In
the previous section we completed the arrangement of bars in the continuous beam, and did the required checks. In this section we will see the curtailment details.
Curtailment of bars
As pointed out earlier, a detailed discussion about 'Development length and curtailment' can be seen here. For our present case, we will be using the recommendations given in SP16. It must be noted that, to use these recommendations, the analysis of the continuous member should be done using the 'method of coefficients'. We have indeed used it in the analysis of our slab, and the results thus obtained were used in the design.
We have already seen that the BM progressively decreases while we move away from a particular section ( support section as well as midspan section). So the steel can also be decreased at greater distances away from the concerned sections.
We will first see the curtailment of top bars at supports. We will take support B as an example. The fig.8.25 given below shows the details.
Fig.8.25
Curtailment of top bars
In the above fig., we can see three types of bars. The top most bars travel uninterrupted. The bottom most layer is curtailed at a distance of 0.15l (from the face of the support) on either sides of the support. So they have a length of 0.15l1 + 0.15l2 + width of the support. This is the most important portion. Just the availability of a length of '0.15l1 + 0.15l2 + width of the support' is not good enough. We must ensure that 0.15 times the respective spans is available on both sides. All the steel (shown in the fig. as Ast) required to resist the hogging moment at this support must be completely present within this distance.
Beyond this 0.15l distance, this full capacity is not required. So the bottom most layer is curtailed at this point. The remaining bars continue their journey. The total area of cross section of these continuing bars should be greater than or equal to 60% of Ast.
The second stage of curtailment is applied to these continuing bars. Thus we see that the middle layer is curtailed at a distance of 0.25l (from the face of the support) on either sides of the support. So they have a length of 0.25l1 + 0.25l2 + width of the support. As in the case of the bottom most layer, the availability of a length of '0.25l1 + 0.25l2 + width of the support' is not good enough. We must ensure that 0.25 times the respective spans is available on both sides. After this curtailment, the remaining bars will and should continue their journey uninterrupted. And the total area of cross section of these continuing bars should be greater than or equal to 20% of Ast.
The above fig.8.25 is applicable where a large quantity of steel is provided at the support. In such cases, the steel will be provided in different layers as in the fig. We can progressively curtail the various layers, while ensuring that the area requirements and length requirements are satisfied.
In our problem, there is only one layer. And there are only three bars (two at the sides and one in the middle) in that layer. Among these three, the two bars at the sides should continue uninterrupted from one end support to the other end support. This is for acting as 'stirrup suspenders'. But we can curtail the middle bar. Because, the two remaining bar will give an area greater than 0.2Ast. This is shown in the steps below:
• Ast (3-#16) = 603.19 mm2
• 2-#16 gives 402.12 mm2
• This is greater than 0.2Ast (0.2 x 603.19 =120.64 mm2)
So the curtailment at support B in our beam is as shown in fig.8.22 of the previous section.
Next we will see the curtailment of bottom bars in the end span AB. Fig.8.26 below gives the details:
Fig.8.26
Bottom bars in span AB
We can see two types of bars. The green colored bars are curtailed at a distance of 0.1l1 from the end support and 0.15l1 from the intermediate support. Both these distances are measured from the center lines of the respective supports. So there is a distance of l1 -0.1l1 -0.15l1 =0.75l1 at the midspan region. This is the 'important distance' as far as the sagging moment in the left end span of a continuous beam is concerned. (In the right end span, a mirror image is applicable). Within this distance, no curtailment is allowed. All bars required to resist the sagging moment should be completely present within this distance.
After curtailment at the above mentioned points, the continuing bars should compulsorily extend into the supports on either sides. Also, these continuing bars should have an area greater than or equal to 30% of the steel at midspan.
In our problem, there is only one layer. And there are only three bars (two at the sides and one in the middle) in that layer. Among these three, the two bars at the sides should continue uninterrupted from one end support to the other end support. This is for holding the stirrups. But we can curtail the middle bar. Because, the two remaining bar will give an area greater than 0.3Ast. This is shown in the steps below:
• Ast (3-#16) = 603.19 mm2
• 2-#16 gives 402.12 mm2
• This is greater than 0.3Ast (0.3 x 603.19 =180.96 mm2)
So the curtailment in midspan AB in our beam is as shown in fig.8.21 of the previous section.
Next we will see the curtailment of bottom bars in the intermediate span BC. Fig.8.27 below gives the details:
Fig.8.27
Bottom bars in span BC
This is similar to the curtailment in an end span. The only difference is that the point of curtailment is at a distance of 0.15l from both supports. All other details are the same.
In our problem, there are only 2 bars in the span BC. These bars are compulsorily required through out the length of ABCDE, for holding the stirrups. So the problem of curtailment of any bars does not arise in this span of our problem.
