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Estimating Costing and Valuation
Diploma –Vth Semester
Topic: Detail Estimate
R.C.C beam
Subject Code:3350604
Prepared by:
Y A .Challawala
Department of Civil Engineering
Government Polytechnic Dahod
Example-1 R.C.C Beam
Consider a beam of clear length of 4m, 300mm wide by 450mm depth. It consists of 2-12
diameter bars at top, and 2-16 diameter and 1 – 12 diameter bars at the bottom. Diameter of
stirrup is 8mm spaced at 180mm center to center. Clear cover to reinforcement provided is
40mm.
Calculate following quantities:
(a) Reinforcement in Beam
(b) Form work for R.C.C beam
(c) Cement concrete (1:2:4)
Solution:
(a) Formwork for R.C.C beam
Sr
no
Item Description No Length Width Height Quantity Total
1 Formwork in beam
Bottom 1 4 0.3 1
1.2
Sides 2 4.46 1 0.45 4.0
Ends 2 - 0.3 0.75 0.45
Total quantity 5.65 sq.m
➢Following are the steps to calculate the quantity of steel for
R.C.C slab/beam:
➢ 1- Prepare a bar bending schedule in order to classify different
shapes of bars
➢ (bent up bar, straight anchor bar, bar, curtail bar, etc) and
diameters.
➢ 2- List down all the shapes of bars from the drawing.
➢ 3- Count the number of bars of each of those shapes.
➢ 4- Then calculate the cutting length of each of those bars
Cutting length of bar
Cutting length of bar
= (length of the member—deduction for cover on both sides) + development length.
➢5- Then calculate the unit weight of each dia bars by the following formula:
W = d2/162 x L
Where d = diameter in mm and weight
w in kg,
L= total length in m
➢6- Then calculate the weight of reinforcement bar
Weight of reinforcement bar = no of bars x cutting length x unit weight
➢7. Add all the weight to get the total steel quantity.
NOTE :
➢1- Minimum % Of Steel Are:
1-For Beam (Tension reinforcement): As = 0.85bd/fy of gross cross-section area.
2- For Slab = 0.12% of total area
3- For Column = 0.8% of cs area
Trusted by millions of students. faculty, and professionals worldwide.
➢2- Maximum % Of Steel Are:
1- For Beam = 4 % of cross-section area.
2- For Slab = 4% of cross-section area.
3- For column = 6% of cross-section area.
➢3- Development Length is usually specified in the drawings, but if not then you
can calculate it as, Dev L = Depth - 2 times cover
➢4- Binding wire = 10 grams per kg of reinforcement.
➢5- No. of stirrups = (length of member - 2 x cover ) / spacing + 1
Typical bar bending schedule:
(b) Form work for R.C.C beam
Now we will calculate the length of reinforcement based on shapes of
reinforcement required for reinforced concrete beam in above example.
We will start with bottom reinforcement, B1.
Bar shape of B1 is as shown below:
6D 6D
Length of B1
= clear distance between walls + 2 x width of walls – 2 x bar cover + 2 x bend
length
Bend length = 6 x 16 = 96 consider as 100mm
Bend length is calculated as 6 x diameter of bar for reinforcement conforming
to IS: 1786-1961
Length of B1 = 4000 + 2 x 230 – 2 x 40 + 2 x100 = 4580mm
Length of bar B2 is calculated based on shape of this bar. This bar bends up
near the support as shown below:
Length of bar B2:
L=A + B + C = 4000 + 2 x 230 – 2 x 40 + 2 (0.42H)
H = 450 – 2 x 40 – 2 x 8 – 2 x 12/2 = 342mm
B2 = 4000 + 2 x 230 – 2 x 40+ 2(0.42×342) = 4667.3 = 4670 mm
Length of Bar T1
= 4000 + 2 x 230 -2 x 40 = 4380mm
Length of Stirrups S1:
Stirrups are spaced at 180mm centre to centre. Stirrups are provided between
walls/support for a beam.
No. of stirrups required for given beam = 4000 + 1 = 24
180
Length a = 450 – 2 x40 – 8 = 362mm
Length b = 300 – 2 x 40 – 8 = 212mm
Therefore, length of 1 stirrup
S1 = 2 x (a + b) + 24D
= 2 x (212 +362) + 24*8= 1340 mm
Where 24D is the hook length as per IS 2502
Bar Bending Schedule for RCC Beam:
No.
Bar
Mark
Bar
dia
(mm)
No. of
bars
Length
(mm)
Weight
of bars
(kg)
Bar Shape
1 B1 16 2 4580 14.5
2 B2 12 1 4670 4.15
3 T1 12 2 4380 7.80
4 S1 8 24 1340 12.70
Total quantity 39.15 kg
(c) Concreting (1:2:4) for R.C.C beam
Sr
no
Item Description No Length Width Height Quantity Total
1 Concreting (1:2:4) for
R.C.C beam
1 4.46 0.3 0.45 0.6021
Total quantity 0.6021
sq.m
Thank you
Any Question???
Example-1 R.C.C Beam
Consider a beam of clear length of 5m, 300mm wide by 450mm depth. It
consists of 2-12 diameter bars at top, and 2-16 diameter and 2 – 12
diameter bars at the bottom. Diameter of stirrup is 8mm spaced at
180mm center to center. Clear cover to reinforcement provided is 50mm.
