In the previous section, we completed the analysis and design of flanged sections. All the beam sections, rectangular or flanged, that we considered so far were 'Singly reinforced'. In this section we will discuss about 'Doubly reinforced sections'.
Suppose we are required to design a beam, spanning between given supports. The loads acting on the beam are also given. We know the procedure to design the beam. If the span is greater, and/or the loads are of greater magnitudes, we will have to provide a beam of higher strength. This increase in strength is usually achieved by increasing the depth and/or width.
It may be noted that the increase in strength can also be achieved by using higher grades of concrete and/or steel.
But there are certain situations where the increase in the depth and/or width is restricted. This restriction may be due to architectural considerations, or due to the requirement for more head room below the beam. In such situations, the strength of the beam can be increased by making it 'doubly reinforced'. They are called 'doubly reinforced' because, main steel bars are provided at both the top and bottom of the beam.
Doubly reinforced beams are also provided in situations where reversal of moments is likely to occur. For example, in a multi storeyed frame subjected to lateral loads, the clock wise moments in the members may change to anti clock wise moments, and vice versa, when the direction of lateral loads changes. When the moment is reversed, the position (top or bottom ) of the beam, where steel is required will also be reversed. So a suitably designed doubly reinforced beam, which has steel at both top and bottom, will be able to resist both clock wise moments and anti clock wise moments.
The compression steel also help to reduce long term deflection due to shrinkage.
Consider a beam which is subjected to sagging moment. That is., the beam is bending downwards. So tension steel is provided at the bottom portion of the beam. If this beam to be made 'doubly reinforced', we have to provide steel at the top portion. As the beam is subjected to sagging moment, we know that the top portion of this beam will be in compression. So the steel that we provide at the top will also be in compression. So the top steel is usually called as 'compression reinforcement'.
Now consider a beam section which is subjected to hogging moment. That is., at the section, the beam is bending upwards. (This happens at the interior supports of continuous beams. See video here.) So, at the section, the tension steel is provided at the top portion of the beam. If this beam to be made 'doubly reinforced', we have to provide steel at the bottom portion. As the beam section is subjected to hogging moment, we know that the bottom portion of this beam section will be in compression. So the steel that we provide at the bottom will also be in compression. So here, the bottom steel will be called as 'compression reinforcement'.
As these bars are in compression, there is a possibility for them to buckle. In order to prevent that, the compression steel should always be enclosed by closed stirrups. The closed stirrups helps also to confine the concrete properly, thus giving some ductility.
We are given the factored moment Mu. We have the maximum values of width b and effective depth d that can be provided. With this b and d, we assume a singly reinforced section. Now we calculate the Limiting moment of resistance Mulim. For this we have derived Eq.3.26 (same as the equation given in cl G 1.1(c) of the code).
Eq.3.26
We know that Mulim of a section is the 'full potential' that the section can offer. In other words, it is the maximum applied factored moment that the section can carry. If the moment is increased beyond that, the concrete in the extreme compression fibre will crush.
We know that corresponding to Mulim, there is a quantity of steel called Ast,lim. We derived Eq.3.29 for this.
Eq.3.29
If we provide this Ast,lim, to the section, the following conditions will be obtained at the ultimate state:
• When the beam section is at the point of impending failure (ie., at the ultimate state), it will be offering a resistance of Mulim
• At this ultimate state, the depth of NA will be xu,max (see 3.31)
• The tension steel would have yielded
• The strain in the extreme compression fibre will be 0.0035, and so it would be about to fail
• The compressive force Cu in concrete will be equal to the tensile force Tu in steel.
So Mulim is self contained in itself. The details of Mulim can be represented as given in the following fig.11.1:
Fig.11.1
Limiting Moment of resistance
We know how to calculate the limiting moment of resistance of the rectangular section based on the above fig. But we want the section to offer more than this Mulim. Because the load that we will apply on the beam will produce a bending moment greater than Mulim.
It is clear that if we apply a moment greater than Mulim, the section will fail by the crushing of concrete in the extreme fibre. To prevent this crushing, we can give some steel in the compression zone, so that, some of the compressive force will be taken up by the steel. In that case the concrete will not experience the heavy crushing force, and thus, it's failure can be prevented. This will mean that we can then apply more load at the beam section.
But it is not that simple. We have to do some calculations:
• The exact position at which this new steel is to be provided should be known.
• The quantity required for this new steel should be known.
• The extra compressive force in the new steel should be balanced by some tensile force so that equilibrium is maintained.
To do the calculations, it is convenient to make a 'separation'. A separation into two parts of the quantity: 'The resisting moment'.
