Showing posts with label Top bars at supports. Show all posts
Showing posts with label Top bars at supports. Show all posts

Sunday, January 24, 2016

Chapter 15.11 - Curtailment of top bars at simple supports

In the previous section we saw the arrangements of top bars at various types of supports. In this section we will see the top bars at simple supports.

When the beam or slab is simply supported, the bending moment at the support is equal to zero. So theoretically there is no need to provide top steel at the supports. A similar situation arises in continuous beams and slabs also, when the end span is simply supported at the discontinuous end. But the code requires us to provide top steel in these cases also. This is because, if a partial fixity is induced at the support, there will be a negative (hogging) moment. A presentation showing an example of partial fixity is shown here. In such cases top steel will be required to resist the hogging moment.

First we will see the details of the simply supported end of a slab. We have to follow the clause D -1.6 of the code in this case. This clause gives us the requirements regarding the two parameters: (i) quantity and (ii) length.

(i) Quantity: The area to be provided for the top bars is equal to half of the area of bottom steel required at the midspan.
(ii) Length: The length of each top bar should be equal to 0.1l . Where l is the effective span.

The above requirements can be satisfied by providing bars as shown in the following figs.

Fig.15.57
Top bars at the simple supports of a slab
Top bars are required at simple supports to resist possible negative or hogging moment

Now we will see the simply supported end of a beam. For this we have to follow cl.22.5.2 of the code. This clause gives us the procedure to determine the amount of steel that has to be provided:

• First we calculate the moment Wl⁄24 . (15.8)
Where W is the total factored load on the span, and l the effective span.
• Then we determine the amount of steel required to resist this moment. This much steel has to be provided as top steel at the simple support of the beam. This is the 'quantity'.

Now we need the length for which this much steel is to be provided. We know that for a beam, there will always be two bars acting as ‘stirrup hangers’ at the top. We must provide these two bars in such a way that their total area will be greater than that required to resist the moment in 15.8. So the ‘area requirement’ will be satisfied. We also know that these bars being 'stirrup hangers', will extend from support to support. That is., they are provided for the full length of the beam. So the ‘length requirement’ if any, will also be satisfied. But we have to provide enough embedment [Ld (unique value)] for these two bars into the support. These details are shown in the fig.15.58 below:

Top bars at the simple supports of a Beam
Top bars are required at simple supports to resist any negative or hogging moment that may arise due to partial fixity.

The above fig.15.58 can be used for a single span simply supported beam, and also for the discontinuous end of a multi-span continuous beam. If this discontinuous end is simply supported.

So we have completed the discussion about top bars. That is., bars provided to resist the hogging moment. These bars are provided at the supports.

We have seen the method of curtailment of Top bars at various types of supports. It will be better to compile the discussions that we had about this topic in a compact form. The fig.15.59 below shows the five types of supports at which we discussed the top bars.

Support conditions



The descriptions about each of the above supports and the links to the figs., explaining the curtailment of top bars at that support are given below:


(i) Support of a single span simply supported member [Figs.15.57 and 15.58 shown above]

(ii) End support (which is simply supported) of a multi span continuous member [Figs.15.57 and 15.58 shown above]

(iii) End support of a multi span continuous beam which is part of a frame. [Figs.15.55 and 15.56]

(iv) Intermediate support of a continuous beam or slab [Figs.15.51 and 15.52]

(v) Intermediate support of a continuous beam which is part of a frame. [Figs.15.53 and 15.54]

Just as we did the final check for development length in the case of bottom bars using fig.15.50, we have to do the final check for top bars also. Here we will consider each of the above five cases separately.

(i) For this case, the Ld requirement to prevent the bar from pulling out from the support, is shown in the fig.15.58 itself.
(ii) For this case also fig.15.58 can be used.
(iii) For this case, the Ld requirement to prevent the bar from pulling out from the support, is shown in the Figs.15.55 and 15.56 itself.

In general, the above three cases are end supports, and so it is easy to give the required Ld at the design stage itself.

(iv) For this case, Ld requirement is shown in the fig.15.60 below:
Development length requirements for top bars of continuous beams

The details shown in the above fig. can be explained as follows:
At the support, the beam will be bending upwards due to the hogging moment. So the bars will be in tension, and they will try to 'contract' which is same as 'pulling into' the support. So all the bars should have a minimum Ld from the face of the support, and this length should be embedded into the span. This will prevent the bars from pulling into the support. 

Similarly, at the cut-off section XX, the remaining bars will be under a greater tension. So each of these remaining bars should also be given the required Ld into the span.

(v) For this case, Ld requirement can be shown just by modifying the support conditions of the previous case (iv). This is shown in the fig.15.61 below. The explanations are also the same.

Fig.15.61
Development length requirements for top bars of continuous beams in a frame



This completes the compilation for Top bars. Next we will make a similar compilation for Bottom bars:


For bottom bars also, the supports encountered are those shown in the fig.15.59 above. So we will discuss the five cases shown in that fig.

