Project on Concrete Mix Design for various grades of Concrete
AIM : MIX DESIGN FOR M35 GRADE OF CONCRETE WITH ADDITION OF FLY ASH
INTRODUCTION : FLY ASH IS USED IN CONCRETE AS A REPLACEMENT TO ORDINARY PORTLAND CEMENT (OPC)
- STIPULATIONS FOR PROPORTIONING
| Grade Designation | M35 | |
| Type of Cement used | Ordinary portland cement (OPC) | 43 Grade ---Conforming to IS 8112 |
| Type of mineral admixture | Fly ash | Conforming to IS 3812 (Part 1) |
| Maximum nominal size of aggregate | 20 mm | |
| Exposure condition | Severe | For Reinforced Concrete |
| Minimum Cement content | 320 kg/m^3 | From table 5 of IS 456:2000 - Exposure condition |
| Maximum Water content | 0.45 | From table 5 of IS 456:2000 - Exposure condition |
| Method of concrete placing | Pumping | |
| Workability | 100 mm | Slump value for pumpable concrete |
| Type of aggregate | Crushed angular aggregate | |
| Maximum Cement content | 450 kg/m^3 | |
| Chemical admixture used | Superplasticizer | Conforming to IS 9103 |
- TEST DATA OF MATERIALS
| Specific Gravity of Cement | 3.15 | |
| Specific Gravity of Fly ash | 2.2 | |
| Specific Gravity of Aggregate | 2.74 | Fine Aggregate |
| 2.74 | Coarse Aggregate | |
| Water Absorption | 0.1% | Fine Aggregate |
| 0.5% | Coarse Aggregate |
STEP 1. CALCULATE TARGET MEAN STRENGTH OF MIX PROPORTIONING (F'CK) :
Formula for, Target mean strength = F'ck = Fck + 1.65 X S
where , fck = compressive strength of concrete
S = Standard Deviation --------------From IS 1062
Therefore, F'ck = 35 + 1.65 X 5 = 43.25 N/mm^2
STEP 2. SELECTION OF WATER CEMENT RATIO :
- Value of w/c ratio is considered from three criteria
- From graph 1 of compressive strength vs water/cement ratio , w/c = 0.35
- From grapgh 2 of compressive strenth in 28 days vs w/c ratio , w/c = 0.40
- From Table 5 of IS 456 2000 for severe exposure , w/c = 0.45
- From above and based on experience, smallest value is assumed. Therefore, w/c = 0.35
STEP 3. SELECTION OF WATER CONTENT
- From Table 2 of IS 10262 : 2009 , Maximum water content for 25 to 50 mm Slump = 186 kg = 186 litre
- Therefore, Estimated water content for 100 mm slump = 186 + (6/100 X 186) = 197 litre
- As superplasticizer is used, the water content can be reduced upto 20% and based on the trials water reduction of 29% is achieved with plasticizer
- Hence, The water content = 197 X 0.71 = 140 litre = 140 kg
STEP 4. CEMENT AND FLYASH CONTENT
- w/c ratio assumed is 0.35
- Therefore, Cementitious material i.e. (Cement + Flyash) = C = 140/0.35 = 400 kg/m^3
- From Table 5 of IS 456 minimum cement content for severe exposure condition is 320 kg/m^3
- Therefore, 400 kg/m^3 > 320 kg/m^3 ,HENCE OK
- As we are using Fly ash in this mix the Cementitios material is increased by 10%
- Therefore, Cementitious material content = 400 X 1.10 = 440 kg/m^3
- So Water content is 140 kg
- Therefore w/c = 140/440 = 0.318 ----------Final value of water/cement ratio
- Now, Fly ash @ 30% of Total cementitious material = 440 X 30% = 132 kg/m^3
- Cement OPC = 440 - 132 = 308 kg/m^3
- Cement saved by using flyash = 400 - 308 = 92 kg/m^3
From above calculations value obtaimned are :
| w/c ratio | 3.18 |
| Cement content | 308 kg/m^3 |
| Fly ash content | 132 kg.m^3 |
STEP 5. VOLUME OF COARSE AGGREGATE AND FINE AGGREGATE
- From Table 3 of IS 10262 : 2009 , volume of coarse aggregate corresponding to 20 mm size aggregate and fine agggregate zone 1 water cement ratio od 0.5 = 0.6
