In
the previous section we saw the final table 8.3 which gives the required spacing at the important points. We also said that the spacing in that table cannot be given as such. In this section, we will see the steps for obtaining a good arrangement of bars. We will discuss this based on the figs below:
Fig.8.1
Bar types:a & b
Let us take midspan AB. Two types of bars: Bar type:'a' & Bar type:'b' are provided as shown in the fig.8.1. 'a' has a bent-up at the left side, while 'b' has a bent-up at the right side. Both of them have similar shapes. They look like mirror images. But as we will soon see, their lengths are not the same. The point at which the bent-up is done, is also not the same. If we provide any one type of bars 'a' or 'b', the reinforcement requirements at midspan AB will not be satisfied. Both of them should be provided together. Plus, they must be provided alternately. This is shown in the 3D view given below:
Fig.8.2
3D view of bent up bars
In the fig.8.1, they are shown separately, only for clarity.
[In the sectional elevations in fig.8.1 above, we see some lengths marked off as 0.15l1, 0.3l2, 0.25l1 etc., These are the points of bent-up and points of curtailment. We will see their details after this discussion]
A plan view is shown in the fig. below:
Fig.8.3
Plan view of bars of slab
In this fig., the two bars are arranged alternately. At the site, this is achieved by two simple steps:
• Lay all the bars (prior to bending) at an uniform spacing of s1
• Give the bent-up at left side and right side for alternate bars.
The bars placed at a spacing of s1 will be resisting the sagging moment at midspan AB. So the spacing s1 mentioned above should be less than or equal to 155.27mm (row 3 of table 8.3 of the previous section). For convenience, the table 8.3 is shown again below:
Table 8.3:
So the spacing s1 is obtained from the required spacing at midspan AB. We may want to change this statement later. So let us note it as - - - (1).
In this arrangement, the spacing between any two adjacent 'a' bars will be 2s1. The spacing between any two adjacent 'b' bars will also be 2s1.
Now we look at the details at supports:
At support A, we need top bars. We get bars at top because of the bent-up in 'a'. These bars 'a' constitute 50% of those at midspan. All of them are bent-up at support A. This means that the top bars at support A have an area equal to 50% of that at the midspan. This satisfies the cl.D-1.6 of the code. So the 'a' bars will do two jobs:
• They will 'take part' in resisting the sagging moment at midspan AB
• They will 'solely' resist the hogging moment due to partial fixity at support A
Note the terms 'take part' and 'solely'. These are used because, at midspan, they will be working along with 'b' bars to resist sagging moment. But at the top portion in support A, there are no other types of bars.
At support B also we need top bars. We do get some bars at top because of the bent-up in 'b'. These bars 'b' constitute 50% of the total bars at midspan. All of 'b' are bent-up at support B. This means that the top bars at support B, now have an area equal to 50% of that at the midspan AB. So, if the hogging moment at support B is (numerically) less than or equal to 'half of the sagging moment' at midspan AB, then these top bars are sufficient. But in reality, the hogging moment at support B is generally greater than (numerically) the full sagging moment at midspan AB. We can see that this is true in our problem also. From table 8.3, 29.76>28.60. This means that 50% steel that is obtained by bending up 'b' is no where near to what is actually required. This is shown in the fig.8.4 below:
Fig.8.4
Insufficient steel at support
So we need more steel. How can this be achieved? The answer is to get some steel from the adjacent span BC. Just as we like what we see at midspan AB, at midspan BC also, there will be two types of bars. One set with the bent-up on the left side, and the other set with the bent-up on the right side. We want the set with the bent-up on the left side. Let us 'bring them in' to span AB. This is shown in the fig.8.5 below:
Fig.8.5
Bar type:c
We are talking about the 'Bar type: c' in the above fig. They are originally provided to take part in resisting the sagging moment at midspan BC. But after the bent-up, their left side become top bars. Then they travel towards the left, cross the support B, and travels further into span AB for a sufficient distance. So at support B, we get two sets of top bars:
1. 'b' from span AB and
2. 'c' from span BC.
This is shown in the plan view of support B given below:
Fig.8.6
Plan view at support B
Let us see if these two bars 'b' and 'c' can together satisfy the steel requirements at support B: Recall the statement that we marked as (1) above: 'The spacing s1 is obtained from the required spacing at midspan AB'.
• So the 'b' bars which have a spacing of 2s1 will be able to resist a BM which has a magnitude equal to half the sagging moment at midspan AB.
• Similarly, the 'c' bars which is coming from BC, and which has a spacing 2s2 will be able to resist a BM which has a magnitude equal to half the sagging moment at midspan BC.
