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 = Ø10mm@ 150mm / 0.492126ft / 6”
Shear Wall mesh = Ø12mm@ 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 - Ø 10mm @ 150mm / 6"
Read about:
HOW
TO CALCULATE WEIGHT OF STEEL BARS IN FEET
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:
Read about:
HOW
TO CALCULATE WEIGHT OF STEEL BARS IN FEET
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). Ø = 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 Ø10mm 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). Ø = 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 Ø12mm 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 |
Ø10mm |
Kgs |
62.50 |
165.52 |
22 |
10345 |
|
|
|
|
|
|
|
|
|
2 |
Ø12mm |
Kgs |
62.50 |
411.92 |
38 |
25745 |
|
|
|
|
|
|
|
|
|
|
|
|
|
|
Total |
36090 |
|
|
|
|
|
Add
10% wastage |
3609.0 |
|
|
|
|
|
|
Add
5% Contingencies |
1804.5 |
|
|
|
|
|
|
Total |
41503.5 |
|
If you find
This information is helpful, please share it.
Thanks! For reading the article...
No comments:
Post a Comment