Project on Concrete Mix Design for various grades of Concrete

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

I. Stipulations for proportioning:

a) Grade designation: M35

b) Type of cement: OPC 43 grade conforming to IS 8112

c) Type of mineral admixtures: Fly ash conforming to IS 3812

d) Max nominal size of aggregate: 20 mm

e) Minimum cement content: 320 Kg/m^3(from Table-5 of IS 456:2000 exposure condition)

f) Max water-cement ratio= 0.45(from Table-5 of IS 456:2000 exposure condition)

g) Workability= 100mm(slump value for pumpable concrete)

h) Exposure condition: Severe(for reinforced concrete)

i) Method of concrete placing: Pumping

j) Degree of supervision: good

k) Type of aggregate: crushed angular aggregate

l) Maximum cement content= 450 Kg/m^3

m) Chemical admixture: super plasticizer 

 

II Test data for materials:

a) Cement used=OPC 43 grade conforming to IS 8112

b) Specific gravity=3.15

c) Fly ash: conforming to IS 3812(part 1)

d) Specific gravity of fly ash=2.2

e) Chemical admixture: super plasticizer conforming to IS 9103

f) Specific gravity Coarse aggregate=2.74 Specific gravity Fine aggregate=2.74

g) Water absorption of Coarse aggregate= 0.5%, Water absorption of Fine aggregate= 1.0%

h) Free surface moisture for Coarse aggregate= NIL, Free surface moisture for Fine aggregate=NIL

i) Sieve analysis- Coarse aggregate conforming to table 2 of IS 383 and Fine aggreagte conforming to grading zone 1 of table 4 of IS 383

 

III. Target Strength for mix proportioning:

           f'ck=fck+1.65S

where f'ck=Target average compressive strength at 28 days

           fck= characteristic compressive strength at 28 days

             S= Standard deviation

From table 1 of IS 10262:2009, standard deviation,S=5 N/mm^2

Therefore,Target strength=35+1.65*5= 43.25 N/mm^2

 

IV. Selection of Water-cement ratio:

From Table 5 of IS 456: 2000 for severe exposure, maximum water-cement ratio=0.45

Based on experience,adopted water-cement ratio is 0.40

0.40<0.45, hence O.K

 

V. Selection of water content:

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

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

                                                              =197 litres

As plasticizer is used, water content can be reduced by 20% and above

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

Hence arrived water content=197*0.71=140 litres

 

VI. Calculation of Cement content:

Water-cement ratio=0.40

Cementitious material(cement+ash)= 140/0.40=350 Kg/m^3

From table 5 of IS 456:2000 min cement content for'severe' exposure condition= 320  Kg/m^3

350Kg/m^3>320Kg/m^3.Hence OK

 

VII. Proportioning of mix containing fly ash:

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

a) The percentage of fly ash needs to be decided based on project requirement and quality of materials.

b) In some cases increase in cementitious material can be warranted.Increase in cementitious material content and its percentage may be based on experience and trial(this example is with increase of 10% cementitious material content).

Cementitious material content= 350*1.10=385 Kg/m^3

Water content= 140 Kg/m^3

Water-cement ratio=140/385=0.364

Fly ash @30% total cementitious material content=385*30%=115 Kg/m^3

Cement(OPC)=385-115=270 Kg/m^3

Saving of cement while using fly ash=350-270 =80 Kg/m^3

Fly ash being utilized=115 Kg/m^3

 

VIII. Proportion of Coarse aggregate and Fine aggregate:

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

In the present case the water cement ratio = 0.40

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

As water-cement ratio is lower by 0.10, the proportion of volume of coarse aggregate is increased by 0.02(at the rate of-/+ 0.01 for every -/+ change in w/c ratio)

Therefore,corrected volume of Coarse aggregate for w/c ratio of 0.4=0.62

For pumpable concrete these values are reduced by 10%

Therefore,volume of Coarse aggregate = 0.62*0.90 = 0.56

Volume of Fine aggregate content = 1-0.56 = 0.44

                                                 

IX. Mix Calculations

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

a) Volume of concrete = 1 m^3

b) Volume of cement = mass of cement/specific gravity of cement*1/1000

                               = 270/3.15*(1/1000)

                               = 0.086 m^3 

c) Volume of Fly ash  = mass of cement/specific gravity of cement*1/1000

                               = 115/2.2*(1/1000) 

                               = 0.052 m^3                            

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

                           =140/1*(1/1000)

                           =0.140 m^3 

e) Volume of chemical admixture=mass of chemical admixture/Specific gravity of admixture*1/1000

                                               =7.6/1.145*(1/1000)

                                               =0.006 m^3

f) Volume of aggregate(Fine aggregate and Coarse aggregate)=((a-(b+c+d+e))

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

                                                                                         =0.716 m^3

g) Mass of Coarse aggregate = f*Volume of coarse aggregate*Specific gravity of coarse aggregate*1000

                                          = 0.716*0.56*2.74*1000

                                          = 1098 Kg/m^3 

h) Mass of Fine aggregate = f*volume of fine aggregate*specific gravity of fine aggregate*1000

