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)

---

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

---

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 9 (cont..9) Limiting moment of resistance of flanged sections

In the previous sections we saw a number of solved examples demonstrating the procedure of analysis of flanged beams. In this section, we will see the 'Limiting moment of resistance' of flanged sections.

We have learned about the limiting moment of resistance M_ulim of a rectangular section here. It is the maximum resisting moment (the full potential) that the beam section can offer at the ultimate state, with out being over reinforced. We have derived Eq.3.25 for calculating it. 

Eq.3.25
M_u,lim = 0.362 f_ck b x_u,max( d - 0.416 x_u,max )

To calculate M_ulim of a rectangular beam section, we do not require the area of steel. We only need the width of the beam and the effective depth. Also we need the grade of steel and grade of concrete. The grade of steel and effective depth are required to determine x_u,max from Table 3.4. This table is shown again below:

Table 3.4

Similar to the rectangular sections, for flanged sections also, we will want to calculate the M_ulim on several occassions. The data given will be: the cross sectional dimensions, the grade of steel and the grade of concrete.

First of all, how can we define the 'limiting moment of resistance of a flanged section'? We can provide an answer similar to the answer for a rectangular section: M_ulim of a flanged section is the maximum resisting moment that the flanged section can offer at the ultimate state. Of course, the section should not be over reinforced. 


In the discussion about M_ulim, we have seen (Eq.3.31) that, when the moment of resistance offered by a section at the ultimate state is exactly equal to it's M_ulim,  the depth of NA at this ultimate state is equal to x_u,max. 

Normally, for calculating a resisting moment, we need to calculate x_u. But for calculating M_ulim, we do not need to calculate it because we now know that for M_ulim, x_u = x_u,max. So our work is very much reduced. 

But there exists another problem: For a rectangular section, we can calculate M_ulim directly, just by putting x_u = x_u,max. For a flanged section also, we can put x_u = x_u,max in the equation for M_uR, and thus obtain M_ulim. But for flanged sections, we have three cases and three corresponding equations for M_uR. We have to determine which equation to use. This problem can be further detailed as follows:


Imagine that we have a flanged section. It is set up in such a way that, when it is loaded to it's ultimate state, the resistance that it will offer will be equal to M_ulim. (This set up is done by giving A_st = A_st,lim.) Also at this ultimate state, the depth of it's NA will be equal to x_u,max. This x_u,max will determine the position of NA. Where is this NA situated? In the flange or in the web?

• If it is in the flange, we can use M_uR = C_u x z Where C_u is given by Eq.3.7  

Eq.3.7
C_u  = 0.362 f_ck b x_u  and z = d - 0.416x_u

• If it is in the web, we can use either Eq.9.11 or Eq.9.17 depending upon the height of the rectangular portion of the stress block. 

Eq.9.11

Eq.9.17

We can use the appropriate equation only if we know the correct position of the NA.

So our next task is to determine the appropriate case. We will consider the cases one by one. The first case is when x_u,max lies in the flange. We will name it as Case 1_lim: D_f ≥ x_u,max

Case 1_lim: D_f ≥ x_u,max
In this case, all the concrete in the web, and some concrete below the NA in the flange is in tension. This condition is shown in the fig. below. It is similar to the case 1 that we discussed earlier.

Fig.9.34
x_u,max lies in the flange

So it will behave like a rectangular section of width b_f. The eq.3.7 given above can be used to calculate C_u and then multiply it with z. This will give M_ulim. The only changes to be done are these:
• Change x_u to x_u,max
• Change b to b_f

The following point may be noted about the 'Titles' that are given to this case and the former 'Case 1: D_f ≥ x_u' : We can see that the titles are similar. The only differences are that a subscript 'lim' has been added and x_u has been changed to x_u,max in our present case. 

We will now see the next case. We will call it as case 2_lim(1): D_f < x_u,max AND D_f ≤ 0.43x_u,max

case 2_lim(1): D_f < x_u,max AND D_f ≤ 0.43x_u,max:

This case arises when 
• The NA lies in the web (indicated by D_f < x_u,max) and 
• The depth of the rectangular portion is greater than or equal to D_f (indicated by D_f ≤ 0.43x_u,max). 
This is shown in the fig. below: It is similar to the case 2(1) that we discussed earlier 

Fig.9.35
x_u,max lies in the web

x_u,max can be readily calculated using the table 3.4 shown above. So first we check whether x_u,max > D_f. Then we check if D_f is less than or equal to 0.43x_u,max. If both these conditions are satisfied, then our beam section belongs to 'case 2_lim(1)'. We can use Eq.9.11 given above to calculate M_ulim. The only changes to be done are: 
• Change x_u to x_u,max.
• Change M_uR to M_ulim

Now we will calculate A_st,lim. This is the quantity of steel that has to be provided so that when the beam section reaches the ultimate state, the resisting moment that it will be offering is equal to the above calculated M_ulim. For calculating it, we equate the compressive force to tensile force. 

The compressive force is obtained by putting x_u = x_u,max in Eq.9.12. So we get: C_uw + C_uf = C_u = 0.362f_ck b_w x_u,max + 0.447f_ck D_f (b_f -b_w)

The tensile force = 0.87f_yA_st,lim. Equating the two, we get 

Eq.9.19

In this case also the 'Title' is similar to the earlier case 2(1): D_f < x_u AND D_f ≤ 0.43x_u. The only differences are that a subscript 'lim' has been added and x_u has been changed to x_u,max in our present case.

We will now see the next case. We will call it as case 2_lim(2): D_f < x_u,max AND D_f ≤ 0.43x_u,max

case 2_lim(1): D_f < x_u,max AND D_f  > 0.43x_u,max:

This case arises when 
• The NA lies in the web (indicated by D_f < x_u,max) and 
• The depth of the rectangular portion is less than D_f (indicated by D_f > 0.43x_u,max). 
This is shown in the fig. below: It is similar to the case 2(2) that we discussed earlier 

Fig.9.36
x_u,max lies in the web

x_u,max can be readily calculated using the table 3.4 shown above. So first we check whether x_u,max > D_f. Then we check if D_f is greater than to 0.43x_u,max. If both these conditions are satisfied, then our beam section belongs to 'case 2_lim(2)'. We can use Eq.9.17 given above to calculate Mulim. The only changes to be done are: 
• Change x_u to x_u,max.
• Change M_uR to M_ulim

Now we will calculate the A_st,lim for this case. But we do not need to do detailed derivations. Because this case is similar to the previous case. There is no change in the web portion. The only change is in the depth of the 'equivalent rectangular block' in the flange portion. D_f in the previous case is changed to y_f in the present case. So we can use the above Eq.9.19, just by changing D_f in that equation to y_f.

In this case also the 'Title' is similar to the earlier case 2(2): D_f < x_u AND D_f > 0.43x_u. The only differences are that a subscript 'lim' has been added and x_u has been changed to x_u,max in our present case.

So now we know how to calculate the M_ulim of any given flange section. We will add these three new cases to the flow chart we prepared earlier (Fig.9.17). Thus we will get:

Fig.9.37
Classification of flanged beams including limiting conditions

The new cases with subscript 'lim' are to be considered when we are given a flanged beam and asked to determine it's M_ulim. 

In the next section, we will determine the M_ulim of the flanged beam in each of the four solved examples that we did earlier in this chapter.


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