In
the previous chapter we completed the analysis of the continuous slabs and beams. In this chapter we will discuss the design of continuous beams and continuous one way slabs.
We will consider the design of continuous slabs first. For this, the first step is to setup a preliminary depth for the slab. This is for calculating the self wt. But for continuous slabs and beams, this is all the more important (unless we are using moment coefficients) because, only with the knowledge of the dimensions of cross section, we can do the 'detailed structural analysis'.
We have already seen (in the chapter 6: design of simply supported one-way slabs) that the 'control of deflection' is the criterion for deciding the depth in the case of slabs. We have derived the expressions which will enable us to use this criterion in the case of one-way slabs:
6.1: (l/d)actual ≤ 20 x 1.25 ⇒ (l/d)actual ≤ 25. So d that we actually provide in the final slab should be greater than or equal to l/25. This is for simply supported slabs.
6.2: (l/d)actual ≤ 26 x 1.25 ⇒ (l/d)actual ≤ 32.5. So d that we actually provide in the final slab should be greater than or equal to l/32.5.≈ l/32 . This is for continuous one-way slabs.
So using 6.2, we can easily fix up a preliminary depth for the continuous slab. But we have to consider a few points before we use this expression 6.2.:
In continuous systems, the end spans will be subjected to greater sagging moment than the interior spans. So we have to use the properties of end spans for fixing up the depth. (The depth so derived from the end span is provided for all the spans. This is for uniformity). But the end span is not 'perfectly continuous'. It is continuous only at it's interior support. At the end support, it is discontinuous or simply supported. So we cannot use the value '26'. The solution for this problem is to use the average of simply supported condition and continuous condition. That is., average of 20 and 26, which is equal to 23. Thus we get l/d basic = 23. Thus for the end span we get:
8.1: (l/d)actual ≤ 23 x 1.25 ⇒ (l/d)actual ≤ 28.75. So d that we actually provide in the final slab should be greater than or equal to l/28.75.≈ l/29 . This is for end spans in continuous one-way slabs.
From the above expression, what we get is the effective depth d. To this we must add Cc and half of Ф to get the total depth D. Ф can be assumed to be equal to 10mm.
We must do all the design checks to ensure that the final section is adequate. If the section is found to be inadequate, the whole process should be repeated with improved dimensions.
After fixing the preliminary depth, we can start the design process. The design process is same as that for a simply supported slab that we saw earlier in chapter 6: We designed it as a beam of width 1000 mm and total depth D. First we calculated d required. Then we compared it with the d obtained from the preliminary dimensions. If the required value was less, we proceeded to find the steel required to resist the sagging BM at midspan. Here, in the case of continuous one-way slabs also, we have to do the same. But we have to do this for each span. That is., we have to calculate the steel required to resist the sagging BM at the midspan of each span. Plus, we have to do this for the hogging BM at each of the supports also.
After obtaining the steel at all midspans and supports, we arrange the bars in such a way that the maximum and minimum spacing requirements between the bars are satisfied. We have to design the distribution steel also. Then we do the final checks like area of minimum steel required, check for pt,lim, check for deflection etc., These steps will become more clear when we see an actual solved example:
Solved example 8.1:
We will do the design of the continuous slab, of which we did the structural analysis in the previous chapter 7. We have obtained the BM and SF at all the important points. We will use the results obtained by using the 'method of coefficients' fig.7.14 and 7.15. Those results are reproduced in the table 8.1 below:
Table 8.1:
Now we can start the design. As we have discussed at the beginning of this chapter, the first step is to obtain the preliminary depth of the slab. But in this problem, the total depth (200mm) was already given. However, we will check if the depth of 200mm satisfies the expression 8.1:
(l/d)actual ≤ 29
where l is the effective span of the end span AB.
So the d that is provided in the slab should be greater than or equal to l/29. l/29 =4500/29 =155.17mm.
Assuming =10mm, D =155.17 +30 +5 = 190.17. So 200mm is satisfactory.
Though we found it satisfactory, after the design, we must do all the necessary checks and confirm that the slab section with depth 200mm is adequate for the given problem. If it is found that the section is not adequate, it should be redesigned.
The next step is to determine the steel required to resist the BM at various points. First we will do this for the midspan AB. From the table 8.1 above, we can see that the BM at midspan AB is 28.60kNm. The pdf file given below shows the detailed step of design:
Steel at midspan AB
So we got the spacing required for #10 bars as 155.27mm at midspan AB.
The next pdf file given below shows the detailed steps of the design for steel at support B:
Steel at support B
So we got the spacing required for #10 bars as 148.88mm at Support B .In a similar way we can obtain the required spacing at other points also. These are shown in the table 8.2 below.
Table 8.2:
The above table 8.2 is some what incomplete. This is because, we have not considered the end supports A and D. We know that they are simply supported ends and so the BM will be zero. We will indeed get zero as the BM at these supports when we do a detailed structural analysis of the slab ABCD using any of the methods like Kani's method, slope deflection method, moment distribution method etc., So theoretically, steel is not required at these supports.
But we have to consider the partial fixity that may be introduced at a future stage. We learned about it when we discussed about simply supported one way slabs. And we applied cl.D-1.6 of the code in the solved example of a simply supported one way slab. The same is applicable here also. So the table 8.2 should be modified as given below:
Table 8.3:
One row for support A has been added at the beginning, and another row for support D has been added at the end. Now the table is complete. Note that only half the steel at midspan AB is given at support A. (505.82÷2 =252.91) So double the spacing at midspan AB is sufficient at support A. (155.27×2 =310.54). Similar is the case with support D.
