Project on Concrete Mix Design for various grades of Concrete

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

1. The concrete mix design for M50 grade of concrete without fly ash

Equations:

• Target mean strength: f’ck =fck + 1.65 s

where,

f’ck = target mean strength

fck = characteristics compressive strength at 28 days, and

s = standard deviation.

Procedure and calculation:

1. Target mean strength

f’ck =fck + 1.65 s

= 50N/mm²

S= 5N/mm²

f’ck = 50+(1.65*5)

=58.25 N/mm²

2. Selection of water/cement ratio

Given M50

Assume:

Exposure Condition

Therefore,

Extreme = w/c ratio 0.40

M50 = w/c ratio 0.40

Water cement ratio = 0.40

Minimum water cement content = 360kg/m³

3. Selection of water content

Assume:

Nominal size of aggregate = 20mm

Minimum water content = 186 litres

Slump range from 25 to 50

As superplasticizers (20% reduced)

186*80/100 = 148.8 (100-20=80)

Water content = 149 litres

4. Calculation of cement content

Water cement ratio = 0.40

Water content = 149 litres

Cement content = 149/0.40 = 372.5 (373kg/m³)

Minimum cement content = 360 kg/m³

5. Volume of coarse aggregate and fine aggregate

Fine aggregate are sieved according to zone 1 table 4 IS 383

Coarse aggregate are sieved according to table 2 IS 383

According,

Table 3 IS 10262:2009

Fine aggregate, (zone 1)

Therefore,

Volume of coarse aggregate = 0.60

Water cement ratio = 0.40

(+/- 0.001 for every 0.005)

So Volume of Coarse aggregate = 0.62

Volume of fine aggregate = 1-0.62 = 0.38

6. Mix calculation

• Volume of concrete = 1m³
• Volume of cement = (373/3.15)*1/1000 = 0.118m³
• Volume of water = (149/1)*1/1000 = 0.149 m³
• Volume of admixture = (7.6/1.145)*1/1000 = 0.006 m³
• Volume of coarse aggregate and fine aggregate = a-(b + c +d)
• = 1-(0.118+0.149+0.006)= 0.727m³
• Mass of coarse aggregate =0.72*0.62*2.74*1000 = 1235.02 kg/m³
• Mas of fine aggregate = 0.72*0.38*2.74*1000 = 756.9 kg/m³

2. The concrete mix design for M35 grade of concrete with fly ash

Procedure and calculation:

1. Target mean strength

f’ck =fck + 1.65 s

= 35N/mm²

S= 5N/mm²

f’ck = 50+(1.65*5)

=43.25 N/mm²

2. Selection of water/cement ratio

Given M35

Assume:

Exposure Condition

Therefore,

Extreme = w/c ratio 0.40

M35 = w/c ratio 0.45

Water cement ratio = 0.40

Minimum water cement content = 360kg/m³

3. Selection of water content

Assume:

Nominal size of aggregate = 20mm

Minimum water content = 186 litres

Slump range from 25 to 50

As superplasticizers (20% reduced)

186*80/100 = 148.8 (100-20=80)

Water content = 149 litres

4. Calculation of cement content (Cement + fly ash)

Water cement ratio = 0.40

Water content = 149 litres

Cement content = 149/0.40 = 372.5 (373kg/m³)

Minimum cement content = 360 kg/m³

360 < 370 kg/m³

Increase by 10%

370*0.10+370 = 407 kg/m³

W = 149kg

Water cement ratio = 149/407 = 0.366

Fly ash 30% total cementitious content

407*30/100 = 122 kg/m³

Cement (Ordinary Portland cement) = 407-122 = 285 kg/m³

Saving of cement by using Fly ash = 370-285 = 85 kg/m³

Fly ash = 122 kg/m³

5. Volume of coarse aggregate and fine aggregate

Fine aggregate are sieved according to zone 1 table 4 IS 383

Coarse aggregate are sieved according to table 2 IS 383

According,

Table 3 IS 10262:2009

Fine aggregate, (zone 1)

Therefore,

Volume of coarse aggregate = 0.60

Water cement ratio = 0.40

(+/- 0.001 for every 0.005)

So Volume of Coarse aggregate = 0.62

Volume of fine aggregate = 1-0.62 = 0.38

6. Mix calculation

• Volume of concrete = 1m³
• Volume of cement = (285/3.15)*1/1000 = 0.090 m³
• Volume of fly ash = (122/2.2)*1/1000 = 0.055 m³
• Volume of water = (149/1)*1/1000 = 0.149 m³
• Volume of admixture = (7.6/1.145)*1/1000 = 0.006 m³
• Volume of coarse aggregate and fine aggregate = a-(b + c +d + e)
  • = 1-(0.99+0.055+0.149+0.006)
  • = 0.7m³
• Mass of coarse aggregate =0.7*0.62*2.74*1000 = 1189.16kg/m³
• Mas of fine aggregate = 0.7*0.38*2.74*1000 = 728kg/m³