Project on Concrete Mix Design for various grades of Concrete
1) i) Mix Proportioning of Concrete using Fly ash as part of replacement of OPC
1. Stipulations for proportioning
a) Grade designation: M35
b) Type of cement: OPC 43 grade conforming to 15 8112
c) Type of mineral admixture: Fly ash conforming to IS 3812 (Part d) Maximum nominal size of aggregate: 20 mm
e) Minimum cement content: 360 kg/m^3 (From Table 5 of 15 456: 2000- Exposure condition)
f) Maximum water-cement ratio : 0.40 (From Table 5 of 15 456: 2000-Exposure condition) 8) Workability: 25-50 mm (slump value for pumpable concrete)
h) Exposure condition: Extreme (for reinforced concrete)
h) Method of concrete placing: Pumping
i) Degree of supervision: Good
i) Type of aggregate: Crushed angular aggregate
k) Maximum cement content: 450 kg/m^3
1) Chemical admixture type: Superplasticizer
II. Test data for materials:
a) Cement used: OPC 43 grade conforming to IS 8112
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) Chemical admixture: Superplasticizer conforming to IS 9103
f)
Specific gravity of:
• Coarse aggregate :2.74
• Fine aggregate
:2.74
d)
Water absorption:
Coarse aggregate :0.5%
Fine aggregate
: 1.0%
e) Free Surface moisture:
• Coarse aggregate
: Nil
Fine aggregate
: Nil
Target Strength for Mix proportioning:
f'ck = fck + 1.65 S
Where, f'ck Target average compressive strength fck Characteristic compressive strength at 28 days
at 28 days
S = Standard deviation
From Table 1 of IS 10262: 2009, standard deviation, S = 5 N/mm^2
Therefore, Target strength =35 + ( 1.65 x 5) = 43.25 N/mm^2
IV. Selection of Water-cement ratio:
From Table 5 of IS 456: 2000 for severe exposure, maximum water-cement ratio = 0.45 Based on experience, adopt water-cement ratio as 0.40
0.40 < 0.45, Hence O.K
V. Selection of Water content
From Table 2 of IS 10262: 2009, maximum water content for 20mm aggregate = 186 litre (for 25mm to 50mm slump range)
As superplasticizer is used, the water content can be reduced up to 20%
Hence the arrived water content =
186 x 80/100 = 148.8 (100-20 = 80)
Water content = 149 litres
VI. Calculation of Cement Content:
Water-cement ratio = 0.4
Cementitious content = c=149/0.40 = 372.5 = 370 kg/m^3
Minimum cement content = 360 kg/m^3
360< 370 kg/m^3 ( hence okay )
Now, to proportion a mix containing
fly ash the following steps are suggested:
a) The percentage of fly ash needs to be decided based on project requirement and quality of materials b) In some cases increase in cementitious material can be warranted. Increase in cementitious material content and its percentage may be based on experience and trial
Increase by 10 %
370 x 0.10 + 370 = 407 kg/m^3
W= 149 kg
Water cement ratio = 149/407 = 0.366
Fly ash at 30 % total cementitious material content
= 407 x 30/100 = 122 kg/m^3
Cement (ordinary Portland cement ) = 407-122 = 285 kg/m^3
Saving of cement while using fly ash = 370 – 285 = 85 kg/ m^3
Fly ash = 122 kg/m^3
VOLUME OF COARSE AGGREGATES AND FINE AGGREGATES
Fine aggregates are sieved according to zone 1 of table 4 IS 383
Coarse aggregates are sieved according to table 2 IS 383
According ,
Table 3 IS 10262: 2009
Nominal size of aggregates = 20 mm
Fine aggregates ( zone 1)
Therefore ,
Volume of coarse aggregate = 0.60
Water cement ratio = 0.4
( +/- 0.01 for every +/- 0.005)
So, volume of coarse aggregate = 0.62
Volume of fine aggregate = 1-0.62 = 0.38
Mix Calculations:
