Project on Concrete Mix Design for various grades of Concrete
MIX PROPORTIONING OF CONCRTE
M35 GRADE WITH FLY ASH
1 STIPULATIONS FOR PROPORTIONS
a. Grade designation : M35
b. Type of cement : OPC 53 grade conforming to IS 12269 (1987)
c. Type of mineral admixture : Fly ash conforming to IS 3812 (Part 1)
d. Maximum nominal size of aggregate : 20mm
e. Minimum cement content and maximum
Water –cement ratio to be adopted and/ or : Severe (for reinforced concrete)
Exposure conditions are per table 3 and
Table 5 of IS 456 : 2000
f. Workability : 100mm (slump)
g. Method of concrete placing : Pumping
h. Degree of supervision : Good
i. Type of aggregate : Crushed angular aggregate
j. Maximum cement content (OPC content) : As per IS 456
2 TEST DATA OF MATERIALS
a. Cement : OPC 53 grade conforming to IS 12269
b. Specific gravity of cement : 3.15
c. Fly ash : Conforming to IS 3812 (part 1)
d. Specific gravity of fly ash : 2.2
e. Specific Gravities of
f. Coarse aggregate (at SSD condition) : 2.7 (based on saturated surface dry condition)
g. Fine aggregate (at SSD condition) : 2.6 (based on saturated surface dry condition)
h.Water absorption:
i. Coarse Aggregate :0.5%
ii. Fine Aggregate : 1%
i. Moisture content of Aggregate
[As per IS 2386 (Part 3)]
i. Coarse aggregate : Nil
ii. Fine aggregate : Nil
j. Grading of Fine Aggregate : ZONE 1 conforming to Table 9 of IS 383
3 TARGET MEAN STRENGTH
f’ck = fck+ 1.65 S
or
f’ck = fck + X
Whichever is higher
Where,
f’ck = target average compressive strength at 28 days,
fck = characteristic compressive strength at 28 days,
S = standard deviation, and
From Table 2, standard deviation S = 5 N/mm2
From Table 1, X = 6.5.
Therefore, target strength using both equations, that is,
a) f’ck = fck +1.65 S
= 35 + 1.65 x 5 = 43.25 N/mm2
b) f’ck = fck + X
= 35 + 6.5 = 41.5 N/mm2
The higher value is to be adopted. Therefore, target strength will be 48.25 N/mm2
as 43.25 N/mm2 > 41.5 N/mm2.
4 SELECTION OF WATER- CEMENT RATIO
From the Figure 1 of 10262: 2019, Water-cement ratio required for the target strength of 43.25 N/mm2 is taken as 0.45 for OPC 53 grade curve. Hence maximum water cement ratio is also 0.45 for ‘Very Severe’ exposure for reinforced concrete as per Table 5 of IS: 456-2000
0.45 w/c from both references, hence O.K.
5 SELECTION OF WATER CONTENT
From Table 4 of IS:10262 - 2019,
Water content= 186 kg (for 50 mm slump) for 20 mm aggregate. Estimated water content for 100 mm slump (increasing at the rate of 3 percent for every 25 mm slump).
As here, for 100mm slump, 6 percent of increase in water content is necessary.
= 186 + (6% of 186L)
= 186 + 186 x (6/100)
= 186 + 11.16
= 197.16 kg or L
As super plasticizer is used, the water content may be reduced. Based on trial data, the water content reduction of 23 percent is considered while using super plasticizer at the rate of 1.0 percent by weight of cement.
Hence, the arrived water content = 197.16L x 0.77
= 151.81kg /L
6 CALCULATION OF CEMENT CONTENT
Water- cement ratio = 0.45
Cement content = Water content / Water – cement ratio
= 151.81/ 0.45 = 337.36 kg/m3
From Table 5 of IS 456, minimum cement content for ‘severe’ exposure condition = 320 kg/m3
As 337.36 kg/m3 > 320 kg/m3, hence okay with the cement content
As fly ash is used as admixture, 10 % of the cementitious material content is increased
a) Cementitious content = 337.36 x 1.1 = 371.1 kg/m3
b) Water content = 151.81 kg/m3
c) water – cementitious ratio
= 151.81/371.1 = 0.41
d) Flyash @ 30 % of total cementitious material content,
= 371.1 x 30%
= 111.33 kg/m3
e) Cement (OPC) = 371.1- 111.33 = 259.77 kg/m3
From table 5 of IS 456, minimum cementitious content for ‘very severe’ exposure condition is 340 kg/m3. Hence okay
Hence, we save = 337.372 – 259.77 = 77.6 kg of cement, by using fly-ash @30% replacement.
