In
the previous section we derived Eq.9.11 for calculating MuR of flanged section for: 'case 2(1): Df < xu AND Df ≤ 0.43xu'. To use this equation we want the value of xu. So now we will derive an expression for xu.
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
Cuw = 0.362fck bw xu
• The compressive force in the flange given by Eq.9.10
Eq.9.10
Cuf = 0.447fck Df (bf -bw)
So the total compressive force =
Eq.9.12
Cuw + Cuf = Cu = 0.362fck bw xu + 0.447fck Df (bf -bw)
The tensile force in steel is given by [stress in steel x Area of steel]. Thus:
Eq.9.13
Tu = fst Ast
Equating 9.12 and 9.13 we get:
Eq.9.14
So we obtained the equation for xu. 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 MuR 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 0.87fy at the ultimate state. In that case, we can directly use Eq.9.14 to calculate xu, by putting fst = 0.87fy
• 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 fst = 0.87fy. 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): Df < xu AND Df ≤ 0.43xu' . 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 xu,max by xu in the equations given in G-2.2. So we move upwards from G-2.3 to G-2.2. If we substitute xu,max by xu 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 xu, the the depth of NA should be less than xu,max. So that, it is an under reinforced section. This condition also says that xu must be greater than Df which means that the NA lies in the web.
• The second condition is that Df / xu does not exceed 0.43. That is., Df / xu ≤ 0.43. This can be written as Df ≤ 0.43xu. Which is same as saying: The depth of rectangular portion of the stress block is greater than or equal to Df.
So both the conditions are same as those conditions of the case that we named as: 'case 2(1): Df < xu AND Df ≤ 0.43xu'
Thus we conclude that if we are given a beam section which falls under 'case 2(1): Df < xu AND Df ≤ 0.43xu', we can use the first part (The second part is related to the case when Df / xu exceeds 0.43) of G-2.3 of the code to find MuR. This completes the discussion about: 'case 2(1): Df < xu AND Df ≤ 0.43xu'
Next we consider the other sub case in case 2. We will call it:
case 2(2): Df < xu AND Df > 0.43xu
So this case is related to the situation in which two conditions are satisfied:
• The NA lies in the web (indicated by the relation: Df < xu) and
• The depth of the rectangular stress block is less than the depth of flange (indicated by the relation: Df > 0.43xu)
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
Cuw = 0.362fck bw xu
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 (bf -bw).
So we have to get the depth and the thickness. The code gives us the guidelines for fixing these:
• The depth is equal to 0.447fck.
• The thickness (denoted as yf)is given by Cl.G-2.2.1 of the code as
Eq.9.15
yf = 0.15xu + 0.65Df
The code imposes a restriction that the maximum value that yf can take is Df. So we must always check that:
9.16
yf ≤ 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, yf is shown to be equal to Df. However, if on calculating Eq.9.15 above, we get a value less than or equal to Df, it is OK. But if the value obtained is greater than Df, then we must take yf to be equal to Df.
In the next section, we will see how the beam section is analysed with this new stress block.
PREVIOUS
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
Cuw = 0.362fck bw xu
• The compressive force in the flange given by Eq.9.10
Eq.9.10
Cuf = 0.447fck Df (bf -bw)
So the total compressive force =
Eq.9.12
Cuw + Cuf = Cu = 0.362fck bw xu + 0.447fck Df (bf -bw)
The tensile force in steel is given by [stress in steel x Area of steel]. Thus:
Eq.9.13
Tu = fst Ast
Equating 9.12 and 9.13 we get:
Eq.9.14
So we obtained the equation for xu. 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 MuR 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 0.87fy at the ultimate state. In that case, we can directly use Eq.9.14 to calculate xu, by putting fst = 0.87fy
• 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 fst = 0.87fy. 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): Df < xu AND Df ≤ 0.43xu' . 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 xu,max by xu in the equations given in G-2.2. So we move upwards from G-2.3 to G-2.2. If we substitute xu,max by xu 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 xu, the the depth of NA should be less than xu,max. So that, it is an under reinforced section. This condition also says that xu must be greater than Df which means that the NA lies in the web.
• The second condition is that Df / xu does not exceed 0.43. That is., Df / xu ≤ 0.43. This can be written as Df ≤ 0.43xu. Which is same as saying: The depth of rectangular portion of the stress block is greater than or equal to Df.
So both the conditions are same as those conditions of the case that we named as: 'case 2(1): Df < xu AND Df ≤ 0.43xu'
Thus we conclude that if we are given a beam section which falls under 'case 2(1): Df < xu AND Df ≤ 0.43xu', we can use the first part (The second part is related to the case when Df / xu exceeds 0.43) of G-2.3 of the code to find MuR. This completes the discussion about: 'case 2(1): Df < xu AND Df ≤ 0.43xu'
Next we consider the other sub case in case 2. We will call it:
case 2(2): Df < xu AND Df > 0.43xu
So this case is related to the situation in which two conditions are satisfied:
• The NA lies in the web (indicated by the relation: Df < xu) and
• The depth of the rectangular stress block is less than the depth of flange (indicated by the relation: Df > 0.43xu)
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
Cuw = 0.362fck bw xu
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 (bf -bw).
So we have to get the depth and the thickness. The code gives us the guidelines for fixing these:
• The depth is equal to 0.447fck.
• The thickness (denoted as yf)is given by Cl.G-2.2.1 of the code as
Eq.9.15
yf = 0.15xu + 0.65Df
The code imposes a restriction that the maximum value that yf can take is Df. So we must always check that:
9.16
yf ≤ 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, yf is shown to be equal to Df. However, if on calculating Eq.9.15 above, we get a value less than or equal to Df, it is OK. But if the value obtained is greater than Df, then we must take yf to be equal to Df.
In the next section, we will see how the beam section is analysed with this new stress block.
PREVIOUS
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