In
the previous section we completed the analysis of an over reinforced T-beam. In this section we will analyse an L-beam section.
Solved Example 9.3:
Calculate the Ultimate moment of resistance MuR of the L-beam section given in fig.9.27 below:
Fig.9.27
Section of L-beam
Assume fck = 25 N/mm2 and Fe 415 steel.
Solution:
We will first write the following data:
bf =1000 mm, Df =100 mm, D =500 mm, bw =325 mm, d =420 mm, Ast =3436.12 mm2 (7-25ะค)
For Fe 415 steel, xu,max / d = 0.4791 (Table 3.4)
So xu,max for our beam section = 0.4791 x 420 =201.22 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. 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 =100 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 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.9.28 below:
Fig.9.28
Stress distribution when xu = Df
In the above fig., xu = 100 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 100 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 = 905000 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 = 1240611.126 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.9.28 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 do a test to check whether the beam falls into the category of 'case 2(1): Df < xu AND Df ≤ 0.43xu'. For doing this test, we have to make the following assumption:
• The height (0.43xu) of the rectangular portion of the stress block (coloured in red) has the smallest value possible within case 2(1).
This smallest possible height of the red block is equal to Df. So we put 0.43xu = Df =100 mm. Which gives xu = 100/0.43 = 232.56 mm.- - - (2)
We need to get a good understanding about this assumption. We can see a '≤' symbol in the name of this case. This means that 0.43xu can have a value greater than or equal to Df. The smallest value possible is Df. In the above assumption, we are giving this smallest possible value to 0.43xu. In other words, we are giving the smallest possible value of 100 mm for the 'height of the red block'. In case 2(1), the red block cannot get any smaller than this. In this situation, the stress distribution will be as shown in the fig. below:
Fig.9.29
Stress distribution when 0.43xu = Df
In the above fig., 0.43xu is given the smallest possible value in case 2(1)
• Also, when we assume the above value for 0.43xu, it automatically implies that we are 'assuming the section to be over reinforced'. This is because, the value of 232.56 mm (obtained in (2) above) for xu is greater 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
Eq.9.12
Cuw + Cuf = Cu = 0.362fck bw xu + 0.447fck Df (bf -bw) = 1418863.68 N
But the calculation of Tu is not so easy. This is because xu is greater than xu,max. We have to first calculate the stress in steel. For this, we will have to use the strain compatibility method. We have earlier used this method for the analysis of an over reinforced 'rectangular section'. It can be seen here.
Eq.3.19
= 0.002821
From the table for Fe 415 steel we get:
So we will get Tu = fst x Ast = 352.33 x 3436.12 =1210648.16 N.
What we did in the above strain compatibility method can be simply explained as follows:
• We have a value of xu =232.56 mm. Unfortunately, this is greater than xu,max.
• So if 232.56 mm is indeed the value of xu at the ultimate state, steel will not yield at the ultimate state. If steel does not yield, we cannot say that the stress in it is 0.87fy.
• Instead, we have to calculate the stress by using strain compatibility. That is., we first calculate the strain corresponding to the xu value of 232.56 mm. This we do using Eq.3.19 above.
• Then we calculate the stress corresponding to this strain. This we do by using linear interpolation between values from the table for Fe 415 steel.
We find that Cu is now larger than Tu. This will not give equilibrium. So we want a decreased value for Cu.
Here, we must do some simple calculations to decide which way to proceed. We have earlier discarded Case 1. We are now in case 2(1). We have to check whether there is any possibility by which we can remain in this case 2(1):
• We have just now seen that xu must be reduced to a value below 232.56 mm. This is to reduce Cu. This means that point 'A' in the above fig.9.20 must move upwards.
• But when 'A' moves upwards, xu will surely decrease, but along with that, '0.43xu' will also decrease. That is., point 'B' will also move upwards. This will mean that the height of the red block will be lesser than Df. So we will no longer be in case 2(1).
• This means that if we are to remain in case 2(1), we can never decrease xu. That is., 'A' must remain in the same position.
• Keeping 'A' in the same position, can 'B' be lowered?
• In that case we will still be in case 2(1) because the ht. of red block will be greater than Df.
• But then another problem arises: If 'B' is lowered, thus increasing the ht. of red block from it's present value of 100 mm, it would mean that xu has also increased because the ht. of red block = 0.43xu. In other words, lowering of 'B' is possible only if we increase xu.
• This increment in xu will further increase Cu.
• Thus we can conclude that we cannot remain in Case 2(1) under any circumstances.
So we have discarded two cases. The only remaining option is case 2(2). The calculations and discussions that we did so far in this problem were important 'Tests'. Tests to arrive at a conclusion about the case which our beam section will belong to. At the end of those tests, we conclude that our beam section belongs to 'case 2(2): Df < xu AND Df > 0.43xu'.
