Chapter 15.9 - Requirements at Simple supports and Continuous supports

In the previous section we saw the development length required at the inflection points in continuous beams. In this section we will see some code requirements that has to be satisfied at simple supports and at inflection points.

We know that in a simply supported beam, even after curtailment, some of the bottom bars will surely continue into the supports. In fact, it is compulsory to continue some bars and embed a portion of these continuing bars into the supports. Now we will see the details of the portion that we have to compulsorily embed inside the support. Details about this are given in the cl.26.2.3.3(a) of the code. 

This clause gives requirements about three parameters. They are : 
(i) Length of the embedment. 
(ii) quantity of the bars which are embedded. 
(iii) Position of the bars which are embedded. 

We will see each of these parameters in detail:
(i) Length of embedment: Each of the continuing bars that get embedded inside the support should have an embedment length not less than L_d ⁄ 3 . If it is not possible to obtain this much embedment, bends can be provided.

(ii) Quantity of embedment: The continuing bars which get embedded inside the support of a simply supported beam should have a total area of A_st ⁄ 3 . Where A_st is the steel provided at the midspan to resist the maximum sagging moment. 

An example demonstrating this requirement about 'quantity' is given below:
• Let there be 3- 16# and 2- 12# at midspan (of a simply supported beam) giving an area of 829.38 mm². 
• One third of this is equal to 276.46 mm²
• Let 1 -16# and 2 -12# be curtailed at various suitable points. So that only 2 -16# goes into the support. 
• Area of these continuing bars = 402.12 mm². This is greater than 276.46 mm². Hence OK.

(iii) Position of continuing bars: The bars that we intend to embed in the support, 'should extend along the same face into the support'. This can be best explained by the example of bent-up bars. We know that in some cases, one or two bars are 'bent up' near the support region. At the midspan region, these bars are at the bottom face of the beam. After the point of bent, they deviate from the bottom face, and move towards the top face of the beam. So they will finally become top bars at the support, and will get embedded at the top face in the support. Such bars may satisfy (i) and (ii) above. But they do not satisfy (iii). So the bars that are compulsorily required to be embedded must not be bent up. In fact, they must not take any deviation, and 'must continue along the same face into the support'.

The figs. below shows the above requirements for a simply supported beam.

Fig.15.42
Bars having simple extension

Fig.15.43
Bars having a standard 90 degree bend

In the above fig., the anchorage value from B to D is 8Φ. So the total embedded length = A’B + 8Φ. This length should be greater than or equal to L_d ⁄ 3  

Fig.15.44
Bars having an extension beyond the bend

In the above fig., the anchorage value from B to D is 8Φ. So the total embedded length = A’B + 8Φ + DE. This length should be greater than or equal to L_d ⁄ 3 

---

Earlier we saw the application of the expression:
L_d (unique value) ≤ M_uR ⁄ V_u + L₀
The final arrangement of bars should be in such a way that, each bottom bar at the simple support satisfies the above expression. 

Once this is checked, we must proceed to check the code requirements given in cl.26.2.3.3(a). That is., we must compare the final layout of bars with the fig.15.42, 43 or 44, which ever is applicable. 

• If it is found that the embedded length is less than L_d ⁄ 3 then, more length should be provided in such a way that each bar has an embedded length greater than or equal to L_d ⁄ 3. 
    ♦ If the length provided is found to be greater than that indicated in the figs., no reduction should be made.

• If it is found that the embedded area is less than A_st ⁄ 3, then more bars should be brought into the support to increase the area. 
    ♦ If the area provided is found to be greater than that indicated in the figs., no reduction should be made.

• Finally, the position of bars should also be checked to ensure that 'the bars continue along the same face into the support'.

---

The clause gives the requirements of the three parameters related to continuous beams also. These can be summarized as follows:

(i) Length of embedment : For continuous beams, this is same as that of simply supported members, which is L_d ⁄ 3

(ii) Quantity of embedment: For continuous beams, this is equal to A_st ⁄ 4

(iii) Position of continuing bars: For continuous beams, this requirement is same as that of simply supported members. That is., 'bars must extend along the same face'.

For continuous beams, the bars will not require hooks or  90 degree bends, as enough space is available to extend the bars. So the above requirements can be shown in a single fig. as given below:

Fig.15.45
Extension required for bottom bars of continuous beams

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Earlier we saw the application of the expression:
L_d (unique value) ≤ M_uR ⁄ V_u + L₀
The final arrangement of bars should be in such a way that, each bottom bar at the continuous support satisfies the above expression. 

Once this is checked, we must proceed to check the code requirements given in cl.26.2.3.3(a). That is., we must compare the final layout of bars with the fig.15.45.

