Project on Concrete Mix Design for various grades of Concrete
Calculate the Concrete Mix Design for M35 grade concrete with fly ash
AIM : To calculate the mix design for M35 grade concrete with fly ash
INTRODUCTION : Concrete mix design may be defined as the art of selecting suitable ingredients of concrete and determining their relative proportion with the object of producing concrete of certain minimum strength and durability as economically as possible.
DATA FOR CONCRETE MIX DESIGN OF M35 GRADE CONCRETE WITH FLY ASH
Grade of concrete - 35N/mm^2
Type of cement - OPC 53 Grade confirming to IS-8112
Maximum size of aggregate - 20mm
Type of aggregate - Crushed angular aggregate
Slump workability - 75 mm
Exposure - Extreme (For reinforced concrete)
Method of concrete placing - Non pumpable
Maximum cement content - As per IS 456:2000
Chemical admixture - Superplasticizer
Zone of fine aggregate - Zone I
Type of Mineral admixture - Fly ash confirming to IS3812(part-1)
TEST DATA FOR MATERIALS
The following data are collected by performing standatd laboratory test
Specific gravity of cement - 3.15
Specific gravity of coarse aggregate - 2.74
Specific gravity of fine aggregate - 2.56
Specific gravity of fly ash - 2.2
STEP AND CALCULATION
1 - TAREGT STRENGTH FOR MIX PROPORTIONING
f'ck = fck + 1.65S
Where,
f'ck = Target mean strength
fck = Charactersitics compressive strength at 28 days
S = Standard deviation
fck=35 N/mm^2
From Table 1-IS 10262:2009
S= 5 n/MM^2
f'ck = 35+1.65*5
f'ck = 43.25 N/mm2
2. CALCULATION OF WATER CEMENT RATIO
Exposure condition : Extreme (for reinforced concrete)
From table 5 - IS456:2009
Water-cement ratio- 0.4
Based on experience , W/C ratio= 0.35
0.35< 0.4 hence ok
3. SELECTION OF WATER CONTENT
From table 2 - IS10262:2009
Maximum water content for 20mm size aggregate = 186 lt ( 25 to 50mm slump)
For slump value of 75 mm
Maximum water content = 186 +( 3/100)*186 = 191 lt
As, superplasticizer is used , water content can be reduced upto 30%
Bsed on trial , 29% of water is reduced from mix
The arrived water content = 135 lt
4. CALCULATION OF CEMENT AND FLY ASH CONTENT
W/C ratio = 0.35
Cementitious material(cement+fly ash) = 135/0.35=386 Kg/m^3
From table 5 -IS456:2009 Minimun cement content for extreme condition= 360Kg/m^3
so, 386 Kg/m^3 > 360 Kg/m^3 hence ok
Increase the content by 10%
386 X 1.10 = 425 kg/m^3
So W/C ratio = 135/425 = 0.317
Fly ash at 30% of the total cementitious material
Fly ash content = 425 X 30% = 127 kg/m^3
Cement ( Ordinary portland cement) = 425-127=298 Kg/m^3
Saving of cement whil using fly ash = 386-298=88 Kg/m^3
Fly ash = 127 Kg/m^3
5. VOLUME FINE AND COARSE AGGREGATE
From Table 3 IS10262:2009
Volume of coasre aggregate = 0.6
(for 20mm Nominal size aggregate and fine aggregate confirming to zone 1)
Water-Cement ratio =0.35
(+/- 0.01 For every +/- 0.05 in w/c ratio)
So, Volume of coarse aggregate = 0.63
Volume of Fine aggregate = 1-0.63=0.37
6. MIX CALCULATION
Mix calculation for unit volume of concrete
a. Volume of concrete = 1 m^3
b. Volume of cement = ( Mass of cement/Specific gravity of cement) X 1/1000
= ( 298/3.15) X 1/1000
= 0.094 m^3
c. Volume of fly ash = ( Mass of fly ash/Specific gravity of fly ash) X 1/1000
= ( 127/2.2) X 1/1000
= 0.057 m^3
d. Volume of water = ( Mass of water/ Specific gravity of water) X1/1000
= ( 135/1) X1/1000
= 0.135 m^3
e. Volume of admixture = ( Mass of admixture/Specific gravity of admixture) X1/1000
(2% of cementitious material) = ( 7.7/1.145) X 1/1000
= 0.006 m^3
f. Volume of all in aggregate = (a-(b+c+d+e))
= (1-(0.094+0.057+0.135+0.006))
= 0.708 m^3
g. Mass of coarse aggregate = Volume of aggregate X volume of coarse aggregate X Specific gravity of CA X 1000
= 0.708 X 0.63 X 2.74 X 1000
= 1222.14 Kg
