In the previous section, we derived the two equations which relate the extra bending moment ΔMu, with the force [Cus in the compression steel Asc] and the force [ΔTu in the extra tension steel ΔAst]:
Eq.11.4
ΔMu = (fsc - 0.447fck)Asc (d - d'). and
Eq.11.5
ΔMu = 0.87fy ΔAst (d - d')
ΔMu in the above equations is a known quantity. It is the extra moment capacity that the beam must acquire when it become 'doubly reinforced'. So we can obtain ΔMu using the following equation:
Eq.11.6
ΔMu = Mu - Mulim.
Thus in Eqs.11.4 & 11.5, Asc and ΔAst are the unknowns. We can use 11.5 to find ΔAst. But Asc should not be calculated using 11.4. We will see the reason for this in a short while.
First we will see ΔAst. Substituting 11.6 in 11.5 and rearranging, we get:
Eq.11.7
When we add this ΔAst to Ast,lim, we will get the total tension steel Ast provided in the beam section. So we can write:
Eq.11.8
Ast = Ast,lim + ΔAst
Once we calculate Ast we can fix the number and diameter of bars for all the bars on the 'tension side' of the beam. We must choose the number of bars and their diameters in such a way that, the total area obtained is as close as possible, and at the same time, a little greater than what is obtained from Eq.11.8.
When the tension bars are finalized, the d may differ from what we had assumed. In that case, Mulim, Ast,lim, and Ast should be recalculated. This is one of the reasons why we should not use Eq.11.4 for the calculation of Asc. The d that we use in it may change after we calculate Ast. The second reason is as follows:
We get a value for Ast from our calculations. Based on that value, we try different combinations of 'numbers and diameters' of bars to make up the Ast. We can provide only discrete number of bars to make up a certain Ast. If we use bundled bars, then each bar in a bundle should be discrete. For example, we can provide '3' sepatate bars of 20 mm bars giving an Ast of 942 mm2. We can not provide numbers like 31⁄2 or 31⁄4 or 31⁄8 of 20 mm bars. This is shown in the fig. below:
Fig.11.4
Fractions of bar area cannot be provided
Also only a certain diameters are available in the market. So it is clear that a combination giving the exact Ast as obtained in the calculations cannot be achieved. Out of all the combinations which give values close to Ast, some will be less than Ast, and others will be greater than Ast. We must never use those combinations which are less. Always select a combination which give an area greater than but close to Ast.
Based on the above discussion, we can now get a better understanding about the 'quantity' of ΔAst. That is., we subtract Ast,lim from Ast (the final Ast calculated from the actual bars provided) to get ΔAst actually provided, and it will be greater than the calculated ΔAst.
We can denote this ΔAst which is actually provided as: ΔAst(pd). We must ensure that this ΔAst(pd) is also self contained in itself. That is., there is equilibrium between the force in ΔAst(pd) and the force in Asc. So it is clear that it is the force in ΔAst(pd) that we must equate to the force in Asc, and not ΔAst.
So we can write: (fsc - 0.447fck)Asc = 0.87fy ΔAst(pd). Rearranging this we get:
Eq.11.9
Thus we obtained the equation for Asc. Now we have to discuss an important point about this Asc. As in the case of Ast, Asc also cannot be given the exact value obtained in 11.9. We must choose the combination which give an area greater than but close to Asc.
So in the final beam section we will be giving an Asc which is a little higher than what is obtained from Eq.11.9. So the force in the Asc will also be greater than what is obtained in the calculations. That is., we calculated the Asc in such a way that it will balance the force in ΔAst(pd). But now it is greater the calculated Asc. So the force will also be greater. Then what about the equilibrium? The answer is that, to get equilibrium at the ultimate state, the compressive force in concrete will be reduced by the upward shifting of NA, thus reducing concrete area. So the depth of NA will become less than xu,max. This will ensure that the steel Ast on the tension side will yield at the ultimate state, and thus, the beam section will indeed be an 'under reinforced' one.
