Distribution bars
In the previous section we completed the layout of main bars of the slab ABCD. In this section we will design the distribution bars. We know that the quantity of distribution bars required depends only on Ag, the gross area of cross section of the slab, which is obtained as Ag = b xD = 1000mm x D. So even when our continuous slab has many spans, the area of distribution bars required will be the same for all spans. This is because, all spans in our slab has the same depth D, which is equal to 200mm.
Let us use #8 of Fe415 grade steel. So the area required = 0.0012Ag = 0.0012 x bD = 0.0012 x1000 x200 = 240mm2
Now, spacing required can be determined using Eq.5.3
Where Φ= 8mm and Ast = 240mm2
Thus we get spacing s =209.44mm. Let us provide #8 @200 mm c/c
The actual area provided is given by Eq.5.2
Where s is the actual spacing provided, which is equal to 200mm. So we get Ast provided =251.33mm2
Maximum spacing allowable between distribution bars of the slab :
(cl 26.3.3.b(2) of the code)
According to this clause, the spacing should not be more than the smallest of the following:
1. Five times the effective depth of the slab = 5 x d = 5 x165 =825mm
2. 450mm
So the spacing should not be more than 450mm. Thus the spacing of 200mm is OK
Check whether the minimum required distribution steel is provided:
The spacing of 200mm which is actually provided, is less than the required spacing of 209.44mm. So the area will be greater than the minimum required.
Thus we have completed the design of distribution bars also. We can now do the various checks.
In the previous section we completed the layout of main bars of the slab ABCD. In this section we will design the distribution bars. We know that the quantity of distribution bars required depends only on Ag, the gross area of cross section of the slab, which is obtained as Ag = b xD = 1000mm x D. So even when our continuous slab has many spans, the area of distribution bars required will be the same for all spans. This is because, all spans in our slab has the same depth D, which is equal to 200mm.
Let us use #8 of Fe415 grade steel. So the area required = 0.0012Ag = 0.0012 x bD = 0.0012 x1000 x200 = 240mm2
Now, spacing required can be determined using Eq.5.3
Thus we get spacing s =209.44mm. Let us provide #8 @200 mm c/c
The actual area provided is given by Eq.5.2
Maximum spacing allowable between distribution bars of the slab :
(cl 26.3.3.b(2) of the code)
According to this clause, the spacing should not be more than the smallest of the following:
1. Five times the effective depth of the slab = 5 x d = 5 x165 =825mm
2. 450mm
So the spacing should not be more than 450mm. Thus the spacing of 200mm is OK
Check whether the minimum required distribution steel is provided:
The spacing of 200mm which is actually provided, is less than the required spacing of 209.44mm. So the area will be greater than the minimum required.
Thus we have completed the design of distribution bars also. We can now do the various checks.
The pdf file given below gives the detailed steps involved in the various checks:
Solved example 8.1 - final checks
Curtailment of bars
Now we can discuss about 'curtailment of bars' in our slab. We know that in a continuous beam or slab, there will be hogging moment at support, and sagging moment at midspan. This can be seen in the BM diagram of a continuous member.
We earlier saw the BM diagram for our present slab in fig.7.5. On either side of the support, the hogging moment is decreasing progressively. That is., at greater distances away from the support, there will be lesser hogging moments.
We have designed the top steel at supports for the full design moment. But now we see that at sections away from the support, the BM decreases, and so the full steel is not necessary. So we can reduce the steel. A detailed discussion about 'Development length and curtailment' can be seen here. But for our problem, we will use 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 will learn about the curtailment of top bars by taking support B of our continuous slab as an example. The fig.8.10 given below shows the details.
Fig.8.10
Curtailment of top bars
Here, bar types 'b' from span AB, and bar type 'c' from span BC are working together to resist the hogging moment at support B. These bars are shown separately only for clarity. In reality, they are at the same level, as indicated by the 0mm distance in the fig.
