Project on Concrete Mix Design for various grades of Concrete
AIM:
To calculate the Concrete Mix Design for M35 grade Concrete with fly ash
Equations:
- Target mean strength: f’ck= fck +1.65 s
f’ck = target mean strength
fck= compressive strength at 28 days of concrete
s = standard deviation (IS 10262:2009, table -1)
- Volume of concrete = 1 m^3
- Volume of water = (mass of water /specific gravity of water) x1 /1000.
- Volume of cement = (mass of cement /specific gravity of cement) x1 /1000
- Volume of chemical admixtures = (mass of chemical admixture / specific gravity of chemical admixture) x 1/1000
- Volume of fly ash = (mass of fly ash/specific gravity of fly ash) x 1/1000
- Volume of aggregates = (volume of concrete (volume of cement +volume of fly ash + volume of water + volume of admixture)
- Mass of coarse aggregates = volume of aggregate x volume of coarse aggregate x specific gravity of coarse aggregate x1000
- Mass of fine aggregate = volume of aggregate x volume of fine aggregate x specific gravity of fine aggregate x1000
Codes For Concrete Mix Design:
- IS 8112
- IS 10262: 2009
- IS 456: 2000
- IS 3812 (part 1)
- IS 9103
Objectives:
- Produce concrete as economically as possible
- Achieve the desired workability, durability, and minimum strength of M35 grade concrete
Calculation:
- Target Mean Strength
F’ck = fck + 1.65 5
fck= 35 N/mm2 (M35 )
Table 1- IS 10262:2009
s = 5 N/mm^2
f ‘ck = 35 + (1.65 x 5)
= 43.25 N/mm2
- Selection Of Water Cement Ratio:
Given:
M35
Assume:
Exposure condition: Extreme (for reinforced concrete)
Table 5 – IS 456: 2000
Therefore,
Extreme= w/c ratio= 0.40
M35 = w/c ratio= 0.45
Lower the water-cement ratio better the workability
0.40 < 0.45
Water cement ratio = 0.40
- Selection Of Water Content:
Assume:
Nominal size of aggregates= 20 mm
table 2 -IS 10262: 2009
- Minimum water content = 186 liters
- Slump ranges from 25 to 50 mm
As superplasticizers (20 % reduced) (100-20 = 80)
= 186 x 80/100 = 148.8
Water content = 149 liters
- Calculation Of Cementitious Content (Cement + Fly Ash):
Water cement ratio = w/c= 0.40
Water content = W = 149 kg
Cementitious content = c=149/0.40 = 372.5 = 370 kg/m3
Minimum cement content = 360 kg/m3 (Table 5 – IS 456: 2000)
360< 370 kg/m^3 (it’s okay), Increase by 10 %
370 x 0.10 + 370 = 407 kg/m^3
W = 149 kg
Water cement ratio = 149/407 = 0.37
Fly ash at 30 % total cementitious material content
= 407 x 30/100 = 122 kg/m3
Cement (ordinary Portland cement) = 407-122 = 285 kg/m3
Fly ash = 122 kg/m3
- Volume Of Coarse Aggregates and Fine Aggregates:
Fine aggregates are sieved according to zone 1 of table 4 IS 383
Coarse aggregates are sieved according to table 2 IS 383
Referring to,
Table 3 IS 10262: 2009
Nominal size of aggregates = 20 mm
Fine aggregates (zone 1)
Therefore,
Volume of coarse aggregate = 0.60
Water cement ratio = 0.4
(+/- 0.01 for every +/- 0.005)
So, the volume of coarse aggregate = 0.62
Volume of fine aggregate = 1-0.62 = 0.38
- Mix Calculation:
Volume of concrete = 1 m3
Volume of cement = (285 / 3.15) x1/1000= 0.090 m3
Volume of fly ash = (122/2.2) x1/1000= 0.055 m3
Volume of water = (149/1) x 1/1000= 0.149 m3
Volume of admixture = (7.6/ 1.145) x1/1000= 0.006 m3
Volume of coarse aggregate and fine aggregate = a – (b + c + d+
e = 1-( 0.090 + 0.055 +0.149+ 0.006) = 0.7 m^3
Mass of coarse aggregate = 0.7 x 0.62 x 2.74 x1000 = 1189.16 kg/m3
