Chapter 15.15 - Curtailment of Bottom bars when Moment coefficients are used

In the previous section we were discussing the method of curtailment when Moment coefficients are used for the analysis of continuous beams. In this section, we continue the discussion.

We will see an example in which there is a lesser number (3-16#) of top bars at the support. The arrangement is shown in the fig.15.86 below:

Fig.15.86
Arrangement after curtailment

In this case, there are only three bars. Only one of them can be curtailed because two bars have to continue as stirrup hangers. So there will be only one stage of curtailment. The calculations are shown in the fig. below:

Fig.15.87
Calculations of areas after curtailment

We can see that the two 16 mm dia. bars give an area of 66.67%. This is greater than the required 60%. So the middle 16 mm bar can be curtailed at 0.15 l₁. There is no need to take it upto 0.25 l₁. However, the above arrangement should be finalized only after the development length checks using the values of l₁ and l₂

Let us consider one more example in which 2-16# and 1-20# are given. In this case also, there can be only one stage of curtailment, and that is of the 20 mm dia. bar. The two 16 mm dia. bars have to continue as stirrup hangers. The calculations are given below:

Fig.15.88
Calculations of areas after curtailment

We can see that after curtailment, only 56.14% remains. But we want 60%. So we must extend the bar up to 0.25 l₁. In other words, the curtailment should be done only at 0.25 l₁. And also, the development length checks should be done using the values of l₁ and l₂ . The arrangement is shown below:

Fig.15.89
Arrangement after curtailment

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So we have completed the discussion about the curtailment of top bars when 'Moment coefficients are used for design of steel'. But in the discussions, we did not consider the case when the end support of the continuous beam is simply supported. In such a support condition, there will not be any bending moment, and the arrangement shown in fig.15.58  that we discussed earlier can be provided.

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Now we will discuss about the bottom bars. We know that when Moment coefficients are used, the sagging BM at midspans are also calculated using coefficients. So there will not be any BM diagrams, and we will have to follow the same fig.8.15 of SP 34.

Fig.15.90
Curtailment of bottom bars when Moment coefficients are used

From the fig.15.90, we can see that there is only one stage of curtailment for the bottom bars. The distance of this curtailment is 0.1 l at end supports, and 0.15 l₁ at intermediate supports. These distances are measured from the 'imaginary vertical lines within the supports' which mark the effective spans. The area of remaining bars which continue towards the supports is same for both end supports and continuous supports, and is equal to 0.3A_st .

We can work out a quick example to demonstrate the area requirements. Let there be 3-16# as bottom bars in a beam. The middle one can be curtailed. The other two has to continue because, there must be two corner bars at the bottom for the stirrups. So the percentage of 2-16# is equal to 66.67. (calculations are same as that in fig.15.87 above) This is greater than the required 30%.

Let us see another example with 2-16# + 1-20# . The middle 20 mm bar can be curtailed. So the percentage of 2-16# is equal to 56.14. (calculations are same as that in fig.15.88 above) This is greater than the required 30%.

So we will get the curtailment details of ‘bottom bars within the span’ from the above fig.15.90. But we have to work out the following separately:
(i) amount of bottom bars that has to be embedded in the supports.
(ii) length of bottom bars that has to be embedded in the supports.
(iii) Development length requirements for bottom bars at supports.

(i) and (ii) above were discussed before, when we saw the details about cl.26.2.3.3(a) of the code. We saw the figs.15.42 to 15.45 in chapter 15.9, which are based on this clause. So, using those figs., we can calculate (i) and (ii) .

(iii) was also discussed before. The procedure for both simple supports and continuous supports were discussed. M_uR is the ‘Ultimate moment of resistance’ of the section, and V_u is the factored shear force at the support. The shear forces are determined by using the coefficients from Table 13. 

Thus we are now in a position to do the curtailment design of a continuous beam when it is designed using ‘Moment coefficients’. The very same procedure can be used also for a continuous slab, when it is designed using moment coefficients. For this we must use fig.9.5 of SP 34.

