Project on Concrete Mix Design for various grades of Concrete
1. Design mix calculation for M35 grade concrete with fly ash
Given,
- Grade of concrete=35N/mm^2
- Type of cement = OPC 43 grade
- Maximum nominal size of aggregate = 20mm
- Minimum cement content = 320 kg/m^3
- Maximum W/C ratio = 0.40
- Wokability = 100mm (Slump value for pumpable concrete)
- Exposure = severe
- Type of aggregate = 20mm Crushed angular
Test data,
- Cement used = 43 grade
- specific gravity of cement = 3.15
- specific gravity of Fly ash = 2.2
- Specific Gravity , Fine Aggregate = 2.74 , Course Aggregate = 2.74
- Water absortion , Fine aggregate = 1% , Course aggregate = 0.5%
i. To find target mean strength
f'ck = fck + 1.65(S)
f'ck = 35 + 1.65(5) = 43.25N/mm^2
ii. To water cement ratio
W/C ratio = 0.4 {From IS 456 2000}
iii. Selection of water content
From IS 10262 2009 Table 2 , maximum content for 20mm size aggregate is 186lts
Estimated water content for 100mm slump = 186 + .06*186 = 197lts
As Fly ash is used the water content is reduced by 29%
Water content = 197 * 0.71 * 197 = 140lts
iv. Calculation of cement content
water/cement = 0.4
Cement content = 140/0.4 = 350 kg/m^3 + Fly ash
To proportion the fly ash with cement:
- The % of fly ash needs to be decided based on the project requirments.
- In Concrete the increase in cementatious material can be varied. Increase in cementatious material and its increase in it's persentage may be based on experience
Cementatious material of concrete = 350 * 1.10 = 385 kg/m^3
water content = 140lts
So, w-c ratio = 140/385 = 0.364
Fly ash at 30% total cementatious content = 385 * 30% = 115kg/m^3
Cement = 385 -115 = 270 kg/m^3
Saving cement = 350 - 270 = 80kg/m^3
Fly ash being utilized = 115kg/m^3
v. proportion of volume of CA & FA
From Table 3 of IS 10262 2009, volume of CA corresponding to 20mm size & FA (Zone 1) for W/C ratio 0.35
CA & FA = .6
Therefore, Volume of CA needs to be increased and FA should be decreased.
As W/C ratio is lowered by 0.10, the proportion of volume of CA is increased by 0.02
Therefore, Corrected W/C ratio = 0.62
Volume of Course aggregate = .62 * .9 = 0.56
Volume of Fine Aggregate = 1 - 0.56 = 0.44
vi. Mix Calculation
- Volume of cement = 1m^3
- Volume of cement = 270/3.15 *1/1000 = 0.086m^3
- Volume of Fly ash = 115/2.2 *1/1000 = 0.052m^3
- Volume of water = 140/1000 = 0.140m^3
- Volume of Chemical admixture = 7.6/1/145 *1/1000 = 0.006m^3
- Volume of aggregate = 1 -(0.086+0.052+0.140+0.006) = 0.716m^3
- Mass of Course aggregate = 0.716 * 0.56*2.74*1000 = 1098kg/m^3
- Mass of Fine aggregate = 0.716 * 0.44*2.75*1000 = 863 kg/m^3
vii. Mix Proportion
- Cement = 270 kg/m^3
- Fly ash = 115 kg/m^3
- Water = 140lts
- Fine aggregate = 863 kg/m^3
- Course aggregate = 1098 kg/m^3
- Water-cement ratio = .364
MIX RATIO = 1 : 2.24 : 2.85
2. Design mix calculation for M50 grade concrete without fly ash
Given,
- Grade of concrete=50N/mm^2
- Type of cement = OPC 43 grade
- Maximum nominal size of aggregate = 20mm
- Minimum cement content = 320 kg/m^3
- Maximum W/C ratio = 0.40
- Wokability = 100mm (Slump value for pumpable concrete)
- Exposure = severe
- Type of aggregate = 20mm Crushed angular
Test data,
- Cement used = 43 grade
- specific gravity of cement = 3.15
- Chemical admixture = Super plasticizer conforming IS 9103
- Specific Gravity , Fine Aggregate = 2.74 , Course Aggregate = 2.74
- Water absortion , Fine aggregate = 1% , Course aggregate = 0.5%
i. To find target mean strength
f'ck = fck + 1.65(S)
f'ck = 50 + 1.65(5) = 58.25N/mm^2
ii. To water cement ratio
W/C ratio = 0.35 {From IS 456 2000}
iii. Selection of water content
From IS 10262 2009 Table 2 , maximum content for 20mm size aggregate is 186lts
Estimated water content for 100mm slump = 186 + .06*186 = 197lts
As super plasticizer is used the water content is reduced by 29%
Water content = 197 * 0.71 * 197 = 140lts
iv. Calculation of cement content
W/C ratio = .35
Cement = 140/.35 = 400 kg/m^3
v. proportion of volume of CA & FA
From Table 3 of IS 10262 2009, volume of CA corresponding to 20mm size & FA (Zone 1) for W/C ratio 0.35
CA & FA = .6
Therefore, Volume of CA needs to be increased and FA should be decreased.
As W/C ratio is lowered by 0.10, the proportion of volume of CA is increased by 0.02
Therefore, Corrected W/C ratio = 0.62
Volume of Course aggregate = .62 * .9 = 0.56
Volume of Fine Aggregate = 1 - 0.56 = 0.44
vi. Mix Calculation
- Volume of concrete =1m^3
- Volume of cement = 400/(3.15*1000) = 0.127m^3
- Vlume of water = 140/(1*1000) = 0.140m^3
- Volume of Chemical Admixture = 7.6/(1.145*1000) = 0.006m^3
- Volume of total aggregate = a-(b+c+d) = 1-(0.127+.140+.006) = 0.728m^3
- Mass of CA = .728 *0.56 *2.74 *1000 = 1117.043 kg/m^3
- Mass of FA = .728 *.44*2.74*1000 = 877.67 kg/m^3
vii. Mix Proportion
- Cement = 400 kg/m^3
- Water = 140 kg/m^3
- Fine aggregate = 877.67 kg/m^3
- Course aggregate = 1117.043 kg/m^3
- Chemical Admixture = 7.6 kg/m^3
- W/C ratio = .35
- Mix Ratio = 1 : 2.19 : 2.8