Project on Concrete Mix Design for various grades of Concrete
1. The concrete design mix of M35 grade of concrete using Flyash
1. STIPULATIONS FOR PROPORTIONING
a. Grade of concrete = M35
B. Cement type = opc 43 grade
C. Type of mineral admixture flyash conforming to IS 3812 to part 1
D. Maximum nominal size of aggregates 20 mm
E. Minimum cement content 320 kgs/m3
F. Maximum water cement ratio 0.45
G. Workability 100 mm slump for pumpable concrete
H. Exposure condition extreme
I. An extreme degree of supervision good
J. Method of concrete placing = pumping
K. Type of aggregates = crushed angular aggregates
L. Maximum cement content 450kg/m3
M. Chemical admixture type super plasticiser
Step 2
Test data for materials
A.Cement used obcy 43 grade spellspell
B.Specific gravity of cement 3.15
C.Fly ash
D.Specific gravity of flyash = 2.2
E.Chemical admixture super plasticiser
F. Specific gravity
Coarse aggregates =2.74
Find aggregates = 2.74
G. Water absorption
Coarse aggregate 0.5%
Fine aggregate = 1%
H. Free surface moisture
Coarse aggregate = Nil
Fine aggregates = Nil
Step 3
Target mean strength
f'ck=fck+1.65 * s
=35+ 1.85 * 5
=43.25 N/mm2
Step 4:
Selection of water cement ratio
From table 5 IS 456-2000 for severve condition
Maximum W/C ratio = 0.40
Based on experience adopted W/C rario = 0.40
0.40<0.45 Hence ok
Step 5:
Selecttion of water ceontent
FRom table 2 IS 10262-2009 water content for 20mm coarse aggregates - 186 for 25 to 50mm slump ranges
est water content for 100mm slump
= 186+6/100 * 186
=197 kg or lit
Super plastisizer is used the water content can be reduced 20% and above based on trails
Water content reduction of 29% is achived with plastisizers
Hence arrived water content = 197 * 0.71 = 140 lit
Step 6:
Calculation of cement content
W/C ratio = 140/0.4 = 350kg/m3
cement content = 320kg/m3
from IS 456-2000 table 5 cement content for severe expousure condition
350>320 kg/m3
Hence ok
To proportion mix containing flyash the following steps are suggested
a.The percentage of flyash needs to be decided based upon the project requirement and quality of materials
b.In some cases increase the cementitious material can be warned . Increase in cementitious material content and its percentage is based on experience and trail
In this increase of 10% cemnetitious material content is considered
Cementitious material content = 350 * 1.1 = 385kg/m3
water content = 140kg/m3
W/C ratio = 140/385 = 0.364
Fly ash 30% total cementitious material content = 385 * 30 = 115kg/m3
cement = 385-115 = 270kg/m3
saving of cement while using fly ash = 385-270 = 80kg/m3
Fly ash can be utilized = 115kg/m3
Step 7
Proportion of volume of coarse aggregates and fine aggregate content
From table 3 of IS 10262-2009 Vol of CA corresponding to 20mm sizs of aggregates and FA ( zone 1 )
In the present case the W/C ratio is 0.40
As W/C ratio is lower by 0.10 the proportion of vol of CA is increased by 0.02 at the rate of -/+0.01 for every +/-0.05 change in W/C ratio
Therefore corrected vol of CA for W/C ratio of 0.40 = 0.64
for Pumpable concrete therse values are reduced by 10%
Vol of CA = 0.64 * 0.9 = 0.576
Vol of FA = 1= 0.576 = 0.424
Step 8
Mix calculations
a.VOl of concrete = 1m3
b.Vol of cement = (Mass of cement / sp.gravity of cement ) * 1/1000
= 270/3.15 *1/1000
= 0.086m3
c, Vol of flyash = (115 / 2.2 )* 1/ 1000
= 0.052 m3
d. Vol of water = (140 / 1 ) *1/1000
