Project on Concrete Mix Design for various grades of Concrete
ASSIGNMENT 5
Aim:
To prepare the detail procedure for M35 concrete mix design calculation as per IS 10262:2000
Procedure:
Step 1: Stipulation for proportioning:
- Grade designing :
- Type of cement:
- Maximum nominal size of aggregate:
- Minimum cement content:
- Maximum water cement ratio:
- Workability:
- Exposure condition :
- Method of concrete placing:
- Degree of supervision:
- Type of aggregate:
- Maximum cement content:
- Chemical admixture:
Step 2: Test data for materials:
- Cement used:
- Specific gravity of cement:
- Chemical admixture:
- Specific gravity of materials:
- Water absorption:
- Free surface moisture:
Step 3: Selection of water content
Step 4: selection of cement content
Step 5: Volume of coarse and Fly ash.
Step 6: Mix calculation:
- Volume of concrete:
- Volume of cement:
- Volume of water:
- Volume of admixture:
- Volume of aggregate (coarse and fly ash):
- Mass o coarse aggregate:
- Mass of fly ash:
- Mix proportions:
- Mix ratio:
- Specific gravity of coarse aggregate:
- Water absorption:
Results
Procedure for design the M35 grade concrete mix design as follows the code procedures.
Step 1: Stipulation for proportioning:
- Grade designing : M35
- Type of cement: OPC 43 grade conforming IS 8112
- Maximum nominal size of aggregate: 20 mm
- Minimum cement content: 340 kg /m3 from table 5 IS 456:2000
For usage of fly ash add 1.1% of cementetious material = 340x1.1=374 kg / m3.
- Maximum water cement ratio: 0.45
- Workability: 100 mm for pump able concrete
- Exposure condition : very Severe for RCC
- Method of concrete placing: Pumping
- Degree of supervision: good
- Type of aggregate: crushed angular aggregate
- Maximum cement content:
If can be calculated from water cement ratio minimum cement content taken in Table 5 IS 456:2000 =0.45
Cementetious material content:
Water content= 140 kg/m3
So water cement ratio is 140/413.33 = 0.33
Fly ash @ 30% total cementetious material content cement (OPC) = 413.33 x 30% = 123.99 kg/m3
Cement (OPC) = 413.33-123.99 = 289.34 kg/m3
Saving of cement while using fly ash = 340-289.34= 50.66
Fly ash being utilized : 124 kg/m3
- Chemical admixture: super plasticizer.
Step 2: Test data for materials:
- Cement used: OPC 43 grade
- Specific gravity of cement: 3.15
- Chemical admixture: super plasticizer for IS 9103
- Specific gravity of coarse aggregate- 2.74 and fly ash- 2.2
- Water absorption: coarse aggregate-0.5% and fly ash-0.5%
- Free surface moisture: coarse aggregate-Nil and fly ash- Nil
Step 3: Selection of water content
From table-2 IS 10262:2009 maximum water content for 25 mm to 50 mm slump = 186 kg = 186 liters.
Assume 3% of water increasing 186 +(6/100)x186 = 197 liters.
Reduction of 29% for super plasticizers = 197x0.71 = 140liters
Step 4: selection of cement content
It can be calculated from water cement ratio minimum cement content taken in Table – 5 IS 456:2000
For W= 186 kg water cement ratio = 0.45
C= W/0.45 = 186/0.45 = 413.33 kg
Step 5: Volume of coarse and fly ash
This taken from Table – 3 from IS 10262:2009
For 20mm aggregate take 0.60 volume based on dry condition
Step 6: Mix calculation:
Target mean strength: For M35 concrete used Fck value taken as 35 ‘S ‘means standard deviation
F’ck = Fck + 1.65 S
F’ck = 35 + (1.65x5)=43.25 N/mm2
Target mean strength = 43.25 N/mm2
- Volume of concrete: 1 m3
- Volume of cement:
Volume of cement = Mass of cement / Specific gravity of cement x (1/1000)
Volume of cement from, table – 5 IS 456 : 2000 =340 kg/m3 very severe hence ok.