So it follows that the 'b' and 'c' which now work together at support B will be able to resist a BM equal to 0.5 xBM at midspan AB + 0.5 xBM at midspan AB. We will write this sum in a simpler form, and mark it as (2):
0.5(BM at midspan AB + BM at midspan BC) - - - (2)
But we find (from the values in table 8.3) that this sum in (2) is less than the hogging moment at B.
0.5(28.60 +17.75) = 23.175 <29.76
This means that the total capacity of 'b' and 'c' put together is not sufficient to resist the hogging moment at B.
Any way, this situation is encountered generally. That is., the hogging moment at a support is greater than the sum of half the sagging moments on the spans on either sides of that support. This can be shown in the form of the diagram given below:
Fig.8.7
In fact, we do not need to work out half of the sagging moments. The calculation of 'halves' can be avoided if we use a very simple mathematical theorem:
If P and Q, both are less than R, then '0.5(P + Q)' will be less than R. We will write it in the form of an expression, and mark it as (3):
IF P<R AND Q<R
THEN 0.5(P + Q) < R - - -(3)
Here, P and Q are the full magnitudes of the sagging moments on two adjacent spans. If both of them are less than R, the hogging moment at the support between them, then half of their sum will indeed be less than the hogging moment. So looking at a final table (like table 8.3 above), we can say (with out any calculations) whether the contributions from the spans will 'work' at the support. In our case, 28.6 and 17.75 are both less than 29.76. So it will not 'work'.
Though this situation generally occur, we must check each problem to see if it is true for that particular problem.
But we can solve this problem. We just need to increase the capacity of 'b' and 'c'. We can achieve this increase just by 'decreasing the spacing'. We don't need to change the shape of bars or any other parameter in the figures that we saw until now. Just a 'decrease in spacing' is enough.
So what is the new 'decreased value'? To find this, we have to do our calculations in a sort of 'reverse' manner. Recall that in (1) we have stated that the spacing s1 is obtained from the required spacing at midspan AB. Then half of it was given to support A. Similarly, s2 is obtained from the required spacing in midspan BC. Then half of it was given to support B. We will now reverse this procedure: We will first fix up the spacing at the support, and then give half the bars to either spans.
From table 8.1, we see that the spacing required at support B is 148.88mm. Rounding of to the nearest lower multiple of 10, we get 140mm. In fig.8.6 above, we put the 'spacing at support B' = 140mm. As this is a uniform spacing, 2s1 for bars 'b' on the left side, and 2s2 for bars 'c' on the right side, both will be equal to 2 x140 = 280mm.
When the spacing of 'b' is 280mm, spacing of 'a' should also become 280mm. So the spacing at midspan where both 'a' and 'b' are present will be equal to 140mm. Similarly, when spacing of 'c' is equal to 280, the other set of bars (let us call it 'd') will also be 280mm. So the spacing a midspan BC will also become 140mm. Thus the spacing at midspan in both spans will become equal to 140mm. In AB, this 'decrease in spacing' is of a lower extent. Because 140mm is comparable with 155.27, which is the spacing actually required in AB. But in BC, we are giving 140mm in place of 253.78. This will lead to the provision of an 'unwanted quantity' of bars. But for uniformity and safety, the only option is to give 140mm.
We have had such a long discussion in order to show the application of (3) above. Next time when we get the final table (as table 8.3 above), we can straight away check the magnitudes of the moments, and if (3) is satisfied, we can use the 'required spacing of top bars at the support' to start the arrangement of bars.
Now we have to design the support C and midspan CD. Proceeding as before, we find (from table 8.3) that the hogging moment at support C is numerically greater than the sagging moments in both BC and CD. So the support moment is the criterion. We have to provide a spacing of 160.05mm for the top bars at C. We will round it off to the nearest lower multiple of 10mm, which is 160mm. From here, working towards the spans on either side, we get a spacing of midspan bars of both BC and CD as 160mm. But we have already fixed the spacing of bars in BC as 140mm. So the bars from BC which contribute towards the top bars at support C will be having a spacing of 140mm. This means that all the bars of the support C and span CD will also be having a spacing of 140mm.
This spacing will lead to the provision of some unwanted quantity of bars in CD. This is just like what we saw in the case of BC. But this situation cannot be avoided because we want to ensure both uniformity and safety. So we get a spacing of 140mm at all midspans and intermediate supports. At the end supports, the spacing will be 2 times this, which is equal to 280mm. The following table shows the final spacing of bars.
Table 8.4
The final sectional elevation of the slab, showing the details of all slabs is given below:
Fig.8.8
Sectional elevation
We will see the design of distributor bars in the next section.