We can see a distance of 0.15l (from the face of the support) marked off on either side. So there is a particular horizontal length equal to 0.15l1 + 0.15l2 + width of the support. Within this length, no curtailment is allowed. In other words, within this length, all bars (which are intended to resist the design hogging moment) should be compulsorily present. So the quantity of steel in this length is denoted as Ast,sup.B. 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. This is indicated by the '≥' sign. In this length, all of 'b' and 'c' are working together. But beyond this distance, some bars can be allowed to leave. We see that on the left side, 'b' bars are leaving. So after this distance, on the left side, only 'c' is present. The 'b' bars have taken deviation and left. They will become bottom bars at the midspan region. The 'c' bars remain as top bars, and continue to do their work. We can say that, the group does not need 'b' beyond 0.15l because, the moment is of a lower magnitude in that region.
But there is a restriction on the quantity of bars that can be curtailed in this way: 50% of the bars required to resist the full hogging moment must be compulsorily present beyond the 0.15l length. Thus, in the fig., the quantity of steel beyond 0.15l is denoted as 0.5Ast,sup.B. So all of 'c' must continue. This will ensure the required 50%. There is a restriction on the length also. All the 'c' must compulsorily extend a distance of another 0.15l. So they will have a length of 0.30l from the face of the support.
A mirror image of the above arrangement happens on the right side of the support. There, 'c' bars will be leaving the group of top bars.
The above fig.8.10 is applicable to interior supports. Now we will see the curtailment of bottom bars. We will take the example of span BC, which is an interior span. Fig.8.11 given below shows the details:
Fig.8.11
Curtailment of bottom bars
Here, the bars 'c' and 'd' are working together to resist the sagging moment at midspan BC. These bars are shown separately only for clarity. In reality, they are at the same level, as indicated by the 0mm distance in the fig.
We can see a distance of 0.25l2 (from the center of supports) marked near either supports. So there will be a portion (of length l2 -0.50l2 =0.50l2) at the middle of the slab. This is the important portion as far as the 'sagging moment in an interior span' is concerned. All the bars which are intended to resist the sagging moment at midspan should be completely present in this portion. Thus, in the fig., the quantity of steel is denoted as Ast,mid.BC. Beyond this portion, on either sides, the BM is of lower magnitude. So 'c' is allowed to leave on the left side, and 'd' is allowed to leave on the right side. They are bent up and will become top bars on the respective supports.
Like in the case of top bars, here also, there are some restrictions: Half the area of bars at the midspan should be compulsorily present after curtailment on either sides. Thus, in the fig., the quantity of steel after curtailment is denoted as 0.5Ast,mid.BC. There is a restriction on length also: All the bars remaining after curtailment should be compulsorily extended into the supports on either sides.
So we have completed the discussion on the curtailment of bars at interior supports and spans. From the above figs.8.10 and 8.11, one point can be noted: Each bent-up bar will be marked with three lengths. They are 0.15l and 0.30l at the top and 0.25l at the bottom. The lengths at top should be measured from the respective face of support. The length at bottom should be measured from the center line of support.
Now we will see the curtailment at an end span. Together, we will see the curtailment at an end support also. Fig.8.12 below shows the details.
Fig.8.12
Curtailment details in end span
First we will see the details of top bars at the end support A. The length marked here is 0.1l from the face of the support. So we must ensure that the top horizontal portion of all the 'a', which are bent-up bars, have a length of 0.1l from the face of the support. The area of these bars should not be less than half of that provided at midspan for the sagging moment.
Next we will see the bottom bars. This is similar to what we saw in fig.8.11 above, except that the distance from the center line of end support is 0.15l instead of 0.25l. So the 'important' portion in the midspan region has a length of l1 -0.15l1 -0.25l1 =0.60l2 . All the bars should be present in this region. Beyond this region, the bars should have an area of half of that provided at midspan, and they must extend into supports on either sides.
This completes the details of the arrangement using 'bent-up bars'. In the next section we will see how the same reinforcement requirements of this slab can be satisfied with 'straight bars'.
Solved example 8.1 - final checks
Curtailment of bars
Now we can discuss about 'curtailment of bars' in our slab. We know that in a continuous beam or slab, there will be hogging moment at support, and sagging moment at midspan. This can be seen in the BM diagram of a continuous member.
We earlier saw the BM diagram for our present slab in fig.7.5. On either side of the support, the hogging moment is decreasing progressively. That is., at greater distances away from the support, there will be lesser hogging moments.
