Project on Concrete Mix Design for various grades of Concrete
AIM : Concrete Mix Design for M35 grade concrete with fly ash
PROCEDURE :
STEP 1
1) Stipulation for proportioning :
a) Grade designation : M35
b) Type of cement : OPC 53 grade confirmimg to IS 8112
C)Maximum nominal size : 20 mm
d) maximum cement content : 350 kg/m^3(from table 5 of IS 456 -2000 -Very severe condition )
e) Maximum water content : 0.45 ( fro m table 5 of IS 456 -2000 - Very severe condition )
f) Workability : 100 mm (sump valvue )
g) Exposure condition : Very severe ( reinforced concrete)
h ) Method for concrete placing : Pumping
i) Type of aggregate : Crushed angular aggregate
j) Degree of supervision : Good
k) Maximum cement content : 450 kg/m^3
l) Chemical Admixture : Super plasticizer
STEP 2: TEST DATA FOR MATERIAL :
a) Cement used : OPC 53 grade { confirming to IS 8112 }
b) Specific gravity of cement : 3.15
c) Chemical admixture : Superplasticizer
d) Specific gravity of : Coarse aggregate : 2.74
: Fine aggregate : 2.74
e)Water absorption : Coarse aggregate : 0.5 %
: Fine aggregate : 1.0 %
f)Free surface moisture : Coarse aggregate : Nil
: Fine aggregate : Nil
g) Sieve analysis
: Coarse aggregate :
STEP 3 : TARGET STRENGTH FOR MIX PROPORTIONING :
F'ck = fck + 1.65 S
Where ,
f'ck = Target average compressive strength at28 days
S = Standard deviation
FROM table 1 , of IS 10262: 2009 ,standard deviation =S= 5 N/mm^2
fck = 35 N/mm^2
Therefore , Target strength = fck +1.65 S
f'ck = 35+ 1.65 x 5
=43.25 n/mm^2
STEP 4 : SELECTION OF WATER CEMENT RATION :
From table 5 of IS 456 - 2000 FOR very severe exposure maximum water cement ratio = 0.45
Based on experience , adopted water cement ratio = 0.45
Hence OK .
STEP 5 : SELECTION OF WATER CONTENT :
FROM table 5 of IS 10262 :2009
Maximum water content for 20 mm aggregate = 186 litre ..............( For 25 to 50mm sump range)
Estimated water content for 100mm slump = 186 + 6/100 x 186
= 197 litre.
As super plasticizer is used , the water content can be reduced up to 20% and above based on trials water content
Hence , the arrived water content = 197 - [197 x (29/100)]
= 139.87 =140 litre
STEP 6 : CALCULATION FOR CEMENT FLY ASH CONTENT :
Water cement ratio =0.4
Cement content = 140/0.4
= 350 kg/m^2 = 350 kg/m^2
Hence ok .
NOW, to proportion a mix containing fly ash ; -
a) Decide the percentage fly ash to be used based on project requirement and quality of materials .
b) The cementeties content has to be increased in order to attein good workability and strength
c) We are increasing cementitious content by 10 %
final cementitious content = 350 x 1.10
= 385 kg/m3
385 kg /m3 > 350 kg/m3 ..........Hence ok .
cementitious material content = 385 kg/m3
new water cement ratio = 140/385
=0.3
Fly ash content is taken as 30 % of total
Cementitious content = 385 x 0.3 = 115 kg / m^3
Cement require = 385-115 =270 kg/ m^3
Cement quantity saved due to use of fly ash = 350 - 270
= 80 kg / m^3
Fly ash utilized = 115 kg/m^3
STEP 7 :PROPORTION OF VOLUME OF COARSE AGGREGATE AND FINE AGGREGATE :
From table 3 ( IS 102262 : 2009 ) volume of coarse aggregate & fine aggregate (zone II ) FOR WATER - CEMENT ratio =0.6
In present case the water ratio is 0.45
Therefore , coarse aggregate is increased by 0.02
Therefore , corrected proportion of volume of coarse agregate for water cement ratio 0.45 =0.62
In case of the coarse aggregate is not angular one then also volume of coarse aggregate may be required to be increased suitably based on experience
For pumpable concrete these values should be reduced by 10 %
Therefore , volume of coarse aggregate = 0.62 x 0.9 = 0.56
Volume of fine aggregate = 1- 0.56 = 0.44
STEP 8 :MIX CALCULATION :
MIx calculation as per unit volume of concrete :-
a) VOLUME OF CONCRETE = 1 m ^3
b) VOLUME OF CEMENT = [ Mass of cement /specific gravity of cement ] x [ 1/ 1000 ] = [270/3.15] x [1/1000]
=0.052
c) VOLUME OF FLY ASH = [ Mass of fly ash /specific gravity of fly ash ] x [1/1000] =[155/2.2] x [1/1000]
= 0.052
d) VOLUME OF WATER = [ Mass of water / specific gravity of water ] x [1/1000] =[140/1 ] x [ 1/ 1000]
= 0.14 m^3
e) VOLUME OF CHEMICAL ADMIXTURE =
[Mass of admixture / specific gravity of admixture] x [1/ 1000] = [7/1.145] x [ 1/100]
=0.007 m^3
f) VOLUME OF ALL IN AGGREGATE ( FINE & COARSE AGGRGATE )
= [a- ( b+c+d+e) = [1 - ( 0.085+0.052+0.14+0.007] = 0.7
g) MASS OF COARSE AGGREGATE =
[F X volume of fine aggregate x specific gravity of coarse aggregate x 1000]
= 0.7 x 0.56 x 2.74 x 1000 = 862 kg
STEP 9 : MIX PROPORTION FOR TRAIL NUMBER 1 :
CEMENT : 270 kg/m^3
FLY ASH : 115
WATER : 140 kg/m^3
FINE AGGREGATE : 862 kg/m^3
COARSE AGGREGATE : 1058 kg/m^3
CHEMICAL ADMIXTURE : 7 KG/M^3
WATER CEMENT RATIO : 0.45
MIX PROPORTION BY WEIGHT 1 : 2.23 : 2.74
AIM : Concrete Mix Design for M50 grade concrete without Fly ash.
