Project on Concrete Mix Design for various grades of Concrete

Challenge 5

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

1. Aim – To find out the mix proportion for the M35 grade of concrete with fly ash by performing mix design.

Introduction - The mix design is performed by the structural engineer to determine the proportion of the concrete mix. The production of concrete with appropriate properties will be a most economical one. It is performed, so that the concrete should have minimum compressive strength required for the structural consideration, adequate workability, max water/cement ratio to attain adequate durability and water content to avoid shrinkage.

Procedure

1. Stipulation for proportioning

Grade designation – M35

Type of cement – OPC 43 grade conforming to IS 8112

Type of mineral admixture – fly ash (IS 3812 past I)

Max nominal size of aggregate – 20 mm

Min cement content – 320 kg/m3 (Table 5 IS 456:2000)

Max w/c ratio - 0.45 (Table 5 IS 456:2000)

Workability – 100 mm (slump value for pumped concrete)

Exposure condition – Sever (For RC)

Method of concrete placing – pumping

Degree of supervision – good

Type of agg – crushed angular agg

Max cement content – 450 kg/m3

Chemical admixture – super plasticizer

1. Test data for materials

Specific gravity, cement – 3.15, FA – 2.74, CA – 2.74

Specific gravity, fly ash – 2.20

Water absorption, CA – 0.5 %, FA – 1.0 %

Free surface moisture – CA, FA - Nil

• Target strength for mix proportioning

f’ck = fck + 1.65 S

From table 1 of IS 456:2000,

Std deviation (S)= 5 N/mm2

Target strength (f’ck) = 35 + 1.65 * 5 = 43.25 N/mm2

1. Selection of w/c ratio

As per table 5 IS 456:2000,

Assume w/c = 0.4

From Fig 1 of IS 10262:2009,

W/C ratio for 43.25 N/mm2, we can use curve 2 = 0.35

Therefore, w/c = 0.35

From table 2 max water content for 20 mm agg = 186 lit (for 25 mm to 50 mm slump)

For 100 mm slump = 186 + 6/100 * 186 = 197 lit

As super plasticizer is used, the water content can be reduced up to 20 % and above

Based on trials, water content reduction of 25 % is achieved with plasticizers

Hence, arrived water content = 197 * 0.71 = 140 lit

1. Calculation of cement content

w/c ratio = 0.35

Cement content = 140/0.35 = 400 kg/m3

From table 5 of IS 456 min cement content for severe exposure condition = 320 kg/m3

Hence it is ok.

Now, to proportion a mic containing fly ash the following steps are taken,

Consider the cementitious content is increase by 10 %

Cementitious material content = 320*1.1 = 352 kg/m3

Water content = 140 kg/m3

w/c ratio = 140/352 = 0.398

fly ash @ 30 % total cementitious material content = 352 * 30% = 106 kg/m3

cement (OPC) = 352 -106 = 246 kg/m3

saving of cement while using fly ash = 320 – 246 = 74 kg/m3

fly ash being utilized = 106 kg/m

chemical admixture = 2% of cementitious material

= 320 * 2/100

= 6.2 kg/m3

1. Proportion of volume of coarse agg and fine agg

From table 3 of IS 10260:2009 volume of coarse aggregate corresponding to 20 mm agg and FA,

Considering zone 1 for w/c ratio of 0.50 = 0.60

In the present case, w/c ratio is 0.35

The volume of coarse agg needs to be increased to decrease the fine agg content as the rate of +/- 0.01 for every +/- 0.05 change in w/c ratio.

New volume of CA (X) = [{(0.5-0.35) *0.01}/0.05] +0.6

X = 0.03+0.6 = 0.63

Corrected volume of CA for the w/c ratio of 0.35 = 0.63

For pumping concrete these values are reduced by 10 %

Volume of CA = 0.63 * 0.9 = 0.567

Volume of FA = 1-0.567 = 0.433

• Mix calculation
  1. Volume of concrete = 1 kg/m3
  2. Volume of cement = 320/ (3.15 * 1000) = 0.102 m3
  3. Volume of fly ash = 106/ (2.2 * 1000) = 0.048 m3
  4. Volume of water = 140/ (1 * 1000) = 0.140 m3
  5. Volume of chemical admixture = 6.2/ (1.145 * 1000) = 0.0054 m3
  6. Volume of agg (FA and CA) = 1-(0.102+0.140+0.0054) = 0.753 m3
  7. Mass of CA = 0.753 * 0.56 * 2.74 * 1000 = 1155 kg/m3
  8. Mass of FA = 0.753 * 0.44 * 2.74 * 1000 = 907 kg/m3
• Mix proportions