This completes the discussion about curtailment of bars in continuous beams when coefficients are used for calculating BM and SF. With this we have completed the discussion on the design of continuous slabs and beams. In the next chapter we will discuss about the analysis and design of flanged sections.
Curtailment of bars
As pointed out earlier, a detailed discussion about 'Development length and curtailment' can be seen here. For our present case, we will be using the recommendations given in SP16. It must be noted that, to use these recommendations, the analysis of the continuous member should be done using the 'method of coefficients'. We have indeed used it in the analysis of our slab, and the results thus obtained were used in the design.
We have already seen that the BM progressively decreases while we move away from a particular section ( support section as well as midspan section). So the steel can also be decreased at greater distances away from the concerned sections.
We will first see the curtailment of top bars at supports. We will take support B as an example. The fig.8.25 given below shows the details.
Fig.8.25
Curtailment of top bars
Beyond this 0.15l distance, this full capacity is not required. So the bottom most layer is curtailed at this point. The remaining bars continue their journey. The total area of cross section of these continuing bars should be greater than or equal to 60% of Ast.
The second stage of curtailment is applied to these continuing bars. Thus we see that the middle layer is curtailed at a distance of 0.25l (from the face of the support) on either sides of the support. So they have a length of 0.25l1 + 0.25l2 + width of the support. As in the case of the bottom most layer, the availability of a length of '0.25l1 + 0.25l2 + width of the support' is not good enough. We must ensure that 0.25 times the respective spans is available on both sides. After this curtailment, the remaining bars will and should continue their journey uninterrupted. And the total area of cross section of these continuing bars should be greater than or equal to 20% of Ast.
The above fig.8.25 is applicable where a large quantity of steel is provided at the support. In such cases, the steel will be provided in different layers as in the fig. We can progressively curtail the various layers, while ensuring that the area requirements and length requirements are satisfied.
In our problem, there is only one layer. And there are only three bars (two at the sides and one in the middle) in that layer. Among these three, the two bars at the sides should continue uninterrupted from one end support to the other end support. This is for acting as 'stirrup suspenders'. But we can curtail the middle bar. Because, the two remaining bar will give an area greater than 0.2Ast. This is shown in the steps below:
• Ast (3-#16) = 603.19 mm2
• 2-#16 gives 402.12 mm2
• This is greater than 0.2Ast (0.2 x 603.19 =120.64 mm2)
So the curtailment at support B in our beam is as shown in fig.8.22 of the previous section.
Next we will see the curtailment of bottom bars in the end span AB. Fig.8.26 below gives the details:
Fig.8.26
Bottom bars in span AB
We can see two types of bars. The green colored bars are curtailed at a distance of 0.1l1 from the end support and 0.15l1 from the intermediate support. Both these distances are measured from the center lines of the respective supports. So there is a distance of l1 -0.1l1 -0.15l1 =0.75l1 at the midspan region. This is the 'important distance' as far as the sagging moment in the left end span of a continuous beam is concerned. (In the right end span, a mirror image is applicable). Within this distance, no curtailment is allowed. All bars required to resist the sagging moment should be completely present within this distance.
After curtailment at the above mentioned points, the continuing bars should compulsorily extend into the supports on either sides. Also, these continuing bars should have an area greater than or equal to 30% of the steel at midspan.
In our problem, there is only one layer. And there are only three bars (two at the sides and one in the middle) in that layer. Among these three, the two bars at the sides should continue uninterrupted from one end support to the other end support. This is for holding the stirrups. But we can curtail the middle bar. Because, the two remaining bar will give an area greater than 0.3Ast. This is shown in the steps below:
• Ast (3-#16) = 603.19 mm2
• 2-#16 gives 402.12 mm2
• This is greater than 0.3Ast (0.3 x 603.19 =180.96 mm2)
So the curtailment in midspan AB in our beam is as shown in fig.8.21 of the previous section.
Next we will see the curtailment of bottom bars in the intermediate span BC. Fig.8.27 below gives the details:
Fig.8.27
Bottom bars in span BC
This is similar to the curtailment in an end span. The only difference is that the point of curtailment is at a distance of 0.15l from both supports. All other details are the same.
In our problem, there are only 2 bars in the span BC. These bars are compulsorily required through out the length of ABCDE, for holding the stirrups. So the problem of curtailment of any bars does not arise in this span of our problem.
This completes the discussion about curtailment of bars in continuous beams when coefficients are used for calculating BM and SF. With this we have completed the discussion on the design of continuous slabs and beams. In the next chapter we will discuss about the analysis and design of flanged sections.
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