Project on Concrete Mix Design for various grades of Concrete

                                                                          PROJECT

1.AIM:- To find the mix design for M50 grade of concrete without fly ash.

PROCEDURE:- 

 Step 1:- Stipulations for proportioning 

1. Grade designation : M50
2. Tyoe of cement : OPC 43 grade conforming to IS 8112
3. Maximum nominal size of aggragate : 20mm
4. Minimmum cement content; 320 kg/m^3 (from table 5 IS 456:2000- SEVERE EXPOSURE)
5. Maximum cement content : 0.45 (from table 5 IS 456:2000- SEVERE EXPOSURE)
6. Workability: 100mm  ( slump value for pumpable concrete)
7. Exposure condition: Severe  ( for reinforced concrete)
8. Method of concrete placing: Pumping
9. Degree of suoer vission: Good
10. Type of aggregate: Crushed angular aggregate.
11. Maximum cement content: 450 kg/m^3
12. Chemical admixture type: Superplastizer conforming to IS 9103

Step 2 :- Test data for material

1. Cement used : OPC 43 grade conforming to IS 8112
2. Specific gravity of cement: 3.15
3. Chemical admixture: Superplastizer conforming to IS 9103

  4.Specific gravity of:

1. Coarse aggregate : 2.74
2. Fine aggregate     : 2.74

     5.Water absorption:

• Coarse aggregate : 0.5%
• Fine aggregate     : 1.0%

     6. Free surface moisture

• Coarse aggregate : Nil
• Fine aggregate     : Nil 

     7. Sieve analysis

• Coarse aggregate : conforming to table 2 of IS 383
• Fine aggregate     : Conforming to grading zone 1 of table 4 of IS 383

 STEP 3. Target strength for Mix proportioning; f'ck = fck+1.65S

      where, f'ck- target average compressive strength at 28 dyas

                 fck- Characteristic compressive strength at 28 days

                  S - Standard deviation.

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

                                                                         fck = 50 N/mm^2

    therefore, target strength, f'ck = 50 +1.65*5= 58.25 N/mm^2

 STEP 4. Selection of water cement ratio;

   From table 5 of IS 456:2000 for severe exposure, maximum w/c ratio=0.45

   from fig 1 of IS 1026:2019 W/C ratio for respective f'ck is 0.35

   0.35<0.45, hence ok.

STEP  5. Seletion of water content.

from table 2 of IS 10262;2009, maximum water content for 20mm aggregate= 186 litres (for 25mm to 50mm slump range)

Estinmated water content for 100mm slump= 186+6/100*186 = 197 litres.

A s superplasticer is used, the water content reduction of 29% is achived plasticizer.

Hence the arrived water content= 197*0.71 = 140 litres

STEP 6. Calculation of cement ratio;

Water cement ratio = 0.35

Cement content  = 140/0.35 = 400 kg/m^3

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

Here it is 400 kg/m3> 320 kg/m3, hence, O.K

STEP 7. PROPORTION OF VOLUME OF COARSE AGGREGATE AND FINE AGGREGETECONTENT

From Table 3. From IS 10262:2009 volume of coarse aggregate corresponding to 20 mm size aggregate and fine aggregate (Zone I) for water-cement ratio of 0.50 =0.60.
In the present case water-cement ratio is 0.35. Therefore. volume of coarse aggregate is required to be increased to decrease the fine aggregate content. As the water-cement ratio is lower by 0.15. the proportion of volume of coarse aggregate is increased by 0.03 (at the rate of -/+ 0.01 for every ± 0.05 change in water-cement ratio).
Therefore. corrected proportion of volume of coarse aggregate for the water-cement ratio of 0.35 = 0.63

For pumpable concrete these values are reduced by 10%.

therefor,volume of concrete = 0.63*0.9= 0.567

therefor volume of fine aggregate= 1- 0.567 = 0.433

STEP 8. MIX CALCULATIONS.

