Project on Concrete Mix Design for various grades of Concrete
Q.CALCULATE THE CONCRETE MIX DESIGN FOR M35 GRADE CONCRETE WITH FLY ASH &M50 GRADE CONCRETE WITHOUT FLY ASH?
ANS-Calculate the concrete mix design for m35grade concrete with fly ash
1.stipulation for proportioning
a) grade designaton-M35
b)type of cement-opc43 grade confirming to IS8112
types of mineral admixture-fly ash confirming to IS3812(part1)
c)maximum nominal size of aggregate-20mm
d)minimum cement content-320kg\m3
e) maximum water cement ratio-0.45
f)workability-100mm(slump for pumpable concrete)
g)exposure condition-severe for reinforcement concrete
h)mathod of concret placing-pumping
i)degree of supervision-good
j)type of aggregate-crushed angular aggregate
k)maximum cement content-450m3
l)chemical admixture type- super plasticizer
2.TEST DATA FOR MATERIAL
a)cement used :OPC43 grade confirming IS8112
b)specific gravity of cement3.15
c)fly ash confirming to 3812 part1
d)specific gravity of fly ash-2.2
c)chemical admixture:super plasticizer
d)specific gravity of
1)coarse aggregate-2.74
2)fine aggregate:2.74
e)water absorption:
1)coarse aggregate:0.5%
2)fine aggregate(m.sand):2.5%
f)free moisture
1)coarse aggregate:0.5%
2)fine aggregate:1.0%
g)sieve analysis:
1)coarse aggregate:confirming to all in aggregate of table 2 of IS383
2)fine aggregate:confirming to grading zone llof table 4 of IS383
3.To calculate the target mean strength:
f'ck=fck+1.65s
Where,
f'ck is the target average compressive strength at 28 day's
fck is the characteristics compressive strength at 28 day's
S is the standard deviation
As per the table1of IS:10262-2009,the standard deviation,s is5.0n/mm2
therefore,target strength=35+1.65x5=43.25n/mm2
4.SELECTION OF WATER CEMENT RATIO
maximum water cement ratio =0.45from the table 5 of IS456 2000 for very sever exposur maximum water cement ratio is 0.40 ,0.40<0.45 hence value is ok
5.CALCULATION OF WATER CONTENT
from table 2of IS10262:2009,maximum water content for 20mm aggregate =186 lit(for 25 to50mm siump range )estimate water content for 100mm slump
=186+(6/186)=197lit
based on trials with super plasticizer water content reduction of 30%has been achived hence the arrived water content=197-[197x(20/100)]=158lit
6.CALCULATION OF CEMENT CONTENT
w/c ratio =0.40
cementecious material(cement+fly ash)
cement content=(140/0.4)=350kg/m3
from table 5of IS456,minimumcement content for moderate explousure conditions 320kg/m3
preparing the mix containing fly ash the following steps are follow
percentage of fly ash need to be deside
in this design mix we increase the 10% of cementeciuos material content
cementesious material content =350x1.10=385kg/m3
water content =140kg/m3
so w/c ratio =(140/385)=0.364
fly ash @30% of cementecious material content-385x30%=115kg/m3
cement OPC-385-115kg,m3
saving of cement with using fly ash=350-270=80kg/m3
fly ash being utilised=115kg/m3
7. PROPORTION OF VOLUME OF COURSE AGGRIGAT AND FINE AGGRIGATE CONTENT
from table 3 of ( is 10262:2009) vollume of course aggrih=gate corresponding to 20 mm size aggrigate and fine aggrigtate zone1 nd of water cement ratio 0.6 in the present case water cement ratio 0.4 thererfore volumeof course aggrigate is required to be increase to decrease the fine aggrigate content . As the water cement ratio is lower by 0.06 the proportion of volume of courseaggrigate is increase by 0.02
Therefore, corrected proportion of volume of course aggrigate for gthe water cement ratio of 0.4-0.64
For pumpable concrete this valuees should be reduce up to 10 % .Therefore volum,e of course aggrigate = 0.62 x 0.9 =0.56 .
Volume of finer aggrigate content 1-0.56=0.44
8. MIX CALCULATION
The mix calculatoion per unit volume of concrete shall be as follow
a) Volume of concrete =1 m3
b) Volume of cement = (Mass of cement / Specific gravity of cement) x (1/1000)
= (270/3.15)x(1/1000)=0.0086m3
= 0.0086m3
c) volume of fly ash=(mass of fly ash/specific gravity of fly ash)(1/1000)
(115/22)x(1/1000)=0.052m3
=0.052m3
c) Volume of water = ( Mass of water / Specific gfravitty of water) x(1/ 1000)
= (140/1)(1/1000)=0.140m3
= 0.140m3
d) Volume of chemical admixture = (mass of chemical admixture/specific gravity of admixture)x(1/1000)
(7.6/1.145)x(1/1000)=0.006m3
e) volume of all in aggrigate = (volume of concrete-(volume of cement+volume of fly ash+volume of water+volume of chemical admixture+1
=( 1-(0.086+0.052+0.140+0.006)).
= 0.716m3
f) Mass of course aggrigate = volume of aggr x Volume of course aggrigate x specific gravity of fine aggrigate x 1000 .
