Project on Concrete Mix Design for various grades of Concrete

AIM:

To calculate the concrete mix design for M35 grade concrete with fly ash.

STIPULATIONS FOR PROPORTIONING

1. Grade designation:M35
2. Type of cement:OPC 43grade
3. Type of mineral admixture:Fly ash 
4. Maximum nominal size of aggregrate:20mm
5. Workability:120mm
6. Method of concrete placing:pumping
7. Type of concrete:crushed angular aggregrate

TEST DATA FOR MATERIALS

1. Cement used:OPC43grade
2. Specific gravity of cement:3.15
3. Fly ash:conforming to IS3812
4. Specific gravity of fly ash:2.2

Chemical admixture:Superplasticizer conforming to IS9103

PROCEDURE:

1.Target mean strength of concrete(f'ck):

f'ck=fck+1.65S

where,

fck=Compressive strength of concrete

S=Standard deviation

From table 1,Standard deviation,S=5N/mm^2

      =35+1.65(5)

       =43.3N/mm^2

2.Selection of water cement ratio:

Maximum water cement ratio =0.45

3.Selection of water content:

Water content =186Kg(for 50mm slump) for 20mm aggregrate.

Estimated water content for 120mm slump(increasing at the rate of 35% for every 25mm slump)

                    =186+8.4/100*186

                    =201.62Kg

As superplasticizer is used the water content may be reduced.

Based on trial data the water content reduction of 23 percent is considered while using superplasticizer at the rate of 1 percent by weight of cement.

Hence the arrived water content =201.62*0.77=155.25Kg =155Kg.

4.Calculation of cement content:

Water cement ratio=0.45

cement content=155/0.45

                      =344.45Kg/m^3=345Kg/m^3.

To propotion a mix containing fly ash the following steps are suggested:

1. Decide the percentage fly ash to be used based on project requirement and quantity of fly ash.
2. In certain situations,increase in cementitious maerial content may be warranted.The decision on increase in cementitious material content and its percentage may be based on experience and trials.

Cementitious material content 

                   =345*1.10=380Kg/m^3

water content=155Kg/m^3

so water cementitious ratio

            =155/380

            =0.407

Fly ash @30% of total cementitious material content =380*30%=114Kg/m^3

cement=380-114=266Kg/m^3

From table 5 of IS456,minimum cementitious content for M35 grade =340Kg/m^3

380Kg/m^3>340Kg/m3,hence ok

5.Proportion of volume of coarse aggregrate and fine aggregrate content:

From table5,volume of coarse aggregrate correspomding to 20mm size aggregrate and fine aggregrate(zoneII)for water cement ratio=0.62

In present case water cement ratio is 0.407.Volume of coarse aggregrate is required to be inscreased to decrease the fine aggregrate content.As the water cement ratio is lower by 0.173 the proportion of volume of coarse aggregrate is increased by 0.034.

For pumped concrete these values may be reduced by upto 10%.

Volume of coarse aggregrate=0.6546*0.9=0.59m^3.

Volume of fine aggregrate content=1-0.59=0.41m^3.

6.Mix calculation:

1. Total volume=1m^3
2. Volume of cement=mass of cement/specific gravity of cement*1/1000

                                   =266/3.15*1/1000=0.084m^3

     c.Volume of fly ash =mass of fly ash/specific gravity of fly ash*1/1000

                                  =114/2.2*1/1000=0.052m^3

     d.Volume of water=mass of water/specific gravity of water*1/1000

                                =155/1*1/1000=0.155m^3

    e.Volume of chemical admixture=mass of chemical admixture/specific gravity of chemical admixture*1/1000

                                                 =4.74/1.145*1/1000=0.004m^3

    f.Volume of all aggregrate=(a-(b+c+d+e))

                                        =(1-(0.084+0.052+0.155+0.004))=0.705m^3

    g.Mass of coarse aggregrate=f*volume of coarse aggregrate*specific gravity of coarse aggregrate*1000

                                             =0.705*0.59*2.74*1000=1140Kg

    h.Mass of fine aggregrate=f*volume of fine aggregrate*specific gravity of fine aggregrate*1000

                                         =0.705*0.41*2.65*1000=766Kg

7.Mix proportion:

Cement =266kg/m^3

Fly ash:114kg/m^3

water:155kg/m^3

Fine aggregrate:766kg/m^3

coarse aggregrate:1140kg/m^3

Chemical admixture:4.74kg/m^3

Water cement ratio=0.407

Result:

Mix ratio=1:2.88:4.3

AIM:

To calculate the concrete mix design for M50 grade concrete without fly ash.

