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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I am Thomas Britto here to share my experiences in the civil engineering field to all my readers.Today many students are struggling to buy books at high prices. So I decided to start a blog and share my experience and knowledge with all my readers.
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