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Calculate the Concrete Mix Design for M35 grade concrete with fly ash. Aim of the project:- To design a concrete mix of M35 grade with the use of fly ash as a replacement for cement with the help of indian standard codes. Introduction of the project:- Data collection:- Grade designation=M35 Type of cement=43 grade conforming…
MD ABU ATHAR
updated on 27 May 2022
Calculate the Concrete Mix Design for M35 grade concrete with fly ash.
Aim of the project:- To design a concrete mix of M35 grade with the use of fly ash as a replacement for cement with the help of indian standard codes.
Introduction of the project:-
Data collection:-
Grade designation=M35
Type of cement=43 grade conforming to IS8112
Maximum nominal size of aggregate=20mm
minimum cement content=320kg/m3
Maximum water cement ratio=0.45(from table 5 of IS 456:2000 based on exposure condition, here the adopted exposure condition is severe for RCC)
workability=100mm (for pumpable concrete)
Method of concrete placing=pumping
Type of aggregate=crushed angular aggregate
maximum cement content=320kg/m3
Chemical admixture type= superplasticizer
Test data for materials:-
chemical admixture=1.145
coarse aggregate=2.74
fine aggregate=2.74
f. water absorption
coarse aggregate=0.5%
fine aggregate=1.0%
g. free surface moisture
coarse aggregate & fine aggregate=nil
Concrete Mix Design Procedure
f’ck=fck+1.65S
From table 1 of IS 10262:2009, for M35 grade of concrete
S= standard deviation=5.0N/mm2
F’ck=35+1.65*5
=43.25 N/mm2
From table 5 of IS 456:2000 for severe exposure adopted for RCC maximum w/c ratio =0.45
but, based on experience adopted w/c ratio is =0.40
and, 0.40<0.45,hence,ok
From table 2 of IS 10262:2009,maximum water content for 20mm aggregate size=186 lit.(for 20 to 50mm slump range)
but, our target slump is 100mm(code specifies that for each 25mm increment in slump the water content should be increased by 3%)
so, the estimated water content for 100mm slump
=186+(6/100*186)=197lit.
As superplasticizer is used the water content can be reduced upto 30%
Based on trials water content reduction achieved is 29%
so, =197*0.71=140lit.
water cement ratio=0.40
cementitious material=(cement +fly ash)=140/0.4= 350kg/m3
from table 5 of IS 456:2000 minimum cement content for severe exposure condition =320kg/m3
here, 350>320,hence,ok.
now,proportioning a mix containing fly ash
based on the project requirement, experience and trials, fly ash replacement @30% of cementitious material is recommended
fly ash content=0.3*350=105kg.
cement content=350-105=245kg.
As the chemical admixture is also used=mass of chemical admixture=2% of cementitious material
so,350*(2/100)=7kg.
from table 3 of IS10262:2009,
volume of coarse aggregate corresponding to 20mm size aggregate and fine aggregate (zone 1) for w/c ratio of 0.5=0.6
but, in the present case the w/c ratio =0.40
therefore in this case, the volume of coarse aggregate needs to be increased to decrease the fine aggregate content
As the w/c ratio is lower by 0.10 the proportion of volume of coarse aggregate needs to be increased by 0.02(i.e., @-/+0.01 for every +/-0.05 change in w/c ratio)
therefore , the corrected volume of coarse aggregate for the w/c ratio of 0.40=0.62
but, for pumpable concrete these values are reduced by 10%, therefore
volume of coarse aggregate=0.62*0.9=0.56
volume of fine aggregate=1-0.56=0.44.
=(7/1.145)*(1/1000)=0.00611m3
f. Volume of aggregate(F.A&C.A)=(a-(b+c+d+e)) =(1-(0.077+0.042+0.140+0.00611))=0.73489m3
g. mass of coarse aggregate=f.*volume of coarse aggregate*sp.gravity of CA*1000=1128kg/m3
h.mass of fine aggregate=f.*volume of FA*sp.gravity of FA*1000=0.73489*0.44*2.74*1000=886kg./m3
MIX PROPORTION OF THE ABOVE TRIAL
cement=245 kg./m3
fly ash=105 kg./m3
water=140kg/m3
Fine aggregate=886kg./m3
Coarse aggregate=1128kg./m3
Chemical admixture=7kg./m3
TRIAL MIX RATIO=1:2.53:3.22
RESULT:-
The mix ratio suggests that for 1 bag of cement that is 50 kg. 126.5kg. of fine aggregate will be used and 161kg. of coarse aggregate shall be used. The water content for the mix is 140lits.
