aggregate *specific gravity of coarse aggregate *1000

Mass of fine aggregate = volume of aggregate * volume of fine aggregate *specific gravity of fine aggregate *1000

 

Objective :

The objective of this project is to produce M35 grade concrete economically.

To achieve desired workability, durability, and minimum strength of M35 grade concrete.

Here we use IS 456:2000 and IS 10262:2000 as a standard codebook for reference.

 

Procedure and calculation :

1. Target mean strength is

f’ck=fck+1.65s

       = 35 N/mm^2

From table 1 of IS 10262:2009

S=5N/mm^2

2. Calculation of water-cement ratio :

Assuming extreme weather conditions

From table 5 of IS 456:2000

For extreme conditions w/c ratio =0.40

Minimum cement content from table 5 of IS 456:2000 is 360 kg/m^3

3. Selection of water-cement ratio :

 Assuming the nominal size of aggregate i.e 20mm

Therefore from table 2 of IS 10262:2009

Minimum water content = 186 liters

Slump range from 25 to 50mm

As superplasticizers reduce 20% water content

= 186*(100-20)/100 = 148.8

Approximately 149 liters

 

4. Calculation of cementitious material :

Water cement  ratio = 0.40

Water content W= 149 kg

Cementatious content = 149/0.40= 372.5=370kg/m^3

From table 5 we can see that the minimum cement content to be 360 kg/m^3

Therefore 360<370

So we increase by 10%

370*0.10 +370= 407 kg/m^3

w/c = 149/407

        = 0.366

Fly ash at 30 % total cementitious material content = 407*30/100= 122kg/m^3

Cement (OPC) = 407-122= 285 kg/m^3

Saving of cement while using fly ash = 370-285= 85 kg/m^3

Fly ash = 122kg/m^3

5. The volume of Coarse aggregates and Fine aggregates :

Fine aggregates are sieved through Zone 1 table 4 of IS 383

Coarse aggregates are sieved through table 2 of IS 383

According to table 3 of IS 10262:2009

The nominal size of aggregates is 20mm.

Therefore the volume of coarse aggregate is 0.60

Water cement ratio =0.4

(+/-0.01 for every +/-0.005)

So, the volume of coarse aggregate is 0.62

Volume of fine aggregate =1-0.62 = 0.38

6. Mix calculations :

a.Volume of concrete = 1m^3

b.Volume of cement = (285/3.15)*1/1000= 0.090m^3

c.Volume of fly ash = ( 122/2.2)*1/1000= 0.055m^3

d.Volume of water = (149/1)*1/1000 = 0.149m^3

e.Volume of  admixture = (7.6/1.145)*1/1000= 0.006m^3

1. Volume of corse aggregate and fine aggregate = a-(b+c+d+e)

                                                                                        = 1-(0.090+0.055+0.149+0.006)

                                                                                        = 0.7m^3

1. Mass of coarse aggregate = 0.7*0.62*2.74*1000= 1189.16 kg/m^3
2. Mass of fine aggregate = 0.7*0.38*2.74*1000=728kg/m^3

 

7. Results and calculation :

Cement = 285kg /m^3

Flyash = 122kg/m^3

Water = 149kg

Coarse aggregate =1189kg/m^3

Fine aggregate =728kg/m^3

Chemical admixture = 7.6 kg/m^3

Water cement ratio = 0.366

Therefore trial mix ratio = 285:728:1189 = 1.2.55:4.17 ( C:FA:CA )

We have to check for durability and strength of this mix if not again a trial ratio mix needs to be done.

2 Answer )

 

 

Aim of the project :

To calculate M50 grade concrete design without using fly ash.

The following equations need to be considered :

                                                                            f’ck= fck+1.65s

here f’ck is the mean target strength for concrete mix design

fck is characteristic compressive strength at 28 days

s  is the standard deviation

The volume of cement =1m^3

The volume of cement = (mass of cement /specific gravity of cement ) *1/1000

The volume of water = (mass of water/specific gravity of water )*1/1000

The volume of chemical admixture =   (mass of chemical admixture/specific gravity of admixture) *     *1/1000

The volume of aggregates = (volume of concrete –(volume of cement +volume of fly ash+volume of water +volume of admixture

Mass of coarse aggregate = volume of aggregate *volume of coarse aggregate *specific gravity of coarse aggregate *1000

Mass of fine aggregate = volume of aggregate * volume of fine aggregate *specific gravity of fine aggregate *1000

Objective :

The objective of this project is to produce M35 grade concrete economically.

To achieve desired workability, durability, and minimum strength of M35 grade concrete.

Here we use IS 456:2000 and IS 10262:2000 as a standard codebook for reference.

 

Procedure and calculation :

1. Target mean strength is

f’ck=fck+1.65s

fck = 50N/mm^2

From table 1 of IS 10262

S=5N/mm^2

F’ck=50+(1.65*5)

         = 58.25 N/mm^2

2. Selection of water-cement Ratio :

Assuming exposure condition: Extreme

Therefore from table 5 IS 456:2000

Extreme w/c ratio = 0.40

Lower the w/c ratio better the workability.

Minimum cement content = 360kg/m^3

3. Selection of water content :

Assuming nominal size of aggregates = 20mm

From table 2 of IS 10262:2009

MinimumWater content = 186 liter

As superplasticizers would reduce the water content by 20%

    = 186*(100-20)/100 = 148.8

Water content = 149 liter

4. Calculation of cement content :

 Water cement ratio = w/c ratio = 0.40

Water content = w= 149kg

Cement content c= 149/0.40=372.5 = 373 kg/m^3

360<373kg/m^3

Hence okay.

5. The volume of coarse aggregate and fine aggregate:

According to table 3 IS 102062:2009

The nominal size of aggregate is 20mm

Therefore the volume of coarse aggregate = 0.60

Water cement ratio =0.4

For every ( +/-0.01for every +/-0.005)

So the volume of coarse aggregate =0.62

Volume of fine aggregate = 1.0.62= 0.38

6. Mix calculation :

 

6. Volume of concrete = 1m^3
7. Volume of cement = (373/3.15)*1/1000= 0.118m^3
8. Volume of water = (149/1)*1/1000 = 0.149m^3
9. Volume of admixture = (7.6/1.145)*1/1000 = 0.006
10. Volume of coarse aggregate and fine aggregate = a-(b+c+d) = 1-( 0.118+0.19+0.006)= 0.727m^3
11. Mass of coarse aggregate = 0.727*0.62*2.74*1000= 1235.02kg/m^3
12. Mass of fine aggregate =0.727*0.38*2.74*1000= 756.9kg/m^3

 

7.Results and calculations :

cement = 373kg/m^3

water = 149 kg

coarse aggregate = 1235.02kg/m^3

Fine aggregate = 756.9kg/m^3

Chemical admixture = 7.6kg/m^3

Water cement ratio = 0.40

Therefore trial mix ratio : 373:756.9:1235.02  = 1:2.02:3.31  ( C: FA : CA )

Therefore thid trial mix should satisfy the strength and durability of M50concrete grade mix.

If  doesn't met the required conditions we have to increase the details by 10% and should be checked.