Prof. Kamble A.S.: Sub: Quantity Surveying, Contracts & Tenders. BE Civil

Ahmednagar Jilha Maratha Vidya Prasarak Samaj’s
Shri Chhatrapati Shivaji Maharaj

College of Engineering, Nepti, Ahmednagar.
Department of Civil Engineering

Present
Sub: Quantity Surveying , Contracts & Tenders.

BE Civil
Prof. Kamble A.S.

Bar Bending Schedule for RCC Footing
Length of Footing = X = 2m
Breadth of Footing = Y = 1.6m
Height of the footing (Thickness) = h =0.3m
Diameter of the Main reinforcement bars = dm= 12mm
Diameter of Distribution Reinforcement Bars = d =12mm
Spacing of Main reinforcement bars = Sm = 150mm c/c
Spacing of Distribution Reinforcement bars = Sd = 150mm c/c
Cover for reinforcement = C = 50mm
From the given details, the following values are calculated:
• Solution
1. Calculation of Number of Main Reinforcement (X- Bars)
Nm = (Y / Spacing of Main Reinforcement) + 1
Nm = (1.6/.15)+1
Nm= 12nos
2. Calculation of Number of Distribution Reinforcement (Y-Bars)
Nd = (X/Spacing of Distribution Reinforcement) +1
Nd = (2/.15)+1
Nd= 14nos
3. Cutting Length of Main Reinforcement (X-Bars)
Cm1 = [X-2C] +2[h-2C]-2[2dm]
Cm1 = [2-(2x.05)]+2[.3-(2 x .05)]-2 (2 x .012)
Cm1 = 1.9+.4-.048
Cm1=2.252m
Total Cutting Length of Main Reinforcement
Cm= Nm {[X-2C] +2[h-2C]-2[2dm]}
Cm = 11 x 2.252
Cm = 24.772m
4. Cutting Length of Distribution Reinforcement (Y-Bars)
Cd1 = [Y-2C] +2[h-2C]-2[2dm]
Cd1 = [1.6-(2x.05)]+2[.3-(2 x .05)]-2 (2 x .012)
Cd1 = 1.5+.4-.048
Cd1 = 1.852m
Total Cutting Length of Main Reinforcement
Cd = Nd x Cd1
Cd = 14 x 1.852
Cd = 25.928m
5. Estimation of Steel Quantity
W = D2L/162;
Total Steel Quantity for Main Reinforcement,
Wm = (12 x 12 x 24.772)/162
Wm= 22kg
Total Steel Quantity for Distribution Reinforcement,
Wd=(12x12x25.928)/162
Wd =23.05kg
∴ Summary Bar Bending Schedule of RCC footing is
Dia: of
No. of Length of Total Length of Total Weight of Bar
Bar Type Bar
Bar Bar(m) Bar(m) Kg
(mm)
Main
Reinforcement 12 11 2.252 24.772 22.02
(X-Bars)
Distribution
Reinforcement 12 14 1.852 23.05 23.05
(Y-Bars)
45.07
Bar Bending Schedule for RCC Column
• Given Data
Slab Thickness – 125 mm & 100 mm
Floor height – 3000 mm or 3 m
Ground Floor Level – 3300 mm
Footing Height – 300 mm
Development Length – 50d
Column has 6 numbers of 20 mm dia bars
8mm stirrups @ 150 mm C/C
Footing Clear Cover – 40 mm
Slab Clear Cover – 25 mm

• Step 1 – Find the length of Vertical Bar
Length of Vertical bar = Development length (Ld) + Height of Ground Level + Floor Heights (1,2,3)+ Slab
Thicknesses + Overlap Length (Det.B)
= (2X50×20)+(3300+300 - (40+20+20))+ (3×3000)+((3×125)+100) +(50×20)
= 15995 mm= 15.99m
• Step 2 -Cutting Length of Stirrups
Length of One Hook = 9d
Cutting length of Stirrup = Perimeter of stirrup + Number of Bends + Number of Hooks
= 2(a+b) + 3 numbers of 90 degree bends + 2 numbers of hooks
= 2(500+200)+(3 x2d)+(2x9d)= 2×700+3x2x20+2x9x20
Cutting length of Stirrup = 1880 mm= 1.88m
Step 4 – Number of Stirrups
Number of stirrups required = (Total length of Column/spacing of stirrups)+1
= (3300+125+3000+125+3000+125+3000+100+ 300 - (40+20+20))/150 + 1= (12995/150)+1
Number of stirrups = 87 no.s

