Project on Concrete Mix Design for various grades of Concrete
QUESTION: Calculate the Concrete Mix Design for M35 grade concrete with fly ash & M50 grade concrete without Fly ash
Answer: Calculation of the concrete mix design of M35 grade concrete with fly ash.
AIM:
The aim of the project is to design M35 concrete mix design with fly ash.
OBJECTIVE OF THE PROJECT:
The objective of the project is to find the proportion of each ingredient used in concrete to make the M35 concrete with fly ash.
GIVEN DATA:
|
Grade of concrete |
M35 |
|
Type of cement |
OPC 43 grade conforming to IS 8112 |
|
Exposure condition |
Very severe |
|
Min. cement content |
340 kg/m^3 (Table 5, IS 456:2000) |
|
Mineral admixture |
Fly ash conforming to IS 3812 part 1 |
|
Type of aggregate |
Crushed angular aggregate |
|
Workability |
100mm slump value |
|
Max. w/c ratio |
0.45 (Table 5, IS 456:2000) |
|
Supervision |
Good |
|
Max. nominal size of aggregates |
20mm |
|
Concrete placing by |
Pumping |
|
Chemical admixture |
Super plasticizer conforming to IS 9103 |
|
Zone of aggregates |
Zone 1 |
Specific gravity of
- Cement - 3.15
- Coarse aggregates - 2.74
- Fine aggregates - 2.74
- Flyash - 2.2
- Superplasticizer - 1.145
Water absorption of
- Coarse aggregate - 0.5%
- Fine aggregate - 1%
Free surface moisture of aggregates - Nil
Sieve analysis of aggregates confronting to IS 383
Body of the content:
Step 1- To find the target strength (f'ck)
From IS 10262:2019, the formula for target strength f'ck is given by
f'ck = fck + 1.65 S
Where, fck - Characteristic compressive strength at 28 days.
S - Standard deviation
From table 2 of IS 10262:2019, S = 5 N/mm^2 for M35 grade concrete.
Therefore, f'ck = 35 + 1.65 * 5
f'ck = 43.25 N/mm^2
Step 2 - To find the water-cement ratio
The maximum w/c ratio for different exposure condition is specified on IS 456:2000 Table 5. For very severe conditions the maximum w/c ratio to be used is 0.45 as per table 5 of IS 456:2000
But based on experience 0.40 w/c ratio is OK. Hence w/c ratio is adopted as 0.40
Step 3 - Selection of water content
As per table 4 of IS 10262:2019, the maximum water content for cubic concrete with 20mm aggregate is 186kg.
This is for slump value till 50mm. In our concrete, the slump value is 100mm, so additional 3% water has to be added for every 25mm slump increased. Here we have to add 6% extra.
Water content for 100mm slump value = 186 + 6%186
= 197.16 kg
There is addition of superplasticizer which can further reduce the water content by upto 30%. Based on trials, the water content was able to be reduced upto 29%.
Water content required after adding superplasticizer = 197.16 * 0.71
Adopted water content = 140 kg
Step 4 - Calculation of Cement and flyash content
Water-cement ratio = 0.4
Water content = 140 kg
Cementitious material (cement + flyash) content = 140/0.4
= 350 kg/m^3
This is greater than minimum cement content. Since it also contains flyash as mineral admixture, the cementitious material content is increased by 10%.
Therefore, total content of cementitious material = 1.1 * 350
= 385 kg/m^3
Water cement ratio used is = 140/385
w/c ratio = 0.364
Flyash is used 30% of the cementitious material, i.e) flyash content = 0.3*385
= 115.5 kg/m^3
Cement content (OPC) in the cementitious content = 385 - 115.5
= 269.5 kg/m^3
Cement saved by using flyash = 350 - 269.5
= 80.5 kg/m^3
Mass of chemical admixture (super plasticizer) = 2% cementitious material
= (2 * 385) / 100
= 7.7 kg/m^3
Step 5 - Volume of coarse aggregate and fine aggregate
From table 5 of IS 10262:2019, the volume of coarse aggregate corresponding to 20mm coarse aggregate in zone 1 with w/c ratio 0.5 is = 0.6
But here, the w/c ratio taken is 0.4. As the w/c ratio is decreased by 0.1, the volume of coarse aggregates has to be increased by 0.02.