The mix calcualtions per unit volume of concrete shall be as follows;

   a) Volume of concrete =  1 m^3

   b) Volume of cement =  Mass of concrete/(specific gravity of cement)*(1/1000) 

                               = (400/3.15)*(1/1000)

                               = 0.127 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= (maas of chemical admixture)/(specific gravity of admixture)*(1/1000)

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

  e) Volume of aggregate (FA & CA) = (a-(b+c+d)) = (1-(0.127+0.140+0.006))

                                                    = 0.727 m^3

 f) Mass of coarse aggregate  = e* volume of coarse aggregate*specific gravity of coarse aggregate*1000

                                            =  0.727*0.567*2.74*1000

                                            =1129.453 kg/m^3

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

                                           = 0.727*0.433*2.74*1000

                                           = 862.527 kg/m^3

STEP 9. Mix proportions for trial number 1 :

       Cement           = 400  kg/m^3

       water              = 140  kg/m^3

fine aggregate        = 862.527 kg/m^3

coarse aggregate    = 1129.453 kg/m^3

chemical admixture =     7  kg/m^3 

water cement ratio  =    0.35

STEP 9: The slump shall be measured and the water content and dosage of admixture shall be adjusted for achieving the required slump on trial.The mix proportions shall be reworked for the actual water ocontent and checked for durability requirement.

STEP 10: Two more trials having variations of 10% of water cement ratio in clause  shall be carried out and a graph between three water cement ratio and corresponding strengths shall be plotted to worked out the mix proportions for the given target strength for field   trials. However durabiltyt shall be met out.

RESULTS

Mix proportions for M50 grade of concrete.

       Cement            = 400  kg/m^3

       water               = 140  kg/m^3

fine aggregate         = 862.527 kg/m^3

coarse aggregate     = 1129.453 kg/m^3

chemical admixture  =     7  kg/m^3 

water cement ratio   =    0.35

2.  AIM:- To find the mix design for M35 grade of concrete with fly ash.

Step 1:- Stipulations for proportioning 

a) Grade designation                                  :-  M35

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

c) Type of mineral admixture                       :-  Fly ash conforming to IS 38I2 (Part I)

d) Maximum nominal size of aggregate        :-  20mm

e) Minimum cement content                       :-  320 kg/m'

f) Maximum water-cement ratio                  :-  0.45

G) Workability                                           :-  100 mm (slump)

h) Exposure condition                                :-  Severe (for reinforced concrete)

j) Method of concrete placing                     :-  Pumping

k) Degree ofsupervision                            :-   Good

m) Type of aggregate                               :-   Crushed angular aggregate

n) Maximum cement (OPC) content           :-   450 kg/m'

p) Chemical admixture type                      :-   Superplasticizer

Step 2 :- Test data for material

1. Cement used : OPC 43 grade conforming to IS 8112
2. Specific gravity of cement: 3.15
3. fly ash                              :- Conforming to IS 3812 (Part I)
4. specific gravity of fly ash    :- 2.2
5. Chemical admixture: Superplastizer conforming to IS 9103

    4.Specific gravity of:

1. Coarse aggregate : 2.74
2. Fine aggregate     : 2.74

     5.Water absorption:

• Coarse aggregate : 0.5%
• Fine aggregate     : 1.0%

     6. Free surface moisture

• Coarse aggregate : Nil
• Fine aggregate     : Nil 

     7. Sieve analysis

• Coarse aggregate : conforming to table 2 of IS 383
• Fine aggregate     : Conforming to grading zone 1 of table 4 of IS 383

STEP 3. Target strength for Mix proportioning; f'ck = fck+1.65S

      where, f'ck- target average compressive strength at 28 dyas

                 fck- Characteristic compressive strength at 28 days

                  S - Standard deviation.

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

                                                                         fck = 35 N/mm^2

    therefore, target strength, f'ck = 35+1.65*5= 43.25 N/mm^2

 STEP 4. Selection of water cement ratio;

   From table 5 of IS 456:2000 for severe exposure, maximum w/c ratio=0.45

   from fig 1 of IS 1026:2019 W/C ratio for respective f'ck is 0.37

   0.37<0.45, hence ok.

