In
the previous section we saw how to fix up the preliminary dimensions of the beam section. We also saw the checks that have to be done immediately after fixing up these dimensions.
We can now take up the actual design of the beam. After fixing up the preliminary cross sectional dimensions, we can do the structural analysis, and calculate the total factored BM that the beam will have to resist. This factored BM is denoted as Mu. The beam section that we provide will have a unique value of 'resisting moment' that it can offer at ultimate state. We have learned about it in the previous section. It is called the 'Ultimate Moment of Resistance' MuR of the section. So MuR should be greater than or equal to Mu. Thus we can write one condition that the design should satisfy:
4.7: MuR ≥ Mu
Another condition that the design should satisfy is that, the beam that we design should be under reinforced. This is to ensure that , if the beam is over loaded and fails, the failure will be a 'tension failure'. So there will be enough warning about the failure. We know that in an under reinforced section, xu, the depth of NA will be less than xu,max. So we can write the second condition:
4.8: xu < xu,max
We can write the steps for designing a beam which satisfies the above conditions:
Step 2: If we make a section with width = b and effective depth =d obtained from eq.4.9, we will be having a section whose Mu,lim = Mu. Effective depth can never be less than the d that is obtained from eq.4.9, because, then the Mu,lim will be less than Mu. So in this step we give a 'final' form to our section. For this we assume a diameter Φ for the reinforcing bars, and a diameter Φl for the links. Then we will get
Eq.4.5 (based on fig.4.8)
D = d + Φ/2 + Φl + Cc
Another easier method is to use Table 4.2. By using that table, we are assuming that Φ = 20mm and Φl = 10mm. Then
D = d + effective cover
This D should be less than the preliminary value of D that we have used earlier. This is shown in the fig.4.9 below:
Fig.4.9
Calculated required depth should be less than the preliminary depth
If
the calculated value is larger, then it is obvious that the section with the
preliminary dimensions will not be capable to resist Mu. Then we must
go back and do the structural analysis again with new improved
preliminary values of cross sectional dimensions. If
the required value is lesser as shown in the fig.4.9 above, then we can discard the required value and
finalize the section with the preliminary values. In this situation, the 'preliminary dimensions' will hereafter be called the 'Final dimensions' of the beam section
We can think in this way: Suppose we give a certain area of steel. Corresponding to this area of steel and the final values of b and d, the beam will have a certain value of MuR. This MuR should be greater than or equal to Mu. So one condition is that: the area of steel which we are going to provide should enable the section to attain an MuR value which is greater than or equal to Mu. (condition written in 4.7 above)
We have one more condition. The final section should be under reinforced. This is because, if over loading occurs, and the beam fails, the failure should be a 'tension failure'. So we have a second condition: xu of the section should be less than xu,max. (condition written in 4.8 above)
With these two conditions, we can calculate the required area of steel. Let us see how this is done:
Let us simplify it further:
We can see that it is a quadratic equation. We can write it as:
Eq.4.10:
A X2 + B X + C = 0
Where A = 0.1506 bd2fck ; B = -0.362 bd2fck ; C = Mu and X = xu/d
A, B and C contain only the known values b, d, fck and Mu. So if we are using a spread sheet program, it will calculate A, B and C just when the input data are given. The program will then directly give the two solutions of the quadratic equation. That is., we will get two values for xu/d. Out of these, one will be greater than 1, and the other will be less than 1. We must choose the lesser value because xu is always less than d. If we choose the value which is greater than 1, when we multiply it with d, we will get a value for xu which is greater than d.
This step comes to an end with the calculation of xu . So step 3 can be summarized as:
• Calculate the values of A, B and C and then solve xu
Step 4: With the calculation of xu, we can split the section into an upper part and a lower part. The compressive force in the upper part is given by the same eq.3.7 that we saw in step 3 above:
Eq.3.7:
Cu = 0.362 fck b xu
The tensile force in the steel steel in the lower part is given by :
Tu = 0.87 fy Ast
Where Ast is the area of steel. The stress in steel is equal to 0.87 fy because, it is an under reinforced section, and so the steel would have yielded at the ultimate state.
For equilibrium, Cu = Tu. So, equating the two, we get
Eq.4.11:
So step 4 can be summarized as:
• Calculate Ast the area of steel by using Eq.4.11
The area of steel calculated above is the "Area of steel required". For beams, this has to be expressed in the form of Diameter of bars and number of bars of that particular diameter.
