Chapter 15.8 - Development length requirements at inflection points

In the previous section we saw the development length required (at the support) for the bottom bars of a simply supported beam. Now we will see the bottom bars of a continuous beam. Fig.15.38 below is a part elevation showing an intermediate span CD of a continuous beam.

Fig.15.38

Intermediate span of a continuous beam

The fig. also shows the bending moment diagram. The points of inflection are marked as R and S. [The bending moment at the 'points of inflection' is equal to zero. So the portion between the R and S is similar to the portion between the supports of a simply supported beam] The bottom bars provided for the sagging moment between R and S will be under tension, and so will be stretched. They will be trying to contract to their original length. So we can say that, the ends of the bars will be trying to pull inwards (towards the center of the span) from R and S. 

• In the case of simply supported beams, we calculated the minimum length that is available to prevent the pull out from the support.

• In the above continuous beam, our aim is to calculate the minimum length that is available to prevent the contraction of the bottom bars between R and S. Or in other words, the 'pulling out' from R and S.

So both cases are similar. Just as we did in the case of a simply supported beam, here also we can take a small segment pq of the beam for analysis. This small segment is at a distance of l_x from the point of inflection R as shown in the fig.15.39 below:

Fig.15.39

Small segment pq of the beam

The forces on the segment pq will be exactly similar to what we saw in fig.15.30 for the simply supported beam. So the calculations will also be the same, and the available length which will resist the pull out will be derived as M_uR ⁄ V_u. 

• In the case of simply supported beam, V_u is the maximum shear force that occur between the supports. This maximum occurs at the supports.

• In the case of the above continuous beam, V_u is the maximum shear force that occurs between R and S. From the topic on 'Analysis of continuous beams', we know that this maximum occurs exactly at R or S as shown in fig.13.40 below:

Fig.15.40

Shear force diagram

Values immediately to the left of R and to the right of S in the above shear force diagram are higher. But those values are related to the hogging moment at supports. They are not related to the bottom bars. So we can ignore those higher values.

Thus the length M_uR ⁄ V_u  is a part of 'that length of the bar which is within R and S'. This is the least available length that will help to prevent the pull out. 

But the length L₀ if any, beyond R and S will also contribute to prevent the pull out. So, just as the simply supported beam, the total length available to resist the pull out is equal to M_uR ⁄ V_u, + L₀ . It must be greater than or equal to L_d (unique value). So we can write:

L_d (unique value) ≤ M_uR ⁄ V_u + L₀
This expression is same as 15.6 that we derived earlier for simply supported beams.

[It may be noted that, in the fig.15.38 above, the BM diagram shown is that for a uniform loading. There is continuity between sagging and hogging parts. But if Live loads are present, we must consider the ‘envelope’. This topic was discussed in a previous section of this chapter, with the help of fig.15.17. In such a case, the points of inflection to be taken are those marked as  p₀^s in fig.15.17] 

But this L₀ in the case of a point of inflection has an important difference from that in a simply supported beam. According to cl.26.2.3.3 (c) of the code, L₀ at a point of inflection cannot exceed the larger of the following:

(i) d

(ii) 12Φ

This requirement can be detailed as follows:
At an intermediate support, there is enough space to extend the bottom bars to a longer distance. But only a certain length (which is the larger of d or 12Φ) measured from the point of inflection will be eligible to be considered as L₀. This is shown in fig.15.41 below:

Fig.15.41

Restriction on L₀ at point of inflection

In a simply supported beam, there is no such restrictions. The bar can be extended to any distance. We can also give bends (fig.15.36) and even extension beyond bends (fig.15.37). But such measures to increase L₀ should satisfy cover requirements. So it may not be possible to extend the bars to the required length. So in effect, there are restrictions on the availability of L₀ in the cases of both simply supported beams and continuous beams.

We have to satisfy the relation:

L_d (unique value) ≤ M_uR ⁄ V_u + L₀ 

To increase the right side, we can increase L₀ . But we have seen that L₀ has upper limits. Even with the upper limiting value, it may not be possible to satisfy the relation. In such a situation, we can increase M_uR. For this, we will have to reduce the curtailments done to the bottom bars so that more bars reach the support. This method is effective even though it will result in increased costs.

