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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Chapter 14 (cont..2) - Bends and Hooks to improve anchorage

In the previous section we saw some situations that require proper provision of development lengths. Now we will see some methods to improve the anchorage.

Bends and Hooks to improve anchorage

We have seen the importance of providing sufficient development length for the bars. The provision of this extra length will require space. In many situations, there will not be any space restrictions. Some examples are:
• Bottom bars at the midspan region:
There will be enough space for providing this extra length for the bottom bars at the midspan regions of simply supported and continuous beams. These bars are at the mid span, and there indeed will be enough space on either sides.
• Top bars at interior supports:
These bars are at the interior supports, and so there indeed will be enough space on either sides.

But this may not be so for the top and bottom bars near supports. The length of the beam is coming to an end at the support. So we may not be able to extend the bar to a sufficient distance. When the end of the beam is supported on a column, the bars of the beam will be, and should be, extended into the column. But then also the dimensions of the column may be such that it is not possible to provide the required development length. In such cases Bends or Hooks are provided at the end of bars to improve the anchorage. Fig.14.10 below shows the method of providing a bend in the bar.

Fig.14.10
90^o bend at the end of a bar

AB is the straight portion of the bar. The bend starts at B and extends upto C. OB =OC =r. We can see that the bend BC is a part of a perfect circle. But it has an inner radius and an outer radius. So we can say that it is a part of a 'perfect ring'. It has an inner radius of r and an outer radius of (r + φ). Inner radius r is the one which is important to us. The center of the ring is obtained by drawing the perpendicular BO to AB such that BO = r.

The value of r should not be less than
•  4φ for deformed bars and
•  2φ for plain mild steel bars.

Angle COB is exactly 90^o. So this is a 90^o bend, and the bend is 1/4 of the ring.
Thus we have obtained the bend. But this is not a 'Standard 90^o bend' specified by the code. For that, we have to extend the bar even further from C. But this time, in a straight line. This is shown in fig.14.11 below. The amount of extension required for this extension CD is 4φ.

Fig.14.11
Standard 90^o bend

So we have obtained the standard 90^o bend. In this we have extended the original bar from B to D. How much development length does this bend give? The code (cl.26.2.2.1) says that the anchorage value (that is the development length) contributed by a bend is equal to 4 times the diameter of the bar for each 45^o bend, subject to a maximum of 16 times the diameter of the bar. That is., each 45^o in the bend will give a length of 4φ. But we cannot obtain an indefinite length just by giving more angle. The maximum development length that can be obtained in this way is 16φ. In the 90^o bend that we saw above, there are two 45 degrees. So the anchorage value is equal to 8φ. It should be noted that this contribution of 8φ  is inclusive of the extension CD. That is., we will get this 8φ, only if we provide an extension CD. So 4φ from the portion CD cannot be added to 8φ as if to get a total anchorage of 12φ. The total contribution of the portion from B to D of the ‘standard 90^o bend’ shown in fig.14.11 is equal to 8φ. However, if we provide some length beyond D, that length will be considered as an additional contribution to anchorage.

Now we will see the details about hooks. Fig.14.12 below shows the method of providing a hook in the bar.

Fig.14.12
Hook at the end of a bar

Just as in the case of a 90^o bend, here, the portion BCD is a part of a ring having inner radius r. Angle BOD is exactly 180^o. So it is a 180^o bend, and it is 1/2 of the ring. In fact, DC is an exact mirror image of BC. Thus we have the shape of a 'hook' from B to D.

But this is not a 'Standard U-type hook’ specified by the code. For that, we have to extend the bar even further from D, in a straight line. This is shown in fig.14.13 below. The amount of extension required for this extension DE is 4φ.

Fig.14.13
Standard U-type hook

So we have obtained the standard U-type hook. In this we have extended the original bar from B to E. How much development length does this hook give? As we saw earlier in the case of 90^o bend, the code (cl.26.2.2.1) says that the anchorage value (that is the development length) contributed by a bend is equal to 4 times the diameter of the bar for each 45^o bend, subject to a maximum of 16 times the diameter of the bar. In the 180^o hook that we saw above, there are four 45 degrees. So the anchorage value is equal to 16φ. It should be noted that, as in the case of a bend, this contribution of 16φ is inclusive of the extension DE. That is., we will get this 16φ, only if we provide an extension DE. So 4φ from the portion DE cannot be added to 16φ as if to get a total anchorage of 20φ. The total contribution of the portion from B to E of the ‘standard U-type hook’ shown in fig.14.13 is equal to 16φ.


In the next section we will see how a standard 90 degree bend can be used in a practical situation.