∴ Summary Bar Bending Schedule of RCC Column is
Diameter of Bar Numbers Cutting Length Total Length Wt. of Steel

Vertical Bar 20 6 15995 mm 95.94 m 236.88 kg
Stirrups 8 85 1880 mm 159.8m 63.19 kg
Bar Bending Schedule for RCC Beam
As you can see in the figure, the beam has clear span of 3metre consists of 2 numbers of 16 mm dia at bottom and
2 numbers of 12mm dia bars at top with 8mm dia stirrups at 150mm Clear Cover. Assuming Clear Cover of 25 mm at
both ends and sides of the beam.
• Given Data
Clear Span of Beam = 3000 mm
Development Length Ld = 50d (assumption)
Clear Cover on any ends = 25 m
Bottom – 2 numbers of 16⌀
Top – 2 numbers of 12⌀
Stirrups – 8⌀ @ 150mm clear cover

Step 1 – Find cutting length of top bar
Cutting length of top bar = Clear Span of Beam + Development length (Anchorage) Ld on 2 sides – Clear Cover on 2
ends
= 3000+(2 x 50d) – (2 x 25) = 3000 + (2 x 50 x 12) – 50
= 4150 mm
• Step 2 – Find cutting length of bottom bar
Cutting length of bottom bar = Clear Span of Beam + Development length (Anchorage) Ld on 2 sides – Clear Cover on
2 ends
= 3000+(2 x 50d) – (2 x 25) = 3000 + (2 x 50 x 16) – 50
= 4550 mm
Step 3 – Find out Number of Stirrups
Number of Stirrups required = (Clear Span of Beam/Spacing Stirrups)+ 1
= (3000 / 150) +1 = 21 No.s
Step 4 – Find out cutting length of Each Stirrup
Let’s split the sides as (a, b) for easy calculation.

Length of One Hook = 9d
Cutting length of Stirrup = Perimeter of stirrup + Number of Bends + Number of Hooks
= 2(a+b) + 3 numbers of 90 degree bends + 2 numbers of hooks
= 2(300+500) + (3 x 2d)+(2 x 9d) = 1600 + (3 x 2 x 8) + (2 x 9 x 8) = 1792 mm
∴ Summary Bar Bending Schedule of RCC beam section is

Top Bar 12 2 4150 mm 8300 mm 7.33 kg
Bottom Bar 16 2 4550 mm 9100 mm 14.38 kg
Stirrups 8 21 1792 mm 37632 mm 14.86 kg
Bar Bending Schedule for RCC Slab
Suppose we have a one-way slab, which has a length 5 m or width 2 m (clear span). The Main bars will be 12 mm
in diameter with 100 mm c/c spacing. The Distribution bars will be 8 mm in diameter with 125 mm c/c spacing.
The Clear cover will be 25 mm (Top or Bottom) and the thickness of the slab is 150 mm.
• GIVEN DATA.
Length = 5 m (5000 mm).
Width = 2 m (2000 mm).
Main Bar = 12 mm @ 100 mm c/c.
Distribution Bar = 8 mm @ 125 mm c/c.
Clear cover = 25 mm from (Top and Bottom).
Thickness = 150 mm
• SOLUTION
The quantity is done in two steps.
STEP 1. (CALCULATION OF BARS NO’S)
First, calculate the number of bars required (main and distribution both).
FORMULA = (Total length – Clear cover)/center to center spacing + 1

Main bar = (5000 – (25+25))/100 + 1
= 4950 / (100 + 1) = 51 Bars
Distribution bar = (2000 – (25+25))/125 + 1
= 1950 / )125 + 1) = 17 Bars
STEP 2. (CUTTING LENGTH)
MAIN BAR:
FORMULA = (L) + (2 x Ld) + (1 x 0.42D) – (2 x 1d)
# Where
L = Clear Span of the Slab
Ld = Development Length which is 40 d (where d is diameter of bar)
0.42D = Inclined length (Bend length)
1d = 45° bends (d is diameter of bar)
First calculate the length of D
D = (Thickness) – 2 (Clear cover at Top, BOTTOM) – Diameter of the bar.
= 150 – 2(25) -12
D = 88 mm
By putting Values.
Cutting length = 2000 + (2 x 40 x 12) + (1 x 0.42 x 88) – (2 x 1 x 12)
Cutting length = 2000 + 960 + 36.96 – 24 =2972.96 mm ~ 2973 mm or 2.973 m
DISTRIBUTION BAR:
= Clear Span + (2 x Development Length (Ld))
= 5000 + (2 x 40 x 8) = 5640 mm or 5.64 m
∴ Summary Bar Bending Schedule of RCC Slab structure is