This completes the discussion on the design procedure. We can now write the procedure as various steps in sequential order:
Step 1: Calculate Mulim. (Eq.3.26)
Step 2: Calculate Ast,lim. (Eq.3.29)
Step 3: Calculate ΔMu from the relation: Mu = Mulim + ΔMu
Step 4: Calculate ΔAst using Eq.11.7. Then Ast = Ast,lim + ΔAst
Step 5: Provide suitable combination of bars to obtain the above Ast. The area so obtained must be greater than but close to Ast.
Step 6: Check whether actual d provided is same as the assumed d. If there is change, the above steps should be repeated.
Step 7: Find the area of this combination of bars actually provided.
Step 8: Subtract Ast,lim from this to obtain ΔAst(pd)
Step 9: Calculate d'/d and use it to determine fsc from Table 11.1
Step 10: Use Eq.11.9 to calculate Asc
Step 11: Provide a suitable combination of bars which give an area greater than but close to Asc.
A doubly reinforced beam designed by the above process will have all the required properties:
• The ultimate moment of resistance MuR of the section will be greater than the applied factored moment Mu
• The depth of NA xu at the ultimate state will be less than xumax.
However we must always do an analysis of the newly designed doubly reinforced section, to verify that these conditions are satisfied. For this we must learn how to analyse a doubly reinforced rectangular beam section. This will be discussed in the next chapter.
At present we will look into the other essential aspects of the design procedure.
The basic concepts of the process of design of singly reinforced rectangular sections were discussed earlier. There, in the introduction part, we discussed about:
1.concrete cover
2.Minimum distance to be provided between the bars of beams
3.Maximum spacing allowable between bars of beams
4.Beams with overall depth greater than 750 mm
5.Minimum area of flexural reinforcements in beams
6.Maximum allowable area of flexural reinforcements in beams
7.Deflection control of singly reinforced beams and
8.Guide lines for fixing up the dimensions of beams
Links to each of the above items can be seen here.
All the above eight topics are applicable to Doubly reinforced beams also. However, some minor modifications have to be made to some of them. We will now look at the required modifications:
No.6: Maximum allowable area of flexural reinforcements in flanged beams:
We have seen earlier that, according to cl 26.5.1(b), area of the tension steel Ast provided should not be more than 4% of the total cross sectional area of the beam ie., Ast ≯ 0.04bwD The code specifies an allowable limit for the compression steel also. According to cl 26.5.1.2, area of compression steel provided should not be more than 4% of the total cross sectional area of the beam ie., Asc ≯ 0.04bwD
From the above, we can see that, if both and are provided in a beam at their maximum allowable limits, the area of steel will take up about 8% of the total cross sectional area of the beam. This is rather excessive. In such cases, the high steel areas should be avoided by using improved grades of steel and concrete. Increasing the size of the beam will also result in a lesser quantity of required steel.
No.7: Deflection control
We have seen the modification factors that have to be applied to the basic l/d ratio. There we mentioned that the modification factor kc will be discussed when we take up the design of doubly reinforced sections.
So we have to learn about this new factor kc. For this we look at cl. 23.2.1(d) of the code. According to this clause, we must calculate kc from fig. 5 of the code, and then the basic l/d ratio must be multiplied by this kc.
Another method to determine kc is to use the following formula given by SP 24:
Where pc is the percentage of compression reinforcement given by
Now the final equation for a doubly reinforced rectangular beam section can be written as follows:
Thus we can write the deflection control equations in the final form:
11.10
For doubly reinforced beams with span less than 10m,
(l/d)actual ≤ [(l/d)basic] kt kc
11.11
For doubly reinforced beams with span greater than 10m,
(l/d)actual ≤ [(l/d)basic] α kt kc
If the beam is a cantilever with span greater than 10 m, actual deflection calculations should be made.
No.8: Guide lines for fixing up the dimensions of beams
We have seen at the beginning of this chapter that, generally, we make the decision to make a beam 'doubly reinforced', when we find that the cross sectional dimensions of the beam cannot be increased further to provide the required bending moment capacity as a singly reinforced beam. This means that the dimensions are already fixed when we start to design it as a singly reinforced beam. So we do not have to learn any new methods to fix up the preliminary dimensions for doubly reinforced beams.