For trial no 1

Cement = 320 kg/m3

Fly ash = 106 kg/m3

Water = 140 kg/m3

FA = 907 kg/m3

CA = 1154 kg/m3

Chemical admixture = 6.2 kg/m3

w/c ratio = 0.35

Trial mix = 1:2.12:2.7

2. Aim – To find out the mix proportion for the M50 grade of concrete without fly ash by performing mix design.

Introduction - The mix design is performed by the structural engineer to determine the proportion of the concrete mix. The production of concrete with appropriate properties will be a most economical one. It is performed, so that the concrete should have minimum compressive strength required for the structural consideration, adequate workability, max water/cement ratio to attain adequate durability and water content to avoid shrinkage.

Procedure

1. Stipulation for proportioning

Grade designation – M50

Type of cement – OPC 43 grade conforming to IS 8112

Max nominal size of aggregate – 20 mm

Min cement content – 320 kg/m3 (Table 5 IS 456:2000)

Max w/c ratio - 0.45 (Table 5 IS 456:2000)

Workability – 100 mm (slump value for pumped concrete)

Exposure condition – Sever (For RC)

Method of concrete placing – pumping

Degree of supervision – good

Type of agg – crushed angular agg

Max cement content – 450 kg/m3

Chemical admixture – super plasticizer

1. Test data for materials

Specific gravity, cement – 3.15, FA – 2.74, CA – 2.74

Water absorption, CA – 0.5 %, FA – 1.0 %

Free surface moisture – CA, FA - Nil

• Target strength for mix proportioning

f’ck = fck + 1.65 S

From table 1 of IS 456:2000,

Std deviation (S)= 5 N/mm2

Target strength (f’ck) = 50 + 1.65 * 5 = 58.25 N/mm2

1. Selection of w/c ratio

As per table 5 IS 456:2000,

Assume w/c = 0.4

From Fig 1 of IS 10262:2009,

W/C ratio for 58.25 N/mm2, we can use curve 2 = 0.291

Therefore, w/c = 0.291

From table 2 max water content for 20 mm agg = 186 lit (for 25 mm to 50 mm slump)

For 100 mm slump = 186 + 6/100 * 186 = 197 lit

As super plasticizer is used, the water content can be reduced up to 20 % and above

Based on trials, water content reduction of 25 % is achieved with plasticizers

Hence, arrived water content = 197 * 0.71 = 140 lit

1. Calculation of cement content

w/c ratio = 0.291

Cement content = 140/0.291 = 481 kg/m3

From table 5 of IS 456 min cement content for severe exposure condition = 481kg/m3

Hence it is ok.

chemical admixture = 2% of cementitious material

= 481 * 2/100

= 9.62 kg/m3

1. Proportion of volume of coarse agg and fine agg

From table 3 of IS 10260:2009 volume of coarse aggregate corresponding to 20 mm agg and FA,

Considering zone 1 for w/c ratio of 0.50 = 0.60

In the present case, w/c ratio is 0.291

The volume of coarse agg needs to be increased to decrease the fine agg content as the rate of +/- 0.01 for every +/- 0.05 change in w/c ratio.

New volume of CA (X) = [{(0.5-0.291) *0.01}/0.05] +0.6

X = 0.042+0.6 = 0.642

Corrected volume of CA for the w/c ratio of 0.291 = 0.642

For pumping concrete these values are reduced by 10 %

Volume of CA = 0.642 * 0.9 = 0.578

Volume of FA = 1-0.578 = 0.422

• Mix calculation
  1. Volume of concrete = 1 kg/m3
  2. Volume of cement = 481/ (3.15 * 1000) = 0.137 m3
  3. Volume of water = 140/ (1 * 1000) = 0.140 m3
  4. Volume of chemical admixture = 9.62/ (1.145 * 1000) = 0.0084 m3
  5. Volume of agg (FA and CA) = 1-(0.137+0.140+0.0084) = 0.715 m3
  6. Mass of CA = 0.715 * 0.56 * 2.74 * 1000 = 1097 kg/m3
  7. Mass of FA = 0.715 * 0.44 * 2.74 * 1000 = 862 kg/m3
• Mix proportions

For trial no 1

Cement = 481 kg/m3

Water = 140 kg/m3

FA = 862 kg/m3

CA = 1097 kg/m3

Chemical admixture = 9.62 kg/m3

w/c ratio = 0.291

Trial mix = 1:1.79:2.28