Curtailment of Bundled bars
When bundled bars are to be curtailed, all the bars in the bundle should not be curtailed at a single section. Each bar must be curtailed at sections which are at least 40Φ apart. This is shown in the fig.15.85 below:

Fig.15.91
Curtailment of individual bars in a bundle

First, the whole bundle is given an extension of L_a beyond the theoretical cut-off point. Then the individual members of the bundle are curtailed one at a time, and the distance between the sections are greater than or equal to 40Φ. [In the fig., the bars of the bundle are shown separately. This is only for clarity. In actual case, all the bars of a bundle will be in contact]

Another point to note is that while curtailing bundled bars, the bars which are closer to the Neutral axis should be curtailed first. So for a bundled bar at the top of a beam, the bars which are at the bottom in the bundle should be curtailed first. Similarly, for a bundled bar at the bottom of a beam, the bars which are at the top in the bundle should be curtailed first. This is shown in the figs. below:

Fig.15.92
Bundled bar at top of beam section

Fig.15.93
Bundled bar at bottom of beam section

After fixing up the final curtailments, we have to check for development lengths. This is necessary for bundled bars also. We have seen that this check is done based on fig.15.50 (bottom bars) and fig.15.60 (top bars). Now, while checking this for bundled bars, the increased value of L_d should be used based on fig.14.9. This increased value should be available on both sides of a section. 

As an example, let us consider the curtailment involving a bundle of 3 nos. of 12 mm bars. (Assume that the bars are in tension, the bars are of grade Fe 415 deformed steel, and the grade of concrete is M20)  From fig.14.9 we see that L_d of each bar of a bundle of 3 bars should be increased by 20%. We have also seen that for the above assumptions, L_d of 12 mm bars is equal to 564.14 mm. Increasing this by 20%, we get 564.14 x 1.2 = 676.97 = 677 mm. 

• If our bundle belongs to the ‘category of curtailed bars’ in fig.15.50, then the bundle as a whole should have a distance greater than 677 mm on both sides of CL, and then only the curtailment of individual bars can begin. (The individual bars should be curtailed at sections which are more than 40 x 12 = 480 mm apart.)  

• If it belongs to the ‘category of continuing bars’, then the distance on both sides of the theoretical cut-off section should be greater than 677 mm. These requirements are shown in the figs. below:

Fig.15.94
Development length requirements of a 'curtailed' bundled bar

Fig.15.95
Development length requirements of a 'continuing' bundled bar

In fig.18.94, the diameter of the curtailed bar should be used in the calculation of L_a.

So we have completed the discussion about the curtailment of bars. We will see some solved examples.

Solved example 15.1
In this example we will design the curtailment and layout of the bars of the beam that we designed in solved example 4.3

In that example, the final section was designed. Now, The design of curtailment and layout is given now as the Solved example 15.1

Solved example 15.2
In this example we will do the ‘shear design’ of the beam that we designed just above in solved example 15.1. While doing this, we will be considering the points that have to be checked while doing the shear design of a beam in which curtailment of tensile bars have been done. 

In the next chapter we will discuss about the design of Stairs.

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Chapter 15.14 - Curtailment of bars when Moment coefficients are used

In the previous section we saw the anchorage requirements and curtailment details of bent-up bars. In this section we will see a general scheme.

Curtailment when moment coefficients are used 
So far we have discussed about the curtailment of bars based on Bending moment diagrams. When we analyse the structure, we will be able to draw the BM diagrams based on the results of the analysis. These diagrams are used for the design of steel at various sections. These diagrams are also used for determining the sections at which curtailment of steel can be done. But in chapter 7 (cont..3) we saw a method of designing steel 'without doing actual analysis and drawing the BM diagram'. In this method (if the structure satisfies certain conditions), moment coefficients given in Table 12 of the code are used to find the Bending moments . So in this method, there will not be any BM diagrams to calculate the theoretical points of curtailment.