                                      = 0.716*0.44*2.74*1000

                                      = 863 Kg/m^3

 

X. Mix Calculation:

a) Cement = 270 Kg/m^3

b) Fly ash = 115 Kg/m^3

b) Water = 140 Kg/m^3

c) Fine aggregate = 863 Kg/m^3

d) coarse aggregate = 1098 Kg/m^3

e) Chemical admixture = 7 Kg/m^3

f)  w/c ratio = 0.364

Trial mix ratio = 1:2.24:2.85

 

Aim: Calculate the Concrete Mix design for M50 grade concrete without Fly ash

I. Stipulations for proportioning:

a) Grade designation: M50

b) Type of cement: OPC 43 grade conforming to IS 8112

c) Max size of aggregate: 20 mm

d) Minimum cement content: 320 Kg/m^3(from Table-5 of IS 456:2000 exposure condition)

e) Max water-cement ratio= 0.45(from Table-5 of IS 456:2000 exposure condition)

f) Workability= 100mm(slump value for pumpable concrete)

g) Exposure condition: Severe(for reinforced concrete)

h) Method of concrete placing: Pumping

i) Degree of supervision: good

j) Type of aggregate: crushed angular aggregate

k) Maximum cement content= 450 Kg/m^3

l) Chemical admixture= super plasticizer 

 

II Test data for materials:

a) Cement used: OPC 43 grade conforming to IS 8112

b) Specific gravity=3.15

c) Chemical admixture: Super plasticizer conforming to IS 9103

d) Specific gravity Coarse aggregate=2.74 Specific gravity Fine aggregate=2.74

e) Water absorption of Coarse aggregate= 0.5%, Water absorption of Fine aggregate= 1.0%

f) Free surface moisture for Coarse aggregate= NIL, Free surface moisture for Fine aggregate=NIL

g) Sieve analysis- Coarse aggregate conforming to table 2 of IS 383 and Fine aggreagte conforming to grading zone 1 of table 4 of IS 383

 

III. Target Strength for mix proportioning:

           f'ck=fck+1.65S

where f'ck=Target average compressive strength at 28 days

           fck= characteristic compressive strength at 28 days

             S= Standard deviation

From table 1 of IS 10262:2009, standard deviation,S=5 N/mm^2

Therefore,Target strength=50+1.65*5=58.25 N/mm^2

 

IV. Selection of Water-cement ratio:

From Table 5 of IS 456: 2000 for severe exposure, maximum water-cement ratio=0.45

Based on experience,adopted water-cement ratio is 0.40

0.40<0.45, hence O.K

 

V. Selection of water content:

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

Estmated water contentfor 100 mm slump=186+6/100*186

                                                             =197 litres

As plasticizer is used, water content can be reduced by 20% and above

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

Hence arrived water content=197*0.71=140 litres

 

VI. Calculation of Cement content:

      water-cement ratio=0.40

      Cement content= 140/0.4=350 Kg/m^3

From table 5 of IS 456:2000 min cement content for'severe' exposure condition= 320 Kg/m^3

350 Kg/m^3>320 Kg/m^3.Hence OK

 

VII. Proportion of Coarse aggregate and Fine aggregate:

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

In the present case the water cement ratio = 0.40

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

As water-cement ratio is lower by 0.10, the proportion of volume of coarse aggregate is increased by 0.02(at the rate of-/+ 0.01 for every -/+ change in w/c ratio)

Therefore,corrected volume of Coarse aggregate for w/c ratio 0.4=0.62

For pumpable concrete these values are reduced by 10%

Therefore,volume of Coarse aggregate = 0.62*0.90 = 0.56

Volume of Fine aggregate content = 1-0.56 = 0.44

                                                 

VIII. Mix Calculations

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

a) Volume of concrete=1 m^3

b) Volume of cement=mass of cement/specific gravity of cement*1/1000

                              =350/3.15*(1/1000)

                              =0.111 m^3 

c) Volume of water=mass of water/specific gravity of water*1/1000

                           =140/1*(1/1000)

                           =0.140 m^3 

d) Volume of chemical admixture=mass of chemical admixture/Specific gravity of admixture*1/1000

                                               =7.6/1.145*(1/1000)

                                               =0.006 m^3

e) Volume of aggregate(Fine aggregate and Coarse aggregate)=((a-(b+c+d))

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

                                                                                         =0.743 m^3

f) Mass of Coarse aggregate=e*Volume of coarse aggregate*Specific gravity of coarse aggregate*1000

                                         =0.743*0.56*2.74*1000

                                         =1140 Kg/m^3 

g) Mass of Fine aggregate=e*volume of fine aggregate*specific gravity of fine aggregate*1000

                                     =0.743*0.44*2.74*1000

                                     =896 Kg/m^3

IX. Mix Calculation:

a) Cement=350 Kg/m^3

b) Water=140 Kg/m^3

c) Fine aggregate=896 Kg/m^3

d) coarse aggregate=1140 Kg/m^3

e) Chemical admixture=7 Kg/m^3

f)  w/c ratio=0.40

Trial mix ratio=1:2.56:3.26