Project on Concrete Mix Design for various grades of Concrete
1 ) Calculate the Concrete Mix Design for M35 grade concrete with fly ash ?
Solution:
Step 1: Target mean strength:
f’ck = fck + 1.65 s
Fck = 35
S = 5 N/mm^2 (As per table 1 of IS 10262:2009)
= 35 + 1.65 * 5
= 43.25 N/mm^2
Step 2: Water cement ratio
For M35
Exposure condition: Extreme
Therefore.
Extreme = 0.45 (As per table 5 of IS 456:2000)
w/c ratio = 0.40 (As per graph)
hence 0.40 < 0.45
for the better results we will consider the lower value
therefore=w/c ratio = 0.40
Minimum cement content = 360 kg / m^3 (As per table 5 of IS 456:2000)
Step 3: Selection of water content
Nominal size of aggregate = 20 mm (As per table 2 of IS 10262:2009)
Minimum water content = 186 litres or kg
Slump ranges from 25 to 50 mm
As superplasticizers (20 % reduced)
= 186 * 80/100 = 148.8
Water content = 149 litres
Step 4: calculation of cementitious content (Cement + Fly Ash)
Water cement ratio w/c =0.40
Water content W = 149 kg
Cementitious content = c = 149/ 0.40 = 372.5 = 370 kg/m^3
Minimum cement content = 360 kg / m^3
Therefore 360 < 370 kg/m^3 (hence okay)
Increase by 10 %
370 * 0.10 + 370 = 407 kg/m^3
W = 149 kg
Water cement ratio = 149 / 407 = 0.366
Fly ash at 30% total cementitious material content
= 407 * 30/100 = 122 kg/m^3
Cement = 407 – 122 = 285 kg/m^3
Saving of cement using fly ash = 370 – 285 = 85 kg/m^3
Fly ash = 122 kg/m^3
Step 5: Volume of coarse aggregates and fine aggregates
( Table 4 and table 2 of 383 are used for fine and coarse aggregates )
As per table 3 IS 10262 : 2009
Nominal size of aggregates = 20 mm
Select zone 1 which is used for construction
Volume of coarse aggregates = 0.60
w/c Cement ratio = 0.4
(As per the codes +/- 0..01 for every 0.005 )
Volume of coarse aggregates = 0.62
Volume of fine aggregates = 1 - 0.62 = 0.38
Step 6: Mix Calculation
Volume of concrete = 1m^3
Volume of cement = (Mass of cement / Specific gravity of cement) * 1 / 1000
= (285/3.15) * (1/1000) = 0.090 m^3
Volume of fly ash = (Mass of Fly Ash / Specific gravity of cement) * 1 / 1000
= (122/2.2) * (1/1000) = 0.055 m^3
Volume of water = (Mass of Water / Specific gravity of cement) * 1 / 1000
(149/1) * 1/1000 = 0.006 m^3
Volume of admixtures = (Mass of admixtures / Specific gravity of cement) * 1 / 1000 (7.6 / 1.145) * 1/1000 = 0.006 m^3
Volume of coarse aggregates and fine aggregates = a – ( b + c + d+ e )
= 1 – (0.090+ 0.055+0.149+00.006 )
= 0.7 m^3
Mass of course aggregates = 0.7 *0.62 * 2.74 * 1000 = 1189.16 kg/m^3
Mass of fine aggregates = 0.78 * 0.38 * 2.74 * 1000 = 728 kg/m^3
Result
Cement = 285 kg/m^3
Fly Ash = 122 kg/m^3
Water = 149 kg
Water cement ratio = 0.36
Coarse aggregate = 1189 kg/m^3
Fine aggregate = 728 kg/m^3
Chemical admixtures = 7.6 kg/m^3
Mix ratio = 285 : 728 : 1189
= 1 : 2.55 : 4.17
2 ) Calculate the Concrete Mix Design for M50 grade?
Ans: Step 1: Target mean strength:
f’ck = fck + 1.65 s
Fck = 50
S = 5 N/mm^2 (As per table 1 of IS 10262:2009)
= 50 + 1.65 * 5
= 58.25 N/mm^2
Step 2: Water cement ratio
For M35
Exposure condition: Extreme
Therefore.
Extreme = 0.45 (As per table 5 of IS 456:2000)
w/c ratio = 0.40 (As per graph)
hence 0.40 < 0.45 Hence ok
for the better results we will consider the lower value
therefore=w/c ratio = 0.40
Minimum cement content = 360 kg / m^3 (As per table 5 of IS 456:2000)
Step 3: Selection of water content
Nominal size of aggregate = 20 mm (As per table 2 of IS 1026:2009)
Minimum water content = 197 litres or kg
Slump ranges from 25 to 50 mm
As superplasticizers (30 % reduced)
Based on trails water content reduction of 29 % is achieved with plasticizer
= 197 * 71/100 = 140
Water content = 140 litres
Step 4: calculation of cementitious content (Cement + Fly Ash)
Water cement ratio w/c =0.40
Water content W = 149 kg
Cementitious content = c = 140/ 0.40 = 350= 350 kg/m^3
Minimum cement content = 320 kg / m^3
Therefore 350 < 320 kg/m^3 (hence okay)
Step 5: Volume of coarse aggregates and fine aggregates
( Table 4 and table 2 of 383 are used for fine and coarse aggregates )
As per table 3 IS 10262 : 2009
Nominal size of aggregates = 20 mm
Select zone 1 which is used for construction w/c ratio = 0.6
Actual w/c Cement ratio = 0.4
There is 0.2 diff
So reduce by 10 %
(As per the codes +/- 0..01 for every 0.005 )
Volume of coarse aggregates = 0.64*0.9 = 0.576
Volume of fine aggregates = 1 – 0.576 = 0.424
Step 6: Mix Calculation
Volume of concrete = 1m^3
Volume of cement = (Mass of cement / Specific gravity of cement) * 1 / 1000
= (350/3.15) * (1/1000) = 0.111 m^3
Volume of fly ash = (Mass of Fly Ash / Specific gravity of cement) * 1 / 1000
= (140/2.2) * (1/1000) = 0.140 m^3
Volume of water = (Mass of Water / Specific gravity of cement) * 1 / 1000
(149/1) * 1/1000 = 0.006 m^3
Volume of admixtures = (Mass of admixtures / Specific gravity of cement) * 1 / 1000 (7.6 / 1.145) * 1/1000 = 0.006 m^3
Volume of coarse aggregates and fine aggregates = a – ( b + c + d+ e )
= 1 – (0.111+ 0.140+0.006 )
= 0.743 m^3
Mass of course aggregates = 0.743 *0.576 * 2.74 * 1000 = 1173 kg/m^3
Mass of fine aggregates = 0.743 * 0.424 * 2.74 * 1000 = 863 kg/m^3
Result
Cement = 350kg/m^3
Water = 140 kg
Water cement ratio = 0.40
Coarse aggregate = 1173 kg/m^3
Fine aggregate = 863 kg/m^3
Chemical admixtures = 7.6 kg/m^3
Mix ratio = 350 : 728 : 1173
= 1 : 2.08 : 3.35