The
following table 4.3 gives the areas of standard bar sizes:
For any diameter Ф, the area Ab of the bar can be obtained as
Ab = πФ2/4
So for a chosen bar diameter, the number of bars required to provide the area of tension steel will be given by Ast / Ab . This ratio must be rounded off to the nearest higher whole number. In some cases, it may be economical to make a combination of two different bar diameters (close to each other) in order to get an area which is just higher than the area Ast.
After fixing the number and diameter of bars, we must calculate Ast,p which is the actual area of tension steel provided in the beam. Ast,p should always be greater than Ast.
This completes the design part. Now we have to learn some details like the 'rules for distributing the calculated steel in the beam'. Also we have to see the details about the checks that have to be done after completing the design part. This includes 'minimum steel that has to be provided in a beam section', 'Check for deflection' etc.,. We will learn about them in the next section.
We can now take up the actual design of the beam. After fixing up the preliminary cross sectional dimensions, we can do the structural analysis, and calculate the total factored BM that the beam will have to resist. This factored BM is denoted as Mu. The beam section that we provide will have a unique value of 'resisting moment' that it can offer at ultimate state. We have learned about it in the previous section. It is called the 'Ultimate Moment of Resistance' MuR of the section. So MuR should be greater than or equal to Mu. Thus we can write one condition that the design should satisfy:
4.7: MuR ≥ Mu
Another condition that the design should satisfy is that, the beam that we design should be under reinforced. This is to ensure that , if the beam is over loaded and fails, the failure will be a 'tension failure'. So there will be enough warning about the failure. We know that in an under reinforced section, xu, the depth of NA will be less than xu,max. So we can write the second condition:
4.8: xu < xu,max
We can write the steps for designing a beam which satisfies the above conditions:
Step
1: We want a beam section which will 'work' at it's 'full potential' to resist Mu. We know how to calculate the 'full potential' of a
section. It is called the 'Limiting Moment of Resistance' Mu,lim. We
already have a section with preliminary cross sectional dimensions.
Let us forget the depth D of that section for a while. We will need D in Step 2. But for now, we are discarding it. But we need it's
b. We want a section with the same width b, and whose Mu,lim is equal
to Mu. So we write the Mu,lim of a section with width b:
Eq.3.26:
Eq.3.26:
Let us put the quantity on the right side equal to 'K'. So we get
Mu,lim = Kbd2
Now we equate Mu,lim to Mu:
Mu,lim = Mu = Kbd2
Mu,lim = Kbd2
Now we equate Mu,lim to Mu:
Mu,lim = Mu = Kbd2
In
this equation, the only unknown is d. We can bring it to the left side and write:
Eq.4.9:
So Step 1 can be summarized as :
Eq.4.9:
So Step 1 can be summarized as :
• Write Eq.4.9 and get the effective depth d of the section whose width = b and Mu,lim = Mu.
Step 2: If we make a section with width = b and effective depth =d obtained from eq.4.9, we will be having a section whose Mu,lim = Mu. Effective depth can never be less than the d that is obtained from eq.4.9, because, then the Mu,lim will be less than Mu. So in this step we give a 'final' form to our section. For this we assume a diameter Φ for the reinforcing bars, and a diameter Φl for the links. Then we will get
Eq.4.5 (based on fig.4.8)
D = d + Φ/2 + Φl + Cc
Another easier method is to use Table 4.2. By using that table, we are assuming that Φ = 20mm and Φl = 10mm. Then
D = d + effective cover
This D should be less than the preliminary value of D that we have used earlier. This is shown in the fig.4.9 below:
Fig.4.9
Calculated required depth should be less than the preliminary depth
When
we finalize the section in this way, one important point should be
noted. The finalized D is greater than the required D obtained from eq.4.5 above. So the
final d will also be greater than that calculated using.4.9. So we
must calculate the final d as:
Eq.4.6:
d = D - (Φ/2 + Φl + Cc)
Eq.4.6:
d = D - (Φ/2 + Φl + Cc)
It
is very important to make sure that the final d calculated using eq.4.6 above is greater than d calculated in step 1. Because if it is less,
it would mean that the final section will have a Mu,lim less than Mu.