= 0.716 x 0.56 x 2.74 x1000
= 1098kg/m3
g) Massof fine aggrigate = volume of aggre x Volume of fine aggrigate x specific gravity of fine aggrigate x 1000 .
=0.716x0.44x2.74x1000
= 863 kg / m3 .
9) MIX PROPORTION .
Cement = 270kg m3
fly ash=115kg/m3
Water =140kg/m3
fine aggregate=863kg/m3
coarse aggregate=1098kg/m3
chemical admixture==7kg/m3
Water cement ratio = 0.364
Mix proportiomn by weight = 1:2.24:1.85
2.calculation of concrete mix design for M50 grade of concrete without fly ash
1.Data
type of cement-OPC43 grade confirming to IS8112
maximum nominal size of aggregate-20mm
minimum cement content-320kg\m3
maximum water cement ratio-0.45
workability-100(slump value for pumpable concrete)
exposure condition-severe for reinforced concrete(for renforced concret)
mathod of concret placing-pumping
degree of supervision-good
type of aggregate-crushed angular aggregate
maximum cement content-450kg/m3
chemical admixture type- super plasticizer
2.TEST DATA FOR MATERIAL
a)cement used :OPC43 grade confirming IS8112
b)specific gravity of cement3.15
c)chemical admixture:super plasticizer
d)specific gravity of course aggrigate and fine aggregate=2.74
e)water absorption:
1)coarse aggregate:0.5%
2)fine aggregate(m.sand):1.0%
f)free moisture
1)coarse aggregate:nil
2)fine aggregate:nil
g)sieve analysis:
1)coarse aggregate:confirming to all in aggregate of table 2 of IS383
2)fine aggregate:confirming to grading zone lof table 4 of IS383
3.TARGET STRENGTH FOR MIX PROPORTIONING
f'ck=fck+1.65s
Where,
f'ck is the target average compressive strength at 28 day's
fck is the characteristics compressive strength at 28 day's
S is the standard deviation
As per the table1of IS:10262-2009,the standard deviation,s is5.0n/mm2
therefore,target strength=50+1.65x5=58.25n/mm2
4.SELECTION OF WATER CEMENT RATIO
adopted maximum water cement ratio =0.4from the table 5 of IS456:2000 for very sever exposur maximum water cement ratio is 0.40 ,0.45.....henc
from table 2 of IS10262:2009,maximum water content for 20mm aggregate =186 lit(for 25 to50mm siump range )
estimate water content for 100mm slump
=186+(6/186)=197lit
as per the super plasticizer water content reduction of 20% and above based on trial water content reduced of 29%is achived with plasticizier
hence,the arrived water content 197x0.71=140lit
6.CALCULATION OF CEMENT CONTENT
adopted cement ratio =0.4
cement content=140/0.4=350kg/m3
from table 5of IS456,minimum cement content for moderate explousure conditions 320 kg/m =350 kg/m3 >340 kg/m3 hence ok
7. PROPORTION OF VOLUME OF COURSE AGGRIGAT AND FINE AGGRIGATE CONTENT
from table 3 of ( is 10262:2009) vollume of course aggrih=gate corresponding to 20 mm size aggrigate and fine aggrigtate zone 1 of water cement ratio 0.6
in the present case water cement ratio 0.40
thererfore volumeof course aggrigate is required to be increase to decrease the fine aggrigate content
. As per water cement ratio is lower by 0.10 the proportion of volume of courseaggrigate is increase by 0.02
Therefore, corrected proportion of volume of course aggrigate for gthe water cement ratio of 0.40=0.62
For pumpable concrete this valuees should be reduce up to 10 % .Therefore volum,e of course aggrigate = 0.62 x 0.9 =0.56
Volume of finer aggrigate content 1-0.56=0.44
8. MIX CALCULATION
The mix calculatoion per unit volume of concrete shall be as follow
a) Volume of concrete =1 m3
b) Volume of cement = (Mass of cement / Specific gravity of cement) x (1/1000)
= 350/ 3.15x1/ 1000
= 0.111m3
c) Volume of water = ( Mass of water / Specific gfravitty of water) x(1/ 1000)
= (140/1)(1/1000)
= 0.140m3
d) Volume of chemical and mixture = (mass of chemical admixture/specific gravity of chemical admixture)x(1/1000)
=(7.5/1.145)(1/1000)=0.006m3
e) volume of all in aggrigate = ( volume of concrete(voiume of cement+volume of watre+volume of chemical admixture)).
=( 1-(0.111+0.140+0.006)).
= 0.743m3
f) Mass of course aggrigate = volume of aggregate x Volume of course aggrigate x specific gravity of fine aggrigate x 1000 .
= 0.743 x 0.56 x 2.74 x1000
= 1140 kg /m3
g) Massof fine aggrigate = volume of aggregate x Volume of fine aggrigate x specific gravity of fine aggrigate x 1000 .
0.743 x0.44 x 2.74x1000
= 896 kg / m3 .
9) MIX PROPORTION .
Cement = 350kg/ m3
Water=140kg/m3
fine aggregate=896kg/m3
course aggregate=1140kg/m3
Chemical admixture = 7 kg /m3 (0.4% by the weight of cement )
Water cement ratio = 0.4
RESULT=1:2.56:3.26