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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Chapter 9 (cont..8)

In the previous section we completed the analysis of an L-beam section. In all the solved examples that we saw so far in this chapter, the effective width b_f of the sections were already given. We will now see an example in which we have to calculate b_f ourselves. The following presentation shows the basic details of a structure. Using those details, we can calculate the b_f of the T beams in that structure.

Solved example 9.4:



Assume f_ck = 20  N/mm² and Fe 415 steel.

Solution:
We will first write the following data:
D_f =110 mm, D =510 mm, b_w =250 mm, d =464 mm, A_st =603.19 mm² (3-16Ф).

Clear span of beam = cs =3500 mm
Widths of supports = ws =230 mm
c/c spacing of beam s₁,s₂ =4250 mm

Now we can start the analysis procedure:
Our first task is to calculate the effective flange width b_f. For this we can use Eq.9.5 

b_f is the lesser of (a) and (b). But we have to calculate l₀ first. For simply supported beams, l₀ is the effective span.

Effective span is the lesser of the following:
(1) Clear span + effective depth = 3500 + 464 = 3964 mm
(2) c/c distance between supports = 3500 + 230/2 + 230/2 =3730 mm

So l₀ = effective span =3730 mm. Substituting all the values in Eq.9.5 we get: 

(a) =1531.67 ; (b) =4250 mm. b_f = Lesser of the two = 1531.67 ≈ 1532 mm
So our aim is to analyse the beam section shown in the fig. below:

Fig.9.31
Section of T-beam

  
For Fe 415 steel, x_u,max / d = 0.4791 (Table 3.4)
So x_u,max for our beam section = 0.4791 x 464 =222.30 mm - - - - (1) 

We know that the given beam section may fall into any of the three cases shown in fig.9.17. To find x_u and M_uR, we first need to know the particular case into which our beam will fall. For this, we have to do some tests. 

The first test will be to check whether the beam falls into the category of 'Case 1: D_f ≥ x_u'. For doing this test, we have to make the following assumption:

• x_u has the maximum value possible within case 1. 

This maximum possible value is equal to D_f. So we put x_u = D_f =110 mm

We need to get a good understanding about this assumption. We can see a '≥' symbol in the name of this case 1. This means that x_u can have a value less than or equal to D_f. The maximum value possible is D_f. In the above assumption, we are giving this maximum possible value to x_u. In this situation, the stress distribution will be as shown in the fig.9.32 below:

Fig.9.32
Stress distribution when x_u = D_f

In the above fig., x_u = 110 mm. So the NA coincides with the bottom edge of the flange.

• Also, when we assume the above value for x_u, it automatically implies that: 'we are assuming the section to be under reinforced'. This is because, the value of 110 mm for x_u is less than x_u,max which is obtained in (1)

If the above two assumptions that we make are true, at ultimate state, the compressive force in concrete will be equal to

C_u = 0.362 f_ck b_f x_u = 1220084.8 N

Note that the above equation is obtained by changing b to b_f in Eq.3.7  

And the tensile force in steel (steel would have yielded at ultimate state if it is an under reinforced section) will be equal to

T_u = 0.87f_y A_st = 217781.75 N 

From the above results, we find that C_u is larger than T_u. This is not allowable. So we want a decreased value for C_u. Let us assess this situation:

• If we want a decreased value of C_u, then concrete area should be decreased.
• This will imply that the position of NA should be raised, which is same as saying: value of x_u should decrease.
• We have given the maximum possible value of x_u in Case 1.
• If we now decrease x_u, then, Point 'A' in the above fig.9.32 will remain inside the flange.
• This will imply that our beam belongs to case 1.

• Also, the value of 110 mm for x_u is already lesser than x_u,max. We are now going to decrease it further. So the beam section will indeed be under reinforced.

Thus the 'tests' are complete. We can proceed to do the analysis. We know that 'Case 1' is same as a beam of rectangular cross section. We have earlier seen a solved example demonstrating the calculation of xu of an under reinforced rectangular section here. Let us now follow the same procedure and calculate the x_u of our present beam:  

x_u is given by 

= 19.635 mm < x_u,max. Hence OK

Note that the above equation is obtained by replacing b by bf in Eq.3.17

Now we have to calculate M_uR. This is given by:
M_uR = T_u x z. 
Where T_u = 0.87f_y x A_st = 217781.75 mm
z = d - 0.416x_u = 464 - 0.416 x 19.64 = 455.83 mm

Thus we get M_uR = 99271454.87 Nmm = 99.27 kNm

This completes the analysis of the given T-beam. In the next section, we will see the limiting moment of resistance of T-beams and L-beams.

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