Therefore, corrected volume of coarse aggregates for w/c ratio 0.4 is 0.62
For pumpable concrete, these values are further reduced by 10%.
Volume of coarse aggregate = 0.9*0.62 = 0.56
Volume of fine aggregate = 1 - 0.56 = 0.44
Step 6 - Mix calculation
a) Volume of concrete = 1 m^3
b) Volume of cement = Mass of cement / ( specific gravity of cement * 1000)
= 269.5 / (3.15*1000)
= 0.086 m^3
c) Volume of flyash = Mass of flyash / (specific gravity of flyash * 1000)
= 115.5/(2.2*1000)
= 0.053 m^3
d) Volume of water = Mass of water /( specific gravity of water * 1000)
= 140/(1*1000)
= 0.140 m^3
e) Volume of superplasticizer = Mass of superplasticizer / (specific gravity of superplasticizer * 1000)
= 7.7/(1.145*1000)
= 0.007 m^3
f) Volume of aggregates = a - (b+c+d+e)
= 1 - ( 0.086 + 0.053 + 0.14 + 0.007 )
= 0.714 m^3
g) Mass of coarse aggregate = f * volume of coarse aggregate * specific gravity of coarse aggregate * 1000
= 0.714*0.56*2.74*1000
= 1095.5 kg/m^3
h) Mass of fine aggregate = f * volume of fine aggregate * specific gravity of fine aggregate * 1000
= 0.714*0.44*2.74*1000
= 860.8 kg/m^3
Step 7 - Mix proportions
The quantity of materials needed for concrete is given below,
Cement =269.5 kg/m^3
Flyash =115.5 kg/m^3
Water =140 kg/m^3
Fine aggregates =860.8 kg/m^3
Coarse aggregtes =1095.5 kg/m^3
Superplasticizer =7.7 kg/m^3
Water-cement ratio =0.364
Mix ratio for M35 concrete with flyash = 1 : 2.24 : 2.85
Conclusion
Indian standard code books like IS 456:2000, IS 10262:2019 were used for the completion of the project.
The final mix ratio for M35 concrete with flyash is given by 1 : 2.24 : 2.85
-----------------------------------------------------------------------------------------------
Answer: Calculation of the concrete mix design of M50 grade concrete.
AIM:
The aim of the project is to design M50 concrete mix design.
OBJECTIVE OF THE PROJECT:
The objective of the project is to find the proportion of each ingredient used in concrete to make the M50 concrete.
GIVEN DATA:
|
Grade of concrete |
M50 |
|
Type of cement |
OPC 43 grade conforming to IS 8112 |
|
Exposure condition |
Extreme |
|
Min. cement content |
360 kg/m^3 (Table 5, IS 456:2000) |
|
Type of aggregate |
Crushed angular aggregate |
|
Workability |
100mm slump value |
|
Max. w/c ratio |
0.40 (Table 5, IS 456:2000) |
|
Supervision |
Good |
|
Max. nominal size of aggregates |
20mm |
|
Concrete placing by |
Pumping |
|
Chemical admixture |
Superplasticizer conforming to IS 9103 |
|
Zone of aggregates |
Zone 1 |
Specific gravity of
- Cement - 3.15
- Coarse aggregates - 2.74
- Fine aggregates - 2.74
- Superplasticizer - 1.145
Water absorption of
- Coarse aggregate - 0.5%
- Fine aggregate - 1%
Free surface moisture of aggregates - Nil
Sieve analysis of aggregates confronting to IS 383
Body of the content:
Step 1- To find the target strength (f'ck)
From IS 10262:2019, the formula for target strength f'ck is given by
f'ck = fck + 1.65 S
Where, fck - Characteristic compressive strength at 28 days.
