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)
*0PC 43 GRADE CEMENT(IS8112)
*20MM SIZE AGGREGATE,ZONE 1 FINE AGGRGATE
*SEVERE ENVIRONMENT CONDITION
*ADMIXTURE SELECTED AS PER IS3812-PART-1(FLY ASH)
*SLUMP VALUE 100MM
*S.G OF CA AND FA =2.74
*S.G OF CEMENT = 3.15
*S.G OF FLY ASH = 2.2
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.
W/C RATIO
0.4 as per IS 120262 : 2019 (Comparing 28days compressive strength vs water cement ratio from the graph)
There fore, the least values of the obtained values is to choosed
Hence W/C RATIO = 0.4
STEP 3
WATER CONTENT:
As per table-2 IS10262:2009 as per code maximum water content for 20mm aggregate = 186liters for (25mm to 50mm slump range , for additional slump 3% may be added for every 25mm slump).
there fore , estimated water content for 100mm slump
=186+186x(6/100)
=186+5.58=197lit
as superplastrizres are used then the water content can be reduced by 30%
=197-197x(30/100)=140lit
STEP 4
CEMNT CONTENT:
W/C=0.4
Cement content =140/0.4=350KG/m^3
It is premissiabe as the minimum is 320kg and maximum is 450kg as per table-5 IS 456:2000.
The % of fly ash needs to bee decided based on project requirement and quality , adding 10% cement to cementitious material .
cementitious material content = 350 x 1.10 = 385kg/m^3
As fly ash is to be added it is preffered to add 30% of fly based on trail and error results.
385x30%=115.5kg
cement= 385 - 115.5 =269.5kg
fly ash=115.5kg
STEP 5
PROPORTION VLOUME OF COARSE AND FINE AGGREGATE VOLUME :
*From Table 3 of IS 10262 : 2009 volume of coarse corresponding to 20mm size of aggrgates and fine aggregates zone 1 , W/C ratio is given as 0.6 , in present case W/C ratio obtained is 0.4 , there fore the volume of coarse aggregates needs to be increased to decrease the fine aggregate content.
*As W/C is lower by 0.10,the proportion of volume of coarse agregate is increased by 0.02 (at the rate of -/+ 0.01) for average + or - 0.05 change in W/C ratio
There fore corrected , volume of course aggregate=0.62
volume of fine agg =1-0.62=0.38
volume of materials per unit volume of concrete:1m^3
STEP 6
MIX CALCULATIONS:
(a) volume of cement=(mass of cement/specific graviy of cement)x1/1000
=269.5/3.15x0.001
=0.085m^3
(b) volume of fly ash=(mass of fy ash/specific graviy of fly)x1/1000
=115.5/2.2x0.001
=0.052m^3
(c) volume of water =140/1x0.001
=0.14
(d) total volume of aggregate =1-(a+b+c)
=1-(0.085+0.052+0.14)
=0.723m^3
mass of CA= (dxvolume of CA)X(specific gravity of CA)X1000
=0.723x0.62x2.74x1000
=1228.23KG
mass of FA= (dxvolume of FA)X(specific gravity of FA)X1000
=0.723x0.38x2.74x1000
=752.78m^3
STEP 7
MIX CONTENT:
CEMENT=269.5KG
FLY ASH=105KG
C.A=1228.23KG/M^3
F.A=752.78KG/M^3
WATER=140KG/M^3
RESULT MIX RATO:
1:2.79:4.5
M50 DESIGN MIX PROCEDURE:(REQUIRED DATA IS ASSUMED)
GIVEN DATA:(ASSUMED)
*0PC 43 GRADE CEMENT(IS8112)
*20MM SIZE AGGREGATE,ZONE 1 FINE AGGRGATE
*SEVERE ENVIRONMENT CONDITION
*ADMIXTURE SELECTED AS PER IS3812-PART-1(FLY ASH)
*SLUMP VALUE 100MM
*S.G OF CA AND FA =2.74
*S.G OF CEMENT = 3.15
*S.G OF FLY ASH = 2.2
*MAX CEMENT CONTENT = 350 KG
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.
W/C RATIO
0.4 as per IS 120262 : 2019 (Comparing 28days compressive strength vs water cement ratio from the graph)
There fore, the least values of the obtained values is to choosed
Hence W/C RATIO = 0.4
STEP 3
WATER CONTENT:
As per table-2 IS10262:2009 as per code maximum water content for 20mm aggregate = 186liters for (25mm to 50mm slump range , for additional slump 3% may be added for every 25mm slump).
there fore , estimated water content for 100mm slump
=186+186x(6/100)
=186+5.58=197lit
as superplastrizres are used then the water content can be reduced by 30%
=197-197x(30/100)=140lit
STEP 4
CEMNT CONTENT:
W/C=0.4
Cement content =140/0.4=350KG/m^3
It is premissiabe as the minimum is 320kg and maximum is 450kg as per table-5 IS 456:2000.
cementitious material content = 350KG/M^3
STEP 5
PROPORTION VLOUME OF COARSE AND FINE AGGREGATE VOLUME :
*From Table 3 of IS 10262 : 2009 volume of coarse corresponding to 20mm size of aggrgates and fine aggregates zone 1 , W/C ratio is given as 0.6 , in present case W/C ratio obtained is 0.4 , there fore the volume of coarse aggregates needs to be increased to decrease the fine aggregate content.
*As W/C is lower by 0.10,the proportion of volume of coarse agregate is increased by 0.02 (at the rate of -/+ 0.01) for average + or - 0.05 change in W/C ratio
There fore corrected , volume of course aggregate=0.62
volume of fine agg =1-0.62=0.38
volume of materials per unit volume of concrete:1m^3
STEP 6
MIX CALCULATIONS:
(a) volume of cement=(mass of cement/specific graviy of cement)x1/1000
=350/3.15x0.001
=0.111m^3
(b) volume of water =140/1x0.001
=0.14
(c) total volume of aggregate =1-(a+b)
=1-(0.111+0.14)
=0.749m^3
mass of CA= (cxvolume of CA)X(specific gravity of CA)X1000
=0.749x0.62x2.74x1000
=1272.40KG
mass of FA= (cxvolume of FA)X(specific gravity of FA)X1000
=0.749x0.38x2.74x1000
=779.85m^3
MIX CONTENT:
CEMENT=350KG
C.A=1272.40KG/M^3
F.A=779.85KG/M^3
WATER=140KG/M^3
RESULT MIX RATO:
1:2.28:3.63