Project on Concrete Mix Design for various grades of Concrete

Concrete mix design for M50 grade concrete:

1. Data collection

Grade designation = M50

Assume,

Type of cement = OPC 43 grade conforming to IS 8112

Maximum nominal size of aggregate = 20 mm size of aggregate

• Workability = 100 mm slump for pumpable concrete

Exposure condition = Severe ( For reinforced concrete )

Chemical admixtures = Superplasticizers

specific gravity of cement = 3.15

specific gravity of coarse aggregate = 2.74

specific gravity of fine aggregate = 2.74

2. Calculation of target mean strength of concrete

f'ck = fck + 1.65s

Where , Fck = 50 N/ mm^2

• From table - 1, IS 10262:2009, Let we take standard deviation at M50 grade,

S = 5.0 N /mm^2

Then, f'ck = 50 + ( 1.65 x 5 )

= 58.25 N/mm^2

3. Calculation of Water cement ratio

• Let we take the graph from code

Compressive strength vs w/c ratio graph

Assume , w/c = 0.30

• W/C ratio for graph - 2

28 days compressive strength vs 7 days strength of cement

w/c = 0.39

• From table - 5, IS 456:2000, it can be taken depending upon the exposure condition

w/c = 0.40 ( for severe )

let we take about the lower value of three value is adopted as w/c

finally take , w/c = 0.30

4. Calculation of water content

• From table - 2, IS 10262:2009

Maximum water content for 20 mm aggregate = 186 litre ( from 25mm to 50mm )

• Estimate the water content for 100mm slump,

= 186 + 6%

= 186 + ( 6/100 x 186)

W = 197 litre.

As superplasticizer are used , the water content can be reduced at 20%, and based on trials water content reduction of 29% is achived with plasticizer,

Hence arrived water content = 197 x 0.71

Finally, W = 140 litre.

5.Calculations of cement content

W/C = 0.30

140 / C = 0.30

C = 140 / 0.30

C = 467 kg/ m ^3.

6. Proportion of volume of coarse and fine aggregate

Assume, Zone- II

THEN,

• From table - 3, IS 10262:2009
• Nominal size of aggregate to 20 mm in size, then volume of coarse aggregate per unit volume of total aggregate for ZONE -II Of fine aggregate,

w/c = 0.62

• In the present case of water cement ratio is

w/c = 0.30

Therefore, the volume of coarse aggregate needs to be increased to decrease the fine aggregate.

0.30 = 0.62

For pumpable concrete, values are reduced by 10%

Volume of coarse aggregate = 0.62 x 0.9

= 0.56

volume of fine aggregate = 1 - 0.56

= 0.44

7. Mix Calculation

The mix per unit volume of concrete,

a) Volume of concrete = 1 m^3

b ) Volume of cement = ( mass of cement / specific gravity of cement ) x 1/1000

= 467/ 3.15 x 1/1000

= 0.148 m^3

c ) Volume of water = ( mass of water / specific gravity of water ) x 1/1000

= 140 / 1 x 1/1000

= 0.140 m^3

f ) volume of chemical admixtures = ( mass of chemical admixtures / specific gravity of ch. admix ) x 1/1000

= 7.6/1.145 x 1/1000

= 0.006 m^3

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

= 1 - ( 0.148 + 0.140 + 0.006 )

= 0.706 m^3

f ) Mass of coarse aggregate = volume of aggregate x volume of coarse agg x specific gravity of CA x 1000

= 0.706 x 0.56 x 2.74 x 1000

= 1084 kg/m^3

e ) Mass of fine aggregate = Volume of aggregate x volume of FA x specific gravity of FA x 1000

= 0.706 x 0.44 x 2.74 x 1000

= 851 kg/m^3

8. Mix proportion of concrete

• Cement = 467 kg/m^3
• water = 140 kg/m^3
• FA = 851 kg / m^3
• CA = 1084 kg / m^3
• Chemical admixtures = 7 kg/m^3
• water cement ratio = 0.30

