Project on Concrete Mix Design for various grades of Concrete
1.Concrete Mix Design for M35 grade Concrete with Fly Ash:
Step 1: Stipulation of Concrete design mix:
Grade of Concrete = M35
Type of Cement = OPC 43 grades (confirming to IS 8112)
Admixture = Fly ash
Size of Aggregate = 20 mm
Partial replacement= 30% (by weight of cement)
Max Water cement ratio =0.45 (as per IS 456 code, table 5)
S.G of cement = 3.15
S.G of Aggragates =2.74
S.G of Fly ash = 2.90
Exposure condition = Severe
Step 2: Target Mean strength
Fck= fck+1.65S
were,
Fck = Target mean strength at 28 days
fck = Characteristic compressive strength
S= Standard Deviation
S=5 As per IS-10262 table 1
therefore Fck=35+1.65*5 =43.25 N/mm2
Step 3: water cement ratio:
maximum water cement ratio is 0.45 (As per Table 5 IS 456)
Step 4: Maximum Water Content:
186 L for 50 mm slump (as per Table 2 IS10262)
for 100 mm slump we need to increase the water content.
(Clause 4.2 of IS 10262 suggest increases of 3% of Water content to get 25 mm extra slump so for 100 mm slump we need to add 6% of water)
=186+0.06*186
=197.16 liter =197 liter
Step 5: Calculation of cement and fly ash content:
W/C ratio=0.45
Total Cement+ fly ash requirement = 197/0.45 = 437.55 kg
Minimum cement content for severe exposure=340 kg
438>340 hence Okay.
so adopted cement content is 438 kg.
Fly ash is 30% against the cement content
so, =438*0.30
=131.4 kg
Fly ash content = 131 kg
Cement Content = 306 kg
Step 6: Coarse aggregate and Fine aggregate
Coarse Aggregate ratio =0.62 (as per IS10262 table 3 for W/C ratio 0.5)
for 0.45 W/C ratio =0.63 (adding 0.01 for every decrease 0.05)
Final fine aggregate ratio =1-0.63 =0.37
Volume of C.A = 0.63 cum
Volume of F.A = 0.37 Cum
Step 7: Mix Calculation:
a) Volume of Concrete = 1 cum
b) Total volume of Cement=306/3.15*1000 =0.09 cum
c)Total Volume of Fly ash =131/2.50*1000 =0.052 cum
d)Volume of Water =197/1*1000 = 0.197 cum
Total Aggregates requirement=a-(b+c+d) = 0.661 cum
C.A = 0.661*0.63*2.74*1000=1141 kg
F. A= 0.661*0.37*2.74*1000=670.12 kg
Step 8: Mix ratio:
Ratio=1:2.18:3.72
2.Concrete Mix design for M50 Concrete:
Data:
Step 1: Stipulations for concrete design mix
Grade of Cement = M50
Type of cement = OPC 53 grade
Specific gravity of cement=3.15
Nominal size of aggregate =20 mm
Exposure Condition = Severe Condition
Max Water Cement ratio=0.45 (as per IS 456, Table 5 exposure condition)
Workability =100 mm
Type of aggregate = Angular
Maximum Cement Content =450kg/m3
Step 2: Test data for Materials
Cement = OPC 53 Grade
Specific Gravity=3.15
Specific Gravity of coarse aggregate =2.74
Specific gravity of fine aggregate =2.64
water absorption of coarse aggregate =0.5%
water absorption of fine aggregate =1%
Step 3: Target Mean strength:
Fck=fck+1.65XS
S=5 as per IS10262 Table 1
Therefore= 50+1.65*5=58.25 N/mm
Target Mean strength=58.25 N/mm2 (Target average strength at 28 days)
Step 4: Selection of Water cement Ratio:
The max water cement ratio =0.45 (as per IS 456 table 5)
Max Water Cement ratio is =0.45
Based on experience W/C Ratio is 0.44
0.44<0.45, Hence OK
Adopted water cement ratio =0.44
Step 5: Selection of water content:
Max Water content for 20 mm =186 liter for (25 to 50 mm slump range) as per table 2 of IS 10262
Clause 4.2 suggests increasing 3% of Water content for every extra slump from 25mm
so,
Estimated water content for 100 mm slump
186+(0.06*186) =197 liter.
Step 6: Calculation of Cement content:
Adopted water cement ratio =0.44
Cement content=197/0.44=447.72 kg/m3
from table 5 IS 456 Min cement content for severe condition is 320 kg/m3
448>320, Hence ok
Cement Required =448 kg/m3
Step 7: Volume of Coarse Aggregate and Fine aggregate:
W/C Ratio of 0.50 is 0.62(from table 3 IS 10262)
But For W/C Ratio =0.44
as the water cement ratio is lower by 0.03. the proportions of volume of Coarse aggregates are increases by 0.02(at the rate of -/+ 0.01 for every 0.05 change in water cement ratio)
therefore =0.62+0.01=0.632
for pumpable concrete these values should be reduced up to 10% therefore 0.0632*0.9=0.568
Volume of Aggregate =1-0.56=0.431
Volume of F.A =0.431
Volume of C. A=0.56
Step 8: Mix Calculations
a) Volume of concrete = 1 cum
b) Volume of Cement = Mass of Cement / (S.G of Cement*1000) =0.142 cum
c) Volume of Water =Mass of water / (S.G of Water*1000) = 0.197 cum
d)Volume of all aggregate =1-(a+b) = 1- (0.142+0.197) =0.661
e) Volume of Coarse Aggregate = 0.661*0.56*2.74*1000 =1014.23 kg
f) Volume of Fine aggregate =0.661*0.43*2.75*1000 = 781.63 kg
Step 9: Ratio
Mix Ratio = 1: 1.74: 2.26