Project on Concrete Mix Design for various grades of Concrete
Aim: To calculate concrete mix design of M35 grade of concrete using fly ash content.
step1: given data:
grade of concrete: M35,therefore, fck=35 N/mm^2
therefore, target mean strength(ft)= fck+1.65(s)
value of standard deviation(s)=5 ....from IS 10262:2009 table1.
therefore, ft= 35+1.65(5)= 43.25N/mm^2
step2: selection of water cement ratio:
assume severe condition, so wc ratio for sever condition according to environmental exposure condition=0.45
from IS 10262:2019, graph of page4,wc ratio=0.4
taking the smaller value of wc so,final water cement ratio=0.4
step3: calculation of water content:
From table 2 of IS 10262:2009, the maximum nominal size of aggregate is 20 mm.
From table, For 20mm nominal maximum size of aggregate the water content =186 kg (that is 186 litres)
( for 25mm to 50mm slump range)
But we take for 100 mm slump..... (For pumpable concrete)
therefore, water content= 186+(6%of186)
= 186+(0.06*186)= 197kg=197litres.
(For additional slump value 3% is added for every 25 slump,hence added 6% for 100mm slump value)
A superplasticizer is added as an admixture.
So when a superplasticizer is added the water content should be reduced up to 30% and above
Based on trial, reduction in water content of 29 percentage is achieved.Hence 71 percentage of water content should be considered.
Final water content=71% of 197= 0.71*197=140litres.
Step4: cement content calculation:
cement content=watercontent/water cement ratio.
= 140/0.4= 350kg/m^3
From IS 456: 2000
Minimum cement content is equals to 320 kg per metre cube, a higher cement content is ok.
hence,350>320 therefore ok
cement content=350kg/m^3
consider 10% cementitious material is added
therefore cementitious material content= 350*(110/100)= 385kg/m^3
water content=140lit
therefore water cement ratio=140/385=0.364
therefore, fly ash in 30% of total cementitious material content= 385*0.3=115kg/m^3
therefore, cement (OPC) used from cementitious material=385-115=270 kg/m^3
therefore amount of cement saved by using fly ash= 350-270= 80 kg/m^3
fly ash being utilized= 115 kg/m^3
Step5: Calculation of volume of coarse aggregates and fine aggregates:
from IS 10262:2019 table5:
For water cement ratio of 0.5 the volume of coarse aggregate for 20 mm size aggregate of zone 1 is 0.6
But in present case water cement ratio= 0.4
Therefore the volume of coarse aggregate needs to be increased to decrease the fine aggregate content.
Here, cement ratio is less by 0.1
Therefore the volume of coarse aggregate has to be increased by 0.02
[At the rate of -/+0.01 for every -/+0.05 changing water cement ratio]
Therefore the corrected volume of coarse aggregate of water cement ratio of 0.4= 0.62
For pump concrete the values are reduced by 10 percentage.
volume of coarse aggregate= 0.9*0.62= 0.56
therefore,Volume of fine aggregate content= 1 - 0.56= 0.44
Step6:Mix calculation:
volume of concrete= 1m^3
volume of cement=270/(3.15*1000)= 0.086m^3
volume of fly ash= 0.052 m^3
volume of water= 140/1000=0.140m^3
volume of admixture= 7.6/(1.145*1000)= 0.006m^3
volume of aggregate= 1-(0.086+0.052+0.140+0.006)
=0.716m^3
mass of coarse aggregate= 0.716*0.56*2.74*1000= 1098.6304 kg
mass of fine aggregates=863.2096 kg
Step7:
mix proportion for trial1:
cement=270kg/m^3
fly ash= 115 kg/m^3
water=140lit
fine aggregates=863.2096 kg/m^3
coarse aggregates=1098.6304 kg/m^3
admixture=7.6kg/m^3
water cement ratio=0.4
trial mix= 1: 3.19 :4.066
Aim:To calculate the Concrete Mix Design for M50 grade concrete without Fly ash
Step1:
given data:
grade of concrete: M50,therefore, fck=50 N/mm^2
therefore, target mean strength(ft)= fck+1.65(s)
value of standard deviation(s)=5 ....from IS 10262:2009 table1.
therefore, ft= 50+1.65(5)= 58.25N/mm^2
step2: selection of water cement ratio:
assume severe condition, so wc ratio for sever condition according to environmental exposure condition=0.45
from IS 10262:2019, graph of page4,wc ratio=0.35
taking the smaller value of wc so,final water cement ratio=0.35
step3: calculation of water content:
From table 2 of IS 10262:2009, the maximum nominal size of aggregate is 20 mm.
From table, For 20mm nominal maximum size of aggregate the water content =186 kg (that is 186 litres)
( for 25mm to 50mm slump range)
But we take for 100 mm slump..... (For pumpable concrete)
therefore, water content= 186+(6%of186)
= 186+(0.06*186)= 197kg=197litres.
(For additional slump value 3% is added for every 25 slump,hence added 6% for 100mm slump value)
A superplasticizer is added as an admixture.
So when a superplasticizer is added the water content should be reduced up to 30% and above
Based on trial, reduction in water content of 29 percentage is achieved.Hence 71 percentage of water content should be considered.
Final water content=71% of 197= 0.71*197=140litres.
Step4: cement content calculation:
cement content=watercontent/water cement ratio.
= 140/0.35= 400kg/m^3
From IS 456: 2000
Minimum cement content is equals to 320 kg per metre cube, a higher cement content is ok.
hence,400>320 therefore ok.
cement content=400kg/m^3
Step5: Calculation of volume of coarse aggregates and fine aggregates:
from IS 10262:2019 table5:
For water cement ratio of 0.5 the volume of coarse aggregate for 20 mm size aggregate of zone 1 is 0.6
But in present case water cement ratio= 0.35
Therefore the volume of coarse aggregate needs to be increased to decrease the fine aggregate content.
Here, cement ratio is less by 0.15
Therefore the volume of coarse aggregate has to be increased by 0.03
[At the rate of -/+0.01 for every -/+0.05 changing water cement ratio]
Therefore the corrected volume of coarse aggregate of water cement ratio of 0.35= 0.63
For pump concrete the values are reduced by 10 percentage.
volume of coarse aggregate= 0.9*0.63= 0.567
therefore,Volume of fine aggregate content=1-0.567=0.433
Step6:Mix calculation:
volume of concrete= 1m^3
volume of cement=400/(3.15*1000)= 0.12698 m^3
volume of water= 140/1000=0.140m^3
volume of admixture= 7.6/(1.145*1000)= 0.006m^3
volume of aggregate= 1-(0.12698+0.140+0.006)
=0.72702m^3
mass of coarse aggregate= 0.72702*0.567*2.74*1000= 1129.4837 kg/m^3
mass of fine aggregates=862.55 kg/m^3
Step7:
mix proportion for trial1:
cement=400kg/m^3
water=140lit
fine aggregates=862.55kg/m^3
coarse aggregates=1129.4837kg/m^3
admixture=7.6kg/m^3
water cement ratio=0.35
trial mix= 1: 2.156: 2.823