= 0.140 m3
e Vol of chemical admixtues = (7.6 / 1.145 ) * 1/1000
= 0.006m3
f. Vol of aggregates ( FA & CA ) = (a-(b+c+d+e))
=(1-(0.086+0.052+0.140+0.006))
=0.716 m3
g. mass of CA = f * Vol of CA * sp.gravity of CA * 10000
= 0.716 * 0.576 * 2.74 *1000
=1130kg/m3
h. Mass of FA = 0.716 * 0.424 * 2.74 *1000
=831kg/m3
Cement = 270kg/m3
flyash = 115kg/m3
water = 140 kg/m3
fine aggregates = 831 kg/m3
coarse aggregates = 1130 kg/m3
chemical admixture = 7kg/m3
W/C ratio = 0.364
Mix ratio = 1: 2.15: 2.94
MIX DESIGN OF M50 GRADE OF CONCRETE
Step 1
Stipulation for design
a.Grade designation : M 50
b.Type of cement : OPC 53 grade confirming to IS 8112
c.Maximum nominal size of aggregate : 20 mm
d.Minimum cement content : 360 kg/m3
e.Maximum water-cement ratio : 0.35
f.Workability : 100 mm (slump)
g.Exposure condition : Extreme (for reinforced concrete)
h.Method of concrete placing : Pumping
i.Degree of supervision : Good
j.Type of aggregate : Crushed angular aggregate
k.Maximum cement content : 450 kg/m3
l.Chemical admixture type : Super Plasticizer
Step 2
The following data are collected by performing standard tests in a laboratory.
Specific gravity of cement : 3.12
Specific gravity of Coarse Aggregate : 2.74
Specific gravity of Fine Aggregate : 2.56
Step 3
Target mean strength
f'ck = fck+1.65 *s
= 50 + 1.65 * 5
= 58. 25 N/mm2
Step 4
Selection of water cement ratio
From table 5 IS 456-2000 for severve condition
Maximum W/C ratio = 0.35
Based on experience adopted W/C rario = 0.35
Step 5
Selecttion of water ceontent
FRom table 2 IS 10262-2009 water content for 20mm coarse aggregates - 186 for 25 to 50mm slump ranges
est water content for 100mm slump
= 186+6/100 * 186
=197 kg or lit
Super plastisizer is used the water content can be reduced 20% and above based on trails
Water content reduction of 27% is achived with plastisizers
Hence arrived water content = 197 * 0.73 = 143 lit
Step 6
Water-cement ratio = 0.35
Cement content = 143/0.35 = 410 kg/m3
From Table 5 of IS 456-2000, minimum cement content for ‘Extreme exposure’ condition = 360 kg/m3
410 kg/m3 > 360 kg/m3, hence, O.K.
Step 7
From Table 3 of IS 10262-2009 . volume of coarse aggregate corresponding to 20 mm size aggregate and fine aggregate (Zone II) for a water-cement ratio of 0.50 =0.62.
Therefore. the corrected proportion of the volume of coarse aggregate for the water-cement ratio of 0.35 = 0.65
For pumpable concrete, these values should be reduced by 10 percent.
Therefore, volume of coarse aggregate = 0.65 x 0.9 =0.585
Volume of fine aggregate content =1 – 0.585 = 0.415
Step 8
Mix Calculations
a) Volume of concrete = 1m3
b) Volume of cement = (Mass of Cement/ Its Specific Gravity ) X (1/1000)
= (410/3.12) X (1/1000)
= 0.131 m3
c) Volume of Water = (Mass of water/specific gravity of water) X (1/1000)
= (143/1) X (1/1000)
= 0.143 m3
d) Volume of Superplasticizer (2%) = (Mass of Superplasticizer / Specific gravity of admixture) X (1/1000)
= (8.2/1.145) X (1/1000)
= 0.007 m3
e) Volume of all aggregate = (a – (b + c + d))
= 1 – (0.131 + 0.143+0.007)
= 0.719 m3
e) Mass of coarse aggregate
= 0.719 x 0.585 x 2.74 x 1000
= 1152.48 kg
f) Mass of fine aggregate
= 0.719 x 0.415 x 2.56 x 1000
= 763.86 kg
Mix Proportions
Cement = 410 kg/m3
Water = 143 kg/m3
Super plasticizer = 7 kg/m3
Fine aggregates = 763.86 kg/m3
Coarse aggregate = 1152.48 kg/m3
W/c = 0.35
Mix ratio 1: 1.86 : 2.81