PROCEDURE :
1) Stipulation for proportioning :
a) Grade designation : M 50
b) Type of cement : OPC 43 grade confirmimg to IS 8112
C)Maximum nominal size : 20 mm
d) maximum cement content : 320 kg/m^3(from table 5 of IS 456 -2000 -Exposure condition )
e) Maximum water content : 0.45 ( from table 5 of IS 456 -2000 - Exposure condition )
f) Workability : 100 mm (sump valvue )
g) Exposure condition : severe ( reinforced concrete)
h ) Method for concrete placing : Pumping
i) Type of aggregate : Crushed angular aggregate
j) Degree of supervision : Good
k) Maximum cement content : 450 kg/m^3
l) Chemical Admixture : Super plasticizer
STEP 2: TEST DATA FOR MATERIAL :
a) Cement used : OPC 43 grade { confirming to IS 8112 }
b) Specific gravity of cement : 3.15
c) Chemical admixture : Superplasticizer
d) Sieve analysis
: Coarse aggregate :
Fine aggregate : confirming to grading zone l pf table 4 to IS 383
STEP 3 : TARGET STRENGTH FOR MIX PROPORTIONING :
F'ck = fck + 1.65 S
Where ,
f'ck = Target average compressive strength at28 days
S = Standard deviation
FROM table 1 , of IS 10262: 2009 ,standard deviation =S= 5 N/mm^2
fck = 50N/mm^2
Therefore , Target strength = fck +1.65 S
f'ck = 50+ 1.65 x 5
= 58.25 n/mm^2
STEP 4 : SELECTION OF WATER CEMENT RATION :
From table 5 of IS 456 - 2000 FOR severe exposure maximum water cement ratio = 0.45
Based on experience , adopted water cement ratio = 0.40
0.40 < 0.45 Hence OK .
STEP 5 : SELECTION OF WATER CONTENT :
FROM table 5 of IS 10262 :2009
Maximum water content for 20 mm aggregate = 186 litre ..............( For 25 to 50mm sump range)
Estimated water content for 100mm slump = 186 + 6/100 x 186
= 197 litre.
As super plasticizer is used , the water content can be reduced up to 20% and above based on trials water content reduction of 29% is achieved with plasticizer .
Hence , the arrived water content = 197 x 0.71
= 140 litre
STEP 6 : CALCULATION FOR CEMENT CONTENT :
Water cement ratio =0.4
Cement content = 140/0.4
= 350 kg/m^2 > 320 kg/m^2
From table 5 of IS 456 minimum cement content for severe exposure condition is 320 kg/m^3
Hence ok .
STEP 7 :PROPORTION OF VOLUME OF COARSE AGGREGATE AND FINE AGGREGATE :
From table 3 ( IS 102262 : 2009 ) volume of coarse aggregate & fine aggregate (zone I ) FOR WATER - CEMENT ratio =0.6
In present case the water ratio is 0.40
Therefore , coarse aggregate is increased by 0.02
Therefore , corrected proportion of volume of coarse agregate for water cement ratio 0.40 =0.62
In case of the coarse aggregate is not angular one then also volume of coarse aggregate may be required to be increased suitably based on experience
For pumpable concrete these values should be reduced by 10 %
Therefore , volume of coarse aggregate = 0.62 x 0.9 = 0.56
Volume of fine aggregate = 1- 0.56 = 0.44
STEP 8 :MIX CALCULATION:
MIx calculation as per unit volume of concrete :-
a) VOLUME OF CONCRETE = 1 m ^3
b) VOLUME OF CEMENT = [ Mass of cement /specific gravity of cement ] x [ 1/ 1000 ] = [350/3.15] x [1/1000]
=0.111 m^3
c) VOLUME OF WATER = [ Mass of water / specific gravity of water ] x [1/1000] =[140/1 ] x [ 1/ 1000]
= 0.14 m^3
d) VOLUME OF CHEMICAL ADMIXTURE =
[Mass of admixture / specific gravity of admixture] x [1/ 1000] = [7/1.145] x [ 1/100]
=0.006 m^3
e) VOLUME OF ALL IN AGGREGATE ( FINE & COARSE AGGRGATE )
= [a- ( b+c+d+e) = [1 - ( 0.111+0.14+0.006] = 0.743 m^3
f) MASS OF COARSE AGGREGATE =
[F X volume of fine aggregate x specific gravity of coarse aggregate x 1000]
= 0.743 x 0.56 x 2.74 x 1000 = 1140 kg
g) MASS OF FINE AGGREGATE =
[F X volume of fine aggregate x specific gravity of fine aggregate x 1000]
= 0.743 x 0.44 x 2.74 x 1000 = 896 kg
STEP 9 : MIX PROPORTION FOR TRAIL NUMBER 1 :
CEMENT : 350 kg/m^3
WATER : 140 kg/m^3
FINE AGGREGATE : 896 kg/m^3
COARSE AGGREGATE : 1140 kg/m^3
CHEMICAL ADMIXTURE : 7 KG/M^3
WATER CEMENT RATIO : 0.4
MIX PROPORTION BY WEIGHT 1 : 2.56 : 3.25