Main bar 12 51 2.973m 151.623m 134.776 kg
Distribution Bar 8 17 5.64m 95.88m 37.87 kg

Prof. Kamble A.S.: Sub: Quantity Surveying, Contracts & Tenders. BE Civil

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Ahmednagar Jilha Maratha Vidya Prasarak Samaj’s

Shri Chhatrapati Shivaji Maharaj

Department of Civil Engineering

Sub: Quantity Surveying , Contracts & Tenders.

Prof. Kamble A.S.

3. Cutting Length of Main Reinforcement (X-Bars)

Cm1 = [X-2C] +2[h-2C]-2[2dm]

Cm1 = [2-(2x.05)]+2[.3-(2 x .05)]-2 (2 x .012)

Cm= Nm {[X-2C] +2[h-2C]-2[2dm]}

4. Cutting Length of Distribution Reinforcement (Y-Bars)

Cd1 = [Y-2C] +2[h-2C]-2[2dm]

Cd1 = [1.6-(2x.05)]+2[.3-(2 x .05)]-2 (2 x .012)

Total Cutting Length of Main Reinforcement

5. Estimation of Steel Quantity

Total Steel Quantity for Main Reinforcement,

∴ Summary Bar Bending Schedule of RCC footing is

Slab Thickness – 125 mm & 100 mm

Floor height – 3000 mm or 3 m

Ground Floor Level – 3300 mm

Footing Height – 300 mm

Development Length – 50d

Column has 6 numbers of 20 mm dia bars

8mm stirrups @ 150 mm C/C

Footing Clear Cover – 40 mm

Slab Clear Cover – 25 mm

= (2X50×20)+(3300+300 - (40+20+20))+ (3×3000)+((3×125)+100) +(50×20)

= 15995 mm= 15.99m

• Step 2 -Cutting Length of Stirrups

Length of One Hook = 9d

Cutting length of Stirrup = Perimeter of stirrup + Number of Bends + Number of Hooks

= 2(a+b) + 3 numbers of 90 degree bends + 2 numbers of hooks

= 2(500+200)+(3 x2d)+(2x9d)= 2×700+3x2x20+2x9x20

Cutting length of Stirrup = 1880 mm= 1.88m

Step 4 – Number of Stirrups

Number of stirrups required = (Total length of Column/spacing of stirrups)+1

= (3300+125+3000+125+3000+125+3000+100+ 300 - (40+20+20))/150 + 1= (12995/150)+1

Number of stirrups = 87 no.s

Diameter of Bar Numbers Cutting Length Total Length Wt. of Steel

Clear Span of Beam = 3000 mm

Development Length Ld = 50d (assumption)

Clear Cover on any ends = 25 m

Bottom – 2 numbers of 16⌀

Top – 2 numbers of 12⌀

Stirrups – 8⌀ @ 150mm clear cover

= 3000+(2 x 50d) – (2 x 25) = 3000 + (2 x 50 x 12) – 50

= 3000+(2 x 50d) – (2 x 25) = 3000 + (2 x 50 x 16) – 50

Step 3 – Find out Number of Stirrups

Number of Stirrups required = (Clear Span of Beam/Spacing Stirrups)+ 1

= (3000 / 150) +1 = 21 No.s

Step 4 – Find out cutting length of Each Stirrup

Let’s split the sides as (a, b) for easy calculation.

Cutting length of Stirrup = Perimeter of stirrup + Number of Bends + Number of Hooks

= 2(a+b) + 3 numbers of 90 degree bends + 2 numbers of hooks

= 2(300+500) + (3 x 2d)+(2 x 9d) = 1600 + (3 x 2 x 8) + (2 x 9 x 8) = 1792 mm

∴ Summary Bar Bending Schedule of RCC beam section is

Diameter of Bar Numbers Cutting Length Total Length Wt. of Steel

Length = 5 m (5000 mm).

Width = 2 m (2000 mm).