In the next section we will see a solved example.
PREVIOUS
Eq.11.4
ΔMu = (fsc - 0.447fck)Asc (d - d'). and
Eq.11.5
ΔMu = 0.87fy ΔAst (d - d')
ΔMu in the above equations is a known quantity. It is the extra moment capacity that the beam must acquire when it become 'doubly reinforced'. So we can obtain ΔMu using the following equation:
Eq.11.6
ΔMu = Mu - Mulim.
Thus in Eqs.11.4 & 11.5, Asc and ΔAst are the unknowns. We can use 11.5 to find ΔAst. But Asc should not be calculated using 11.4. We will see the reason for this in a short while.
First we will see ΔAst. Substituting 11.6 in 11.5 and rearranging, we get:
Eq.11.7
When we add this ΔAst to Ast,lim, we will get the total tension steel Ast provided in the beam section. So we can write:
Eq.11.8
Ast = Ast,lim + ΔAst
Once we calculate Ast we can fix the number and diameter of bars for all the bars on the 'tension side' of the beam. We must choose the number of bars and their diameters in such a way that, the total area obtained is as close as possible, and at the same time, a little greater than what is obtained from Eq.11.8.
When the tension bars are finalized, the d may differ from what we had assumed. In that case, Mulim, Ast,lim, and Ast should be recalculated. This is one of the reasons why we should not use Eq.11.4 for the calculation of Asc. The d that we use in it may change after we calculate Ast. The second reason is as follows:
We get a value for Ast from our calculations. Based on that value, we try different combinations of 'numbers and diameters' of bars to make up the Ast. We can provide only discrete number of bars to make up a certain Ast. If we use bundled bars, then each bar in a bundle should be discrete. For example, we can provide '3' sepatate bars of 20 mm bars giving an Ast of 942 mm2. We can not provide numbers like 31⁄2 or 31⁄4 or 31⁄8 of 20 mm bars. This is shown in the fig. below:
Fig.11.4
Fractions of bar area cannot be provided
Also only a certain diameters are available in the market. So it is clear that a combination giving the exact Ast as obtained in the calculations cannot be achieved. Out of all the combinations which give values close to Ast, some will be less than Ast, and others will be greater than Ast. We must never use those combinations which are less. Always select a combination which give an area greater than but close to Ast.
Based on the above discussion, we can now get a better understanding about the 'quantity' of ΔAst. That is., we subtract Ast,lim from Ast (the final Ast calculated from the actual bars provided) to get ΔAst actually provided, and it will be greater than the calculated ΔAst.
We can denote this ΔAst which is actually provided as: ΔAst(pd). We must ensure that this ΔAst(pd) is also self contained in itself. That is., there is equilibrium between the force in ΔAst(pd) and the force in Asc. So it is clear that it is the force in ΔAst(pd) that we must equate to the force in Asc, and not ΔAst.
So we can write: (fsc - 0.447fck)Asc = 0.87fy ΔAst(pd). Rearranging this we get:
Eq.11.9
Thus we obtained the equation for Asc. Now we have to discuss an important point about this Asc. As in the case of Ast, Asc also cannot be given the exact value obtained in 11.9. We must choose the combination which give an area greater than but close to Asc.
So in the final beam section we will be giving an Asc which is a little higher than what is obtained from Eq.11.9. So the force in the Asc will also be greater than what is obtained in the calculations. That is., we calculated the Asc in such a way that it will balance the force in ΔAst(pd). But now it is greater the calculated Asc. So the force will also be greater. Then what about the equilibrium? The answer is that, to get equilibrium at the ultimate state, the compressive force in concrete will be reduced by the upward shifting of NA, thus reducing concrete area. So the depth of NA will become less than xu,max. This will ensure that the steel Ast on the tension side will yield at the ultimate state, and thus, the beam section will indeed be an 'under reinforced' one.
