Project on Concrete Mix Design for various grades of Concrete

Answer:

For M35 grade concrete with Fly ash:

• STIPULATIONS FOR PROPORTIONING:

a) Grade designation: M35

b) Type of cement: OPC 43 Grade conforming IS 8112

c) Maximum nominal size of aggregate: 20mm

d) Minimum cement content: 340 kg/m3 (IS 456:2000)

e) Maximum water-cement ratio: 0.40 (Table 5 of IS 456:2000)

f) Workability: 75 mm slump

g) Exposure condition: Extreme (For Reinforced Concrete)

h) Method of concrete placing: Pumping

j) Degree of supervision: Good

k) Type of aggregate: Crushed Angular Aggregates

m) Maximum cement content : 450 kg/m3

n) Chemical admixture type: Super Plasticizer

• TEST DATA FOR MATERIALS:

a) Cement used: OPC 43 Grade conforming IS 8112

b) Specific gravity of cement: 3.15

c) Chemical admixture: Super Plasticizer conforming to IS 9103

d) Specific gravity of

1) Coarse aggregate 20mm: 2.74

2) Fine aggregate: 2.74

e) Water absorption:

1) Coarse aggregate: 0.5 %

2) Fine aggregate (M.sand): 1.0 %

f) Free (surface) moisture:

1) Coarse aggregate: Nil

2) Fine aggregate: Nil

g) Sieve analysis:

1) Coarse aggregate: Conforming to all aggregates of Table 2 of IS 383

2) Fine aggregate: Conforming to Grading Zone I of Table 4 of IS 383

• TARGET STRENGTH FOR MIX PROPORTIONING:

f’ck =fck + 1.65 s

where

f’ck = target average compressive strength at 28 days,

fck = characteristics compressive strength at 28 days, and

s = standard deviation.

From Table I of IS 10262:2009, Standard Deviation, s = 5 N/mm2. Therefore, target strength = 35 + 1.65 x 5 = 43.25 N/mm2.

• SELECTION OF WATER-CEMENT RATIO:

From Table 5 of IS 456 for Extreme maximum, Water Cement Ratio is 0.40

Based on Graph (IS 10262-2019), the adopted water-cement ratio is 0.38

Hence the lower value is adopted i.e., 0.38

• SELECTION OF WATER CONTENT:

From Table 2 of IS 10262:2009, maximum water content for 20 mm aggregate = 186 litres (for 25 to 50 mm slump range)

Estimated water content for 75 mm slump = 186+ { 186 x (3/100) } = 191.58 litre.

(As Superplasticizer is used, the water content can be reduced up to 30% and above.)

Based on trials with Super plasticizer water content reduction of 30% has been achieved, Hence the arrived water content = 191.58 x {(100-30)/100} = 197*0.70 = 134.1 litre.

• CALCULATION OF CEMENT CONTENT:

Water-cement Ratio = 0.38

Cement Content = 134.1/ 0.38 = 352.89 kg/m3

From Table 5 of IS 456, the Minimum cement content for ‘Extreme’ exposure conditions is 340kg/m3

= 352.89 kg/m3 > 340 kg/m3, hence ok.

As Fly Ash is added, therefore Cementanious material will be = Cement content increased by 110%

Cementanious material = 352.89 x 1.1 = 388 kg.

Fly Ash content = 30 % of Cementanious material

= 388 x 30%

=116.4 kg

New Cement content = Cementanious material - Fly Ash content

= 388 - 116.4 = 272.1 Kg

Cement Saved = Previous Cement Content - New Cement content

=352.88 - 272.1 = 81 kg

I.e, around one and a half bags of cement.

• PROPORTIO0N OF VOLUME OF COARSE AGGREGATE AND FINE AGGREGATE CONTENT:

From Table 3 of (IS 10262:2009) Volume of coarse aggregate corresponding to 20 mm size aggregate and fine aggregate (Zone I) for the water-cement ratio of = 0.60.

