Project on Concrete Mix Design for various grades of Concrete
1Ans:Mix Design for M35 Grade concrete with fly ash
Data
Grade of Concrete = M35
Compressive Strength = 35N/mm^2
Course Aggregate (Maximum Nominal Size) = 20mm
Type of Cement = OPC 43 Grade pertaining to IS 8112:1989
Minimum Cement Content = 320Kg/m^3 (From Table5 of IS 456:2000 Environment condition Severe)
Maximum Water Cement Ratio = 0.45 (From Table5 of IS 456:2000)
Workability = 75mm (Slump value for Pumpable concrete)
Environment Exposure condition = Severe reinforced concrete
Method of concrete placing = Pumpable
Degree of Supervision = Good
Type of Aggregate = Crushed Angular Aggregate
Chemical Admixture Type = Superplasticizer pertaining to IS 9103
Material Test Data
Specific Gravity of Cement = 3.15
Specific Gravity of Course Aggregate = 2.74
Specific Gravity of Fine Aggregate = 2.74
Water Absorption Course Aggregate = 0.5% & Fine Aggregate = 1%
Free Surface Moisture, Course Aggregate = Nil & Fine Aggregate = Nil
Seive Analysis, Course Aggregate Table2 of IS 383 & Fine Aggregate Zone II Table 4 of IS 383
Step1: Calculating Target Mean Strength(Ft)
Ft = Fck + 1.65σ
Where,
Ft - Target Mean strength of concrete
Fck - Compressive Strength of concrete
σ - Standard deviation (from table 1 IS 10262:2009)
Ft = Fck + 1.65σ
= 35 + 1.65 x 5 (σ = 5N/mm^2 from Table 2 IS 10262:2009)
= 35 + 8.25
Ft = 43.25N/mm^2
Step2: Selection of Water-Cement ratio
From Table 5 of IS 456:2000 for exposure severe condition & reinforced concrete is 0.45
From Page 4 of IS 10262 graph the curve 2 the water cement ratio comes to 0.38
Lesser value we adopt W/C ratio = 0.38
Step3: Selection of water content
From Table 2 of IS:10262:2009, for 20 mm maximum size aggregate and sand grading for zone II –
maximum water content =186 litre (for 25 to 50 mm slump range)
Estimated Water Content for 75mm slump = 186 + (3% of 186)
= 192litres
As superplasticizer is used the water content can be reduced upto 30% based on trials
Hence, Arrived Water Content = 0.70 x 197
Water Content = 134litres or kg/cu.m
Step4: Calculation of Cement Content
Water-cement ratio = 0.38
Cement content = 134/0.38 =352 kg/m^3
From Table 5 of IS 456, minimum cement content for ‘severe’ exposure condition = 320 kg/m^3
352kg/cu.m > 320kg/cu.m, Hence OK
Step4 a: Fly Ash ( Mineral Admixtures used)
Increase of 10% for Cementitious material ( Cement + fly ash)
cement content + fly ash = 352 x (110/100) = 388kg/cu.m
Water content is 134litres
Therefore, w/c ratio = 134/352 = 0.38 hence OK
Fly Ash at 30% total cement content = (388 x 30%) = 116kg/cu.m
Fly ash content = 116kg/cu.m
OPC cement = 388-116 = 272Kg/cu.m
Step5: Calculation of Volume of Course aggregate and Fine aggregate
From Table 3 of IS 10262:2009,
For Nominal maximum size of aggregate = 20 mm,
Volume of coarse aggregate for water-cement ratio of = 0.62(Corresponding Zone II)
In this case W/C ratio lower by 0.10. So, 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).
Corrected Volume of course aggregate for W/C ratio 0.38 = 0.03 + 0.62 = 0.65
For pump or worked around congested reinforcing steel required more workable concrete the coarse aggregate proportion may be reduced up to 10%.
