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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Chapter 15.13 - Curtailment of Bent-up bars

In the previous section we saw the anchorage requirements of bent-up bars at simply supported ends and at end supports in frames. In this section we will see the case (iv) and (v) of fig.15.59.

Case (iv): Intermediate supports in continuous beams and slabs

The following fig.15.73 shows such a support.

Fig.15.73
Bent up bar at continuous support

In the fig.15.67, the bar provided as bottom bar in span BC is bent-up at Q. It then becomes top steel at support C and continues as top steel into the span CD. Now, when the span BC bends downwards due to the applied loads, the bar will experience tension. So it will try to pull out from support C. To prevent this, we know that adequate development length should be provided, and from the discussion that we had about the 'development length of bent-up bars', we can say that the length RST should be greater than or equal to L_d. This will prevent the bar in span BC from pulling out from the support C.

This same bar will experience another tension also: At the support, the beam is bending upwards, so the bar is being pulled from span BC as well as from span CD. So the bar will try to pull into the support C from both sides. That is., from span BC and from span CD.

We will first consider span BC. The portion inside BC has a bent shape. So a bar with a bent shape is preventing the 'pulling into support C'. This situation requires a different method of measurement for L_d. It is shown in the fig. below:

Fig.15.74
Method of measuring L_d when the bent up bar is under hogging moment

From the fig.15.74, we can see that the procedure is simpler. We do not need to mark the point R. The measurement is taken from the bottom bend point Q towards the left side. The available length towards the left of Q should be greater than or equal to L_d. If this is satisfied, the bar will not 'contract' into support C from the span BC.

In a similar way, the bar will try to 'contract' into the support C from span DC also. To prevent this we must provide L_d from the face of the support towards the right side.

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Now we will discuss about the performance of the bent-up bar in resisting the hogging moment at the support C. We have seen figs.15.51 and 52 which give the requirements of top bars at an intermediate support in a continuous beam. Fig.15.51 is shown here again for easy comparison.

Fig.15.51
Curtailment of 'straight' top bars

We can see that a bar is curtailed at a distance of L_a from the theoretical cut-off point. The other bars continue beyond the point of inflection, (two of them will continue even beyond the next support to act as stirrup hangers and also to take part in resisting the hogging moment at the next support. In a general simple case, there will be three bars at the top, and the middle one will be curtailed. If the hogging moment is of a larger magnitude, more than three bars will have to be provided in layers). Here we concentrate our attention on the bar which is cut-off at a distance L_a from the theoretical cut-off point. This actual cut-off point is close to the support. We can think of using a bent-up bar in the place of this bar. This is shown in the fig.15.75 below:

Fig.15.75
Bent up bar in place of curtailed bar at top of continuous support

From the fig., we can see that the horizontal portion ST can be extended to the desired point in the right side of support C. But it is not possible to do so in the left side, and thus it may not have the required length in the left side. So the bar will perform well on the right side of the support C. And any contribution that it makes to resist the hogging moment on the left side of C should be ignored. But this arrangement will create a sort of asymmetry because, on the right of support C, three bars (two stirrup hangers and one bent-up bar) are available, while on the left side, only the two stirrup hangers are available to take up the hogging moment. This is solved by bringing in the bent-up bar of span CD as top bar into span BC as shown in the fig. below:

Fig.15.76
Bent up bar at a support from adjacent spans

In the above fig., the two bars are shown at different levels only for clarity. In the actual beam, they are provided at the same level as indicated by the ‘0 mm’ clearance between the bars.

Now we will see the curtailment requirements when bent-up bars are provided. Earlier we have seen fig.15.51 (shown again above) which showed the curtailment details when straight bars are used. This fig. is based on the cl.26.2.3.4 of the code. The same requirements apply here also. So we can draw a fig. for bent-up bars by making suitable modifications to fig.15.47. This is given in fig.15.77 below:

Fig.15.77
Curtailment of top bars at an intermediate support when bent up bars are used

We can see that the bent-up bar from span CD is not included in the calculation of A_st1. This is because, it's top horizontal portion does not have the required length inside span CD. So it's contribution cannot be taken into account in span CD. But it's contribution can be taken into account in the span BC.