We have designed the top steel at supports for the full design moment. But now we see that at sections away from the support, the BM decreases, and so the full steel is not necessary. So we can reduce the steel. A detailed discussion about 'Development length and curtailment' can be seen here. But for our problem, we will use 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 will learn about the curtailment of top bars by taking support B of our continuous slab as an example. The fig.8.10 given below shows the details.
Fig.8.10
Curtailment of top bars
Here, bar types 'b' from span AB, and bar type 'c' from span BC are working together to resist the hogging moment at support B. These bars are shown separately only for clarity. In reality, they are at the same level, as indicated by the 0mm distance in the fig.
We can see a distance of 0.15l (from the face of the support) marked off on either side. So there is a particular horizontal length equal to 0.15l1 + 0.15l2 + width of the support. Within this length, no curtailment is allowed. In other words, within this length, all bars (which are intended to resist the design hogging moment) should be compulsorily present. So the quantity of steel in this length is denoted as Ast,sup.B. 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. This is indicated by the '≥' sign. In this length, all of 'b' and 'c' are working together. But beyond this distance, some bars can be allowed to leave. We see that on the left side, 'b' bars are leaving. So after this distance, on the left side, only 'c' is present. The 'b' bars have taken deviation and left. They will become bottom bars at the midspan region. The 'c' bars remain as top bars, and continue to do their work. We can say that, the group does not need 'b' beyond 0.15l because, the moment is of a lower magnitude in that region.
But there is a restriction on the quantity of bars that can be curtailed in this way: 50% of the bars required to resist the full hogging moment must be compulsorily present beyond the 0.15l length. Thus, in the fig., the quantity of steel beyond 0.15l is denoted as 0.5Ast,sup.B. So all of 'c' must continue. This will ensure the required 50%. There is a restriction on the length also. All the 'c' must compulsorily extend a distance of another 0.15l. So they will have a length of 0.30l from the face of the support.
A mirror image of the above arrangement happens on the right side of the support. There, 'c' bars will be leaving the group of top bars.
The above fig.8.10 is applicable to interior supports. Now we will see the curtailment of bottom bars. We will take the example of span BC, which is an interior span. Fig.8.11 given below shows the details:
Fig.8.11
Curtailment of bottom bars
We can see a distance of 0.25l2 (from the center of supports) marked near either supports. So there will be a portion (of length l2 -0.50l2 =0.50l2) at the middle of the slab. This is the important portion as far as the 'sagging moment in an interior span' is concerned. All the bars which are intended to resist the sagging moment at midspan should be completely present in this portion. Thus, in the fig., the quantity of steel is denoted as Ast,mid.BC. Beyond this portion, on either sides, the BM is of lower magnitude. So 'c' is allowed to leave on the left side, and 'd' is allowed to leave on the right side. They are bent up and will become top bars on the respective supports.
Like in the case of top bars, here also, there are some restrictions: Half the area of bars at the midspan should be compulsorily present after curtailment on either sides. Thus, in the fig., the quantity of steel after curtailment is denoted as 0.5Ast,mid.BC. There is a restriction on length also: All the bars remaining after curtailment should be compulsorily extended into the supports on either sides.
So we have completed the discussion on the curtailment of bars at interior supports and spans. From the above figs.8.10 and 8.11, one point can be noted: Each bent-up bar will be marked with three lengths. They are 0.15l and 0.30l at the top and 0.25l at the bottom. The lengths at top should be measured from the respective face of support. The length at bottom should be measured from the center line of support.
Now we will see the curtailment at an end span. Together, we will see the curtailment at an end support also. Fig.8.12 below shows the details.
Fig.8.12
Curtailment details in end span
Next we will see the bottom bars. This is similar to what we saw in fig.8.11 above, except that the distance from the center line of end support is 0.15l instead of 0.25l. So the 'important' portion in the midspan region has a length of l1 -0.15l1 -0.25l1 =0.60l2 . All the bars should be present in this region. Beyond this region, the bars should have an area of half of that provided at midspan, and they must extend into supports on either sides.
This completes the details of the arrangement using 'bent-up bars'. In the next section we will see how the same reinforcement requirements of this slab can be satisfied with 'straight bars'.
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