Calculate the Concrete Mix Design for M50 grade concrete without Fly ash.
Aim of the project:- To design a concrete mix of M35 grade with the use of fly ash as a replacement for cement with the help of indian standard codes.
Introduction of the project:-
DATA COLLECTION:-
Grade designation= M50
Type of cement =OPC 43 grade conforming to IS 8112
Maximum nominal size of aggregate=20mm
Minimum cement content=320kg/m3 (from table 5 of IS 456:2000 based on the exposure condition)
Maximum water-cement ratio=0.45 (from table 5 of IS 456:2000 based on the exposure condition)
workability=100mm (slum value for pumpable concrete)
method of concrete placing=pumping
Exposure condition = severe (for RCC structures)
Degree of supervision=good
Type of aggregate=crushed angular aggregate
maximum cement content=450kg./m3
chemical admixture - superplasticizers
TEST DATA FOR MATERIALS:-
Cement used-OPC 43 grade conforming to IS 8112
Specific gravity of cement=3.15
Chemical admixture-superplasticizer conforming to IS 9103
Specific gravity
water absorption
free moisture content
sieve analysis
CONCRETE MIX DESIGN PROCEDURE
f’ck=fck+1.65S
From table 1 of IS 10262:2009, for M50 grade of concrete
S= standard deviation=5.0N/mm2
F’ck=50+1.65*5
=58.25 N/mm2
From table 5 of IS 456:2000 for severe exposure codition
maximum water-cement ratio=0.45
but, based on experience, adopted w/c ratio=0.40
0.40<0.45,hence ,ok.
From table 2 of IS 10262:2009,maximum water content for 20mm aggregate =186lit.(for 25 to 50mm sdlump range)
but, here workability required is 100mm so, the code has suggested to increase the water content by 3% for every 25mm increment of slump value
Estimated water content for 100mm slump
186+((6/100)*186)=197 lit.
Code hass also suggested that if the plasticizer is used water content can be reduced by 20%
Based on trials water content reduction of 29% is achieved
so,the arrived water content=197*0.71=140 lit.
CALCULATION OF CEMENT CONTENT:-
W/C ratio =0.40
Cement content=140/0.4=350kg/m3
From table 2 of IS 456:2000, minimum cement content suggested for severe condition is =320kg./m3
since, 350>320kg./m3 hence,ok
As the chemical admixture is also used=mass of chemical admixture=2% of cementitious material
so,350*(2/100)=7kg.
PROPORTION OF VOLUME OF COARSE AGGREGATE AND FINE AGGREGATE CONTENT
From table 3 of IS 10262:2009,
volume of coarse aggregate corresponding to 20mm size aggregate and fine aggregate (zone 1) for w/c ratio of 0.5=0.60
but, in the present situation we have,the w/c ratio =0.40
therefore, the code has suggested to increase the volume of coarse aggregate to decrease the volume of fine aggregate content.
As the w/c ratio is lower by 0.10 the proportion of volume of coarse aggregate needs to be increased by 0.02(i.e., @-/+0.01 for every +/-0.05 change in w/c ratio)
so,the corrected volume of coarse aggregate for the w/c ratio of 0.40=0.62
for pumpable concrete, these values are reduced by 10%
therefore,volume of coarse aggregate =0.62*0.9=0.56
volume of fine aggregate=1-0.56=0.44
MIX CALCULATIONS:-
The mix calculation per unit volume of concrete
=(7/1.145)*(1/1000)=0.006m3
e. volume of aggregate(FA&CA)=(a-(b+c+d))
=(1-(0.111+0.140+0.006))=0.743m3
f. mass of coarse aggregate=e.*volume of coarse aggregate*sp.gravity of CA*1000=0.743*0.56*2.74*1000
=1140kg./m3
g.mass of fine aggregate=e.*volume of FA*sp.gravity of FA*1000=0.743*0.44*2.74*1000=896kg./m3
MIX PROPORTIONING
cement=350 kg./m3
water=140kg/m3
Fine aggregate=896kg./m3
Coarse aggregate=1140kg./m3
Chemical admixture=7kg./m3
hence, THE ARRIVED MIX RATIO
1:2.56:3.26
RESULT:-
The mix ratio suggests that for 1 bag of cement that is 50 kg. 128kg. of fine aggregate will be used and 163kg. of coarse aggregate shall be used. The water content for the mix is 140lits.
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