Project on Concrete Mix Design for various grades of Concrete
Aim : To calculate the concrete mix design for M35 grade concrete with fly ash.
Objective: The main objective is to produce concrete as economically as possible to achive desired durability, workability and strenght of M35 grade concrete.
Codes for concrete mix design:
IS 10262:2009
IS 456:2000
Procedure and Calculation:
Step 1:
Target mean strenght f'ck = fck+ 1.65 s
fck = Characteristic compressive strenght
s = Standard deviation (Table 1 IS 10262:2009)
fck = 35 N/mm2
s = 5
f'ck = 35+1.65*5
f'ck = 43.25 N/mm2
Step 2:
Water cement ratio = 0.45 (For severe condition, from table 5 IS 10262:2009)
on expriecnce water cement ratio = 0.4
0.4<0.45, hence ok.
Step 3:
Calculate the water content
Maximum water content for 20mm aggregate = 186 litres (Table 2 IS 10262:2009)
for 25mm to 50mm slump range.
For 75mm slump = 186+(3/100)*186
= 191.58 litres
As superplaticizer is used the water content is reduced upto 30%
Based on trials water content reduction of 29% is achived with plasticizer
Water content = 191.58*0.71
= 136.02 litres
Step 4:
Calculation of cement content
Water cement ratio = 0.4
Cement content = 136.02/0.4
= 340.05 kg/m3
In table 5 IS 456:2000 the cement content for severe exposure = 320 kg/m3
340.05 kg/m3> 320 kg/m3 hence ok.
To proportion the mix containing flyash the following steps:
* The percentage of flyash need to be decided based on project requirement and quality materials.
* In some cases increase in cementatious materials can be warrned. increase in cementatious material content and its percentage may be used on experience and trial.
In this increase of 10% cementatious material content is considered.
Cementatious material content = 340.05*1.10
= 374.056 kg/m3
Water content = 136.02 litres
So, water cement ratio = 136.02/374.056
= 0.36
Fly ash at 30% total cementatious material content = 374.056* (30/100)
= 112.216 kg/m3
Cement(OPC) = 374.056 - 112.216
= 261.84 kg/m3
Saving of cement while using flyash = 340.05-112.217
= 227.883 kg/m3
Step 5:
Proportion of volume of coarse aggregate and fine aggregate content
From table 3 IS 10262:2009, volume of coarse aggregate corresponding to 20mm size aggregate and fine aggregate zone 1 for water cement ratio = 0.6
In present case the water cement ratio = 0.4
Therefore the volume of coarse aggregate needs to be increased to decrease the fine aggregate content.
As the waer cement ratio is lowered by 0.1 the proportion of the volume of coarse aggregate is increased by 0.2 (at the ratio of +/-0.01 for every +/-0.05 changes in water cement ratio).
Hence the corrected value of coarse aggregate for the water cement ratio = 0.64
If coarse aggregate is not of angular type, then also the volume of coarse aggregate may be required to be increased suitably, based on experience.
For pumpable concrete, these are reduced by 10%
volume of coarse aggregate = 0.64*0.9
= 0.576
volume of fine aggregate = 1-0.576
= 0.424.
Step 6:
Mix calculations
Mix calculation per unit volume of concrete
A] Volume of concrete = 1m3
B] Volume of cement = (mass of cement/specific gravity of cement)*1/1000
= (261.84/3.15)*1/1000
= 0.083m3
C] Volume of flyash = (mass of flyash/specific gravity of flyash)*1/1000
= (261.839/2.2)*1/1000
= 0.098m3
D] Volume of water = (mass of water/specific gravity of water)*1/1000
= (136.02/1)*1/1000
= 0.136m3
E] Volume of chemical admixture = (mass of admixture/specific gravity of admixture)*1/1000
= (7.4/1.14)*1/1000
= 0.006m3
F] Volume of aggregate(fine and coarse) = A-(B+C+D+E)
= 1-(0.083+0.098+0.136+0.006)
= 0.677m3
G] Mass of coarse aggregate = volume of aggregate*volume of coarse aggregate*specific gravity of coarse aggregate*1000
= 0.677*0.576*2.74*1000
= 1068.468 kg/m3
H] Mass of fine aggregate = volume of aggregate*volume of fine aggregate*specific gravity of fine aggregate*1000
= 0.677*0.424*2.74*1000
= 786.511 kg/m3
Step 7:
Mix proportion
Cement = 261.84 kg/m3
Flyash = 261.839 kg/m3
water = 136.02 kg/m3
Fine aggregate = 786.511 kg/m3
Coarse aggregate = 1068.468 kg/m3
Chemical admixture = 7.4 kg/m3
Water cement ratio = 0.4
Mix ratio = 1:3:4.08
Result: The mix design for M35 grade of concrete with flyash is 1 : 3 : 4.08.
