Project on Concrete Mix Design for various grades of Concrete
DESIGN MIX FOR M35 GRADE OF CONCRETE WITH FLY-ASH
A)Collection of data
- grade of concrete-M35
- type of cement & grade of cement -OPC 43grade
- maximum nominal size of aggregate-20mm
- exposure condition-very severe
- minimum cement content-340kg/m^3
- workability in terms of slump-80mm
- transportation time-30min
- method of placing-pumping
- standard deviation-5N/MM^2
- type of coarse aggregate-crushed angular aggregate
- maximum cement content-450kg/m^3
- water cement ratio-0.45
- fine aggregate-zone2
B)Test data
1.Target strength
f'ck=fck+1.65*s
f'ck-target mean compressive strength at 28days
fck-characteristic compressive strength
s-standard deviation
as per IS:10262-2019,Table-2 s=5N/MM^2
then f'ck=35+1.65*5
=35+8.25
=43.25N/MM^2
2.water cement ratio
Exposure-very severe
as per IS:456-2000,Table-3
W/C ratio-0.45
but as we are using admixtures the W/C ratio can reduced
so the resultant W/C ratio-0.45-0.05=0.4
3.water content
as per IS:10262-2019,Table-4
the minimum water content for 20mm aggregates is '186litres' for 50mm slump value
we have 80mm slump
for every increase in 25mm slump additional 3% to be added to the provided value
water content=186+3% of 186
=186+5.58
=191.58~192litres
as we are using super plasticizers the water content may reduced to 20%
water content=192-20% of 192
=192-38.4
=153.6~154litres
4.calculation of cement content
W/C = water content/cement content
0.4=154/cement content
cement content=154/0.4
=385kg/m^3
385kg/m^3>340kg/m^3
but the value should not exceed the value of maximum cement content
W/C = water content/cement content
0.4=154/cement content
cement content=154/0.4
=385kg/m^3
385kg/m^3>340kg/m^3
but the value should not exceed the value of maximum cement content
Now ,to proportion containing flyash the following precedure is suggested
a)the percentage of flyash based on the project requirement and quality of material content
b)in some cases based on trial usage of flyash the cementious material percentage to be increased by 10%
cement content=385+10% of 385
=385+38.5
=423.5kg/m^3~423kg/m^3
then the water-cement ratio=154/423
=0.364~0.36
Flyash used is 30% of the cementitious material
Flyash content=30% of 423
=126.9~127kg/m^3
the reduced cement content =423-127
=296kg/m^3
saving of cement while using flyash=385-296
=89kg/m^3
fly-ash content=127kg/m^3
5.aggregate proportion b/w coarse aggregate & fine aggregate
as per IS:10262-2009,Table-3
the volume of coarse aggregate per unit volume of total aggregate given as per nominal size of aggregate of 20mm is under zone2 is 0.66 for W/C ratio of 0.30
as fine aggregate content should be less than that of coarse aggregate content
so we have to increase the coarse aggregate content
for every increase in W/C ratio by 0.05 increase in the volume by 0.01
as we have W/C ratio 0.36 but as per IS code the given W/C ratio is 0.30
so the volume of coarse aggregates to be increased for 0.06 decrease in the W/C ratio given as per IS code
at the rate of +/- 0.01 for every increase in the W/C ratio by 0.05
Therefor,coorected volume of coarse aggregate for W/C ratio 0.36=0.67
For pumpable concrete these values are reduced by 10%
then the volume of coarse aggregate =0.67-10% of 0.67
=0.603
volume of fine aggregtes=1-0.603
=0.397
6.mix calculation per unit volume of concrete
a)volume of concrete=1m^3
b)volume of cement=mass of cement divided by specific gravity of cement*1/1000
=296/3.15*1/1000
=0.093m^3
c)volume of flyash=mass of flyash 'divided by' specific gravity of flyash*1/1000
=127/2.2*1/1000
=0.057m^3
d)volume of water=mass of water 'divided by' specific gravity of water*1/1000
=154/1*1/1000
=0.154m^3
e)volume of chemical admixture=mass of chemical admixture 'divided by' specific gravity of chemical admixture*1/1000
=7.6/1.145*1/1000
=0.0063m^3
f)volume of aggregate C.A & F.A
=(a-(b+c+d+e))
=(1-(0.093+0.057+0.154+0.0063)
=(1-0.3103)
=0.689
g)mass of coarse aggregate=f*volume of C.A*specific gravity of C.A*1000
=0.689*0.603*2.74*1000
