I am performing this project to find a design mix for M35 with fly ash & M50 concrete without fly ash.

INTRODUCTION

Concrete is one of the most widely used and versatile building materials. It comprises different ingredients that have various functions, concrete as a building material is made from cement, aggregate that is rock and sand, admixtures(Optional) and water in definite proportion. The properties of concrete mainly depend on the mixing of these ingredients, concrete act as a Binder material that binds the materials together and cement in combination with aggregate acts as a strong building material all these ingredients help in modifying the properties of the fresh as well as harden concrete, in short, these materials are like members of the family and they perform their role in such a manner that they contribute to the overall success of the concrete as a whole family.

MATERIALS

1. Cement
2. Fine aggregate
3. Coarse aggregate
4. Water
5. Admixture(Superplasticizers)
6. Fly ash.

DATA PROVIDED FOR M50 GRADE CONCRETE

Grade Of Concrete	M50
Type of cement	43
Maximum nominal size of aggregate	20mm
Minimum cement content	320Kg/m^3
Maximum water-cement ratio	0.45
Workability	True slump at 100 mm
Environment exposure	Severe
Method of concrete placing	Pumping
Degree of supervision	Good
Type of aggregate	Crush angular
Maximum cement content	450kg/m3
Chemical admixtures	Superplasticizers
Specific gravity of cement	3.15
Specific gravity of coarse aggregate	2.74
Specific gravity of fine aggregate	2.74
Sieve analysis of fine aggregate	Zone 1 (table 4, IS 383)
Sieve analysis of coarse aggregate	Table 2, IS 383

PROCEDURE FOR M50 DESIGN MIX WITHOUT FLY ASH

1. The first step in our design mix is to find the target mean strength as target mean strength acts as a benchmark to our design mix. The Target mean strength of concrete mix is using the following formula from IS 10262-2009,

f’ck= fck + 1.65×S

fck=Characteristic compressive strength of concrete at 28 days

S= Standard deviation, (values from table 1)

Step 1 :

Target mean strength

From IS 10262-2009, Table1

f’ck = fck+1.65*S

f’ck = 50+1.65*5

f’ck = 58.25Nmm2Nmm2

2. Based on reinforced concrete or plain cement concrete select the water-cement ratio for the given Exposure condition

Step 2 :

Find the Water cement ratio based on Environment exposure:

As per IS 456-2000, table-5:

w/c = 0.45, Based on experience take 0.4

Minimum cement content =320kgmm2kgmm2

3. Selection of water content is taken from table 2, IS 10262:2009. The maximum water content for 25mm to 50 mm slum values is given, For additional slump, 3% may be added for every 25mm slump. Also, make sure that the aggregate is dry.

Step 3 :

selection of water content: from table 2, IS 10262:2009

Water content is 186

True slump at 100 mm

= 186+186*0.06

=197lit

Use of superplasticizers reduces water content by 30%

After observation =29%

=197*0.71

= 140lit

4. Next is the selection of cement content. We can take it from the water-cement ratio, Minimum cement content can be taken from Table 5 IS 456-2000.

C=WwcC=Wwc

W= Water content

w/c= water-cement ratio

Step 4 :

Selection of cement content

Cement content of the concrete mix=Wwc=Wwc

= 1400.4

=350 kg

350>320[minimum cement coment]

So ok

5. Volume of coarse aggregate and fine aggregate are taken from- table 3, of IS10262-2009.

Step 5 :

Volume of Agg(CA&FA)

Table 3 of IS 10262:2009

FA= zone 1 ; 0.6

But value of w/c =0.4

So Increase in volume of coarse aggregate and decrease in volume of water Content

as the water-cement ratio is decreased by 0.10 so increase by 0.02

Water cement ratio of coarse aggregate= 0.4=0.62

[for pumpable concrete value decreases by 10%]

Volume of CA=0.62*0.9

=0.56

Volume of FA=1-0.56

=0.44

6. Perform mix calculations for all the materials.

Step 6:

Mix calculation

a. Volume of concrete=1m^3

b. The volume of cement =massofcementspecificgravityofcement×11000massofcementspecificgravityofcement×11000

= 350/3.15 * 1/1000 =0.111m^3

c. The volume of water =massofwaterspecificgravityofwater×11000massofwaterspecificgravityofwater×11000

=140/1 * 1/1000 = 0.140m3m3

d. The volume of chemical admixture =massofadmixturespecificgravityofadmixture×11000massofadmixturespecificgravityofadmixture×11000

=7.6/1.145 * 1/1000 = 0.006m3m3

e. Volume of agg(FA&CA)

=(a-(b+c+d))

=(1-(0.111+0.140+0.006))

=0.743

f. Mass of CA

=0.743*0.56*2.74*1000

=1140Kg

g. Mass of FA

=0.743*0.44*2.74*1000

=896Kg

State all the mix proportions we got after calculating mix calculations of material

Step 7 :

Cement = 350kg

Water = 140kg

FA = 896kg

CA = 1140kg

Chemical admixtures = 7kg

Water cement ratio = 0.4

Trail mix ratio = 1 : 2.56 : 3.26

[Using this mix ratio perform experiments to find if the concrete meets the target mean strength, if yes go ahead with this design mix if no alter few values and Calculator new design mix till the target mean strength is reached.]

