Project on Concrete Mix Design for various grades of Concrete
Concrete Mix Design for M35 grade concrete with fly ash
Ans:
AIM : The main aim of the concrete mix design is to find out the desired proportion of each ingredients which are cement, coarse aggregate, fine aggregate, water etc. to obtain the required properties of resulting mix with addition of the fly ash for the some extent of the concrete.
introduction :
Each ingredient used in the concrete mix design is tested for its quality. The aggregates with good strength, shape, specific gravity and free from organic content are used. The water used also is of good quality generally potable water is used.Concrete with good quality improves its properties such as strength, durability etc. The design mix which is prepared from suitable ingredients in respective proportions itself improves the concrete properties.Fly ash or pulverished fuel ash (pfa) is a finely divided powder thrown out as a waste material at the thermal power plants using pulverized coal for raising steam in the boilers. In the building industry, the use of fly ash a part replacement of cement in mortar and concrete at the construction site has been made all over the world including India and is well known.
Given grade designation = M35
Assume
• OPC 43 grade cement
• 20 mm maximum nominal size of aggregate
• 320 kg/m^3 minimum cement content for exposure condition
• 0.45 water content for exposure condition
• Workability 100mm
• Exposure condition = severe
• Ft = fck + 1.65S (targeted compressive strength)
Ft= 35 +1.65*5 = 43.25 N/mm^2
0.4 is the water content based on experience
0.45 from graph IS 10262:2009


Since 0.4<0.45<0.45, 0.4 is used as the water cement ratio
• From table 2 of IS 10262:2009, maximum water content of 20 mm aggregate is 186 litres for 25 to 50mm slump
For 100 mm slump water content =
186+(6*186/100) = 197 litre
As super plasticizer is used, water content is reduced by 20 %and above
Based on the trials water content reduction of 29% is achieved by the plasticizer
= 197*0.71 = 140 litre
• Cement content
w/c = 0.4
cement content = 140/0.4 = 350 kg/m^3
from table 5 of IS 456 minimum cement content for severe condition is 320 kg/m^3
350> 320 hence OK
320 – cement
30 – fly ash
• From table 3 of IS 456 volume of course aggregate corresponding to 20 mm size aggregate and FA (zone 1) for w/c ratio 0.6

In present case the w/c ratio is 0.4
Therefore, volume of CA needs to be increased to decrease the FA content so, the corrected volume of the CA for w/c ratio of 0.4 is 0.6
For pumpable concrete these values are reduced by 10 %
Therefore, volume of CA = 0.6*0.9=0.54
Volume of FA = 1-0.54 = 0.46
• Volume of cement
= (mass of cement/sp.gravity of cement) *(1/1000)
=320/3.15 * 1/1000 = 0.101 m^3
• Volume of water
= (mass of water/sp.gravity of water) *(1/1000)
=140/1 *1/1000 = 0.140 m^3
• Volume of fly ash
= (mass of fly ash/sp. gravity of fly ash) *(1/1000)
= 30/3 *1/1000 = 0.01 m^3
• Volume of CA & FA
= (Volume of concrete- (Volume of cement+ Volume of water+ Volume of fly ash))
=(1-(0.101+0.14+0.01)) = 0.749 M^3
• Mass of CA
= Volume of CA & FA* Volume of CA* sp. gravity of CA*1000
=0.749*0.54*2.74*1000
=1108 kg/m^3
• Mass of FA
= Volume of CA & FA* Volume of FA* sp. gravity of FA*1000
=0.749*0.46*2.74*1000
=944 kg/m^3
• Mix proportion
Cement =320 kg/m^3
Water = 140 kg/m^3
FA = 944 kg/m^3
CA= 1108 kg/m^3
Fly ash = 30 kg/m^3
w/c = 0.4
• Trial mix ratio = 1: 2.95: 3.46
therfore for the 50 kg of the cement use 147.5 kg fine aggregate (i.e.,50kg*2.95) and 173kg course aggregate (i.e.,50kg*3.46) and water cement ratio of the 0.4.in the 50 kg cement, mass of fly ash is also included.
M50 grade concrete without Fly ash
ans:
AIM: The main aim of the M50 concrete mix design is to find out the desired proportion of each ingredients which are cement, coarse aggregate, fine aggregate, water etc. to obtain the required properties of resulting mix to attain the maximum compressive strength of 50N/mm^2 in 28 days.
introduction:Each ingredient used in the concrete mix design is tested for its quality. The aggregates with good strength, shape, specific gravity and free from organic content are used. The water used also is of good quality generally potable water is used.Concrete with good quality improves its properties such as strength, durability etc. The design mix which is prepared from suitable ingredients in respective proportions itself improves the concrete properties.
Given grade designation = M50Assume
• OPC 53 grade cement
• 20 mm maximum nominal size of aggregate (zone 1)
• 320 kg/m^3 minimum cement content for exposure condition
• 0.45 water content for exposure condition
• Workability 75mm
• Exposure condition = severe
• Ft = fck + 1.65S (targeted compressive strength)
Ft= 50 +1.65*5 = 58.25 N/mm^2
0.4 is the water content based on experience
0.4 from graph IS 10262:2009


Since, 0.4 is used as the water cement ratio
• From table 2 of IS 10262:2009, maximum water content of 20 mm aggregate is 186 litres for 25 to 50mm slump
For 75 mm slump water content (increase 3 %) =
186+(3*186/100) = 191.58 litre
• Cement content
w/c = 0.4
cement content = 191.58/0.4 = 478.95 kg/m^3
from table 5 of IS 456 minimum cement content for severe condition is 320 kg/m^3

478.95> 320 hence OK
• From table 3 of IS 456 volume of course aggregate corresponding to 20 mm size aggregate and FA (zone 1) for w/c ratio 0.6

In present case the w/c ratio is 0.4
Therefore, volume of CA needs to be increased to decrease the FA content so, the corrected volume of the CA for w/c ratio of 0.4 is 0.6
For pumpable concrete these values are reduced by 10 %
Therefore, volume of CA = 0.6*0.9=0.54
Volume of FA = 1-0.54 = 0.46
• Volume of cement
= (mass of cement/sp.gravity of cement) *(1/1000)
=478.95/3.15 * 1/1000 = 0.152 m^3
• Volume of water
= (mass of water/sp.gravity of water) *(1/1000)
=191.58/1 *1/1000 = 0.191 m^3
• Volume of CA & FA
= (Volume of concrete- (Volume of cement+ Volume of water+ Volume of fly ash))
=(1-(0.152+0.191)) = 0.657 M^3
• Mass of CA
= Volume of CA & FA* Volume of CA* sp. gravity of CA*1000
=0.657*0.54*2.74*1000
=972.09 kg/m^3
• Mass of FA
= Volume of CA & FA* Volume of FA* sp. gravity of FA*1000
=0.657*0.46*2.74*1000
=828.08 kg/m^3
• Mix proportion
Cement =478.95 kg/m^3
Water = 191.58 kg/m^3
FA = 828.08 kg/m^3
CA= 972.09 kg/m^3
w/c = 0.4
• Trial mix ratio = 1: 1.72: 2.02
therfore for the 50 kg of the cement use 86 kg fine aggregate (i.e.,50kg*1.72) and 101kg course aggregate (i.e.,50kg*2.02) and water cement ratio of the 0.4.