Chapter 9 (cont..4)

In the previous section we saw that a new 'Equivalent rectangular stress block' is applied to the flanges. We saw fig.9.16 which gives the stress distribution in the flange. So the case has in effect become similar to the previous case: 'case 2(1): D_f < x_u AND D_f ≤ 0.43x_u'. This is because in that case also, we have full stress block acting in the web portion, and rectangular stress block acting in the flange region. The only difference is in the thickness of this rectangular stress block. 
• In the previous case, the thickness is equal to D_f, the depth of the flange.
• In the present case, the thickness is equal to y_f  

So all the steps in the derivation of expressions are the same. We will indeed obtain the same expressions with just D_f changed to y_f.

So we can write the equation for M_uR as:
Eq.9.17

This is same as Eq.9.11, with just D_f changed to y_f.

Similarly, we can write the equation for x_u as:
Eq.9.18

This is same as Eq.9.14, with D_f changed to y_f.

Just as it was mentioned in the previous case, here also, we will have to use strain compatibility method if the section given to us is an 'Over reinforced section'. The procedure will become more clear when we discuss a solved example.

So now we know how to analyse a section if it falls under 'case 2(2): D_f < x_u AND D_f > 0.43x_u' . Let us see how this case is related to the provisions given in the code.

The clause which is related to our present discussion is G-2.3. The second part of this clause states the following:
The moment of resistance may be calculated by substituting  x_u,max by x_u in the equations given in G-2.2.1. So we move upwards from G-2.3 to G-2.2.1. If we substitute x_u,max by x_u in the equation given in G-2.2.1, we get the same Eq.9.17 above.

But according to this clause, to use this equation, the section has to satisfy the following two conditions:

• The first condition is that x_u, the the depth of NA should be less than x_u,max. So that, it is an under reinforced section. This condition also says that x_u must be greater than D_f which means that the NA lies in the web.

• The second condition is that D_f / x_u must exceed 0.43. That is., D_f / x_u > 0.43.   This can be written as D_f > 0.43x_u. Which is same as saying: The depth of rectangular portion of the stress block is less than D_f. 

So both the conditions are same as those conditions of the case that we named as: 'case 2(2): D_f < x_u AND D_f > 0.43x_u'

Thus we conclude that if we are given a beam section which falls under 'case 2(2): D_f < x_u AND D_f > 0.43x_u', we can use the second part of G-2.3 of the code to find M_uR. This completes the discussion about this case.

So we have discussed about the different possible cases that a flanged beam section may fall into. We will show a flow chart like presentation of the cases that we have learned so far in this chapter.

Fig.9.17
Classification of flanged beams

The above flow chart will have to be modified by adding more details, when we discuss about the 'Limiting moment of resistance' of these flanged sections. But before we discuss about it, we will see some solved examples that demonstrates the process of analysis in the different cases.

Solved Example 9.1:
Calculate the Ultimate moment of resistance M_uR of the T-beam section given in fig.9.18 below:

Fig.9.18
Section of T-beam

Assume f_ck = 20  N/mm² and Fe 250 steel.

Solution:
We will first write the following data:
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)

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 mm - - - - (1) 

Now we can start the analysis procedure:
We know that the given beam section may fall into any of the three cases shown in fig.9.17 above. 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, within 'Case 1', 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.9.19 below:

Fig.9.19
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 = 756580 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 = 803555.93 N 

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

• If we want an increased value of C_u, then concrete area should be increased.
• This will imply that the position of NA should be lowered, which is same as saying: value of x_u should increase.
• But we have already given the maximum possible value of x_u in Case 1.
• If we further increase x_u, then, Point 'A' in the above fig.9.19 will fall in the web, and it will no longer be Case 1 
• This will imply that our beam is beyond the limits of case 1.

Thus we discard Case 1 and try 'Case 2(1)'. We will see this in the next section.

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Chapter 9 (cont..3)

In the previous section we derived Eq.9.11 for calculating MuR of flanged section for: 'case 2(1): D_f < x_u AND D_f ≤ 0.43x_u'. To use this equation we want the value of x_u. So now we will derive an expression for x_u.

For this we use the equilibrium condition. That is., the compressive force in concrete is equal to the tensile force in steel. The compressive force is the sum of two quantities: 
• The compressive force in the web given by Eq.9.9
Eq.9.9
C_uw = 0.362f_ck b_w x_u  

• The compressive force in the flange given by Eq.9.10
Eq.9.10
C_uf = 0.447f_ck D_f (b_f -b_w)

So the total compressive force =

Eq.9.12
C_uw + C_uf = C_u = 0.362f_ck b_w x_u + 0.447f_ck D_f (b_f -b_w) 

The tensile force in steel is given by [stress in steel x Area of steel]. Thus:
Eq.9.13
T_u = f_st A_st

Equating 9.12 and 9.13 we get:
Eq.9.14

So we obtained the equation for x_u. But there are two 'possible directions' in which we can proceed using this equation. Let us assess this situation:

• We are on our way to calculate M_uR of the section.
• The section may be 'under reinforced' or 'over reinforced'. But we don't know which is our case yet.
• If the section is an 'under reinforced' one, the stress in steel would be 0.87f_y at the ultimate state. In that case, we can directly use Eq.9.14 to calculate x_u, by putting f_st = 0.87f_y
• On the other hand, if the section that we are having, is an 'over reinforced' one, then steel would not have yielded at the ultimate state. So we cannot put f_st = 0.87f_y. In such a case, we will have to use strain compatibility method, just as we did in the case of over reinforced rectangular sections. 