DATA PROVIDED FOR M35 GRADE CONCRETE

Grade Of Concrete	M35
Type of cement	43
Maximum nominal size of aggregate	20mm
Minimum cement content	320Kg/m^3
Maximum water-cement ratio	0.45
Workability	True slump at 100 mm
Environment exposure	Severe
Method of concrete placing	Pumping
Degree of supervision	Good
Type of aggregate	Crush angular
Maximum cement content	450kg/m^3
Chemical admixtures	Superplasticizers, fly ash
Specific gravity of cement	3.15
Specific gravity of fly ash	2.2
Specific gravity of coarse aggregate	2.74
Specific gravity of fine aggregate	2.74
Sieve analysis of fine aggregate	Zone 1 (table 4, IS 383)
Sieve analysis of coarse aggregate	Table 2, IS 383

PROCEDURE FOR M35 DESIGN MIX WITH THE USE OF FLY ASH

1. The first step in our design mix is to find the target mean strength as target mean strength acts as a benchmark to our design mix. The Target mean strength of concrete mix is using the following formula from IS 10262-2009,

f’ck= fck + 1.65×S

fck=Characteristic compressive strength of concrete at 28 days

S= Standard deviation, (values from table 1)

Step 1 :

Target mean strength

From IS 10262-2009, Table1

f’ck = fck+1.65*S

f’ck = 35+1.65*5

f’ck = 43.25 Nmm2Nmm2

2. Based on reinforced concrete or plain cement concrete select the water-cement ratio for the given Exposure condition

Step 2 :

Find the Water cement ratio based on Environment exposure:

As per IS 456-2000, table-5:

w/c = 0.45, Based on experience take 0.4

Minimum cement content =320kgm2kgm2

3. Selection of water content is taken from table 2, IS 10262:2009. The maximum water content for 25mm to 50 mm slum values is given, For additional slump, 3% may be added for every 25mm slump. Also, make sure that the aggregate is dry.

Step 3 :

selection of water content: from table 2, IS 10262:2009

Water content is 186

True slump at 100 mm

= 186+186*0.06

=197lit

Use of superplasticizers reduces water content by 30%

After observation =29%

=197*0.71

= 140lit

4. Next is a selection of cement content. We can take it from the water-cement ratio, Minimum cement content can be taken from Table 5 IS 456-2000.

Step 4 :

Selection of cement content

Cement content of the concrete mix=W/w/c

= 1400.4

=350 kg

350>320

So ok

Now total Cementitious materials( cement + fly ash) = 350

There is an increase of 10% of cementitious materials due to the addition of fly ash = 350*1.1

=385kg/m^3

Water content = 140kg/m^3

So water-cement ratio=140/385

=0.364

Fly ash at 30% cementaious materials

Fly ash = 385*0.3

= 115kg/m^3

Cement = 385-115

= 270kg/m^3

Cement saved = 80kg/m^3

5. Volume of coarse aggregate and fine aggregate are taken from- table 3 ,of IS10262-2009.

Step 5 :

Volume of Agg(CA&FA)

Table 3 of IS 10262:2009

FA= zone 1 ; 0.6

Value of w/c =0.4

So Increase in volume of coarse aggregate and decrease in volume of water Content

as the water-cement ratio is decreased by 0.10 so increase by 0.02

Water cement ratio of coarse aggregate= 0.4=0.62

[for pumpable concrete value decreases by 10%]

Volume of CA=0.62*0.9

=0.56

Volume of FA=1-0.56

=0.44

6. Perform mix calculations for all the materials.

Step 6:

Mix calculation

a. Volume of concrete=1m^3

b. The volume of cement =massofcementsπcificgravityofcement×11000massofcementsπcificgravityofcement×11000

= 270/3.15 * 1/1000 =0.0086m^3

c. The volume of fly ash =massofflyashspecificgravityofflyash×11000massofflyashspecificgravityofflyash×11000

=115/2.2 *1/1000 = 0.052m^3

d. The volume of water =massofwaterspecificgravityofwater×11000massofwaterspecificgravityofwater×11000

=140/1 * 1/1000 = 0.140

e. The volume of chemical admixture =massofadmixturesπcificgravityofadmixture×11000massofadmixturesπcificgravityofadmixture×11000

=7.6/1.145 * 1/1000 = 0.006m^3

f. Volume of agg(FA&CA)

=(a-(b+c+d+e))

=(1-(0.086+0.052+0.140+0.006))

=0.716m^3

g. Mass of CA

=0.716*0.56*2.74*1000

=1098Kg

h. Mass of FA

=0.716*0.44*2.74*1000

=863Kg

7. State all the mix proportions we got after calculating mix calculations of material

Step 7 :

Cement = 270kg/m^3

Fly ash= 115kg/m^3

Water = 140kg/m^3

FA = 863kg

CA = 1098kg

Chemical admixtures = 7kg/m^3

Water cement ratio = 0.36

Trail mix ratio = 1 : 2.24 : 2.85

[Using this mix ratio perform an experiments to find if the concrete meets the target mean strength, if yes go ahead with this design mix, if no alter few values and Calculator new design mix till the target mean strength is reached.]

RESULTS

By performing this project we have gained the design mix for M35 concrete with fly ash and M50 concrete without fly ash.

1. Trail mix ratio for M35 grade concrete with fly ash = 1 : 2.24 : 2.85 [We save 80kg/m^3 of cement]
2. Trail mix ratio for M50 grade concrete without fly ash = 1 : 2.56 : 3.26

CONCLUSION

I would like to conclude my project by stating that every design mix varies depending on the aggregates present, the state of aggregates and many other factors. Using this mix ratio perform experiments to find if the concrete meets the target mean strength, if yes go ahead with this design mix if no alter few values and Calculator new design mix till the target mean strength is reached.

REFERENCES

• IS 10262-2009,2019
• IS 456-2000
• IS 383-1970,2016
• SKILL LYNC COURSE