Project on Concrete Mix Design for various grades of Concrete
Calculate the Concrete Mix Design for M50 grade concrete without fly ash
Aim:
To Calculate the Concrete Mix Design for M50 grade concrete without fly ash.
Introduction:
In order to calculate the concrete mix design,
- Calculate the Target mean strength of the concrete
- To select the Water Cement ratio
- To calculate the Water Content
- To calculate the Cement Content
- Volume of Coarse aggregate and Fine Aggregate per unit volume
- To calculate Mix Calculation
- To calculate the Mix Proportions
Procedure:
M50
Step 1 : To find Target strength of concrete
F' ck = F ck + 1.65S
To find Target strength of concrete ?
Standard Deviation - IS 10262 : 2009, Table 1
= 50 + 1.65(5.0)
= 58.25 N/mm^2
F' ck = 58.25 N/mm^2
Step 2 : To calculate the Water Cement ratio
IS 10262 : 2019 , Table no 4
Water cement ratio = 0.34
IS 456 : 2000 , Table 5
Assume weather ( Severe ) = 0.45
so the lower value of water cement ratio is 0.34. reducing the W/c ratio which gives more strength to concrete.
Step 3 : To calculate the Water Content:
To find Water Content , IS 10262 2019, Table no 4, Page 5
Aggegate nominal size = 20mm
=186 kg
Here we have slump value is 100.
for every 25 increase in slump 3% water content must be added. we have slump value of 100 .
6% of water content added
= 186+6/100X186
= 197.2kg
Here we are adding super plasticizer to reduce water content .
we can reduce water content from 20 to 30%
Here i reduced 30%
= 197.2 X 0.7
= 138 litres
Water content is 138 litres.
Step 4 : To calculate the Cement Content:
Water content /Cement = water cement ratio
138/ cement = 0.34
Cement = 405 kg
Step 5 : To calculate the Volume of Coarse aggrefate and fine aggregate per unit volume:
IS 10262 : 2009 , Table 3
Nominal size of aggregate 20mm, for zone 1
= 0.60
Volume of Coarse Aggregate = 0.60
This formula mentioned in clause if water cement ratio is below 0.50
= [0.5- water cement ratio/0.05 X 0.01] + volume of aggregate nominal size according to zone
= [0.5-0.34/0.05 X 0.01] +0.60
= 0.632
Volume of coarse aggregate is 0.632 . but we are pumping concrete 10% should be reduced in coarse aggregate.
= 0.632 X 0.9
= 0.568
volume of Coarse aggregate = 0.568 .
To find Fine Aggregate ?
Fine Aggregate = 1 - 0.568
Fine Aggregate = 0.432
Step 6 : Mix Calculation :
(a) Volume of concrete = 1m^3 is taken
(b) volume of Cement = Mass of cement /specific gravity of cement X 1/1000
= 405/3.15 X 1/1000
= 0.128m^3
(c) Volume of water = Mass of water /specific gravity of waterX 1/1000
= 138/1 X 1/1000
= 0.138m^3
(d) Volume of Chemical Admixture = Mass of Chemical Admixture /specific gravity of Chemical Admixture X 1/1000
= 8.1/1.145 X 1/1000
= 0.007m^3
(e) Volume of Aggregate = (a- (b+c+d))
= (1-(0.128+0.138+0.007)
= 0.727 m^3
Mass of Coarse Aggregate = e X volume of coase aggregate X specific gravity of coarse aggregate X 1000
= 0.727 x 0.568 x 2.74 x 1000
= 1131 Kg/m^3
Mass of fine Aggregate = 0.727 x 0.432 x 2.74 x 1000
= 860 kg/m^3
Step 7 : Mix proportion :
Mix proportion
Cement = 405 kg/m^3
Water = 138 kg/m^3
Fine Agregate = 860 kg/m^3
Coarse Aggregate = 1131 kg/m^3
Chemical Admixture = 8.1
Water cement ratio = 0.34
Mix Proportion = 1 : 2.12 : 2.79
Calculate the Concrete Mix Design for M35 grade concrete with fly ash
Aim:
To Calculate the Concrete Mix Design for M35 grade concrete with fly ash.