The descriptions about each of the supports and the links to the figs., explaining the curtailment of bottom bars at that support are given below:


(i) Support of a single span simply supported member [Figs.15.42, 43 and 44]

(ii) End support (which is simply supported) of a multi span continuous member. [Same as above. Figs.15.42, 43 and 44]

(iii) End support of a multi span continuous beam which is part of a frame. [Figs.15.48 and 15.49]

(iv) Intermediate support of a continuous beam or slab [Figs.15.45]

(v) Intermediate support of a continuous beam which is part of a frame. [Figs.15.47]

We have to do a final check for development length in the case of bottom bars using fig.15.50. This completes the compilation for Bottom bars. In addition to the above, the application of 15.4 and 15.6 should be considered where ever necessary for both Top and Bottom bars. The largest length obtained should be provided in the final design.


In the next section we will discuss about 'Bent-up bars'.

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Wednesday, September 2, 2015

Chapter 8 (cont..6) Arrangement of bars in continuous beam

In the previous section we formed table.8.8 which gives the number and dia. of bars that we intend to give at various supports and midspans. In this section we will see the arrangement of bars that will give these required steel at the various points. First we will take midspan AB. The following fig. gives the details:

Fig.8.21
Bottom bars at midspan AB

Bottom bars are provided at the mid span of a continuous beam to resist the sagging moment

The above fig.8.21 shows the bars required to resist the sagging moment at midspan AB. 
• Bar type: 'a' has a shorter length. It is provided in the middle of the cross section of the beam. This can be seen in section XX. There is only one bar of this type. 
• Bar type : 'b' are longer. They extend from support to support. They are provided at the sides of the cross section of the beam. This can be seen in section YY. There are two bars of this type. 

These three bars should be provided together. They work together to resist the sagging moment at midspan AB. 

We can see that the two 'b' bars extend from support to support. In fact they extend from the first end support A to the last end support E. This is because, we compulsorily require two bars at the sides (for holding the stirrups), through out the length of ABCDE. 

Thus we have successfully provided the required three bars at the midspan AB. Out of the three bars, 'a' have a shorter length. This is because all the three bars are not required at sections away from the midsection. We will learn more about this in the next section about curtailment of bars. Now we will see the top bars at supports A & B. The fig. below gives the details:

Fig.8.22
Top bars at supports A & B


In a continuous beam, top bars are provided at supports to resist hogging moment

The above fig.8.22 shows the top bars at supports A and B. 
• Bar type: 'c' has a shorter length. It is provided in the middle of the cross section of the beam. This can be seen in section XX. There is only one bar of this type. 
• Bar type : 'd' are longer. They extend from support to support. They are provided at the sides of the cross section of the beam. This can be seen in section YY. There are two bars of this type. 

At support B, these three bars should be provided together. They work together to resist the hogging moment at support B

We can see that the two 'd' bars extend from support to support. In fact they extend from the first end support A to the last end support E. This is because, we compulsorily require two bars (to act as 'stirrup suspenders') at the sides, through out the length of ABCDE. These 'stirrup suspenders' serve another purpose also: They act as top bars at support A. 

Thus we have successfully provided the required three top bars at the support B, and also the required two top bars at support A. Out of the three bars, 'c' have a shorter length. This is because all the three bars are not required at sections away from the support. We will learn more about this in the next section about curtailment of bars. Now we will see the bottom bars at midspan BC. The fig. below gives the details:

Fig.8.23
Bottom bars at midspan BC

The above fig.8.23 shows only one type of bar, which is 'b'. They are given at the sides as shown in section XX. There are two numbers of this type. Recall that, these are the same bars given at the bottom of span AB. There we said that these two bars will continue upto the other end-support E. This is for holding the stirrups properly. Thus we meet them again in our present span BC. Also note that we need only two no. bottom bars of 16 mm dia in BC. So these bars will serve two purposes: To hold the stirrups, and to resist the sagging moment. 

At this point of our discussion, we have completed the following types of supports and spans:
• An end support (support A)
• An intermediate support (support B)
• An end span (span AB)
• An intermediate span (span BC)  

Based on this we can now give a satisfactory arrangement at other spans and supports of our beam ABCDE. Because CD is an intermediate span, DE is an end span, C and D are intermediate supports and E is an end support. The fig. showing the complete details of the whole beam is given below:

Fig.8.24
Sectional elevation of the continuous beam ABCDE
Sectional elevation showing the reinforcement details of a continuous beam

So we have completed the design of the continuous beam. We will now see the various checks. The pdf file given below gives the detailed steps involved in the various checks:

Solved example 8.2 final checks

In the next section we will discuss about the curtailment of bars in a continuous beam.




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Saturday, August 29, 2015

Chapter 8 (cont..5) Design of a continuous beam

In the previous section we completed the design of a continuous slab. In this section we will see the design of a continuous beam. For this, the first step is to setup a preliminary cross sectional dimensions of the beam. This is for calculating the self wt. But for continuous slabs and beams, this is all the more important (unless we are using moment coefficients) because, only with the knowledge of the dimensions of cross section, we can do the 'detailed structural analysis'. At the beginning of this chapter, we saw the guidelines for fixing up the depth of a continuous one way slab. 