=1138.37kg
h)mass of fine aggregate=f*volume of F.A*specific gravity of F.A*1000
=0.689*0.397*2.74*1000
=749.48kg
FINAL TRIAL MIX:-
CEMENT-296kg
FLYASH-127kg
C.A-1138.37
F.A-749.48
WATER CONTENT-154kg
CHEMICAL ADMIXTURE-7kg
W/C RATIO-0.36
MIX PROPORTION FOR M35=CEMENT:F.A:C.A
=296:749.48:1138.37
=1:2.532:3.845
CONCRETE MIX DESIGN FOR M50 GRADE CONCRETE WITHOUT FLY-ASH
1.STIPULATIONS FOR PROPORTIONING
a) Grade designation: M50
b) Type of cement: OPC 53GRADE
c) Maximum nominal size of aggregate: 20MM
d) Minimum cement content: 320kg/m^3
e) Maximum water-cement ratio:0.4
f) Workability: 100MM
g) Exposure condition: EXTREME
h) Method of concrete placing: PUMPING
j) Degree of supervision: GOOD
k) Type of aggregate: CRUSHED ANGULAR
l) Maximum cement content:450kg/m^3
m) Chemical admixture type: SUPER PLASTICIZERS
2. TEST DATA FOR MATERIALS
a) Cement used: OPC 53grade
b) Specific gravity of cement:3.15
c) Chemical admixture: super plasticizers
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: 1%
f) Free (surface) moisture:
1) Coarse aggregate: nill
2) Fine aggregate: nill
g) Sieve analysis:
1) Coarse aggregate : Confirming of Table-2 of IS-383
2) Fine aggregate: Confirming of grading zones of Table-4 of IS-383
B)Test data
1.Target strength
f'ck=fck+1.65*s
f'ck-target mean compressive strength at 28days
fck-characteristic compressive strength
s-standard deviation
as per IS:10262-2019,Table-2 s=5N/MM^2
then f'ck=50+1.65*5
=50+8.25
=58.25N/MM^2
2.water cement ratio
as per IS:10262-2019,Fig-1
W/C ratio-0.34
3.water content
as per IS:10262-2019,Table-4
the minimum water content for 20mm aggregates is '186litres' for 50mm slump value
we have 100mm slump
for every increase in 25mm slump additional 3% to be added to the provided value
water content=186+6% of 186
=186+11.16
=197.16~197litres
as we are using super plasticizers the water content may reduced to 20%
water content=197-20% of 197
=197-39.4
=157.6~158litres
4.calculation of cement content
W/C = water content/cement content
0.34=157/cement content
cement content=157/0.34
=461.76
461kg>320kg
but the value should not exceed the value of maximum cement content
5.aggregate proportion b/w coarse aggregate & fine aggregate
as per IS:10262-2009,Table-3
the volume of coarse aggregate per unit volume of total aggregate given as per nominal size of aggregate of 20mm is under zone2 is 0.66 for W/C ratio of 0.30
as fine aggregate content should be less than that of coarse aggregate content
so we have to increase the coarse aggregate content
for every increase in W/C ratio by 0.05 increase in the volume by 0.01
as we have W/C ratio 0.34 but as per IS code the given W/C ratio is 0.30
so the volume of coarse aggregates to be increased for 0.06 decrease in the W/C ratio given as per IS code
at the rate of +/- 0.01 for every increase in the W/C ratio by 0.05
Therefore,corrected volume of coarse aggregate for W/C ratio 0.34=0.669
For pumpable concrete these values are reduced by 10%
then the volume of coarse aggregate =0.669-10% of 0.669
=0.669-0.0669
=0.6021
volume of fine aggregtes=1-0.6021
=0.3979
6.mix calculation per unit volume of concrete
a)volume of concrete=1m^3
b)volume of cement=mass of cement divided by specific gravity of cement*1/1000
=461.76/3.15*1/1000
=0.146kg/m^3
c)volume of water=mass of water 'divided by' specific gravity of water*1/1000
=158/1*1/1000
=0.158kg/m^3
d)volume of chemical admixture=mass of chemical admixture 'divided by' specific gravity of chemical admixture*1/1000
=7.6/1.145*1/1000
=0.0063kg/m^3
e)volume of aggregate C.A & F.A
=(a-(b+c+d))
=(1-(0.146+0.158+0.0063))
=(1-0.3103)
=0.6897kg/m^3
f)mass of coarse aggregate=e*volume of C.A*specific gravity of C.A*1000
=0.6897*0.6021*2.74*1000
=1137.83kg
g)mass of fine aggregate=e*volume of F.A*specific gravity of F.A*1000
=0.6897*0.3979*2.74*1000
=751.942kg
FINAL TRIAL MIX:-
CEMENT-461.76kg
C.A-1137.83
F.A-751.942
WATER CONTENT-158kg
CHEMICAL ADMIXTURE-7kg
W/C RATIO-0.34
MIX PROPORTION FOR M50=CEMENT:F.A:C.A
=461.76:751.942:1137.83
=1:1.628:2.464