Project on Concrete Mix Design for various grades of Concrete
1) Calculate the concrete mix design for M35 grade concrete with fly ash
Given ,
M35 grade concrete withfly ash
Mix Design
a) target mean strength
grade of concrete = M35
fck = 35N/mm2 ( FROM TABLE 1 OF IS 10262: 2019 )
s = 5 N/mm2
f'ck = fck + 1.65 S
= 35 + ( 1.65 x 5 )
43.25 N/mm2
b) obtain the w/c ratio :
from the graph for the corresponding 28 days compressive strength which is the target mean strength abtain W/C ratio
w/c ratio = 0.37
based of given exposure condition , from table 5 of IS 456
w/c = 0.45
the lower value is taken w/c ratio = 0.37
1.c) selection of water content
,
max water content = 186 lit
considering slump as 100 mm = 186 + (6/100) 186
= 197.16
1.d) calculation of cement content :
w/c ratio = 0.37
cement content = 197.16 /0.37 = 532.86 kg/m3
= 533 kg / m3
cementitious material content
= 533 x 1.10 = 586.3
= 586 kg/m3
water content = 197 kg/m3
so, water cementitious ratio
= 197/ 586 =0.336
fly ash @ 30% of total cementittious material content
= 586X 30% = 175.8 = 176
cement (OPC) = 586 - 176 = 410 kg/m3
table 5 of IS 456 , minimum cement cementitious content exposure condition
= 320 kg/m3
586 kg / m3 > 320 kg / m3 , hence ok
1.e) proportion of volume of CA and FA
the proportionate volume of CA to 20mm size aggregate and FA (zone II) for water ratio of 0.50 = 0.62
in our case W/C ratio is 0.37 , volume of CA is required to increase and decrease FA as , W/C ratio is lower by 0.13
volume of coarse aggregates is increased by 0.026
volume of coarse aggregate = 0.62 + 0.026 = 0.646
volume of fine aggregate content = 1 - 0.646 = 0.354
1.f) mix calculation
a ) volume of concrete = 1m3
b) volume of entrapped air in wet concrete = 0.01m3
c ) volume of cement = (mass of cement / sp. gravity of cement) x (1/1000)
= ( 410 / 3.15 ) X 1 / 1000
= 0.130 m3
d ) volume of fly ash= ( mass of fly ash / specific grvity of fly ash
X 1 / 1000
= 176 / 2.2 X 1 / 1000
= 0.08 m3
e )volume of water = ( mass of water / spec gravity of water )
X 1 / 1000
= 197 / 1 X 1 / 1000
= 0.197 m3
f ) volume of chemical admixture ( superplasticizer ) @ 1.0 % by mass of cementitious material
= ( mass of chemical admixture / spec gravity of admixture ) X 1 / 1000
= 5.86 / 1.145 X 1 / 1000
0.005 m3
g ) volume of all in aggregate = [(a-b)-(c + b +e + f)]
= (1 - 0.01 ) - ( 0.130 + 0.01 + 0.197 + 0.005)
= 0.648 m3
h) mass of coarse aggregate
= g X volume oF CA X specific gravity of CA X 1000
= 0.648 X 0.64 X 2.74 X 1000
= 1136.33 = 1136 kg
j) mass of FA
= g X volume of FA X specific gravity of FA X 1000
= 0.648 X 0.35 X 2.65 X1000
=601.02 = 601 kg
Cement = 410 kg/m3
fly ash = 176 kg/m3
water = 197 kg/m3
fine aggregate = 601 kg/m3
coarse aggregate = 1136 kg/m3
chemical admixture = 5.86 kg/m3
free water - cementitious material ratio
= 0.336
concrete mix degsin for M50 grade concrete
Given ,
M50 grade concrete without fly ash
Mix Design
a) target mean strength
grade of concrete = M50
fck = 50N/mm2 ( FROM TABLE 1 OF IS 10262: 2019 )
s = 5 N/mm2
f'ck = fck + 1.65 S
= 50 + ( 1.65 x 5 )
58.25 N/mm2
b) obtain the w/c ratio :
from the graph for the corresponding 28 days compressive strength which is the target mean strength abtain W/C ratio
w/c ratio = 0.28
based of given exposure condition , from table 5 of IS 456
w/c = 0.35
the lower value is taken w/c ratio = 0.28
1.c) selection of water content
,
max water content = 186 lit
considering slump as 120 mm = 186 + (8.4/100) X 186
= 201.624 = 202kg/m3
As superplasticizer (Polycarboxylate ether based) is
used, the water content can be reduced by 30 percent.
Hence, the reduced water content
= 202 - 30%
= 141.4 = 141 kg/m3
1.d) calculation of cement content :
w/c ratio = 0.28 ,
cement content = 141 /0.28 = 503.57 kg/m3
= 503kg / m3
the cementitious material content
= 503 X 1.10 = 553 kg / m3
revised w/cm = 141/ 553 = 0.254
1 e ) proportion of volume of CA and FA
from
From Table 10, volume of coarse aggregate
corresponding to 20 mm size aggregate and fine
aggregate grading Zone II 0.5= 0.66
in our case the W/C ratio is 0.28 whuch is lower by 0.22, the volume of CA is increased by 0.039
vol of CA = 0.039
vol of FA = 1 - 0.039 = 0.961
1 f )
1a ) volume of concrete = 1m3
b) volume of entrapped air in wet concrete = 0.01m3
c ) volume of cement = (mass of cement / sp. gravity of cement) x (1/1000)
= 503 /3.15 X 1 / 1000 = 0.159m3
d )volume of water = ( mass of water / spec gravity of water
= 141 /1 X 0.001
= 0.141 m3
e)
Volume of chemical admixture
(superplasticizer) (@ 0.5 percent by mass of
cementitious material)
= mass of chemical admixture / specific gravity of admixture
X 1 / 1000
= 553 X 0.5% / 1.08 X 0.001
= 0.0025 m3
f) volume of all in aggregates
= [ (a -b ) - ( c + d + e )]
[ ( 1 - 0.01 ) - ( 0.159 +0.141 + 0.0025 )]
= 1 - 0.01 - 0.3025
= 0.686 m3
g ) Mass of coarse aggregate
= f × Volume of coarse aggregate × Specific
gravity of coarse aggregate × 1 000
= 0.686 X 0.036 X 2.74 X 1000
= 676 kg
h )
) Mass of fine aggregate
= f × volume of fine aggregate × Specific
gravity of fine aggregate × 1 000
= 0.686 X 0.961 X 2.65 X 1000
= 174 kg/m3
cement = 553 kg/m3
water = 141 kg/m3
fine aggregates = 174 kg/m3
coarse aggregates = 676 kg/m3
chemical admixture = 0.0025m3
w/cm = 0.254