But the tests does not help us to conclude whether the section is over reinforced or under reinforced. This is because:
• We have seen that 0.43xu will be less than Df which is equal to 100mm
• So xu will be less than 100/0.43 = 232.56 mm
• We can say that the section is under reinforced only if we obtain a statement that 'xu will be less than xu,max which is equal to 201.22 mm'
• But we have not obtained such a statement during our 'tests'
• So at present we cannot conclude whether our beam section is under reinforced or over reinforced.
But we can try to obtain such a statement. This is by assuming xu = xu,max = 201.22 mm. Based on this assumption, we will get 0.43xu = 0.43 x 201.22 = 86.525 mm. The stress distribution in this condition is shown in the fig. below:
Fig.9.30
Stress distribution based on the assumption that xu = xu,max
In the above fig.,
• xu = xu,max. So the steel would have yielded at ultimate state.
• And 0.43xu < Df. So it is case 2(2).
The compressive force is given by:
Cuw + Cuf = Cu = 0.362fck bw xu + 0.447fck yf (bf -bw)
This is obtained by changing Df to yf in Eq.9.12
Where, yf = 0.15xu + 0.65Df = 0.15 x 201.22 + 0.65 x 100 = 95.18 mm (Eq.9.15)
Thus we get Cu = 1309798.845
Tu = 0.87fy x Ast = 1240611.126
Here, Cu is greater than Tu. So the xu must be decreased further. This means that xu will be less than xu,max, and so, our beam section is indeed an under reinforced one.
So now we know that our beam section belongs to case 2(2), and that it is an under reinforced one. We can use the equations directly.
xu is given by
Eq.9.18
Where fst = 0.87fy (section is under reinforced)
And, yf = 0.15xu + 0.65Df = 0.15xu + 65 (Eq.9.15)
So we get xu =184.23 mm < xu,max Hence OK
and yf = 0.15 x 184.23 + 0.65 x 100 = 92.63 mm < Df Hence OK
0.43xu = 0.43 x 184.23 = 79.22 mm < Df. So the section indeed belongs to case 2(2)
MuR is given by
Eq.9.17
So we get MuR = 447165077.31 Nmm = 447.16 kNm
This completes the analysis of the given beam section. In the next section we will see another example.
PREVIOUS
Solved Example 9.3:
Calculate the Ultimate moment of resistance MuR of the L-beam section given in fig.9.27 below:
Fig.9.27
Section of L-beam
Solution:
We will first write the following data:
bf =1000 mm, Df =100 mm, D =500 mm, bw =325 mm, d =420 mm, Ast =3436.12 mm2 (7-25ะค)
For Fe 415 steel, xu,max / d = 0.4791 (Table 3.4)
So xu,max for our beam section = 0.4791 x 420 =201.22 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. 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 =100 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 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.9.28 below:
Fig.9.28
Stress distribution when xu = Df
In the above fig., xu = 100 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 100 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 = 905000 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 = 1240611.126 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.9.28 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 do a test to check whether the beam falls into the category of 'case 2(1): Df < xu AND Df ≤ 0.43xu'. For doing this test, we have to make the following assumption:
• The height (0.43xu) of the rectangular portion of the stress block (coloured in red) has the smallest value possible within case 2(1).
This smallest possible height of the red block is equal to Df. So we put 0.43xu = Df =100 mm. Which gives xu = 100/0.43 = 232.56 mm.- - - (2)
We need to get a good understanding about this assumption. We can see a '≤' symbol in the name of this case. This means that 0.43xu can have a value greater than or equal to Df. The smallest value possible is Df. In the above assumption, we are giving this smallest possible value to 0.43xu. In other words, we are giving the smallest possible value of 100 mm for the 'height of the red block'. In case 2(1), the red block cannot get any smaller than this. In this situation, the stress distribution will be as shown in the fig. below:
Fig.9.29
Stress distribution when 0.43xu = Df
In the above fig., 0.43xu is given the smallest possible value in case 2(1)
• Also, when we assume the above value for 0.43xu, it automatically implies that we are 'assuming the section to be over reinforced'. This is because, the value of 232.56 mm (obtained in (2) above) for xu is greater 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
Eq.9.12
Cuw + Cuf = Cu = 0.362fck bw xu + 0.447fck Df (bf -bw) = 1418863.68 N
But the calculation of Tu is not so easy. This is because xu is greater than xu,max. We have to first calculate the stress in steel. For this, we will have to use the strain compatibility method. We have earlier used this method for the analysis of an over reinforced 'rectangular section'. It can be seen here.
