During site construction, Reinforcement Chair or sometimes called as Rebar Spacer, Steel Chair or simply Chair is commonly seen in raft or mat foundations. This is commonly shaped as Z or inverted U type reinforcement to evenly separate the top and bottom mesh reinforcement or to provide enough space between the meshes. In the installation of the main mesh reinforcement during construction, steel chairs play a vital role to keep the spacing evenly and hold the mesh reinforcement in place before and after concreting. That is why Steel chairs should be designed also accordingly and not taken for granted.
This article will show you how to design reinforcement chairs. And Reinforcement chair design can be further understood by showing an illustrative example given in the image below. Let us design the reinforcement chairs needed as per the typical raft details that are given below:
Summarizing the image above, the following are given:
- Raft thickness: 2000mm
- Main Mesh Reinforcement:
- Bottom: T32-200mm both ways
- Top: T25-200mm both ways
The first thing to consider is how much the weight of the rebar reinforcement in the raft which is tabulated as:
- T25: 0.0378 kN/m length
- T32: 0.0632 kN/m length
An additional 1.5kN/m2 Live Load to be considered to account for any construction loads.
If you are wondering rebar weight with a unit of kN/m is derived using the formula:
Calculate for Weight of Rebar that the rebar chair will carry:
- Top Reinforcement: (0.0378 kN/m/0.20) x 2 layers = 0.378kN/m2
- Bottom Reinforcement: (0.0632 kN/m/0.20) x 2 layers = 0.632kN/m2
Assumed Diameter of Reinforcement Chair to use:
Let us use say T25 for chair reinforcement
Check Buckling Capacity of T25 Rebar using the Euler’s Formula:
Maximum Buckling Force is given by:
- Pcr: allowable buckling force
- E: Modulus of Elasticity of Rebar can be taken as 200, 000 MPa or 200×106 kPa
- I: Rebar’s area moment of inertia in m4 with the formula:
and I is equal to 1.917 x10-8 m4
- K: effective length factor = 1.0
- L: unsupported length= 2.0m
Substituting to the above equation:
- Pcr = 9.46kN
- Pcr = 0.75×9.46kN= 7.1kN
- is the strength reduction factor as specified by table 21.2.1 given below by the ACI 318-14 code.
Solving for the Number of Reinforcement Chair required per square meter:
Since the chair reinforcements support only the top layer, then we will consider the weight of top mesh reinforcement as dead load.
No of Bars/m2 = Factored Load / Pcr
Factored Load = 1.2Dead Load + 1.6 Live Load
= 1.2 x 0.378kN/m2 + 1.6 x 1.5kN/m2
= 2.85 kN/m2
No of Bars/m2 = 2.85 kN/m2/7.1kN= 0.40 bar/ m2 ;
Hence, the reinforcement chair required is 1 T25 reinforcement chair per square meter.
An Image view of Reinforcement Chair
Since the reinforcement chair supports only the top mesh reinforcement, the bottom mesh reinforcement needs as well as support to maintain its concrete cover. Likewise, a spacer is required. There are so many types of spacers that can be used during site construction, but the most commonly used at the site is the concrete spacer. But how do we check if the dimension of the concrete spacer is enough so it can withstand the load of the raft reinforcement it carries? That will be tackled in the next article-“Design of Concrete Spacer” so stay tuned or subscribe to our newsletter for any updates and notification.
What do you think of the above this article? Tell us your thoughts! Leave your message in the comment section below. Feel free to share this article, subscribe to our newsletter and follow us on our social media pages.
18,926 total views, 36 views today