This completes the discussion on the design procedure. We can now write the procedure as various steps in sequential order:
Step 1: Calculate Mulim. (Eq.3.26)
Step 2: Calculate Ast,lim. (Eq.3.29)
Step 3: Calculate ΔMu from the relation: Mu = Mulim + ΔMu
Step 4: Calculate ΔAst using Eq.11.7. Then Ast = Ast,lim + ΔAst
Step 5: Provide suitable combination of bars to obtain the above Ast. The area so obtained must be greater than but close to Ast.
Step 6: Check whether actual d provided is same as the assumed d. If there is change, the above steps should be repeated.
Step 7: Find the area of this combination of bars actually provided.
Step 8: Subtract Ast,lim from this to obtain ΔAst(pd)
Step 9: Calculate d'/d and use it to determine fsc from Table 11.1
Step 10: Use Eq.11.9 to calculate Asc
Step 11: Provide a suitable combination of bars which give an area greater than but close to Asc.
A doubly reinforced beam designed by the above process will have all the required properties:
• The ultimate moment of resistance MuR of the section will be greater than the applied factored moment Mu
• The depth of NA xu at the ultimate state will be less than xumax.
However we must always do an analysis of the newly designed doubly reinforced section, to verify that these conditions are satisfied. For this we must learn how to analyse a doubly reinforced rectangular beam section. This will be discussed in the next chapter.
At present we will look into the other essential aspects of the design procedure.
The basic concepts of the process of design of singly reinforced rectangular sections were discussed earlier. There, in the introduction part, we discussed about:
1.concrete cover
2.Minimum distance to be provided between the bars of beams
3.Maximum spacing allowable between bars of beams
4.Beams with overall depth greater than 750 mm
5.Minimum area of flexural reinforcements in beams
6.Maximum allowable area of flexural reinforcements in beams
7.Deflection control of singly reinforced beams and
8.Guide lines for fixing up the dimensions of beams
Links to each of the above items can be seen here.
All the above eight topics are applicable to Doubly reinforced beams also. However, some minor modifications have to be made to some of them. We will now look at the required modifications:
No.6: Maximum allowable area of flexural reinforcements in flanged beams:
We have seen earlier that, according to cl 26.5.1(b), area of the tension steel Ast provided should not be more than 4% of the total cross sectional area of the beam ie., Ast ≯ 0.04bwD The code specifies an allowable limit for the compression steel also. According to cl 26.5.1.2, area of compression steel provided should not be more than 4% of the total cross sectional area of the beam ie., Asc ≯ 0.04bwD
From the above, we can see that, if both and are provided in a beam at their maximum allowable limits, the area of steel will take up about 8% of the total cross sectional area of the beam. This is rather excessive. In such cases, the high steel areas should be avoided by using improved grades of steel and concrete. Increasing the size of the beam will also result in a lesser quantity of required steel.
No.7: Deflection control
We have seen the modification factors that have to be applied to the basic l/d ratio. There we mentioned that the modification factor kc will be discussed when we take up the design of doubly reinforced sections.
So we have to learn about this new factor kc. For this we look at cl. 23.2.1(d) of the code. According to this clause, we must calculate kc from fig. 5 of the code, and then the basic l/d ratio must be multiplied by this kc.
Another method to determine kc is to use the following formula given by SP 24:
Where pc is the percentage of compression reinforcement given by
Thus we can write the deflection control equations in the final form:
11.10
For doubly reinforced beams with span less than 10m,
(l/d)actual ≤ [(l/d)basic] kt kc
11.11
For doubly reinforced beams with span greater than 10m,
(l/d)actual ≤ [(l/d)basic] α kt kc
If the beam is a cantilever with span greater than 10 m, actual deflection calculations should be made.
No.8: Guide lines for fixing up the dimensions of beams
We have seen at the beginning of this chapter that, generally, we make the decision to make a beam 'doubly reinforced', when we find that the cross sectional dimensions of the beam cannot be increased further to provide the required bending moment capacity as a singly reinforced beam. This means that the dimensions are already fixed when we start to design it as a singly reinforced beam. So we do not have to learn any new methods to fix up the preliminary dimensions for doubly reinforced beams.
In the next section we will see a solved example.
PREVIOUS
