Project on Concrete Mix Design for various grades of Concrete
Project 1- To Calculate Concrete mix design for M35 concrete with fly ash.
Aim- To calculate concrete mix design for M35 concrete with fly ash.
Assume- Type of cement OPC 43 grade.
Lets take site conditions severe and workability 120mm and method of placing- pumping.
Step1- To find target mean strength (f'ck) in MPa-
F'ck = fck+1.65S = 35+1.65*5= 43.25MPa. (S=5 from the table)
Step2- To calculate the water cement ratio -
From IS 10262:2019 Table1
Maximum water content for severe exposure is 0.36. (Taking lesser value)
Step3- To calculate water content- Adding 6% for each 25mm additional slump.
for 50mm slump water content is 186 kg.
therefore for 120mm slump water content is
186+6%*186+6%*186+6%*186= 219.48 kg
Step4- To find cement content-
It is calculated by formula= Water content/Water cement ratio
219.48/0.36 = 609.66kg/m^3
Fly ash content= 30%*610 = 183kg/m^3
Cement content= 610-183=427kg/m^3
Also maximum cement content according to IS code should be less than 450kg/m^3.
Hence it is okkk.
Step5- Volume of coarse and fine aggregates-
Lets take nominal size of aggregates=20mm
So volume of aggregates according to the table for 20mm aggregates is 0.62
So volume of fine aggregates is 1-0.62= 0.38
Step6- Mix Calculations- Lets take total volume = 1m^3 (unit volume of concrete)
Volume of air = 0.01m^3
Volume of water= 219.48/1000 = 0.219m^3
Volume of cement = 427/3.15*1000 = 0.135m^3
Volume of fly ash = 183/2.2*1000 = 0.083m^3
Volume of chemical admixture = 1% of (427+183) = 6.1kg = 6.7/1.145*1000 = 0.005m^3
Total volume of aggregates = 1-(a+b+c+d+e) = 0.548m^3
Mass of coarse aggregate = Total volume of aggregate*% of aggregate*Specific gravity of coarse aggregate*1000
= 0.548*0.62*2.74*1000= 930kg/m^3
Mass of fine aggregates = 0.548*0.38*2.65*1000= 551.836kg/m^3
Therefore,
Cement= 427kg/m^3
Water= 219kg/m^3
fly ash= 183kg/m^3
coarse aggregates=930kg/m^3
fine aggregates= 552kg/m^3
Project 2 - Calculation of mix design for M50 grade concrete without fly ash-
Aim- To calculate mix design for M50 grade of concrete.
Assume-
Type of cement OPC 43 grade.
Lets take site conditions severe and workability 120mm and method of placing- pumping.
Step1- To find target mean strength (f'ck) in MPa-
F'ck = fck+1.65S
f'ck= 50+1.65*5=58.25MPa
Step2- To calculate the water cement ratio -
From IS 10262:2019 Table1
Maximum water content for severe exposure is 0.36. (Taking lesser value)
Step3- To calculate water content- Adding 6% for each 25mm additional slump.
for 50mm slump water content is 186 kg.
therefore for 120mm slump water content is
186+6%*186+6%*186+6%*186= 219.48 kg
Step4- To find cement content-
It is calculated by formula= Water content/Water cement ratio
219.48/0.36 = 609.66kg/m^3
Sand content = 30% of 610 =
183kg/m^3
Cement content= 610-183=427kg/m^3
Step5- Volume of coarse and fine aggregates-
Lets take nominal size of aggregates=20mm
So volume of aggregates according to the table for 20mm aggregates is 0.62
So volume of fine aggregates is 1-0.62= 0.38
Step6- Mix Calculations- Lets take total volume = 1m^3 (unit volume of concrete)
Volume of air = 0.01m^3
Volume of water= 219.48/1000 = 0.219m^3
Volume of sand = 183/2.65*1000 = 0.069m^3
Volume of cement = 427/3.15*1000 = 0.135m^3
Volume of chemical admixture = 1% of (427+183) = 6.1kg = 6.7/1.145*1000 = 0.005m^3
Total volume of aggregates = 1-(a+b+c+d+e) = 0.548m^3
Mass of coarse aggregate = Total volume of aggregate*% of aggregate*Specific gravity of coarse aggregate*1000
= 0.548*0.62*2.74*1000= 930kg/m^3
Mass of fine aggregates = 0.548*0.38*2.65*1000= 551.836kg/m^3
Therefore,
Cement= 427kg/m^3
Water= 219kg/m^3
Sand= 183kg/m^3
coarse aggregates=930kg/m^3
fine aggregates= 552kg/m^3