In the previous section, we saw that the given beam section does not belong to 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 =110 mm. Which gives xu = 110/0.43 = 255.81 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 110 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.9.20
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 under reinforced'. This is because, the value of 255.81 mm (obtained in (2) above) 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
Eq.9.12
Cuw + Cuf = Cu = 0.362fck bw xu + 0.447fck Df (bf -bw) = 1194837.91 N
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 now greater 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 find that Cu is greater than Tu, which means, we have to reduce the value of xu. 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 255.81 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 110 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'.
We also get another important information about the beam section: We have seen that xu must have a value less than 255.81 mm. But this value of 255.81 is already less than xu,max which is equal to 276.12 mm (1). So our section is indeed an under reinforced section.
Based on the above conclusions, we can proceed to calculate the MuR as if we 'already know that the section belongs to case 2(2)'. The stress distribution in the beam section will be as shown in the fig. below:
Fig.9.21
Actual stress distribution
In fact, we do not require the above stress distribution for calculating MuR. We know that the beam section belongs to case 2(2), and we have all the required equations.
xu is given by
Eq.9.18
Where fst = 0.87fy (section is under reinforced)
And, yf = 0.15xu + 0.65Df = 0.15xu + 71.5 (Eq.9.15)
So we get xu =127.50 mm < xu,max Hence OK
and yf = 0.15 x 127.50 + 0.65 x 110 = 90.625 mm < Df Hence OK
0.43xu = 0.43 x 127.50 = 54.825 mm < Df. So the section indeed belongs to case 2(2)
MuR is given by
Eq.9.17
So we get MuR = 379296054.089 Nmm = 379.3 kNm
This completes the analysis of the given beam section. In the next section we will see another example.
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• 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 =110 mm. Which gives xu = 110/0.43 = 255.81 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 110 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.9.20
Stress distribution when 0.43xu = Df
• Also, when we assume the above value for 0.43xu, it automatically implies that we are 'assuming the section to be under reinforced'. This is because, the value of 255.81 mm (obtained in (2) above) 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
Eq.9.12
Cuw + Cuf = Cu = 0.362fck bw xu + 0.447fck Df (bf -bw) = 1194837.91 N
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 now greater 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 find that Cu is greater than Tu, which means, we have to reduce the value of xu. 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 255.81 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 110 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'.
We also get another important information about the beam section: We have seen that xu must have a value less than 255.81 mm. But this value of 255.81 is already less than xu,max which is equal to 276.12 mm (1). So our section is indeed an under reinforced section.
Based on the above conclusions, we can proceed to calculate the MuR as if we 'already know that the section belongs to case 2(2)'. The stress distribution in the beam section will be as shown in the fig. below:
Fig.9.21
Actual stress distribution
In fact, we do not require the above stress distribution for calculating MuR. We know that the beam section belongs to case 2(2), and we have all the required equations.
xu is given by
Eq.9.18
Where fst = 0.87fy (section is under reinforced)
And, yf = 0.15xu + 0.65Df = 0.15xu + 71.5 (Eq.9.15)
So we get xu =127.50 mm < xu,max Hence OK
and yf = 0.15 x 127.50 + 0.65 x 110 = 90.625 mm < Df Hence OK
0.43xu = 0.43 x 127.50 = 54.825 mm < Df. So the section indeed belongs to case 2(2)
MuR is given by
Eq.9.17
So we get MuR = 379296054.089 Nmm = 379.3 kNm
This completes the analysis of the given beam section. In the next section we will see another example.
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thank you
ReplyDeleteThank you for your help in understanding the subject.
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