Chapter 14.4 - Anchorage at the ends of stirrups

In the previous section we saw one method to give the required anchorage at the end of stirrups. Now we will see the other methods.

Method 2: Using 180^o bend 

In this, we use a type of bending similar to the standard U-type hook that we discussed earlier. As before, the hook is given at both the ends of the bar of the stirrup. This is shown in the fig.14.19 below:

Fig.14.19

Anchorage of stirrups: Method 2

Here, the extension DE beyond the bent portion is the same 4φ that we saw earlier in the case of the ‘standard U-type hook’. [Note that, in the Method 1, where we used standard 90^o bend at the ends of stirrups, the length of DE was 8φ.] The resulting final shape of the stirrup is shown in fig.14.20 below.

Fig.14.20

Resulting shape of stirrup

Method 3: Using 135^o bend

In this case we use a type of bending that we have not discussed before. We know that the ‘standard 90^o bend’ has an angle of 90^o, and a ‘standard U-type hook’ has an angle of 180^o. The new type which we are going to see has an angle which is the average of these two values. That is : (90 + 180) / 2 = 135^o. This is shown in the fig.14.21 below:

Fig.14.21

135^o bend in a bar

Here angle DOB = 135^o. So the portion BCD is the exact 3/8 of a ring (∵ 135/360 = 3/8) having inner radius r. As usual, we must extend it beyond D. Here, for the purpose of anchoring stirrup ends, this extension required is equal to 6φ  . Note that 6φ is the average of 8φ (Method 1) and 4φ (Method 2). Fig.14.22 shows the extension.

Fig.14.22

135^o bend for anchoring the stirrup

Note that angle EDO is 90^o. We will apply this bend to the two ends of the stirrup bar as shown below:

Fig.14.23

Anchorage of stirrups: Method 3

The resulting final shape of the stirrup is shown in fig.14.24 below:

Fig.14.24

Resulting shape of stirrup

This completes the discussion about the anchorage to be provided to the stirrups. A few important additional points are given below:

The grip is exerted by the concrete mainly on the straight portion DE coming after the bend, at the ends of the stirrup bar. In the previous section we saw the method of anchorage for stirrups by 90^o bend (Method 1). We discussed it based on fig.14.17. In this method, the straight portion is nearer to the outer surface of a concrete member. (In methods 2 and 3, DE is embedded into the mass of concrete of the beam). So, if the defect known as ‘spalling’ occurs in the concrete member, the outer cover for DE may fall off. (More details about spalling can be seen here) This will reduce the anchorage provided to the ends of the stirrup. In such a situation, if the stirrup bars undergo higher tensions, it may open out due to the non availability of concrete to keep the ends in position. So we must adopt Method 2 or Method 3 in situations where concrete cover around the stirrups is not restrained against spalling.

The next point that we have to note, is the radius of the bends at the four corners of the stirrups. In the figs. 14.17, 14.19, and 14.23, that we saw for the three methods of anchorage, the diameter of the bar of the stirrup is about 10 mm. We have seen that, the radius of the bend should not be less than 4 times the diameter. So a radius of 40 mm is given. Then the diameter of the bend will be 80 mm. This means that the bar shown in blue colour at the corner of the stirrup has a diameter of 80 mm. But bar diameters greater than 36 mm are not generally used in practice. And it is not practical to give such large diameter bends at each corner of a beam. 

The main reason for specifying a large radius for the bend is to reduce the 'bearing stress' on concrete. But in a beam, there will be a longitudinal bar at each corner of the stirrup. So, even if the radius is small, the concrete will not be subjected to much bearing stress at the corners. We will learn more details about bearing stress in a later section.

We used a large value of bending radius in the figs. of stirrups, just for the clarity of showing the bending procedure, and for showing the similarity with the standard bends. Fig.14.25 below shows the correct procedure. A 10 mm dia. bar is given a 135^o bend around a normal diameter bar. The larger radius bend is also shown along side for comparison. In the fig., the dia. of the bar of stirrup is the same in both the cases.

Fig.14.25

Comparison between small and large radius of bend

In the above fig.,

• The stirrup bar AE has the same diameter in both the cases.

• The length of the extension DE is the same in both the cases.

• Angle BOD is equal to 135^o in both the cases.

• BCD in both the cases are exact 3/8 of their corresponding rings.

• The only difference is that, the radius of the ring on the left side is small, and that on the right side is large.

3D views of the stirrups having small diameter of bend is shown in the fig. below:

In the above fig., (a) shows a stirrup with both ends bent through 90^o, (b) shows 180^o and (c) shows 135^o.

In the next section, we will discuss about the 'bearing stress' caused in concrete by the forces in the bends.

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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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