Project on Concrete Mix Design for various grades of Concrete

Mix proportioning of concrete using fly ash as part of replacement of OPC for M35 grade concrete

1. Stipulations for proportioning

1. Grade designation : M35
   1. TYPE OF CEMENT :OPC 43 grade conforming to IS 8112
2. Type of mineral admixture :Fly ash conforming to IS 3812 (Part I)
3. Maximum nominal size of aggregate :20 mm
4. Min Cement content = 320 kg/m3
5. Max w/c ratio = .45 (From table 5 of IS 456:2000- exposure condition)
6. Workability = 100 mm(slimp value for pumbable concrete)
7. Exposure condition :Severe (for reinforced concrete)
8. Method of concrete placing: Pumbing
9. Degree of supervision :Good
10. Type of aggregate :Crushed angular aggregate
11. Max cement content = 450 kg/m3
12. Chemical admixture type : Superplasticizers

B. Test data for materials

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 1)
4. SPECIFIC GRAVITY OF FLY ASH =2.2
5. CHEMICAL ADMIXTURE :Superplasticizers conforming to IS 9103
6. Specific gravity of

Corase aggregate :2.74

Fine aggregate =2.747.

7.Water absorption:

Coarse aggregate :0.5%

Fine aggregate = 1.0%

8.Free surface moisture:

Coarse agrregate :Nil

Fine aggreagte :Nil

9.Sieve Analysis

Coarse agrregate & Fine aggregate

C.Target strength for Mix proportioning

f'Ck= fck + 1.65 S

f'ck = Target avg compressive strength @ 28 days

fck = characteristic compressive @ 28 days

S = Standard deviation

For 35 grade concrete ,

f'ck = 35+1.65*5

= 43.25N/mm2

C. Selection of W/C ratio

Frm the graph w/c ratio M35 grade ,0.35 is the value

Based on experience,

adopted w-c ratio is

0.35<0.45,Hence ok.

D.Selection of water content

From table 2 of IS 10262:2009,

Max W/C for 20 mm aggregate = 186 l(25mm to 50mmm slump range)

Estimate water content for 100mm slump

=186 +0.06*186

= 197l

As super plasticizer is used, water content can be reduced to be 20% .Based on 29% in wet phase

Hence arrived water content =197*0.71 = 140 l

D.Calculation of cement content

Water - cement Ratio = 0.35

Cementitous material (Cement+fly ash content )= 140

Cement content = 140/0.35 = 400 kg/m3

From table 5 of IS 456:2000 MIN cement content for severe exposure condition = 340 kg/m3

400 kg/m3 > 340 kg /m3

Hence ok.

E.Proportion of volume of coarse aggregate & fine aggregate content :

From table 3 of IS 10262 :2009, Vol of coarse aggregate corresponding to 20 mm the coarse & fine aggregate (zone 1 ) for water cement ratio = 0.6 for M35 grade concrete

In the present case the w/c ratio is 0.35

therefore the 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 -1/+ 0.01 for every +- o.05 change in water cement ratio)

Therefore the corrected value of coarse aggregate for the w-c ratio of 0.35 = 0.64

For pumpable concrete these values are reduced by 10%

Volume of coarse aggregate = 0.64*0.9

= 0.576

Volume of fine aggregate = 1-0.576 = 0.424

F. Mix calculations

Now to proportion a mix containing fly ash

a)Percentage of fly ash needs to be decided based on project requirement & quality of materials

b)In some cases increase in cementitious material can be warranted.Increase in cementitious material cement & its % may be based on experience & trial (eg : with increase of 10 % cementitious material content )

Cementitious material content = 400*1.1

= 440 kg/m3

Water content = 140 kg/m3

so w/c ratio = 0.318

Fly ash @ 30 % total cementitious material content

= 440* 30 %

=132 kg/m3

Cement (opc) = 440-132

= 308 kg/m3

Saving of cement while using fly ash = 400 -308

= 92 kg/m3

Fly ash being utilized = 132 kg/m3

Mix calculation per unit volume of concrete shall be as follows :

a)Volume of concrete = 1m3

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

=(308/3.15)*(1/1000)

=0.097 m3

c)Volume of fly ash =(132/2.2)*(1/1000)

=0.06 m3

d)Volume of water = (Mass of water /Specific gravity of water)* (1/1000)

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

=0.140 m3

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

=(7/1.143)*(1/1000)

=0.006 m3

f) Volume of aggregate (FA& CA) =(a- (b+c+d)

=(1- (0.097+0.06+0.14+0.006))

=0.697 m3

g)Mass of coarse aggregate = e* Vol of coarse aggregate * S.G of coarse aggregate *1000

= 0.697 * 0.576* 2.76 * 1000

= 1108.06kg/m3

h)Mass of fine aggregate = c* Vol of fine aggregate * SG of fine aggregate * 1000

