Project on Concrete Mix Design for various grades of Concrete
Calculate the Concrete Mix Design for M35 grade concrete with fly ash?
Ans.
Concrete mix design for M35 gade of concrete
Given data:
Gade of concrete = M35
Add mixture used= Fly ash
Assume,
type of cement = OPC 43 grade Confirming to IS 8112
Minimum cement content = 320 kg/m^3
Max size of aggregate=20mm
Max w/c ratio = 0.45 ( From table 5 IS 456:2000-Exposure condition)
Chemical admixture type = Superplasticizer
workability = 100mm
exposure condition = Severe
Method of concrete = Pumping
Type of aggregate = Crushed angular aggregate
Max cement content = 450kg/m^3
Specific gravity:
Cement=3.15
F A = 2.74
C A = 2.74
Fly ash = 2.2
Water absoprption:
C A = 0.5%
F A = 1%
Step1: Target mean strength for mix proportion:
Ft= Fck+1.65*S
Ft = Traget avarege compressive strength at 28 days.
Fck= Compressive strength at 278 days.
S = Stander deviation.
From Table 1 IS 10262:2009. Standerd deviation S=5N/mm^2
Ft=35+1.65*5
Ft= 43.25N/mm^2
Step 2: Selection of w/c ratio:
From table 5 of IS456-2000 for Severe exposure, Max W/c ratio = 0.45
Based pn the experience adopt W/c ratio as 0.40
0.40<0.45 Hence O.K.
Step 3: Selection of water content:
From table 2 of IS 10262:2009, max water content for 20mm aggregate is 186kg=186lit.
The slump range for pumping concrete is 100mm
therefor, Water content=186+186*6/100
=197ltr
As superplasticizer used, the water content can be reduced upto 30%
Based on the trails water content reduction of 28% is achived with plasticizer.
Hence arrived water content = 197*0.72 = 142ltr.
Step 4: Calculation of cement content:
W/c ratio = 0.40
Cementitious materials(cement+flyash)content = 142/0.4
= 355kg/m^3
From table 5 of IS 456 maximum cement content for sevre exposure condition=320kg/m^3.
Hence, 355kg/m^3 > 320kg/m^3.
Now proportion of mix containing fly ash the following steps are suggested,
a) The percentage of flyash needs to be decided based on project requirement and quality of materials,
b) In some casesincrease in cementaseous matireals can be warrented. Increase in cementaseous material content and its percentage may be based on experience and trails
Cementacious matireals content = 355*1.10 = 390.5kg/m^3
Water content = 142 Kg/m^3
Water-cement ratio = 142/390.5 = 0.363
Flyash @ 30% total cemetaceous content = 390.5*0.3 = 117.15Kg/m^3
Cement ( OPC ) = 390.5-117.15 = 273.35Kg/m^3
Saving of cement while using fly ash = 350-273.35 = 81.65Kg/m^3
Flyash being utilised = 117.15Kg/m^3
Step 5: Proportion of volume of CA nd FA content:
From table 3 of IS10262-2009, Volume of CA corresponding to 20mm size of aggregate and fine aggregate (Zone I) for W/c Ratio of 0.5=0.6
In the present case of the W/cRatio is 0.40
- Therefore, the volume of CA is needs to icrease by decreasing volume of FA content
- As w/c ratio is lower by 0.10, the proportion of volume of CA is increased by 0.02.