The slump shall be measured and the water content and dosage of admixyure shall be adjusted for achieving the required slump on trial . The mix proportion shall be reworked for the actual water content and checkedd for durability requirements.
Aim: To calculate the M50 grade of concrete without fly ash.
Objective: The main objective is to produce concrete as economically as possible to achive desired durability, workability and strenght of M35 grade concrete.
Codes for concrete mix design:
IS 10262:2009
IS 456:2000
Prosedure and calculation
Step 1:
To find target mean strenght for proportioning
f'ck = fck+1.65s
fck = characteristic compressive strenght
s = standard deviation
f'ck = 50+1.65*5
= 58.25 N/mm2
Step 2:
selection of water cement ratio
From Table :5 of IS 456-2000, by exposure condition
Here severe condition is considered , Maximum water-cement ratio=0.45
Based on experience water- cement ratio =0.4
0.4<0.45, hence ok.
Step 3:
Selection of water content
Maximum water content for 20mm aggregate =186 litres ( Table 2 of IS 1026:2009) for 25mm to 50mm slump range
For 100 mm slump ,= 186+(6/100)*186 =197 litres
As superplaticizer is used, the water content is reduced upto 30%
Based on trials water content reduction of 29% is achieved with plasticizer
Hence arrived water content = 197*0.71 = 140 litres
Step 4:
Calculation of cement content
Water - cement ratio = 0.4
Cement content = 140/0.4
= 350 kg/m3
In table 5 of IS 456-2000, the cement content for severe exposure = 320 kg/m3
350 kg/m3 > 320 kg/m3, hence ok
Step 5:
Proportion of volume of coarse aggregate and fine aggregate content
From table 3 of IS 1026:2009 ,volume of coarse aggregate corresponding to 20 mm size aggregate and fine aggregate (Zone 1 ) for water cement ratio =0.6
In present case the water - cement ratio is 0.4
Therefore the volume of coarse aggregate needs to be increased to decrease the fine aggregate content .
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 the corrected value of coarse aggregate for the water-cemnt ratio = 0.64
If coarse aggregate is not of angular type, then also the volume of coarse aggregate may be required to be increased suitably ,based on experience
For pumpable concrete, these are reduced by 10 %
Volume of coarse aggregate = 0.64*0.9 = 0.576
Volume of fine aggregate = 1-0.576 =0.424
Step 6:
Mix calculation
For unit volume of concrete
A] Volume of concrete = 1m3
B] Volume of cement = (mass of cement/specific gravity of cement)*1/1000
= (350/3.15)*1/1000
= 0.111 m3
C] volume of water = (mass of water/specific gravity of water)*1/1000
= (140/1)*1/1000
= 0.140 m3
D] Volume of chemical admixture = (mass of admixture/specific gravity of admixture)*1/1000
= (7.6/1.145)*1/1000
= 0.006 M3
E] Volume of aggregate ( fine and coarse ) = A-(B+C+D)
= 1-(0.111+0.140+0.006)
= 0.743 m3
F] Mass of coarse aggregate = volume of aggregate*volume of coarse aggregate*specific gravity of coarse aggregate*1000
= 0.743*0.576*2.74*1000
= 1173 kg/m3
G] Mass of fine aggregate = volume of aggregate*volume of fine aggregate*specific gravity of fine aggregate*1000
= 0.743*0.424*2.74*1000
= 863 kg/m3
Step 7:
Cement = 350 kg/m3
Water = 140 kg/m3
Fine aggregate = 863 kg/m3
Coarse aggregate = 1173 kg/m3
Chemical admixture = 7 kg/m3
Water cement ratio = 0.4
Mix ratio = 1 : 2.46 : 3.35
Result: The mix ratio for M 50 grade concrete is 1:2.46:3.35.
The slump shall be measured and the water content and dosage of admixyure shall be adjusted for achieving the required slump on trial . The mix proportion shall be reworked for the actual water content and checkedd for durability requirements.