How To Calculate Bar Bending Schedule For Rectangular Sump Reinforcement? | BBS | Quantity Survey | LCETED -lceted LCETED INSTITUTE FOR CIVIL ENGINEERS

## Dec 18, 2021

In this article, we explained about how to calculate Bill of Quantities And Bar bending schedule for Rectangular Sump Reinforcement For a given plan & Section Plan of Sump (water tank below ground level)

To find

(A). Weight of Steel required

(B). No. of Steel bars required, of length 12m/40Ft

(C). Bill of Quantities for Rectangular Sump Reinforcement if Rate of steel/kg=62.50 rupees. (on 2021)

GIVEN DATA

Reinforcement Details :

Bottom and Top Slab mesh = Ã˜10mm@ 150mm / 0.492126ft / 6”

Shear Wall mesh = Ã˜12mm@ 150mm / 0.492126ft / 6”

Size of Sump = 12' x 9' x 6'9"   Thickness of bottom and top slab = 6"

Solution:

(i). Bottom and Top Slab mesh Calculations:

Note: while providing Steel mesh in the bottom and top slab, 1" concrete cover shall be deducted from each side of the Slab.

1. Length of x-bar = 12' - 1" - 1"

= 12' - 0.166'

= 11.833 Ft

2. Length of y-bar = 9' - 1" - 1"

= 9' - 0.1666'

= 8.833 Ft

3. No. of x-bars = [8.833 / 0.492] +1

= 18.953 approximately 19

4. No. of y-bars = [11.833 / 0.492] +1

= 25.05 approximately 26

Formula used to find, no. of bars = [ opposite length / spacing ] + 1

5. Total length of x-bar and y-bar = 11.833 x 19 + 8.833 x 26 = 454.485Ft

Deduction of Cover from top slab mesh:

Length of x-bar = 2'

Length of y-bar = 2'

Spacing of bar – 4”- 0.333

No. of x-bars = [2' / 0.333] +1 = 7

No. of y-bars = [2' / 0.333] +1 = 7

Note: End bars will be deducted from each side of cover therefore, 7 - 2 = 5 bars shall be considered.

Total length of Steel bar for cover = 2 x 5 + 2 x 5 = 20 Ft

Total length of Steel bar required = 454.485- 20 = 434.485 Ft

6. Dia of bar for Bottom and Top Slab mesh = 10 mm

According to thumb rule:

Slab mesh: Dia - Ã˜ 10mm @ 150mm / 6"

HOW TO CALCULATE WEIGHT OF STEEL BARS IN METER

7. Weight required for bottom and top slab mesh = weight / Ft x Total length x no. of Slabs

= 0.188 x 434.485 x 2 = 163.36 kgs

(ii). Shear Wall along horizontal axis:

1. Length of x-bar = 12' - 1" - 1" + 50D x 2

= 12' - 0.1666' + 50 x 0.03937 x 2

= 15.77 Ft

2. Length of y-bar = 6.75' + 3" + 3" + 16D x 2

D = 12mm = ? Ft

1” = 25.4mm

=12 / 25.4

= 0.472"

= 0.472 /12

= 0.03937Ft

= 6.75 + 0.5' + 16 x 0.03937 x 2

= 8.50 Ft

3. No. of x-bars

L= 6.75' - 1" - 1"

L= 6.75'- 0.16661’

L= 6.583 Ft

L = [6.583 / 0.492] +1

L = 14.38 approximately 15

4. No. of y-bars

L =12'-1"-1"

L = 11.833Ft

L = [11.833 / 0.492 ] +1

L = 25.05 approximately 26

5. Total length of Steel bar = 15.77 x 15 + 8.50 x 26 = 457.55 Ft

(iii). Shear wall along vertical axis:

1. Length of x-bar = 9' - 1" - 1"

= 9' - 0.1666'

= 8.833 Ft

2. Length of y-bar = 6.75' + 3" + 3" + 16D x 2

12mm = ? Ft

1” = 25.4mm

=12 / 25.4

= 0.472"

= 0.472 /12

= 0.03937Ft

= 6.75’ + 0.5' + 16 x 0.03937 x 2

= 8.50 Ft

3. No. of x-bars

L= 6.75'- 1" - 1"

L= 6.75'- 0.16661’

L= 6.583 Ft

L = [6.583 / 0.492] +1

L = 14.38 approximately 15

4. No. of y-bars

L=9'-1"-1"

L= 8.833Ft

= [8.833 / 0.492 ] +1

= 18.95 approximately 19

5. Total length of x-bar = 8.833 x 15 + 8.50 x 19 = 293.995 Ft

Total length of Steel bars = Shear wall along horizontal axis + shear wall along vertical axis

Total length of Steel bars = 457.55 + 293.995 = 751.545 Ft

6. Dia of the bar for Shear wall mesh = 12 mm

According to the thumb rule:

HOW TO CALCULATE WEIGHT OF STEEL BARS IN METER

7. Total weight required = Weight per feet x Total length x no. of walls

= 0.271 x 751.545 x 2

= 407.33 kgs

No. of Steel bars required of length 12m/40Ft :

(a). Ã˜ = 10mm

Required wt. of steel = 163.36 kgs

1. Standard length of each steel bar = 40 Ft.

2. Dia of steel bar = 10mm

3. Weight of 10mm Steel bar in kgs / Ft  = 0.1881 kgs/Ft

4. weight of each steel bar of length 40' = 7.524 kgs/nos

5. No. of Steel bars of length 40' required= wt of steel required / wt of each steel bar

= 163.36 / 7.524

= 21.711 or approximately equal to 22 bars of Ã˜10mm required.

For 21.711 steel bars the required weight of steel = 163.36 kgs (required wt of steel)

For 22 Steel bars the required wt of steel = 7.524 x 22 = 165.52 kgs (Actual wt of steel)

No. of Steel bars required of length 12m/40Ft:

(b). Ã˜ = 12mm

Required wt. of steel = 407.33 kgs

1. Standard length of each steel bar = 40 Ft.

2. Dia of steel bar = 12mm

3. Weight of Steel bar in kgs / Ft = 0.271 kgs/Ft

4. Weight of each steel bar of length 40' = 10.84 kgs

5. No. of Steel bars of length 40' required = wt of steel required / wt of each steel bar

= 407.33 / 10.84

= 37.57 or approximately equal to 38 bars of Ã˜12mm required.

For 37.57 steel bars the required weight of steel = 407.33 kgs (required wt of steel)

For 38 Steel bars the required wt of steel = 10.84 x 38 = 411.92 kgs (Actual wt of steel)

 S.no. Description Unit Rate/unit in Rupees Total Quantity Total Amount in Rupees Remark Dia of bar Kgs Actual weight of Steel required in kgs Required bars of 40' length 1 Ã˜10mm Kgs 62.50 165.52 22 10345 2 Ã˜12mm Kgs 62.50 411.92 38 25745 Total 36090 Add 10% wastage 3609.0 Add 5% Contingencies 1804.5 Total 41503.5

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