The mix calculation per unit volume of concrete shall be as follows:
a) Volume of Concrete = 1 m^3
b) Volume of Cement =
Mass of cement
Specific gravity of cement 1000
=(285 / 3.15) x1/1000= 0.090 m^3
c) Volume of Fly ash=
Mass of fly ash
Specific gravity of fly ash 1000
( 122/2.2) x1/1000=0.055 m^3
Mass of water
d) Volume of water
= Specific gravity of water 1000
( 149/1)x 1/1000=0.149 m^3
e) Volume of chemical admixture
Mass of chemical admixture Specific gravity of admixture
( 7.6/ 1.145)x1/1000=0.006 m^3
f) Volume of aggregate (FA & CA)
a – ( b + c + d+
e) = 1-( 0.090 + 0.055 +0.149+ 0.006 )= 0.7 m^3
g) Mass of Coarse aggregate = fx Volume of coarse aggregate x Specific gravity of
coarse aggregate X 1000
- 0.7 x 0.62 x 2.74 x1000 = 1189.16 kg/m^3
h) Mass of fine aggregate
= fx Volume of fine aggregate x Specific gravity of fine
aggregate X1000
- 0.7 x 0.38 x 2.74 x1000 = 728 kg/m^3
IX. Mix proportions for Trial Number
Cement = 285 kg/m^3
Fly ash = 122 kg/m^3
Water = 149 kg
Coarse aggregate = 1189 kg/m^3
Fine aggregate = 728 kg/m^3
Chemical admixture = 7.6 kg/m^3
Water cement ratio = 0.366
Trial mix ratio : 285 :728 : 1189 = 1: 2.55 : 4.17 ( C: FA: CA)
ii) M50 grade without flyash:
Target Strength for Mix proportioning:
f'ck = fck + 1.65 S
Where, f'ck Target average compressive strength fck Characteristic compressive strength at 28 days
at 28 days
S = Standard deviation
From Table 1 of IS 10262: 2009, standard deviation, S = 5 N/mm^2
Therefore, Target strength =
f ‘ck = 50 + ( 1.65 x 5)
= 58.25 N/mm^2
SELECTION OF WATER CEMENT RATIO :
Given :
M50
Assume:
Exposure condition : Extreme ( for reinforced concrete)
Table 5 – IS 456 : 2000
Therefore,
Extreme= w/c ratio= 0.40
M50 = w/c ratio= 0.40
Lower the water cement ratio better the workability
0.40
Water cement ratio = 0.40
Minimum cement content = 360 kg/m^3
SELECTION OF WATER CONTENT
Assume :
Nominal size of aggregates= 20 mm
Minimum water content = 186 litres or kg
Slump ranges from 25 to 50 mm
As superplasticizers (20 % reduced)
= 186 x 80/100 = 148.8 (100-20 = 80)
Water content = 149 litres
4.CALCULATION OF CEMENT CONTENT
Water cement ratio = w/c= 0.40
Water content = W = 149 kg
Cement content = c=149/0.40 = 372.5 = 373 kg/m^3
Minimum cement content = 360 kg/m^3
360< 373 kg/m^3 ( hence okay )
5.VOLUME OF COARSE AGGREGATES AND FINE AGGREGATES
Fine aggregates are sieved according to zone 1 of table 4 IS 383
Coarse aggregates are sieved according to table 2 IS 383
According ,
Table 3 IS 10262: 2009
Nominal size of aggregates = 20 mm
Fine aggregates ( zone 1)
Therefore ,
Volume of coarse aggregate = 0.60
Water cement ratio = 0.4
(+/- 0.01 for every +/- 0.005)
So, volume of coarse aggregate = 0.62
Volume of fine aggregate = 1-0.62 = 0.38
6.MIX CALCULATION :
- Volume of concrete = 1m^3
- Volume of cement = (373/3.15)x1/1000 = 0.118 m^3
- Volume of water = ( 149/1)x1/1000= 0.149 m^3
- Volume of admixture = ( 7.6/1.145)x1/1000 = 0.006
- Volume of coarse aggregate and fine aggregate = a – ( b + c + d) = 1-( 0.118 + 0.149 + 0.006 )= 0.727 m^3
- Mass of coarse aggregate = 0.727 x 0.62 x 2.74 x1000 = 1235.02 kg/m^3
- Mass of fine aggregate = 0.727 x 0.38 x 2.74 x1000 = 756.9 kg/m^3
RESULTS AND CONCLUSION :
TRIAL MIX:
Cement = 373 kg/m^3
Water = 149 kg
Coarse aggregate = 1235.02 kg/m^3
Fine aggregate = 756.9 kg/m^3
Chemical admixture = 7.6 kg/m^3
Water cement ratio = 0.40
Trial mix ratio : 373 :756.9 : 1235.02 = 1: 2.02: 3.31 ( C: FA: CA)