Project on Concrete Mix Design for various grades of Concrete
Design mix for M35 grade concrete with fly ash
Aim: To Design mix for M35 concrete with fly ash content in different percent with replacement of cement
Procedure:
Preliminary data required for mix design
*Exposure conditions of the structure:
The general environment to which the concrete will be exposed during its service life, is categorised into five class to severity ,as per IS 456.
The exposer condition limit the minimum cement content ,maximum water-cement ratio and minimum grade of concrete.
*Minimum thickness of member:
Size of aggregate should not be more than one -fourth of minimum thickness of member,mostly 20 mm nominal size aggregate is suitable for most works.
*Cement grade:opc 53 grade
*Workability:
Placing condition of concrete governs its workability low slump of 25-75mm to high slump of 100-150mm.
Stipulation for proportionin concrete incrediants
1.characteristic compressive strength required in the field at 28 days grade design mix M35
2.Type of cement : opc 53 grade confirming to IS 12269
3.Type of fly ash: class f
4.maximum nominal size of aggregate : 20mm
5. shape of coarse aggregate : angular
6.workability required:100mm slump
7.Type of exposer the structure will be subjected to as defined in IS 456 :Moderate
8.Method of concrete placing : pumable concrete
Test data of material
1.cement used grade confirming used to IS 12269:opc 53
2.fly ash: class f
3.specific gravity of cement :3.15
4.specific gravity of flyash :2.2
5.specific gravity of fine aggregate: 2.70
6.specific gravity of coarse aggregate: 2.80
7.water absorption of coarse aggregate : 0.4%
8.water absortion of fine aggregate :1.0%
9.seive analysis of fine aggregate : confirmin to zone 1 of IS 383
Mix design of M35 concrete with flyash
step1: determing target strength for mix design
F’ck = fck + 1.65 x S
Where,
F’ck = Target average compressive strength at 28 days
Fck = Characteristic compressive strength at 28 days
S = Assumed standard deviation in N/mm2 = 5 (as per table -1 of IS 10262- 2009)
= 35 + 1.65 x 5.0 = 43.25 N/mm2
Note : Under control conditions if Target average compressive strength is achieved then at field ,the probability of getting compressive strength of 35 MPa is very high
step 2: selection of water cement ratio
From Table 5 of IS 456, Maximum water-cementitious (w/cm)ratio = 0.5
Note : Do not start with w/cm ratio above 0.50,even though the other desired results like Strength, workability could be achieved.
step 3: selection of water content
From Table 2 of IS 10262- 2009,
Maximum water content for 20 mm aggregate = 186 Kg (for 25 to 50 slump)
We are targeting a slump of 100mm, we need to increase water content by 3% for every 25mm above 50 mm i.e. increase 6% for 100mm slump
i.e. Estimated water content for 100 Slump = 186+(6/100) X 186 = 197 litre
Water content = 197 liters
step 4: calculation of cement content
Water-Cement Ratio = 0.50
Water content from Step – 3 i.e. 197 liters
Cement Content = Water content / “w-cm ratio” = (197/0.50) = 394 kgs
From Table 5 of IS 456,
Minimum cement/cementitious content for moderate exposure condition = 300 kg/m3
394 kg/m3 > 300 kg/m3, hence, OK.
As per clause 8.2.4.2 of IS: 456
Maximum cement content = 450 kg/m3, hence ok too.
step 5: proportion of volume of coarse aggregate and fine aggreate
From Table 3 of IS 10262- 2009, Volume of coarse aggregate corresponding to 20 mm size and fine aggregate (Zone I) = 0.60
Note 1: In the present case water-cement ratio is 0.5.So there will be no change in coarse aggregate volume i.e. 0.60 .
Note 2: Incase the coarse aggregate is not angular one, then volume of coarse aggregate may be required to be increased suitably based on experience.
step 6: Estimation of concrete mix calculation
Mix 1 with 15 % of fly ash
The mix calculations per unit volume of concrete shall be as follows:
a) Volume of concrete = 1 m3
As cement is partially replaced by flyash, initially we will go for 15% replacement.