Now we want to set up the preliminary dimensions of a continuous beam. Unlike for a slab (width of a strip is taken as 1000 mm), for a beam, we want to set up both the preliminary width and depth.  We have already seen the procedure for doing this here, when we discussed about the design of beams. Based on that, we can use a range of 10 to 16 for l/D for the depth.

After fixing the preliminary dimensions, we can start the design process. The design process is same as that for a simply supported beam that we saw earlier here : First we calculated d required. Then we compared it with the d obtained from the preliminary dimensions. If the required value was less, we proceeded to find the steel required to resist the sagging BM at midspan. Here, in the case of continuous beams also, we have to do the same. But we have to do this for each span. That is., we have to calculate the steel required to resist the sagging BM at the midspan of each span. Plus, we have to do this for the hogging BM at each of the supports also.

After obtaining the steel at all midspans and supports, we arrange the bars in such a way that the various code requirements are satisfied. Then we do the final checks like area of minimum steel required, check for pt,lim, check for deflection etc., These steps will become more clear when we see an actual solved example:

Solved example 8.2:

We will do the design of the continuous beam, of which we did the structural analysis in the previous chapter 7. We have obtained the BM and SF at all the important points. We will use the results obtained by using the 'method of coefficients' (fig.7.22 and 7.23). Those results are reproduced in the table 8.6 below:
Table 8.6
BM
Span AB 55.49
Supp. B -63.69
Span BC 41.44
Supp. C -53.55
Span CD 38.89
Supp. D -61.76
Span DE 55.49
Now we can start the design. As we have discussed at the beginning of this chapter, the first step is to obtain the preliminary dimensions of the beam. But in this problem, these are already given (b =230 mm & D =400 mm). However, we will check if the depth of 400 mm satisfies the 'general rule':

We will use the largest span, which is the end span. So l = 4230 mm. D = 400 mm. Thus we get l/D = 4230/400 =10.575. This falls between 10 and 16. So the value of D is satisfactory.

Though we found it satisfactory, after the design, we must do all the necessary checks and confirm that the beam section with overall depth 400 mm is adequate for the given problem. If it is found that the section is not adequate, it should be redesigned.


The next step is to determine the steel required to resist the BM at various points. First we will do this for the midspan AB. From the table 8.6 above, we can see that the BM at midspan AB is 55.49 kNm. The pdf file given below shows the detailed steps of design:

Steel at midspan AB

So we got the steel as 3-#16 at midspan AB. The next pdf file given below shows the detailed steps of the design for steel at support B:

Steel at support B

So we got the steel as 3-#16 at Support B. In a similar way we can obtain the required spacing at other points also. These are shown in the table 8.7 below:
Table 8.7
BM d req. Ast req. dia. No. Ast pr.
Span AB 55.49 263.62 481.39 16 3 603.19
Supp. B -63.69 282.42 562.9 16 3 603.19
Span BC 41.44 228.58 349.16 16 2 402.12
Supp. C -53.55 259.81 462.61 16 3 603.19
Span CD 38.89 221.38 325.84 16 2 402.12
Supp. D -61.76 279.02 543.42 16 3 603.19
Span DE 55.49 263.62 481.39 16 3 603.19

The above table 8.2 is some what incomplete. This is because, we have not considered the end supports A and E. We know that they are simply supported ends and so the BM will be zero. We will indeed get zero as the BM at these supports when we do a detailed structural analysis of the beam ABCDE using any of the methods like Kani's method, slope deflection method, moment distribution method etc., So theoretically, steel is not required at these supports.

But we have to consider the partial fixity that may be introduced at a future stage. We learned about it when we discussed about simply supported one way slabs. And we applied cl.D-1.6 of the code in the solved example of a simply supported one way slab. But for beams, we have to apply cl.22.5.2. According to this clause, we have to design the supports A and E for a hogging moment whose magnitude is given by Wl/24. Where W is the total load, and l is the effective span. So we will first calculate this moment:

Here W is the total load. The total load is obtained as W =wl. where w is the total factored load per unit length. So the hogging moment can be obtained as wl2/24. This is equal to = 26.22kNm (w =24.94 +10.23 & l =4.23 m). With this value of the hogging moment, we have to do a complete design just as we do at other supports B, C and D. The results are: d required = 181.93mm, Ast required = 215.88 mm2. These results are applicable at support E also because spans AB and DE are exact mirror images. So the table 8.7 can be modified by adding these details:

Table 8.8
BM d req. Ast req. dia. No. Ast pr.
Supp. A -26.22 181.93 215.88 16 2 402.12
Span AB 55.49 263.62 481.39 16 3 603.19
Supp. B -63.69 282.42 562.9 16 3 603.19
Span BC 41.44 228.58 349.16 16 2 402.12
Supp. C -53.55 259.81 462.61 16 3 603.19
Span CD 38.89 221.38 325.84 16 2 402.12
Supp. D -61.76 279.02 543.42 16 3 603.19
Span DE 55.49 263.62 481.39 16 3 603.19
Supp. E -26.22 181.93 215.88 16 2 402.12

One row for 'support A' has been added at the beginning, and another row for 'support E' has been added at the end. Now the table is complete. In the next section we will see the arrangement of bars according to the numbers given in the above table.




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