In the previous section, we derived the 'most probable' value of d. Now we begin the actual design process.
We will now fix fck: It is given that the beam is subjected to 'moderate' exposure conditions. So we have to use M25 concrete. Thus fck = 25 N/mm2.
Step A:
Calculations for Mulim:
For Fe415 steel, xu,max / d = 0.4791 (Table 3.4)
So xu,max for our beam section = 0.4791 x 623 =298.48 mm - - - (1)
Here xu,max > Df so the section belongs to either case 2lim(1) or case 2lim(2)
0.43xu,max = 0.43 x 298.48 = 128.35 > Df. So the beam belongs to case 2lim(1)
So we can use Eq.9.11
• Change xu to xu,max.
• Change MuR to Mulim
Substituting all the known values, we will get Mulim = 1108.6 kNm
We can calculate Ast,lim using
Eq.9.19
Substituting all the known values, we get Ast,lim = 5649.1 mm2.
We obtained Mulim = 1108.6 kNm. So, the given T-beam section has dimensions and concrete grade fck such that, it will require an applied moment of 1105.06 kNm, to bring it to a state of impending failure, provided we give it a steel quantity of 5649.1 mm2 of Fe415 grade.
The resisting moment will be equal to the applied moment. So we can describe the situation in another way: If we give an area of steel equal to 5649.1 mm2, to the beam section, it will offer a resistance of 1108.6 kNm at the ultimate state. This is the Mulim, and unless we make it doubly reinforced, or improve the concrete and/or steel grades, we can not expect a higher resistance, even if we give more tension steel.
But we do not want it to offer us this much moment capacity. Our requirement is only 750 kNm. Based on the above results, we can say that, the beam section will indeed be able to resist 750 kNm, if we provide the appropriate quantity of steel. So our next aim is to find this 'appropriate' quantity of steel. For that, first we have to calculate xu.
Step B:
Here we analyse the section to calculate xu:
• Assume that xu has the maximum value possible within case 1.
This maximum possible value is equal to Df. So we put xu = Df =100 mm
• 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 equal to 298.48 mm (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 = 1267000 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 = 1315756.46 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, the NA will fall in the web.
• 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 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 rectangular portion of the stress block'. In case 2(1), this height cannot get any smaller than this.
• 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 232.56 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 = 0.362fckbwxu +0.447fckDf (bf -bw) = 1500750 N
And the tensile force in steel (steel would have yielded at ultimate state if it is an under reinforced section) will be same as before:
Tu = 0.87fy Ast = 1315756.46 N
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.
• But when xu is decreased, '0.43xu' will also decrease. This will mean that the height of the rectangular portion of the stress 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.
• But xu has to be decreased for obtaining equilibrium.
• 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 step B 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'.
Also, when xu is further reduced from the value of 232.56 mm, the section will indeed be 'under reinforced'.
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 equation 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 =113.35 < xu,max Hence OK
and yf = 0.15 x 113.35 + 0.65 x 100 = 82.00 mm < Df Hence OK
0.43xu = 0.43 x 113.35 = 48.74 < Df. And xu > Df. So the section indeed belongs to case 2(2).
The equation that we used above to calculate xu is Eq.9.18. The Eq.9.17 gives us MuR. Let us calculate that also at this stage. It will be required at a later stage in this problem.
Eq.9.17
Substituting all the known values, we will get MuR = 763.88 kNm - - - (3)
Thus we complete Step B. At the end of this step we found out the case into which the section belongs.
Step C:
In the previous step we calculated a value of xu. This value enabled us to put the beam section into the 'most probable case' into which it will belong. That is., if we provide 3 -32# and 2 -28#, of Fe 415 grade steel, at the ultimate state, the beam section will have the following properties:
• The depth of NA will be equal to 113.35 mm
• The effective depth is equal to 623 mm
• The resistance it will be offering at it's ultimate state will be equal to 763.88 kNm
• It will belong to case 2(2)
The above properties are the most probable properties because 3-32# and 2-28# is the most probable quantity of steel that the beam will require to resist the ultimate moment of 750 kNm. If we actually provide 3-32# and 2-28#, the capacity of the beam will be 763.88 kNm, found in (3). We do not need this much. We need only 750 kNm. So we want to know the exact steel which will give the capacity of 750 kNm.
For this we want to know the value of xu at the following condition:
• A moment of 750 kNm is applied at the section.
• When this moment is applied, the section reaches it's ultimate state.
In this step we try to find this xu. We have seen the Eq.10.9 which is used to calculate this xu.
Eq.10.9
The solution steps are shown here. We get xu = 107.66 mm. This is the actual value of xu if the effective depth d of the given section is 623 mm. And, if a moment of 750 kNm is applied at the section. Also, when this moment is applied, the section will reach it's ultimate state. Based on this information, we can calculate Ast. This we do in the next step.
Step D:
In this step, we calculate Ast using the value of xu obtained in the previous step. For this we use Eq.10.10. It's evaluation is also shown in the pdf file given in the previous step.
Thus we get Ast = 3572.45 mm2. This required value is a little less than the assumed value of 3644.25 mm2 (given by 3-32# + 2-28#). So we can safely provide 3 - #32 + 2 - #28.
When we provide this same steel, d will remain the same. So the section will have the same MuR calculated in (3). Thus the section is safe. We can also see that Ast is less than Ast,lim.
This completes the design procedure of the given beam section. The bars can be provided as we saw earlier in fig.10.13.
