Project on Concrete Mix Design for various grades of Concrete

1. Develop the Concrete Mix Design for M35 grade concrete with fly ash & M50 grade concrete without Fly ash.

CASE - 1

Aim: To develop the Concrete Mix Design for M35 grade concrete with fly ash

Introduction: Before starting the mix design process, we need to gather the following information:

Grade of Designation = M35

Type of cement = OPC 43Grade conforming to IS 269

Type of mineral admixture = Fly ash conforming to IS 3812 (Part1)

Chemical admixture type = Superplasticizer- normal

Fine aggregate confirming = Zone 2

Maximum size of aggregate = 20mm (Crushed Angular)

Minimum cement content = 320kg/cum

Workability = 75mm slump

Test data of Materials

Specific gravity of cement = 3.15

Chemical admixture = Superplasticizer conforming to IS 9103

Specific gravity of fine aggregate = 2.65

Specific gravity of coarse aggregate = 2.74

Water absorption of fine aggregate = 1%

Water absorption of coarse aggregate = 0.5%

Total moisture content of fine aggregate = 5%

Total moisture content of coarse aggregate = 2%

Free (surface) moisture Fine aggregate = Total moisture content – Water absorption = 5.0 – 1.0 = 4%

Free (surface) moisture Coarse aggregate = Total moisture content – Water absorption = 2.0 – 0.5 =1.5%

Procedure:

Step 1 - Target mean compressive strength: In this case, M35 grade concrete has a target mean compressive strength of 35 MPa (megapascals).

f’ck = fck + 1.65 S (or) f’ck = fck + X (whichever is higher)

where,

f’ck = target average compressive strength at 28 days

fck = characteristic compressive strength at 28 days

S = standard deviation

X = factor based on grade of concrete.

From Table 2, standard deviation, S = 5 N/mm2.

From Table 1, X = 6.5

Therefore, target strength using both equations, that is,

1. a) f’ck = fck+1.65 S

= 35 + 1.65 × 5 = 43.25 N/mm2

6. b) f’ck = fck + 5

= 35 + 6.5 = 41.5 N/mm2

The higher value is to be adopted. Therefore, target strength will be 43.25 N/mm2 as 43.25N/mm2 > 41.5 N/mm2.

Step 2 - Air Entrapped: From Table 3 of IS 10262, the approximate amount of entrapped air to be expected in normal (non-air-entrained) concrete is 1.0 percent for 20 mm nominal maximum size of aggregate.

Step 3 - Selection of Water-Cement Ratio: From Fig. 1 of IS 10262, the free water-cement ratio required for the target strength of 43.25 N/mm2 is 0.38 for OPC 43 grade curve. This is lower than the maximum value of 0.45 prescribed for ‘severe’ exposure for reinforced concrete as per Table 5 of IS 456. 0.38 < 0.45, hence O.K.

Step 4 - Selection of Water-Content: From Table 4 of IS 10262, water content = 186 kg (for 50 mm

slump) for 20 mm aggregate. Estimated water content for 75 mm slump (increasing at the rate of

percent for every 25 mm slump) = 186+(3*186)/100 = 191.58 kg

As superplasticizer is used, the water content may be reduced. Based on trial data, the water Content reduction of 25 percent is considered while using superplasticizer at the rate 1.0 percent by weight of cement.

Hence the arrived water content = 191.58 × 0.75 = 143.68 kg ≈ 144 kg.

Step 4 - Selection of Cement-Content: Water-cement ratio = 0.38

Cement content = 144/0.38 = 378.94 kg/m3 ≈ 379 kg/m3

To proportion a mix containing fly ash the following steps are suggested, Assuming with increase of 20 percent cementitious material content.

Cementitious material content = 379 × 1.20 = 454.8kg/m3 ≈ 455kg/m3

Water content = 144 kg/m3

So, water-cementitious ratio =144/455 = 0.316

Fly ash @ 25 percent of total cementitious material content = 455 × 25% = 113.75 ≈ 114kg/m3

Cement (OPC) = 455 – 114 = 341.25 kg/m3

From Table 5 of IS 456, minimum cementitious content for ‘severe’ exposure condition = 320kg/m3 455kg/m3 > 320 kg/m3, hence O.K.

Step 4 – Proportion of volume of Coarse aggregate and Fine aggregate content:

From Table 5 of IS 10262, volume of coarse aggregate corresponding to 20 mm size aggregate and fine aggregate (Zone II) for water-cement ratio of 0.50 = 0.62.

