How to Calculate Steel Quantity from Drawing
Estimation of steel quantity is an essential skill of any civil engineer. Every civil engineer must know the method of calculation steel quantity from drawing.
Data requires for estimation of steel quantity:
1) Plan of Section or Element. (Slab, Beam, Column & Foundation)
2) Section and Elevation Details.
3) All dimensions must be clear and co-related.
Read More: BUILDING CONSTRUCTION STEP BY STEP PROCESS
Steel Quantity Calculation for Slab:
Let, we take one example of slab for steel estimation.
Following Quantity we have to calculate from drawing.
- Cement Concrete for Slab.
- Centering and Shuttering for Slab.
- Weight of 12mm dia. Bars.
- Weight of 8mm dia. Bars.
- Prepare Bar Bending Schedule.
(Note: Weight of Steel bars 12mm – @ 0.9 kg / m
8 mm – @ 0.4 kg/m,
How to Calculate Steel Quantity from Drawing
Estimation Sheet:
Sr. No. | Item Description | No. | Length (m) | Width/ Breadth (m) | Height/ Wt (m) | Quantity |
1 | Cement Concrete for Slab (1:1.5:3): | |||||
L = 4 + 0.23 + 0.23 = 4.46 m | 1 | 4.46 | 3.46 | 0.12 | 1.85 cu m | |
B = 3 + 0.23 + 0.23 = 3.46 |
(See drawing in to in dimension (without wall) of slab is given 3 m x 4 m. For calculating Quantity of slab concrete we have to calculate out to out dimension of slab.)
2 | Centering & Shuttering for Slab: | ||||||
Bottom | 1 | 4.00 | 3.00 | – | 12.00 | cu m | |
Sides | 2 | 4.46 | – | 0.15 | 1.34 | cu m | |
( Take 30mm Extra Above Slab Thickness) | 2 | 3.46 | – | 0.15 | 1.04 | cu m | |
Total | 14.38 | cu m |
(Formwork Quantity is calculated in square meter. There is one bottom portion and 4 side shuttering. For bottom we have in to in dimension of slab 3m x 4m. For Side formwork there is 2 different dimensions, 2 with 4 m side and 2 with 3 m side. For total length we have to add wall width of both sides to get total length.)
Read More: ESTIMATION OF A BUILDING WITH PLAN
How to Calculate Steel Quantity from Drawing
3 | 12 mm dia main Steel bars @ 150 mm c/c Alternate bent up. | ||||||
L = 3 + 0.23 + 0.23 + (2 x 9 x 0.012) – (2 x 0.05) | |||||||
L = 3.58 (Straight Length) (For finding out nos. of bar we have to calculate straight length of main bar without adding extra length due to bent up) | |||||||
Span = 4 + 0.23 + 0.23 – (2 x 0.05) = 4.36 m | |||||||
No. of Bars = (4.36/0.15) + 1 = 30 nos. | |||||||
Extra Length of Bent up: |
Extra Length = 0.45 X | |||||||
X = 0.12(Slab Thickness) – 2 x 0.02 (Top & Bottom Cover) – 0.012/2(Harf dia top) – 0.012/2( Half dia bottom) | |||||||
X = 0.12- 2 x 0.02 – 0.012/2 – 0.012/2 = 0.068 m | |||||||
L
= 3.58 + 0.45X L = 3.58 + 0.45 x 0.068 3.61 | 30 | 3.61 | @ | 0.90 | 97.47 | kg |
(In case of bent up bars are provided on both sides, we have to add 0.45 X twice in length)
4 | 8 mm dia. Distribution steel @ 180 mm c/c | ||||||
Bars at Bottom: | |||||||
Hook
Length = 9d = 9 x 0.008 = 0.072 < 0.075 ( Mini. Hook length = 0.075) | |||||||
L = 4 + 0.23 + 0.23 + 2 x 0.075(Both side hook(9d) | |||||||
L = 4.51 m | |||||||
Width of Slab = 3 + 0.23 + 0.23 – 2 x 0.05 (Both Side Cover) = 3.36 m | |||||||
No. of Bars = (3.36 / 0.18) + 1 = 20 nos. | |||||||
Bars at Top: | |||||||
Width
of Slab at one end for bent up at Top | |||||||
L= 0.23 + 0.45 – 0.068 – 0.05(cover) | |||||||
(When bent up bar bent to top at the end of bar, there is also distribution steel is provide. we have to add this bars in total distribution steel bar nos.) | |||||||
L = 0.562 m | |||||||
Nos. of Bars at one End: = (0.562/0.18) + 1 = 5 nos. | |||||||
Nos.
of bars at both side = 2 x 5 = 10 | |||||||
Total Bars = 20 + 10 + 30 | 30 | 4.51 | @ | 0.40 | 54.12 | kg |
Free Download Above Excel Sheet :
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