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): Df < xu AND Df ≤ 0.43xu'. 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 Df, the depth of the flange.
• In the present case, the thickness is equal to yf
So all the steps in the derivation of expressions are the same. We will indeed obtain the same expressions with just Df changed to yf.
So we can write the equation for MuR as:
Eq.9.17
This is same as Eq.9.11, with just Df changed to yf.
Similarly, we can write the equation for xu as:
Eq.9.18
This is same as Eq.9.14, with Df changed to yf.
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): 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 second 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-2.2.1. So we move upwards from G-2.3 to G-2.2.1. If we substitute xu,max by xu 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 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 must 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 less than Df.
So both the conditions are same as those conditions of the case that we named as: 'case 2(2): Df < xu AND Df > 0.43xu'
Thus we conclude that if we are given a beam section which falls under 'case 2(2): Df < xu AND Df > 0.43xu', we can use the second part of G-2.3 of the code to find MuR. 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 MuR of the T-beam section given in fig.9.18 below:
Fig.9.18
Section of T-beam
Assume fck = 20 N/mm2 and Fe 250 steel.
Solution:
We will first write the following data:
bf =950 mm, Df =110 mm, D =600 mm, bw =300 mm, d =520 mm, Ast =3694.51 mm2 (6-28ะค)
For Fe250 steel, xu,max / d = 0.531 (Table 3.4)
So xu,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 xu and MuR, 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: Df ≥ xu'. For doing this test, we have to make the following assumption:
• xu has the maximum value possible within case 1.
This maximum possible value is equal to Df. So we put xu = Df =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', xu can have a value less than or equal to Df. The maximum value possible is Df. In the above assumption, we are giving this maximum possible value to xu. In this situation, the stress distribution will be as shown in the fig.9.19 below:
Fig.9.19
Stress distribution when xu = Df
In the above fig., xu = 110 mm. So the NA coincides with the bottom edge of the flange.
• Also, when we assume the above value for xu, it automatically implies that: 'we are assuming the section to be under reinforced'. This is because, the value of 110 mm for xu is less than xu,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
Cu = 0.362 fck bf xu = 756580 N
Note that the above equation is obtained by changing b to bf 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
Tu = 0.87fy Ast = 803555.93 N
From the above results, we find that Cu is less than Tu. This is not allowable. So we want an increased value for Cu. Let us assess this situation:
• If we want an increased value of Cu, then concrete area should be increased.
• This will imply that the position of NA should be lowered, which is same as saying: value of xu should increase.
• But we have already given the maximum possible value of xu in Case 1.
• If we further increase xu, 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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• In the previous case, the thickness is equal to Df, the depth of the flange.
• In the present case, the thickness is equal to yf
So all the steps in the derivation of expressions are the same. We will indeed obtain the same expressions with just Df changed to yf.
So we can write the equation for MuR as:
Eq.9.17
This is same as Eq.9.11, with just Df changed to yf.
Similarly, we can write the equation for xu as:
Eq.9.18
This is same as Eq.9.14, with Df changed to yf.
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): 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 second 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-2.2.1. So we move upwards from G-2.3 to G-2.2.1. If we substitute xu,max by xu 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 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 must 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 less than Df.
So both the conditions are same as those conditions of the case that we named as: 'case 2(2): Df < xu AND Df > 0.43xu'
Thus we conclude that if we are given a beam section which falls under 'case 2(2): Df < xu AND Df > 0.43xu', we can use the second part of G-2.3 of the code to find MuR. 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 MuR of the T-beam section given in fig.9.18 below:
Fig.9.18
Section of T-beam
Solution:
We will first write the following data:
bf =950 mm, Df =110 mm, D =600 mm, bw =300 mm, d =520 mm, Ast =3694.51 mm2 (6-28ะค)
For Fe250 steel, xu,max / d = 0.531 (Table 3.4)
So xu,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 xu and MuR, 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: Df ≥ xu'. For doing this test, we have to make the following assumption:
• xu has the maximum value possible within case 1.
This maximum possible value is equal to Df. So we put xu = Df =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', xu can have a value less than or equal to Df. The maximum value possible is Df. In the above assumption, we are giving this maximum possible value to xu. In this situation, the stress distribution will be as shown in the fig.9.19 below:
Fig.9.19
Stress distribution when xu = Df
In the above fig., xu = 110 mm. So the NA coincides with the bottom edge of the flange.
• Also, when we assume the above value for xu, it automatically implies that: 'we are assuming the section to be under reinforced'. This is because, the value of 110 mm for xu is less than xu,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
Cu = 0.362 fck bf xu = 756580 N
Note that the above equation is obtained by changing b to bf 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
Tu = 0.87fy Ast = 803555.93 N
From the above results, we find that Cu is less than Tu. This is not allowable. So we want an increased value for Cu. Let us assess this situation:
• If we want an increased value of Cu, then concrete area should be increased.
• This will imply that the position of NA should be lowered, which is same as saying: value of xu should increase.
• But we have already given the maximum possible value of xu in Case 1.
• If we further increase xu, 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.
PREVIOUS
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