Project on Concrete Mix Design for various grades of Concrete

Calculate the Concrete Mix Design for M35 grade concrete with fly ash & M50 grade concrete without Fly ash

1. Concrete mix design for M35 grade with Fly ash

We have to calculate the mix proportion for M35 grade with Fly ash.Which ratio of cement:Fine aggregate : coarse aggregate

M35 Grade indicates that the m is mixed design and 35 indicates the compressive strenght of concrete after 28 days .

1. Mix design OF M35 grade with Fly ash

AIM:- Mix design of M35 grade with Fly ash

Method /procedure;

1.Data collection

• Grade of concrete = M35
• Cement type = OPC
• Type of mineral admixture = fly ash
• Max. nominal size of aggregate = 20 mm
• Aggregate type = angular
• We take ,exposure condition = very severe ( r/f concrete)
• From table 5 of IS456:2000 the min cement content= 340 kg/ m3
• W/C ratio= 0.45 as per table 5 of IS456:2000
• Workability= 75 mm
• Method of concrete= pumping concrete
• Max cement content = 450 Kg/m3
• Admixture= superplasticizer

2. Tested data for materials

• Grade of cement= 43 grade
• Specific gravity of cement= 3.15
• Specific gravity of coarse aggregate= 2.74
• Specific gravity of fine aggregate = 2.74
• Specific gravity of admixture(superplasticizer)=1.14
• Specific gravity of fly ash=2.2
• Water absorption of coarse aggregate= 0.5%
• Water absoption of fine aggregate =1%
• Free surface moisture for;

Coarse aggregate = NIL

Fine aggregate = NIL

• Sieve analysis of aggregate is done with the help of zone 1 of table 4 of IS 383

3. Calculation of targeted strenght

f'ck = fck+1.65*S

f'ck = 35+1.65*5 ................as S is a standard deviation ,take it from table 8 of IS 10262:2009

f'ck = 43.25 N/mm2

4. Selection of w/c ratio

Maximum w/c ratio we have= 0.45 ......as per the IS 456:2000

Experience w/c ratio =0.4

0.4<0.45 ..ok to have.

5. Water content calculation

For 20 mm aggregate size the max water content is 186 .......as per table 2 o IS 10262:2009

For 75 mm slump = 186 +(3/100)*186

=191.58 litre

Because we used admixture as a superplasticizer , water content reduced by 30%

In trials 29% water reduced occurs mostly;

Water content = 191.58*0.71

=136.02 litre

6. Calculation of cement content

We have w/c ratio=0.4

(cement + fly ash) i.e cementitious material =136.02/0.4=340.025 Kg/m3

as per IS456:2000 the cement content for very severe condition is 340 kg/m3

340.05>340 hence,its okay

Increase in cementitious materials the increase in precentage in content based on the trials we take,

In that case we considered a 10% increase;

Cementitious materials content = 340.02*1.1

= 374.022 Kg/m3

Water content we calculate before is 136.02 litre

so ,the w/c ratio = 136.02/374.022

=0.363

Fly ash is 30% of total cementitious materials i.e = 374.022*30/100

=112.20 kg/ m3

Then, OPC cement is = 374.022-112.20 = 261.82 kg/m3

because of fly ash use we save cement i.e = 340.022-261.82 = 78.2 kg/m3

Superplasticizer is 0.6% of total cementitious materials = 374.02*0.6/100 = 2.24 kg/m3

7. Coarse and Fine aggregates

From table 3 of IS10262:2009 fro 20 mm aggregate size the w/c ratio is 0.6

in present case we have w/c ratio is 0.45

So as the water cement ratio is decresed by 0.1 we have increase the volume of aggregate by 0.02

Corrected value of coarse aggregate for w/c ratio is = 0.6*(0.15*0.05)*0.01

= 0.63

for pumpable concrete , these values reduced by 10%

volume of coarse aggregate = 0.63*0.9

= 0.567

volume of fine aggregate = 1-0.567

= 0.433

8. Mix calculations

Calculations for unit vomume of concrete are as below;

1. Volume of concrete = 1 m3
2. Volume of cement = ( mass of cement/specific gravity of cement)*1/1000

= (261.82/3.15)*1/1000

= 0.083 m3

C.Volume of water = ( mass of water/specific gravity of water)*1/1000

= (136.02/1)*1/1000

= 0.136 m3

D. Volume of fly ash = ( mass of fly ash/specific gravity of fly ash)*1/1000

= (112.20/2.2)*1/1000

= 0.051 m3

E. Volume of admixture = ( mass of admixture/specific gravity of admixture)*1/1000

= (2.24/1.14)*1/1000

= 0.0019 m3

F. Volume of CA and FA aggregates = A-(B+C+D+E)

= 1-(0.083+0.136+0.051+0.0019)

= 0.7281 m3

G. Weight of coarse aggregate = F* volume of coarse aggregate*specific gravity of CA*1000

= 0.7281*0.567*2.74*1000

= 1131.16 KG

H. Weight of Fine aggregate = F* volume of fine aggregate*specific gravity of FA*1000

= 0.7281*0.433*2.74*1000

= 863.83 KG

9. Mix proportions

• Cement = 261.82 kg
• Water = 136.02 litre
• Fly ash = 112.20
• Coarse aggregate = 1131.16 kg
• Fine aggregate = 863.83 kg
• Water cement ratio = 0.363
• Admixture weight = 2.24

Therefore the mix ratio of M35 grade with fly ash is 261.82:863.83:1131.16

i.e 1: 3.3 : 4.32

10. The slump shall be measured and the water content and dosage of admixture shall be adjusted for achieving therequired slump .trail may be rework and then check the durability.