Earlier in Step 4 we got cement content of 394 kg/cum ,15% of it replaced by fly ash i.e. ~59 kg of flyash and cement content of 335 kg/cum, keeping cementitious content of 335 as arrived in STEP 4.
b) Volume of cement = (Mass of cement / Specific gravity of cement) x (1/1000)
= (335/3.15) x (1/1000) = 0.106 m3
c) Volume of flyash = (Mass of flyash / Specific gravity of cement) x (1/1000)
= (59/2.2) x (1/1000) = 0.026m3
d) Volume of water = (Mass of water / Specific gravity of water) x (1/1000)
= (197/1) x (1/1000) = 0.197 m3
e) Total Volume of Aggregates = 1- (b+c+d) =1- (0.106+0.026+0.197) = 0.671 m3
f) Mass of coarse aggregates = d X Volume of Coarse Aggregate X Specific Gravity of Coarse Aggregate X 1000 = 0.671 X 0.60 X 2.80 X 1000 = 1127.28 kgs/m3
g) Mass of fine aggregates = d X Volume of Fine Aggregate X Specific Gravity of Coarse Aggregate X 1000 = 0.671 X 0.40 X 2.70 X 1000 = 724.68 kgs/m3
Concrete mix proportion
Cement = 355 kg/m3
Fly Ash = 39 kg/m3
Water = 197 kg/m3
Fine aggregates = 724.68 kg/m3
Coarse aggregate = 1127.28. kg/m3
Water-cementitious ratio = 0.50
Mix 2 with 10% of fly ash
The mix calculations per unit volume of concrete shall be as follows:
a) Volume of concrete = 1 m3
As cement is partially replaced by flyash, initially we will go for 10% replacement.
Earlier in Step 4 we got cement content of 394 kg/cum ,10% of it replaced by fly ash i.e. ~39 kg of flyash and cement content of 355 kg/cum, keeping cementitious content of 394 as arrived in STEP 4.
b) Volume of cement = (Mass of cement / Specific gravity of cement) x (1/1000)
= (355/3.15) x (1/1000) = 0.112m3
c) Volume of flyash = (Mass of flyash / Specific gravity of cement) x (1/1000)
= (39/2.2) x (1/1000) = 0.017m3
d) Volume of water = (Mass of water / Specific gravity of water) x (1/1000)
= (197/1) x (1/1000) = 0.197 m3
e) Total Volume of Aggregates = 1- (b+c+d) =1- (0.112+0.017+0.197) = 0.674 m3
f) Mass of coarse aggregates = d X Volume of Coarse Aggregate X Specific Gravity of Coarse Aggregate X 1000 = 0.674 X 0.60 X 2.80 X 1000 = 1132.32 kgs/m3
g) Mass of fine aggregates = d X Volume of Fine Aggregate X Specific Gravity of Coarse Aggregate X 1000 = 0.674 X 0.40 X 2.70 X 1000 = 754.88 kgs/m3
Concrete mix proportion
Cement = 355 kg/m3
Fly Ash = 39 kg/m3
Water = 197 kg/m3
Fine aggregates = 754.88 kg/m3
Coarse aggregate = 1132.32 kg/m3
Water-cementitious ratio = 0.50
Mix 3 with 5% of fly ash
The mix calculations per unit volume of concrete shall be as follows:
a) Volume of concrete = 1 m3
As cement is partially replaced by flyash, initially we will go for 5% replacement.
Earlier in Step 4 we got cement content of 394 kg/cum ,5% of it replaced by fly ash i.e. ~20 kg of flyash and cement content of 355 kg/cum, keeping cementitious content of 374 as arrived in STEP 4.
b) Volume of cement = (Mass of cement / Specific gravity of cement) x (1/1000)
= (374/3.15) x (1/1000) = 0.118m3
c) Volume of flyash = (Mass of flyash / Specific gravity of cement) x (1/1000)
= (20/2.2) x (1/1000) = 0.009m3
d) Volume of water = (Mass of water / Specific gravity of water) x (1/1000)
= (197/1) x (1/1000) = 0.197 m3
e) Total Volume of Aggregates = 1- (b+c+d) =1- (0.118+0.009+0.197) = 0.676 m3
f) Mass of coarse aggregates = d X Volume of Coarse Aggregate X Specific Gravity of Coarse Aggregate X 1000 = 0.676 X 0.60 X 2.80 X 1000 = 1135.68 kgs/m3
g) Mass of fine aggregates = d X Volume of Fine Aggregate X Specific Gravity of Coarse Aggregate X 1000 = 0.676 X 0.40 X 2.70 X 1000 = 730.08 kgs/m3
Concrete mix proportion
Cement = 374 kg/m3
Fly Ash = 20 kg/m3
Water = 197 kg/m3
Fine aggregates = 730.08 kg/m3
Coarse aggregate = 1135.68 kg/m3
Water-cementitious ratio = 0.50
Result:
| TRIAL MIX | CEMENT CONTENT | FLY ASH CONTENT | MIX PROPORTION |
| MIX 1 | 85% | 15% | 1:1.84:2.86 |
| MIX 2 | 90% | 10% | 1:1.92:2.87 |
| MIX 3 | 95% | 5% | 1:1.96:2.88 |
Conclution:
*In mix 1,mix2,mix 3 addition of fly ash 15%,10%,5% replacement of cement respectively.
