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 Mulim 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
Mu,lim = 0.362 fck b xu,max( d - 0.416 xu,max )
To calculate Mulim 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 xu,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 Mulim 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: Mulim 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 Mulim, 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 Mulim, the depth of NA at this ultimate state is equal to xu,max.
Normally, for calculating a resisting moment, we need to calculate xu. But for calculating Mulim, we do not need to calculate it because we now know that for Mulim, xu = xu,max. So our work is very much reduced.
But there exists another problem: For a rectangular section, we can calculate Mulim directly, just by putting xu = xu,max. For a flanged section also, we can put xu = xu,max in the equation for MuR, and thus obtain Mulim. But for flanged sections, we have three cases and three corresponding equations for MuR. 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 Mulim. (This set up is done by giving Ast = Ast,lim.) Also at this ultimate state, the depth of it's NA will be equal to xu,max. This xu,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 MuR = Cu x z Where Cu is given by Eq.3.7
Eq.3.7
Cu = 0.362 fck b xu and z = d - 0.416xu
• 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 xu,max lies in the flange. We will name it as Case 1lim: Df ≥ xu,max
Case 1lim: Df ≥ xu,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
xu,max lies in the flange
So it will behave like a rectangular section of width bf. The eq.3.7 given above can be used to calculate Cu and then multiply it with z. This will give Mulim. The only changes to be done are these:
• Change xu to xu,max
• Change b to bf
The following point may be noted about the 'Titles' that are given to this case and the former 'Case 1: Df ≥ xu' : We can see that the titles are similar. The only differences are that a subscript 'lim' has been added and xu has been changed to xu,max in our present case.
We will now see the next case. We will call it as case 2lim(1): Df < xu,max AND Df ≤ 0.43xu,max
case 2lim(1): Df < xu,max AND Df ≤ 0.43xu,max:
This case arises when
• The NA lies in the web (indicated by Df < xu,max) and
• The depth of the rectangular portion is greater than or equal to Df (indicated by Df ≤ 0.43xu,max).
This is shown in the fig. below: It is similar to the case 2(1) that we discussed earlier
Fig.9.35
xu,max lies in the web
xu,max can be readily calculated using the table 3.4 shown above. So first we check whether xu,max > Df. Then we check if Df is less than or equal to 0.43xu,max. If both these conditions are satisfied, then our beam section belongs to 'case 2lim(1)'. We can use Eq.9.11 given above to calculate Mulim. The only changes to be done are:
• Change xu to xu,max.
• Change MuR to Mulim
Now we will calculate Ast,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 Mulim. For calculating it, we equate the compressive force to tensile force.
The compressive force is obtained by putting xu = xu,max in Eq.9.12. So we get: Cuw + Cuf = Cu = 0.362fck bw xu,max + 0.447fck Df (bf -bw)
The tensile force = 0.87fyAst,lim. Equating the two, we get
Eq.9.19
In this case also the 'Title' is similar to the earlier case 2(1): Df < xu AND Df ≤ 0.43xu. The only differences are that a subscript 'lim' has been added and xu has been changed to xu,max in our present case.
We will now see the next case. We will call it as case 2lim(2): Df < xu,max AND Df ≤ 0.43xu,max
case 2lim(1): Df < xu,max AND Df > 0.43xu,max:
This case arises when
• The NA lies in the web (indicated by Df < xu,max) and
• The depth of the rectangular portion is less than Df (indicated by Df > 0.43xu,max).
This is shown in the fig. below: It is similar to the case 2(2) that we discussed earlier
Fig.9.36
xu,max lies in the web
xu,max can be readily calculated using the table 3.4 shown above. So first we check whether xu,max > Df. Then we check if Df is greater than to 0.43xu,max. If both these conditions are satisfied, then our beam section belongs to 'case 2lim(2)'. We can use Eq.9.17 given above to calculate Mulim. The only changes to be done are:
• Change xu to xu,max.
• Change MuR to Mulim
Now we will calculate the Ast,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. Df in the previous case is changed to yf in the present case. So we can use the above Eq.9.19, just by changing Df in that equation to yf.
In this case also the 'Title' is similar to the earlier case 2(2): Df < xu AND Df > 0.43xu. The only differences are that a subscript 'lim' has been added and xu has been changed to xu,max in our present case.
So now we know how to calculate the Mulim 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 Mulim.
In the next section, we will determine the Mulim of the flanged beam in each of the four solved examples that we did earlier in this chapter.