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So we have completed case (iv) of fig.15.59. The next case (v) is similar to case (iv). To show a final diagram showing the curtailment details, all we need to do is to change the support condition in the above fig.15.77. This is shown below:

Fig.15.78
Bent up bars at the intermediate supports of a frame

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So we have seen the final case (v) also of fig.15.59. It may be noted that for the cases (iii), (iv) and(v), Moment envelope should be used instead of Bending moment diagram where ever applicable.

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Shape of Bent-up bars

We have seen the usage of bent-up bars in various situations. Now we will see an important aspect that we have to consider in the 'actual making' of a bent-up bar. In the figs. that we saw above, the bends are all shown as ‘sharp’, as in the fig. below:

Fig.15.79
Sharp bends shown in illustrations

But this is used only for illustration purpose. In an actual beam, such sharp bends should be avoided. This can be achieved by introducing curved portions in between straight segments as shown in the fig. below:

Fig.15.80
Actual method for forming a bent-up bar

Each of the curved portions at the top and bottom bend points is a part of a perfect ring having inner radius r and outer radius r + Φ. More details about the formation of such bent bars are given in IS 2502-1963: Code of practice for bending and fixing of bars for concrete reinforcement.

So we have completed the discussion about bent-up bars. In the next section, we will see the method of curtailment when the continuous members are designed using moment coefficients.

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Chapter 15.12 - Anchorage for Bent-up bars at supports

In the previous section we saw the compilation of the various requirements for the curtailment of Top and Bottom bars. In all those cases, we have used 'straight bars'. In this section we will see the use of 'bent-up' bars.

We have seen that some of the tensile bars can be curtailed at sections beyond which they are no longer required (after giving the required extensions). Instead of curtailing them at those sections, they can be bend upwards. The angle of bend is usually 45 to 60 degrees. After bending, these bars continue upwards up to the top face of the beam. (the required concrete cover should be provided at the top face). At the top face, these bars are given a bend once again so that they become horizontal again. After becoming horizontal, they continue towards the support.

The bending up of the bars is done near the supports. So we will discuss it at the same five cases that we saw earlier in fig.15.59 in the previous section.

In the first three cases, the bent up bars will not have space for required extension. So they must be given a standard 90^o bend, so as to obtain the required L_d. In the last two cases, required space will be available for extension. So when they are extended, they will have enough length to contribute towards 'serving as a part of the negative top steel' required at the support. We will discuss each of the cases in detail.

Case (i) Simply supported ends: We know that a bar should be extended beyond the theoretical cut-off point by distance greater than or equal to L_a. We discussed this based on fig.15.25 to 27. Based on that, we must begin the sloping portion of the bar only at a distance of L_a from the theoretical cut-off point. But, when we provide a bent-up bar, it will intercept the diagonal crack. as shown in the fig.15.62 below:

Fig.15.62

Bent-up bar intercepting the crack

Because of this type of interception, the bent-up bars contribute towards reducing the spread of diagonal cracks. [This type of a contribution is not obtained from straight bars]. So we need not extend the bar to a distance of full L_a before beginning the sloping portion. According to cl.8.3 (b) of SP 34, for extension purpose, the bent-up bar may be considered to be effective up to the point where the bar crosses half of the effective depth of the beam. 

This can be explained as follows:
The bent-up bar originally travels through the bottom side of the beam. This is at the midspan region.  Near the support region, it is bent upwards. Once it is bent, the bar is no longer available at the bottom side. So we are tempted to think in this way:
• a bottom straight bar no longer contributes any thing at the bottom side after the cut-off point. 
• in the same way, the bent-up bar no longer contributes any thing at the bottom side, after the point of bent-up. 

Based on this, we will want to bring the point of bend up to the beginning of the crack as shown in the fig.15.63(b) below:

Fig.15.63
Extension for bent-up bar

Fig.(a) shows the original point r of the bent-up. But as we are required to provide an extension of L_a, the point of bent up is given at q as shown in the fig.(b). But this is not required. The tension in the bent-up bar is effective to resist the 'external applied moment M_u,x' shown in fig.15.26, even after the point of bent-up. Now the question arises: For how much 'more distance' is it effective? The answer is given by the above mentioned cl.8.3(b) of SP 34. 