Based on experience. adopt water-cement ratio as 0.40.
0.40 < 0.45, hence, O.K

STEP  5. Seletion of water content.

from table 2 of IS 10262;2009, maximum water content for 20mm aggregate= 186 litres (for 25mm to 50mm slump range)

Estinmated water content for 100mm slump= 186+6/100*186 = 197 litres.

A s superplasticer is used, the water content reduction of 29% is achived plasticizer.

Hence the arrived water content= 197*0.71 = 140 litres

Based on experience. adopt water-cement ratio as 0.40.
0.40 < 0.45, hence, O.K.

STEP 6. CALCULATION OF CEMENT AND FLY ASH CONTENT

Water-cement ratio (see note under 4.1)  :- 0.40

Cementitious material (cement + fly ash) content = 140/0.40 = 350 kg/m^3

From Table 5 of IS 456, minimum cement content for 'severe' exposure conditions 320 kg/m^3,

350 kg/m^3 > 320 kg/m^3, hence. O.K

Now, to proportion a mix containing fly ash the following steps are suggested

a) Decide the percentage fly ash to be used based on project requirement and quality of materials

b) In certain situations increase in cementitious material content may be warranted, The decision on increase  in cementitious material content and    its percentage may be based on experience and trial.

Cementitious material content  -  350 x 1.10 = 385 kg/m^3

Water Content                           -   140 kg/m^3

So. water-cement ratio             -    140/385  = 0.364

Fly ash @ 30% of total cementitious material content  = 385 x 30% =115 kg/m^3

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

Saving ofcement while using fly ash                              = 350 - 270 = 80 kg/m^3, and

Fly ash being utilized                                                       = 115 kg/m^3

STEP 7. PROPORTION OF VOLUME OF COARSE AGGREGATE AND FINE AGGREGETECONTENT

From Table 3. From IS 10262:2009 volume of coarse aggregate corresponding to 20 mm size aggregate and fine aggregate (Zone I) for water-cement ratio of 0.50 =0.60.

In the present case water-cement ratio is 0.35. Therefore. volume of coarse aggregate is required to be increased to decrease the fine aggregate content. As the water-cement ratio is lower by 0.15. the proportion of volume of coarse aggregate is increased by 0.03 (at the rate of -/+ 0.01 for every ± 0.05 change in water-cement ratio).

Therefore. corrected proportion of volume of coarse aggregate for the water-cement ratio of 0.40 = 0.62

For pumpable concrete these values are reduced by 10%.

therefor,volume of concrete = 0.62*0.9= 0.56

therefor volume of fine aggregate= 1- 0.56 = 0.44

STEP 8. MIX CALCULATIONS.

The mix calcualtions per unit volume of concrete shall be as follows;

   a) Volume of concrete =  1 m^3

   b) Volume of cement =  Mass of concrete/(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= (maas of chemical admixture)/(specific gravity of admixture)*(1/1000)

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

  e) Volume of aggregate (FA & CA) = (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* vol of fine aggregate*specific gravity of fine aggregate*1000

                                           = 0.743*0.44*2.74*1000

                                           = 896  kg/m^3

STEP 9. Mix proportions for trial number 1 :

       Cement           =  350 kg/m^3

       water              = 140  kg/m^3

fine aggregate        = 896 kg/m^3

coarse aggregate    = 1140 kg/m^3

chemical admixture =     7  kg/m^3 

water cement ratio  =    0.4

STEP 9: The slump shall be measured and the water content and dosage of admixture shall be adjusted for achieving the required slump on trial.The mix proportions shall be reworked for the actual water ocontent and checked for durability requirement.

STEP 10: Two more trials having variations of 10% of water cement ratio in clause  shall be carried out and a graph between three water cement ratio and corresponding strengths shall be plotted to worked out the mix proportions for the given target strength for field   trials. However durabiltyt shall be met out.

RESULTS

Mix proportions for M35 grade of concrete.

Cement                          =  350 kg/m^3

water                             = 140  kg/m^3

fine aggregate                =  896 kg/m^3

coarse aggregate            = 1140 kg/m^3

chemical admixture         =     7  kg/m^3 

water cement ratio          =    0.4