In the present case, the water-cement ratio is 0.38.

Therefore, the 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 the volume of coarse aggregate is increased by 0.02 (at the rate of -/+ 0.01 for every ± 0.05 change in water-cement ratio).

By using the formula [{(0.5-w/c ratio)/0.5}*0.01]+ 0.60

Therefore, the corrected proportion of the volume of coarse aggregate for the water-cement ratio of 0.38 = 0.62

(NOTE – If the coarse aggregate is not angular type, then also the volume of coarse aggregate may be required to be increased suitably based on experience).

For pumpable concrete, these values should be reduced up to 10%.

Therefore, volume of coarse aggregate = 0.62 x {(100-10)/100} = 0.62 x 0.9 = 0.56.

The volume of fine aggregate content = 1 – 0.56 = 0.44.

• MIX CALCULATIONS:

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] / {[Specific Gravity of Cement] x 1/1000}

= (272.1/3.15) x 0.001

= 0.08 m3

c) Volume of Fly Ash = [Mass of Fly Ash] / {[Specific Gravity of Cement] x 1/1000}

=(116.4/2.2) x 0.001

=0.05 m3

d) Volume of water = [Mass of water] / {[Specific Gravity of water] x 1/1000}

= (134.1/1) x 0.001

= 0.13m3

e) Volume of chemical admixture = [Mass of chemical admixture ] / {[Specific Gravity of admixture] x 1000}

= [{7.76/1.145} x 1000]

= 0.0067m3

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

= [1-(0.08+0.05+0.13+0.006)]

= 0.73m3

g) Mass of coarse aggregate= e x Volume of Coarse Aggregate x Specific Gravity of Fine Aggregate x 1000

= 0.73x 0.56 x 2.74 x 1000

= 1120.11 kg/m3

h) Mass of fine aggregate= e x Volume of Fine Aggregate x Specific Gravity of Fine Aggregate x 1000

= 0.73x 0.44x 2.74 x 1000

= 880 kg/m3

• MIX PROPORTIONS:

Cement = 272 kg/m3

Water = 134.1 l/m3

Fine aggregate = 880 kg/m3

Coarse aggregate = 1120kg/m3

Chemical admixture = 7.76 kg/m3

Water-cement ratio = 0.38

Trail Mix Ratio = 272:880:1120

= 1:3.23:4.11

For M50 grade of concrete without Fly Ash:

• STIPULATIONS FOR PROPORTIONING:

a) Grade designation: M50

b) Type of cement: OPC 43 Grade conforming IS 8112

c) Maximum nominal size of aggregate: 20mm

d) Minimum cement content: 400 kg/m3 (IS 456:2000)

e) Maximum water-cement ratio: 0.40 (Table 5 of IS 456:2000)

f) Workability: 75 mm slump

g) Exposure condition: Extreme (For Reinforced Concrete)

h) Method of concrete placing: Pumping

j) Degree of supervision: Good

k) Type of aggregate: Crushed Angular Aggregates

m) Maximum cement content : 450 kg/m3

n) Chemical admixture type: Super Plasticizer

• TEST DATA FOR MATERIALS:

a) Cement used: OPC 43 Grade conforming IS 8112

b) Specific gravity of cement: 3.15

c) Chemical admixture: Super Plasticizer conforming to IS 9103

d) Specific gravity of

1) Coarse aggregate 20mm: 2.74

2) Fine aggregate: 2.74

e) Water absorption:

1) Coarse aggregate: 0.5 %

2) Fine aggregate (M.sand): 1.0 %

f) Free (surface) moisture:

1) Coarse aggregate: Nil

2) Fine aggregate: Nil

g) Sieve analysis:

1) Coarse aggregate: Conforming to all aggregates of Table 2 of IS 383

2) Fine aggregate: Conforming to Grading Zone I of Table 4 of IS 383

• TARGET STRENGTH FOR MIX PROPORTIONING:

f’ck =fck + 1.65 s

where

f’ck = target average compressive strength at 28 days,

fck = characteristics compressive strength at 28 days, and

s = standard deviation.

