Chapter 15.6 - Development length required at simple support

In the previous section we were discussing the free body diagram of small segment pq in a simply supported beam. We will now continue the discussion and learn more details about dT.

dT is the difference in the tensions at the two ends of the small segment of the bar in pq. [If there are N bars of the same diameter Φ, then dT is the total of the difference in tension in all the N bars.]  This small bar segment is shown in the fig.15.32 below:

Fig.15.32

Free body diagram of the bar segment PQ

From the fig. we can see that the tension at q is greater. So pq must move towards the right. But another force is coming into play here. This force is the one due to the bond stress τ_b between steel and concrete, and it helps to keep pq in equilibrium. That is., the difference in tensions is neutralised by the bond force. We will try to calculate the magnitude of this force:

•The surface area of the bar (having diameter Φ) within the segment pq = лΦdx. 

• This area multiplied by the bond stress τ_b will give the bond force. 

• So the bond force is equal to лΦdxτ_b

• If there are N bars of the same diameter, then the total surface area = NлΦdx 

• and bond force is equal to NлΦdxτ_b 

• This force is the one which neutralizes dT. So we can write:

dT= NлΦdxτ_b . 

• In the previous section we have derived dM =dT z. So substituting for dT we get we get dM = NлΦdxτ_b z

• This can be rearranged as dM⁄dx  = NлΦτ_b z

But from the basic lessons on strength of materials, we know that dM⁄dx is the shear force V at a section XX through the small segment pq of length dx. This is shown in the fig.15.33 below. [pq has a length dx. But this dx is infinitesimal. So the load effects at section XX is same as the load effects at the position of pq]

Fig.15.33

Section XX passing through pq

So we can write:

V = NлΦτ_b z which is same as

τ_b =  V ⁄NлΦ z  - - - (4)

So we have derived the expression for the bond stress acting on the segment pq. 

We will take a small deviation in our discussion here. We are going to focus our discussion on the development length requirements of segment pq. We have earlier analysed the development length requirements of a top bar in a cantilever beam. We did this in chapter 14 on bond stress, and we derived Eq.14.6. The very same procedure can be  adopted here also. But we will write the required steps again:

Consider fig.15.29 . 
• Let  the tensile stress in the bar at section XX be σ_s.

• Then, the total tensile force in all the 'N' number of bars at section XX will be equal to (NлΦ²⁄4 ) σ_s - - - (5)

• This force tries to pull out the bars from the support. The pulling out force is resisted by the bond force developed along the length l_x. Where l_x is the length of bar to the left of the section.

• The surface area on which the bond stress acts = NлΦl_x - - -(6)
• So, if τ_b is the bond stress between steel and concrete, then the force resisting the pull out = ( NлΦl_x) τ_b - - - (7)

• Equating (5) and (7) we get

 - - - (8)

• Now we go back to the point where we took the deviation and get the expression for τ_b. From (4) we get the value of τ_b. Substituting this in (8), we get

Where A_b = NлΦ²⁄4, the total area of the bars.

- - - (9)

So we have obtained the expression for the length which will prevent the bars from being pulled out. Let us now see the changes that have to be made to the above expression at the state of impending failure, that is., at the ultimate state. At the ultimate state, the stress σs in steel will be equal to 0.87f_y. So we can write the above expression as:

- - - (10)

But A_b multiplied by 0.87f_y is equal to the ultimate force in the tensile steel (∵ Force =Area x Stress). And this force multiplied by the lever arm z is the 'ultimate moment of resistance of the section' having area of tensile steel A_b. In other words, it is the M_uR of the beam at section XX. So we can write :

At ultimate state,

 - - - (11)

So, when we calculate M_uR ⁄ V , we will get l_x, which is the length that prevents the bars from pulling out at the ultimate state. But this length should be sufficient to develop enough grip so that the pulling out can be prevented. As we are considering the pulling out at ultimate state, the length that prevents the pull out should be the unique value that we discussed in chapter 14. So the length obtained from (11) should be greater than or equal to the unique value. If it is less than that, enough grip cannot be developed, and a bond failure will occur causing a pull out.

The next point that we must note is that, while evaluating (11) to get the available length, we must get the ‘least amount of length’ that can possibly occur. [This is because, even the least possible value of l_x should be greater than or equal to the unique value of L_d]. If l_x is to be as small as possible, V in the denominator in (11) should be as large as possible. We know that the largest possible value of V is the factored shear at support. Let us denote it as V_u.

So it is M_uR ⁄ V_u that has to be determined. And, even this least possible value should be greater than or equal to L_d. This can be written as

L_d (unique value) ≤ M_uR ⁄ V_u - - - (15)

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If we do a dimensional analysis of (15), the two sides will tally as shown below:

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We must note that the expression (for calculating the least available length) has M_uR in the numerator. This become significant when curtailment is done near support. If curtailment is done near support, M_uR at that section will be low. As M_uR is in the numerator, the available length will also become low. So we must take this aspect into account while doing curtailment.

---

Thus we obtained the least length that is available. We have obtained this length by considering the BM diagram which extends from the point of zero moment in one support to the point of zero moment in the other support. So this length will be a part of the bars which falls within these points of zero moments. But if some extension of the bars is available beyond these points of zero moments, then that length of extension will also contribute towards preventing the pull out. In the next section, we will discuss about this contribution.

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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

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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.

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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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