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 

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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 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'.

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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 15.7 - Development length requirements at simple supports

In the previous section we saw the least possible length (on which bond stress will be exerted to prevent pulling out) that is available at a simple support. This length is a portion of that 'length of the bar which is coming between the points of zero moments'. If any length is available beyond these points of zero moments, then concrete will exert grip on that length also. So that length will also assist in preventing the pull out.

Let us denote the extension beyond the point of zero moment as L₀. This L₀ can be given in a number of different ways. We will now see each of these methods in detail.

Fig.15.34

Simple extension

In the above fig.15.34, the bar is given a simple extension (that is., with out any bends or hooks) beyond the point of zero moment. But the bar cannot continue indefinitely. It can be continued upto a maximum point where the concrete cover C_c at the end becomes just sufficient. C_c is specified by the code. Details can be seen here. 

In many of the worked out examples that we did until now, 'moderate' exposure conditions were assumed, and so C_c was taken as 30 mm for the bottom bars. This much cover is to be given from the outer surface of the stirrups. So the main bars get even more cover. But in our present case, the cover is taken from the end surface of the beam. Diameter of stirrups does not come into the picture. So we can assume 30 mm from the end of the bar to the end surface of the beam. However, it is important to consider the exposure conditions and other code requirements while fixing up C_c for each design problem. 

From the fig.15.34, we can easily calculate L₀ if the width of support and C_c are known. 

In some cases, the CL of the support may not coincide with the Point of zero moment. This happens when the effective span is different from the c/c distance between the supports. In such cases necessary adjustments should be made in the calculation of L₀.

Now we consider the case when the bar is given a standard 90^o bend at the support as shown in the fig.15.35 below:

Fig.15.35

Extension with a standard 90^o bend

We can see that a small portion of the 'original straight portion' of the bar extends beyond the CL of the support. This is denoted as BA' in the fig.15.36 below. In an actual beam, depending upon the width of the support and C_c, there may or may not be such a small portion. In the above fig.15.35, we have to calculate the length of this portion. 

Before that, let us see the other features of this arrangement: We can see that the bend portion begins at the end of the straight portion. We can also see that after the bend portion, a straight length of 4Φ is given to make it a standard 90^o bend. We know that the radius of the bend is 4Φ. So the calculations can be done based on the fig.15.36 given below:

Fig.15.36

Calculation of L₀

From the fig., we can see that

• C_c + Φ + r + BA’ = half of the width of support = ws/2

• But r = 4Φ. So we can write

• C_c + 5Φ + BA’ = ws/2. From this we will get BA’.

• Now, the anchorage value of the bend from B to D is 8Φ.

• So the value of L₀  = BA’ + 8Φ

Finally we consider the case when an extension is given to the bar even beyond point D in the above fig. This is shown in the fig.15.37 below:

Fig.15.37

Extension beyond point D

In this case, the calculation of BA’ is same as the previous one. Also, the value from B to D is 8Φ.

• So we get  L₀  = BA’ + 8Φ + DE

So now we know how to calculate L₀. This length L₀ will also help to prevent the pull out. So the total length available to us is equal to M_uR ⁄ V_u , + L₀, and this much available length should be greater than or equal to L_d. So we can write:

15.6

L_d (unique value) ≤ M_uR ⁄ V_u + L₀

This is the same expression given in the cl.26.2.3.3 (c) of the code. In the code, the ultimate moment of resistance M_uR is denoted as M1, and the factored shear force V_u is denoted as V.

Next, we will see a modification that has to be made to the above expression, when the simply supported end of the beam satisfies a 'special condition'. In the topic on design for shear, we discussed about the critical sections for shear design. There we saw that, when a 'transverse compression' is present at a support, the shear strength of the beam is increased near the support region. Figs. 13.62 to 13.65 of that section were used to illustrate the support conditions were such compression will occur. The same is applicable for our present discussion also. That is., when such a compression is available, the ends of the bars of the beam which are extended into the support will experience a confining effect. So a greater force will be necessary to pull the bars out. This is in effect, having a longer length of bars to resist the pull out. So the code allows us to increase MuR ⁄ Vu by 30 per cent. Thus, the expression that we can use when the ends of bars are under transverse compression is:

15.7:

In the next section, we will discuss about the development length requirements of the bottom bars of continuous beams.

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Chapter 14 (cont..3) - Bends and Hooks for bars in compression

In the previous section we saw how to improve the anchorage by using bends and hooks. Now we will see some practical situations where bends and hooks can be used.