7 SELECTION AND CALCULATION OF COARSE AGGREGATE
From Table 5 of IS 10262: 2019, Volume of coarse aggregate corresponding to 20 mm size aggregate and fine aggregate (Zone I) for water-cement ratio of 0.50 = 0.6
In the present case water-cementitious ratio is 0.41
Therefore, volume of coarse aggregate is required to be increased to decrease the fine aggregate content. As the water-cement ratio is lower by 0.09, the proportion of volume of coarse aggregate is increased by 0.018 (at the rate of 0.01 for every ± 0.05 change in water-cement ratio). Therefore, corrected proportion of volume of coarse aggregate for the water-cementitious ratio of 0.41 = 0.60 + 0.018 = 0.618
For pumpable concrete, these values may be reduced by up to 10 percent (From IS 10262: 2019, Para 5.5.2). So here, 10 percent reduction is considered.
Therefore, volume of coarse aggregate = 0.618 × 0.9 = 0.5562, say 0.556 m3
8 CALCULATION OF FINE AGGREGATE
Volume of fine aggregate content = Volume of All aggregates – Volume of coarse aggregate
= 1 – 0.556 = 0.444 m3
9 MIX CALCULATIONS
The mix calculations per unit volume of concrete shall be as follows:
a) Total Volume of concrete = 1m3
b) Volume of cement = (Mass of cement/ Specific gravity of cement) x (1/1000)
= (259.77/3.15) (1/1000) = 0.082m3
c) Volume of water = (Mass of water / specific gravity of water) x (1/1000)
= (151.81/1) (1/1000) =0.152m3
d) Volume of fly ash = (Mass of fly ash / specific gravity of fly ash) x (1/1000)
= (111.33/2.2) (1/1000) = 0.0506 m3
e) Volume of Chemical admixture (superplasticizer) @ 1.0 percentage (0.01) by mass of cementitious material
= (mass of chemical admixture / specific gravity of admixture) x (1/1000)
= ((371.1 x 0.01) (1/1000) = 0.003711m3
f) Volume of all in aggregates = (Volume of concrete) – (Volume of water + volume of fly ash + volume of chemical admixture)
= 1– 0.082- 0.152 – 0.0506 – 0.003711 =0.712m3
g) Mass of fine aggregate =
=0.712 x 0.444 x 2.6 x 1000 = 821.933 kg
h) Mass of coarse aggregate
= 0.712 x 0.556 x 2.7 x 1000 = 1068.85 kg
10 MIX PROPORTIONS
Cement = 259.77 kg/m3
Fly Ash = 111.33 kg/m3
Water = 152 kg/m3
Fine Aggregate (SSD) = 821.93 kg/m3
Coarse Aggregate (SSD) = 1068.85 kg/m3
Chemical admixture = 3.711 kg/m3
Free water- cementitious materials ratio = 0.41
(Mass of Cement / Mass of cement): (Mass of Fine Aggregate / Mass of cement): (Mass of Coarse aggregate / Mass of cement)
(259.77 /259.77): (821.93/259.77): (1068.85/ 259.77)
MIX PROPORTIONS = 1: 3.164 : 4.11
MIX PROPORTIONING OF CONCRTE
M50 GRADE
1 STIPULATIONS FOR PROPORTIONS
a. Grade designation : M50
b. Type of cement : OPC 53 grade conforming to IS 12269 (1987)
c. Maximum nominal size of aggregate : 20mm
d. Minimum cement content and maximum
Water –cement ratio to be adopted and/ or : Severe (for reinforced concrete)
Exposure conditions are per table 3 and
Table 5 of IS 456 : 2000
e. Workability : 100mm (slump)
f. Method of concrete placing : Pumping
g. Degree of supervision : Good
h. Type of aggregate : Crushed angular aggregate
i. Maximum cement content (OPC content) : As per IS 456
2 TEST DATA OF MATERIALS
a. Cement : OPC 53 grade conforming to IS:12269
b. Specific gravity of cement : 3.09
e. Specific Gravities of
Coarse aggregate (at SSD condition) : 2.74 (based on saturated surface dry condition)
Fine aggregate (at SSD condition) : 2.68 (based on saturated surface dry condition)
h.Water absorption:
i. Coarse Aggregate :0.5%
ii. Fine Aggregate : 1%
i. Moisture content of Aggregate
[As per IS 2386 (Part 3)]
i. Coarse aggregate : Nil
ii. Fine aggregate : Nil
j. Grading of Fine Aggregate : ZONE II conforming to Table 9 of IS 383
3 TARGET MEAN STRENGTH
f’ck = fck+ 1.65 S
or
f’ck = fck + X
Whichever is higher
Where,
f’ck = target average compressive strength at 28 days,
fck = characteristic compressive strength at 28 days,
S = standard deviation, and
From Table 2, standard deviation S = 5 N/mm2
From Table 1, X = 6.5.