The final doubly reinforced section will be having a capacity to resist the large bending moment that will be applied on it. We call this capacity as the 'Ultimate moment of resistance' MuR. We are going to separate this capacity into two parts.
• First part is the 'full potential' that we can extract from the section if the section is not 'doubly reinforced'. That is., the first part is the Mulim. It is self contained in itself. So we can calculate it and keep it aside.
• The second part is the extra capacity (denoted as ΔMu) that the beam section gets, by giving some steel in the compression zone. We will treat this as a separate part and do the calculations. At the end of those calculations, this part will also become self contained in itself.
When we add the two parts, we will get the total capacity MuR of the doubly reinforced section. Thus we can write
MuR = Mulim + ΔMu
This MuR should be 'greater than or equal to' the applied external factored moment Mu at the section. Note the inequality 'greater than or equal to' in this statement. When we design a new beam, we first equate MuR to Mu. After we obtain the complete designed section, we do an analysis and ensure that MuR is greater than Mu. Thus the inequality will be satisfied.
The calculations related to the second part can be done using the fig. given below:
Fig.11.2
Extra Moment capacity acquired
The fig. shows the stresses, strains, and the resultant forces, at the ultimate state. The new steel given in the compression zone have an area of Asc. We want the force in that steel. For that, we want the stress in it. The stress calculation is some what lengthy. We will see it step by step:
Step 1: In this step, we calculate the strain (denoted as εsc ) in Asc. For this we use the strain diagram. Strain diagram is a straight line because plane sections remain plane even after bending. (See fig.3.6). Consider the portion above the NA, in the strain diagram. This portion has two triangles. The following points will help us to distinguish between the two triangles:
• altitude of the larger triangle = xu,max
• base of the larger triangle = 0.0035
• altitude of the smaller triangle = (xu,max - d') d' gives us the position of the compression steel. It is the distance of the c.g (of the group of bars constituting the Asc) from the top most compression fibre.
• base of the smaller triangle = εsc
These two triangles are similar. So we get:
Rearranging this we get:
Eq.11.1:
Thus we get the strain in the compression steel. With this, we can calculate fsc , the stress. We will see it in the next section.
PREVIOUS
Suppose we are required to design a beam, spanning between given supports. The loads acting on the beam are also given. We know the procedure to design the beam. If the span is greater, and/or the loads are of greater magnitudes, we will have to provide a beam of higher strength. This increase in strength is usually achieved by increasing the depth and/or width.
It may be noted that the increase in strength can also be achieved by using higher grades of concrete and/or steel.
But there are certain situations where the increase in the depth and/or width is restricted. This restriction may be due to architectural considerations, or due to the requirement for more head room below the beam. In such situations, the strength of the beam can be increased by making it 'doubly reinforced'. They are called 'doubly reinforced' because, main steel bars are provided at both the top and bottom of the beam.
Doubly reinforced beams are also provided in situations where reversal of moments is likely to occur. For example, in a multi storeyed frame subjected to lateral loads, the clock wise moments in the members may change to anti clock wise moments, and vice versa, when the direction of lateral loads changes. When the moment is reversed, the position (top or bottom ) of the beam, where steel is required will also be reversed. So a suitably designed doubly reinforced beam, which has steel at both top and bottom, will be able to resist both clock wise moments and anti clock wise moments.
The compression steel also help to reduce long term deflection due to shrinkage.
Consider a beam which is subjected to sagging moment. That is., the beam is bending downwards. So tension steel is provided at the bottom portion of the beam. If this beam to be made 'doubly reinforced', we have to provide steel at the top portion. As the beam is subjected to sagging moment, we know that the top portion of this beam will be in compression. So the steel that we provide at the top will also be in compression. So the top steel is usually called as 'compression reinforcement'.
Now consider a beam section which is subjected to hogging moment. That is., at the section, the beam is bending upwards. (This happens at the interior supports of continuous beams. See video here.) So, at the section, the tension steel is provided at the top portion of the beam. If this beam to be made 'doubly reinforced', we have to provide steel at the bottom portion. As the beam section is subjected to hogging moment, we know that the bottom portion of this beam section will be in compression. So the steel that we provide at the bottom will also be in compression. So here, the bottom steel will be called as 'compression reinforcement'.
As these bars are in compression, there is a possibility for them to buckle. In order to prevent that, the compression steel should always be enclosed by closed stirrups. The closed stirrups helps also to confine the concrete properly, thus giving some ductility.
How to decide whether to use a Singly reinforced or Doubly reinforced section?