In such a case, we can do the curtailment based on the fig.8.15 of SP 34: Handbook of Concrete Reinforcement And Detailing. Based on that fig., the method of curtailment of the top bars at the end support of a continuous beam (when the beam is framing into a column at that end support) can be shown as in the fig.15.81 below:

Fig.15.81
Curtailment of top bars at end support when moment coefficients are used

In the above fig., A_st is the maximum top steel provided at the support. This much steel is not required at regions away from the support. So curtailments can be done. The first curtailment is done at a distance of 0.15 l₁ from the face of the support. The remaining bars continue towards the next support. The area of these remaining bars should be greater than or equal to 60% of A_st. 

Out of these continuing bars, some can be curtailed at a distance of 0.25 l₁ from the face of the support. The area of the remaining bars after this curtailment should be greater than or equal to 20% of A_st . These bars continue uninterrupted to the last end support, and will take part in resisting the hogging moments at various supports.

So we know the lengths at which curtailments are done and the areas that should remain after each curtailment. One more important aspect that we have to consider is that of development length. Each and every bar in the above fig requires an embedment in the column, This embedment should be greater than or equal to L_d. We know that the bars will be pulled from both ends, and so the development length should be provided on both sides of the section. For those bars which are curtailed at 0.15 l₁, the only length available for anchorage at the right side of the section (that is., section through the face of the support) is 0.15 l₁. So this 0.15 l₁ must be greater than or equal to 'L_d of those bars which are curtailed at 0.15 l₁'. For example if l₁ = 3250 mm, 0.15 l₁ = 487.5 mm. This is less than 564.14 mm. (L_d of 12 mm dia. bars of Fe 415 grade steel when M20 concrete is used).

If the condition that 0.15 l₁ ≥ L_d is satisfied, our first impression will be that 0.25 l₁ will be naturally greater than L_d. But this need not be true if the bars curtailed at both the sections are not of the same diameter.

For example,
• Let the bars curtailed at 0.15 l₁ have a diameter of Φ1 and a development length of L_d1
• Let the bars curtailed at 0.25 l₁ have a diameter of Φ2 and a development length of L_d2
• Then, if 0.15 l₁ ≥ L_d1
    ♦ 0.25 l₁ ≥ L_d1 will be true. But
    ♦ 0.25 l₁ ≥ L_d2 need not be true.
So, if the bars are of different diameters, we must check that 0.25 l₁ is also greater than the L_d of those bars which are curtailed at 0.25 l₁.

This completes the discussion on the above fig.15.81. Next we will see the curtailment at an intermediate support. This is shown in fig.15.82 below:

Fig.15.82
Curtailment of top bars at intermediate support when moment coefficients are used

From the above fig., we can see that the length requirements and area requirements are exactly same as those at the end support. The development length requirements should also be checked in the same way that we discussed for end support, based on fig.15.81 that we saw earlier.

Now we will see how the rules related to area can be applied to an actual beam. Let there be 3-20# and 2-16# as top bars at an intermediate support. A sectional view is shown in the fig.15.83 below:

Fig.15.83
Top bars at a support

We must remember a general rule in curtailment: The bars in the layer closer to the NA of the beam are curtailed first. So when curtailment of top bars at a support is done, the bars in the bottom most layer of the bar group is curtailed in the first stage. In the second stage, the bars in the second bottom most layer is curtailed. Similarly, when the curtailment of bottom bars in a span is done, the bars in the top most layer of the bar group is curtailed in the first stage. In the second stage, the bars in the second top most layer is curtailed.

So, in the first stage, the two 16 mm bars can be curtailed, and at the second stage, one 20 mm bar can be curtailed. The calculations are shown in the fig.15.84 below:

Fig.15.84
Calculations of the areas after curtailments

From the calculations we can see that:
After the first curtailment, 3-20# bars remain. This much steel gives an area which is 70.09% of the A_st at support. This is greater than the required 60%. Hence OK

After the second curtailment, 2-20# bars remain. This much steel gives an area which is 46.73% of the A_st at support. This is greater than the required 20%. Hence OK

The 2-20# which remain after the second and final curtailment will continue as stirrup hangers, and also will take part in resisting the hogging moments at various supports. The fig. showing this arrangement is given below:

Fig.15.85
Arrangement after curtailment

However, the above arrangement should be finalised only after the development length checks using the values of l₁ and l₂.

In the next section we will see more such curtailments.

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