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Chapter 14 (cont..1) - Situations which demand Development length and Anchorage

In the previous section we discussed about the bond stress between concrete and steel. We saw the importance of providing the required development length. We did our discussions based on a cantilever beam. Now we will see some other situations.

Consider a doubly reinforced cantilever beam as shown in fig.14.5 below:

Fig.14.5

Load applied on a doubly reinforced cantilever beam

In the previous section, we have seen how the top steel in the above beam can be made safe from being pulled out. Now we will see the bottom steel. The bottom steel is in compression. So they will be pushed in. The concrete should have sufficient grip on the bar for preventing this from happening. For this, there should be sufficient embedded length. Here the required embedment is the L_d for compression steel. We can calculate it using the same Eq.14.6 used for tension steel. The only difference is that, if deformed bars are being used, we must increase the bond stress in table 14.1, first by 60 per cent, and then by 25 per cent.

Next we consider a situation encountered in a simply supported beam. Fig.14.6 shows the region near the left support of a simply supported beam carrying a UDL. The bending moment diagram is also shown.

Fig.14.6

Simply supported beam carrying a  UDL

The beam is reinforced with 2 types of bars – bar 'a' and bar 'b' . Bar 'a' continues uninterrupted from one support to other. But bar 'b' is curtailed at section AA. This is because the bending moment is decreasing progressively towards the support, and so all the bars provided near the midspan region need not be provided at the regions near supports. We can find the section AA beyond which bar 'b' is no longer required. We will discuss the method to determine the position of AA in a later section. At present we are more concerned about another matter: 

We cannot cut off the bar 'b' at the exact section AA as shown in the above fig. The reason is as follows: We know that bar 'b' is carrying tensile loads. That means it is being stretched. So it will have a tendency to regain it's original length, and so it will try to shorten. It will try to shorten back to it's original length. If this shortening happens, it will mean that the end of the bar is moving away from section AA, towards the mid span region of the beam. So the bar will no longer be available at AA. To prevent this from happening, we must extend the bar 'b' even beyond AA to a certain distance as shown in fig.14.7 below. This distance should be such that, it is sufficient for the concrete to exert the required grip on the bar, to prevent it from shortening back to it's original length. In other words, the bar 'b' must be given sufficient ‘anchorage’ beyond section AA.

Fig.14.7

Extension of the bar

Curtailment of bar is done for economy in the design. In addition to the above discussed points, some other aspects also have to be considered while doing bar curtailment. So we will discuss about them as a separate topic in a later section.

Next we look at a portion of a continuous beam shown in fig.14.8. Here top tensile steel is provided for the hogging moment at the support, and bottom tensile steel is provided for the sagging moment at midspan regions. (A video demonstrating the requirement of top bars at continuous supports can be seen here)

Fig.14.8

Continuous beam carrying UDL

From the bending moment diagram, we can see that the hogging moment progressively decreases on either sides of the support (Value at C decreases towards B on the left side, and towards D on the right side). This indicates that the steel provided for this moment can be curtailed at some distance away from the support. But while doing this curtailment, we must take care of the anchorage requirements beyond the section.

The same is true for the bars provided for the sagging moments also in such continuous beams. As we move from the midspan regions towards either supports, the sagging moment progressively decreases. So here also curtailment can be used for economy, provided anchorage requirements are satisfied.

Yet another point that we have to consider in the above fig. is the bar arrangement related to the points of inflection. In the fig., A,B,D and E are points of inflection. The moments change signs at these points. The bottom steel provided for the region between A and B is not required in the region from B to C, because there, the moment is changing signs from – ve to + ve. But we cannot stop the bars at exact A or B. Anchorage and other requirements like shear, prevention of cracks etc., that are specified by the code should be satisfied.

As mentioned earlier, we will discuss about them as a separate topic ‘curtailment of bars’ in a later section. At present, we have obtained an understanding that anchorage and development length requirements is one of the many important points that have to be considered while finalizing the arrangement of bars in a structural member. We will now focus our attention on the other aspects of bond and development length.

Development length of bundled bars

In the case of bundled bars, the development length required will have a higher value than individual bars. This is because, when bars are bundled together, the area of contact between each bar and the concrete will be reduced. So the grip that concrete exert on the steel will also be reduced. This can be compensated by providing a greater development length, so that more portion of steel will come in contact with the concrete. The quantity by which the length has to be increased is specified by the code:

When two bars are in a bundle: 
• First calculate the L_d for a single bar.
• Then increase this L_d by 10 per cent. That is., new 'increased L_d' = L_d x1.1
• Give this increased L_d for each of the two bars in the bundle.

When three bars are in a bundle:
• First calculate the L_d for a single bar. 
• Then increase this L_d by 20 per cent. That is., new 'increased L_d' = L_d x1.2
• Give this increased L_d for each of the three bars in the bundle.

