Construction Technology_Concrete Mix Design_Project 1
1. Concrete Mix Design for M35 grade concrete with fly ash
Answer:
- AIM: To calculate Concrete Mix Design for M35 grade concrete with fly ash content as per IS 10262:2009 and IS 456:2000
- PROCEDURE:
Given:
- Grade of concrete: M35
- Type of mineral admixture: Fly ash conforming to IS 3812 (Part 1)
Assuming the following data:
- Exposure condition: Moderate
- Size of aggregate: 20mm
- Slump value: 100mm
- Admixture: Superplasticizer
- Coarse aggregate: Table 2 of IS 383
- Fine aggregate: Zone I ( Table 1 of IS 383)
STEP 1: Target mean strength for mix proportioning
f'ck = fck + 1.65 s
where, f'ck = target average compressive strength at 28 days,
fck = characteristics compressive strength at 28 days, and
s = standard deviation.
From Table-1 of IS 10262:2009, s = 5.0 Therefore, f'ck = 35 + (1.65 x 5) = 43.25 N/mm2.
STEP 2: Selection of Water-cement ratio
From Table-5 of IS 456:2000, maximum water-cement ratio=0.5
Based on experience, adopt water cement ratio=0.45
0.45<0.5, hence, ok.
STEP 3: Selection of water content
From Table-2 of IS 10262:2009, maximum water content for 20mm aggregate =186 kg or 186 lit (for 25 to 50mm slump range)
Here slump taken is 100mm. For every 25mm slump 3 % of water content is increased
Estimated water content for 100mm slump= 186 + ((6/100) x 186) = 197 lit
As superplasticizer is used, the water content can be reduced up 30 percent.
Based on trials with superplasticizer water content reduction of 29 percent has been achieved, Hence, the
arrived water content = 197 x 0.71 = 140 litres.
Step 4: Calculation of Cement and Fly Ash content
Water-cement ratio = 0.45
Cementitious material (cement+fly ash) content = 140/0.45 = 311 kg/m3
From Table-5 of IS 456:2000, minimum cement content for "moderate" exposure condition = 300 kg/m3
311 kg/m3 > 300 kg/m3 , hence OK.
Due to fly ash, increase in cementitious material by 10%
Cementitious material content = 311 x 1.10 = 342 kg/m3
Water content = 140 kg/m3
So, water-cement ratio = 140/342 = 0.41
Fly ash @ 30% of total cementitious material content = 342 x 30% = 103 kg/m3
Cement (OPC) = 342-103 = 239 kg/m3
Saving of cement using fly ash = 311-239 = 72 kg/m3
Fly ash being utilised = 103 kg/m3
STEP 5: Volume of Coarse aggregate and Fine aggregate content
Fine aggregate (Zone I)
Water-cement ratio = 0.45
Size of aggregate = 20mm
From Table-3 of IS 10262:2009, volume of coarse aggregate corresponding to fine aggregate (zone I) for water-cement ratio of 0.5=0.6
The present water cement ratio is = 0.45
The water cement ratio is lower by 0.15
For every 0.05 difference in w/c ratio ,0.01 of volume of coarse aggregate is increased.
Therefore ,the volume of coarse aggregate = 0.63
As the slump is 100mm ( pumpable concrete) so reduced by 10 % (100-10=90)
The volume of coarse aggregate = 0.63 * 90% = 0.56
volume of fine aggregate = 1- 0.56= 0.44
STEP 6: Mix Calculations
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) = (239/3.15) x (1/1000) = 0.0759 m3 (specific gravity of cement = 3.15)
c) Volume of fly ash = (Mass of fly ash/Specific gravity of fly ash) x (1/1000) = (103/2.2) x (1/1000) = 0.0468 m3 (specific gravity of fly ash = 2.2)
d) Volume of water = (Mass of water/Specific gravity of water) x (1/1000) = (140/1 x (1/1000) = 0.14 m3 (specific gravity of water = 1)
e) Volume of admixture = (Mass of admixture/Specific gravity of admixture) x (1/1000) = (6.22/1.145) x (1/1000) = 0.0054 m3 (specific gravity of admixture= 1.145)
f) Volume of aggregate (FA and CA) = (a-(b+c+d+e)) = (1-(0.0759+0.0468+0.14+0.0054)) = 0.7319 m3
g) Mass of coarse aggregate = e x Volume of CA x Specific gravity of CA x 1000 = 0.7319 x 0.56 x 2.7 x 1000 = 1107 kg/m3
h)Mass of fine aggregate = e x Volume of FA x Specific gravity of FA x 1000 = 0.7319 x 0.44 x 2.7 x 1000 = 870 kg/m3
STEP 7: Mix Proportions
CEMENT: 239 kg/m3
FLY ASH : 103 kg/m3
FINE AGGREGATE : 870 kg/m3
COARSE AGGREGATE : 1107 kg/m3
WATER CONTENT (W) = 140 litre
WATER CEMENT RATIO: 0.41
ADMIXTURE: 6.22 kg/m3
STEP 8: Mix Ratio
Cement : Fine Aggregate : Coarse Aggregate
1 : 3.64 : 4.63
- RESULT:
The Mix Design for M35 grade concrete with fly ash as per IS code is as shown above. The mix ratio obtained from the design is 1 : 3.64 : 4.63 .
