Project on Concrete Mix Design for various grades of Concrete
Calculation of concrete mix design for M35 grade concrete with fly ash
AIM
To calculate the concrete mix design for M35 grade concrete with fly ash.
EQUATIONS
- Target Mean Strength, f'ck = fck + 1.65 S
where, f'ck = target mean strength
fck = compressive strength at 28 days of concrete
S = standard deviation (IS 10262:2009, Table 1)
- Volume of cement = (mass of cement / specific gravity of cement) * (1 / 1000)
- Volume of water = (mass of water / specific gravity of water) * (1 / 1000)
- Volume of chemical admixture = (mass of chemical admixture / specific gravity of chemical admixture) * (1 / 1000)
- Volume of fly ash = (mass of fly ash / specific gravity of fly ash) * (1 / 1000)
- Volume of concrete = 1 m^3
- Volume of aggregates = Volume of concrete - (volume of cement + volume of fly ash + volume of water + volume of admixture)
- Mass of coarse aggregates = volume of aggregate * volume of coarse aggregate * specific gravity of coarse aggregate * 1000
- Mass of fine aggregate = volume of aggregate * volume of fine aggregate * specific gravity of fine aggregate * 1000
CODES FOR CONCRETE MIX DESIGN
- IS 10262:2009
- IS 456:2000
- IS 8112
- IS 9103
- IS 3812 (Part 1)
OBJECTIVES
- The main objective is to produce concrete as economically as possible.
- To achieve desired workability, durability and minimum strength of M35 grade concrete.
PROCEDURE AND CALCULATION
1. TARGET MEAN STRENGTH
f'ck = fck + 1.65 S
fck = 35 N/mm^2 (M35)
From table 1 - IS 10262:2009
S = 5 N/mm^2
f'ck = 35 + (1.65 * 5)
= 43.25 N/mm^2
2. SELECTION OF WATER CEMENT RATIO
For M35,
Assume,
Exposure condition: Extreme (for reinforced concrete)
From Table 5 - IS 456:2000
Therefore,
Extreme = w/c ratio = 0.40
M35 = w/c ratio = 0.45
Lower the water-cement ratio better the workability
0.40 < 0.45
Water cement ratio = 0.40
Minimum cement content = 360 kg/m^3
3. SELECTION OF WATER CONTENT
Assume,
Nominal size of aggregate = 20mm
From Table 2 - IS 10262:2009
Minimum water content = 186 litres or kg (for 25 to 50mm slump range)
As Superplasticizers (20% reduced)
= 186 * 80/100
= 148.8 litre
Water content = 149 litre
4. CALCULATION OF CEMENTITIOUS CONTENT (CEMENT + FLY ASH)
Water cement ratio = w/c = 0.40
Water content = W = 149 kg
Cementitious content = C = 149/0.40 = 372.5 kg = 370 kg/m^3
Minimum cement content = 360 kg/m^3
360 < 370kg/m^3 (Hence OK)
Increase by 10%,
370 * 0.10 + 370 = 407 kg/m^3
W = 149 kg
Water cement ratio = 149/407 = 0.366
Fly ash at 30% total cementitious material content
= 407 * 30/100 = 122 kg/m^3
Cement (Ordinary Portland Cement) = 407 - 122 = 285 kg/m^3
Saving cement while using fly ash = 370 - 285 = 85 kg/m^3
Fly ash =122 kg/m^3
5. VOLUME OF COARSE AGGREGATES AND FINE AGGREGATES
Fine aggregates are sieved according to zone 1 of Table 4 - IS 383
Coarse aggregates are sieved according to Table 2 - IS 383
According,
From Table 3 - IS 10262:2009
Nominal size of aggregates = 20mm
Fine aggregates zone 1
Therefore,
Volume of coarse aggregate = 0.60
Water cement ratio = 0.4
(+/- 0.01 for every +/- 0.005)
So, volume of coarse aggregate = 0.62
Volume of fine aggregates = 1 - 0.62 = 0.38
6. MIX CALCULATION
- Volume of concrete = 1 m^3
- Volume of cement = (285 / 3.15) * (1 / 1000) = 0.090 m^3
- Volume of fly ash = (122 / 2.2)* (1 / 1000) = 0.055 m^3
- Volume of water = (149 / 1)* (1 / 1000) = 0.149 m^3
- Volume of admixture = (7.6 / 1.145)* (1 / 1000) = 0.006 m^3
- Volume of coarse aggregate and fine aggregate = a - (b + c + d + e)
= 1 - (0.090 + 0.055 + 0.149 + 0.006) = 0.7 m^3
- Mass of coarse aggregate = 0.7 * 0.62 * 2.74 * 1000 = 1189.16 kg/m^3
- Mass of fine aggregate = 0.7 * 0.38 * 2.74 * 1000 = 728 kg/m^3
RESULTS AND CONCLUSION
TRIAL MIX
Cement = 285 kg/m^3
Fly ash = 122 kg/m^3
Water = 149 kg
Coarse aggregate = 1189 kg/m^3
Fine aggregate = 728 kg/m^3
Chemical admixture = 7.6 kg/m^3
Water cement ratio = 0.366
Trial mix ratio = 285:728:1189
= 1:2.55:4.17 (C:FA:CA)
- This trial mix ratio should match the workability, durability and minimum strength of M35 grade concrete.