Mass of fine aggregate = 0.7 x 0.38 x 2.74 x1000 = 728 kg/m3
Results And Conclusion:
Trial Mix:
- Cement = 285 kg/m3
- Fly ash = 122 kg/m3
- Water = 149 kg
- Coarse aggregate = 1189 kg/m3
- Fine aggregate = 728 kg/m3
- Chemical admixture = 6 kg/m3
- Water cement ratio = 366
Trial mix ratio: 285 :728 : 1189 = 1: 2.55 : 4.17 ( C: FA: CA)
Aim:
To calculate M50 grade without fly ash
Equations:
- Target mean strength: f’ck= fck +1.65 s
f’ck = target mean strength
fck= compressive strength at 28 days of concrete
s = standard deviation (IS 10262:2009, table 1)
- Volume of cement = (mass of cement /specific gravity of cement) x1/1000
- Volume of water = (mass of water/specific gravity of water ) x1 /1000
- Volume of chemical admixtures = ( mass of chemical admixture/ specific gravity of chemical admixture )x 1/1000
- Volume of concrete = 1 m^3
- Volume of aggregates = ( volume of concrete -( volume of cement + volume of water + volume of chemical admixture) )
- Mass of coarse aggregates = volume of aggregate x volume of coarse aggregate x specific gravity of coarse aggregate x1000
- Mass of fine aggregate = volume of aggregate x volume of fine aggregate x specific gravity of fine aggregate x1000
Codes For Concrete Mix Design:
- A .IS 8112
- IS 10262: 2009
- IS 456: 2000
- IS 3812 (part 1)
- IS 9103
Objectives:
- Produce concrete as economically as possible
- Achieve the desired workability, durability, and minimum strength of M35 grade concrete
Calculation:
- Target Mean Strength
F’ck = fck + 1.65 5
fck= 50 N/mm2 (M50 )
Table 1- IS 10262:2009
s = 5 N/mm2
f ‘ck = 50 + (1.65 x 5)
= 58.25 N/mm^2
- Selection Of Water Cement Ratio:
Given:
M50
Assume:
Exposure condition: Extreme ( for reinforced concrete)
Table 5 – IS 456: 2000
Therefore,
Extreme= w/c ratio= 0.40
M50 = w/c ratio= 0.40
Water cement ratio = 0.40
- Selection Of Water Content:
Assume:
Nominal size of aggregates= 20 mm
Table 2 -IS 10262: 2009
Minimum water content = 186 liters or kg
Slump ranges from 25 to 50 mm
As superplasticizers (20 % reduced) (100-20 = 80)
= 186 x 80/100 = 148.8
Water content = 149 liters
- Calculation Of Cement Content:
Water cement ratio = w/c= 0.40
Water content = W = 149 kg
Cement content = c=149/0.40 = 373 kg/m3
Minimum cement content = 360 kg/m3
360 < 373 kg/m3 (okay)
- Volume Of Coarse Aggregates and Fine Aggregates:
Fine aggregates are sieved according to zone 1 of table 4 IS 383
Coarse aggregates are sieved according to table 2 IS 383
Referring,
Table 3 IS 10262: 2009
Nominal size of aggregates = 20 mm
Fine aggregates (zone 1)
Therefore,
Volume of coarse aggregate = 0.60
Water cement ratio = 0.4
(+/- 0.01 for every +/- 0.005)
So, the volume of coarse aggregate = 0.62
Volume of fine aggregate = 1-0.62 = 0.38
- Mix Calculation:
Volume of concrete = 1m3
Volume of cement = (373/3.15) x1/1000 = 0.118 m3
Volume of water = (149/1) x1/1000= 0.149 m3
Volume of admixture = (7.6/1.145) x1/1000 = 0.006
Volume of coarse aggregate and fine aggregate = a – (b + c + d) = 1-( 0.118 + 0.149 + 0.006) = 0.727 m3
Mass of coarse aggregate = 0.727 x 0.62 x 2.74 x1000 = 1235.02 kg/m3
Mass of fine aggregate = 0.727 x 0.38 x 2.74 x1000 = 756.9 kg/m3
Results And Conclusion:
Trail Mix:
- Cement = 373 kg/m3
- Water = 149 kg
- Coarse aggregate = 02 kg/m3
- Fine aggregate = 9 kg/m3
- Chemical admixture = 6 kg/m3
- Water cement ratio = 40
Trial mix ratio: 373 :756.9: 1235.02 = 1: 2.02: 3.31 ( C: FA: CA)