Fig.8.1
Bar types:a & b
Let us take midspan AB. Two types of bars: Bar type:'a' & Bar type:'b' are provided as shown in the fig.8.1. 'a' has a bent-up at the left side, while 'b' has a bent-up at the right side. Both of them have similar shapes. They look like mirror images. But as we will soon see, their lengths are not the same. The point at which the bent-up is done, is also not the same. If we provide any one type of bars 'a' or 'b', the reinforcement requirements at midspan AB will not be satisfied. Both of them should be provided together. Plus, they must be provided alternately. This is shown in the 3D view given below:
Fig.8.2
3D view of bent up bars
 |
[In the sectional elevations in fig.8.1 above, we see some lengths marked off as 0.15l1, 0.3l2, 0.25l1 etc., These are the points of bent-up and points of curtailment. We will see their details after this discussion]
A plan view is shown in the fig. below:
Fig.8.3
Plan view of bars of slab
In this fig., the two bars are arranged alternately. At the site, this is achieved by two simple steps:
• Lay all the bars (prior to bending) at an uniform spacing of s1
• Give the bent-up at left side and right side for alternate bars.
The bars placed at a spacing of s1 will be resisting the sagging moment at midspan AB. So the spacing s1 mentioned above should be less than or equal to 155.27mm (row 3 of table 8.3 of the previous section). For convenience, the table 8.3 is shown again below:
Table 8.3:
| BM | d req. | Ast req. | S req. | |
| Supp. A | 252.91 | 310.54 | ||
| Span AB | 28.60 | 90.76 | 505.82 | 155.27 |
| Supp. B | -29.76 | 92.59 | 527.55 | 148.88 |
| Span BC | 17.75 | 71.79 | 309.32 | 253.78 |
| Supp. C | -27.61 | 89.53 | 490.47 | 160.05 |
| Span CD | 24.92 | 85.06 | 440.32 | 178.28 |
| Supp. D | 220.16 | 356.56 |
So the spacing s1 is obtained from the required spacing at midspan AB. We may want to change this statement later. So let us note it as - - - (1).
In this arrangement, the spacing between any two adjacent 'a' bars will be 2s1. The spacing between any two adjacent 'b' bars will also be 2s1.
Now we look at the details at supports:
At support A, we need top bars. We get bars at top because of the bent-up in 'a'. These bars 'a' constitute 50% of those at midspan. All of them are bent-up at support A. This means that the top bars at support A have an area equal to 50% of that at the midspan. This satisfies the cl.D-1.6 of the code. So the 'a' bars will do two jobs:
• They will 'take part' in resisting the sagging moment at midspan AB
• They will 'solely' resist the hogging moment due to partial fixity at support A
Note the terms 'take part' and 'solely'. These are used because, at midspan, they will be working along with 'b' bars to resist sagging moment. But at the top portion in support A, there are no other types of bars.
At support B also we need top bars. We do get some bars at top because of the bent-up in 'b'. These bars 'b' constitute 50% of the total bars at midspan. All of 'b' are bent-up at support B. This means that the top bars at support B, now have an area equal to 50% of that at the midspan AB. So, if the hogging moment at support B is (numerically) less than or equal to 'half of the sagging moment' at midspan AB, then these top bars are sufficient. But in reality, the hogging moment at support B is generally greater than (numerically) the full sagging moment at midspan AB. We can see that this is true in our problem also. From table 8.3, 29.76>28.60. This means that 50% steel that is obtained by bending up 'b' is no where near to what is actually required. This is shown in the fig.8.4 below:
Fig.8.4
Insufficient steel at support
So we need more steel. How can this be achieved? The answer is to get some steel from the adjacent span BC. Just as we like what we see at midspan AB, at midspan BC also, there will be two types of bars. One set with the bent-up on the left side, and the other set with the bent-up on the right side. We want the set with the bent-up on the left side. Let us 'bring them in' to span AB. This is shown in the fig.8.5 below:
Fig.8.5
Bar type:c
1. 'b' from span AB and
2. 'c' from span BC.
This is shown in the plan view of support B given below:
Fig.8.6
Plan view at support B
Let us see if these two bars 'b' and 'c' can together satisfy the steel requirements at support B: Recall the statement that we marked as (1) above: 'The spacing s1 is obtained from the required spacing at midspan AB'.
• So the 'b' bars which have a spacing of 2s1 will be able to resist a BM which has a magnitude equal to half the sagging moment at midspan AB.
• Similarly, the 'c' bars which is coming from BC, and which has a spacing 2s2 will be able to resist a BM which has a magnitude equal to half the sagging moment at midspan BC.
So it follows that the 'b' and 'c' which now work together at support B will be able to resist a BM equal to 0.5 xBM at midspan AB + 0.5 xBM at midspan AB. We will write this sum in a simpler form, and mark it as (2):
0.5(BM at midspan AB + BM at midspan BC) - - - (2)
But we find (from the values in table 8.3) that this sum in (2) is less than the hogging moment at B.
0.5(28.60 +17.75) = 23.175 <29.76
This means that the total capacity of 'b' and 'c' put together is not sufficient to resist the hogging moment at B.
