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.15.73Case (iv): Intermediate supports in continuous beams and slabs
The following fig.15.73 shows such a support.
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 Ld. 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 Ld. It is shown in the fig. below:
Fig.15.74
Method of measuring Ld 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 Ld. 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 Ld from the face of the support towards the right side.
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
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 Ast1. 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.
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
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.
Bent up bars at the intermediate supports of a frame
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.
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.
Copyright©2016 limitstatelessons.blogspot.com - All Rights Reserved
No comments:
Post a Comment