Project on Concrete Mix Design for various grades of Concrete
1. AIM:- CALCULATION OF DESIGN MIX FOR M35 GRADE OF CONCRETE USING FLY ASH.
I. STIPULATIONS FOR PROPORTION OF MIX DESIGN
| GRADE OF CONCRETE | M35 |
| GRADE OF CEMENT | OPC 53 |
| EXPOSURE CONDITION | SEVERE |
| WORKABILITY | 125mm |
| MAX. NOMINAL SIZE OF AGGREGATE | 20mm |
| SHAPE OF AGGREGATE | CRUSHED ANGULAR AGGREGATES |
| METHOD OF PLACING | PUMPING |
| TYPE OF ADMIXTURE | FLY ASH |
| CEHMICAL ADMIXTURE | SUPERPLASTICIZER |
| MIN. CEMENT CONTENT | 320kg/m^3 (REINFORCED CONCRETE) |
| MAX. WATER CEMENT RATIO | 0.45 TABLE 5 IS 456:2000 |
| MAX. CEMENT CONTENT (OPC) | 450kg/m^3 |
II. TEST DATA MATERIALS:
Values of materials while working in field may range from its standard value,so trial and laboratory tests are performed,the values recieved from test for M35 grade of concrete are:
| CEMENT USED | OPC 53 CONFIRMING TO IS 8112 |
| SP. GRAVITY OF CEMENT | 3.15 |
| FLY ASH | CONFIRMING TO IS 3812 PART1 |
| SP. GRAVITY OF FLY ASH | 2.4 |
| CHEMICAL ADMIXTURE | CONFORMING TO IS 9103 |
| SP. GARVITY OF COARSE AGGREGATE | 2.74 IS 2386 |
| SP. GARVITY OF FINE AGGREGATE | 2.72 IS 2386 |
| SP. GRAVITY OF ADMIXTURE | 1.145 IS 2386 |
| VOL. OF COARSE AGGREGATE | ZONE II CONFIRMING IS 383 TABLE2 |
III. TARGET MEAN STRENGTH OF CONCRETE(F'ck):
Target means strength of concrete can calculated by formula F'ck= (Fck+1.65xS) where 'S' is Standard deviation,'Fck' is compressive strength of concrete after 28 days.
value of 'S' is taken from Table1 IS 10262:2009 according to the grade of concrete. Now,for M35, S=5.0 and Fck=35 N/mm^2.
so F'ck=35 + 1.65 x 5.0 =43.25 N/mm^2
IV. WATER CEMENT RATIO CALCULATION:
Now for water cement calculation we refer to Table5 IS 456:2000 ,according to the SEVERE exposure condition maximum water cement ratio for reinforcing concrete is 0.45 , from the trial batcehs we found water content suitable of 0.4 (since 0.4< 0.45 Hence OK.)
Now adopted w/c=0.4
V. SELECTION OF WATER CONTENT:
| MAX. NOMINAL SIZE OF AGGREGATE | MAX WATER CONTENT |
| 20mm | 186Ltr |
The above value is from Table2 IS 10262:2009 , Above water-content is for 25 to 50mm slump but the slump we require is 125mm. hence we need to increase the water content to get the required slump for good workability. Clause 4.2 of IS10262 suggests an increase of 3% of water content from its original value for every increase of 25mm. Here we have a 75mm increase in the water content hence we have to increase the water content by 9%.
186+0.09*186= 202.74 or 203 Ltr (approx)
Since superplasticizers are used water content can be reduced to get good strength of concrete. Percentage of reduction of water depends on the type of superplasticizers used in the concrete. Superlasticizer reduce water content by 30%,but in trials we found it get reduced to 29%.
203*0.71=144 Ltr
VI. CALCULATION OF CEMENT AND FLY ASH CONTENT:
Cement content is now calculated from the water-cement ratio :
w/c=0.4 i.e. 144/c=0.4
Now Cement content c=144/0.4= 360kg/m^3 , Now the cement content is 360kg/m^3> 320kg/m^3 (greater than minimum cement content required ) Hence OK.
Since Fly ash is used,the amount of Cementitious content (Cement+Fly ash) shall be increased to get good workability and strength. the percentage of increase depending on experience and research. Here we increase the cementitious content by 1.1 %.
Now ,New cement content is :- 360*1.1 =396kg/m^3 .