The spacing in the above table cannot be given as such in the final slab. They must be rounded off to the next higher multiple of 5 or 10mm. Also, while using 'bent-up' bars, the spacing at any one midspan or support in the slab can have some 'influence' on the spacing in an adjacent midspan or support. We must take this 'influence' into consideration. Because, by taking it into consideration, we will get an 'arrangement of bars' which will be uniform and convenient for placement. We will see more details about this in the next section.
We will consider the design of continuous slabs first. For this, the first step is to setup a preliminary depth for the slab. This is for calculating the self wt. But for continuous slabs and beams, this is all the more important (unless we are using moment coefficients) because, only with the knowledge of the dimensions of cross section, we can do the 'detailed structural analysis'.
We have already seen (in the chapter 6: design of simply supported one-way slabs) that the 'control of deflection' is the criterion for deciding the depth in the case of slabs. We have derived the expressions which will enable us to use this criterion in the case of one-way slabs:
6.1: (l/d)actual ≤ 20 x 1.25 ⇒ (l/d)actual ≤ 25. So d that we actually provide in the final slab should be greater than or equal to l/25. This is for simply supported slabs.
6.2: (l/d)actual ≤ 26 x 1.25 ⇒ (l/d)actual ≤ 32.5. So d that we actually provide in the final slab should be greater than or equal to l/32.5.≈ l/32 . This is for continuous one-way slabs.
So using 6.2, we can easily fix up a preliminary depth for the continuous slab. But we have to consider a few points before we use this expression 6.2.:
In continuous systems, the end spans will be subjected to greater sagging moment than the interior spans. So we have to use the properties of end spans for fixing up the depth. (The depth so derived from the end span is provided for all the spans. This is for uniformity). But the end span is not 'perfectly continuous'. It is continuous only at it's interior support. At the end support, it is discontinuous or simply supported. So we cannot use the value '26'. The solution for this problem is to use the average of simply supported condition and continuous condition. That is., average of 20 and 26, which is equal to 23. Thus we get l/d basic = 23. Thus for the end span we get:
8.1: (l/d)actual ≤ 23 x 1.25 ⇒ (l/d)actual ≤ 28.75. So d that we actually provide in the final slab should be greater than or equal to l/28.75.≈ l/29 . This is for end spans in continuous one-way slabs.
From the above expression, what we get is the effective depth d. To this we must add Cc and half of Ф to get the total depth D. Ф can be assumed to be equal to 10mm.
We must do all the design checks to ensure that the final section is adequate. If the section is found to be inadequate, the whole process should be repeated with improved dimensions.
After fixing the preliminary depth, we can start the design process. The design process is same as that for a simply supported slab that we saw earlier in chapter 6: We designed it as a beam of width 1000 mm and total depth D. First we calculated d required. Then we compared it with the d obtained from the preliminary dimensions. If the required value was less, we proceeded to find the steel required to resist the sagging BM at midspan. Here, in the case of continuous one-way slabs also, we have to do the same. But we have to do this for each span. That is., we have to calculate the steel required to resist the sagging BM at the midspan of each span. Plus, we have to do this for the hogging BM at each of the supports also.
After obtaining the steel at all midspans and supports, we arrange the bars in such a way that the maximum and minimum spacing requirements between the bars are satisfied. We have to design the distribution steel also. Then we do the final checks like area of minimum steel required, check for pt,lim, check for deflection etc., These steps will become more clear when we see an actual solved example:
Solved example 8.1:
We will do the design of the continuous slab, of which we did the structural analysis in the previous chapter 7. We have obtained the BM and SF at all the important points. We will use the results obtained by using the 'method of coefficients' fig.7.14 and 7.15. Those results are reproduced in the table 8.1 below:
Table 8.1:
| BM | |
| Span AB | 28.60 |
| Supp. B | -29.76 |
| Span BC | 17.75 |
| Supp. C | -27.61 |
| Span CD | 24.92 |
(l/d)actual ≤ 29
where l is the effective span of the end span AB.
So the d that is provided in the slab should be greater than or equal to l/29. l/29 =4500/29 =155.17mm.
Assuming =10mm, D =155.17 +30 +5 = 190.17. So 200mm is satisfactory.
Though we found it satisfactory, after the design, we must do all the necessary checks and confirm that the slab section with depth 200mm is adequate for the given problem. If it is found that the section is not adequate, it should be redesigned.
The next step is to determine the steel required to resist the BM at various points. First we will do this for the midspan AB. From the table 8.1 above, we can see that the BM at midspan AB is 28.60kNm. The pdf file given below shows the detailed step of design:
Steel at midspan AB
So we got the spacing required for #10 bars as 155.27mm at midspan AB.
The next pdf file given below shows the detailed steps of the design for steel at support B:
Steel at support B
So we got the spacing required for #10 bars as 148.88mm at Support B .In a similar way we can obtain the required spacing at other points also. These are shown in the table 8.2 below.
Table 8.2:
| BM | d req. | Ast req. | S req. | |
| 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 |
But we have to consider the partial fixity that may be introduced at a future stage. We learned about it when we discussed about simply supported one way slabs. And we applied cl.D-1.6 of the code in the solved example of a simply supported one way slab. The same is applicable here also. So the table 8.2 should be modified as given 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 |
One row for support A has been added at the beginning, and another row for support D has been added at the end. Now the table is complete. Note that only half the steel at midspan AB is given at support A. (505.82÷2 =252.91) So double the spacing at midspan AB is sufficient at support A. (155.27×2 =310.54). Similar is the case with support D.
The spacing in the above table cannot be given as such in the final slab. They must be rounded off to the next higher multiple of 5 or 10mm. Also, while using 'bent-up' bars, the spacing at any one midspan or support in the slab can have some 'influence' on the spacing in an adjacent midspan or support. We must take this 'influence' into consideration. Because, by taking it into consideration, we will get an 'arrangement of bars' which will be uniform and convenient for placement. We will see more details about this in the next section.