• If it is found that the embedded length is less than L_d ⁄ 3 then, more length should be provided in such a way that each bar has an embedded length greater than or equal to L_d ⁄ 3. 
    ♦ If the length provided is found to be greater than that indicated in the figs., no reduction should be made. 

• If it is found that the embedded area is less than A_st ⁄ 4, then more bars should be brought into the support to increase the area. 

    ♦ If the area provided is found to be greater than that indicated in the figs., no reduction should be made.

• Finally, the position of bars should also be checked to ensure that 'the bars continue along the same face into the support'.

---

So we have discussed the sub clause (a) of 26.2.3.3. It is related to the bottom bars at simple supports and continuous supports. In the next section, we will discuss the sub clause (b) which is related to the bottom bars at various supports in a lateral load resisting frame.

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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 10 (cont..1) - Deflection control and preliminary dimensions of flanged beams

In the previous section, we have seen the conditions for ensuring integral action between the slab and the beam. We also saw different types of flanged beams and their performances. Now we will see the basic requirements that a newly designed 'flanged beam' should satisfy.

The basic concepts of the process of design of singly reinforced rectangular sections were discussed earlier. There, in the introduction part, we discussed about:
1.concrete cover 
2.Minimum distance to be provided between the bars of beams
3.Maximum spacing allowable between bars of beams
4.Beams with overall depth greater than 750 mm
5.Minimum area of flexural reinforcements in beams
6.Maximum allowable area of flexural reinforcements in beams
7.Deflection control of singly reinforced beams and
8.Guide lines for fixing up the dimensions of beams


Links to each of the above items can be seen here.

All the above eight topics are applicable to Flanged beams also. However, some minor modifications have to be made to some of them. We will now look at the required modifications:

No.5: Minimum area of flexural reinforcements in flanged beams

We have seen that the minimum area of flexural reinforcement required in the case of a rectangular section is given by Eq.4.1

In the case of flanged sections, the width b should be replaced by b_w, the width of the web. So we get

Eq.10.1

No.6: Maximum allowable area of flexural reinforcements in flanged beams

Similar to the above,  the width b should be replaced by b_w, the width of the web. So we get

10.2
A_st ≯ 0.04b_wD 

No.7: Deflection control of singly reinforced flanged beams

We have seen the modification factors that have to be applied to the basic l/d ratio. There we mentioned that the reduction factor k_f will be discussed when we take up the design of flanged sections. In fact, we have to do two things: 
• we have to see some modifications to the factor k_t and 
• we have to learn about the new factor k_f

So we now look at cl 23.2.1(e) of the code. When we do the deflection calculations for a singly reinforced 'flanged beam', we have to determine the values of three quantities: Modification factor α, Modification factor k_t, and the reduction factor  k_f. The final equation can be written as:

(l/d)_actual  ≤  [(l/d)_basic] α k_t k_f 

Modification factor α can be determined by the same procedure as for a rectangular section. It may be noted that we do not need to calculate α if the span is less than 10 m, because then it's value is equal to 1.

Let us first see k_t: To calculate k_t, we have to first calculate the percentage of tension reinforcement by using the equation:

Where A_st,p is the actual area of steel provided

But in our present case of flanged beams , the effective width of flange, b_f has to be used instead of b . So the equation becomes:

The rest of the procedure for calculating k_t  is same as that for a rectangular beam.

Now we come to the factor  k_f.  For it's calculation, first we calculate the ratio of 'web width' to 'flange width'. That is, the ratio: b_w / b_f  . Then we use this ratio to obtain k_f from the fig 6 of the code.

Thus we can write the deflection control equations in the final form:

10.3
For singly reinforced flanged beams with span less than 10m,

(l/d)_actual  ≤  [(l/d)_basic] k_t k_f

10.4
For singly reinforced flanged beams with span greater than 10m,

(l/d)_actual  ≤  [(l/d)_basic] α k_t k_f

If the beam is a cantilever with span greater than 10 m, actual deflection calculations should be made.

So we have Seen the code procedure for deflection control of a flanged beam. But this method has been found to give results that are not normally expected. So it is recommended that for calculations regarding deflection control of a flanged beam, the over hanging portions should be ignored, and it should be considered as a rectangular section of width b_w and effective depth d . That is., when we have to do the deflection control calculations of a singly reinforced flanged section, we must consider it as a rectangular section of width b_w, and effective depth d, and then do the calculations.

No.8: Guide lines for fixing up the dimensions of beams

Fig.10.11 below shows the view of a part of a structure. It shows a slab supported over some beams and masonry walls. If the slab is cast integrally with the beams, and the condition of the arrangement of slab bars mentioned in the previous section is satisfied, they can be considered as T-beams. As these T-beams are supported over a number of masonry walls, they are 'continuous T-beams.'