In a similar way, when there are 4 bars in a bundle, the increase must be 33 per cent. This method of increment can be diagrammatically represented as shown in the fig. below:

Fig.14.9

Development length for bundled bars

Development length when area of steel provided is greater than area of steel required

As we have seen in the design of beams and slabs, the calculated area cannot be given as such. Bars are available only in certain diameters. We have to choose the proper combination of bar diameters and the number of bars to get the required area. And care should be taken to see that area provided is not less than the area required. This will usually result in an upward rounding. That is., the area provided will usually be greater than area required. In such situations we can provide a modified value of development length denoted as L_dm. The expression for obtaining L_dm  given in clause 25.2.1 of SP 24 is shown below:

Eq.14.7

So we first determine L_d as usual, and multiply it by the ratio (As,required / As,provided).

In the next section we will discuss about bends, hooks etc., that are given to bars for improving anchorage.

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Chapter 14 - Bond and Development length

In the previous section we completed the shear design of beams and slabs. Now we will discuss about 'Bond and development length'.

Consider the cantilever beam shown in the fig.14.1

Fig.14.1

Load applied on a cantilever beam

Fig. shows a cantilever beam. It is cantilevering from a supporting member. The supporting member may be a RCC column or a RCC wall. When a load is applied on the beam, the beam bends. When this happens, the reinforcing bar experiences a tensile force. It is as if the bar is being pulled out from the supporting member. We have to analyse the mechanism by which this 'pulling out' is prevented.

The concrete in the supporting member exerts a grip on the reinforcing bar. This grip is formed due to the following reasons:

• When freshly placed concrete sets, various chemical reactions are taking place in it, and many products are formed as a result of these reactions. These products have gum like properties that creates an adhesion between concrete and steel.

• Frictional force is developed between concrete and steel. This is due to the surface roughness of the steel bar and the gripping force that concrete exerts on steel when it shrinks.

• The deformed bars have surface protrusions that project in a direction perpendicular to the axis of the bar. This will create a mechanical interlocking between steel and concrete. Thus preventing the movement of the bar.

Out of the above three reasons, the last one, which is the mechanical interlocking, will not be available when we use plain bars with no protrusions. So we must always use deformed bars for reinforcements.

So now we know the reasons for the formation of the gripping force. We will now try to find the magnitude of this force. Concrete of the supporting member is able to exert this force only on a 'certain length' of the bar. This length on which the gripping force can be exerted, is the 'length which is embedded' in the supporting member. The supporting member will not be able to exert a grip on that portion of the bar which is inside the cantilevering portion. The embedded length is shown as L in the fig.14.1. Now, the force is exerted on this length, over the 'perimeter surface area' of the bar.

• The perimeter is equal to πΦ 

• So the perimeter surface area is given by πΦL.

This is shown in fig.14.2 below:

Fig.14.2

Area on which force is exerted

When the tensile force in the bar tries to pull it out, a stress will be developed on this surface. This stress is the result of the gripping force exerted by the concrete. This stress is called the 'anchorage bond stress', and is denoted by u_a. 

• This stress will be parallel to the axis of the bar. 

• It will be acting in a direction opposite to the force which tries to pull the bar out. 

But the magnitude of this stress will not be uniform through out the length L. It will be maximum at the edge of the supporting member, and zero at the end of the bar. This is shown in fig.14.3 below:

Fig.14.3

Variation of u_a

Let the average of all the values of u_a from zero to the maximum value be denoted as u_av. For design purposes, we assume that this average value is acting uniformly over the length L. This is shown in the next fig.14.4:

Fig.14.4

Average bond stress u_av

If we multiply this bond stress u_av by the area over which it acts, we will get the force which resist the pulling out of the bar. So we get

Eq.14.1

Force resisting the pulling out =(πΦL)u_av  

Now we will see the details of the force that tries to pull the bar out. We know that the top bar of the cantilever shown in the fig. will be in tension, and it is this tension in the bar which tries to pull it out. 

• Let  σ_s be the stress in the bar. 

• Multiplying this stress by the area of the bar will give us the tensile force in the bar. 