= 0.697* 0.424*2.74*1000

= 809.746 kg/m3

G. Mix proportion for trial 1 :

Cement = 308 kg/m3

Fly ash = 132 kg/m3

Water = 140 kg/m3

Fine aggregate = 1108.06 kg/m3

Coarse aggregate = 809.746 kg/m3

Chemical admixture = 7kg/m3

w/c ratio =0.35

Trial mix ratio = 1:2.63:3.6

2. Mix Design for M50 grade without flyash

1. Stipulations for proportioning

1. Grade designation : M50
   1. TYPE OF CEMENT :OPC 43 grade conforming to IS 8112
2. Type of mineral admixture :Fly ash conforming to IS 3812 (Part I)
3. Maximum nominal size of aggregate :20 mm
4. Min Cement content = 320 kg/m3
5. Max w/c ratio = .45 (From table 5 of IS 456:2000- exposure condition)
6. Workability = 100 mm(slimp value for pumbable concrete)
7. Exposure condition :Severe (for reinforced concrete)
8. Method of concrete placing: Pumbing
9. Degree of supervision :Good
10. Type of aggregate :Crushed angular aggregate
11. Max cement content = 450 kg/m3
12. Chemical admixture type : Superplasticizers

B. Test data for materials

1. Cement used :OPC 43 grade conforming to IS 8112
2. Specific gravity of cement :3.15
3. SPECIFIC GRAVITY OF FLY ASH =2.2
4. CHEMICAL ADMIXTURE :Superplasticizers conforming to IS 9103
5. Specific gravity of

Corase aggregate :2.74

Fine aggregate =2.747.

6.Water absorption:

Coarse aggregate :0.5%

Fine aggregate = 1.0%

7.Free surface moisture:

Coarse agrregate :Nil

Fine aggreagte :Nil

8.Sieve Analysis

Coarse agrregate & Fine aggregate

C.Target strength for Mix proportioning

f'Ck= fck + 1.65 S

f'ck = Target avg compressive strength @ 28 days

fck = characteristic compressive @ 28 days

S = Standard deviation

f'ck = 50 + 1.65*5

= 58.25 N/mm2

D.Selection of W/C ratio

Frm the graph w/c ratio M50grade ,0.4 is the value (from table 5 of IS 456:2000 UNDER SEVERE EXPSURE CONDITION MAXIMUM W/C RATIO =0.4

Based on experience,

adopted w-c ratio is

0.4<0.45,Hence ok.

Selection of water content

From table 2 of IS 10262:2009,

Max W/C for 20 mm aggregate = 186 l(25mm to 50mmm slump range)

Estimate water content for 100mm slump

=186 +0.06*186

= 197l

As super plasticizer is used, water content can be reduced to be 20% .Based on 29% in wet phase

Hence arrived water content =197*0.71 = 140 l

VI)Calculation of cement conent

water - cement ratio =0.4

cement content = (140/0.4)= 350 kg/m3

Proportion of volume of coarse aggregate & fine aggregate content :

From table 3 of IS 10262 :2009, Vol of coarse aggregate corresponding to 20 mm the coarse & fine aggregate (zone 1 ) for water cement ratio = 0.6 for M50 grade concrete

In the present case the w/c ratio is 0.4

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

As water - cement ratio s lower by 0.10 ,the proportion of volume of coarse aggregate is increades by 0.02

(at the rate of -1/+ 0.01 for every +- o.05 change in water cement ratio)

Therefore the corrected value of coarse aggregate for the w-c ratio of 0.4 = 0.62

For pumpable concrete these values are reduced by 10%

Volume of coarse aggregate = 0.62*0.9

= 0.56

Volume of fine aggregate = 1-0.56 = 0.44

Mix calculation per unit volume of concrete shall be as follows :

a)Volume of concrete = 1m3

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

=(350/3.15)*(1/1000)

=0.111 m3

c)Volume of water = (Mass of water /Specific gravity of water)* (1/1000)

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

=0.140 m3

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

=(7/1.143)*(1/1000)

=0.006 m3

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

=(1- (0.111+0.14+0.006))

=0.743m3

f)Mass of coarse aggregate = e* Vol of coarse aggregate * S.G of coarse aggregate *1000

= 0.743 * 0.56* 2.76 * 1000

= 1140kg/m3

g)Mass of fine aggregate = c* Vol of fine aggregate * SG of fine aggregate * 1000

= 0.743* 0.44*2.74*1000

= 896 kg/m3

G. Mix proportion for trial 1 :

Cement = 350 kg/m3

Water = 140 kg/m3

Fine aggregate = 896 kg/m3

Coarse aggregate = 1140 kg/m3

Chemical admixture = 7kg/m3

w/c ratio =0.4

Trial mix ratio = 1:2.56 :3.36