Project on Concrete Mix Design for various grades of Concrete
grades of concrete : M.5
characteristuc compressive strength required in fieled at 25 days : 35mPa
types of exposes : moderate
1. target mean strngth
ft= fck + 1.65*s
= 35+1.65*5
=43.25N/mm2
2.dtermination of w/c ratio
depend upon exposer condition : moderate
as per table 5 IS 456:2000 (page no. 20)
max w/c ratio =0.5
based on expince adopt w/c ratio = 0.5-0.05
=0.45
.. determonation of water content
from table -2 is IS 10262:2009
we gwt water content for 20mm c.aggregate
= 186litres
now ,
target of slum is 15-175mm
as per Is 10262:2009
increse 5% for every assifience 25mm slump
estimate water content for 175mm skup =
186+(15/100*186)
=213liters
as admixture is used the water content can reduced by 5% and above the actual water to be used =
=213.9-(15/100*213.9)
=181.82ltrs
4. calcilation of cement content
as per IS 456:2000
modarate exposed condition
cement =300kg/m3
w/c ratio = 045
water used =181082l
:.cement content =water content / water cement rato
=181/045=
=404kg/m3
5. calculate of volume course and fine aggregate
from table 3 of IS 1062:2009 of c.a errespectivally to 20mm size agg.
fine agg. zone 3 for water content ratio 0.5=0.64
now w/c ratio =0.45
it is less by (0.5-0.45_=0.65
concrete proporstion of volume of course aggregate
= 064+0.01
=0.65
IS10262:20009 cause 404
by reducing 10% we get
=0.65(10/100*0.65)
=0.505
vol. of fine aggregate -1-0.585=0.415
6.deszign of mix calculation
volume of concrete =1m3
vol. of cememt = mass of cement /specific gravty of cement *1000
= 409/293*1000
=0.138m3
vol. of water =mass of water /s.p gravity of water *1000
=181.82/1*1000
= 0.182
vol. of admixture = 102/100*404
=1.21
vol. of A.d = 2/100*1
= 0.02
mass of c.a= 0.656*0.585*2.61*1000
=1082kg
mass of f.a = 0.656*0415*2.65*1000
= 721.4kg/m33
cements = 104kg
grade of concrete =M50
type of cemnet 43 grade
fine agg = zone ll
sp gravity
cement =3.15
f agg =2.61
c agg =2.65
w/c ratio= 0.45
1. target strength
ft= fck +1.65*s
=50+1.65*5
= 58.25mPa
w/c = 0.35
3. calculation of water
water content for 20mm size aggrgate = 180kg/m3
water content =180*0.8
= 144kg /m3
4. calculation of cements content
w/c ratio =0.35
water content / cum of concrete = 144kg
cement content = 144/0.35
= 411.4kg/m3
cement = 412 kg /m3
calculation of course and fine aggrgate
volume of concrete = 1m3
vol. of admixture =4.99/(1.145*1000)
= 0.004
total wt of C.A = 01308+0.1440+0.0043
= 0.2791m3
vol. of c.a = 1-0.2791
= 0.7209
vol. of f.agg = 0.7209-02379
=0.2379
vo. of c.a = 0.7209-0.2379
= 0.4830
wt of f.a = 0.2379*2.61*1000
=620.914kg/m3
wt of c.a = 0.4830x2.65x1000
= 1282.365kg /m3
considered
cement = 412x1.025
=422 kg
water = 144x1.025
= 147.6kg
f.a = 621kg
c.a = 706kg