Project on Concrete Mix Design for various grades of Concrete
concrete mix design for M35 with Fly ash:
Stipulations for propotioning
Grade =M35
Type of cement= OPC 43 grade conforming to 8112
Type of mineral admixture= fly ash Conforming to IS 3812 (part 1)
Max nominal size of aggragate= 20mm
Min cement content= 320kg/m3 from table 5 of IS 456:2000 for severe condtion .
Maximum water-cement ratio=.45 (from table 5 of IS 456:2000)
Workability=100mm (slump value for pumpable concrete)
Exposure condition= Severe (For reinforced concrete)
Max Cement content=450 kg/m3
Chemical admixture type= Superplaticizer
Test Data for materials:
a) Cement used: OPC 43 grade conforming to IS 8112
b)Specific Gravity of Cement=3.15
c)Fly Ash = Conforming to IS 3812 (Part 1)
d)Specific gravity of Fly Ash=2.2
e) Specific Gravity of FA and CA =2.74
F) Water Absorption: CA=.5%
FA=1.0%
g) Free Surface moisture= Nil For both FA and CA
h) Fine Aggregate= Conforming to grading Zone 1 foa table 4 of IS 383
Procedure for finding out the Concrete mix design for M35 grade.
- Target mean strength: f'ck = fck +1.65s
f'ck= 35+1.65(5), here 5 comes from the table 8 of IS 10262:2009
=43.25 N/mm2
2. Selection of w/c ratio
from table 5 of IS 456: 2000 for severe exposure, max w/c ratio =.45,
based on experience adopt w/c=.4
.4<.45, Hence OK
3. Selection of water content
From table 2 of IS 10262:2009, max water content for 20mm aggregate=186l
Asuming the workability as 100mm , estimated water content = 186+.06*186=191.58= 197.16l = 197 litre approx.
Assuming platicizer are used , Water content can be reduced to 30%
Based on trails water content reduction of 29% is achieved with plasticizer.
Therefore water content =192* .71 =140L
4. calculation of water content:
water cement ratio=0.4
Cementious material (Cement =fly ash) content= 140/.4= 350kg/m3
therefore cement content=350kg/m3
From table 5 of IS456:2000, min cement content for severe condition is 320kg/m3
since 320<350, Hence ok.
Now , to proportion a mix containing fly ash the following steps are suggested:
a) The % fo fly ash needs to be decided based on project requiremnent and quality of materials
b)In Some cased increase in cementious material can be warranted. Increase in cementious material content and its % may be based on the experience and trial ( This eg is with increase of 10% cementitious material content).
Cementious material content= 350*1.1=385kg/m3
Water content=140kg/m3
w/c=140/385= .364
fly ash @ 30% total cementious material content=385*30%=115kg/m3
cement =385-115=270kg/m3
Saving of Cement while using Fly ash=350-270= 80kg/m3
Fly ash being utilized =115kg/m3
5. Proportioning of volume of CA and FA
From table 3 of IS 10262:2009, voulme of CA corresponding to 20mm size aggregate and FA Zone 1 for water cement content ratio of .5 to .6
In the present case w/c =.4
Therefore the vol. of CA needs to be increased to decrease the FA content.
