In the previous section, we calculated the effective flange width bf. So the section can be drawn as shown in fig.10.22 below, and we can begin the actual design process.
Fig.10.22
Section of T-beam
We will fix fck now: 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 323 =154.75 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 154.75 = 66.54 < Df. So the beam belongs to case 2lim(2)
So we can use Eq.9.17
• Change xu to xu,max.
• Change MuR to Mulim
yf = 0.15xu,max + 0.65Df =101.21 mm
Substituting all the known values, we will get Mulim = 522.33 kNm
We can calculate Ast,lim using
Eq.9.19
• Change Df to yf
Substituting all the known values, we will get Astlim = 5356.1 mm2
We obtained Mulim =522.33 kNm. So, the given T-beam section has dimensions and concrete grade fck such that, it will require an applied moment of 522.33 kNm, to bring it to a state of impending failure, provided we give it a steel quantity of 5356.1 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 5356.1 mm2, to the beam section, it will offer a resistance of 522.33 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 78.9 kNm. Based on the above results, we can say that, the beam section will indeed be able to resist 75 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 =120 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 120 mm for xu is less than xu,max which is equal to 154.75 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 = 1797330 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 = 275629.18 N
From the above results, we find that Cu is greater than Tu. This is not allowable. So we want a decreased value for Cu. Let us assess this situation:
• If we want a decreased value of Cu, then concrete area should be decreased.
• This will imply that the position of NA should be raised, which is same as saying: value of xu should decrease.
• This will imply that we will remain in case 1.
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 1: Df ≥ xu'.
Also, when xu is further reduced from the value of 120 mm, the section will indeed be 'under reinforced'.
So now we know that our beam section belongs to case 1, and that it is an under reinforced one. This is the simplest case in which the beam behaves as a rectangular section of width bf. We can use the equation directly.
Eq.3.17
xu = 0.87fyAst / 0.362fckb
• Change b to bf
Substituting all the known values, we get xu = 18.403. This is less than Df. So the section indeed belongs to case 1.
Let us calculate MuR also at this stage. It will be required at a later stage in this problem.
MuR = Tu x z = 0.87fyAst x (d -0.416xu) = 86.9 kNm - - - (3)
Thus we complete Step B.
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 -18#, 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 18.4 mm
• The effective depth is equal to 323 mm
• The resistance it will be offering at it's ultimate state will be equal to 86.9 kNm
• It will belong to case 1
The above properties are the most probable properties because 3-18# is the most probable quantity of steel that the beam will require to resist the ultimate moment of 78.9 kNm. If we actually provide 3-18#, the capacity of the beam will be 86.9 kNm, found in (3). We do not need this much. We need only 78.9 kNm. So we want to know the exact steel which will give the capacity of 78.9 kNm.
For this we want to know the value of xu at the following condition:
• A moment of 78.9 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.4.10 which is used to calculate this xu.
Eq.4.10:
A X2 + B X + C = 0
Where A = 0.1506 bd2fck ; B = -0.362 bd2fck ; C = Mu and X = xu/d
• Change b to bf
Solving this quadratic equation, we get xu/d = 0.0516. So xu = 0.0516 x 323 =16.67 mm
Ast can be obtained from Eq.4.11
Eq.4.11
• Change b to bf
Thus we get Ast = 691.53 mm2. - - - (4)
This required value is a less than the assumed value of 763.41 mm2 (given by 3-#18). So we can safely provide 3 - #18.
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 beam section. We will now do the various checks:
We did the three checks which are to be done immediately after fixing up the preliminary dimensions here. In those checks, the first two will give the same results because the parameters related to them have not changed. The third check for 'slenderness' involves 'd' which has changed from 320 to 323 mm. We will get the same result (calculation steps are not shown here): It is not a slender beam. So we will proceed to the other checks:
(1) Spacing between the bars have already been checked here
(2) Check for minimum area of flexural reinforcement (Details here):
As = 0.85bd / fy
• Change b to bw
We get As = 152.16 mm2 < 763.41 mm2. Hence OK
(3) Check for maximum area of flexural reinforcement (Details here):
0.04bwD = 3404 mm2 > 763.41 mm2. Hence OK
(4) Check whether pt is less than pt,lim:
We have seen that Ast provided is less than Ast,lim. So pt will be less than pt,lim.
(5) Check for deflection control (Details here):
(l/d)actual ≤ [(l/d)basic] kt
For simply supported spans, (l/d)basic = 20
To find kt:
• Ast required = 691.53 mm2 from (4) • Ast,p = 763.41 mm2
• fst = 218.04 N/mm2 • pt = 1.028 • From Fig.4 of the code, kt = 1.07
• So l/d = 20 x kt = 21.4 • (l/d) actual = 4230/323 = 13.1
13.1 < 21.4 Hence OK
This completes all the checks on the designed beam section. The bars can be provided as we saw earlier in the fig.10.21
In the next section, we will see a new topic, which is 'The design of Doubly reinforced rectangular beam sections'.