Imagine a horizontal plane through half the effective depth of the beam. Any bent-up bar in that beam will have a point of intersection with the plane. This point of intersection will be in the 'sloping portion'. According to the clause, the tension in the bar at this point of intersection is effective in resisting M_u,x.  Based on this information, we can calculate the required extension. The calculations are based on fig.15.64 below:

Fig.15.64

Calculation of extension

The point of intersection between the bent-up bar and the horizontal plane through d/2 is our 'point of interest'. We want that point to be at the beginning of the crack. The above fig. shows exactly that. q' is the point of intersection, and it is on section YY which passes through the beginning of the crack. To this finalised position, we can add two triangles stq and str. With the help of those triangles, we calculate the required extension.

• sr is original position of the bent-up bar.

• tanα = st⁄tr = (d/2)⁄tr  . So tr = d/2 cotα.

• Similarly tanθ = st⁄qt = (d/2)⁄qt  . So qt = d/2 cotθ.

If we assume that the diagonal crack occurs at an angle θ = 45^o and the bent up is also done at an angle of α = 45^o, then we will get tr = qt = d/2.  (∵ tan45 = cot45 = 1). So, if we extend the horizontal portion of the bar beyond the section XX by a distance d⁄2 , and thus start the sloping portion from the point t, the point s will move horizontally to q’. So we can conclude that the additional extension that has to be given beyond the theoretical point of bent-up is L_a = tr = d ⁄2.

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Now we will discuss about the development length requirements of bent-up bars. The fig.15.65 below shows a bent-up bar PQRSTU , provided in a beam.

Fig.15.65

Development length for bent up bar

The bend is given at Q. So the sloping portion begins at Q , and the angle of the bend is shown as α . [Bend can be given at Q only if the bar is no longer required to take up any sagging moment, and L_a =d/2 is given]. But from the discussion that we had based on fig.15.50, development length L_d should be provided on both sides of Q. As the portion on the left side of Q is having a sloping shape, we need a standard method to measure L_d on a sloping portion. 

This method is shown in the above fig.15.65 and can be explained as follows: The point R is marked on the bar. It is the point of intersection of the bar with a horizontal plane through d/2. The length available from R to the end of the bar is measured. This length should be greater than or equal to L_d . If at an end support, enough space is not available, then a standard 90^o bend can be given at T. 

If the point Q is a little more away from the support, it may be possible to avoid the 90^o bend. This is shown in the fig.15.66 below:

Fig.15.66

Development length of bent up bar with out standard 90^o bend at the end

For the bar B in the above fig., the point Q at which the sloping portion begins, is further away from the support. So there is enough space to extend the bar, and a 90^o bend can be avoided at the end.

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As we are discussing about the bent-up bars at this stage, it will be better if we look at the performance of these bars as 'top bars' also. The above figs.15.65 and 15.66 can be used for this purpose. Earlier, we saw the method of providing top bars at a simple support of a beam using 'straight bars'. We discussed it based on fig.15.58 in the previous section. There we saw the quantity of bars to be provided. Now, in our present case, looking at the above fig.15.65, we can say that the portion STU can take part in resisting the hogging moment at the support. But we must consider the length of the horizontal portion ST. This length depends on the position of Q, and angle α. So it may not be always possible to get a sufficient length for ST to resist the hogging moment. Because of this, it is better to ignore any contribution made by the bent-up bar towards resisting the hogging moment at a simple support of a beam. Instead, we must use the two stirrup hanger bars to satisfy this requirement. The hanger bars should be given required anchorage at supports.

Now let us look at the Bar B in fig.15.66. This bar cannot make any contribution towards resisting the hogging moment because it is not anchored into the support. Even if we decide to extend it and give the required anchorage within the support, the horizontal length at the top may not be sufficient always. So in this case also, we must ignore any contribution made by the Bar B towards resisting the hogging moment at a simple support of a beam. Instead, we must use the two stirrup hanger bars to satisfy this requirement. But these bent-up bars perform well as a bottom bars to resist the sagging moment, and also assists in resisting the diagonal tension crack.