STIPULATIONS FOR PROPORTIONING

1. Grade designation:M50
2. Type of cement:PPC conforming toIS1489
3. Maximum nominal size of aggregrate:20mm
4. Exposure condition:severe
5. Workability:75mm slump
6. Method of concrete placing:chute
7. Degree of site control:good
8. Type of aggregrate:crushed angular aggregrate
9. Maximum cement content:450kg/m^3
10. Chemical admixture type:superplasticizer-normal

TEST DATA FOR MATERIAL

1. Cement used :ppc
2. Specific gravity of cement:2.88
3. chemical admixture:superplasticizer
4. specific gravity of
5. water absorption

PROCEDURE:

1.Target strength of mix proportion

f'ck=fck+1.65S

Where,

fck=compressive strength at 28 days

S=standard deviation

From table 2 standard deviation=5

f'ck=50+1.65(5)=58.25N/mm^2

2.Selection of water cement ratio

From fig.1,the free water cement ratio required for target strength of 58.25N/mm^2 is 0.39 for OPC grade curve.This is lower than the maximum value of 0.45 precribed for severe exposure for reinforced concrete .

0.39<0.45,hence ok.

3.Selection of water content:

From table 4,water content=186kg(for 50mm slump) for 20mm aggregrate.

Estimated water content for 75mm slump=186+3*186/100=191.58kg

As superplasticizer is used,the water content may be reduced.Based on the trial data the water content reduction of 23% is considered while using superplasticizer at the rate 1%by weight of cement 

Hence the water content=191.58*0.77=148kg

4.Calculation of cement content:

Water cement ratio=0.39

Cement content=148/0.39=380kg/m^3

From table 5 of IS456,minimum cement content for severe exposure condition=320kg/m^3

380kg/m^3>320kg/m^3,hence ok.

5.Proportion of volume of coarse aggregrate and fine aggregrate

From table 5 the proportionate volume of coarse aggregrate corresponding 20mm aggregrate for water cement ratio=0.62

In the present case of water cement ratio is 0.39.therefore volume of coarse aggregrate is required to increased to decreased the fine aggregrate content.As the water cement ratio is lower by 0.14 the proportion pf volume of coarse aggregrate is increased by 0.028 0.39=0.62+0.028=0.648

Volume of fine aggregrate content =1-0.648=0.352

6.Mix calculation

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

a.Total volume=1m^3

b.volume of cement=mass of cement/specific gravity of cement*1/1000

                             =380/2.88*1/1000=0.132m^3

c.volume of water=mass of water/specific gravity of water*1/1000

                          =148/1*1/1000=0.148m^3

d.volume of cement admixture=mass of chemical admixture/specific gravity of chemical admixture*1/1000

                                            =4.12/1.145*1/1000=0.003m^3

e.volume of all in aggregrate=(a-(b+c+d))

                                          =(1-(0.132+0.148+0.003)=0.717m^3

f.mass of coarse aggregrate=e*volume of coarse aggregrate*specific gravity of coarse aggregrate*1000

                                         =0.717*0.648*2.74*1000=1273kg

g.mass of fine aggregrate=e*volume of fine aggregrate*specific gravity of fine aggregrate*1000

                                     =0.717*0.352*2.65*1000=669kg

Mix proportion:

cement=380kg/m^3

water=148kg/m^3

fine aggregrate=669kg/m^3

coarse aggregrate=1273kg/m^3

chemical admixture=4012kg/m^3

water cement ratio=0.39

RESULT

Mix ratio=1:1.76:3.35