Chapter 15 - Curtailment of bars

In the previous section we completed the discussion on splices. In this section we will discuss about 'Curtailment of bars'

In a simply supported beam, the maximum bending moment occurs at the midspan or near the midspan, and this bending moment is sagging in nature. So the steel is provided at the bottom side of the beam. (If it is designed as a doubly reinforced beam, then both bottom and top steel will have to be provided.) Similarly, in the case of a continuous beam, top steel has to be provided for the hogging moment at the support. (Here also, if it is designed as a doubly reinforced beam, then steel will have to be provided at the bottom side also.) Also in a continuous beam, bottom steel has to be provided for the sagging moment at the midspan region. So there are three critical points in a beam where bending moment is considered:
• Midspan of a simply supported beam
• Support of a continuous beam
• Midspan of a continuous beam.
The bending moment will be maximum at these points.

But if we look at the bending moment diagram, we can see that the BM decreases progressively, as we move away from these points. This means that the sections of the beam, away from these points will be experiencing lesser bending moments. So these sections need not be as strong as the section at the critical points. This means that, we can reduce the quantity of steel progressively, as we move away from the critical points.

But bars which are no longer needed after a particular section, cannot be 'just stopped' at that section. They must be continued for some more distance. The code gives us the guidelines for doing this. We will discuss about them in the following sections.

Theoretical bar cut-off points

Consider the bending moment diagram of a simply supported beam shown in the fig.15.1. below:

Fig.15.1
Bending moment diagram of a simply supported beam

Let the factored BM at the mid span be M_u,CL, and the steel provided at this section be A_st,CL. And, let the externally applied factored BM at a section XX be M_u,XX , and the steel provided at this section be A_st,XX . The steel provided at XX is different from that at CL because, some of the bars which constituted the total of A_st,CL at the mid span, are curtailed at XX so that only remaining bars continue to the support A. So the applied BM M_u,XX at section XX, is resisted by the continuing bars. The position of XX at which some bars are curtailed should be such that Mu,XX at this section will be less than or equal to the ultimate moment of resistance of the beam section at XX. That is., the beam section should be able to resist the applied factored BM at XX, by using the remaining bars. Or in other words, the condition M_u,XX  ≤ M_uR,XX should be satisfied. This can be illustrated with the help of an example.

Let the beam shown in fig 15.1 above have a span of 4m, and carry a udl of 25 kN/m including self weight. Let the width of the beam section be 230 mm and the total depth be 450 mm.

The maximum factored BM at midspan is given by (1.5 x 25 x 42) /8 = 75kNm. Assuming M20 concrete, and Fe 415 steel, the beam can be designed for this BM, and the result obtained is : A_st,CL = 597.51 mm², at an effective depth d of 404 mm.

For the sake of our discussion, let us assume that exactly the same amount of steel is provided at the same effective depth of 404 mm. Then, if we analyse the beam section, by the methods that we discussed in  chapter 3, we will get the ultimate moment of resistance M_uR,CL at CL, exactly equal to 75 kNm. 

Now let it be decided to curtail one third of the bars. We have to calculate the position of the suitable section to do this curtailment. This calculation can be done as follows:
When one third of steel is curtailed, two third will remain. So the remaining area = 2/3 x 597.51 = 398.34 mm². We will find the M_uR of a section having this much area of steel at the same effective depth. That is., we want the M_uR of a section having a tensile steel of 398.34 mm² at an effective depth of 404 mm. Using the methods of analysis that we discussed in the chapter 3, this can be obtained as 52.94 kNm

Now we find the position of the section at which the applied factored bending moment is equal to 52.94 kNm:
The factored reactions R_A and R_B = 75 kN.
The equation for BM at any section on the beam can be written as
Eq.15.1:
M_u,XX = 75 x – (1.5 x 25 x²)/2 where x is the distance of the section from the support A.
We can put M_u,XX = 52.94 in the above equation, and obtain x = 0.9153 and 3.0847 m. These two values are symmetric (∵ 0.9153 + 3.0847 = 4.0m) about the section CL passing through the mid span. So we will concentrate on one single value, 0.9153. We must place section XX at a distance of 0.9153 from support A. This position of XX is the 'Theoretical point of curtailment'. If there are three bars of the same diameter at the midspan, one of them can be stopped at this section. The other two will continue upto support A. This is shown in the fig.15.2 below:

Fig.15.2
One bar curtailed at XX

We can superimpose the graph of M_uR over the above diagram. This is shown in the fig. below:

Fig.15.3
Graph of M_uR superimposed

From the superimposition, it is clear that at the point of curtailment, the applied moment is equal to the M_uR (the 'capacity') of the beam at that point. So at the point of curtailment, that is, at section XX, the remaining two bars will be fully stressed. 

We have to note an important point at this time. In the above fig. which shows the superimposition, some M_uR appears to be 'available in excess'. This is marked in the fig. below:

Fig.15.4
Seemingly excess M_uR

The excess M_uR is in the region between XX and CL. In this region, no curtailment has been made. Can we curtail the bar in this region so as to make the design more economical? The answer is 'NO'. 

Such a curtailment will result in undesirable effects. We can explain this by trying to make a curtailment in this region.  Let us curtail the bar at a section AA instead of at XX. This AA is in between XX and CL as shown in the fig. below.

Fig.15.5
Undesirable effect due to curtailment between XX and CL

We can see in the graph of M_uR that, the 'block with a height of 52.94' is now further extended to the right upto AA. This is because, after AA, there are only two bars, and the 'capacity' that the two bars offer is 52.94 kNm. 

But now, some portion of M_u is exposed. This means that, the 'capacity' of the beam is less than the applied moment in the exposed region. This can never be allowed.

So the one bar which was curtailed at XX cannot be stopped any earlier before reaching XX. That is., it cannot be curtailed at any point between XX and CL. The bar should be curtailed only at the point were the applied factored BM falls to such a low value that, the remaining bars can take up the load.

When we decide to curtail a portion of bars, the remaining bars will play the main role in deciding the position of the section at which curtailment can be made.

In the next section, we will see the example of a beam in which curtailment is made at more than one section to the left of CL.

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