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 cement : (mass of cement / specific gravity of cement) X 1/1000
- Volume of water : (mass of water / specific gravity of water) X 1/1000
- 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 concrete : 1 cu m
- Volume of aggregates : (volume concrete -(volume of cement + volume of fly ash + volume of water + volume of admixture)
- Mass of coarse aggregates: volume aggregate X volume of coarse aggregate X specific gravity of coarse aggregate X 1000
- 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 10262: 2009
- IS 456: 2000
- IS 8112
- IS 9103
- IS 3812(PART 1)
OBJECTIVES:
- The main objective is to produce concrete as economically as possible
- To achieve desired workability, durability, and minimum strength of M35 grade concrete
PROCEDURE AND CALCULATION:
1 TARGET MEAN STRENGTH
F'CK=fck + 1.65 s
fck=35N/MM SQ(M35)
TABLE 1- IS 10262 : 2009
THIS TABLE REQUIRED TO, s=5N/mm sq
f'ck=35+(1.65 X 5)
=43.25N/mm sq
2SELECTION 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
Minimum cement content = 360kg / cu m
3.SELECTION OF WATER CONTENT
Assume :
Nominal size of aggregates = 20 mm
Table 2 - IS 10262 : 2009
Minimum water content = 186 litres or kg
Slump ranges from 25 to 50 mm
As superplasticizers (20% reduced)
=186 X 80/100 = 148.8 (100-20=80)
Water content =149 litres
4.CALCULATION OF CEMETITIOUS CONTENT(CEMENT + FLY ASH)
Water cement ratio = w/c=0.40
Water content = W=149 kg
Cementitious content = C = 149/0.40=372.5 = 373 kg/ cu m
Minimum cement content = 360 kg/ cu m
360<373 kg/ cu m ( hency ok)
Increse by 10%
370 X 0.10 + 370 = 407 KG/ CU M
W = 149 kg
Water cement ratio = 149/407 = 0.366
Fly ash at 30% total cementitious material content = 407 X 30/100 = 122 kg/cu m
Cement (ordinary portland cement) = 407-122 = 285 kg/cu m
Saving of cement while using fly ash = 370-285=85kg/cu m
Fly ash =122 kg/ cu m.
5. 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
According,
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 =0.40
(+/-0.01 for every =/- 0.005).
So, the volume of coarse aggregate = 0.62
Volume of fine aggregate = 1-0.62 = 0.38
6. MIX CALCULATION :
- Volume of concrete = 1 cu m
- Volume of cement = (285/3.15) X 1/1000
- Volume of fly ash = (122/2.2) X 1/1000 = 0.055 CU M
- Volume of water = (149/1) x 1/1000=0.146 cu m
- Volume of addmixture = (7.6/1.145) x 1/1000=0.006 cu m
- Volume of coarse aggregate and fine aggregate = a-(b+c+d+e)=1-(0.090+0.055+0.149+0.006)=0.7 cu m
- Mass of coarse aggregate = 0.7 X 0.62 X 2.74 x 1000 =1189.16kg /cu m
- Mass of fine aggregate =0.7 X 0.38 X2.74 X1000 =728 KG/CU M
RESULTS AND CONCLUSION :
TRIAL MIX :
Cement = 285 kg/cu m
Fly ash 122 kg/ cu m
Water =149 kg
Coarse aggregate =1189 kg/cu m
Fine Aggregate =728 kg / cu m
Chemical admixture = 7.6 kg / cu m
Water cement ratio = 0.366
Trial mix ratio : 285 : 728: 1189, =1:2.55:4.17(C:FA:CA)
- This trial mix ratio should match the workability, durability, and minimum strength of M35 grade concrete.
- If it does not satisfy the above criteria two or more trials for an increase of 10% should be done.
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(IS10262 : 2009,TABLE-1)
- Volume of cement : (mass of cement / specific gravity of cement) x 1/1000
- Volume of water : (mass of water / specific gravity of water) x 1/1000
- Volume of chemical admixtures : ( mass of chemical admixture / specific gravity of chemical admixture) x 1/1000
- Volume of cement : 1 cu m
- 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 X 1000
- Mass of fine aggregate: volume of aggregate X volume of fine aggregate X specific gravity of fine aggregate X 1000
CODES FOR CONCRETE MIX DESIGN:
- IS 10262:2009
- IS 456:2000
- IS 8112
- IS 9103
- IS 3812( PART 1)
OBJECTIVES:
- The main objective is to produce concrete as economically as possible
- To achieve desired workability, durability, and minimum strength of M50 GRADE concrete
PROCEDURE AND CALCULATION:
1. TARGET MEAN STRENGTH
F'ck = fck+1.65 s
fck =50N/mm sq (M50)
Table 1-IS 10262:2009
s=5 n/mm sq
f'ck=50+(1.65X5)
=58.25 N/MM SQ
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
Lower the water cement ratio better the workability
0.40
Water cement ratio = 0.40
Minimum cement content = 360kg / cu m
3. selection of water content
Assume:
Nominal size of aggregates = 20 mm
Table 2 - IS 10262 : 2009
Minimum water content = 186 litres or kg
slump ranges from 25 to 50 mm
As superplasticizers (20% reduced)
= 186 x 80/100 148.8 (100-20=80)
water content = 149 litres
4.CALCULATION OF CEMENT CONTENT
Water cement ratio =w/c 0.40
Water content =w =149 kg
Cement content =C= 149/0.40 = 372.5=373 kg / cu m
Minimum cement content = 360 kg / cu m.
360<373 kg / cu m( HENCE OKAY)
5. VOLUME OF COARSE AGGREGATES AND FINE AGGREGATES
fine aggregates are sieved according to zone 1 of table 4 IS 383
Coarse aggregate is sieved according to table 2 IS 383
According,
Table 3 IS 10262: 2009
Nominal size of aggregates = 20 mm
Fine aggregates(zone1)
Therefore,
Volume of coarse aggregate = 0.60
Water cement ratio = 0.40
(+/-0.01 for every +/- 0.005)
So, the volume of coarse aggregate = 0.62
Volume of fine aggregate = 1-0.62=0.38
6. MIX CALCULATION :
- Volume of concrete =1 cu m
- Volume of cement = (373/3.15) X 1/1000=0.118 CU M
- Volume of water = (149/1) X 1/1000=0.149 CU M
- Volume of admixture = (7.6/1.145) x 1/1000 = 0.006
- Volume of coarse aggregate and fine aggregate =a-(b+c+d) = 1-(0.118+0.49+0.006)=0.727 cu m
- Mass of coarse aggregate =0.727 x 0.62 X 2.74 X 1000 = 1235.02 KG/CU M
- Mass of fine aggregate =0.727 X 0.38 X 2.74 X 1000 =756.9 KG / CU M
RESULTS AND CONCLUSION :
TRIAL MIX :
Cement =373 kg / cu m
Water = 149 kg
Coarse aggregate = 1235.02 kg / cu m
Fine aggregate = 756.9 kg / cu m
Chemical admixture = 7.6 kg / cu m
Water cement ratio =0.40
Trial mix ratio : 373:756.9:1235.02, 1:2.02:3.31(C:FA:CA)
- This trial mix ratio should match the workability, durability, minimum strength of M50 grade concrete
- If it does not satisfy the above criteria or more trials for an increase of 10% should be done.