Project on Concrete Mix Design for various grades of Concrete
Q.1 Calculate the Concrete Mix Design for M35 grade concrete with fly ash
Solution -1. - stipulations for proprotioning
Grade of concrete - M35
type of cement - opc 43 grade conforming to IS 8112
type of mineral admixture - fly ash conforming to IS 3812 ( part 1 )
maximum cement content - 320 kg?m^3 from table 5 of IS 456: 2000 exposure condition
workability - 100 mm( slump value for pumpable concrete )
exposure condition - severe ( for reinforced concrete )
method of concrete placing - pumping
degree of supervision - good
type of aggregate - crushed angular aggregate
chemical admixture type - superplasticizer
2. 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
d- specific gravity of fly ash - 2.2
e- chemical admixture - superplasticizer conforming to IS 9103
f- specific gravity of coarse aggregate - 2.74
specific gravity of fine aggragate - 2.74
d- water absorption of coarse aggragate - 0.5%
water absorption of fine aggregate - 1.0%
e- free surface mositure coarse aggregate - nil
free surface of fine aggregate - nil
d- sieve analysis
3. Target strength for mix proportioning
f'ck = fck + 1.65 s
standard deviation from table 1 IS 10262:2009
S= 5 N/mm^2
35+1.65*5
=43.25
4. Selection of water-cement ratio
from table-5 of IS 456:2000 for severe exposure , maximum water-cement ratio=0.45,based on experiance , adopt water- cement ratio as 0.40
0.40<0.45,hence ok
5. Selection of water content
from table 2 of IS 10262:2009,maximum water content for 20mm aggregate = 186 litre
(for 25mm-50 mm slump range)
EStimated water content for 100mm slump
=186+6/100*186
=197 litre
As superplasticizer is used , the water content can be reduced up to 30% ,Based on trials water content reduction of 29% is acheived with plasticizer.
Hence, the arrived water content= 197*0.71
=140 litre
6. Calculation of cement content
water cement ratio = 0.4
cementitious material ( cement + fly ash )content = water /cement 140/0.4
350 kg/m^3
From table of IS 456 minimum cement content for severe exposure condition = 320 kg/m^3
350 kg/m^3 > 320 kg/m^3, hence ok
Now, to proportion a mix containing fly ash the following steps are suggested.
a. - The percentage of fly ash needs to be decided based on project requirment and quality of materials.
b. - cementitious material content - 350*1.10 = 385 kg/m^3
water content = 140 kg/m^3
so, water-cement ratio= 140/385= 0.364
Fly ash @30% total cementitious material content = 385*30% = 115 kg/m^3
cement (opc) = 385-115=270 kg/m^3
saving of cement while using fly ash = 350-270= 80kg/m^3
Fly ash being utilized = 115 kg/m^3
7. Proportion of volume of coarse aggregate and fine aggregqate content
from table 3 of IS 10262:2009, volume of coarse aggregate corresponding to 20 mm size water- cement ratio of 0.5= 0.6
In the present case the water cement ratio is 0.40
therefore, the volume of coarse aggregate needs to be increased to decreased the fine aggregate content.
formula , 0.5 - w/c /0.05 * 0.01 + volume of aggregate
= 0.5 - 0.4 / 0.05 * 0.01 + 0.6
= 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 content
= 1 - 0.56
= 0.44
8. - Mix calculations
The mix calculation per unit volume of concrete shall be as follows
a. - volume of concrete = 1 m^3
b. - volume of cement = mass of cement / specific gravity of cement * 1/1000
270 / 3.15 * 1/1000 = 0.086 m ^3
c. - volume of fly ash = mass of fly ash / specific gravity of fly ash * 1/1000
= 115 / 2.2 * 1/1000
= 0.052 m ^3
d. - volume of water = mass of water / specific gravity of water * 1/1000
= 140 / 1 * 1/1000
= 0.140 m^3
e. - volume of chemical admixture = mass of chemical admixture / specific gravity of admixture * 1/1000
= 7.6/1.145 * 1/1000
= 0.006 m^3
f. - volume of aggregate ( FA and CA ) = (a- (b+c+d+e)
= (1- (0.086+0.052+0.140+0.006)
= 0.716 m^3
g. - mass of coarse agggregate = f * volume of CA * specific gravity CA * 1000
= 0.716 * 0.56 * 2.74 * 1000
= 1098 kg / m ^3
h. - mass of fine aggregate = f * volume FA * specific gravity FA * 1000
= 0.716 * 0.44 * 2.74 * 1000
= 863 kg /m^3
9. Mix proprotions for trails no. 1
cement - 270 kg/m^3
fly ash = 115 kg/m^3
water = 140 kg/m^3
fine aggregate = 863 kg/m^3