Therefore, the corrected w/c ratio of 0.4=0.62
For Pumpable concrete these values are reduced by 10%
Therefore, Volume of CA=0.62*0.9 = 0.56
Volume of FA = 1-0.56 = 0.44
Step 6: Mix Calculations:
a) Volume of concrete =1m^3
b) Volume of cement = Mass of cement/(specific gravity of cement*1000)
= 273.35/(3.15*1000)
= 0.0867m^3
c) Volume of flyash = Mass of flyash/(specific gravity of flyash*1000)
= 117.15/(2.2*1000)
= 0.053
d) Volume of chemical admixture = Mass of chemical admixture/(specific gravity of chemical admixture*1000)
= 7.6/(1.145*1000)
= 0.006m^3
e) Volume of water = Mass of water/(specific gravity of water*1000)
= 142/(1*1000)
= 0.142m^3
f) Volume of aggregate (CA&FA) = a-b-c-d-e
= 1-0.0867-0.053-0.006-0.142
= 0.712m^3
g) Mass of CA = Volume of aggregate x Volume of CA x Specific gravity of CA x 1000
= 0.712 x 0.56 x 2.74 x 1000
= 1092 Kg/^3
h) Mass of FA = Volume of aggregate x Volume of FA x Specific gravity of FA x 1000
= 0.712 x 0.44 x 2.74 x 1000
= 858 Kg/m^3
Step 7: Mix proportions for Trail numer 1:
Cement = 273.35 Kg/m^3
Fly ash = 117.15 Kg/m^3
Water = 142 Kg/m^3
F A = 858 Kg/m^3
C A = 1092 Kg/^3
Chemical admixture = 7 Kg/m^3
W/c ratio = 0.363
Trail mix ratio = 1:2.19:2.79
Calculate the Concrete Mix Design for M50 grade concrete without Fly ash
Ans:
Concrete mix design for M35 gade of concrete
Given data:
Gade of concrete = M50
Assume,
type of cement = OPC 43 grade Confirming to IS 8112
Minimum cement content = 320 kg/m^3
Max size of aggregate=20mm
Max w/c ratio = 0.45 ( From table 5 IS 456:2000-Exposure condition)
Chemical admixture type = Superplasticizer
workability = 100mm
exposure condition = Extreme
Method of concrete = Pumping
Type of aggregate = Crushed angular aggregate
Max cement content = 450kg/m^3
Specific gravity:
Cement=3.15
F A = 2.74
C A = 2.74
Fly ash = 2.2
Water absoprption:
C A = 0.5%
F A = 1%
Step1: Target mean strength for mix proportion:
Ft= Fck+1.65*S
Ft = Traget avarege compressive strength at 28 days.
Fck= Compressive strength at 278 days.
S = Stander deviation.
From Table 1 IS 10262:2009. Standerd deviation S=5N/mm^2
Ft=50+1.65*5
Ft= 58.25N/mm^2
Step 2: Selection of w/c ratio:
From table 5 of IS456-2000 for Severe exposure, Max W/c ratio = 0.4
Based on the experience adopt W/c ratio as 0.38
0.38<0.4 Hence O.K.
Step 3: Selection of water content:
From table 2 of IS 10262:2009, max water content for 20mm aggregate is 186kg=186lit.
The slump range for pumping concrete is 100mm
therefor, Water content=186+186*6/100
=197ltr
As superplasticizer used, the water content can be reduced upto 30%
Based on the trails water content reduction of 28% is achived with plasticizer.
Hence arrived water content = 197*0.72 = 142ltr.
Step 4: Calculation of cement content:
W/c ratio = 0.38
Cementitious materials(cement+flyash)content = 142/0.38
= 373 kg/m^3
From table 5 of IS 456 maximum cement content for sevre exposure condition=320kg/m^3.
Hence, 373 kg/m^3 > 320kg/m^3.
Step 5: Proportion of volume of CA nd FA content:
From table 3 of IS10262-2009, Volume of CA corresponding to 20mm size of aggregate and fine aggregate (Zone I) for W/c Ratio of 0.5=0.6
In the present case of the W/cRatio is 0.38
- Therefore, the volume of CA is needs to icrease by decreasing volume of FA content
- As w/c ratio is lower by 0.10, the proportion of volume of CA is increased by 0.02.
Therefore, the corrected w/c ratio of 0.38=0.624
For Pumpable concrete these values are reduced by 10%
Therefore, Volume of CA=0.624*0.9 = 0.561
Volume of FA = 1-0.561 = 0.439
Step 6: Mix Calculations:
a) Volume of concrete =1m^3
b) Volume of cement = Mass of cement/(specific gravity of cement*1000)
= 373/(3.15*1000)
= 0.1184m^3
c) Volume of chemical admixture = Mass of chemical admixture/(specific gravity of chemical admixture*1000)
= 7.6/(1.145*1000)
= 0.006m^3
d) Volume of water = Mass of water/(specific gravity of water*1000)
= 142/(1*1000)
= 0.142m^3
e) Volume of aggregate (CA&FA) = a-b-c-d
= 1-0.1184-0.006-0.142
= 0.7336m^3
f) Mass of CA = Volume of aggregate x Volume of CA x Specific gravity of CA x 1000
= 0.7336 x 0.561 x 2.74 x 1000
= 1128 Kg/^3
g) Mass of FA = Volume of aggregate x Volume of FA x Specific gravity of FA x 1000
= 0.7336 x 0.439 x 2.74 x 1000
= 882 Kg/m^3
Step 7: Mix proportions for Trail numer 1:
Cement = 373 Kg/m^3
Water = 142 Kg/m^3
F A = 882 Kg/m^3
C A = 1128 Kg/^3
Chemical admixture = 7 Kg/m^3
W/c ratio = 0.38
Trail mix ratio = 1:2.24:2.87