New water cement ratio according to the cementitious content added :- 144/396=0.36
Now from the cementitious content added ,30% contained Fly ash, So the fly ash content is:- 396*0.3= 118.8kg/m^3 or 119kg/m^3 approx
Cement content requirement(OPC)= 396-119 =277kg/m^3 (Total quantity of only cement content)
Quantity of cement saved by using, fly ash in the mix = 360-277 =83kg/m^3
VII. CALCULATION OF COARSE AND FINE AGGREGATE:
Table3 IS10262:2009
| NOMINAL MAX. SIZE OF AGGREGATE | VOLUME OF COASRSE AGGREGATE PER UNIT VOLUME OF TOTAL AGGREGATE |
| 20mm | 0.62 (Zone II) |
we have water cement ratio of 0.4 from Table5 of IS 456:2000 for severe exposure condition. But for the mix design, for every increase or decrease of w/c by 0.05 we decrease or increase coarse aggregate volume by 0.01 respectively. Therefore for 0.4 water content the volume of coarse aggregate is 0.64
Because the placing of concrete is done by pumping, the coarse aggregagate may find difficulty to go in between the reinforcement, so the volume of coarse aggregate is to be reduced by 10%.
so, Now the volume of Coarse aggregate is 0.64*0.9= 0.576 or 0.58 approx
Volume of Fine aggregate= 1-0.58=0.42
VIII. MIX DESIGN CALCULATION OF M35 GRADE OF CONCRETE- AS PER IS 10262:2009
a.volume of concrete = 1m^3
b.volume of cement =[mass of cement or OPC/(sp. gravity of cement * 1000)]
= [277/(3.15 * 1000)] = 0.088m^3
c.volume of fly ash =[mass of fly ash/(sp. gravity of fly ash * 1000)]
= [119/(2.4 * 1000)] = 0.049m^3
d.volume of chemical admixture =[mass of admixture/(sp. gravity admixture * 1000)]
Mass of admixture is taken 1% of cement (OPC+fly ash)= (277+119)*0.01 = 3.96kg/m^3
= [3.96/(1.145 * 1000)] = 0.003m^3
e.volume of water =[mass of water/(sp. gravity of water * 1000)]
= [144/(1 * 1000)] = 0.144m^3
f.Total aggregates(coarse+fine)=[a-(b+c+d+e)]
= [1-(0.088+0.049+0.003+0.144)] = 0.716m^3
g.Coarse aggregate =[Total aggregtaes * Sp. gravity of C.A * vol. of C.A *1000]
= [0.716*2.74*0.58*1000] = 1045kg/m^3
h.Fine aggregate=[Total aggregtaes * Sp. gravity of C.A * vol. of C.A *1000]
= [0.658*2.72*0.42*1000] = 751kg/m^3
IX. DESIGN PROPORTION OF M35 CONCRETE
| MATERIALS | QUANTITY |
| CEMENT(OPC) | 277 kg/m^3 |
| COARSE AGGREGATE | 1045 kg/m^3 |
| FINE AGGREGATE | 751 kg/m^3 |
| WATER | 144 kg/m^3 |
| FLY ASH | 168 kg/m^3 |
| CHEMICAL ADMIXTURE | 3.96 kg/m^3 |
| WATER CEMENT RATIO | 0.36 |
DESIGN MIX RATIO= cement: C.A : F.A : water = 1:3.77:2.71:0.52
NOTE- Aggregates should be used in saturated surface dry condition.
2. AIM:- CALCULATION OF DESIGN MIX FOR M50 GRADE OF CONCRETE WITHOUT FLY ASH.
I. STIPULATIONS FOR PROPORTION OF MIX DESIGN
| GRADE OF CONCRETE | M50 |
| GRADE OF CEMENT | OPC 53 |
| EXPOSURE CONDITION | EXTREME |
| WORKABILITY | 150mm |
| MAX. NOMINAL SIZE OF AGGREGATE | 20mm |
| SHAPE OF AGGREGATE | CRUSHED ANGULAR AGGREGATES |
| METHOD OF PLACING | PUMPING |
| CHEMICAL ADMIXTURE | SUPERPLASTICIZER |
| MIN. CEMENT CONTENT | 360kg/m^3 (REINFORCED CONCRETE) |
| MAX. WATER CEMENT RATIO | 0.45 TABLE 5 IS 456:2000 |
| VOL. OF COARSE AGGREGATE | 0.60 |
II. TEST DATA MATERIALS:
Values of materials while working in field may range from its standard value,so trial and laboratory tests are performed,the values recieved from test for M50 grade of concrete are:
| CEMENT USED | OPC 53 CONFIRMING TO IS 8112 |
| SP. GRAVITY OF CEMENT | 3.15 |
| CHEMICAL ADMIXTURE | CONFORMING TO IS 9103 |
| SP. GARVITY OF COARSE AGGREGATE | 2.65 IS 2386 |
| SP. GARVITY OF FINE AGGREGATE | 2.66 IS 2386 |
| SP. GRAVITY OF ADMIXTURE | 1.12 IS 2386 |
| VOL. OF COARSE AGGREGATE | ZONE I CONFIRMING IS 383 TABLE2 |
III. TARGET MEAN STRENGTH OF CONCRETE(F'ck):
Target means strength of concrete can calculated by formula F'ck= (Fck+1.65xS) where 'S' is Standard deviation,'Fck' is compressive strength of concrete after 28 days.