Therefore, target strength using both equations, that is,
a) f’ck = fck +1.65 S
= 50 + 1.65 x 5 = 58.25 N/mm2
b) f’ck = fck + X
= 50 + 6.5 = 56.5 N/mm2
The higher value is to be adopted. Therefore, target strength will be 58.25 N/mm2
as 58.25 N/mm2 > 56.5 N/mm2.
4 SELECTION OF WATER- CEMENT RATIO
From the Figure 1 of 10262: 2019, Water-cement ratio required for the target strength of 58.25 N/mm2 is taken as 0.34 for OPC 53 grade curve. Hence maximum water cement ratio is also 0.4 for ‘Extreme’ exposure for reinforced concrete as per Table 5 of IS: 456-2000
0.34 w/c is choosen from both references, hence O.K.
5 SELECTION OF WATER CONTENT
From Table 4 of IS:10262 - 2019,
Water content= 186 kg (for 50 mm slump) for 20 mm aggregate. Estimated water content for 100 mm slump (increasing at the rate of 3 percent for every 25 mm slump).
As here, for 100mm slump, 6 percent of increase in water content is necessary.
= 186 + (6% of 186L)
= 186 + 186 x (6/100)
= 186 + 11.16
= 197.16 kg or L
As super plasticizer is used, the water content may be reduced. Based on trial data, the water content reduction of 23 percent is considered while using super plasticizer at the rate of 1.0 percent by weight of cement.
Hence, the arrived water content = 197.16L x 0.77
= 151.81kg /L
6 CALCULATION OF CEMENT CONTENT
Water- cement ratio = 0.34
Cement content = Water content / Water – cement ratio
= 151.81/ 0.34 = 446.5 kg/m3
From Table 5 of IS 456, minimum cement content for ‘severe’ exposure condition = 360 kg/m3
As 446.5 kg/m3 > 360 kg/m3, hence okay with the cement content
7 SELECTION AND CALCULATION OF COARSE AGGREGATE
From Table 5 of IS 10262: 2019, Volume of coarse aggregate corresponding to 20 mm size aggregate and fine aggregate (Zone II) for water-cement ratio of 0.50 = 0.62
But water-cement ratio is 0.34
Therefore, volume of coarse aggregate is required to be increased to decrease the fine aggregate content. As the water-cement ratio is lower by 0.16, the proportion of volume of coarse aggregate is increased by 0.032 (at the rate of 0.01 for every ± 0.05 change in water-cement ratio). Therefore, corrected proportion of volume of coarse aggregate for the water-cementitious ratio of 0.34 = 0.62 + 0.032 = 0.652
For pumpable concrete, these values may be reduced by up to 10 percent (From IS 10262: 2019, Para 5.5.2). So here, 10 percent reduction is considered.
Therefore, volume of coarse aggregate = 0.652 × 0.9 = 0.5868, say 0.587 m3
8 CALCULATION OF FINE AGGREGATE
Volume of fine aggregate content = Volume of All aggregates – Volume of coarse aggregate
= 1 – 0.587 = 0.413 m3
9 MIX CALCULATIONS
The mix calculations per unit volume of concrete shall be as follows:
a) Total Volume of concrete = 1m3
b) Volume of cement = (Mass of cement/ Specific gravity of cement) x (1/1000)
= (446.5/3.09) (1/1000) = 0.1443m3
c) Volume of water = (Mass of water / specific gravity of water) x (1/1000)
= (151.81/1) (1/1000) =0.152m3
d) Volume of Chemical admixture (superplasticizer) @ 1.0 percentage (0.01) by mass of cementitious material
= (mass of chemical admixture / specific gravity of admixture) x (1/1000)
= ((446.5x 0.01) (1/1000) = 0.004465m3
f) Volume of all in aggregates = (Volume of concrete) – (Volume of water + volume of chemical admixture)
= 1– 0.1443- 0.152 – 0.004465 =0.7032m3
g) Mass of fine aggregate =
=0.7032 x 0.413 x 2.68 x 1000 = 778.33 kg
h) Mass of coarse aggregate
= 0.7032 x 0.587 x 2.74 x 1000 = 1131 kg
10 MIX PROPORTIONS
Cement = 446.5 kg/m3
Water = 152 kg/m3
Fine Aggregate (SSD) = 778.33 kg/m3
Coarse Aggregate (SSD) = 1131 kg/m3
Chemical admixture = 4.46 kg/m3
(Mass of Cement / Mass of cement): (Mass of Fine Aggregate / Mass of cement): (Mass of Coarse aggregate / Mass of cement)
(446.5 /446.5): (778.33/446.5): (1131/ 446.5)
MIX PROPORTIONS = 1: 1.743 : 2.53