The best solution that the code recommends is to decrease the left side. ie., decrease L_d . Let us see how this can be done:

We have seen the details about L_d , the unique value of development length for a bar of a particular diameter. We have:

In this equation, the diameter Φ is in the numerator. So when Φ increases, L_d also increases and vice versa. This means that, for a larger diameter bar, more length will be required to exert the necessary gripping force to keep it in position with out causing a pull out. And for a bar of lesser diameter, lesser length will be sufficient. 

We have:
L_d (unique value) ≤ M_uR ⁄ V_u + L₀
Thus by using bars of lower diameter, the left side of the above expression can be reduced. By reducing the left side, we have a better chance of satisfying the condition. Thus, if even after providing the maximum allowable value of L₀, the condition cannot be satisfied, we must reduce the diameter of the bars and check again. However, we must remember that we cannot use very low diameter bars. Diameters less than 12 mm are not generally used for bottom bars of beams.

So now we know how to ensure that 'the development length requirements are satisfied at the simple supports and at the points of inflection'. The required length should be provided for all the bars at the support or at the inflection point.

In the next section, we will discuss about the code requirements regarding the curtailment of bars.

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Chapter 15.2 - Curtailment at points of inflection

In the previous section we saw the method of finding the theoretical point of curtailment at more than one section along the length of the beam. In this section we will see the method for the top bars which resist the hogging moment at an intermediate support of a continuous beam.

In fact the method is the same: Find the M_uR obtained from the remaining bars, and put it in the equation to obtain 'x'. So we will discuss only the general scheme adopted for the top bars. This is shown in the fig.15.9 below:

Fig.15.11
Curtailment of top bars

• The first curtailment is given to the bars in the bottom most layer (when counted from the top) among the top bars. When they are curtailed, the steel that remains is A_st2. 
• The second curtailment is given to the second bottom most layer. After this curtailment, the steel that remains is A_st3. 

When we follow this pattern, greater quantities of steel are available at the regions nearer to the support. This indeed must be the case because, stresses are greater near the supports as far as the 'top bars for hogging moments' are concerned.

Another aspect to note is the position of 'points of inflection'. These are the points where the bending moment changes sign. The hogging nature of the bending moment at the support continues upto the points of inflection. So we have to provide top steel upto the point of inflection. Similarly, the sagging nature of the bending moment at the midspan region also continues upto the points of inflection. So we have to provide the bottom steel upto these points. But we can follow the principles of curtailment, and progressively reduce the quantity of steel at the sections near the points of inflection.

We will now discuss another aspect of the point of inflection. In continuous beams, the maximum bending moment at any section will depend upon the position of the loads along the spans. The DL will be present at all times, and it's position will not change. But the LL will change position. In order to find the maximum bending moment at a section, we have to place the LL in those positions that give maximum bending moment at that section, and then analyse the whole continuous beam. This topic is dealt with in 'influence lines' in Structural analysis classes. 

We had a basic discussion about it in the second section of chapter 7, 'Analysis of a continuous beam'. Let us consider the solved example of a continuous beam ABCDE given in that chapter. We do not require the bending moments or final results obtained in that example. We are interested only in the pattern of loading that we adopted for that beam. This pattern was shown in fig.7.19 and 7.20. 

Let us consider the support C, and the span CD. We want 
• the maximum hogging moment at support C and 
• the maximum sagging moment in the span CD. 
For obtaining the maximum possible hogging moment at support C, we must place the LL on the adjacent spans BC and CD, and also on alternate spans. But after filling up BC and CD, there are no alternate spans. So we will filled up BC and CD with LL, and then analysed the whole beam. The resulting BM diagram is shown below in fig.15.12:

Fig.15.12
Maximum hogging moment at support C

But this BM diagram is prepared xxx exclusively for the hogging moment at support C. That is., only the hogging moment values at the support C should be obtained from this diagram. So we will remove the unwanted parts to obtain the fig.15.13 shown below:

Fig.15.13
Diagram with unwanted values removed

Now we will look at the span CD. For maximum sagging moment in this span, the LL should be placed on CD and on alternate spans. So we placed it on CD and on the first span AB. The resulting BM diagram is shown in the fig.15.14 below:

Fig.15.14
Maximum sagging moment in span CD

But this BM diagram is prepared exclusively for the sagging moment in span CD. That is., only the sagging moment values in span CD should be obtained from this diagram. [For this particular beam ABCDE, the above loading arrangement for the maximum sagging moment in CD is same as the loading arrangement for the maximum sagging moment in AB also. But we are not considering AB in our present discussion]. So we will remove the unwanted parts to obtain the fig.15.13 shown below:

Fig.15.15
Diagram with unwanted values removed

The figs.15.15 and 15.13 gives the required values. So we can combine the two figs. as shown below:

Fig.15.16
Required values shown in a single diagram.

The above fig.15.16 gives the maximum values of the hogging moment at support C and the sagging moment in span CD. By using the same procedure, we can fill up the remaining supports and spans in the fig. Then, from that single fig., we can obtain the maximum values at any section along the length of ABCDE. Such a fig. is called the 'moment envelope'. Now let us take a closer look at the fig.15.14. This is shown in the fig.15.17 below:

Fig.15.17
Details of the moment envelope at support C and span CD and the points of inflection

We can see that the point of zero moment in the hogging moment envelope which is marked as p₀^h is different from p₀^s the point of zero moment in the sagging moment envelope. This is a situation that we often see while analysing continuous beams and continuous one-way slabs.

• The top steel bars should be provided for the hogging moment in the region shown in orange colour in the above fig.15.17. This steel should be available upto the points where the orange graph meets the horizontal axis, which are the points of inflection. In fact, as we will soon see in later sections, the bars should be continued even beyond the points of inflection.

• The bottom steel bars should be provided for the sagging moment in the region shown in yellow colour in the above fig.15.17. This steel should be available upto the points where the yellow graph meets the horizontal axis, which are the points of inflection. In this case also, we will soon see in later sections that, the bars should be continued even beyond the points of inflection.

• So the points of inflection play a major role in the final layout of bars. But these points of inflection for the sagging and hogging moments may not coincide. That is., we cannot expect a 'continuity' at the points of inflection as we saw in the fig.15.11 earlier in this section. The meeting point with the horizontal axis may be different for sagging and hogging moments, just as p₀^h and p₀^s in the fig.15.17. This happens when the different positions of LL are considered. This non-coincidence should be considered, and exact positions of various points of inflection should be determined (from the moment envelope) so as to provide a satisfactory lay out of bars.

In the next section, we will discuss another method to determine the theoretical cut-off points.

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Chapter 9 - Effective flange width of T-beams and L-beams

In the previous section we completed the design of continuous slabs and beams. All the beams that we considered until now had a rectangular cross section. In this section, we are discussing about beams with another cross section: The 'flanged beams'.

We know that the slabs rest on walls or beams. Consider the case when a slab rests on a beam. In this case, if two 'conditions' are satisfied, a portion of the slab will act as the 'flange' of the beam. That is., when the beam bends, the slab will also bend in the same direction. This is shown in the fig.9.1 below.

Fig.9.1
Slab acting as flange of a beam 

The beam will become stronger when it becomes a 'flanged beam' because there is greater concrete area in the upper portion to take up the compression. The lateral stability of the beam also increases. But in order to achieve a flanged section, the following two conditions should be satisfied: The first condition is that the slab and beam on which it rests, should be cast together. The second condition is related to the arrangement of the bars of the slab. This is explained in detail at the beginning of chapter 10 - Design of flanged sections.

T-beams and L-beams are the commonly encountered flanged beams. They are mostly seen in framed structures.  Intermediate beams which have slabs on both sides, act as T-beams. The end beams and the beams around staircase or lift openings, which have slabs only on one side, act as L-beams. In both T-beams and L-beams, a portion of the slab acts integrally with the beam. So when the beam bends, this portion of the slab also bends in the same direction as the beam. This portion of the slab is called the Flange and the portion of the beam below the flange is called the Web.