S - Standard deviation
From table 2 of IS 10262:2019, S = 5 N/mm^2 for M50 grade concrete.
Therefore, f'ck = 50 + 1.65 * 5
f'ck = 58.25 N/mm^2
Step 2 - To find the water-cement ratio
The maximum w/c ratio for different exposure condition is specified on IS 456:2000 Table 5. For extreme conditions the maximum w/c ratio to be used is 0.40 as per table 5 of IS 456:2000
Step 3 - Selection of water content
As per table 4 of IS 10262:2019, the maximum water content for cubic concrete with 20mm aggregate is 186kg.
This is for slump value till 50mm. In our concrete, the slump value is 100mm, so additional 3% water has to be added for every 25mm slump increased. Here we have to add 6% extra.
Water content for 100mm slump value = 186 + 6%186
= 197.16 kg
There is addition of superplasticizer which can further reduce the water content by upto 30%. Based on trials, the water content was able to be reduced upto 29%.
Water content required after adding superplasticizer = 197.16 * 0.71
Adopted water content = 140 kg
Step 4 - Calculation of Cement content
Water-cement ratio = 0.4
Water content = 140 kg
Cement content = 140/0.4
= 350 kg/m^3
The minimum cement has to be 360 kg/m^3, so taking 360 kg/m^3 as cement content.
Water cement ratio used is = 140/360
w/c ratio = 0.389
Mass of chemical admixture (super plasticizer) = 2% cementitious material
= (2 * 360) / 100
= 7.2 kg/m^3
Step 5 - Volume of coarse aggregate and fine aggregate
From table 5 of IS 10262:2019, the volume of coarse aggregate corresponding to 20mm coarse aggregate in zone 1 with w/c ratio 0.5 is = 0.6
But here, the w/c ratio taken is 0.4. As the w/c ratio is decreased by 0.1, the volume of coarse aggregates has to be increased by 0.02.
Therefore, corrected volume of coarse aggregates for w/c ratio 0.4 is 0.62
For pumpable concrete, these values are further reduced by 10%.
Volume of coarse aggregate = 0.9*0.62 = 0.56
Volume of fine aggregate = 1 - 0.56 = 0.44
Step 6 - Mix calculation
a) Volume of concrete = 1 m^3
b) Volume of cement = Mass of cement / ( specific gravity of cement * 1000)
= 360 / (3.15*1000)
= 0.114 m^3
c) Volume of water = Mass of water /( specific gravity of water * 1000)
= 140/(1*1000)
= 0.140 m^3
d) Volume of superplasticizer = Mass of superplasticizer / (specific gravity of superplasticizer * 1000)
= 7.2/(1.145*1000)
= 0.006 m^3
e) Volume of aggregates = a - (b+c+d)
= 1 - ( 0.114+0.140+0.006 )
= 0.74 m^3
f) Mass of coarse aggregate = e * volume of coarse aggregate * specific gravity of coarse aggregate * 1000
= 0.74*0.56*2.74*1000
= 1135.5 kg/m^3
g) Mass of fine aggregate = e * volume of fine aggregate * specific gravity of fine aggregate * 1000
= 0.74*0.44*2.74*1000
= 892.14 kg/m^3
Step 7 - Mix proportions
The quantity of materials needed for concrete is given below,
Cement =360 kg/m^3
Water =140 kg/m^3
Fine aggregates =892.14 kg/m^3
Coarse aggregtes =1135.5 kg/m^3
Superplasticizer =7.2 kg/m^3
Water-cement ratio =0.389
Mix ratio for M50 concrete = 1 : 2.48 : 3.15
Conclusion
Indian standard code books like IS 456:2000, IS 10262:2019 were used for the completion of the project.
The final mix ratio for M50 concrete is given by 1 : 2.48 : 3.15