PREVIOUS
We have learned about the limiting moment of resistance Mulim 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
Mu,lim = 0.362 fck b xu,max( d - 0.416 xu,max )
To calculate Mulim 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 xu,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 Mulim 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: Mulim 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 Mulim, 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 Mulim, the depth of NA at this ultimate state is equal to xu,max.
Normally, for calculating a resisting moment, we need to calculate xu. But for calculating Mulim, we do not need to calculate it because we now know that for Mulim, xu = xu,max. So our work is very much reduced.
But there exists another problem: For a rectangular section, we can calculate Mulim directly, just by putting xu = xu,max. For a flanged section also, we can put xu = xu,max in the equation for MuR, and thus obtain Mulim. But for flanged sections, we have three cases and three corresponding equations for MuR. 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 Mulim. (This set up is done by giving Ast = Ast,lim.) Also at this ultimate state, the depth of it's NA will be equal to xu,max. This xu,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 MuR = Cu x z Where Cu is given by Eq.3.7
Eq.3.7
Cu = 0.362 fck b xu and z = d - 0.416xu
• 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
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 xu,max lies in the flange. We will name it as Case 1lim: Df ≥ xu,max
Case 1lim: Df ≥ xu,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
xu,max lies in the flange
So it will behave like a rectangular section of width bf. The eq.3.7 given above can be used to calculate Cu and then multiply it with z. This will give Mulim. The only changes to be done are these:
• Change xu to xu,max
• Change b to bf
The following point may be noted about the 'Titles' that are given to this case and the former 'Case 1: Df ≥ xu' : We can see that the titles are similar. The only differences are that a subscript 'lim' has been added and xu has been changed to xu,max in our present case.
We will now see the next case. We will call it as case 2lim(1): Df < xu,max AND Df ≤ 0.43xu,max
case 2lim(1): Df < xu,max AND Df ≤ 0.43xu,max:
This case arises when
• The NA lies in the web (indicated by Df < xu,max) and
• The depth of the rectangular portion is greater than or equal to Df (indicated by Df ≤ 0.43xu,max).
This is shown in the fig. below: It is similar to the case 2(1) that we discussed earlier
Fig.9.35
xu,max lies in the web
xu,max can be readily calculated using the table 3.4 shown above. So first we check whether xu,max > Df. Then we check if Df is less than or equal to 0.43xu,max. If both these conditions are satisfied, then our beam section belongs to 'case 2lim(1)'. We can use Eq.9.11 given above to calculate Mulim. The only changes to be done are:
• Change xu to xu,max.
• Change MuR to Mulim
Now we will calculate Ast,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 Mulim. For calculating it, we equate the compressive force to tensile force.
The compressive force is obtained by putting xu = xu,max in Eq.9.12. So we get: Cuw + Cuf = Cu = 0.362fck bw xu,max + 0.447fck Df (bf -bw)
The tensile force = 0.87fyAst,lim. Equating the two, we get
Eq.9.19
In this case also the 'Title' is similar to the earlier case 2(1): Df < xu AND Df ≤ 0.43xu. The only differences are that a subscript 'lim' has been added and xu has been changed to xu,max in our present case.
We will now see the next case. We will call it as case 2lim(2): Df < xu,max AND Df ≤ 0.43xu,max
case 2lim(1): Df < xu,max AND Df > 0.43xu,max:
This case arises when
• The NA lies in the web (indicated by Df < xu,max) and
• The depth of the rectangular portion is less than Df (indicated by Df > 0.43xu,max).
This is shown in the fig. below: It is similar to the case 2(2) that we discussed earlier
Fig.9.36
xu,max lies in the web
xu,max can be readily calculated using the table 3.4 shown above. So first we check whether xu,max > Df. Then we check if Df is greater than to 0.43xu,max. If both these conditions are satisfied, then our beam section belongs to 'case 2lim(2)'. We can use Eq.9.17 given above to calculate Mulim. The only changes to be done are:
• Change xu to xu,max.
• Change MuR to Mulim
In this case also the 'Title' is similar to the earlier case 2(2): Df < xu AND Df > 0.43xu. The only differences are that a subscript 'lim' has been added and xu has been changed to xu,max in our present case.
So now we know how to calculate the Mulim 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 Mulim.
In the next section, we will determine the Mulim of the flanged beam in each of the four solved examples that we did earlier in this chapter.
PREVIOUS
No comments:
Post a Comment