2. Concrete Mix Design for M50 grade concrete without Fly ash
Answer:
- AIM: To calculate Concrete Mix Design for M50 grade concrete as per IS 10262:2009 and IS 456:2000
- PROCEDURE:
Given:
- Grade of concrete: M50
Assuming the following data:
- Exposure condition: Moderate
- Size of aggregate: 20mm
- Slump value: 100mm
- Admixture: Superplasticizer
- Coarse aggregate: Table 2 of IS 383
- Fine aggregate: Zone I ( Table 1 of IS 383)
STEP 1: Target mean strength for mix proportioning
f'ck = fck + 1.65 s
where, f'ck = target average compressive strength at 28 days,
fck = characteristics compressive strength at 28 days, and
s = standard deviation.
From Table-1 of IS 10262:2009, s = 5.0 Therefore, f'ck = 50 + (1.65 x 5) = 58.25 N/mm2.
STEP 2: Selection of Water-cement ratio
From Table-5 of IS 456:2000, maximum water-cement ratio=0.5
Based on experience, adopt water cement ratio=0.45
0.45<0.5, hence, ok.
STEP 3: Selection of water content
From Table-2 of IS 10262:2009, maximum water content for 20mm aggregate =186 kg or 186 lit (for 25 to 50mm slump range)
Here slump taken is 100mm. For every 25mm slump 3 % of water content is increased
Estimated water content for 100mm slump= 186 + ((6/100) x 186) = 197 lit
As superplasticizer is used, the water content can be reduced up 20 percent.
Based on trials with superplasticizer water content reduction of 20 percent has been achieved, Hence, the
arrived water content = 197 x 0.8 = 157 litres.
Step 4: Calculation of Cement content
Water-cement ratio = 0.45
Cement content = 157/0.45 = 350 kg/m3
From Table-5 of IS 456:2000, minimum cement content for "moderate" exposure condition = 300 kg/m3
350 kg/m3 > 300 kg/m3 , hence OK.
STEP 5: Volume of Coarse aggregate and Fine aggregate content
Fine aggregate (Zone I)
Water-cement ratio = 0.45
Size of aggregate = 20mm
From Table-3 of IS 10262:2009, volume of coarse aggregate corresponding to fine aggregate (zone I) for water-cement ratio of 0.5=0.6
The present water cement ratio is = 0.45
The water cement ratio is lower by 0.15
For every 0.05 difference in w/c ratio ,0.01 of volume of coarse aggregate is increased.
Therefore ,the volume of coarse aggregate = 0.63
As the slump is 100mm ( pumpable concrete) so reduced by 10 % (100-10=90)
The volume of coarse aggregate = 0.63 * 90% = 0.56
volume of fine aggregate = 1- 0.56= 0.44
STEP 6: Mix Calculations
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) = (350/3.15) x (1/1000) = 0.111 m3 (specific gravity of cement = 3.15)
c) Volume of water = (Mass of water/Specific gravity of water) x (1/1000) = (157/1 x (1/1000) = 0.157 m3 (specific gravity of water = 1)
d) Volume of admixture = (Mass of admixture/Specific gravity of admixture) x (1/1000) = (7/1.145) x (1/1000) = 0.006m3 (specific gravity of admixture= 1.145)
e) Volume of aggregate (FA and CA) = (a-(b+c+d+e)) = (1-(0.111+0.157+0.006)) = 0.726 m3
f) Mass of coarse aggregate = e x Volume of CA x Specific gravity of CA x 1000 = 0.726 x 0.56 x 2.7 x 1000 = 1098 kg/m3
g)Mass of fine aggregate = e x Volume of FA x Specific gravity of FA x 1000 = 0.726 x 0.44 x 2.7 x 1000 = 863 kg/m3
STEP 7: Mix Proportions
CEMENT: 350 kg/m3
FINE AGGREGATE : 863 kg/m3
COARSE AGGREGATE : 1098 kg/m3
WATER CONTENT (W) = 157 litre
WATER CEMENT RATIO: 0.45
ADMIXTURE: 7 kg/m3
STEP 8: Mix Ratio
Cement : Fine Aggregate : Coarse Aggregate
1 : 2.47 : 3.14
- RESULT:
The Mix Design for M50 grade concrete without fly ash as per IS code is as shown above. The mix ratio obtained from the design is 1 : 2.47 : 3.14