Any way, this situation is encountered generally. That is., the hogging moment at a support is greater than the sum of half the sagging moments on the spans on either sides of that support. This can be shown in the form of the diagram given below:
Fig.8.7
If P and Q, both are less than R, then '0.5(P + Q)' will be less than R. We will write it in the form of an expression, and mark it as (3):
IF P<R AND Q<R
THEN 0.5(P + Q) < R - - -(3)
Here, P and Q are the full magnitudes of the sagging moments on two adjacent spans. If both of them are less than R, the hogging moment at the support between them, then half of their sum will indeed be less than the hogging moment. So looking at a final table (like table 8.3 above), we can say (with out any calculations) whether the contributions from the spans will 'work' at the support. In our case, 28.6 and 17.75 are both less than 29.76. So it will not 'work'.
Though this situation generally occur, we must check each problem to see if it is true for that particular problem.
But we can solve this problem. We just need to increase the capacity of 'b' and 'c'. We can achieve this increase just by 'decreasing the spacing'. We don't need to change the shape of bars or any other parameter in the figures that we saw until now. Just a 'decrease in spacing' is enough.
So what is the new 'decreased value'? To find this, we have to do our calculations in a sort of 'reverse' manner. Recall that in (1) we have stated that the spacing s1 is obtained from the required spacing at midspan AB. Then half of it was given to support A. Similarly, s2 is obtained from the required spacing in midspan BC. Then half of it was given to support B. We will now reverse this procedure: We will first fix up the spacing at the support, and then give half the bars to either spans.
From table 8.1, we see that the spacing required at support B is 148.88mm. Rounding of to the nearest lower multiple of 10, we get 140mm. In fig.8.6 above, we put the 'spacing at support B' = 140mm. As this is a uniform spacing, 2s1 for bars 'b' on the left side, and 2s2 for bars 'c' on the right side, both will be equal to 2 x140 = 280mm.
When the spacing of 'b' is 280mm, spacing of 'a' should also become 280mm. So the spacing at midspan where both 'a' and 'b' are present will be equal to 140mm. Similarly, when spacing of 'c' is equal to 280, the other set of bars (let us call it 'd') will also be 280mm. So the spacing a midspan BC will also become 140mm. Thus the spacing at midspan in both spans will become equal to 140mm. In AB, this 'decrease in spacing' is of a lower extent. Because 140mm is comparable with 155.27, which is the spacing actually required in AB. But in BC, we are giving 140mm in place of 253.78. This will lead to the provision of an 'unwanted quantity' of bars. But for uniformity and safety, the only option is to give 140mm.
We have had such a long discussion in order to show the application of (3) above. Next time when we get the final table (as table 8.3 above), we can straight away check the magnitudes of the moments, and if (3) is satisfied, we can use the 'required spacing of top bars at the support' to start the arrangement of bars.
Now we have to design the support C and midspan CD. Proceeding as before, we find (from table 8.3) that the hogging moment at support C is numerically greater than the sagging moments in both BC and CD. So the support moment is the criterion. We have to provide a spacing of 160.05mm for the top bars at C. We will round it off to the nearest lower multiple of 10mm, which is 160mm. From here, working towards the spans on either side, we get a spacing of midspan bars of both BC and CD as 160mm. But we have already fixed the spacing of bars in BC as 140mm. So the bars from BC which contribute towards the top bars at support C will be having a spacing of 140mm. This means that all the bars of the support C and span CD will also be having a spacing of 140mm.
This spacing will lead to the provision of some unwanted quantity of bars in CD. This is just like what we saw in the case of BC. But this situation cannot be avoided because we want to ensure both uniformity and safety. So we get a spacing of 140mm at all midspans and intermediate supports. At the end supports, the spacing will be 2 times this, which is equal to 280mm. The following table shows the final spacing of bars.
Table 8.4
| BM | d req. | Ast req. | S req. | s pr. | Ast pr. | |
| Supp. A | 252.91 | 310.54 | 280 | 280.36 | ||
| Span AB | 28.60 | 90.76 | 505.82 | 155.27 | 140 | 560.71 |
| Supp. B | -29.76 | 92.59 | 527.55 | 148.88 | 140 | 560.71 |
| Span BC | 17.75 | 71.79 | 309.32 | 253.78 | 140 | 560.71 |
| Supp. C | -27.61 | 89.53 | 490.47 | 160.05 | 140 | 560.71 |
| Span CD | 24.92 | 85.06 | 440.32 | 178.28 | 140 | 560.71 |
| Supp. D | 220.16 | 356.56 | 280 | 280.36 |
The final sectional elevation of the slab, showing the details of all slabs is given below:
Fig.8.8
Sectional elevation
We will see the design of distributor bars in the next section.
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