In the present case water-cementitious ratio is 0.316 Therefore, volume of coarse aggregate is required to be increased to decrease the fine aggregate content. As the water-cement ratio is lower by 0.184, the proportion of volume of coarse aggregate is increased by 0.0368 (at the rate of (-or+) 0.01 for every ± 0.05 change in water cement ratio). Therefore, corrected proportion of volume of coarse aggregate for the water-cementitious ratio of 0.316 = 0.62 + 0.0368 = 0.6568.

For pump able concrete these values may be reduced

by up to 10 percent. (see 5.5.2). Here, 10 percent reduction is considered.

Therefore, volume of coarse aggregate = 0.6568 × 0.9 = 0.5911cum say 0.59cum.

Volume of fine aggregate content = 1 – 0.59 = 0.41 m3

Step 4 – Mix Calculations: The mix calculations per unit volume of concrete shall be as followed:

1. Total volume: 1cum
2. Volume of Entrapped air in the wet concrete = 0.01cum
3. Volume of cement = (Mass of cement/Specific gravity of cement) * (1/1000)

= (341.25/3.15) * (1/1000) = 0.108Cum

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

= (114/2.2) * (1/1000) = 0.051Cum

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

= (144/1) * (1/1000) = 0.144Cum

1. Volume of Chemical admixture (@ 1.0 percent by mass of cementitious material)

= (Mass of Admixture/Specific gravity of admixture) * (1/1000)

= (4.55/1.145) * (1/1000) = 0.0039Cum

1. Volume of all in aggregate = (1 - Volume of cement - Volume of fly ash - Volume of water -Volume of Chemical admixture - Volume of Entrapped air in the wet concrete)

= (1-0.108-0.051-0.144-0.0039-0.01) = 0.6831cum

1. Mass of coarse aggregate = (Volume of all in aggregate*volume of coarse aggregate*

Specific gravity of coarse aggregate*1000)

= (0.6831*0.59*2.74*1000) = 1104.299kgs

1. Mass of fine aggregate = (Volume of all in aggregate*volume of fine aggregate*

Specific gravity of fine aggregate*1000)

= (0.6831*0.41*2.65*1000) = 742.188kgs

Step 5 – Mix proportions:

341. Cement = 341.25kg/m3
342. Fly ash = 114kg/m3
343. Water (Net mixing) = 144kg/m3
344. Fine aggregate = 742.188kg/m3
345. Coarse aggregate = 1104kg/m3
346. Chemical admixture = 4.55 kg/m3
347. Free water-cementitious materials ratio = 0.316

CASE - 2

Aim: To develop the Concrete Mix Design for M50 grade concrete without fly ash

Introduction: Before starting the mix design process, we need to gather the following information:

Grade of Designation = M50

Type of cement = OPC 43Grade conforming to IS 269

Chemical admixture type = Superplasticizer- normal

Fine aggregate confirming = Zone 2

Maximum size of aggregate = 20mm (Crushed Angular)

Minimum cement content = 320kg/cum

Workability = 75mm slump

Test data of Materials

Specific gravity of cement = 3.15

Chemical admixture = Superplasticizer conforming to IS 9103

Specific gravity of fine aggregate = 2.65

Specific gravity of coarse aggregate = 2.74

Water absorption of fine aggregate = 1%

Water absorption of coarse aggregate = 0.5%

Total moisture content of fine aggregate = 5%

Total moisture content of coarse aggregate = 2%

Free (surface) moisture Fine aggregate = Total moisture content – Water absorption = 5.0 – 1.0 = 4%

Free (surface) moisture Coarse aggregate = Total moisture content – Water absorption = 2.0 – 0.5 =1.5%

Procedure:

Step 1 - Target mean compressive strength: In this case, M35 grade concrete has a target mean compressive strength of 35 MPa (megapascals).

f’ck = fck + 1.65 S (or) f’ck = fck + X (whichever is higher)

where,

f’ck = target average compressive strength at 28 days

fck = characteristic compressive strength at 28 days

S = standard deviation

X = factor based on grade of concrete.

From Table 2, standard deviation, S = 5 N/mm2.

From Table 1, X = 6.5

Therefore, target strength using both equations, that is,

1. a) f’ck = fck+1.65 S

= 50 + 1.65 × 5 = 58.25 N/mm2

6. b) f’ck = fck + 6.5

= 50 + 6.5 = 56.5 N/mm2

The higher value is to be adopted. Therefore, target strength will be 48.25 N/mm2 as 58.25N/mm2 > 56.5 N/mm2.

