In this article, we explained how to calculate the Bill of Quantities And Bar bending schedule for Staircase (Step by Step procedure of Doglegged Staircase) For a given plan & Section Plan of Doglegged Staircase.
To Find
(i). Weight of Steel required for
a. Main bars,
b. Distribution bars,
c. Extra bars and
d. Hand Rail mesh.
The thickness of Waist Slab and
Landing = 6"
(ii). No. of Steel bars required, of length 40Ft/12m
(iii). Bill of Quantities for Staircase Reinforcement
if Rate of steel /kg = 65 rupees (Jan 2022).
GIVEN DATA
COMPONENTS OF STAIRCASE
Reinforcement Details:
(a). Waist Slab mesh = Ã˜ 10mm@100mm or 4"
(b). Landing mesh = Ã˜ 10mm@100mm or 4"
(c). Extra bar = Ã˜ 8mm@ 100mm or 4"
(d). Handrail mesh = Ã˜ 8mm@ 100mm or 4"
(I). Waist Slab (Slab 1 & Slab 2):
Note:
(a). While providing steel mesh in waist slab, 1"
concrete cover shall be deducted from each side of waist slab's Main Bars and
add the overlap length of 50d on each side of Distribution Bars to join the handrail
mesh
(b). Each Crank Length = 0.42D
Where, D = Depth of Slab Top and Bottom concrete cover
D = 6"  1"  1" = 4" or 0.333 Ft
(c). the formula used to find,
Number
of Main Bars = (Length of Longer Side / Spacing) + 1
Number
of Distribution Bars = (Length of Shorter Side / Spacing) + 1
Given
Dia of main bar = 10mm  0.0328 Ft
1.
Length of Distribution Bars = 4' + 11.18' + 3' + 0.42D x 2 
1"  1"
= 4' + 11.18' + 3' + 0.42 x 0.333 x 2  0.1666
=
18.293 Ft
2. Length
of Main Bars = 3.5' + 50D x 2
= 3.5' + 50 x 0.0328 x 2
=
6.78' Ft
3. Number of Distribution Bars = (Length
of Shorter Side / Spacing) + 1
L = 3.5’  3'6"
L = 3'6"1"1" = 3'4"
Number of Distribution Bars = [3'4" / 4"] + 1
Number of Distribution Bars
= [3.333' / 0.333'] + 1
Number
of Distribution Bars = 11 bars
4. Number
of Main Bars = (Length of Longer Side / Spacing) + 1
L =11.18'1"1" = 11.01'
L = [ 11.01 / 4" ] + 1
Number
of Main Bars = [11.01 / 0.333] + 1
Number
of Main Bars = 34.06 approximately equal to 34 bars
5. Total
Length of Steel bars = (length of Distribution bar) x (no. of Distribution
bars) + (length of Main bar) x (No. of Main bars)
Total
Length of Steel bars = 18.293 x 11 + 6.78 x 34 = 431.743 Ft
6. Dia of the bar for Bottom and Top Slab mesh = 10 mm
According to thumb rule:
Slab mesh: Dia  Ã˜ 10mm @ 150mm / 6"
Read About:
HOW TO CALCULATE WEIGHT OF STEEL BARS IN FEET
HOW TO CALCULATE WEIGHT OF STEEL BARS IN METER
8.
Total weight required for waist slab = weight in Feet x Total
length x no. of waist slab
= 0.188 x 431.743 x 2
Total
weight required for waist slab = 162.335 kgs
(II). Extra Bar
Extra bar = Ã˜ 8mm @ 100mm / 4"  8mm = 0.0262 Ft
Note:
Extra bar is provided at a distance of L / 4 from corner of Slab where L =
length of xbar including crank length
Length
of each extra bar = [L / 4] + overlap length = [ L / 4] + 50 D
1.
Length of each extra bar in Distribution Bar
L = 11.18 + 0.42D x 2
L = 11.18 + 0.42 x 0.333 x 2
L =
11.459 Ft
Length of each extra bar = [L / 4] + overlap length = [
L / 4] + 50 D
Length
of each extra bar = [11.459
/ 4] + 50 x 0.0262 = 4.174 Ft
2. No.
of Extra bar in Distribution Bars = No. of Distribution Bars x 2 = 11 x 2 = 22
3. Total
length of Extra bar = length of Extrabar x no. of Extrabars =
4.174 x 22 = 91.828 Ft
4. weight
of Extra bar = 8 mm ; Extra bar = Ã˜ 8mm @ 100mm / 4"
According to thumb rule:
Read about:
HOW TO CALCULATE WEIGHT OF STEEL BARS IN FEET
HOW TO CALCULATE WEIGHT OF STEEL BARS IN METER
5. Total
weight required for Extra bar = weight of the bar x
Total length x no. of Waist Slabs
= 0.1204 x 91.828 x 2
=
22.112 kgs
(III). Landing
mesh (Landing 1 & Landing 2):
When
calculating the reinforcement for the landing, it is not necessary to find the
length of the distribution bar because it is already added to the with waist
slab distribution bar length.
So,
It is necessary to determine only the length of the main bar.
Width of landing 1 =
3'
Width of landing 2 =
4'
Total
= 3' + 4' = 7'
1.
length of Landing mesh of Main bar = 7'  1"  1"
= 7'  0.1666'
length
of Landing mesh Main bar = 6.833' Feet
2.
No. of Landing mesh Main bar = Number of
Main Bars = (Length of Longer Side / Spacing) + 1 = [ 6.8333 /
4" ] + 1
= [6.8333 / 0.333 ] + 1
No.
of Landing mesh Main bar = 21.52 or approximately equal to 22 bars Total width
3.
Total Length of Landing mesh Main bar = length of main bar x Number of Main
Bars
= 6.883
x 22 = 153.626 Ft
4. Total
weight required for Main bar = weight / Ft x Total length
10mm weight = 0.188kg/ft
= 0.188 x 153.626
= 28.881
kgs
(IV). Hand Rail Mesh:
1.
Length of Xbar = 11.18'  1"  1"
= 11.18'  0.1666'
=
11.013 Ft
2.
Length of Ybar = 3'  1"  1"
= 3'  0.1666'
=
2.833 Ft
3.
No. of x  bar
L = 3'1"1" = 2'10"
No. of x  bar =
[ 2'10" / 4"] + 1
No. of x  bar =
[2.8333 / 0.333'] + 1
No.
of x  bar = 9.508 bars or approximately equal to 10 bars
4.
No. of y  bar
L =11.18'1"1" = 11.01
No. of y  bar =
[11.01 / 4" ] + 1
No. of y  bar = [11.01 / 0.333] + 1
No.
of y  bar = 34.06
Approximately equal to 34 bars
5. Total
Length of Steel bar = length of x bar x no. of x bars + length of y bar x
no. of y bars
= 11.013 x 10 + 2.833 x 34 = 206.452 Ft
6. Dia
of the bar for Hand Rail mesh in both directions = 8 mm
According to the thumb rule:
Read about:
HOW TO CALCULATE WEIGHT OF STEEL BARS IN FEET
HOW TO CALCULATE WEIGHT OF STEEL BARS IN METER
Total
weight of steel required = weight / Ft x Total length x no. of
Handrails = 0.1204 x 206.452 x 4
Total
weight of steel required = 99.427 kgs
Total Weight Of Steel Required