11. if the more than two trials having w/c ratio has 10 % variation shall be carried out and graph between three w/c ratio and corresponding strenght are plotted to work out the mix proportion, however durablity shall be met.

RESULT :- The mix design fo M35 concrete with Fly ash is 1 : 3.3 : 4.32

2. Mix design of M50 concrete

AIM :- Mix design of M50 concrete

Method /procedure;

1.Data collection

• Grade of concrete = M50
• Cement type = OPC 43 grade
• Max. nominal size of aggregate = 20 mm
• Aggregate type = angular
• We take ,exposure condition = severe ( r/f concrete)
• From table 5 of IS456:2000 the min cement content= 320 kg/ m3
• W/C ratio= 0.45 as per table 5 of IS 456 :2000
• Workability= 100 mm (SLUMP)
• Method of concrete= pumping concrete
• Max cement content = 450 Kg/m3
• Admixture= superplasticizer

2. Tested data for materials

• Grade of cement= 43 grade
• Specific gravity of cement= 3.15
• Specific gravity of coarse aggregate= 2.74
• Specific gravity of fine aggregate = 2.74
• Specific gravity of admixture(superplasticizer)=1.14
• Water absorption of coarse aggregate= 0.5%
• Water absoption of fine aggregate =1%
• Free surface moisture for;

Coarse aggregate = NIL

Fine aggregate = NIL

• Sieve analysis of aggregate is done with the help of zone 1 of table 4 of IS 383

3. Calculation of targeted strenght

f'ck = fck+1.65*S

f'ck = 50+1.65*5 ................as S is a standard deviation ,take it from table 8 of IS 10262:2009

f'ck = 58.25 N/mm2

4. Selection of w/c ratio

Maximum w/c ratio we have= 0.45 ......as per the IS 456:2000

Experience w/c ratio =0.4

0.4<0.45 ..ok to have.

5. Water content calculation

For 20 mm aggregate size the max water content is 186 .......as per table 2 o IS 10262:2009

For 100 mm slump = 186 +(6/100)*186

=197 litre

Because we used admixture as a superplasticizer , water content reduced by 30%

In trials 29% water reduced occurs mostly;

Water content = 197*0.71

=140 litre

6. Calculation of cement content

We have w/c ratio=0.4

Cement content =140/0.4 = 350 Kg/m3

as per IS456:2000 the cement content for very severe condition is 320 kg/m3

350>320 hence okay to have.

7. Coarse and Fine aggregates

From table 3 of IS10262:2009 fro 20 mm aggregate size the w/c ratio is 0.6

in present case we have w/c ratio is 0.40

So as the water cement ratio is decresed by 0.1 we have increase the volume of aggregate by 0.02

Corrected value of coarse aggregate for w/c ratio is = 0.6*(0.20*0.05)*0.01

= 0.64

for pumpable concrete , these values reduced by 10%

volume of coarse aggregate = 0.64*0.9

= 0.576

volume of fine aggregate = 1-0.576

= 0.424

8. Mix calculations

Calculations for unit vomume of concrete are as below;

1. Volume of concrete = 1 m3
2. 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 admixture = ( mass of admixture/specific gravity of admixture)*1/1000

= (7/1.14)*1/1000............7 is the 2% of cementitious materials.

= 0.0061 m3

E. Volume of CA and FA aggregates = A-(B+C+D)

= 1-(0.111+0.140+0.0061)

= 0.7429 m3

F. Weight of coarse aggregate = E* volume of coarse aggregate*specific gravity of CA*1000

= 0.7429*0.567*2.74*1000

= 1154.15 KG

G. Weight of Fine aggregate = E* volume of fine aggregate*specific gravity of FA*1000

= 0.7429*0.424*2.74*1000

= 863.07 KG

9. Mix proportions

• Cement = 350 kg
• Water = 140 litre
• Coarse aggregate=1154.15 KG
• Fine aggregate=863.07 KG
• Water cement ratio=0.40
• Admixture weight= 7

Therefore the mix ratio of M50 grade is 350 : 863.07 : 1154.15

i.e 1 : 2.47 : 3.30

10. The slump shall be measured and the water content and dosage of admixture shall be adjusted for achieving therequired slump .trail may be rework and then check the durability.

11. if the more than two trials having w/c ratio has 10 % variation shall be carried out and graph between three w/c ratio and corresponding strenght are plotted to work out the mix proportion, however durablity shall be met.

RESULT :- The mix design fo M50 concrete is 1 : 2.47 : 3.30

(Note:- For these mix design calculations we take reference from IS 456:2000 ,IS 10262:2009 and IS 383)