Calculate following quantities:
(a). Form work for R.C.C beam
(b). Reinforcement in Beam
(c). Cement concrete (1:2:4)

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RCC beam Detail estimate

  • 1. Estimating Costing and Valuation Diploma –Vth Semester Topic: Detail Estimate R.C.C beam Subject Code:3350604 Prepared by: Y A .Challawala Department of Civil Engineering Government Polytechnic Dahod
  • 2. Example-1 R.C.C Beam Consider a beam of clear length of 4m, 300mm wide by 450mm depth. It consists of 2-12 diameter bars at top, and 2-16 diameter and 1 – 12 diameter bars at the bottom. Diameter of stirrup is 8mm spaced at 180mm center to center. Clear cover to reinforcement provided is 40mm. Calculate following quantities: (a) Reinforcement in Beam (b) Form work for R.C.C beam (c) Cement concrete (1:2:4) Solution:
  • 3. (a) Formwork for R.C.C beam Sr no Item Description No Length Width Height Quantity Total 1 Formwork in beam Bottom 1 4 0.3 1 1.2 Sides 2 4.46 1 0.45 4.0 Ends 2 - 0.3 0.75 0.45 Total quantity 5.65 sq.m
  • 4. ➢Following are the steps to calculate the quantity of steel for R.C.C slab/beam: ➢ 1- Prepare a bar bending schedule in order to classify different shapes of bars ➢ (bent up bar, straight anchor bar, bar, curtail bar, etc) and diameters. ➢ 2- List down all the shapes of bars from the drawing. ➢ 3- Count the number of bars of each of those shapes. ➢ 4- Then calculate the cutting length of each of those bars
  • 5. Cutting length of bar Cutting length of bar = (length of the member—deduction for cover on both sides) + development length. ➢5- Then calculate the unit weight of each dia bars by the following formula: W = d2/162 x L Where d = diameter in mm and weight w in kg, L= total length in m ➢6- Then calculate the weight of reinforcement bar Weight of reinforcement bar = no of bars x cutting length x unit weight ➢7. Add all the weight to get the total steel quantity.
  • 6. NOTE : ➢1- Minimum % Of Steel Are: 1-For Beam (Tension reinforcement): As = 0.85bd/fy of gross cross-section area. 2- For Slab = 0.12% of total area 3- For Column = 0.8% of cs area Trusted by millions of students. faculty, and professionals worldwide. ➢2- Maximum % Of Steel Are: 1- For Beam = 4 % of cross-section area. 2- For Slab = 4% of cross-section area. 3- For column = 6% of cross-section area. ➢3- Development Length is usually specified in the drawings, but if not then you can calculate it as, Dev L = Depth - 2 times cover ➢4- Binding wire = 10 grams per kg of reinforcement. ➢5- No. of stirrups = (length of member - 2 x cover ) / spacing + 1
  • 8. (b) Form work for R.C.C beam Now we will calculate the length of reinforcement based on shapes of reinforcement required for reinforced concrete beam in above example. We will start with bottom reinforcement, B1. Bar shape of B1 is as shown below: 6D 6D Length of B1 = clear distance between walls + 2 x width of walls – 2 x bar cover + 2 x bend length Bend length = 6 x 16 = 96 consider as 100mm Bend length is calculated as 6 x diameter of bar for reinforcement conforming to IS: 1786-1961 Length of B1 = 4000 + 2 x 230 – 2 x 40 + 2 x100 = 4580mm
  • 9. Length of bar B2 is calculated based on shape of this bar. This bar bends up near the support as shown below: Length of bar B2: L=A + B + C = 4000 + 2 x 230 – 2 x 40 + 2 (0.42H) H = 450 – 2 x 40 – 2 x 8 – 2 x 12/2 = 342mm B2 = 4000 + 2 x 230 – 2 x 40+ 2(0.42×342) = 4667.3 = 4670 mm Length of Bar T1 = 4000 + 2 x 230 -2 x 40 = 4380mm
  • 10. Length of Stirrups S1: Stirrups are spaced at 180mm centre to centre. Stirrups are provided between walls/support for a beam. No. of stirrups required for given beam = 4000 + 1 = 24 180 Length a = 450 – 2 x40 – 8 = 362mm Length b = 300 – 2 x 40 – 8 = 212mm Therefore, length of 1 stirrup S1 = 2 x (a + b) + 24D = 2 x (212 +362) + 24*8= 1340 mm Where 24D is the hook length as per IS 2502
  • 11. Bar Bending Schedule for RCC Beam: No. Bar Mark Bar dia (mm) No. of bars Length (mm) Weight of bars (kg) Bar Shape 1 B1 16 2 4580 14.5 2 B2 12 1 4670 4.15 3 T1 12 2 4380 7.80 4 S1 8 24 1340 12.70 Total quantity 39.15 kg
  • 12. (c) Concreting (1:2:4) for R.C.C beam Sr no Item Description No Length Width Height Quantity Total 1 Concreting (1:2:4) for R.C.C beam 1 4.46 0.3 0.45 0.6021 Total quantity 0.6021 sq.m
  • 14. Example-1 R.C.C Beam Consider a beam of clear length of 5m, 300mm wide by 450mm depth. It consists of 2-12 diameter bars at top, and 2-16 diameter and 2 – 12 diameter bars at the bottom. Diameter of stirrup is 8mm spaced at 180mm center to center. Clear cover to reinforcement provided is 50mm. Calculate following quantities: (a). Form work for R.C.C beam (b). Reinforcement in Beam (c). Cement concrete (1:2:4)
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