Mix Design Calculations for conventional concrete
Mix Design Calculations for conventional concrete
Stipulations for proportioning:
Grade of designation : M35
Type of cement : OPC 53 Grade
Maximum normal size of aggregate : 20 mm
Minimum cement content : 340 kg/m3
Maximum water cement ratio : 0.4
Exposure conditions (IS: 456-2000, pg no.18) : Severe
Workability (IS: 456-2000, pg no.17) : Medium
Degree of supervision : Good
Maximum cement content : 400-500 kg/ m3(Assumption)
Test data for material:
Cement used : OPC 53 Grade
Specific Gravity of cement : 3.15
Specific Gravity of coarse aggregate : 2.74
Fine Aggregate : 2.62
Water Absorption : 0.5 -1%
Target strength for mix proportioning:
Fck’ = fck + 1.65S
Target Strength = 35 + 1.65*5
= 43.25 N/ mm2
Selection of Water cement ratio:
Maximum Water cement ratio: 0.45
Based on Experience adopt water cement ratio as 0.40
0.40<0.45(Hence ok)
Maximum water content = 186 litre
= 186 + 6/100 *186
= 197 litre (Every 25 mm increase, add to 3%)
Super Plasticizer is used, the water can be reduced upto 20%
= 197 * 20/200 + 197
= 157.6 litre
Calculation of cement content :
Water cement Ratio = 0.40
Cement Content = 157.6/ 0.40
= 394 kg /m3
Proportion of volume of Coarse aggregate and fine aggregate:
From Table no.3, Coarse aggregate corresponding to 20mm size aggregate and fine aggregate ( zone-II) for present case water cement ratio = 0.40
Therefore increase or decrease water cement ratio = 0.1
Corrected proportion of volume of coarse aggregate for water cement ratio 0.4=0.664
Volume of Coarse aggregate = 0.664* 0.9
= 0.5976
Volume of Coarse aggregate = 1-0.5976
= 0.4024
MIX CALCULATIONS
a)Volume of concrete = 1m3
b)Volume of cement = (Mass of cement) / (specific Gravity of cement)*(1/1000)
= 394 / (3.15*(1/1000))
= 0.125 m3
c)Volume of water = (Mass of water) /(specific Gravity of cement)*(1/1000)
= (157.6)/ (1)*(1/1000)
= 0.158 m3
d)Volume of chemical admixture (super plasticizer) @ 2% by mass of cementitious material
= (Mass of chemical admixture)/( specific Gravity of cement)*(1/1000)
= (7.88/1.145) * (1/1000)
= 0.00688
e)Volume all in aggregate = ( a-(b + c + d))
= 1-(0.125+0.1576+0.0068)
= 0.7105 m3
f)Mass of coarse aggregate = e*vol. Of coarse aggregate * sp.gravity of CA*1000
= 0.7105*0.5976*2.74*1000
=1163.38 kg
g)Mass of fine aggregate = e* vol. Of sand * sp.gravity of FA*1000
= 0.7105*0.4024*2.62*1000
= 749.07 kg
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