Bar Bending Schedule of Doglegged Staircase | BOQ of Doglegged Staircase | BBS of Staircase -lceted LCETED INSTITUTE FOR CIVIL ENGINEERS

## Jan 5, 2022

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 = Ø 10mm@100mm or 4"

(b). Landing mesh = Ø 10mm@100mm or 4"

(c). Extra bar = Ø 8mm@ 100mm or 4"

(d). Handrail mesh = Ø 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 over-lap length of 50d on each side of Distribution Bars to join the hand-rail 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 - Ø 10mm @ 150mm / 6"

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 = Ø 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 x-bar including crank length

Length of each extra bar = [L / 4] + over-lap 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] + over-lap 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 Extra-bar x no. of Extra-bars = 4.174 x 22 = 91.828 Ft

4. weight of Extra bar = 8 mm ; Extra bar = Ø 8mm @ 100mm / 4"

According to thumb rule: 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 X-bar = 11.18' - 1" - 1"

= 11.18' - 0.1666'

= 11.013 Ft

2. Length of Y-bar = 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: 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). Ø=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 Ø10mm 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). Ø = 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 Ø8mm 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 STAIR-CASE 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 Ø10mm Kgs 48.00 225.720 30 10834.560 1 2 Ø8mm 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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