Project on Concrete Mix Design for various grades of Concrete
Given:
- M35 = 35 N/mm^2
- OPC 43 = 43 N/mm^2
- Nominal Size = 20mm
- Minimum cement ratio = 320 kg/m^3
- Maximum Water-Cement Ratio = 0.45
- Workabilty = 100mm
- Exposure = Severe
- Maximum Cement content = 450 kg/m^3
Test Data for material
- Cement used = OPC 43 Grade
- Specific Gravity = 3.15
- Specific Gravity of Coarse aggregate = 2.75 & Fine aggregate = 2.74
- Water absorbtion ; Coarse aggregate = 0.5% , Fine aggregate = 1.0%
- Free surface: Nil
Target Strength for mix proportion
- f'ck = fck+1.65S
f'ck = 35 + 1.65x5 ; Where s= 5 N/mm^2
f'ck = 43.25 kg/m^3
- Max w/c ratio = 0.45
0.40<0.45, Hence ok.
Water content table of ISI 10262:2009
- Maximum water content for 20mm aggregate = 186 Litres for (25mm-60mm) slump range
- Estimate water for 100 mm Slump = 186+(6/100)+186 = 197 Litres
- As superplasticizer is used it can be reduced upto 20%. Based on trials water content reduction is of 29% is adivised.
- Hence, the arrival water content = 197x0.71 = 140 litre
2) Calculation of Cement :
W/C ratio = 0.4
Cement Content =140/0.4 = 350 kg/m^3
From IS 456, Minumum cement content for severe condition =320 kg/m^3
350 kg/m^3 > 320 kg/m^3, Hence Ok.
Volume of Coarse Aggregate = 0.62x0.9 = 0.56
Volume of Fine Aggregate = 1-0.56 = 0.44
As, w/c ratio its lower by 0.10, the proportion volume of coarse aggregate is increased by 0.02 (+/- 0.01 for every -/+ 0.05 change in w/c ratio)
Hence, corrected volume of coarse aggregate for w/c ratio of 0.40 = 0.62
MIX CALCULATION:
A) Volume of Concrete = 1 m^3
B) Volume of cement = (Mass of cement / sp.gr of cement)x(1/1000)
= (350/3.15)x(1/1000)
= 0.11 m^3
C) Volume of water = (Mass of water / sp.gr of water)x(1/1000)
= (140/1)x(1/1000)
= 0.14 m^3
D) Volume of chemical admixture = (Mass of chemical / sp.gr of chemical)x(1/1000)
= (7/1.145)x(1/1000)
=0.006 m^3
E) Volume of Aggregate ( Fine & Coarse) = (a-(b+c+d))
=(1-(0.11+0.14+0.006))
= 0.744 m^3
F) Mass of Coarse aggregate = E x Volume of coarse aggregate x Sp. Gr. of coarse aggregate x 1000
= 0.744 x 0.56 x 2.75 x 1000
= 1140 kg/m^3
G) Mass of Fine Aggregate = E x Volume of Fine aggregate x Sp. Gr. of fine aggregate x 1000
= 0.744 x 0.62 x 2.74 x 1000
= 1263.90 kg/m^3
Given:
- M50 = 50 N/mm^2
- OPC 43 = 43 N/mm^2
- Nominal Size = 20mm
- Minimum cement ratio = 320 kg/m^3
- Maximum Water-Cement Ratio = 0.45
- Workabilty = 100mm
- Exposure = Severe
- Maximum Cement content = 450 kg/m^3
Test Data for material
- Cement used = OPC 43 Grade
- Specific Gravity = 3.15
- Specific Gravity of Coarse aggregate = 2.75 & Fine aggregate = 2.74
- Water absorbtion ; Coarse aggregate = 0.5% , Fine aggregate = 1.0%
- Free surface: Nil
Target Strength for mix proportion
- f'ck = fck+1.65S
f'ck = 50 + 1.65x5 ; Where s= 5 N/mm^2
f'ck = 58.25 kg/m^3
- Max w/c ratio = 0.45
0.40<0.45, Hence ok.
Water content table of ISI 10262:2009
- Maximum water content for 20mm aggregate = 186 Litres for (25mm-60mm) slump range
- Estimate water for 100 mm Slump = 186+(6/100)+186 = 197 Litres
- As superplasticizer is used it can be reduced upto 20%. Based on trials water content reduction is of 29% is adivised.
- Hence, the arrival water content = 197x0.71 = 140 litre
2) Calculation of Cement :
W/C ratio = 0.4
Cement Content =140/0.4 = 350 kg/m^3
From IS 456, Minumum cement content for severe condition =320 kg/m^3
350 kg/m^3 > 320 kg/m^3, Hence Ok.
Volume of Coarse Aggregate = 0.62x0.9 = 0.56
Volume of Fine Aggregate = 1-0.56 = 0.44
As, w/c ratio its lower by 0.10, the proportion volume of coarse aggregate is increased by 0.02 (+/- 0.01 for every -/+ 0.05 change in w/c ratio)
Hence, corrected volume of coarse aggregate for w/c ratio of 0.40 = 0.62
MIX CALCULATION:
A) Volume of Concrete = 1 m^3
B) Volume of cement = (Mass of cement / sp.gr of cement)x(1/1000)
= (350/3.15)x(1/1000)
= 0.11 m^3
C) Volume of water = (Mass of water / sp.gr of water)x(1/1000)
= (140/1)x(1/1000)
= 0.14 m^3
D) Volume of chemical admixture = (Mass of chemical / sp.gr of chemical)x(1/1000)
= (7/1.145)x(1/1000)
=0.006 m^3
E) Volume of Aggregate ( Fine & Coarse) = (a-(b+c+d))
=(1-(0.11+0.14+0.006))
= 0.744 m^3
F) Mass of Coarse aggregate = E x Volume of coarse aggregate x Sp. Gr. of coarse aggregate x 1000
= 0.744 x 0.56 x 2.75 x 1000
= 1140 kg/m^3
G) Mass of Fine Aggregate = E x Volume of Fine aggregate x Sp. Gr. of fine aggregate x 1000
= 0.744 x 0.62 x 2.74 x 1000
= 1263.90 kg/m^3