h. Mass of fine aggregate = Volume of aggregate X volume of coarse aggregate X Specific gravity of CA X 1000
= 0.708 X 0.37 X 2.56 X 1000
= 670.61 Kg
7. RESULTS:
Cement = 298 Kg/m^3
Fly ash = 127 Kg/m^3
Water = 135 Kg/m^3
Fine aggregate = 670.61 Kg/m^3
Coarse aggregate = 1222.14 Kg/m^3
Chemical admixture = 7.7 Kg/m^3
Water-Cement ratio = 0.317
Trai Mix 1 = 1:1.57:2.87
2.Calculate the Concrete Mix Design for M50 grade concrete without fly ash
AIM : To calculate the mix design for M50 grade concrete without fly ash
DATA FOR CONCRETE MIX DESIGN OF M50 GRADE CONCRETE WITH FLY ASH
Grade of concrete - 50 N/mm^2
Tyep of cement - OPC 53 Grade confirming to IS8112
Maximum size of aggregate - 20mm
Type of aggregate - Crushed angular aggregate
Slump workability - 100 mm
Exposure - Extreme (For reinforced concrete)
Method of concrete placing - pumping
Maximum cement content - As per IS 456:2000
Chemical admixture - Superplasticizer
Zone of fine aggregate - Zone I
TEST DATA FOR MATERIALS
The following data are collected by performing standatd laboratory test
Specific gravity of cement - 3.15
Specific gravity of coarse aggregate - 2.74
Specific gravity of fine aggregate - 2.56
Chemical admixture - Superplasticizer
STEP AND CALCULATION
1 - TAREGT STRENGTH FOR MIX PROPORTIONING
f'ck = fck + 1.65S
Where,
f'ck = Target mean strength
fck = Charactersitics compressive strength at 28 days
S = Standard deviation
fck=50 N/mm^2
From Table 1-IS 10262:2009
S= 5 N/MM^2
f'ck = 50 +1.65*5
f'ck = 58.25 N/mm2
2. CALCULATION OF WATER CEMENT RATIO
Exposure condition : Extreme (for reinforced concrete)
From table 5 - IS456:2009
Water-cement ratio- 0.4
Based on experience , W/C ratio= 0.35
0.35< 0.4 hence ok
3. SELECTION OF WATER CONTENT
From table 2 - IS10262:2009
Maximum water content for 20mm size aggregate = 186 lt ( 25 to 50mm slump)
For slump value of 100 mm
Maximum water content = 186 +( 6/100)*186 = 197 lt
As, superplasticizer is used , water content can be reduced upto 30%
Bsed on trial , 29% of water is reduced from mix
The arrived water content = 140 lt
4. CALCULATION OF CEMENT CONTENT
W/C ratio = 0.35
Cement Content = 140/0.35=400 Kg/m^3
From table 5 -IS456:2009 Minimun cement content for extreme condition= 360Kg/m^3
so, 400 Kg/m^3 > 360 Kg/m^3 hence ok
5. VOLUME FINE AND COARSE AGGREGATE
From Table 3 IS10262:2009
Volume of coasre aggregate = 0.6
(for 20mm Nominal size aggregate and fine aggregate confirming to zone 1)
Water-Cement ratio =0.35
(+/- 0.01 For every +/- 0.05 in w/c ratio)
So, Volume of coarse aggregate = 0.63
For pumpable concrete, its value should be reduced by 10%
Therefore the colume of coarse aggregate = 0.63X0.9 =0.56
Volume of Fine aggregate = 1-0.56=0.44
6. MIX CALCULATION
Mix calculation for unit volume of concrete
a. Volume of concrete = 1 m^3
b. Volume of cement = ( Mass of cement/Specific gravity of cement) X 1/1000
= ( 400 /3.15) X 1/1000
= 0.126 m^3
c. Volume of water = ( Mass of water/ Specific gravity of water) X1/1000
= ( 140/1) X1/1000
= 0.140 m^3
d. Volume of admixture = ( Mass of admixture/Specific gravity of admixture) X1/1000
(2% of cementitious material) = ( 8/1.145) X 1/1000
= 0.006 m^3
e. Volume of all in aggregate = (a-(b+c+d))
= (1-(0.126+0.140+0.006))
= 0.728 m^3
f. Mass of coarse aggregate = Volume of aggregate X volume of coarse aggregate X Specific gravity of CA X 1000
= 0.728 X 0.56 X 2.74 X 1000
= 1117.04 Kg
g. Mass of fine aggregate = Volume of aggregate X volume of coarse aggregate X Specific gravity of CA X 1000
= 0.728 X 0.44 X 2.56 X 1000
= 820.01 Kg
7. RESULTS:
Cement = 400 Kg/m^3
Water = 140 Kg/m^3
Fine aggregate = 820.01 Kg/m^3
Coarse aggregate = 1117.04 Kg/m^3
Chemical admixture = 8 Kg/m^3
Water-Cement ratio = 0.35
Trai Mix 1 = 1:2.05:2.79