Chapter 10 - Integral action between slab and beam for flanged beams

In the previous section we completed the analysis of flanged sections. In this section we will consider the design. At the beginning of chapter 9 we saw that two conditions should be satisfied to ensure the integral action between the slab and the beam. There we saw the first condition: The slab and the beam should be cast together. In most cases, the slab and beam are indeed cast together. The form works consists of horizontal board for the bottom side (soffit) of the beam, vertical boards for the sides of the beam, and horizontal boards for the under side of the slab. After tying the reinforcement bars and keeping them in position, fresh concrete is poured from the bottom of the beam. Concrete is poured continuously in such a way that, first the beam is filled up, and then the slab. Necessary compaction is ensured by using a needle vibrator or other such devices.

It may be noted that the stirrups of the beam also play a major role in ensuring integral action between the slab and the beam. These stirrups confine the main bars of the beam. They extend from the beam into the slab, upto the top surface of the slab. They are able to extend upto the top surface because the 'stirrup suspender bars' of the beam are near the top surface of the slab. So sufficient stirrups should be provided for the beam to ensure the integral action. We will learn about the design of stirrups when we discuss 'shear design'.  

We will now see the second condition. Consider a slab and beam system used as the roof of a room, as shown in the plan view below:

Fig.10.1
Main bars of a slab given parallel to beam does not contribute towards flange action

We can see that the main bars extend from wall AB to wall CD. This is because, in the design of this particular slab, the slab loads are assumed to be transferred to the walls AB and CD only. No loads are transferred to the other two walls BC and AD. So no loads are transferred to the beam either. Thus all the main bars of the slab are parallel to the beam. In such a case, even if the slab and beam are cast together, the slab will not make a 'contribution' as the flange of the beam.

For the slab to make the 'contribution', a specific quantity of main bars of the slab should be available in a direction perpendicular to the beam. So in the above case, we must provide some 'special bars' perpendicular to the beam. The details of this are given in the cl.23.1.1.b of the code. The requirements of this clause can be illustrated if we draw the section XX shown in the plan view above. This is shown in the fig.10.2 below:

Fig.10.2
Section XX

We can see that the length of each of the 'special bars' on either side of the beam, should not be less than l/4, where l is the span of the beam. The area of these bars should not be less than 60% of the area provided at the midspan of the slab. Also, these bars should be provided near the top surface of the slab. By providing the specified quantity and length of these 'special bars', at the correct position, we can ensure that the slab will act as the flange.

In the fig.10.1 above, distributor bars will be present in a direction perpendicular to the beam. But they are not available near the top surface of the slab through out the length of the beam. Also their quantity may not satisfy the 60% requirement. So we must provide the 'special bars' even though the distributor bars are present in the perpendicular direction to the beam.

Now consider the slab shown in the fig.10.3 below:

Fig.10.3
Main bars of a slab given perpendicular to beam will contribute towards flange action

In this case, the main bars of the slab are provided in a direction perpendicular to the beam. It is a continuous slab. We have already discussed the analysis and design of such slabs. We know that the slab will experience a hogging moment at the beam. So there will be top bars to resist the hogging moment. And these bars will be perpendicular to the slab. So we do not need any 'special bars'. The top bars that we designed at an intermediate support for a slab can be seen here. It may be noted that even if we provide bent-up bars instead of straight bars, the required effect will be obtained because, the bent-up bars will become top bars at supports. In any case, the required length of l/4, and the quantity of 60% mentioned above should be available.

So now we know how to ensure integral action between slab and beam. We must extend this discussion to isolated T-beams and L-beams also. In those cases, the stirrups of the beam should extend to opposite sides of the beam, and travel upto the edges of the slab. This is shown in the fig.10.4 below.

Fig.10.4
Ensuring integral action between slab and beam for isolated flanged beams 

A 3D view can be seen below:
Fig.10.5
3D view of bars in isolated T-beams

Fig.10.6
3D view of bars in isolated L-beams

From the above figs., we can see that the number of stirrups will be equal to the number of slab bars. The number of stirrups is calculated using shear force considerations. But the number of slab bars (obtained as spacing in 1m) is calculated from bending moment considerations. We must correlate both these and the larger number should be provided. The diameter of these bars should also be checked.

The presentation given below shows some examples of flanged beams, and the method of their performances.





From the above presentation, it is clear that each structure should be examined carefully to determine the locations where we can design a beam as a 'flanged beam'. 

We can now take up the discussion on design procedure. In the next section, we will see the basic requirements like concrete cover, minimum steel etc., 

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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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