The steps for doing this will become more clear when we see a solved example later in this chapter.


So now we know how to analyse a section if it falls under 'case 2(1): D_f < x_u AND D_f ≤ 0.43x_u' . Let us see how this case is related to the provisions given in the code.

The clause which is related to our present discussion is G-2.3. The first part of this clause states the following:
The moment of resistance may be calculated by substituting  x_u,max by x_u in the equations given in G-2.2. So we move upwards from G-2.3 to G-2.2. If we substitute x_u,max by x_u in the equation given in G-2.2, we get this:

We can see that this is the same equation Eq.9.11 that we derived for our present case. But according to this clause, to use this equation, the section has to satisfy the following two conditions:

• The first condition is that x_u, the the depth of NA should be less than x_u,max. So that, it is an under reinforced section. This condition also says that x_u must be greater than D_f which means that the NA lies in the web.

• The second condition is that D_f / x_u does not exceed 0.43. That is., D_f / x_u ≤ 0.43.   This can be written as D_f ≤ 0.43x_u. Which is same as saying: The depth of rectangular portion of the stress block is greater than or equal to D_f. 

So both the conditions are same as those conditions of the case that we named as: 'case 2(1): D_f < x_u AND D_f ≤ 0.43x_u'


Thus we conclude that if we are given a beam section which falls under 'case 2(1): D_f < x_u AND D_f ≤ 0.43x_u', we can use the first part (The second part is related to the case when D_f / x_u exceeds 0.43) of G-2.3 of the code to find M_uR. This completes the discussion about: 'case 2(1): D_f < x_u AND D_f ≤ 0.43x_u'

Next we consider the other sub case in case 2. We will call it:

case 2(2): D_f < x_u AND D_f > 0.43x_u
So this case is related to the situation in which two conditions are satisfied:
• The NA lies in the web (indicated by the relation:  D_f < x_u) and
• The depth of the rectangular stress block is less than the depth of flange (indicated by the relation: D_f > 0.43x_u)

We now need to analyse this case. We have already seen the stress distribution for this case in fig.9.11 and the 3D view in fig.9.12. These figs. are shown again below:

Fig.9.11
Rectangular portion acts only on a small portion at the top of the flange

Fig.9.12
3D view of stress block

From the 3D view, we can see that, just as in the previous case, in the middle web region, the full stress block is acting. So in this case also, it is convenient to consider the web and the over hangs separately. We already saw the method of separation in figs.9.13 and 9.14


As in the previous case, the full stress block is acting in the web region. The compressive force in the web region is same as before, and is given by
Eq.9.9
C_uw = 0.362f_ck b_w x_u

Let us see the compressive force in the flange. We have to get a good understanding of how the stress block is acting on this flange region. From the 3D view in fig.9.12, we can see that the rectangular block is completely acting on the flange, as it is completely within the flange. The force from this portion can be calculated easily. But when we look at the parabolic portion, only a portion of it (in the form of a truncated parabola) is acting on the flange. We have to find:
• The force exerted by this truncated parabolic portion (coloured in blue), and the point of application of that force.
• The force exerted is equal to the volume of the truncated parabola. So we have to find that volume.
• The point of application of the force is at the centroid. So we have to find the centroid of the truncated parabola.
• The force exerted by the rectangular block (coloured in red) and the point of application of that force.
• The resultant of the above two forces and the point of application of the resultant force.

It is not convenient to form a general expression to find the above. So the code allows us to remove the red and blue blocks on the flange (blocks are removed from the flange only. Those in the web region remain untouched), and replace them with a uniform rectangular stress block. This new block is called the 'equivalent rectangular stress block'. This is shown in the fig.9.15 below:

Fig.9.15
Equivalent rectangular stress block

We have removed the combination of 'truncated parabolic' and rectangular stress blocks from the flange. Instead now we have a simple rectangular block (coloured in yellow). As for any other rectangular stress blocks, for this new yellow stress block also, the important parameters are it's depth, width and thickness. The width, is already fixed because it is equal to (b_f -b_w). 

So we have to get the depth and the thickness. The code gives us the guidelines for fixing these:
• The depth is equal to 0.447f_ck. 
• The thickness (denoted as y_f)is given by Cl.G-2.2.1 of the code as
Eq.9.15
y_f = 0.15x_u + 0.65D_f

The code imposes a restriction that the maximum value that y_f can take is D_f. So we must always check that:
9.16
y_f ≤  Df

These dimensions are shown in the fig.9.16 below:

Fig.9.16
Stress distribution in flange based on 'Equivalent rectangular stress block' 

In the 3D view in fig.9.15 above, y_f is shown to be equal to D_f. However, if on calculating Eq.9.15 above, we get a value less than or equal to D_f, it is OK. But if the value obtained is greater than D_f, then we must take y_f to be equal to D_f.

In the next section, we will see how the beam section is analysed with this new stress block.


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