coarse aggregate =1098 kg/m^3
chemical admixture = 7 kg/m^3
W/c ratio = 0.364
C : S : A
1 : 2.24 : 2.85
Solution 2 - M50 grade concrete without Fly ash
1. stipulation for proportioning
a. - Grade designation - M50
b. - type of cement - opc 43 grade conforming to IS 8112
c. - nominal size of aggregate - 20 mm
d. - minimum water cement content - 320 kg/m^3 ( from table 5 of IS 456:2000 - exposure condition )
e. - maximum water cement ratio - 0.45 ( from table 5 of IS 456:2000 - exposure condition )
f. - workability - 100 mm ( slump value for pumpable concrete )
g. - exposure condition - severe ( for reinforced concrete )
h. - method of concrete placing - pumping
I. - degree of supervision - good
j. - type of aggregate - crushed angular aggregate
k. - maximum cement content - 450 kg / m^3
L. - chemical admixture type - superplasticizer
2. Test data for materials
a. cement used - opc 43 grade conforming to IS 8112
b. specific gravity of cement - 3.15
c. chemical admixture - superplasticizer conforming to IS 9103
d. specific gravity of CA - 2.74
S. P. FA. - 2.74
e. water absorption CA. - 0.5%
water absorption FA. - 1.0%
f. free surface moisture CA - nil
free surface moisture FA - nil
d. - sieve analysis
3. Target strength for mix proportioning
f'ck = fck + 1.65s
from table 1 of IS 10262:2009, standard deviation s= 5 n/mm^2
therefore , target strength = 50 + 1.65*5
= 58.25 n/mm^2
4. Selection of water-cement ratio
from table-5 of IS 456:2000 for severe exposure , maximum water-cement ratio=0.45,based on experiance , adopt water- cement ratio as 0.40
0.40<0.45,hence ok
5. Selection of water content
from table 2 of IS 10262:2009,maximum water content for 20mm aggregate = 186 litre
(for 25mm-50 mm slump range)
EStimated water content for 100mm slump
=186+6/100*186
=197 litre
As superplasticizer is used , the water content can be reduced up to 30% ,Based on trials water content reduction of 29% is acheived with plasticizer.
Hence, the arrived water content= 197*0.71
=140 litre
6. Calculation of cement content
water cement ratio = 0.4
cementitious material ( cement + fly ash )content = water /cement 140/0.4
350 kg/m^3
From table of IS 456 minimum cement content for severe exposure condition = 320 kg/m^3
350 kg/m^3 > 320 kg/m^3, hence ok
7. proportion of volume of coarse aggrergate and fine aggregate content :
From table 3 of IS 10262:2009 , volume of coarse aggregate corresponding to 20 mm size aggregate and fine aggregate ( zone 1 ) for water - cement ratio of 0.5 = 0.6
in the present case the water cement ratio is 0.40 , therefore thevolume of coarse aggregate needs to be increased to decreased the fine aggregate content.
as water cement ratio is lower by 0.10 , the proportion of volume of coarse aggregate is increased by 0.02 ( at the rate of -1 + 0.01 for every +/- 0.01 for every + - 0.05 change in water - cement ratio
therefore, corrected volume of coarse aggregate for the water- cement ratio of 0.40= 0.62
( note : if coarse aggregate is not of angular type , then also the volume of coarse aggregate may be required to be increased suitably , based on experience )
for pumpable concrete these values are reduced by 10 %
therefore , volume of coarse aggregate = 0.62 * 0.9
= 0.56
volume of fine aggregate content = 1 - 0.56
= 0.44
8. - Mix calculations
The mix calculation per unit volume of concrete shall be as follows
a. - volume of concrete = 1 m^3
b. - volume of cement = mass of cement / specific gravity of cement * 1/1000
350 / 3.15 * 1/1000 = 0.111 m ^3
c. - volume of water = mass of water / specific gravity of water * 1/1000
= 140 / 1 * 1/1000
= 0.140 m^3
d. - volume of chemical admixture = mass of chemical admixture / specific gravity of admixture * 1/1000
= 7.6/1.145 * 1/1000
= 0.006 m^3
f. - volume of aggregate ( FA and CA ) = (a- (b+c+d+e)
= (1- (0.111+0.140+0.006)
= 0.743 m^3
g. - mass of coarse agggregate = e * volume of CA * specific gravity CA * 1000
= 0.743 * 0.56 * 2.74 * 1000
= 1140 kg / m ^3
h. - mass of fine aggregate = e * volume FA * specific gravity FA * 1000
= 0.743 * 0.44 * 2.74 * 1000
= 896 kg /m^3
9. Mix proprotions for trails no. 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
W/c ratio = 0.4
C : S : A
1 : 2.56 : 3.26