*In this mix design we can observe fly ash content increses coarse aggregate and fine aggregate content decreses
* so we conclude addition of fly ash in concrete must be cost effective.
Design mix for M50 grade concrete without fly ash
Aim: To Design mix for M50 concrete
Procedure:
Preliminary data required for mix design
*Exposure conditions of the structure:
The general environment to which the concrete will be exposed during its service life, is categorised into five class to severity ,as per IS 456.
The exposer condition limit the minimum cement content ,maximum water-cement ratio and minimum grade of concrete.
*Minimum thickness of member:
Size of aggregate should not be more than one -fourth of minimum thickness of member,mostly 20 mm nominal size aggregate is suitable for most works.
*Cement grade:opc 53 grade
*Workability:
Placing condition of concrete governs its workability low slump of 25-75mm to high slump of 100-150mm.
Stipulation for proportionin concrete incrediants
1.Characteristic compressive strength required in the field at 28 days grade design mix M35
2.Type of cement : opc 53 grade confirming to IS 12269
3.Maximum nominal size of aggregate : 20mm
4. Shape of coarse aggregate : angular
5.Workability required:100mm slump
6.Type of exposer the structure will be subjected to as defined in IS 456 :severe
7.Method of concrete placing : pumable concrete
Test data of material
1.Cement used grade confirming used to IS 12269:opc 53
2.Specific gravity of cement :3.15
3.Specific gravity of flyash :2.2
4.Specific gravity of fine aggregate: 2.70
5.Specific gravity of coarse aggregate: 2.80
6.Water absorption of coarse aggregate : 0.4%
7.Water absortion of fine aggregate :1.0%
8.Seive analysis of fine aggregate : confirmin to zone 1 of IS 383
9. Plasticizer: super plasticizer
Mix design of M50 concrete
step1: determing target strength for mix design
F’ck = fck + 1.65 x S
Where,
F’ck = Target average compressive strength at 28 days
Fck = Characteristic compressive strength at 28 days
S = Assumed standard deviation in N/mm2 = 5 (as per table -1 of IS 10262- 2009)
= 50 + 1.65 x 5.0 = 58.25 N/mm2
Note : Under control conditions if Target average compressive strength is achieved then at field ,the probability of getting compressive strength of 50 MPa is very high
step 2: selection of water cement ratio
From Table 5 of IS 456, Maximum water-cementitious (w/cm)ratio = 0.35
Note : Do not start with w/cm ratio above 0.35,even though the other desired results like Strength, workability could be achieved.
step 3; selection of water content
From Table 2 of IS 10262- 2009,
Maximum water content for 20 mm aggregate = 186 Kg (for 25 to 50 slump)
We are targeting a slump of 100mm, we need to increase water content by 3% for every 25mm above 50 mm i.e. increase 6% for 100mm slump
i.e. Estimated water content for 100 Slump = 186+(6/100) X 186 = 197 litre
Water content = 197 liters
when super plastizer is used water content will reduce by 20%
then water content will be 197-(197x(20/100)
=157.6 liters
step 4: calculation of cement content
Water-Cement Ratio = 0.35
Water content from Step – 3 i.e. 157.6 liters
Cement Content = Water content / “w-cm ratio” = (157.6/0.35) = 450 kgs
From Table 5 of IS 456,
Minimum cement/cementitious content for moderate exposure condition = 300 kg/m3
450 kg/m3 > 300 kg/m3, hence, OK.
As per clause 8.2.4.2 of IS: 456
Maximum cement content = 450 kg/m3, hence ok too.
step 5: proportion of volume of coarse aggregate and fine aggreate
From Table 3 of IS 10262- 2009, Volume of coarse aggregate corresponding to 20 mm size and fine aggregate (Zone I) = 0.60
Note 1: In the present case water-cement ratio is 0.35.So there will be no change in coarse aggregate volume i.e. 0.60 .