Eq.3.19
= 0.002821
From the table for Fe 415 steel we get:
| Strain | Stress |
| 0.00276 | 351.80 |
| 0.002821 | 352.33 |
| 0.00380 | 360.90 |
What we did in the above strain compatibility method can be simply explained as follows:
• We have a value of xu =232.56 mm. Unfortunately, this is greater than xu,max.
• So if 232.56 mm is indeed the value of xu at the ultimate state, steel will not yield at the ultimate state. If steel does not yield, we cannot say that the stress in it is 0.87fy.
• Instead, we have to calculate the stress by using strain compatibility. That is., we first calculate the strain corresponding to the xu value of 232.56 mm. This we do using Eq.3.19 above.
• Then we calculate the stress corresponding to this strain. This we do by using linear interpolation between values from the table for Fe 415 steel.
We find that Cu is now larger than Tu. This will not give equilibrium. So we want a decreased value for Cu.
Here, we must do some simple calculations to decide which way to proceed. We have earlier discarded Case 1. We are now in case 2(1). We have to check whether there is any possibility by which we can remain in this case 2(1):
• We have just now seen that xu must be reduced to a value below 232.56 mm. This is to reduce Cu. This means that point 'A' in the above fig.9.20 must move upwards.
• But when 'A' moves upwards, xu will surely decrease, but along with that, '0.43xu' will also decrease. That is., point 'B' will also move upwards. This will mean that the height of the red block will be lesser than Df. So we will no longer be in case 2(1).
• This means that if we are to remain in case 2(1), we can never decrease xu. That is., 'A' must remain in the same position.
• Keeping 'A' in the same position, can 'B' be lowered?
• In that case we will still be in case 2(1) because the ht. of red block will be greater than Df.
• But then another problem arises: If 'B' is lowered, thus increasing the ht. of red block from it's present value of 100 mm, it would mean that xu has also increased because the ht. of red block = 0.43xu. In other words, lowering of 'B' is possible only if we increase xu.
• This increment in xu will further increase Cu.
• Thus we can conclude that we cannot remain in Case 2(1) under any circumstances.
So we have discarded two cases. The only remaining option is case 2(2). The calculations and discussions that we did so far in this problem were important 'Tests'. Tests to arrive at a conclusion about the case which our beam section will belong to. At the end of those tests, we conclude that our beam section belongs to 'case 2(2): Df < xu AND Df > 0.43xu'.
But the tests does not help us to conclude whether the section is over reinforced or under reinforced. This is because:
• We have seen that 0.43xu will be less than Df which is equal to 100mm
• So xu will be less than 100/0.43 = 232.56 mm
• We can say that the section is under reinforced only if we obtain a statement that 'xu will be less than xu,max which is equal to 201.22 mm'
• But we have not obtained such a statement during our 'tests'
• So at present we cannot conclude whether our beam section is under reinforced or over reinforced.
But we can try to obtain such a statement. This is by assuming xu = xu,max = 201.22 mm. Based on this assumption, we will get 0.43xu = 0.43 x 201.22 = 86.525 mm. The stress distribution in this condition is shown in the fig. below:
Fig.9.30
Stress distribution based on the assumption that xu = xu,max
In the above fig.,
• xu = xu,max. So the steel would have yielded at ultimate state.
• And 0.43xu < Df. So it is case 2(2).
The compressive force is given by:
Cuw + Cuf = Cu = 0.362fck bw xu + 0.447fck yf (bf -bw)
This is obtained by changing Df to yf in Eq.9.12
Where, yf = 0.15xu + 0.65Df = 0.15 x 201.22 + 0.65 x 100 = 95.18 mm (Eq.9.15)
Thus we get Cu = 1309798.845
Tu = 0.87fy x Ast = 1240611.126
Here, Cu is greater than Tu. So the xu must be decreased further. This means that xu will be less than xu,max, and so, our beam section is indeed an under reinforced one.
So now we know that our beam section belongs to case 2(2), and that it is an under reinforced one. We can use the equations directly.
xu is given by
Eq.9.18
Where fst = 0.87fy (section is under reinforced)
And, yf = 0.15xu + 0.65Df = 0.15xu + 65 (Eq.9.15)
So we get xu =184.23 mm < xu,max Hence OK
and yf = 0.15 x 184.23 + 0.65 x 100 = 92.63 mm < Df Hence OK
0.43xu = 0.43 x 184.23 = 79.22 mm < Df. So the section indeed belongs to case 2(2)
MuR is given by
Eq.9.17
So we get MuR = 447165077.31 Nmm = 447.16 kNm
This completes the analysis of the given beam section. In the next section we will see another example.
PREVIOUS
Do we need to consider torsion in an L beam due to asymmetry in flange (compared to a T beam)?
ReplyDeleteYes. It is important to check for torsion in L beams. Even in T beams, the check is necessary because, slab on one side may be carrying a greater load, or have greater span.
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