Finally, mix proportion is

= cement : sand : coarse aggregate

= 1 : 1.82 : 2.32

2.Calculation of concrete mix design of M35 grade of concrete with fly ash

1.Data collection

Grade designation = M35

Assume,

Type of cement = OPC 43 grade

Type of mineral admixtures = flyash

Size of aggregate = 20mm

Workability = 100 mm slump for pumpable concrete

Exposure condition = severe ( for reinforcement )

Specific gravity of fly ash = 2.2

Specific gravity of cement = 3.15

Specific gravity of FA = 2.74

Specific gravity of CA = 2.74

2. Calculation of target mean strength of concrete

F’ck = fck + 1.65s

Where, fck = 35 N/mm^2

• From table – 1, IS 10262:2009

S = 5 N/mm ^2

Then , f’ck = 35 + (1.65 x 5 )

= 43.25 N/mm^2

3.Calculation of water cement ratio

• From Table-5, IS 456:2000, Let we taken the water cement ratio based on exposure conditions.

For severe,

W/C = 0.45

4. Calculation of water content

• From table – 2, IS 10262:2009

Maximum water content for 20 mm aggregate

= 186 litres

• Estimate the 100 mm slump of water content

= 186 + 6%

W = 197 litres

As superplasticizer are used, the water content reduction of 29% is achived with plasticizer

Hence arrived water content is

W = 197 X 0.71

W = 140 litres

5. Calculation of cement content

W/C = 0.45

C = 140/0.45

C = 312 kg / m^3

• From table – 5, IS 456 :2000

Maximum cement content of M35 grade ,

C = 340 kg / m^3

Hence , 340 > 312

Finally, C = 340 kg / m^3

Now to proportion of mix fly ash are based on the percentage of fly ash needs to decided the based on project requirement, and its based added to percentage,

This example in which increase the 10% of cementitious material,

C = 340 x 1.10

C = 374 kg / m^3

Fly ash @ 30% of total cementitious material content

Fly ash = 374 x 30%

= 112 kg / m^3

Finally,

Cement (OPC) = 374 – 112

= 262 kg / m^3

6. Proportion of volume of fine and coarse aggregate

ASSUME , Zone – II

Then

• From table – 3, IS 10262 : 2009
• Nominal size of aggregate in 20 mm size

W/C = 0.62

• In prescent case of water cement ratio is

W/C = 0.45

Therefore,

0.45 = 0.62

For pumpable concrete, values are reduced by 10%

• Volume of coarse aggregate

= 0.62 x 0.9

= 0.56

• Volume of fine aggregate

= 1 – 0.56

= 0.44

7. Mix calculation

The mix per unit volume of concrete

a) Volume of concrete = 1 m^3

b ) Volume of cement = ( mass of cement / specific gravity of cement ) x 1/1000

= 262 / 3.15 x 1/1000

= 0.083 m^3

c ) Volume of fly ash = ( mass of fly ash / specific gravity ) x 1/1000

= 112 / 2.2 x 1/1000

= 0.050 m^3

d ) Volume of water = 140/1 x 1/1000

= 0.140 m^3

e ) Volume of chemical admixtures = 7.6/1.45 x 1/1000

= 0.006m^3

f ) Volume of aggregate = a – ( b+c+d+e)

= 1 – ( 0.083 + 0.050 + 0.140 + 0.006)

= 0.721 m^3

g ) Mass of coarse aggregate = 0.721 x 0.56 x 2.74 x 1000

= 1106 kg/m^3

h ) Mass of fine aggregate = 0.721 x 0.44 x 2.74 x 1000

= 870 kg / m^3

8. Mix proportion of concrete

• Cement = 262 kg/m^3
• Water = 140 kg / m^3
• Fly ash = 112 kg /m ^3
• FA = 870 kg/m^3
• CA = 1106 kg/m^3

The mix ratio is

• Cementitious material : sand : coarse aggregate
• 1 : 2.32 : 2.95