Here we will discuss about the method of giving the required L_d for the top bars of a cantilever which is projecting from a column. Consider a cantilever beam shown in fig.14.14 below:

Fig.14.14

Cantilever beam projecting from a column

The top bars of the cantilever should be given the required L_d within the column. The space into which the bar can be extended horizontally is limited because of the limited dimensions of the column. So we can bend the top bars and extend them vertically downwards into the column. The embedded length should be equal to the required L_d as shown in the fig.

As we are using the ‘Limit state method’ for designing the various members, we will be considering the loads at the ‘ultimate state’ ie., the load at the state of impending failure. At this state, the stress in steel will be equal to 0.87fy at the critical section. So it means that it is the ‘unique value’ of L_d that we discussed earlier, which has to be provided.

Another point should be considered while giving such an embedment. We can see that the required L_d consists of a horizontal part and a vertical part. The horizontal part should be given the maximum possible length. By doing this, the full cross sectional strength of the column will be mobilized in resisting the load coming from the cantilever. Thus the column will deflect to a lesser extent. Such an arrangement also gives a greater vertical support (from the concrete in the column) for the bars of the cantilever . So we must avoid the arrangement shown in the fig.14.15 given below:

Fig.14.15

Insufficient horizontal extension

In the figs.14.14 and 14.15 above, a section named as 'critical section' is shown. The significance of critical section can be explained as follows: When we design structural members like beams, slabs etc., we provide steel to take up the stresses induced in the member. The steel resist the external loads by developing stresses within it. We must ensure that these stresses will develop in the steel when the external loads are applied on the member. If there is any slip or displacement for the steel, the required stresses will not develop in it. So we check for the forces that causes such slips and displacements at certain sections called 'critical sections'. And we must ensure that all precautions are taken to prevent any slips at these sections. Such sections are taken at the following points:

• Points of maximum stress. The critical section shown in fig.14.14 and 14.15 are taken at such a point. Because, the maximum stress in this case will be at the face of the support.

• Points within a flexural member where  reinforcement bars are cut off or bent.

• Points of inflection

• Points at simple supports.

At the critical section, we check whether the required embedded length is provided for the bar so that the required stress will develop in it. As mentioned earlier, we will learn more about this in the topic of 'curtailment of bars'

Bends and hooks for compression reinforcement.

We have seen how to calculate the development length in compression (We did this discussion based on a doubly reinforced cantilever beam shown in fig.14.5). Just as in the case of tension bars, for compression bars also, situations can arise where we will need to provide bends or hooks. When bends and hooks are provided for the bars in compression, only their 'projected length' can be considered for the purpose of development length. This is shown in fig.14.16 below:

Fig.14.16

Development length in compression when Bends and Hooks are provided

So we have completed the discussion on Anchorage and Development length, and the use of bends and hooks. The following solved example will demonstrate their application.

Solved example 14.1

Anchorage for stirrups and ties

We have learned about stirrups in chapter 13. There we saw the shapes of various stirrups. We have seen that stirrups are given around the main bars of the beam. A 3D view of a stirrup was shown in fig.13.27. But just enclosing the main bars will not be sufficient. When the tensile force develop in the bar of the stirrup, it may open out. To prevent this from happening, the ends of the stirrup should be properly anchored, so that the concrete can exert sufficient grip at the ends of the stirrup and prevent it from opening out. The code specifies three methods of providing the required anchorage at the ends of the stirrup. We will discuss each of them now.

Method 1: using 90^o bend

In this, we use a type of bending, similar to the standard 90^o bend. We know that one end of the stirrup is at the top horizontal segment, and the other end is at the left vertical segment. The bend is given at both these ends. This is shown in the fig.14.17 below:

Fig.14.17

Anchorage for stirrups: Method 1

The difference between this and the standard 90^o bend is that the extension CD beyond the bent portion should be 8φ instead of 4φ. Thus the length of CD in the above figs. is shown as 8φ. A 3D view of the resulting final shape of the stirrup is shown in fig.14.18 below:

Fig.14.18

Resulting shape of stirrup

[In the above stirrup, the radius of the bends at the four corners seems to be very large. It is indeed very large because we have followed the exact rules for a 'standard 90^o bend', where the radius of bend should not be less than 4φ for deformed bars. In later sections we will see that, for the bends in stirrups, this radius can be reduced. We will learn the reason for this reduction when we discuss 'bearing stress'. At present we have obtained a basic understanding about the method of providing anchorage at the ends of stirrups by using the 'Method 1: using 90^o bend'.]  

We will discuss about the other two methods in the next section.

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