Chapter 9 (cont..10)

In the previous section, we have discussed the method for calculating the M_ulim of any given flanged section. Now we will calculate the M_ulim of the sections in each of the solved examples that we did in this chapter:

Solved example 9.1:
The data for this example are:

b_f =950 mm, D_f =110 mm, D =600 mm, b_w =300 mm, d =520 mm, A_st =3694.51 mm² (6-28Ф), f_y = 250 N/mm², f_ck =20N/mm² 

Calculations:
For Fe250 steel, x_u,max / d = 0.531 (Table 3.4)
So x_u,max for our beam section = 0.531 x 520 =276.12

Here x_u,max > D_f so the section belongs to either case 2_lim(1) or case 2_lim(2)
0.43x_u,max = 0.43 x 276.12 = 118.8 > D_f. So the beam belongs to case 2_lim(1)
So we can use Eq.9.11

• Change x_u to x_u,max.
• Change M_uR to M_ulim

Substituting all the known values, we will get M_ulim = 540.3 kNm 

We can calculate A_st,lim using
Eq.9.19

Substituting all the known values, we get A_st,lim = 5697.85 mm².

Let us compare the above results with those of the solved example 9.1:
The results obtained in 9.1 were:
[x_u = 127.50 mm] < [x_u,max = 276.12 mm] Hence under reinforced
[M_uR = 379.3 kNm] < [M_ulim =540.3 kNm] as expected
[A_st = 3694.51 mm²] < [A_st,lim =5697.85 mm²] as expected.

Solved example 9.2:
The data for this example are:

b_f =1300 mm, D_f =100 mm, D =500 mm, b_w =325 mm, d =420 mm, A_st =4310.27 mm² (7-28Ф),  f_y = 250 N/mm², f_ck =20N/mm²

Calculations:
For Fe415 steel, x_u,max / d = 0.4791 (Table 3.4)
So x_u,max for our beam section = 0.4791 x 420 =201.22 mm

Here x_u,max > D_f so the section belongs to either case 2_lim(1) or case 2_lim(2)
0.43x_u,max = 0.43 x 201.22 = 86.53 < D_f. So the beam belongs to case 2_lim(2)

So we can use Eq.9.17

• Change x_u to x_u,max.
• Change M_uR to M_ulim

Substituting all the known values, we will get M_ulim = 468.2 kNm 

We can calculate A_st,lim using
Eq.9.19

• Change D_f to y_f

Substituting all the known values, we get A_st,lim = 3609.3 mm².

Let us compare the above results with those of the solved example 9.2:
The results obtained in 9.2 were:
[x_u = 260.51 mm] > [x_u,max = 201.22 mm] Hence over reinforced
[M_uR = 509.45 kNm] > [M_ulim =468.2 kNm] as expected
[A_st = 4310.27 mm²] > [A_st,lim =3609.3 mm²] as expected.

Solved example 9.3:
The data for this example are:

b_f =1000 mm, D_f =100 mm, D =500 mm, b_w =325 mm, d =420 mm, A_st =3436.12 mm² (7-25Ф), f_y = 415 N/mm², f_ck =25 N/mm² 

Calculations:
For Fe415 steel, x_u,max / d = 0.4791 (Table 3.4)
So x_u,max for our beam section = 0.4791 x 420 =201.22 mm

Here x_u,max > D_f so the section belongs to either case 2_lim(1) or case 2_lim(2)
0.43x_u,max = 0.43 x 201.22 = 86.53 < D_f. So the beam belongs to case 2_lim(2)

So we can use Eq.9.17

• Change x_u to x_u,max.
• Change M_uR to M_ulim

Substituting all the known values, we will get M_ulim = 466.41 kNm 

We can calculate A_st,lim using
Eq.9.19

• Change D_f to y_f

Substituting all the known values, we get A_st,lim = 3627.82 mm².

Let us compare the above results with those of the solved example 9.3:
The results obtained in 9.3 were:
[x_u = 184.23 mm] < [x_u,max = 201.22 mm] Hence under reinforced
[M_uR = 447.16 kNm] < [M_ulim =466.41 kNm] as expected
[A_st = 3436.12 mm²] < [A_st,lim =3627.82 mm²] as expected.

Solved example 9.4:
The data for this example are:

b_f =1532 mm, D_f =110 mm, D =510 mm, b_w =250 mm, d =464 mm, A_st =603.19 mm² (3-16Ф)., f_y = 415 N/mm², f_ck =20 N/mm² 

Calculations:
For Fe415 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

Here x_u,max > D_f so the section belongs to either case 2_lim(1) or case 2_lim(2)
0.43x_u,max = 0.43 x 222.3 = 95.59 < D_f. So the beam belongs to case 2_lim(2)

So we can use Eq.9.17

• Change x_u to x_u,max.
• Change M_uR to M_ulim

Substituting all the known values, we will get M_ulim = 644.05 kNm 

We can calculate A_st,lim using
Eq.9.19

• Change D_f to y_f

Substituting all the known values, we get A_st,lim = 4442.61 mm².

Let us compare the above results with those of the solved example 9.4:
The results obtained in 9.4 were:
[x_u = 19.635 mm] < [x_u,max = 222.30 mm] Hence under reinforced
[M_uR = 99.27 kNm] < [M_ulim =644.05 kNm] as expected
[A_st = 603.19 mm²] < [A_st,lim =4442.61 mm²] as expected.

With this we have completed the discussion on the analysis of flanged sections. In the next section we will see the design part.


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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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