Project on Concrete Mix Design for various grades of Concrete
AIM:
CALCULATING THE DESIGN MIX FOR M35 CONCRETE WITH FLY ASH AND M50 WITH OUT FLY ASH AS PER IS CODES.
GIVEN DATA:(ASSUMED)
1.0PC 43 GRADE CEMENT(IS8112)
2.20MM SIZE AGGREGATE,ZONE 1 FINE AGGRGATE.
3.EXTREME ENVIRONMENT CONDION
4.ADMIXTURE SELECTED AS PER IS3812-PART-1(FLY ASH)
5.SLUMP VALUE 75MM
M35 DESIGN MIX WITH FLY ASH PROCEDURE:(REQUIRED DATA IS ASSUMED)
STEP:1
f"CK=fCK+1.65(S)
f"ck=Target mean strength
fCK =compressive strengt
S =standed deviation (form IS10262 table-8)
f"ck=35+1.65x5
=43.25N/mm^2
STEP:2
W/C RATIO
0.45 as per IS456:2000 based on exposure conditions.
WATER CONTENT:
186LIT as per table-2 IS10262:2009
as per code for 75mm slump
=186+186x3%
=186+5.58=191.58lit
as superplastrizres are used then the water content can be reduced by 30%
=192-192x30%=134.4lit
CEMNT CONTENT:W/C=0.4
C=134.4/0.4=336KG/m^3 it is premissiabe as the minimum is 320kg and maximum is 450kg as per table-5 IS 456:2000
as fly ash is to be added it is preffered to add 30% of fly based on trail and error results
336x30%=100.8kg
cement=235.2kg
fly ash=100.8kg
AGGREGATE VOLUME :From table-3 IS10262:2009 it is 0.6 for w/c of 0.5 soo the volume os increased by 0.02 as per the codes
volume of coures aggregate=0.62
volume of fine agg =1-0.62=0.38
volume of materials per unit volume of concrete:1m^3
(a) volume of cement=(mass of cement/specific graviy of cement)x1/1000
=236/3.15x0.001
=0.0743m^3
(b) volume of fly ash=(mass of fy ash/specific graviy of fly)x1/1000
=100/2.2x0.001
=0.045m^3
(c) volume of water =135/1x0.001
=0.135
(d) total volume of aggregate =1-(a+b+c)
=1-(0.0743+0.045+0.135)
=0.7457m^3
mass of CA= (dxvolume of CA)X(specific gravity of CA)X1000
=0.7457x0.62x2.74x1000
=1266.7KG
mass of FA= (dxvolume of FA)X(specific gravity of FA)X1000
=0.7457x0.38x2.74x1000
=776.42m^3
MIX CONTENT:
CEMENT=235.2KG
FLY ASH=100.8KG
C.A=1266.7KG/M^3
F.A=776.42KG/M^3
WATER=135KG/M^3
RESULT MIX RATO:
1:2.3125:3.7678
M50 DESIGN MIX PROCEDURE:(REQUIRED DATA IS ASSUMED)
STEP:1
F"CK=FCK+1.65(S)
f"ck=Target mean strength
FCK =compressive strengt
S =standed deviation (form IS10262 table-8)
f"ck=50+1.65x5
=58.25N/mm^2
STEP:2
W/C RATIO
0.45 as per IS456:2000 based on exposure conditions.
WATER CONTENT:
186LIT as per table-2 IS10262:2009
as per code for 100mm slump
=186+186x6%
=186+11.16=197.16lit
as superplastrizres are used then the water content can be reduced by 30%
=197.1-197.16x30%=137.1lit
CEMNT CONTENT:W/C=0.4
C=137.1/0.4=342.7KG/m^3 it is premissiabe as the minimum is 320kg and maximum is 450kg as per table-5 IS 456:2000
AGGREGATE VOLUME :From table-3 IS10262:2009 it is 0.6 for w/c of 0.5 soo the volume os increased by 0.02 as per the codes
volume of coures aggregate=0.62
volume of fine agg =1-0.62=0.38
volume of materials per unit volume of concrete:1m^3
(a) volume of cement=(mass of cement/specific graviy of cement)x1/1000
=342.7/3.15x0.001
=0.1087m^3
(b) volume of water =137.1/1x0.001
=0.137
(c) total volume of aggregate =1-(a+b)
=1-(0.1087+0.135)
=0.7563m^3
mass of CA= (dxvolume of CA)X(specific gravity of CA)X1000
=0.7563x0.62x2.74x1000
=1284.80KG
mass of FA= (dxvolume of FA)X(specific gravity of FA)X1000
=0.7563x0.38x2.74x1000
=787.45m^3
MIX CONTENT:
CEMENT=342.7KG
C.A=1284.80KG/M^3
F.A=787.45KG/M^3
WATER=137.1KG/M^3
RESULT MIX RATO:
1:2.2977:3.7490