Project on Concrete Mix Design for various grades of Concrete
Ans. Data collection
Grade of concrete = M35
Exposure condition = moderate
grade of cement=M33(OPC)
fck=35N/mm^2
std. Deviation for M35= 5N/mm^2
Target mean strength of concrete=ft=fck+1.65*S
ft=35+1.65*5
ft=43.25
Target mean strength of concrete=43.25 N/mm^2
Step1
Using graph 1
Here, 28 days compressive strength of concrete= target mean strength=43.25 N/mm^2
Hence, we assume w/c ratio =0.35
Step2
GRAPH2
28 days compressive strength of concrete =target mean strength=43.25 N/mm^2
Hence, we use curve line E
we assume w/c ratio=0.41
Step3
Exposure condition=Maderate
From the table 5
Here, the limited value of the water-cement ratio is 0.50 for mmoderate environmmentalexposure
Hence, we take w/c ratio =0.50
now compaire all above the three values,
0.35
0.41
0.50
The lower value of the above three value is adopted as w/c ratio for mix design
Hence, our w/c ratio is 0.35
*Sellection of cement content:
-It will be calculated from w/c ratio
-Minimum cement content is taken from table-5 of IS 456:2000
W=186kg
w/c=0.35
C=186/0.35=531.43 Kg
Estimated water content for 100 mm slump
=186+(6/100)X186
=197L
As superplasticizer is used, the water content can be reduced up to 20% and above based on trails water content reduction of 29% is achieved with plasticizer
Hence the arrived water content=197X0.71
=140L
VI. Calculate the cement content
w/c ratio=0.35
Cement content=140/0.35=400Kg/m^3
From Table -5 of IS 456 monimum cement content for 'severe' exposure condition= 320Kg/m^3
400Kg/m^3>320Kg/m^3 hence ok
To proportion a mix containing fly ash, the following steps are to be followed.
1. The % of fly ash need to be decided based on project requrement and quality of materials
2. Increase in cementitous material content is by 10%
cementitious material content=400*1.10=440 Kg/m^3
Water content=140L
w/c ratio=140/440=0.32
Flyash at 30% of cement=440*(30/100)=132 Kg/m^3
cement=440-132=308 Kg/m^3
Seving of cement using flyash=400-308=92 Kg/m^3
vii) Proportion of volume of CA and FA concrete
From table 3 of IS 10262:2009, volume of CA corresponding to 20mm size aggregate and FA (zone1) for w/c ratio=0.60
in the present case the water cement ratio is 0.35
- There fore the volume of CA need to be increased to decrease the FA content
-As w/c ratio is lower by 0.10, the proportion of volume of CA is increased by 0.02 (at the rate of -/+ 0.01 for every +-0.05 charge in w/c ratio
Therefore, corrected volume of CA for the w/c ratio of 0.35=0.62
(Note: if CA is not a angular types then also the volume of CA may be required to be increased suitably based on experience)
For pumpable concrete these value are reduced by 10%
Therefore, volume of CA=0.62X0.9=0.56
volume of FA content=1-0.56=0.44
viii) Mix calculation:
a) volume of concrete=1m^3
b) volume of cement=400/3.15 X 1/1000
=0.13 m^3
c) volume of water=140/1 X 1/1000
=0.140 m^3
d)volume of chemical admixture=7/1.145 X1/1000
=0.006m^3
e) volume of aggregate(FA & CA)=(a-(b+c+d))
=(1-(0.13+0.140+0.006))
=0.743 m^3
f) mass of CA=0.743 X 0.56 X 2.74 X 1000
=1140 Kg/m^3
g) mass of FA = 0.743 x 0.44 x 2.74 x 1000
=896 Kg/m^3
mix proportion for trial noumber 1:
cement = 350 kg/m^3
water = 140 kg/m^3
FA = 896 kg/m^3
CA = 1140 kg/m^3
chemical Adnixture= 7 kg/m^3
w/c ratio = 0.4
Trial mix ratio =1:2.56 : 3.26
FOR M50 GRADE CONCRETE WITHOUT FLY ASH
Data collection--
Grade of concrete =M50
Exposure condition = moderate
Grade of concrete = M33 (OPC)
f'ck= 50 N/mm^2
Std deviation for M50= 5N/mm^2
Target mean strength of concrete =50+1.65X5
=58.25 N/mm^2
From the graf IS 10262 : 2019
Assume w/c ratio =0.27
Selection of cement content:
it can be calculated from w/c ratio
minimum cement content is taken from Table-5 of IS 456:2000
W=186 kg
w/c=0.27
C=186/0.27
=688.89 kg
Estimated water content for 100 mm slump
=186+6/100X186
=197L
Same as the M35 grade of concrete the arrived water content=197X0.71
=140L
vi) calculation of cement content
w/c ratio=0.27
cement content=140/0.27=518.52 Kg/m^3
hence ok from table 5 IS 456
vii) Proportion of volume of CA and FA concrete
From table 3 of IS 10262:2009, volume of CA corresponding to 20mm size aggregate and FA (zone1) for w/c ratio=0.60
in the present case the water cement ratio is 0.27
- There fore the volume of CA need to be increased to decrease the FA content
-As w/c ratio is lower by 0.10, the proportion of volume of CA is increased by 0.02 (at the rate of -/+ 0.01 for every +-0.05 charge in w/c ratio
Therefore, corrected volume of CA for the w/c ratio of 0.27=0.62
(Note: if CA is not a angular types then also the volume of CA may be required to be increased suitably based on experience)
For pumpable concrete these value are reduced by 10%
Therefore, volume of CA=0.62X0.9=0.56
volume of FA content=1-0.56=0.44
viii) Mix calculation:
a) volume of concrete=1m^3
b) volume of cement=518.52/3.15 X 1/1000
=0.164 m^3
c) volume of water=140/1 X 1/1000
=0.140 m^3
d)volume of chemical admixture=7/1.145 X1/1000
=0.006m^3
e) volume of aggregate(FA & CA)=(a-(b+c+d))
=(1-(0.164+0.140+0.006))
=0.69 m^3
f) mass of CA=0.69 X 0.56 X 2.74 X 1000
=1058.74 Kg/m^3
g) mass of FA = 0.69 x 0.44 x 2.74 x 1000
=831.86 Kg/m^3