Fig.10.7
3D View of Continuous T-beams

Let us consider any one of these continuous T-beams. We can analyse it and draw a bending moment diagram. Fig 10.8 below shows a portion of a typical bending moment diagram of a continuous system. It is shown here to get an understanding about the 'nature of bending moments' in continuous systems. And also to see a comparison between the 'values' of the bending moments. A video about continuous systems can be seen here.  

Fig.10.8
Portion of a Bending moment diagram

We can see that at the continuous supports, the Bending moment has values of 19.3 and 18. But at mid span, it has a lower value of 12.9. In this way, the bending moment at continuous supports will normally be higher than the bending moments at mid spans. So the higher bending moments at these supports should be used to fix up the dimensions of the beams.

However, each structure should be examined and analysed carefully to see if this general rule is true for that structure.

It should also be noted that the bending moment at supports are hogging moments. So the beam will experience tension at the fibres above the neutral axis. So the flange, which is at the top portion of the beam will be under tension, and the concrete in the flange will have cracked. So the flange cannot be considered in the design. Thus, at the supports, the beam section should be designed as a rectangular section. Towards the midspan, the moment becomes sagging type. Here the beam will act as a proper flanged beam. But the width of the web b_w and effective depth d are already fixed by considering the moments at the supports. So the only calculation to be done here is that of the tension steel. But the effective width b_f of the flange has to be calculated prior to the calculation of steel. So in the case of continuous T-beams, the preliminary dimensions are fixed by the methods that we discussed for 'rectangular sections'. And these methods should be applied at the supports. We can say that there are no new methods to be learnt for fixing up the preliminary dimensions of a 'continuous T-beam'.

But this is not the case of 'simply supported T-beams'. Here we do have to learn some new methods. Especially for fixing up the 'preliminary depth'. Fig.10.13 below shows the view of some simply supported T-beams. (The slab above the beams is not shown in the fig. for clarity)

Fig.10.13
3D view of Simply supported T-beams

In this case, the bending moment at supports is equal to zero. Towards the mid span, there will be sagging moment in the beam, and the flange will take up compressive force. So it will act as a proper T-beam. The traditional method for fixing up the preliminary width is same as that for a rectangular section.  We saw those details here. Based on that discussion, b_w is selected from general values such as  200, 230, 250, 300 etc.,

Now we come to the depth. We have earlier seen (in the discussion on rectangular beams) that, the strength (to resist bending) of a beam can be increased by increasing  either it's depth or width. So a beam having a large width will have considerable strength. A T-beam or L-beam will indeed be having a larger width, on account of their flange. This will mean that these beams can withstand the bending moments even if they have smaller depths. But in reality, we must not provide smaller depths under any circumstances. This is because, the increased strength obtained from the increased width may be sufficient to resist bending. But the increased strength obtained in this way will not help the beam to resist 'Shear'. Also it will not help the beam to 'control deflection'. That is., even if smaller depth is sufficient to resist 'bending', we do have to provide sufficient depth to resist other effects like 'shear', 'torsion' etc., Also sufficient depth is required to control 'deflection'. 

So the method that we will use to fix up the preliminary depth should take into account the above points. The range of 13 to 16 for l/D ratio, gives satisfactory results for flanged sections. From the value of D, we usually deduct 50 mm to obtain d. We know that the quantity of '50 mm' is obtained by simple calculations on some assumed values: [Concrete cover 30 mm] + [dia. of links 10 mm] + [half the dia. of main bars = half of 20 = 10 mm] = 50 mm. Here the main bars of 20 mm dia. are assumed to be placed in a single layer. But when heavy loads are acting, a large quantity of bars will have to be provided. Such large quantities can be accomodated only if we arrange the bars in 'layers'.

When the bars are arranged in layers, the above quantity of '50 mm' will change significantly. Consider the scenario:
• We start off by simply deducting 50 mm
• The beam actually have a much different 'd' from that calculated using 50 mm
Then the subsequent calculations will not give correct results. This is particularly important for flanged beams because, as we will soon see, there are a lot of calculations involving 'd'. So it is important to get a fairly good approximation of 'd' in the initial stage itself. In the next section we will see how this can be done.


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Chapter 8 (cont..7) curtailment of bars in continuous beams

In the previous section we completed the arrangement of bars in the continuous beam, and did the required checks. In this section we will see the curtailment details.

Curtailment of bars
As pointed out earlier, a detailed discussion about 'Development length and curtailment' can be seen here. For our present case, we will be using the recommendations given in SP16. It must be noted that, to use these recommendations, the analysis of the continuous member should be done using the 'method of coefficients'. We have indeed used it in the analysis of our slab, and the results thus obtained were used in the design.