- In the present case the water cement ratio is 0.35
- Therefore, volume of Coarse aggregate needs to be increased. As water-cement ratio is lower bt o.25 the proportion of volume of coarse aggregate is increased by 0.05 ---(at the rate of -/+0.01 for every +/-0.05 change in water-cement ratio)
- Therefore , corrected volume of Coarse Aggregate for w/c ratio of 0.35 = 0.65
- for pumpable concrete these values are reduced by 10%
- Therefore Volume of Coarse aggregate = 0.65 X 0.9 Volume of Coarse aggreagte = 0.585
- Volume of Fine Aggregate = 1 - 0.585 Volume of Fine Aggregate = 0.415
STEP 6. MIX CALCULATION
| DESCRIPTION | FORMULAE | ANSWERS |
| Volume of Concrete | = 1m^3 | |
| Volume of Cement | = (Mass of Cement/Specific gravity of cement) X (1/1000) |
= (308/3.15) X (1/1000) = 0.097 m^3 |
| Volume of Fly ash | = (Mass of Fly ash/Specific gravity of Fly ash) X (1/1000) |
= (132/2.2) X (1/1000) = 0.06 m^3 |
| Volume of Water | = (Mass of Water/Specific gravity of Water) X (1/1000) |
= (140/1) X (1/1000) = 0.140 m^3 |
| Volume of Chemical Admixture | = (Mass of chemical admixture/Specific gravity of chemical admixture) X (1/1000) |
= (7.6/1.145) X (1/1000) = 0.006 m^3 |
| Volume of Total Aggregates | = 1 - (Volume of Cement + Volume of Water + Volume of Fly ash + Volume of chemical admixture ) |
= 1- (0.097 + 0.06 + 0.140 + 0.006) = 1 - 0.697 = 0.303 m^3 |
| Mass of Coarse Aggregate | = Volume of Total Aggregate X Volume of coarse agggregate X Specific gravity of Coarse aggregate X 1000 |
= 0.303 X 0.585 X 2.74 X 1000 = 486 kg/m^3 |
| Mass of Fine Aggregate | = Volume of Total Aggregate X Volume of Fine agggregate X Specific gravity of Fine aggregate X 1000 |
= 0.303 X 0.415 X 2.74 X 1000 = 345 kg/m^3 |
STEP 7. MIX DESIGN RESULTS
FOR M35 GRADE OF CONCRETE WITH ADDITION OF FLY ASH REQUIRED QUANTITY ARE AS FOLLOW :
- MASS OF CEMENT = 308 kg/m^3
- MASS OF FLYASH = 132 kg/m^3
- MASS OF WATER = 140 kg/m^3
- MASS OF FINE AGGREGATE = 345 kg/m^3
- MASS OF COARSE AGGREGATE = 486 kg/m^3
- MASS OF ADMIXTURE =7.6 kg/m^3
- WATER/CEMENT RATIO = 0.318
MIX RATIO w.r.t CEMENT : SAND (FINE AGGREGATE) : COARSE AGGREGATE IS 1 : 1.12 : 1.57
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AIM : MIX DESIGN FOR M50 GRADE OF CONCRETE
INTRODUCTION : MIX DESIGN FOR M50 GRADE OF CONCRETE IS DONE WITH RESPECTIVE INDIAN STANARDS
- STIPULATIONS FOR PROPORTIONING
| Grade Designation | M50 | |
| Type of Cement used | Ordinary portland cement (OPC) | 43 Grade ---Conforming to IS 8112 |
| Maximum nominal size of aggregate | 20 mm | |
| Exposure condition | Severe | For Reinforced Concrete |
| Minimum Cement content | 320 kg/m^3 | From table 5 of IS 456:2000 - Exposure condition |
| Maximum Water content | 0.45 | From table 5 of IS 456:2000 - Exposure condition |
| Method of concrete placing | Pumping | |
| Workability | 100 mm | Slump value for pumpable concrete |
| Type of aggregate | Crushed angular aggregate | |
| Maximum Cement content | 450 kg/m^3 | |
| Chemical admixture used | Superplasticizer | Conforming to IS 9103 |
- TEST DATA OF MATERIALS
| Specific Gravity of Cement | 3.15 | |
| Specific Gravity of Aggregate | 2.74 | Fine Aggregate |
| 2.74 | Coarse Aggregate | |
| Water Absorption | 0.1% | Fine Aggregate |
| 0.5% | Coarse Aggregate |
STEP 1. CALCULATE TARGET MEAN STRENGTH OF MIX PROPORTIONING (F'CK) :
Formula for, Target mean strength = F'ck = Fck + 1.65 X S
where , fck = compressive strength of concrete