In the case of slabs, it may be possible to use the bent-up bars. Earlier we saw the provision of straight bars as top bars at the supports in fig.15.57. For bent-up bars, this fig. can be modified as shown in fig.15.67 below:

Fig.15.67

Bent up bar for slab

In the above fig., it is shown that:

The distance between the top bend point of the bar and the face of the support should be greater than or equal to 0.1l. (15.7)

So, after fixing up the bottom bend point, we will be able to determine the position of the top bend point, and at that stage, we must check whether condition 15.7 given above can be satisfied. If it can be satisfied, we can bend up the alternate bars of the slab and satisfy the area requirements also as shown in fig.15.67.

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So we have completed the discussion about the performance of the bent-up bars as top bars at 'simple supports'. We have seen case (i). The same procedure can be adopted for [case (ii) of fig.15.59. The simply supported end of continuous members] also when bent-up bars are considered.

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Case (iii) End support of multi span continuous beams which are part of frames:
In this discussion, we will see the details of the performance of bent-up bars at the end support of a frame. In the previous two cases, we were dealing with simple supports. There, hogging moment is zero. So there was not much calculations to be done. But here there will be a hogging moment that has to be considered. First let us see how the bent-up bar will perform as a bottom bar at such a support.

Fig.15.68

Bent up bars at the end support in a frame

The above fig.15.68 shows a bent-up bar provided in a beam which is framing into an end column of a frame. As usual, first, the point Q is determined by extending d/2 from the theoretical point of bend. Then the length RSTU is measured. This length should be greater than or equal to L_d. Then only the bar can effectively perform as a bottom bar and take part in resisting the sagging moment along with other bottom bars.

Now we will see it's performance as a top bar. We have seen figs.15.55 and 56 which give the requirements of top bars at the end support in a frame. Fig.15.55 is shown here again for easy comparison:

Fig.15.55 

Curtailment of 'straight' top bars (end support in framed structures)

We can see that one bar is curtailed at a distance of L_a from the theoretical cut-off point. The other bars continue beyond the point of inflection. (two of them, not shown in the fig., will continue even beyond the next support to act as stirrup hangers and also to take part in resisting the hogging moment at that support). Here we concentrate our attention on the bar which is cut-off at a distance L_a from the theoretical cut-off point. This actual cut-off point is close to the support. We can think of using a bent-up bar in the place of this bar. This is shown in the fig. below:

Fig.15.69

Bent up bar in the place of a curtailed bar at top

In the above fig., the horizontal portion TS is reaching not even up to the theoretical cut-off point. The point S should have reached up to a distance L_a beyond the theoretical cut-off point. Because of this shortage of length, this bent-up bar in the above fig.15.63 cannot be considered to be a member of the ‘bar group which resists the hogging moment’. That is., any contribution that this bar makes towards resisting the hogging moment should be ignored, and other extra straight bars should be provided. We could take the contribution into account if the bar had the shape as shown in the fig.15.70 below:

Fig.15.70

Required shape of the bent up bar

In the fig.15.64, the point S is beyond a distance L_a from the theoretical cut-off point. The bar should also have a length greater than or equal to L_d embedded within the column. So this shape satisfies the length requirements. But it should be noted that it is the 'position of point Q' that we fix up first from sagging moment considerations. The position of S depends upon the position of Q. If it is required that Q is to be a little more to the left, the above ‘ideal shape’ will not be obtained.

If length requirements are satisfied, we must then look into the area requirements. This requirement is same as that for straight bars shown in fig.15.55. Based on that fig., we can prepare another fig. for a beam having a bent-up bar as shown in the fig.15.71 below:

Fig.15.71

Area requirements when bent up bar is used

It should be noted that when the bent-up bar makes contribution as both bottom bar and top bar as in fig.15.71 above, the development length requirements for the sagging portion should also be satisfied according to fig.15.65 shown earlier in this section. But it can be seen that this requirement will be naturally satisfied if the required Ld is provided within the column as shown in the fig.15.65.

It should also be noted that if the ‘ideal shape’ cannot be obtained for the bent-up bar, it’s presence should be ignored from the point of view of top bars. This situation is shown in the fig.15.72 below:

Fig.15.72

Bent-up bar with insufficient length at top

In the fig., it can be seen that the bent-up bar is not included in calculating A_st1. Extra ‘straight bar’ is provided to obtain this area of A_st1.

So we have completed the discussion on case (iii) when bent-up bars are considered.  In the next section, we will discuss about (iv) and (v).

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