Project on Concrete Mix Design for various grades of Concrete
High Rise Building Course
Challenge 5
Project
(Project on Concrete Mix Design for various grades of Concrete)
Student Name : Mayank Bajpai
Technical Support : Sanjana Reddy
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Calculate the Concrete Mix Design for M35 grade concrete with fly ash & M50 grade concrete without Fly ash
Solution:
M35 grade concrete with fly ash
Assumptions:
- Exposure condition: severe and concrete is reinforced
- Maximum nominal size of aggregate: 20 mm
- Workability required: 100 mm
- Specific gravity of aggregates: 2.74
- Specific gravity of cement: 3.14
- No free surface moisture.
- Superplasticiser is used.
Mix Design:
Targeted mean strength after 28 day = f(ck) + 1.65*s
= 35 + 1.65*5 (‘s’ taken from table 2)
= 43.25 N/mm2
So, we will choose OPC 43 grade cement.
From Table 5 of IS:456(2000), for severe exposure condition and reinforced concrete
Maximum water cement ratio is 0.45
Considering Fig. 1 of IS:10262(2019).
Since, no other data regarding cement strength is available. Curve 2 will be used for OPC 43.
w/c ratio = 0.38 (<0.45)
Assuming nominal aggregate size to be 20 mm.
Then, From table 4, IS:10262(2019), maximum water that can be used is 186 kg for 1 m3 of concrete. This much water provides workable concrete with a slump of around 30-50 mm.
Water required for concrete for a slump of 100 mm = 186*(1 + 0.03*(50/25))
= 197.16 Kg
Assuming 30% reduction in water content by use of superplasticizer, water required = 197.16*0.7 = 138 Kg
Cementitious material required = 138/(0.38) = 363 Kg
Minimum cementitious material should be 320 kg as per the Table 5 IS:456(2000)
Assuming 5% increase in cementitious material = 382 kg
Water cement ratio = 138/382 = 0.3613
Assuming 30% fly ash as the total cementitious material = 382*0.3 = 115 kg
OPC 43 = 382 - 115 = 267 kg
Cement saved = 363 - 267 = 96 kg
Fly ash utilized = 115 kg
Assuming Zone-II fine aggregate, from table 5 IS:10262(2019), we obtain ratio of CA to total aggregate for w/c ratio of 0.5 is 0.62.
Thus CA to total aggregate for w/c of 0.361 = 0.62 +0.139/0.05*0.01 = 0.6478
Volume of cement = 267/3.15/1000 = 0.085 m3 (approx)
Volume of fly ash = 115/2.2/1000 = 0.052 m3 (approx)
Volume of water = 138/1000 = 0.138 m3
Volume of chemical admixture = 7.6/1.1.45/1000 = 0.006 m3
Volume of aggregate(CA+FA) = 1 - (0.085 + 0.052 + 0.138 + 0.006) = 0.719 m3
Volume of CA = 0.719*0.6478 = 0.4658 m3
Volume of FA = 0.719 - 0.4658 = 0.2532 m3
Mass of CA = 0.4658*2.74*1000 = 1276 kg
Mass of FA = 0.2532*2.74*1000 = 694 kg
Mix proportion for 1 m3 of concrete:
|
Cement |
267 kg |
|
Fly ash |
115 kg |
|
Water |
138 kg |
|
Fine aggregate |
694 kg |
|
Coarse Aggregate |
1276 kg |
|
Chemical Admixture |
7 kg |
Water cement ratio: 0.361
Mix Ratio: 1:1.817:3.34
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M50 grade concrete without fly ash
Assumptions:
- Exposure condition: severe and concrete is reinforced
- Maximum nominal size of aggregate: 20 mm
- Workability required: 100 mm
- Specific gravity of aggregates: 2.74
- Specific gravity of cement: 3.14
- No free surface moisture.
- Superplasticiser is used.
Mix Design:
Targeted mean strength after 28 day = f(ck) + 1.65*s
= 50 + 1.65*5 (‘s’ taken from table 2)
= 58.25 N/mm2
So, we will choose OPC 53 grade cement.
From Table 5 of IS:456(2000), for severe exposure condition and reinforced concrete
Maximum water cement ratio is 0.45
Considering Fig. 1 of IS:10262(2019).
Since, no other data regarding cement strength is available. Curve 3 will be used for OPC 53.
w/c ratio = 0.34 (<0.45)
Assuming nominal aggregate size to be 20 mm.
Then, From table 4, IS:10262(2019), maximum water that can be used is 186 kg for 1 m3 of concrete. This much water provides workable concrete with a slump of around 30-50 mm.
Water required for concrete for a slump of 100 mm = 186*(1 + 0.03*(50/25))
= 197.16 Kg
Assuming 30% reduction in water content by use of superplasticizer, water required = 197.16*0.7 = 138 Kg
Cementitious material required = 138/(0.34) = 406 Kg
Minimum cementitious material should be 320 kg as per the Table 5 IS:456(2000)
Assuming Zone-II fine aggregate, from table 5 IS:10262(2019), we obtain ratio of CA to total aggregate for w/c ratio of 0.5 is 0.62.
Thus CA to total aggregate for w/c of 0.34 = 0.62 +0.16/0.05*0.01 = 0.652
Volume of cement = 406/3.15/1000 = 0.129 m3 (approx)
Volume of water = 138/1000 = 0.138 m3
Volume of chemical admixture = 7.6/1.1.45/1000 = 0.006 m3
Volume of aggregate(CA+FA) = 1 - (0.129 + 0.138 + 0.006) = 0.727 m3
Volume of CA = 0.727*0.652 = 0.474 m3
Volume of FA = 0.727 - 0.474 = 0.253 m3
Mass of CA = 0.474*2.74*1000 = 1299 kg
Mass of FA = 0.253*2.74*1000 = 693 kg
Mix proportion for 1 m3 of concrete:
|
Cement |
406 kg |
|
Water |
138 kg |
|
Fine aggregate |
693 kg |
|
Coarse Aggregate |
1299 kg |
|
Chemical Admixture |
7 kg |
Water cement ratio: 0.34
Mix Ratio: 1:1.707:3.2
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