Hence,
Volume of coarse aggregate per unit volume of total aggregate = 0.65 x 90% = 0.59
Volume of fine aggregate = 1 – 0.59 = 0.41
Step6: Mix Calculation
a. Volume of Concrete = 1m^3
b. Volume of Cement = (Mass of cement / specific gravity of cement) x (1/1000)
= 272/3.15 x 1/1000
= 0.086cu.m
c. Volume of water = (Mass of water / specific gravity of water) x (1/1000)
= 134/1 x 1/1000
= 0.134cu.m
d. Volume of fly ash = (Mass of fly ash / specific gravity of water) x (1/1000)
= 134/1 x 1/1000
= 0.134cu.m
e. Volume of admixtures = (Mass of admixtures / specific gravity of admixture) x (1/1000)
= 7.76/1.145 x 1/1000 Mass of admixture = 2% x 388 =7.76kg/cu.m
= 0.0067cu.m
f. Volume of total aggregate (CA + FA) = a-[b+c+d+e]
= 1-[0.086 + 0.053 + 0.134 + 0.0067]
= 0.72m^3
f. Mass of Course aggregate(CA) = e x [Vol of CA] x [Specific gravity of CA] x 1000
= 0.72 x 0.59 x 2.74 x 1000
= 1164kg/m^3
f. Mass of Fine aggregate(FA) = e x [Vol of FA] x [Specific gravity of FA] x 1000
= 0.72 x 0.41 x 2.74 x 1000
= 809kg/m^3
Step7: Mix Proportions Trial
Cement Content = 272kg/m^3
Water content = 134litres
Course aggregate = 1164kg/m^3
Fine aggregate = 809kg/m^3
Chemical admixtures = 7.76kg/m^3
water-cement ratio = 0.38
TRIAL MIX PROPORTION Cement : Fine Agrregate : Course Aggregate
1: 2.97 : 4.27
1Ans:Mix Design for M50 Grade concrete without flyash
Data
Grade of Concrete = M50
Compressive Strength = 50N/mm^2
Course Aggregate (Maximum Nominal Size) = 20mm
Type of Cement = OPC 53 Grade pertaining to IS 12269
Minimum Cement Content = 320Kg/m^3 (From Table5 of IS 456:2000 Environment condition Severe)
Maximum Water Cement Ratio = 0.45 (From Table5 of IS 456:2000)
Workability = 75mm (Slump value for Pumpable concrete)
Environment Exposure condition = Severe reinforced concrete
Method of concrete placing = Pumpable
Degree of Supervision = Good
Type of Aggregate = Crushed Angular Aggregate
Chemical Admixture Type = Superplasticizer pertaining to IS 9103
Material Test Data
Specific Gravity of Cement = 3.15
Specific Gravity of Course Aggregate = 2.74
Specific Gravity of Fine Aggregate = 2.74
Water Absorption Course Aggregate = 0.5% & Fine Aggregate = 1%
Free Surface Moisture, Course Aggregate = Nil & Fine Aggregate = Nil
Seive Analysis, Course Aggregate Table2 of IS 383 & Fine Aggregate Zone II Table 4 of IS 383
Step1: Calculating Target Mean Strength(Ft)
Ft = Fck + 1.65σ
Where,
Ft - Target Mean strength of concrete
Fck - Compressive Strength of concrete
σ - Standard deviation (from table 1 IS 10262:2009)
Ft = Fck + 1.65σ
= 50 + 1.65 x 5 (σ = 5N/mm^2 from Table 2 IS 10262:2009)
= 50 + 8.25
Ft = 58.25N/mm^2
Step2: Selection of Water-Cement ratio
From Table 5 of IS 456:2000 for exposure severe condition & reinforced concrete is 0.45
From Page 4 of IS 10262 graph the curve 3 the water cement ratio comes to 0.34
Lesser value we adopt W/C ratio = 0.34
Step3: Selection of water content
From Table 2 of IS:10262:2009, for 20 mm maximum size aggregate and sand grading for zone II –
maximum water content =186 litre (for 25 to 50 mm slump range)
Estimated Water Content for 75mm slump = 186 + (3% of 186)
= 192litres
As superplasticizer is used the water content can be reduced upto 30% based on trials
Hence, Arrived Water Content = 0.70 x 197
Water Content = 134litres or kg/cu.m
Step4: Calculation of Cement Content
Water-cement ratio = 0.38
Cement content = 134/0.34 =394 kg/m^3
From Table 5 of IS 456, minimum cement content for ‘severe’ exposure condition = 320 kg/m^3
394kg/cu.m > 320kg/cu.m, Hence OK
Step5: Calculation of Volume of Course aggregate and Fine aggregate
From Table 3 of IS 10262:2009,
For Nominal maximum size of aggregate = 20 mm,
Volume of coarse aggregate for water-cement ratio of = 0.62(Corresponding Zone II)
In this case W/C ratio lower by 0.10. So, 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).
Corrected Volume of course aggregate for W/C ratio 0.34 = 0.032 + 0.62 = 0.652
For pump or worked around congested reinforcing steel required more workable concrete the coarse aggregate proportion may be reduced up to 10%.
Hence,
Volume of coarse aggregate per unit volume of total aggregate = 0.65 x 90% = 0.59
Volume of fine aggregate = 1 – 0.59 = 0.41
Step6: Mix Calculation
a. Volume of Concrete = 1m^3
b. Volume of Cement = (Mass of cement / specific gravity of cement) x (1/1000)
= 394/3.15 x 1/1000
= 0.125cu.m
c. Volume of water = (Mass of water / specific gravity of water) x (1/1000)
= 134/1 x 1/1000
= 0.134cu.m
d. Volume of admixtures = (Mass of admixtures / specific gravity of admixture) x (1/1000)
= 7.88/1.145 x 1/1000 Mass of admixture = 2% x 394 =7.88kg/cu.m
= 0.00688cu.m
f. Volume of total aggregate (CA + FA) = a-[b+c+d]
= 1-[0.125 + 0.134 + 0.00688]
= 0.734m^3
f. Mass of Course aggregate(CA) = e x [Vol of CA] x [Specific gravity of CA] x 1000
= 0.734 x 0.59 x 2.74 x 1000
= 1187kg/m^3
f. Mass of Fine aggregate(FA) = e x [Vol of FA] x [Specific gravity of FA] x 1000
= 0.734 x 0.41 x 2.74 x 1000
= 825kg/m^3
Step7: Mix Proportions Trial
Cement Content = 394kg/m^3
Water content = 134litres
Course aggregate = 1187kg/m^3
Fine aggregate = 825kg/m^3
Chemical admixtures = 7.88kg/m^3
water-cement ratio = 0.34
TRIAL MIX PROPORTION Cement : Fine Agrregate : Course Aggregate
1: 2.1 : 3