We start out to design a singly reinforced beam as usual. We know that, during the design process, we reach a point, where we calculate the effective depth d required. At this point, we may find that the required depth cannot be provided because of some restrictions, and then we decide to make it a doubly reinforced beam. So at the point where we make this decision, the width and depth of the beam cross section are already known. This means that, in the discussion about 'doubly reinforced beams' we do not have to discuss about 'obtaining the dimensions of the beam section'. The discussion will be related to the 'calculations of steel' only.We are given the factored moment Mu. We have the maximum values of width b and effective depth d that can be provided. With this b and d, we assume a singly reinforced section. Now we calculate the Limiting moment of resistance Mulim. For this we have derived Eq.3.26 (same as the equation given in cl G 1.1(c) of the code).
Eq.3.26
We know that Mulim of a section is the 'full potential' that the section can offer. In other words, it is the maximum applied factored moment that the section can carry. If the moment is increased beyond that, the concrete in the extreme compression fibre will crush.
We know that corresponding to Mulim, there is a quantity of steel called Ast,lim. We derived Eq.3.29 for this.
Eq.3.29
If we provide this Ast,lim, to the section, the following conditions will be obtained at the ultimate state:
• When the beam section is at the point of impending failure (ie., at the ultimate state), it will be offering a resistance of Mulim
• At this ultimate state, the depth of NA will be xu,max (see 3.31)
• The tension steel would have yielded
• The strain in the extreme compression fibre will be 0.0035, and so it would be about to fail
• The compressive force Cu in concrete will be equal to the tensile force Tu in steel.
So Mulim is self contained in itself. The details of Mulim can be represented as given in the following fig.11.1:
Fig.11.1
Limiting Moment of resistance
We know how to calculate the limiting moment of resistance of the rectangular section based on the above fig. But we want the section to offer more than this Mulim. Because the load that we will apply on the beam will produce a bending moment greater than Mulim.
It is clear that if we apply a moment greater than Mulim, the section will fail by the crushing of concrete in the extreme fibre. To prevent this crushing, we can give some steel in the compression zone, so that, some of the compressive force will be taken up by the steel. In that case the concrete will not experience the heavy crushing force, and thus, it's failure can be prevented. This will mean that we can then apply more load at the beam section.
But it is not that simple. We have to do some calculations:
• The exact position at which this new steel is to be provided should be known.
• The quantity required for this new steel should be known.
• The extra compressive force in the new steel should be balanced by some tensile force so that equilibrium is maintained.
To do the calculations, it is convenient to make a 'separation'. A separation into two parts of the quantity: 'The resisting moment'.
The final doubly reinforced section will be having a capacity to resist the large bending moment that will be applied on it. We call this capacity as the 'Ultimate moment of resistance' MuR. We are going to separate this capacity into two parts.
• First part is the 'full potential' that we can extract from the section if the section is not 'doubly reinforced'. That is., the first part is the Mulim. It is self contained in itself. So we can calculate it and keep it aside.
• The second part is the extra capacity (denoted as ΔMu) that the beam section gets, by giving some steel in the compression zone. We will treat this as a separate part and do the calculations. At the end of those calculations, this part will also become self contained in itself.
When we add the two parts, we will get the total capacity MuR of the doubly reinforced section. Thus we can write
MuR = Mulim + ΔMu
This MuR should be 'greater than or equal to' the applied external factored moment Mu at the section. Note the inequality 'greater than or equal to' in this statement. When we design a new beam, we first equate MuR to Mu. After we obtain the complete designed section, we do an analysis and ensure that MuR is greater than Mu. Thus the inequality will be satisfied.
The calculations related to the second part can be done using the fig. given below:
Fig.11.2
Extra Moment capacity acquired
Step 1: In this step, we calculate the strain (denoted as εsc ) in Asc. For this we use the strain diagram. Strain diagram is a straight line because plane sections remain plane even after bending. (See fig.3.6). Consider the portion above the NA, in the strain diagram. This portion has two triangles. The following points will help us to distinguish between the two triangles:
• altitude of the larger triangle = xu,max
• base of the larger triangle = 0.0035
• altitude of the smaller triangle = (xu,max - d') d' gives us the position of the compression steel. It is the distance of the c.g (of the group of bars constituting the Asc) from the top most compression fibre.
• base of the smaller triangle = εsc
These two triangles are similar. So we get:
Rearranging this we get:
Eq.11.1:
Thus we get the strain in the compression steel. With this, we can calculate fsc , the stress. We will see it in the next section.
PREVIOUS
No comments:
Post a Comment