The following presentation gives the basic idea about flanged beams and the effective width of flanges:

   

An actual building under construction is shown below. The T-beams and L-beams can be clearly distinguished

Equivalent flange width

From the above presentation, we have a basic idea about the concept of 'equivalent flange width' b_f . The following properties are associated with b_f:

• It increases when the span of the beam increases

• It increases when the width of the web  increases.

• It increases when the thickness of the flange increases

• It depends on the type of loading (concentrated, distributed etc.,)

• It depends on the type of support (simply supported, continuous etc.,)

Considering the above factors, the cl 23.1.2 of the code gives the following approximate formulae for calculating b_f.

Eq.9.1: for T-beams

Eq.9.2: for L-beams

In the above formulae, we know that b_w and D_f denotes width of the web and depth of the flange respectively. But l₀ is a new term for us. l₀ is defined as 'the distance between points of zero moments in a beam'. When we consider a simply supported T- beam or L-beam, the distance between the points of zero moments in the beam is obviously equal to it's effective span. Because at the supports, the bending moment is zero for a simply supported beam or slab. But this is not the case when we consider a T-beam or L-beam which is continuous over supports, or when it is part of a framed structure. Consider the fig 9.2 below:

Fig.9.2

Portion of the BM diagram of a continuous member

The above fig. shows a portion of the Bending moment diagram drawn for a continuous beam. The moments at supports are of hogging type, taken as positive and so are marked above the X axis. The moments near midspans are of sagging type, taken as negative and so are marked below X axis. Thus there is a transition from positive to negative and then again back to positive when we move along the length of the beam. In the fig, these transitions occur at points P And Q. Each of these points is called a Point of contraflexure, and at these points bending moment is equal to zero. So for the case shown in the fig, l₀ is equal to the distance PQ.

The above method to calculate l₀ is tedious. So the code permits us to use a much simpler procedure to calculate l₀ . This is given as the note below cl. 23.1.2 (c). According to this note, l₀ may be assumed as 0.7 times the effective span in continuous beams and slabs. We have already discussed the methods for calculating the effective spans of continuous members here. 

So now we know how to calculate b_f using eq 9.1 and 9.2. But when we obtain b_f of a beam using the above formulae, there is a possibility that we get a value which is greater than what is actually available for the beam in the structure. Given below is fig 9.3 which shows the top right side portion of the structure that we discussed in the presentation. It shows the details of the T-beam P and the L-beam Q.

Fig.9.3

Maximum possible width of flange

From the fig, we can deduct the following:

Eq.9.3: The maximum share available for the T-beam P (marked with yellow colour) is

Eq.9.4: The maximum share available for the L-beam Q (marked with white colour) is

So the effective flange width calculated using 9.1 should be checked to see that it is not greater than 9.3 and effective flange width calculated using 9.2 should be checked to see that it is not greater than 9.4. 

In other words, the lesser value from the two methods of calculation should be used.

If we denote the center to center (c/c) span on either side of a T-beam as s₁ and s₂ , and the c/c span on the side of a L-beam as s₁ , we can write 9.1 and 9.2 in the general form as:

Eq.9.5: For T-beams, b_f is the lesser of the following two values:

Eq.9.6: For L- beams, b_f is the lesser of the following two values:

Isolated T-beams and L-beams

In some cases, isolated T-beams or L-beams are encountered. Fig 9.4 below shows an example of an isolated T-beam used for a foot bridge. The fig. shows the view from under side of the bridge.

Fig.9.4
Isolated T-beam in a foot bridge

From the fig. we can see that the slab above the beam which acts as the flange, is discontinuous at the sides. The stringer beam of a staircase is another example of an isolated T-beam. A part view of such a staircase is shown in fig.16.84 in the section on transverse stairs. Cl 23.1.2(c) of the code gives the following formulae for calculating for isolated T-beams and L-beams.

Eq.9.7: For isolated T-beams, take lesser of the following two values:

Eq.9.8: For isolated L-beams, take lesser of the following two values:

Where b is the actual width of the flange. Obviously, b_f cannot be greater than b.

So at this stage we are able to calculate the effective width b_f of the flange for different types of T-beams and L-beams. In the next section we will start the discussion on the analysis of these flanged sections.

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