value of 'S' is taken from Table1 IS 10262:2009 according to the grade of concrete. Now for M50, S=5.0 and Fck=50 N/mm^2.
so F'ck=50 + 1.65 x 5.0 =58.25 N/mm^2
IV. WATER CEMENT RATIO CALCULATION:
Now for water cement calculation we refer to Table5 IS 456:2000 ,according to the EXTREME exposure condition maximum water cement ratio for reinforcing concrete is 0.45 , from the trial batches we found water content suitable of 0.35 (since 0.35< 0.45 Hence OK.)
Now adopted w/c=0.35
V. SELECTION OF WATER CONTENT:
| MAX. NOMINAL SIZE OF AGGREGATE | MAX WATER CONTENT |
| 20mm | 186Ltr (25-50mm slump) |
The above value is from Table2 IS 10262:2009 , Above water-content is for 25 to 50mm slump but the slump we require is 150mm. hence we need to increase the water content to get the required slump for good workability. Clause 4.2 of IS10262:2009 suggests an increase of 3% of water content from its original value for every increase of 25mm. Here we have a 75mm increase in the water content hence we have to increase the water content by 12%.
186+0.12*186= 208.32 or 208 kg/m^3 (approx)
Since superplasticizers are used water content can be reduced to get good strength of concrete. Percentage of reduction of water depends on the type of superplasticizers used in the concrete. Superplasticizer reduce water content by 30%,but in trials we found it get reduced to 29%.
208*0.71=147.68 Ltr or 148kg/m^3 (approx)
VI. CALCULATION OF CEMENT CONTENT:
Cement content is now calculated from the water-cement ratio :
w/c=0.35 i.e. 148/c=0.35
Now Cement content c=148/0.35= 422.85kg/m^3 or 423kg/m^3 approx , Now the cement content is 423kg> 360kg Table5 IS 4556:2000 (greater than minimum cement content required ) Hence OK.
VII. CALCULATION OF COARSE AND FINE AGGREGATE:
Table3 IS10262:2009
| NOMINAL MAX. SIZE OF AGGREGATE | VOLUME OF COASRSE AGGREGATE PER UNIT VOLUME OF TOTAL AGGREGATE |
| 20mm | 0.60 (Zone I) |
we have water cement ratio of 0.35 from the trial basis. But for the mix design, for every increase or decrease of w/c by 0.05 we decrease or increase coarse aggregate volume by 0.01 respectively. Therefore for 0.35 water cement ratio, the volume of coarse aggregate is 0.63
Because the placing of concrete is done by pumping, the coarse aggregagate may find difficulty to go in between the reinforcement, so the volume of coarse aggregate is to be reduced by 10%.
so, Now the volume of Coarse aggregate is 0.63*0.9= 0.567 or 0.57 approx
Volume of Fine aggregate= 1-0.57=0.43
VIII. MIX DESIGN CALCULATION OF M35 GRADE OF CONCRETE - AS PER IS 10262:2009
a.volume of concrete = 1m^3
b.volume of cement =[mass of cement /(sp. gravity of cement * 1000)]
= [423/(3.15 * 1000)] = 0.134m^3
c.volume of chemical admixture =[mass of admixture/(sp. gravity admixture * 1000)]
Mass of admixture we take 0.6% of cement= 423*0.006 = 2.54kg/m^3
= [2.54/(1.12* 1000)] = 0.002m^3
d.volume of water =[mass of water/(sp. gravity of water * 1000)]
= [148/(1 * 1000)] = 0.148m^3
e.Total aggregates(coarse+fine)=[a-(b+c+d)]
= [1-(0.134+0.002+0.148)] = 0.716m^3
f.Coarse aggregate =[Total aggregtaes * Sp. gravity of C.A * vol. of C.A *1000]
= [0.716*2.65*0.57*1000] = 1081kg/m^3
g.Fine aggregate =[Total aggregtaes * Sp. gravity of F.A* vol. of F.A *1000]
= [0.716*2.66*0.43*1000] = 819kg/m^3
IX. DESIGN PROPORTION OF M35 CONCRETE
| MATERIALS | QUANTITY |
| CEMENT | 423 kg/m^3 |
| COARSE AGGREGATE | 1081 kg/m^3 |
| FINE AGGREGATE | 819 kg/m^3 |
| WATER | 148 kg/m^3 |
| CHEMICAL ADMIXTURE | 2.54 kg/m^3 |
| WATER CEMENT RATIO | 0.35 |
DESIGN MIX RATIO= cement: C.A : F.A : water = 1:3.0:2.27:0.41
NOTE- Aggregates should be used in saturated surface dry condition.