Project on Concrete Mix Design for various grades of Concrete
Concrete Mix Design for M35 grade concrete with fly ash
Aim Mix design for M35 grade concrete with fly ash
procedure
Step1:
Stipulations for proportioning
a.Grade Designation:M35
b.Type of cement:OPC 43 Grade Cement
c.Type of mineral admixture:Fly Ash
d.Maximum nominal size of agggregate:20mm
e.Maximum Cement Content:320kg/m^3 (table:5 IS456-2000)
f.Maximum Water content:0.45(Table IS 456-2000)
g.Workability:100mm
h.Expoxure condition:Severe(for reinforced concrete)
i.Method of concrete placing:Pumping
j.Degree of supervision:Good
k.Type of Aggregate:Crushed Angular
l.Maximum Cement content:430kg/m^3
m.Chemical admixture type:Super plactizers
Step2
Test materials Data
a.Specific gravity of cement:3.15
b.Specific gravity of Fly ash:2.2
c.Specific gravity of coarse aggregate:2.74
d.Specific gravity of Fine aggregate:2.74
e.water absorption of coarse aggregate:0.5%
f.water absorption of fine aggregate:1%
Step3:
To find mean strength for Mix proportioning
Target mean strength f'ck=Fck+1.65s
=35+1.65X5=43.25N/mm^2
fck-Characteristics compressive strength
s-Standard deviation
Step4:
Slection of water-cement ratio
Maximum water cement ratio=0.45
but compare all table
0.4<0.45 hence OK
Step5:
Calculation of water content
Maximum water content of 20mm aggregate 186liters for slump range 25mm to 50mm slump range
add 3% for every 25%
For 100mm slump (6/100)X186=197 liters
as super plasticizer was used so 30% of water is reduced
based on trails water content reduced upto 29%
so (197X0.71)=140litres approx
Step6:
Calculation of cement content
water cement ratio is 0.4
cementinous material(cement+flyash)=(140/0.4)=350kg/m^3
cement content in exposure condition is 320Kg/m^3
So 350Kg/m^3>320kg/m^3 hence OK
increase 10% for cementinous material (350X1.1)=385kg/m^3
water content is 140kg/m^3
so water cement ratio was (140/385)=0.36
flyash content was 30% in the cementinous material 35-30=(385X30/100)=115kg/m^3
cement=385-115=270kg/m^3
Step7:
Proportioning of Fine aggregates and Coarse aggregates
volume of fine aggregate and coarse aggregate for 20mm in zone 1 is 0.6
As the water- cement ratio is lowered by 0.1 ,the proportion of volume of coarse aggregate is increased by 0.02( at the rate of +/- 0.01 for every +/- 0.05 change in water- cement ratio)
Hence corrected value of coarse aggregate is 0.64
for pumpable concrete reduced 10%
volume of coarse aggregate 0.64X0.9=0.576
volume of fine aggregate 1-0.576=0.424
Step8:
Mix calculation
Mix calculation for unit volume of concrete
Volume of concrete= 1m^3
volume of cement=(Mass of cement/Specific gravity of cement)X1/1000
=(270/3.15)X1/1000=0.086m^3
volume of fly ash=(Mass of fly ash/Specific gravity of Fly ash)X1/1000
=(115/2.2)X1/1000=0.052m^3
volume of water=(Mass of water/Specific gravity of water)X1/1000
=(140/1)X1/1000=0.143m^3
volume of chemical admixture=(Mass of water/Specific gravity of water)X1/1000
=(7.6/1.145)X1/1000=0.006m^3
volume of aggregate(fine and coarse)=a-(b+c+d)
=1-(0.111+0.141+0.006)=0.743m^3
volume of coarse aggregate=e X volume of coarse aggregate X specific gravity of coarse aggregate X 1000
=0.716X0.576X2.74X1000=1130kg/m^3
volume of fine aggregate=e X volume of fine aggregate X specific gravity of fine aggregate X 1000
=0.716X0.424X2.74X1000=831kg/m^3
Step9:
Mix Prportion
Cement=270kg/m^3
Fly Ash=115Kg/m^3
Water=140kg/m^3
Fine aggregate=831Kg/m^3
Coarse aggregate=1130Kg/m^3
Chemical admixture=7Kg/m^3
Water cement Ratio=0.36
Mix Ratio 1:2.15:2.93
Result
Cement=270kg/m^3
Fly Ash=115Kg/m^3
Water=140kg/m^3
Fine aggregate=831Kg/m^3
Coarse aggregate=1130Kg/m^3
Chemical admixture=7Kg/m^3
Water cement Ratio=0.36
Mix Ratio 1:2.15:2.93
CONCRETE MIX DESIGN FOR M50 GRADE CONCRETE WITHOUT FLYASH
Aim
To calculate the concrete mix design for M50 grade concrete without flyash.
