Estimation of Reinforcement Quantity for a drainage section

Accurate estimation of reinforcement bars (rebars) quantity for a given work is usually preceded by accurate detailing of the structural members and accurate preparation of bar bending schedule (BBS). The reinforcement details such as 20 Y12-02-150 mm C/C B can be explained as follows:

1. The quantity (20)
2. Type of bar (High yield steel, Y)
3. Size of bar (12 mm)
4. Bar mark (02)
5. Bar spacing (150 mm) and
6. Position of bar (B)

The BBS shows the actual shape of each of the bars which are usually collated by the bar mark. The schedule also shows the actual length of each component of the bent shapes such as U-bar, L-bars straight bars, reinforcement chairs, etc., and the total quantity required.

The bar mark is usually assigned to bars to differentiate them from other bars based on some features. Two or more bars would have the same bar mark if they have:

1. The same size
2. The same length
3. The same type of bar
4. The same shape

The extraction of the length of bars from different structural members in reinforcement details would also be considered.

1. The concrete cover of the member as would be subtracted from the length, width, and depth of the structural members.
2. The position of the bars such as Top and Bottom reinforcement bars and necessary allowances made for these.
3. The spacing of the bars.

BBS is a very important part of structural drawings and if possible it should be prepared for every type of construction work to avoid guesswork. BBS also helps quantity Surveyors to estimate the required quantity of different types of bars and append prices to them.

For small jobs that may not require the services of a quantity Surveyor, Engineers may be faced with the task of finding the quantity of rebars required for the job. In this article, I will show how to do this using a simple drainage section.

It is required to estimate the quantity of rebars in Ton required to reinforce the drawing section shown below (Figure 1). The length of the drain is 1000 m.

Use Y12@150mm for the main bars and Y10@200mm for the distribution bars. We can see from Figure 2 below the main bars which are represented as U-bar. The distribution bar can be seen as large dots running through the U-bar. The pressure diagram on the walls of the drainage is also represented in Figure 2 and explained in the next paragraph.

Description of Pressure Diagram of Surface Drainage

This was arrived at from the design. In the design of drainage walls, two pressures are considered. These are earth pressure and water pressure. These are well represented in the diagram. We must have heard that pressure increases with depth. That was why the intensity of the pressure increased from zero at the origin downwards. As I mentioned earlier, earth pressure and water pressure are considered in the design of drainage walls. It is expected that the two pressures would cancel out because they are opposite to each other but water pressure only comes during rain but earth pressure is always present.

Besides earth pressure, the lateral pressure from moving vehicles also affects the drainage wall. That is why the reinforcement in drainage walls is placed close to the side of the wall in contact with the earth. The same rule is obtainable with the drainage floor. The drainage floor experiences a gravity load from flowing water and uplift pressure from underground water. Just like the wall, the former is not constant while the latter is. This is not part of the initial aim of this article but it is good we also know it in order to properly monitor similar construction activities at the site.

Estimation of the Quantity of Main Bars (U-bars)

The U-bars would be spaced at 150 mm (0.15 m) and cover a length of 1000 m.

The required number of bars = (total length/spacing) + 1 = (1000 / 0.15) + 1 = 6667 nos

The total length of one U-bar (having removed 50 mm concrete cover) = (2 x 0.65) m + 0.8 m = 2.1 m

The total length of bars required would be (length of one U-bar x number of bars required) = 6667 x 2.1 = 14001 m

The mass per meter of one 12 mm bar = 0.888 kg

Mass of 14001 m of 12 mm bar = 14001 x 0.888 = 12433 kg = (12433/1000) = 12.433 Ton

Estimation of Quantity of Distribution Bars

The distribution bars would be spaced at 200 mm (0.2 m) over the length of one U-bar which is 2.1 m

The required number of bars = (total length/spacing) + 1 = (2.1/0.2) + 1 = 11 nos

The length of one distribution bar (having removed 50 mm concrete cover) = 999.9 m

The total length of bars required would be (length of one bar x number of bars required) = 999.9 x 11 = 10999 m

The mass per meter of one 10 mm bar = 0.616 kg

Mass of 10999 m of 10 mm bar = 10999 x 0.616 = 6776 kg = (6776/1000) = 6.776 Ton