Step 2 - Air Entrapped: From Table 3 of IS 10262, the approximate amount of entrapped air to be expected in normal (non-air-entrained) concrete is 1.0 percent for 20 mm nominal maximum size of aggregate.

Step 3 - Selection of Water-Cement Ratio: From Fig. 1 of IS 10262, the free water-cement ratio required for the target strength of 58.25 N/mm2 is 0.28 for OPC 43 grade curve. This is lower than the maximum value of 0.45 prescribed for ‘severe’ exposure for reinforced concrete as per Table 5 of IS 456. 0.28 < 0.45, hence O.K.

Step 4 - Selection of Water-Content: From Table 4 of IS 10262, water content = 186 kg (for 50 mm

slump) for 20 mm aggregate. Estimated water content for 75 mm slump (increasing at the rate of

percent for every 25 mm slump) = 186+(3*186)/100 = 191.58 kg

As superplasticizer is used, the water content may be reduced. Based on trial data, the water Content reduction of 25 percent is considered while using superplasticizer at the rate 1.0 percent by weight of cement.

Hence the arrived water content = 191.58 × 0.75 = 143.68 kg ≈ 144 kg.

Step 4 - Selection of Cement-Content: Water-cement ratio = 0.28

Cement content = 144/0.28 = 514.28 kg/m3 ≈ 514 kg/m3

From Table 5 of IS 456, maximum cementitious content for ‘severe’ exposure condition = 450kg/m3 514kg/m3 > 450 kg/m3, hence we reduce the cement content to 450kg/m3.

Step 4 – Proportion of volume of Coarse aggregate and Fine aggregate content:

From Table 5 of IS 10262, volume of coarse aggregate corresponding to 20 mm size aggregate and fine aggregate (Zone II) for water-cement ratio of 0.50 = 0.62.

In the present case water-cementitious ratio is 0.28. Therefore, volume of coarse aggregate is required to be increased to decrease the fine aggregate content. As the water-cement ratio is lower by 0.22, the proportion of volume of coarse aggregate is increased by 0.044 (at the rate of (-or+) 0.01 for every ± 0.05 change in water cement ratio). Therefore, corrected proportion of volume of coarse aggregate for the water-cementitious ratio of 0.28 = 0.62 + 0.044 = 0.664.

For pump able concrete these values may be reduced

by up to 10 percent. (see 5.5.2). Here, 10 percent reduction is considered.

Therefore, Volume of coarse aggregate = 0.664 × 0.9 = 0.5976cum say 0.59cum.

Volume of fine aggregate content = 1 – 0.59 = 0.41 m3

Step 4 – Mix Calculations: The mix calculations per unit volume of concrete shall be as followed:

1. Total volume: 1cum
2. Volume of Entrapped air in the wet concrete = 0.01cum
3. Volume of cement = (Mass of cement/Specific gravity of cement) * (1/1000)

= (450/3.15) * (1/1000) = 0.14Cum

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

= (144/1) * (1/1000) = 0.144Cum

1. Volume of Chemical admixture (@ 1.0 percent by mass of cementitious material)

= (Mass of Admixture/Specific gravity of admixture) * (1/1000)

= (4.50/1.145) * (1/1000) = 0.0039Cum

1. Volume of all in aggregate = (1 - Volume of cement - Volume of water - Volume of Chemical

admixture - Volume of Entrapped air in the wet concrete)

= (1-0.14-0.144-0.0039-0.01) = 0.7021cum

1. Mass of coarse aggregate = (Volume of all in aggregate*volume of coarse aggregate*

specific gravity of coarse aggregate*1000)

= (0.7021*0.664*2.74*1000) = 1277.37kgs

1. Mass of fine aggregate = (Volume of all in aggregate*volume of fine aggregate*

Specific gravity of fine aggregate*1000)

= (0.7021*0.41*2.65*1000) = 762.83kgs

Step 5 – Mix proportions:

1. Cement = 450kg/m3
2. Water (Net mixing) = 144kg/m3
3. Fine aggregate = 762.83kg/m3
4. Coarse aggregate = 1277.37kg/m3
5. Chemical admixture = 4.50 kg/m3
6. Free water-cementitious materials ratio = 0.28