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 co lumn 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 t he bar is being pulled out from the supporting member. We have to analyse the me chanism by which this 'pulling out' is prevented.
The concrete in the supporting member exerts a grip on t he reinforcing bar. This grip is formed due to the following reasons: • When freshly placed concrete se ts, various chemical reactions are taking place in it, and many products are formed as a result of o f these reactions. These products have gum like propert ies that creates
an adhesion between concrete and steel. • Frictional force is developed between concrete and steel. This is due to t he 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 ste el and concrete. Thus preventing the
movement of the bar.
Out of the above three reasons, the last one, which is the m echanical 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 t he 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 ua. • This stress will be parallel to the axis of the bar. • It will be acting in a direction opposite to the force which trie s 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 ze ro at the end of the bar. This is shown in fig.14.3 below:
Fig.14.3
Variation of ua
Let the average of all the values of ua from zero to the maximum value be denoted as uav . 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 uav
If we multiply this bond stress uav 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 =(π ΦL)uav
Now we will see the details of the force that tries to pull the bar out. We know that t he top bar of the cantilever shown in the fig. will be in t ension, and it is this tension in the bar which tries to pull it out. • Let
be the stress in the bar.
σ s
• 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 uav 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 uav 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 τbd . 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 τbd . So we can rewrite Eq.14.3 as:
Eq.14.4
The values of τbd given by the code are shown in the table below: Table 14.1 Grade of concrete τbd (N/mm2 )
M20
M25
M30
M35
1.2
1.4
1.5
1.7
M40 and above
1.9
The following points should be noted while using the above values: • The values given in the table ar e 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 ar e used for bars in compression, 25 percent increase can be applied in addition to the 60 percent note d 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 τbd 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 τbd has to increase to keep the bar from pulling out. • But τbd 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 t he 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 Ld . 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 t he 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 = (πΦLd )τbd • The pulling force = stress in steel x Area of steel = (πΦ2 / 4) 0.87fy
Equating and rearranging, we get
Let us consider τbd . It's value will depend on the following: • Whether the bar is plain or deformed. Because for deformed bars, the values for τbd in table 14.1 can be increased by 60% • Whether the bar is in tension or compression. Because when in compression, the values for τbd in Table 14.1 can be increased by 25% • The grade of concrete. Because the values for τbd 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, τbd 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 Ld . This means that if we take any bar in a structural member, it will be having a unique constant value of Ld 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 e xert 'enough grip' to stop the bar from pushing in.
As an example, let us work out the Ld 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 τbd = 1.2 N/mm2. Also from (a), the bar is deformed so τbd = 1.2 x 1.6 = 1.92 N/mm2 Substituting the values in 14.6 we get Ld = 0.87 x 415 x 16 ⁄ 4 x 1.92 = 752.19 mm Similarly, Ld 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 Ld .
Cantilever beam projecting from a column
The top bars of the cantilever should be given the required Ld 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 ex tend them vertically downwards into the column. The embedded length should be equal to the required Ld 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 Ld 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 Ld consists of a horizontal part and a vertic al part. The horizontal part should be given the maximum possible length. By doing this, the full cross sectional strength o f the column will be mobilized in resisting the load coming from the c antilever. Thus the column will deflect to a lesser e xtent. 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 t he external loads are applied on the me mber. If there is any slip or displacement for the steel, the required stresses will not develop in it. So w e 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 1 4.15 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 1 3. 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 o f providing the required anchorage at the e nds of the stirrup. We will discuss each of them now.
Method 1: using 90o bend
In this, we use a type of bending, similar to the standard 90o 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 be low:
Fig.14.17
Anchorage for stirrups: Method 1
The difference between this and the standard 90o 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 90o bend', where the radius of bend should not be less than 4φ for deformed bars. In later sec tions we will see that, for the bends in stirrups, this radius can be reduced. We will learn the reason for this re duction 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'.]
