Project on Concrete Mix Design for various grades of Concrete

Concrete Mix Design for M35 grade with fly ash-

1. Stipulations for proportioning:

• Grade designation : M35
• Type of cement: OPC 43 grade confirming to IS 8112
• Type of mineral admixture: Fly ash confirming to IS 3812 (Part1)
• Maximum nominal size of aggregate: 20mm
• Minimum cement content: 320 kg/m^3 ( Table 5 of IS 456:2000- Exposure condition )
• Maximum water-cement ratio: 0.45 ( Table 5 of IS 456:2000- Exposure condition )
• Workability: 100mm ( slump value for pumpable concrete )
• Exposure condition: severe (for reinforced concrete)
• Chemical admixture: Superplasticizer confirming to IS 9103
• Specific gravity of :

Coarse aggregates - 2.74

Fine aggregates - 2.74

• Water absorption:

Coarse aggregates - 0.5%

Fine aggregates – 1.0%

• Free surface moisture : Nil for coarse and fine aggregates
• Method of concrete placing: pumping
• Degree of supervision: good
• Type of aggregate: Crushed angular aggregate

2. Test data for materials:

• Specific gravity of cement: 3.15
• Fly ash: confirming to IS 3812
• Specific gravity of fly as: 2.2

3. Target mean 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

S = standard deviation

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

Therefore Target mean strength = 35 + (1.65 * 5)

Target mean strength = 43.25 N/mm^2

4. Selection of water cement ratio-

From table 5 of IS 456:2000 for severe exposure condition, maximum water cement ratio is 0.40

0.4 > 0.45 hence OK

5. Selection of water content-

Form table 2 of IS 10262:2009; maximum water content for 20mm aggregate = 186 litre

( for 25mm to 50mm slump range )

Hence estimated water content for 100 slump = 186 + 6/100 * 186 = 197 litre.

As superplasticizer is used, the water content can be reduced upto 30%

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

Hence arrived water content= 197 * 0.71 = 140 litre

6. Calculation of cement content-

w/c ratio = 0.4

cementitious material (cement+fly ash) content = 140/0.4 = 350 kg/m^3

from table 5 of IS 456 min cement content for sever condition is 320kg/m^3

350kg/m^3 > 320kg/m^3 Hence OK

Now, to proportion a mix containg fly ash, the following conditions are suggested-

The percentage fly ash needs to be decided based on project requirements

In some cases increase in cementitious materials can be warrented.

• Cementitious material content = 350 X 1.10 = 385kg/m^3
• Water content = 140kg/m^3
• Water cement ratio = 140/385 = 0.364
• Fly ash @ 30% total cementitious material= 385 X 30% = 115kg/m^3
• Cement OPC = 385 – 115 = 270kg/m^3
• Saving on cement using fly ash= 350-270 = 80kg/m^3
• Fly ash being utilized = 115kg/m^3

7. Volume of coarse and fine aggregates content-

From table 3 if IS 10262;2009 volume of coarse aggregates confirming to 20mm size and fine aggregate (Zone1) for w/c ratio= 0.6

We have w/c ratio = 0.4, hence the vol of aggregates needs to be increased to decrease the fine aggeegate content. As the w/c ratio is lower by 0.10, the proportion of vol of coarse aggregate is increased by 0.02

Therefore corrected volume of coarse aggregate = 0.62

For pumpable concrete the value is reduced by 10%

Hence volume of coarse aggregate = 0.62 * 0.9= 0.56

Volume of fine aggregate= 1 – 0.56 = 0.44

8. Mix calculations-

a) Volume of concrete = 1m^3

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

270/3.15 x 1/1000 = 0.86m^3

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

115/ 2.2 x 1/1000 = 0.052 m^3

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

140/1 x 1/1000 = 0.140m^3

e) Volume of chemical admixtures = (Mass of chemical admixtures/ Specific gravity of chemical a dmixtures) * (1/1000)

7.6/1.145 x 1/1000 = 0.006m^3

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

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

= 0.716 m^3

f) Mass of coarse aggregate = e x Volume of CA x Specific gravity of CA x 1000.

0.716 x 0.56 x 2.74 x 1000 = 1098kg/m^3

g) Mass of fine aggregate = e x Volume of FA x Specific gravity of FA x 1000.

0.716 x 0.44 x 2.74 x 1000 = 863kg/m^3

9. Mix proportions for Trial –

cement	270 kg/m^3
Fly ash	115 kg/m^3
water	140 kg/m^3
Fine aggregate	863 kg/m^3
Coarse aggregate	1098 kg/m^3
Chemical admixture	7 kg/m^3
Water cement ratio	0.364
Trial mix ratio	1 : 2.24 : 2.85

Concrete Mix design for M50 grade concrete-

• Grade Designation = M-50
• Type of cement = O.P.C-43 grade
• Fine Aggregate = Zone-II
• Gravity of

Cement = 3.15

Fine Aggregate = 2.61

Coarse Aggregate (20mm) = 2.65

Coarse Aggregate (10mm) = 2.66

• Minimum Cement (Assume) =400 kg / m3
• Maximum water cement ratio (As per IS coe) = 0.45

Mix Calculation: –

1. Target Mean Strength = 50 + ( 5 X 1.65 ) = 58.25 N/mm^2

= 58.25 N/mm^2

2. Selection of water cement ratio:-

Assume water cement ratio = 0.35

3. Calculation of water: –

Approximate water content for 20mm max. Size of aggregate = 180 kg /m3 (As per Table No. 5 , IS : 10262 ). As plasticizer is proposed we can reduce water content by 20%.

Now water content = 180 X 0.8 = 144 kg /m3

4. Calculation of cement content:-

Water cement ratio = 0.35

Water content per cum of concrete = 144 kg

Cement content = 144/0.35 = 411.4 kg / m^3

Say cement content = 412 kg / m3 (As per code Minimum cement content 400 kg / m3 )

Hence O.K.

5. Calculation for C.A. & F.A

Volume of concrete = 1 m3

Volume of cement = 412 / ( 3.15 X 1000 ) = 0.1308 m^3

Volume of water = 144 / ( 1 X 1000 ) = 0.1440 m^3

Volume of Admixture = 4.994 / (1.145 X 1000 ) = 0.0043 m^3

Total weight of other materials except coarse aggregate = 0.1308 + 0.1440 +0.0043 = 0.2791 m^3

Volume of coarse and fine aggregate = 1 – 0.2791 = 0.7209 m^3

Volume of F.A. = 0.7209 X 0.33 = 0.2379 m3 (Assuming 33% by volume of total aggregate )

Volume of C.A. = 0.7209 – 0.2379 = 0.4830 m^3

Therefore weight of F.A. = 0.2379 X 2.61 X 1000 = 620.919 kg/ m^3

Assume weight of F.A. = 621 kg/ m^3

Therefore weight of C.A. = 0.4830 X 2.655 X 1000 = 1282.365 kg/ m3

Assume weight of C.A. = 1284 kg/ m3

Considering 20 mm: 10mm = 0.55: 0.45

20mm = 706 kg .

10mm = 578 kg .

Hence Mix details per m^3

Increasing cement, water, admixture by 2.5% for this trial

Cement = 412 X 1.025	422 kg
Water = 144 X 1.025	147.6 kg
Fine aggregate	621 kg
Coarse aggregate 20 mm	706 kg
Coarse aggregate 10 mm	578 kg
Admixture = 1.2 % by weight of cement	5.064 kg.
Water: cement: F.A.: C.A.	0.35: 1: 1.472: 3.043