Note 2: Incase the coarse aggregate is not angular one, then volume of coarse aggregate may be required to be increased suitably based on experience.
step 6: Estimation of concrete mix calculation
Mix 1
The mix calculations per unit volume of concrete shall be as follows:
a) Volume of concrete = 1 m3
b) Volume of cement = (Mass of cement / Specific gravity of cement) x (1/1000)
= (450/3.15) x (1/1000) = 0.143 m3
c) Volume of water = (Mass of water / Specific gravity of water) x (1/1000)
= (157.6/1) x (1/1000) = 0.157 m3
d) Total Volume of Aggregates = 1- (b+c) =1- (0.143+0.157) = 0.700 m3
e) Mass of coarse aggregates = d X Volume of Coarse Aggregate X Specific Gravity of Coarse Aggregate X 1000 = 0.700 X 0.60 X 2.80 X 1000 = 1176 kgs/m3
f) Mass of fine aggregates = d X Volume of Fine Aggregate X Specific Gravity of Coarse Aggregate X 1000 = 0.700 X 0.40 X 2.70 X 1000 = 756 kgs/m3
Concrete mix proportion
Cement = 450 kg/m3
Water = 157.6kg/m3
Fine aggregates = 756 kg/m3
Coarse aggregate = 1176kg/m3
Water-cementitious ratio = 0.35
Mix 2
The mix calculations per unit volume of concrete shall be as follows:
a) Volume of concrete = 1 m3
b) Volume of cement = (Mass of cement / Specific gravity of cement) x (1/1000)
= (425/3.15) x (1/1000) = 0.135 m3
c) Volume of water = (Mass of water / Specific gravity of water) x (1/1000)
= (157.6/1) x (1/1000) = 0.157 m3
d) Total Volume of Aggregates = 1- (b+c) =1- (0.135+0.157) = 0.708 m3
e) Mass of coarse aggregates = d X Volume of Coarse Aggregate X Specific Gravity of Coarse Aggregate X 1000 = 0.708 X 0.60 X 2.80 X 1000 = 1189 kgs/m3
f) Mass of fine aggregates = d X Volume of Fine Aggregate X Specific Gravity of Coarse Aggregate X 1000 = 0.700 X 0.40 X 2.70 X 1000 = 792 kgs/m3
Concrete mix proportion
Cement = 425 kg/m3
Water = 157.6kg/m3
Fine aggregates = 792 kg/m3
Coarse aggregate = 1189kg/m3
Water-cementitious ratio = 0.35
Mix 3
The mix calculations per unit volume of concrete shall be as follows:
a) Volume of concrete = 1 m3
b) Volume of cement = (Mass of cement / Specific gravity of cement) x (1/1000)
= (400/3.15) x (1/1000) = 0.127 m3
c) Volume of water = (Mass of water / Specific gravity of water) x (1/1000)
= (157.6/1) x (1/1000) = 0.157 m3
d) Total Volume of Aggregates = 1- (b+c) =1- (0.127+0.157) = 0.716 m3
e) Mass of coarse aggregates = d X Volume of Coarse Aggregate X Specific Gravity of Coarse Aggregate X 1000 = 0.716 X 0.60 X 2.80 X 1000 = 1203 kgs/m3
f) Mass of fine aggregates = d X Volume of Fine Aggregate X Specific Gravity of Coarse Aggregate X 1000 = 0.700 X 0.40 X 2.70 X 1000 = 802 kgs/m3
Concrete mix proportion
Cement = 400 kg/m3
Water = 157.6kg/m3
Fine aggregates = 802 kg/m3
Coarse aggregate = 1176kg/m3
Water-cementitious ratio = 0.35
Result:
| TRIAL MIX | CEMENT CONTENT kg/m3 | MIX PROPORTION |
| MIX 1 | 450 | 1:1.68:2.61 |
| MIX 2 | 425 | 0.94:1.87:2.8 |
| MIX 3 | 400 | 0.89:2:2.95 |
Conclution:
*In mix 1,mix2,mix 3 there was difference in cement content
*In this mix design we can observe cement content increses coarse aggregate and fine aggregate content decresses
*By adding admixer like ggbs ,flyash will reduce the cement content