- If it does not satisfy the above criteria two or more trials for increase of 10% should be done.
Calculation of concrete mix design for M50 grade concrete without fly ash
AIM
To calculate M50 grade concrete without fly ash.
EQUATIONS
- Target mean strength, f'ck = fck + 1.65 S
where, f'ck = target mean strength
fck = compressive strength at 28 days of concrete
S = standard deviation (IS 10262:2009, Table 1)
- Volume of cement = (mass of cement / specific gravity of cement) * (1 / 1000)
- Volume of water = (mass of water / specific gravity of water) * (1 / 1000)
- Volume of chemical admixture = (mass of chemical admixture / specific gravity of chemical admixture) * (1 / 1000)
- Volume of concrete = 1m^3
- Volume of aggregates = volume of concrete - (volume of cement + volume of water + volume of chemical admixture)
- Mass of coarse aggregate = volume of aggregate * volume of coarse aggregate * specific gravity of coarse aggregate * 1000
- Mass of fine aggregate = volume of aggregate * volume of fine aggregate * specific gravity of fine aggregate * 1000
CODES USED FOR CONCRETE MIX DESIGN
- IS 10262:2009
- IS 456:2000
- IS 8112
- IS 9103
- IS 3812 (Part 1)
OBJECTIVES
- The main objective is to produce concrete as economically as possible.
- To achieve desired workability, durability and minimum strength of M50 grade concrete.
PROCEDURE AND CALCULATION
1. TARGET MEAN STRENGTH
f'ck = fck + 1.65 S
fck = 50 N/mm^2 (M50)
From Table 1 - IS 10262:2009
S = 5 N/mm^2
f'ck = 50 + (1.65 * 5)
= 58.25 N/mm^2
2. SELECTION OF WATER CEMENT RATIO
Given,
M50
Assume,
Exposure condition = Extreme (for reinforced concrete)
From Table 5 - IS 456:2000
Therefore,
Extreme = w/c ratio = 0.40
M50 =w/c ratio = 0.40
Lower the water cement ratio better the workability
0.40
Water cement ratio = 0.40
Minimum cement content = 360 kg/m^3
3. SELECTION OF WATER CONTENT
Assume,
Nominal size of aggregates = 20mm
From Table 2 - IS 10262:2009
Minimum water content = 186 litre or kg (for 25 to 50mm slump range)
As superplasticizers (20% reduced)
= 186 * 80 / 100 = 148.8 (100 - 20 = 80)
Water content = 149 litres
4. CALCULATION OF CEMENT CONTENT
Water cement ratio = w/c = 0.40
Water content = W = 149 kg
Cement content = C = 149 / 0.40 = 372.5 = 373 kg/m^3
Minimum cement content = 360 kg/m^3
360 < 373 kg/m^3 (Hence OK)
5. VOLUME OF COARSE AGGREGATES AND FINE AGGREGATES
Fine aggregates are sieved according to zone 1 of Table 4 IS 383
Coarse aggregates are sieved according to Table 2 IS 383
According,
From Table 3 IS 10262:2009
Nominal size of aggregates = 20mm
Fine aggregates zone 1
Therefore,
volume of coarse aggregate = 0.60
Water cement ratio = 0.4
(+/- 0.01 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 = 1 m^3
- Volume of cement = (373 / 3.15) * (1 /1000) = 0.118 m^3
- Volume of water = (149 / 1) * (1 / 1000) = 0.149 m^3
- Volume of admixture = (7.6 / 1.145) * (1 / 1000) = 0.006 m^3
- Volume of coarse aggregate and fine aggregate = a - (b + c + d)
= 1 - (0.118 + 0.149 + 0.006) = 0.727 m^3
- Mass of coarse aggregate = 0.727 * 0.62 * 2.74 * 1000 = 1235.02 kg/m^3
- Mass of fine aggregate = 0.727 * 0.38 * 2.74 * 1000 = 756.9 kg/m^3
RESULTS AND CONCLUSION
TRIAL MIX
Cement = 373 kg/m^3
Water = 149 kg
Coarse aggregate = 1235.02 kg/m^3
Fine aggregate = 756.9 kg/m^3
Chemical admixture = 7.6 kg/m^3
Water cement ratio = 0.40
Trial mix ratio = 373 : 756.9 : 1235.02
= 1 : 2.02 : 3.31 (C : FA : CA)
- This trial mix ratio should match the workability, durability and minimum strength of M50 grade concrete.
- If it does not satisfy the above criteria two or more trials for increase of 10% should be done.