As the w/c is lower by .01, the proportion of volume of CA is increased by .05( as the rate of +- .01 for every +- 0.05 change in w/c)
Therefore the corrected volume for CA for the water cement ratio of .45=.62
For pumpable concrete these values are reduced by 10%
Therefore volume of CA=.62*.9=.56
and Volume of FA=1.56=.44
6. Mix Calculations
The Mix calculations per unit volume of concrete shall be as follows:
Volume of concrete=1m3
volume of cement =mass of cement/(Sp.gravity of cement *1000)
=270/(3.15*1000)=0.086 m3
Volume of Fly Ash= Mass of Fly ash/ (Spe gravity of flyash*1000)
= 115/(2.2*1000)=.052 m3
volume of water= mass of water/(sp gravity of water*1000)
=140/1000=.140m3
Volume of chemical admixture= (mass of chemical admixture/(Spc gravity of admixture*1000))
=7.6/1.145 *1/1000= .006m3
Volume of aggregate=1-(.086+.052+.140+.006)=.716m3
Mass of coarse agg= e*vol of CA*Spec. gravity of CA*1000, (here e= volume of aggregate)
=0716*.56*2.74*1000
=1098kg/m3
mass of FA=e*vol of FA*Spec. gravity of FA*1000
=.716*0.44*2.74*1000=863 kg/m3
Therefore the mix proportion for this trial:
cement=270 kg/m3
Fly Ash=115 Kg/m3
water=140 kg/m3
FA=863 kg/m3
CA=1098 kg/m3
Chemical Admixture = 7kg/m3
W/C =0.364
MIX Ratio=1:2.24:2.85
Two more trials should be calculated having the w/c ratio having variation of 10%. And the graph between b/w the all shall be plotted to work out the mix proportion for the given strength for the field trials.
concrete mix design for M50
Stipulations for proportioning
Grade =M50
Type of cement= OPC 43 grade of cement
Max nominal size of aggragate= 20mm
Min cement content= 320kg/m3 from table 5 of IS 456:2000 for severe condtion .
Workability=100mm (slump value for pumpable concrete)
Exposure condition= Severe (For reinforced concrete)
Max Cement content=427 kg/m3
Chemical admixture type= not used
Procedure for finding out the Concrete mix design for M30 grade.
- Target mean strength: f'ck = fck +1.65s
f'ck= 50+1.65(5), here 5 comes from the table 8 of IS 10262:2009
=58.25 N/mm2
2. Selection of w/c ratio
from table 5 of IS 456: 2000 for severe exposure, max w/c ratio =.45
3. Selection of water content
From table 2 of IS 10262:2009, max water content for 20mm aggregate=186l
Asuming the workability as 75mm , estimated water content = 186+.03*186=191.58= 192 l approx.
Assuming no platicizer are used ,
Therefore water content =192l
4. calculation of water content:
water cement ratio=0.45
therefore water content=192/.45=427kg/m3
From table 5 of IS 456:2000, min cement content for severe condition is 320kg/m3
since 320<427, Hence ok.
5. Propportioning of volume of CA and FA
From table 3 of IS 10262:2009, voulme of CA corresponding to 20mm size aggregate and FA Zone 1 for water cement content ratio of .5 to .6
In the present case w/c =.45
Therefore the vol. of CA needs to be increased to decrease the FA content.
As the w/c is lower by .05, the proportion of volume of CA is increased by .05( as the rate of +- .01 for every +- 0.05 change in w/c)
Therfor the corrected volume for CA for the water cement ratio of .45=.5
For pumpable concrete, these values are reduced by 10%.
Therefore Volume of CA= .5*.9=.45
and For FA=1-.45= .55
6. Mix Calculations
The Mix calculations per unit volume of concrete shall be as follows:
Volume of concrete=1m3
volume of cement =mass of cement/(Sp.gravity of cement *1000)
=320/(3.15*1000)=0.101
volume of water= mass of water/(sp gravity of water*1000)
=427/1000=.427m3
Volume of aggregate=1-(.101+.427)=.472m3
Mass of coarse agg= e*vol of CA*Spec. gravity of CA*1000, (here e= volume of water)
=.472*0.45*2.74*1000
=585.675kg/m3
mass of FA=e*vol of FA*Spec. gravity of FA*1000
=.472*0.55*2.74*1000=711.304kg/m3
Therefore the mix proportion:
cement=320 kg/m3
water=427 kg/m3
FA=585.675 kg/m3
CA=711.304 kg/m3
Chemical Admixture =not used
W/C =.45
Mix Ratio=1:1.83:2.22
Two more trials should be calculated having the w/c ratio having variation of %. And the graph between b/w the all shall be plotted to work out the mix proportion for the given strength for the field trials.