From Table I of IS 10262:2009, Standard Deviation, s = 5 N/mm2. Therefore, target strength = 50 + 1.65 x 5 = 58.25 N/mm2.

• SELECTION OF WATER-CEMENT RATIO:

From Table 5 of IS 456 for Extreme maximum, Water Cement Ratio is 0.40

Based on Graph (IS 10262 – 2019), the adopted water-cement ratio is 0.30

Hence the lower value is adopted i.e., 0.30

• SELECTION OF WATER CONTENT:

From Table 2 of IS 10262:2009, maximum water content for 20 mm aggregate = 186 litres (for 25 to 50 mm slump range)

Estimated water content for 75 mm slump = 186+ { 186 x (3/100) } = 191.58 litre.

(As Superplasticizer is used, the water content can be reduced up to 30% and above.)

Based on trials with Super plasticizer water content reduction of 30% has been achieved, Hence the arrived water content = 191.58 x {(100-30)/100} = 197*0.70 = 134.1 litre.

• CALCULATION OF CEMENT CONTENT:

Water-cement Ratio = 0.30

Cement Content = 134.1/ 0.30 = 447kg/m3

From Table 5 of IS 456, the Minimum cement content for ‘Extreme’ exposure conditions is 340kg/m3

= 447 kg/m3 > 400 kg/m3, hence ok.

• PROPORTIO0N OF VOLUME OF COARSE AGGREGATE AND FINE AGGREGATE CONTENT:

From Table 3 of (IS 10262:2009) Volume of coarse aggregate corresponding to 20 mm size aggregate and fine aggregate (Zone I) for the water-cement ratio of = 0.60.

In the present case, the water-cement ratio is 0.30.

Therefore, the 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 the volume of coarse aggregate is increased by 0.02 (at the rate of -/+ 0.01 for every ± 0.05 change in water-cement ratio).

By using the formula [{(0.5-w/c ratio)/0.5}*0.01]+ 0.60

Therefore, the corrected proportion of the volume of coarse aggregate for the water-cement ratio of 0.30 = 0.60

(NOTE – If the coarse aggregate is not angular type, then also the volume of coarse aggregate may be required to be increased suitably based on experience).

For pumpable concrete, these values should be reduced up to 10%.

Therefore, volume of coarse aggregate = 0.60 x {(100-10)/100} = 0.60 x 0.9 = 0.54.

The volume of fine aggregate content = 1 – 0.54 = 0.46.

• MIX CALCULATIONS:

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] / {[Specific Gravity of Cement] x 1/1000}

= (447/3.15) x 0.001

= 0.141 m3

c) Volume of water = [Mass of water] / {[Specific Gravity of water] x 1/1000}

= (134.1/1) x 0.001

= 0.13m3

d) Volume of chemical admixture = [Mass of chemical admixture ] / {[Specific Gravity of admixture] x 1000}

= [{7.76/1.145} x 1000]

= 0.0067m3

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

= [1-(0.141+0.13+0.0067)]

= 0.722m3

f) Mass of coarse aggregate= e x Volume of Coarse Aggregate x Specific Gravity of Fine Aggregate x 1000

= 0.722x 0.54 x 2.74 x 1000

=1068.27 kg/m3

g) Mass of fine aggregate= e x Volume of Fine Aggregate x Specific Gravity of Fine Aggregate x 1000

= 0.722x 0.46x 2.74 x 1000

= 910 kg/m3

• MIX PROPORTIONS:

Cement = 447 kg/m3

Water = 134.1 l/m3

Fine aggregate = 910 kg/m3

Coarse aggregate = 1068.27kg/m3

Chemical admixture = 7.76 kg/m3

Water-cement ratio = 0.30

Trail Mix Ratio = 447:910:1068.27

= 1:2.03:2.38