S = Standard Deviation --------------From IS 1062
Therefore, F'ck = 50 + 1.65 X 5 = 58.25 N/mm^2
STEP 2. SELECTION OF WATER CEMENT RATIO :
- Value of w/c ratio is considered from three criteria
- From graph 1 of compressive strength vs water/cement ratio , w/c = 0.30
- From grapgh 2 of compressive strenth in 28 days vs w/c ratio , w/c = 0.37
- From Table 5 of IS 456 2000 for severe exposure , w/c = 0.45
- From above and based on experience, smallest value is assumed. Therefore, w/c = 0.30
STEP 3. SELECTION OF WATER CONTENT
- From Table 2 of IS 10262 : 2009 , Maximum water content for 25 to 50 mm Slump = 186 kg = 186 litre
- Therefore, Estimated water content for 100 mm slump = 186 + (6/100 X 186) = 197 litre
- As superplasticizer is used, the water content can be reduced upto 20% and based on the trials water reduction of 29% is achieved with plasticizer
- Hence, The water content = 197 X 0.71 = 140 litre = 140 kg
STEP 4. CEMENT CONTENT
- w/c ratio assumed is 0.35
- Therefore, Cementitious material i.e. Cement = C = 140/0.30 = 466.67 kg/m^3
- From Table 5 of IS 456 minimum cement content for severe exposure condition is 320 kg/m^3
- Therefore, 466.67 kg/m^3 > 320 kg/m^3 ,HENCE OK
STEP 5. VOLUME OF COARSE AGGREGATE AND FINE AGGREGATE
- From Table 3 of IS 10262 : 2009 , volume of coarse aggregate corresponding to 20 mm size aggregate and fine agggregate zone 1 water cement ratio od 0.5 = 0.6
- In the present case the water cement ratio is 0.35
- Therefore, volume of Coarse aggregate needs to be increased. As water-cement ratio is lower bt o.25 the proportion of volume of coarse aggregate is increased by 0.05 ---(at the rate of -/+0.01 for every +/-0.05 change in water-cement ratio)
- Therefore , corrected volume of Coarse Aggregate for w/c ratio of 0.30 = 0.66
- for pumpable concrete these values are reduced by 10%
- Therefore Volume of Coarse aggregate = 0.66 X 0.9 Volume of Coarse aggreagte = 0.594
- Volume of Fine Aggregate = 1 - 0.585 Volume of Fine Aggregate = 0.406
STEP 6. MIX CALCULATION
| DESCRIPTION | FORMULAE | ANSWERS |
| Volume of Concrete | = 1m^3 | |
| Volume of Cement | = (Mass of Cement/Specific gravity of cement) X (1/1000) |
= (466.67/3.15) X (1/1000) = 0.148 m^3 |
| Volume of Water | = (Mass of Water/Specific gravity of Water) X (1/1000) |
= (140/1) X (1/1000) = 0.140 m^3 |
| Volume of Chemical Admixture | = (Mass of chemical admixture/Specific gravity of chemical admixture) X (1/1000) |
= (7.6/1.145) X (1/1000) = 0.006 m^3 |
| Volume of Total Aggregate | = 1 - (Volume of Cement + Volume of Water + + Volume of chemical admixture ) |
= 1- (0.148 + 0.140 + 0.006) = 1 - 0.294 = 0.706 m^3 |
| Mass of Coarse Aggregate | = Volume of Total Aggregate X Volume of coarse agggregate X Specific gravity of Coarse aggregate X 1000 |
= 0.303 X 0.594 X 2.74 X 1000 = 493 kg/m^3 |
| Mass of Fine Aggregate | = Volume of Total Aggregate X Volume of Fine agggregate X Specific gravity of Fine aggregate X 1000 |
= 0.303 X 0.406 X 2.74 X 1000 = 337 kg/m^3 |
STEP 7. MIX DESIGN RESULTS
FOR M35 GRADE OF CONCRETE WITH ADDITION OF FLY ASH REQUIRED QUANTITY ARE AS FOLLOW :
- MASS OF CEMENT = 466 kg/m^3
- MASS OF WATER = 140 kg/m^3
- MASS OF FINE AGGREGATE = 337 kg/m^3
- MASS OF COARSE AGGREGATE = 493 kg/m^3
- MASS OF ADMIXTURE =7.6 kg/m^3
- WATER/CEMENT RATIO = 0.30
MIX RATIO w.r.t CEMENT : SAND (FINE AGGREGATE) : COARSE AGGREGATE IS 1 : 0.72 : 1.05
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