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.
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 90o bend, so as to obtain the required Ld. 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 La. 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 La 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 La 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 La, 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 Mu,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 Mu,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.
• 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 = 45o and the bent up is also done at an angle of α = 45o, 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 La = tr = d ⁄2.
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 La =d/2 is given]. But from the discussion that we had based on fig.15.50, development length Ld 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 Ld 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 Ld . If at an end support, enough space is not available, then a standard 90o 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 90o bend. This is shown in the fig.15.66 below:
If the point Q is a little more away from the support, it may be possible to avoid the 90o bend. This is shown in the fig.15.66 below:
Fig.15.66
Development length of bent up bar with out standard 90o 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 90o bend can be avoided at the end.
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.
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.
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.
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.
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.
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 Ld. 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 La 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 La 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 La 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 La from the theoretical cut-off point. The bar should also have a length greater than or equal to Ld 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 Ast1. Extra ‘straight bar’ is provided to obtain this area of Ast1.
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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