MAKING A MIX DESIGN ACCORDING TO INDIAN STANDARD 10262:2019
For grade M 80 for severe condition as per IS 10262:2019
1) Let start with target mean strength : fck+1.65*standard deviation or fck +8 (whichever greater)
80 +1.65*6 =89.9n/mm2 or 80 +8 =88n/ mm2
we take target mean strength : 89.9 n/mm2
2)water cement ratio from table no. 8 in is 10262 :2019 = 0.28 ( for max. aggregate size 10mm)
0.28 is less than 0.45
Hence ok
3) water content for aggregate MAS 10mm is : 200 kg/m3 (from table number 7) for 50mm slump.
( required slump is =200mm)
for every 25mm slump we increse 3% of water.
so for 200mm slump we require to increase 18% of water
so total water =200+ 200*(18/100)
=200 +36 = 236 kg/m3
we use high water reducing admixture who reduce water up to 27% as per test perform from marsh cone.
so final total water = 236 - 236*(27/100)
=236 - 63.72 =172.28 kg/m3
4) cementitious content = 172/0.28
= 614.28 = 614 kg/m3
614 is more than 320 kg/m3
Hence ok.
But according to experience and trial we increase 20% of cementitious material.
than cementitious material 614+614*(20/100)= 736.8 = 737 kg/m3
and water/cement ratio : 172/737= 0.233
5) From table no 10 of IS 10216:2019
volume of coarse aggregate for second zone of c/sand is : 0.54 ( for w/c = 0.30)
(for every decreament of 0.05 of water cement ratio volume of coarse aggregate increased by 0.01
and for every increment of 0.05 of water cement ratio volume of coarse aggregate decresed by 0.01 here water cement ratio from we make comparison is 0.30 come from table no 10 of 10262)
But in previous our actual water cement ration is : 0.233 ( 4 th step)
here water cement ratio decreased by 0.067 so the coarse aggregate volume increase.
for 0.05 water cement ratio voulme of coarse aggregate increase = 0.01
for 1 water cement ratio volume of coarse aggregate increase = 0.01/0.05= 0.2
for 0.067 water cement ratio volume of coarse aggregate increase by = 0.2*0.067 = 0.0134m3
so the volume of coarse aggregate is: 0.54+0.0134 = 0.5534m3
and fine aggregate voulme = 1-0.5534= 0.4466m3
6) Let cementitious content is = 737 kg
we use 8% as microsilica , 10 % alcofine and 30 percent GGBFS from this calculation is below:
for microsilica = (737*8/100)
= 58.96 kg/m3
for alcofine = (737*10/100)
= 73.7 kg/m3
for GGBFS = ( 737*30/100)
= 221.10 kg/m3
so cement content = 737-(58.96+73.7+221.10)
=737 - 353.76
=383.24 kg/m3
7) Entrapped air % = 1% or admixture is 1% of cementitious =7.37kg/m3
volume of concrete =1 m3
than volume of cement = 383.24 / 3.15*1000 = 0.121m3 ( 3.15=specific gravityof cement
volume of microsilica = 58.96 / 2.25*1000 =0.026 m3 (2.25 = specific gravity of microsilica)
volume of GGBFS = 221.10 / 2.9*1000 = 0.0762 m3
volume of alcofine = 73.7 / 2.9*1000 = 0.0254 m3
volume of admixture = 7.37 / 1.1*1000 = 0 .0067m3
volume of water = 172/1*1000 =0.172m3
volume of all in aggregate = 1-(0.01+0.121+0.026+0.0762+0.0254+0.0067+0.172)
= 1- 0.4273
= 0.5727m3
8) weight of coarse aggregate = 0.5727 *2.9*0.5534*1000 = 919.10 kg /m3 (2.9 s.gravity)
weight of fine aggregate = 0.5727*2.67*0.4466*1000= 683 kg/m3 (2.67 s.gravity)
So final mix design for M80 grade of concrete as per IS 10262 : 2019
cement =383 kg /m3
microsilica = 58.96 kg/m3
GGBFS = 221.10kg/m3
Alcofine = 73.7 kg/m3
10mm = 919.10 kg/m3
c/ sand = 683 kg/m3
water = 172 kg/m3
admixture = 7.37 kg/m3
YIELD = 1.00003 Hence ok
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