Project on Concrete Mix Design for various grades of Concrete

MIX DESIGN FOR M35 GRADE OF CONCRETE WITH FLY ASH :

1. Collected data:

1.  Grade of deisgnation: M35
2.  Type of cement used : OPC-43 grade comfirming to IS 8112
3. Mineral admixture used: Fly ash Confirming to IS 3812 part 1
4. Max. nominal size of agg.: 20mm
5. Minimum cement content: 340 Kg/cub. m (From talble 5 of IS 456-2000 Exposure condition)
6. Max. w/c ratio: 0.45 (From talble 5 of IS 456-2000 Exposure condition)
7. Workability of concrete: 100 mm (Slump value for pumpable concrete)
8. Exposure condition: Very severe (From talble 5 of IS 456-2000 Exposure condition)
9. Method of concrete placing: Pumping
10. Degree of supervision: Good
11. Type of agg.: Crushed angular agg.
12. Chemical admixture type: Super platisizer

2. Test data for materials:

1. Cement used OPC 43 grade confirming IS 8112.
2. Sp. gravity of cement: 3.15 
3. Chemical admixture: confirming to IS 9103
4. Sp. gravity of fly ash: 2.2 
5. Sp. gravity of coarse agg.: 2.74 and fine agg.: 2.74
6. Water absorption coarse agg.: 0.5% and fine agg.: 1%
7. Free surface moisture: coarse and fine agg.: NIL
8. Sieve analysis: Fine agg. confirming to grading zone I of table 4 of IS 383.  

3. Target strength for mix proportioning: 

                       f'ck =fck + 1.65 s
                                                         where,
                                                                   f'ck = target average compressive strength at 28 days,
                                                                   fck = characteristic compressive strength at 28 days, and
                                                                      s = standard deviation.
   From Table I, standard deviation, s =5 N/mm2
Therefore, target strength =35 + 1.65 x 5 =43.25 N/mm2

4. Selection of water cement ratio:

From Table 5 of IS 456, maximum water-cement ratio = 0.45.

According to graph of c.s. of concrete for 28 days and w/c ratio.

w/c = 0.42 

0.42 < 0.45 , hence O.K.

 5. Selection of water content:

From Table 2, maximum water content =186 litre (for 25 to 50 mm slump range)
for 20 mm aggregate
Estimated water content for 100 mm slump =186+((6/100)*186) 

                                                                 =197 litre
As superplasticizer is used, the water content can be reduced up 20 percent and above.
Based on trials with superplasticizer water content reduction of 29 percent has been achieved. Hence, the
arrived water content =197 x 0.71 =140 litre

6. Calculation of cement content:

Cement content = 140/0.42

                         = 333 Kg/m3

From Table 5 of IS 456, minimum cement
content for ' very severe' exposure condition = 340 kg/m3
340 kg/m3 >333 kg/m3 

Here, We have to increase cement content. 

    Cementitious material content = 340 Kg 

                              When we used fly ash then in some case, there is requirement of increase cementitious material content(by 10%).

Cementitious material content = 340*1.1

                                              = 374 Kg

There for w/c, 

                         w/c = 140/374

                                = 0.38

Fly ash @ 30% total cementitous material content = 374*30% = 112 Kg

Saving of cement by using fly ash = 340-262 = 78 Kg

Fly ash being ultilized = 112 Kg.

 7. Proportion of volume of coarse agg. and fine agg. content:

From Table 3. 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 0.40. Therefore. volume of coarse aggregate is required to be increased
to decrease the fine aggregate content. As the water-cement ratio is lower by 0.10. the proportion of volume of
coarse aggregate is increased by 0.02 (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.
NOTE - In case the coarse aggregate is not angular one. then also volume of coarse aggregate may be required 10 be increased
suitably , based on experience.
For pumpable concrete these values should be reduced by 10 percent.
Therefore, volume of coarse aggregate = 0.62 x 0.9 = 0.56.
Volume of fine aggregate content =1 - 0.56 =0.44.

8. Mix calculation:
The mix calculations per unit volume of concrete shall be as follows:
a) Volume of concrete = 1 m3
b) Volume of cement = (Mass of cement/Sp. gravity)*(1/1000)

                                = (262/3.15)*(1/1000)

  volume of cement    = 0.083 m3

c) Volume of Fly ash = (Mass of Fly ash/Sp. gravity)*(1/1000)

                               = (112/2.2)*(1/1000)

                                = 0.051 m3
d) Volume of water = (Mass of water/Sp. gravity)*(1/1000)

                             = (140/1)*(1/1000)

                              =0.140 m3
e) Volume of chemical admixture = (Mass of admixute/Sp. gravity)*(1/1000)

                                                 = (6.8/1.145)*(1/1000)            ..................................(mass of admixture = 2% of cementitous material)

                                                  = 0.006 m3

f) Volume of all in aggregate = (a-(b+c+d+e))

                                            = (1-(0.083+0.140+0.006+0.051))

                                             = 0.72 m3

g) Mass of coarse aggregate = e*volume of coarse agg.*Sp. gravity fine agg.*1000

                                          =0.72*0.56*2.74*1000

                                           =  1105 Kg
i) Mass of fine aggregate = e*volume of fine agg.*Sp. gravity of fine agg.*1000