So
step 2 can be summarized as:
• Using d calculated in step 1, assumed diameters Φ and Φl , find D using eq.4.5
• If
this D is greater than the preliminary value, go back and do the
structural analysis again with new improved values.
• If
this D is less than the preliminary value, finalize the section with
the preliminary values, and calculate final d using eq.4.6
• Make
sure that new d is greater than that calculated in step 1
Step
3: We started off with the calculation of d in step 1. If we give
this exact d, then we will be having a section whose Mu,lim is exactly
equal to Mu. Then the area of steel will be equal to Ast,lim which
corresponds to Mu,lim. But dimensions have changed from those in step1.
We now have a new d, and a section with finalized values for it's cross section. We have to complete the design by giving it the required steel. How can we know the area of this required steel?
We now have a new d, and a section with finalized values for it's cross section. We have to complete the design by giving it the required steel. How can we know the area of this required steel?
We can think in this way: Suppose we give a certain area of steel. Corresponding to this area of steel and the final values of b and d, the beam will have a certain value of MuR. This MuR should be greater than or equal to Mu. So one condition is that: the area of steel which we are going to provide should enable the section to attain an MuR value which is greater than or equal to Mu. (condition written in 4.7 above)
We have one more condition. The final section should be under reinforced. This is because, if over loading occurs, and the beam fails, the failure should be a 'tension failure'. So we have a second condition: xu of the section should be less than xu,max. (condition written in 4.8 above)
With these two conditions, we can calculate the required area of steel. Let us see how this is done:
Lever
arm (Eq.3.8) = z= d - 0.416 xu
So
the moment that the section will offer at ultimate state =
MuR = 0.362 fck b xu (d - 0.416 xu)
This moment should be equal to Mu.
MuR = 0.362 fck b xu (d - 0.416 xu)
This moment should be equal to Mu.
So
we can write:
Mu = 0.362 fck b xu (d - 0.416 xu)
Let
us rearrange this so that it takes a 'familiar' form:
Let us simplify it further:
We can see that it is a quadratic equation. We can write it as:
Eq.4.10:
A X2 + B X + C = 0
Where A = 0.1506 bd2fck ; B = -0.362 bd2fck ; C = Mu and X = xu/d
A, B and C contain only the known values b, d, fck and Mu. So if we are using a spread sheet program, it will calculate A, B and C just when the input data are given. The program will then directly give the two solutions of the quadratic equation. That is., we will get two values for xu/d. Out of these, one will be greater than 1, and the other will be less than 1. We must choose the lesser value because xu is always less than d. If we choose the value which is greater than 1, when we multiply it with d, we will get a value for xu which is greater than d.
This step comes to an end with the calculation of xu . So step 3 can be summarized as:
• Calculate the values of A, B and C and then solve xu
Step 4: With the calculation of xu, we can split the section into an upper part and a lower part. The compressive force in the upper part is given by the same eq.3.7 that we saw in step 3 above:
Eq.3.7:
Cu = 0.362 fck b xu
The tensile force in the steel steel in the lower part is given by :
Tu = 0.87 fy Ast
Where Ast is the area of steel. The stress in steel is equal to 0.87 fy because, it is an under reinforced section, and so the steel would have yielded at the ultimate state.
For equilibrium, Cu = Tu. So, equating the two, we get
Eq.4.11:
So step 4 can be summarized as:
• Calculate Ast the area of steel by using Eq.4.11
Fixing the number of bars and their diameters
The area of steel calculated above is the "Area of steel required". For beams, this has to be expressed in the form of Diameter of bars and number of bars of that particular diameter.
Table 4.3
For any diameter Ф, the area Ab of the bar can be obtained as
Ab = πФ2/4
So for a chosen bar diameter, the number of bars required to provide the area of tension steel will be given by Ast / Ab . This ratio must be rounded off to the nearest higher whole number. In some cases, it may be economical to make a combination of two different bar diameters (close to each other) in order to get an area which is just higher than the area Ast.
After fixing the number and diameter of bars, we must calculate Ast,p which is the actual area of tension steel provided in the beam. Ast,p should always be greater than Ast.
This completes the design part. Now we have to learn some details like the 'rules for distributing the calculated steel in the beam'. Also we have to see the details about the checks that have to be done after completing the design part. This includes 'minimum steel that has to be provided in a beam section', 'Check for deflection' etc.,. We will learn about them in the next section.
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