In the next section we will see another solved example.
PREVIOUS
We will now fix fck: It is given that the beam is subjected to 'moderate' exposure conditions. So we have to use M25 concrete. Thus fck = 25 N/mm2.
Step A:
Calculations for Mulim:
For Fe415 steel, xu,max / d = 0.4791 (Table 3.4)
So xu,max for our beam section = 0.4791 x 623 =298.48 mm - - - (1)
Here xu,max > Df so the section belongs to either case 2lim(1) or case 2lim(2)
0.43xu,max = 0.43 x 298.48 = 128.35 > Df. So the beam belongs to case 2lim(1)
So we can use Eq.9.11
• Change xu to xu,max.
• Change MuR to Mulim
Substituting all the known values, we will get Mulim = 1108.6 kNm
We can calculate Ast,lim using
Eq.9.19
Substituting all the known values, we get Ast,lim = 5649.1 mm2.
We obtained Mulim = 1108.6 kNm. So, the given T-beam section has dimensions and concrete grade fck such that, it will require an applied moment of 1105.06 kNm, to bring it to a state of impending failure, provided we give it a steel quantity of 5649.1 mm2 of Fe415 grade.
The resisting moment will be equal to the applied moment. So we can describe the situation in another way: If we give an area of steel equal to 5649.1 mm2, to the beam section, it will offer a resistance of 1108.6 kNm at the ultimate state. This is the Mulim, and unless we make it doubly reinforced, or improve the concrete and/or steel grades, we can not expect a higher resistance, even if we give more tension steel.
But we do not want it to offer us this much moment capacity. Our requirement is only 750 kNm. Based on the above results, we can say that, the beam section will indeed be able to resist 750 kNm, if we provide the appropriate quantity of steel. So our next aim is to find this 'appropriate' quantity of steel. For that, first we have to calculate xu.
Step B:
Here we analyse the section to calculate xu:
• Assume that xu has the maximum value possible within case 1.
This maximum possible value is equal to Df. So we put xu = Df =100 mm
• 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 equal to 298.48 mm (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 = 1267000 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 = 1315756.46 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, the NA will fall in the web.
• 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 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 rectangular portion of the stress block'. In case 2(1), this height cannot get any smaller than this.
• 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 232.56 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 = 0.362fckbwxu +0.447fckDf (bf -bw) = 1500750 N
And the tensile force in steel (steel would have yielded at ultimate state if it is an under reinforced section) will be same as before:
Tu = 0.87fy Ast = 1315756.46 N
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.
• But when xu is decreased, '0.43xu' will also decrease. This will mean that the height of the rectangular portion of the stress 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.
• But xu has to be decreased for obtaining equilibrium.
• 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 step B 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'.
Also, when xu is further reduced from the value of 232.56 mm, the section will indeed be 'under reinforced'.
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 equation 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 =113.35 < xu,max Hence OK
and yf = 0.15 x 113.35 + 0.65 x 100 = 82.00 mm < Df Hence OK
0.43xu = 0.43 x 113.35 = 48.74 < Df. And xu > Df. So the section indeed belongs to case 2(2).
The equation that we used above to calculate xu is Eq.9.18. The Eq.9.17 gives us MuR. Let us calculate that also at this stage. It will be required at a later stage in this problem.
Eq.9.17
Thus we complete Step B. At the end of this step we found out the case into which the section belongs.
Step C:
In the previous step we calculated a value of xu. This value enabled us to put the beam section into the 'most probable case' into which it will belong. That is., if we provide 3 -32# and 2 -28#, of Fe 415 grade steel, at the ultimate state, the beam section will have the following properties:
• The depth of NA will be equal to 113.35 mm
• The effective depth is equal to 623 mm
• The resistance it will be offering at it's ultimate state will be equal to 763.88 kNm
• It will belong to case 2(2)
The above properties are the most probable properties because 3-32# and 2-28# is the most probable quantity of steel that the beam will require to resist the ultimate moment of 750 kNm. If we actually provide 3-32# and 2-28#, the capacity of the beam will be 763.88 kNm, found in (3). We do not need this much. We need only 750 kNm. So we want to know the exact steel which will give the capacity of 750 kNm.
For this we want to know the value of xu at the following condition:
• A moment of 750 kNm is applied at the section.
• When this moment is applied, the section reaches it's ultimate state.
In this step we try to find this xu. We have seen the Eq.10.9 which is used to calculate this xu.
Eq.10.9
The solution steps are shown here. We get xu = 107.66 mm. This is the actual value of xu if the effective depth d of the given section is 623 mm. And, if a moment of 750 kNm is applied at the section. Also, when this moment is applied, the section will reach it's ultimate state. Based on this information, we can calculate Ast. This we do in the next step.
Step D:
In this step, we calculate Ast using the value of xu obtained in the previous step. For this we use Eq.10.10. It's evaluation is also shown in the pdf file given in the previous step.
Thus we get Ast = 3572.45 mm2. This required value is a little less than the assumed value of 3644.25 mm2 (given by 3-32# + 2-28#). So we can safely provide 3 - #32 + 2 - #28.
When we provide this same steel, d will remain the same. So the section will have the same MuR calculated in (3). Thus the section is safe. We can also see that Ast is less than Ast,lim.
This completes the design procedure of the given beam section. The bars can be provided as we saw earlier in fig.10.13.
In the next section we will see another solved example.
PREVIOUS
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