                                      = 00.72*0.44*2.74*1000

                                        = 868 Kg

9. Mix proportion: 

Cement = 262 Kg

Fly ash = 112 Kg

Water = 140 Kg
Fine aggregate = 868 Kg
Coarse aggregate = 1105 Kg
Chemical admixture = 7 Kg
Water-cement ratio = 0.38

Trial mix ratio:

1 : 2.32 : 2.96

MIX DESIGN FOR M50 GRADE OF CONCRETE:

1. Collected data:

1.  Grade of deisgnation: M50
2.  Type of cement used : OPC-53 grade comfirming to IS 8112
3. Max. nominal size of agg.: 20mm
4. Minimum cement content: 340 Kg/cub. m (From talble 5 of IS 456-2000 Exposure condition)
5. Max. w/c ratio: 0.45 (From talble 5 of IS 456-2000 Exposure condition)
6. Workability of concrete: 100 mm (Slump value for pumpable concrete)
7. Exposure condition: Very severe (From talble 5 of IS 456-2000 Exposure condition)
8. Method of concrete placing: Pumping
9. Degree of supervision: Good
10. Type of agg.: Crushed angular agg.
11. Chemical admixture type: Super platisizer

2. Test data for materials:

1. Cement used OPC 53 grade confirming IS 8112.
2. Sp. gravity of cement: 3.15 
3. Chemical admixture: confirming to IS 9103
4. Sp. gravity of coarse agg.: 2.74 and fine agg.: 2.74
5. Water absorption coarse agg.: 0.5% and fine agg.: 1%
6. Free surface moisture: coarse and fine agg.: NIL
7. Sieve analysis: Fine agg. confirming to grading zone I of table 4 of IS 383.  

3. Target strength for mix proportioning: 

                       f'ck =fck + 1.65 s
                                                         where,
                                                                   f'ck = target average compressive strength at 28 days,
                                                                   fck = characteristic compressive strength at 28 days, and
                                                                      s = standard deviation.
   From Table I, standard deviation, s =5 N/mm2
Therefore, target strength =50 + 1.65 x 5 =58.25 N/mm2

4. Selection of water cement ratio:

From Table 5 of IS 456, maximum water-cement ratio = 0.45.

According to graph of c.s. of concrete for 28 days and w/c ratio.

w/c = 0.42 

0.42 < 0.45 , hence O.K.

 5. Selection of water content:

From Table 2, maximum water content =186 litre (for 25 to 50 mm slump range)
for 20 mm aggregate
Estimated water content for 100 mm slump =186+((6/100)*186) 

                                                                 =197 litre
As superplasticizer is used, the water content can be reduced up 20 percent and above.
Based on trials with superplasticizer water content reduction of 29 percent has been achieved. Hence, the
arrived water content =197 x 0.71 =140 litre

6. Calculation of cement content:

Cement content = 140/0.42

                         = 333 Kg/m3

From Table 5 of IS 456, minimum cement
content for ' very severe' exposure condition = 340 kg/m3
340 kg/m3 >333 kg/m3 

Here, We have to increase cement content. 

For cement content = 340 Kg the w/c ratio is,

                              w/c = 0.41

 7. Proportion of volume of coarse agg. and fine agg. content:

From Table 3. 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 0.40. Therefore. volume of coarse aggregate is required to be increased
to decrease the fine aggregate content. As the water-cement ratio is lower by 0.10. the proportion of volume of
coarse aggregate is increased by 0.02 (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.
NOTE - In case the coarse aggregate is not angular one. then also volume of coarse aggregate may be required 10 be increased
suitably , based on experience.
For pumpable concrete these values should be reduced by 10 percent.
Therefore, volume of coarse aggregate = 0.62 x 0.9 = 0.56.
Volume of fine aggregate content =1 - 0.56 =0.44.

8. Mix calculation:
The mix calculations per unit volume of concrete shall be as follows:
a) Volume of concrete = 1 m3
b) Volume of cement = (Mass of cement/Sp. gravity)*(1/1000)

                                = (340/3.15)*(1/1000)

  volume of cement    = 0.108 m3
c) Volume of water = (Mass of water/Sp. gravity)*(1/1000)

                             = (140/1)*(1/1000)

                              =0.140 m3
d) Volume of chemical admixture = (Mass of admixute/Sp. gravity)*(1/1000)

                                                 = (6.8/1.145)*(1/1000)            ..................................(mass of admixture = 2% of cementitous material)

                                                  = 0.006 m3

e) Volume of all in aggregate = (a-(b+c+d))

                                            = (1-(0.108+0.140+0.006))

                                             = 0.746 m3

t) Mass of coarse aggregate = e*volume of coarse agg.*Sp. gravity fine agg.*1000

                                          =0.746*0.56*2.74*1000

                                           =  1144 Kg
g) Mass of fine aggregate = e*volume of fine agg.*Sp. gravity of fine agg.*1000

                                      = 0.746*0.44*2.74*1000

                                        = 900 Kg

9. Mix proportion: 

Cement = 340 Kg
Water = 140 Kg
Fine aggregate = 900 Kg
Coarse aggregate = 1144 Kg
Chemical admixture = 7 Kg
Water-cement ratio = 0.41

Trial mix ratio:

1 : 2.65 : 3.37