PREVIOUS
Fig.10.22
Section of T-beam
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 323 =154.75 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 154.75 = 66.54 < Df. So the beam belongs to case 2lim(2)
So we can use Eq.9.17
• Change xu to xu,max.
• Change MuR to Mulim
yf = 0.15xu,max + 0.65Df =101.21 mm
Substituting all the known values, we will get Mulim = 522.33 kNm
We can calculate Ast,lim using
Eq.9.19
• Change Df to yf
Substituting all the known values, we will get Astlim = 5356.1 mm2
We obtained Mulim =522.33 kNm. So, the given T-beam section has dimensions and concrete grade fck such that, it will require an applied moment of 522.33 kNm, to bring it to a state of impending failure, provided we give it a steel quantity of 5356.1 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 5356.1 mm2, to the beam section, it will offer a resistance of 522.33 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 78.9 kNm. Based on the above results, we can say that, the beam section will indeed be able to resist 75 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 =120 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 120 mm for xu is less than xu,max which is equal to 154.75 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 = 1797330 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 = 275629.18 N
From the above results, we find that Cu is greater than Tu. This is not allowable. So we want a decreased value for Cu. Let us assess this situation:
• If we want a decreased value of Cu, then concrete area should be decreased.
• This will imply that the position of NA should be raised, which is same as saying: value of xu should decrease.
• This will imply that we will remain in case 1.
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 1: Df ≥ xu'.
Also, when xu is further reduced from the value of 120 mm, the section will indeed be 'under reinforced'.
So now we know that our beam section belongs to case 1, and that it is an under reinforced one. This is the simplest case in which the beam behaves as a rectangular section of width bf. We can use the equation directly.
Eq.3.17
xu = 0.87fyAst / 0.362fckb
• Change b to bf
Substituting all the known values, we get xu = 18.403. This is less than Df. So the section indeed belongs to case 1.
Let us calculate MuR also at this stage. It will be required at a later stage in this problem.
MuR = Tu x z = 0.87fyAst x (d -0.416xu) = 86.9 kNm - - - (3)
Thus we complete Step B.
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 -18#, 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 18.4 mm
• The effective depth is equal to 323 mm
• The resistance it will be offering at it's ultimate state will be equal to 86.9 kNm
• It will belong to case 1
The above properties are the most probable properties because 3-18# is the most probable quantity of steel that the beam will require to resist the ultimate moment of 78.9 kNm. If we actually provide 3-18#, the capacity of the beam will be 86.9 kNm, found in (3). We do not need this much. We need only 78.9 kNm. So we want to know the exact steel which will give the capacity of 78.9 kNm.
For this we want to know the value of xu at the following condition:
• A moment of 78.9 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.4.10 which is used to calculate this xu.
Eq.4.10:
A X2 + B X + C = 0
Where A = 0.1506 bd2fck ; B = -0.362 bd2fck ; C = Mu and X = xu/d
• Change b to bf
Solving this quadratic equation, we get xu/d = 0.0516. So xu = 0.0516 x 323 =16.67 mm
Ast can be obtained from Eq.4.11
Eq.4.11
• Change b to bf
Thus we get Ast = 691.53 mm2. - - - (4)
This required value is a less than the assumed value of 763.41 mm2 (given by 3-#18). So we can safely provide 3 - #18.
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 beam section. We will now do the various checks:
We did the three checks which are to be done immediately after fixing up the preliminary dimensions here. In those checks, the first two will give the same results because the parameters related to them have not changed. The third check for 'slenderness' involves 'd' which has changed from 320 to 323 mm. We will get the same result (calculation steps are not shown here): It is not a slender beam. So we will proceed to the other checks:
(1) Spacing between the bars have already been checked here
(2) Check for minimum area of flexural reinforcement (Details here):
As = 0.85bd / fy
• Change b to bw
We get As = 152.16 mm2 < 763.41 mm2. Hence OK
(3) Check for maximum area of flexural reinforcement (Details here):
0.04bwD = 3404 mm2 > 763.41 mm2. Hence OK
(4) Check whether pt is less than pt,lim:
We have seen that Ast provided is less than Ast,lim. So pt will be less than pt,lim.
(5) Check for deflection control (Details here):
(l/d)actual ≤ [(l/d)basic] kt
For simply supported spans, (l/d)basic = 20
To find kt:
• Ast required = 691.53 mm2 from (4) • Ast,p = 763.41 mm2
• fst = 218.04 N/mm2 • pt = 1.028 • From Fig.4 of the code, kt = 1.07
• So l/d = 20 x kt = 21.4 • (l/d) actual = 4230/323 = 13.1
13.1 < 21.4 Hence OK
This completes all the checks on the designed beam section. The bars can be provided as we saw earlier in the fig.10.21
In the next section, we will see a new topic, which is 'The design of Doubly reinforced rectangular beam sections'.
PREVIOUS