So tensile force in the bar = (πΦ² / 4) σ_s 

We can equate this force to the resisting force given by Eq.14.1. Thus

Eq.14.2

From this we get an expression for u_av as:

Eq.14.3

We can view the above discussion in another way. If we know the maximum value of the bond stress that concrete can carry, we will be able to calculate the minimum length L that should be embedded in concrete so that steel will not be pulled out from the concrete. The maximum value of u_av is given to us by cl.26.2.1.1 of the code. In the code, it is called the design bond stress, and is denoted as τ_bd . So when we do the calculations using the design loads (ie., factored loads), the bond stress that will be induced in concrete should not exceed τ_bd . So we can rewrite Eq.14.3 as:

Eq.14.4

---

The values of τ_bd given by the code are shown in the table below:
Table 14.1

Grade of concrete	M20	M25	M30	M35	M40 and above
τbd(N/mm2)	1.2	1.4	1.5	1.7	1.9

The following points should be noted while using the above values:

• The values given in the table are for plain bars

• For deformed bars, the values shall be increased by 60 percent

• For bars in compression, the values of bond stress for the bars in tension shall be increased by 25 percent. This means that if deformed bars are used for bars in compression, 25 percent increase can be applied in addition to the 60 percent noted above.

---

We can rearrange Eq.14.4 as

Eq.14.5

• If this much L is provided, concrete can exert sufficient grip on the bar so that it will not be pulled out. 

• If L provided is less than this, it means that τ_bd that appears in the denominator will have an increased value. In other words, when L is low, the the perimeter area available is also low. So the stress τ_bd has to increase to keep the bar from pulling out. 

• But τ_bd is the maximum value possible. Any increase in it will cause failure, and so, the bar will be pulled out.

Consider the following scenario about the beam shown in fig.14.1: 

• We design the beam for flexure, and give the required tensile steel. 

• We check the stress σ_s in steel at the working loads. We find that under working loads, this is under allowable limits. 

• We provide a certain amount of embedment for the bar into the supporting member. 

• What if this embedment is less than the required L ?

• When lower loads are applied on the beam section, it may be OK. But when the intended load is applied, the stress in the bar will reach σ_s, and the bar will be pulled out. 

So, even though the reinforcement is capable of carrying σ_s, it will never reach a point at which the stress in it becomes σ_s. It will be pulled out before that because the required grip was not provided. So the beam will not be able to carry the working loads.

So we must provide this required length so that the bar can 'develop the stress' which it is intended to carry. So this length is also called 'Development length', and is denoted by L_d. Thus from 14.5 we get

Eq.14.6

This is the same equation given in cl.26.2.1 of the code.

We have discussed the above scenario in relation to the 'working loads'. We can consider the scenario in relation to the ultimate loads also. When we consider the working loads, the stress in steel σ_s will vary. It will depend on the magnitude of the load. But at the ultimate state, the stress in steel is a constant, which is equal to 0.87f_y.

• The gripping force = perimeter surface area x design bond stress = (πΦL_d)τ_bd 
• The pulling force = stress in steel x Area of steel = (πΦ² / 4) 0.87fy 

Equating and rearranging, we get 

Let us consider τ_bd. It's value will depend on the following:
• Whether the bar is plain or deformed. Because for deformed bars, the values for τ_bd in table 14.1 can be increased by 60%
• Whether the bar is in tension or compression. Because when in compression, the values for τ_bd in Table 14.1 can be increased by 25%
• The grade of concrete. Because the values for τ_bd in table 14.1 depend on the grade of concrete.

In a design problem, the above parameters will be already confirmed:
(a) bars Plain OR Deformed
(b) bars in Tension OR Compression
(c) Grade of concrete 

If we consider a particular bar in a problem, the above three parameters will not vary. They are constants. So for that particular bar, τ_bd is a constant.

The bar under consideration will be having a constant diameter Φ. Thus if we take any particular bar in a structural member, at the ultimate state, all the parameters on the right side of Eq.14.6 above are constants. Thus it will have a constant value of L_d.

This means that if we take any bar in a structural member, it will be having a unique constant value of L_d at the ultimate state. And, this length must be compulsorily provided. If this length is not provided for a bar in tension, the supporting member will not be able to exert 'enough grip' to stop it from pulling out. If the bar is in compression, the supporting member will not be able to exert 'enough grip' to stop the bar from pushing in.

As an example, let us work out the L_d for 16mm dia. bars:
We will first assume the conditions.
(a) The 16 mm dia. bar that we are considering is deformed. (Fe 415)
(b) The bar is in tension
(c) The grade of concrete is M20
From (c) above, we get  τ_bd = 1.2 N/mm². Also from (a), the bar is deformed so τ_bd = 1.2 x 1.6 = 1.92 N/mm²
Substituting the values in 14.6 we get  L_d = 0.87 x 415 x 16 ⁄ 4 x 1.92  = 752.19 mm
Similarly, L_d of 12 mm dia. bars for the same (a), (b) and (c) above is equal to
0.87 x 415 x 16 ⁄ 4 x 1.92  = 564.14 mm

We did the above discussion based on a cantilever beam. In the next section, we will look at some other situations where we must consider the provision of required L_d. 

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