289.34/3.15 x 1/1000 = 0.09 m3
3.Volume of water:
4.Volume of water = Mass of water/ Specific gravity of water x (1/1000)
5.Volume of water = 1/1x(1/1000) = 0.14 m3
6.Volume of coarse aggregate (coarse and fly ash):
7.Volume of aggregate= Mass of aggreagate/ Specific gravity of aggregate x (1/1000)
8.Volume of C.A = 413.33 x(2/100) / 1.145 x (1/1000)=0.0072 m3
9.Volume of aggregate:
10.Volume of aggregate = a-(b+c+d)
= 1-(0.09+0.14+0.0072)
11.Volume of aggregate=0.76 m3
Mass of coarse aggregate:
Mass of coarse aggregate= e x volume of coarse aggregate x specific gravity of coarse aggregate x 1000
= 0.76 x 0.6 x 2.74 x 1000= 1249.44 kg/m3
Mass of fine aggregate:
Mass of fly ash= e x volume of fly ash x specific gravity of fly ash x 1000
= 0.76 x 0.44 x 2.74 x1000 = 916 kg/m3
Volume of chemical added:
7.6/1.145 x (1/1000) = 0.006 m3
Mix Proportion trial I:
Cement=289.34kg/m3
Fly ash=124 kg/m3
Water=197 liters
Fine aggregate=916kg/m3
coarse aggregate=1249.44kg/m3
chemical added=7
water cement ratio=0.33
- Mix ratio:1:3.16:4.31
Result:
Thus the design made up of using IS codes procedures values taken and calculation made by using thumb rules successfully designed.
Procedure for M50 Grade of mix deign Preparation as follows code procedures
Step 1: Stipulation for proportioning:
- Grade designing : M50
- Type of cement: OPC 43 grade conforming IS 8112
- Maximum nominal size of aggregate: 20 mm
- Minimum cement content: 360 kg /m3 from table 5 IS 456:2000.
- Maximum water cement ratio: 0.45
- Workability: 100 mm for pump able concrete
- Exposure condition : very Severe for RCC
- Method of concrete placing: Pumping
- Degree of supervision: good
- Type of aggregate: crushed angular aggregate
- Maximum cement content:
If can be calculated from water cement ratio minimum cement content taken in Table 5 IS 456:2000 =0.45
Cementetious material content:
Water content= 140 kg/m3
So water cement ratio is 140/360 = 0.38
- Chemical admixture: super plasticizer.
Step 2: Test data for materials:
- Cement used: OPC 43 grade
- Specific gravity of cement: 3.15
- Chemical admixture: super plasticizer for IS 9103
- Specific gravity of coarse aggregate- 2.74 and fine aggregate- 2.74
- Water absorption: coarse aggregate-0.5% and fine aggregate-0.5%
- Free surface moisture: coarse aggregate-Nil and fine aggregate- Nil
Step 3: Selection of water content
From table-2 IS 10262:2009 maximum water content for 25 mm to 50 mm slump = 186 kg = 186 liters.
Assume 3% of water increasing 186 +(6/100)x186 = 197 liters.
Reduction of 29% for super plasticizers = 197x0.71 = 140liters
Step 4: selection of cement content
It can be calculated from water cement ratio minimum cement content taken in Table – 5 IS 456:2000
For W= 186 kg water cement ratio = 0.38
C= W/0.45 = 186/0.38 = 489.47 kg
Step 5: Volume of coarse and fly ash
This taken from Table – 3 from IS 10262:2009
For 20mm aggregate take 0.60 volume based on dry condition
Step 6: Mix calculation:
Target mean strength: For M35 concrete used Fck value taken as 35 ‘S ‘means standard deviation
F’ck = Fck + 1.65 S
F’ck = 50 + (1.65x5)=58.25N/mm2
Target mean strength = 58.25 N/mm2
Volume of concrete: 1 m3
Volume of cement:
Volume of cement = Mass of cement / Specific gravity of cement x (1/1000)
Volume of cement from, table – 5 IS 456 : 2000 =340 kg/m3 very severe hence ok.
360/3.15 x 1/1000 = 0.11m3
Volume of water:
Volume of water = Mass of water/ Specific gravity of water x (1/1000)
Volume of water = 1/1 x (1/1000) = 0.14 m3
Volume of coarse aggregate (coarse and fine aggregate):
Volume of aggregate= Mass of aggregate/ Specific gravity of aggregate x (1/1000)
Volume of C.A = 489.47 x (2/100) / 1.145 x (1/1000)=0.008 m3
Volume of aggregate:
Volume of aggregate = a-( b+c+d )
= 1-(0.11+0.14+0.008)
Volume of aggregate=0.74 m3
Mass of coarse aggregate:
Mass of coarse aggregate= e x volume of coarse aggregate x specific gravity of coarse aggregate x 1000
= 0.74 x 0.6 x 2.74 x 1000= 1216.56kg/m3
Mass of fine aggregate:
Mass of fine aggregate= e x volume of fine aggregate x specific gravity of fine aggregate x 1000
= 0.76 x 0.38 x 2.74 x1000 = 719.31 kg/m3
Volume of chemical added:
7.6/1.145 x (1/1000) = 0.006 m3
Mix Proportion trial II:
Cement=360kg/m3
Water=197 liters
Fine aggregate=791.31kg/m3
coarse aggregate=1216.56kg/m3
chemical added=7
water cement ratio=0.38
- Mix ratio:1:2.19:3.37
Result:
Thus the design made up of using IS codes procedures values taken and calculation made by using thumb rules successfully designed.