DIAMETER 
WEIGHT 
1. S1&S2 
10mm 
162.335 
2.
landing 1&2 
10mm 
57.763 

Total 
220.098 
3.
Extra bar 
8mm 
22.112 
4.
Handrail 
8mm 
99.427 

Total 
121.539 
No. of Steel bars required of length
40Ft/12m:
(a). Ã˜=10mm Required wt. of steel = 220.098kgs
1. Standard
length of each steel bar = 40 Ft/12m
2. Dia
of steel bar = 10mm
3. Weight
of 10mm Steel bar in kgs/Feet = 0.1881 kgs/Feet
4. Weight
of each steel bar of length 40'/12m = 0.1881 x 40 = 7.524 kgs
5. No.
of Steel bars of length 40' required = wt of steel required / wt of each steel
bar
= 220.098 / 7.524
= 29.25 or approximately equal to 30 bars of Ã˜10mm required.
For 29.25 steel bars the required weight of steel =
220.098 kgs (required wt of steel) For 30 Steel bars the required wt of steel =
7.524 x 30 = 225.72 kgs (Actual wt
of steel)
(b). Ã˜ = 8mm Required wt. of steel = 121.539kgs
1. Standard
length of each steel bar = 40 Ft/12m
2. Dia
of steel bar = 8mm
3. Weight
of 10mm Steel bar in kgs/Feet = 0.1204 kgs/Feet
4. Weight
of each steel bar of length 40'/12m = 0.1204 x 40 = 4.816 kgs
5. No.
of Steel bars of length 40' required = weight of steel required
/ weight of each steel bar = 121.539 / 4.816 = 25.23 or approximately equal to 26 bars of Ã˜8mm required.
For 25.23 steel bars the required weight of steel =
121.539 kgs (required wt of steel)
For 26 Steel bars the required wt of steel = 4.816 x 26
= 125.216 kgs (Actual wt of steel)
BILL OF QUANTITES
FOR STAIRCASE REINFORCEMENT 

S.no. 
Description Dia of bar 
Unit Kgs 
Rate/unit in Rupees 
Total Quantity 
Total Amount in Rupees 
Remark 

Actual weight of Steel required in kgs 
Required bars of 40' length 

1 
Ã˜10mm 
Kgs 
48.00 
225.720 
30 
10834.560 
1 








2 
Ã˜8mm 
Kgs 
48.00 
125.216 
26 
6010.368 
2 





Total 
16844.928 




Add 10% wastage 
1684.493 





Add 5%
Contingencies 
842.246 





TOTAL 
19371.667 

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