Introduction:
In order to calculate the concrete mix design,
- Calculate the Target mean strength of the concrete
- To select the Water Cement ratio
- To calculate the Water Content
- To calculate the Cement Content
- Volume of Coarse aggregate and Fine Aggregate per unit volume
- To calculate Mix Calculation
- To calculate the Mix Proportions
M35
Step 1 : To find Target strength of concrete
F' ck = F ck + 1.65S
To find Target strength of concrete ?
Standard Deviation - IS 10262 : 2009, Table 1
= 35 + 1.65(5.0)
= 43.25 N/mm^2
F' ck = 43.25 N/mm^2
Step 2 : To calculate the Water Cement ratio
IS 10262 : 2019 , Table no 4
Water cement ratio = 0.38
IS 456 : 2000 , Table 5
Assume weather ( Severe ) = 0.45
so the lower value of water cement ratio is 0.38. reducing the W/c ratio which gives more strength to concrete.
Step 3 : To calculate the Water Content:
To find Water Content , IS 10262 2019, Table no 4, Page 5
Aggegate nominal size = 20mm
=186 kg
Here we have slump value is 75.
for every 25 increase in slump 3% water content must be added. we have slump value of 75 .
3% of water content added
= 186+3/100X186
= 192 kg
Here we are adding super plasticizer to reduce water content .
we can reduce water content from 20 to 30%
Here i reduced 30%
= 192 X 0.7
= 134 litres
Water content is 134 litres.
Step 4 : To calculate the Cement Content:
Water content /Cement = water cement ratio
134/ cement = 0.38
Cement = 352 kg
Fly ash added to the cement based on the requirement of project . Here we added 10 % of fly ash
Cementious cement ( Fly Ash ) = 352 X 1.10
= 387 kg/m^3
Cementious cement = 387 kg/m^3
Water content = 134 kg/m^3
so ,water cement ratio = 134/387
= 0.346
Fly ash 30% cementious material
= 387 X 0.3
= 116 kg/m^3
Cement = 387 - 116
cement = 271 kg/m^3
Total saving of cement by using Fly ash = 352 - 271
= 81 kg/m^3
Total saving of cement by using Fly ash = 81 kg/m^3
The water cement ratio is 0.34 .
Step 5 : To calculate the Volume of Coarse aggrefate and fine aggregate per unit volume:
IS 10262 : 2009 , Table 3
Nominal size of aggregate 20mm, for zone 1
= 0.60
Volume of Coarse Aggregate = 0.60
This formula mentioned in clause if water cement ratio is below 0.50
= [0.5- water cement ratio/0.05 X 0.01] + volume of aggregate nominal size according to zone
= [0.5-0.34/0.05 X 0.01] +0.60
= 0.632
Volume of coarse aggregate is 0.632 . but we are pumping concrete 10% should be reduced in coarse aggregate.
= 0.632 X 0.9
= 0.568
volume of Coarse aggregate = 0.568 .
To find Fine Aggregate ?
Fine Aggregate = 1 - 0.568
Fine Aggregate = 0.432
Step 6 : Mix Calculation :
(a) Volume of concrete = 1m^3 is taken
(b) volume of Cement = Mass of cement /specific gravity of cement X 1/1000
= 271/3.15 X 1/1000
= 0.086m^3
(c) Volume of water = Mass of water /specific gravity of waterX 1/1000
= 134/1 X 1/1000
= 0.134m^3
(d) Volume of Chemical Admixture = Mass of Chemical Admixture /specific gravity of Chemical Admixture X 1/1000
= 5.42/1.145 X 1/1000
= 0.004m^3
(e) Volume of Aggregate = (a- (b+c+d))
= (1-(0.086+0.134+0.004)
= 0.776 m^3
Mass of Coarse Aggregate = e X volume of coarse aggregate X specific gravity of coarse aggregate X 1000
= 0.776 x 0.568 x 2.74 x 1000
= 1207 Kg/m^3
Mass of fine Aggregate = 0.776 x 0.432 x 2.74 x 1000
= 918 kg/m^3
Step 7 : Mix proportion :
Mix proportion
Cement = 271 kg/m^3
Water = 134 kg/m^3
Fine Agregate = 918 kg/m^3
Coarse Aggregate = 1207 kg/m^3
Chemical Admixture = 5.42
Water cement ratio = 0.34
Mix Proportion = 1 : 3.38 : 4.45