We have already seen that the BM progressively decreases while we move away from a particular section ( support section as well as midspan section). So the steel can also be decreased at greater distances away from the concerned sections.

We will first see the curtailment of top bars at supports. We will take support B as an example. The fig.8.25 given below shows the details.

Fig.8.25
Curtailment of top bars

In the above fig., we can see three types of bars. The top most bars travel uninterrupted. The bottom most layer is curtailed at a distance of 0.15l (from the face of the support) on either sides of the support.  So they have a length of 0.15l₁ + 0.15l₂ + width of the support. This is the most important portion. Just the availability of a length of '0.15l₁ + 0.15l₂ + width of the support' is not good enough. We must ensure that 0.15 times the respective spans is available on both sides. All the steel (shown in the fig. as A_st) required to resist the hogging moment at this support must be completely present within this distance. 

Beyond this 0.15l distance, this full capacity is not required. So the bottom most layer is curtailed at this point. The remaining bars continue their journey. The total area of cross section of these continuing bars should be greater than or equal to 60% of A_st.  

The second stage of curtailment is applied to these continuing bars. Thus we see that the middle layer is curtailed at a distance of 0.25l (from the face of the support) on either sides of the support. So they have a length of 0.25l₁ + 0.25l₂ + width of the support. As in the case of the bottom most layer, the availability of a length of '0.25l₁ + 0.25l₂ + width of the support' is not good enough. We must ensure that 0.25 times the respective spans is available on both sides. After this curtailment, the remaining bars will and should continue their journey uninterrupted. And the total area of cross section of these continuing bars should be greater than or equal to 20% of A_st. 

The above fig.8.25 is applicable where a large quantity of steel is provided at the support. In such cases, the steel will be provided in different layers as in the fig. We can progressively curtail the various layers, while ensuring that the area requirements and length requirements are satisfied.

In our problem, there is only one layer. And there are only three bars (two at the sides and one in the middle) in that layer. Among these three, the two bars at the sides should continue uninterrupted from one end support to the other end support. This is for acting as 'stirrup suspenders'. But we can curtail the middle bar. Because, the two remaining bar will give an area greater than 0.2A_st. This is shown in the steps below:
• A_st (3-#16) = 603.19 mm²
• 2-#16 gives 402.12 mm²
• This is greater than 0.2A_st (0.2 x 603.19 =120.64 mm²)
So the curtailment at support B in our beam is as shown in fig.8.22 of the previous section.

Next we will see the curtailment of bottom bars in the end span AB. Fig.8.26 below gives the details:

Fig.8.26
Bottom bars in span AB

We can see two types of bars. The green colored bars are curtailed at a distance of 0.1l₁ from the end support and 0.15l₁ from the intermediate support. Both these distances are measured from the center lines of the respective supports. So there is a distance of l₁ -0.1l₁ -0.15l₁ =0.75l₁ at the midspan region. This is the 'important distance' as far as the sagging moment in the left end span of a continuous beam is concerned. (In the right end span, a mirror image is applicable). Within this distance, no curtailment is allowed. All bars required to resist the sagging moment should be completely present within this distance. 

After curtailment at the above mentioned points, the continuing bars should compulsorily extend into the supports on either sides. Also, these continuing bars should have an area greater than or equal to 30% of the steel at midspan.

In our problem, there is only one layer. And there are only three bars (two at the sides and one in the middle) in that layer. Among these three, the two bars at the sides should continue uninterrupted from one end support to the other end support. This is for holding the stirrups. But we can curtail the middle bar. Because, the two remaining bar will give an area greater than 0.3A_st. This is shown in the steps below:
• A_st (3-#16) = 603.19 mm²
• 2-#16 gives 402.12 mm²
• This is greater than 0.3A_st (0.3 x 603.19 =180.96 mm²)
So the curtailment in midspan AB in our beam is as shown in fig.8.21 of the previous section.

Next we will see the curtailment of bottom bars in the intermediate span BC. Fig.8.27 below gives the details:

Fig.8.27
Bottom bars in span BC

This is similar to the curtailment in an end span. The only difference is that the point of curtailment is at a distance of 0.15l from both supports. All other details are the same.

In our problem, there are only 2 bars in the span BC. These bars are compulsorily required through out the length of ABCDE, for holding the stirrups. So the problem of curtailment of any bars does not arise in this span of our problem.

This completes the discussion about curtailment of bars in continuous beams when coefficients are used for calculating BM and SF. With this we have completed the discussion on the design of continuous slabs and beams. In the next chapter we will discuss about the analysis and design of flanged sections.

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