Introduction
Step1:
Stipulation for proportioning.
a.Grade designation = M50
b.Type of cement = OPC 43 grade confirming to IS8112.
c.Maximum nominaal size of aggregate = 20mm
d.Minimum cement content =360kg/`m^3`.
e.Maximum w/c ratio = 0.4
f.Workability = 100mm
g.Exposure condition = Extreme
h.Method of concret placing = Pumping
i.Degree of supervision = Good
j.Type of aggregate = Crushed angular aggregate
k.Maximum cement content = 450kg/`m^3`
l.Chemical admixture used = Superplasticizers
Step2:
Test data for material
a.Cement used:OPC 43 grade confirming to IS8112
b.Specific gravity of cement = 3.15
c.Chemical admixture = Superplasticizers
d.Specific gravity of coarse aggregate = 2.74; Fine aggregate = 2.74
e.Water absorption ,Coarse aggregate = 0.5%; Fine aggregate = 1%
f.Free surface moisture, Coarse aggregate = Nil; Fine aggregate = Nil.
Step3:
To find mean strength for Mix proportioning
Target mean strength of concrete, ft = fck+1.65S = 50+1.65x5 = 58.25N/`mm^2
Step4:
Slection of water-cement ratio
From table 5 of IS456:2000, the maximum w/c ratio for extreme exposure condition is 0.4. Hence, adopt w/c ratio as 0.35.
Step5:
Calculation of water content
From table 2 of IS10262:2009, the maximum water content corresponding to 20mm nominal size of aggregate is 186 litres for slump value of 25-50mm.
Here, slump value is 100mm, i.e an additional 50mm slump is present.
water content may be increased by about 3% percent for every additional 25mm slump.
Hence, the water content = 186+186x(6/100) = 197 lit.
Since, Superplasticizers are used as admixture, the water content can be reduced by upto 30%.
Let, water be reduced upto 29% i.e, water content = 197x((100-29)/100) = 140 lit.
Step6:
Calculation of cement content
w/c ratio = 0.35
Hence the cement content= 140/0.35= 400kg/`m^3`.
for extreme exposure condition, minimum cement is 360kg/`m^3`, which is less than 400kg/`m^3`. Hence, ok.
Step7:
Proportioning of Fine aggregates and Coarse aggregates
volume coarse aggregate per unit volume total aggregate corresponding to Fine aggreaggte Zone I = 0.6 (corresponding to w/c ratio of 0.5).
Here, the w/c ratio is 0.35, which is lower by 0.15. Hence, coarse aggreagate needs to be increased to decrease the fine aggregate content. So increase the coarse aggregate content by 0.03% (@ of `+-` 0.01% for every `+-`0.05 change in w/c ratio).
Corrected volume of coarse aggregate for w/c ratio of 0.35= 0.63.
For pumpable concrete, values are reduced by 10%.
Hence, volume of coarse aggregate = 0.63x0.9 = 0.567
Volume fine aggregate = 1-0.567 = 0.433.
Step8:
Mix calculation
Volume of concrete = 1m^3
Volume of cement=(Mass of cement/Specific gravity of cement)X1/1000
=(400/3.15)x(1/1000) = 0.127`m^3`
Volume of water=(Mass of water/Specific gravity of water)X1/1000
=(140/1)x(1/1000) = 0.14`m^3`
Volume of chemical admixture=(Mass of water/Specific gravity of water)X1/1000
= (7.6/1.145)x(1/1000) = 0.006`m^3
`
Volume of aggregate=a-(b+c+d)
= 1-(0.127+0.14+0.006) = 0.727`m^3`
Mass of coarse aggregate=e X volume of coarse aggregate X specific gravity of coarse aggregate X 1000
= 0.727x0.567x2.74x1000 = 1129kg/`m^3
Mass of fine aggregate=e X volume of fine aggregate X specific gravity of fine aggregate X 1000
= 0.727x0.433x2.74x1000 = 862kg/`m^3`
Step9:
Mix Prportion
Cement = 400g/`m^3`
Water content = 140kg/`m^3`
Coarse aggregate = 1129kg/`m^3`
Fine aggregate = 862kg/`m^3`
Chemical admixture = 7kg/`m^3`
w/c ratio = 0.35
Mix ratio = 1:2.82:2.15
Result
Cement = 400kg/`m^3`
Water content = 140kg/`m^3`
Coarse aggregate = 1129kg/`m^3`
Fine aggregate = 862kg/`m^3`
Chemical admixture = 7kg/`m^3`
w/c ratio = 0.35
Mix ratio = 1:2.82:2.15