In
the previous section we completed the design of a continuous slab. In this section we will see the design of a continuous beam. For this, the first step is to setup a preliminary cross sectional dimensions of the beam. 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'. At the beginning of this chapter, we saw the guidelines for fixing up the depth of a continuous one way slab.
Now we want to set up the preliminary dimensions of a continuous beam. Unlike for a slab (width of a strip is taken as 1000 mm), for a beam, we want to set up both the preliminary width and depth. We have already seen the procedure for doing this here, when we discussed about the design of beams. Based on that, we can use a range of 10 to 16 for l/D for the depth.
After fixing the preliminary dimensions, we can start the design process. The design process is same as that for a simply supported beam that we saw earlier here : 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 beams 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 various code requirements are satisfied. 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 8.2:
We will do the design of the continuous beam, 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.7.22 and 7.23). Those results are reproduced in the table 8.6 below:
Now we can start the design. As we have discussed at the beginning of this chapter, the first step is to obtain the preliminary dimensions of the beam. But in this problem, these are already given (b =230 mm & D =400 mm). However, we will check if the depth of 400 mm satisfies the 'general rule':
We will use the largest span, which is the end span. So l = 4230 mm. D = 400 mm. Thus we get l/D = 4230/400 =10.575. This falls between 10 and 16. So the value of D is satisfactory.
Though we found it satisfactory, after the design, we must do all the necessary checks and confirm that the beam section with overall depth 400 mm 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 8.6 above, we can see that the BM at midspan AB is 55.49 kNm. The pdf file given below shows the detailed steps of design:
Steel at midspan AB
So we got the steel as 3-#16 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 steel as 3-#16 at Support B. In a similar way we can obtain the required spacing at other points also. These are shown in the table 8.7 below:
Table 8.7
The above table 8.2 is some what incomplete. This is because, we have not considered the end supports A and E. 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 beam ABCDE 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. But for beams, we have to apply cl.22.5.2. According to this clause, we have to design the supports A and E for a hogging moment whose magnitude is given by Wl/24. Where W is the total load, and l is the effective span. So we will first calculate this moment:
Here W is the total load. The total load is obtained as W =wl. where w is the total factored load per unit length. So the hogging moment can be obtained as wl2/24. This is equal to = 26.22kNm (w =24.94 +10.23 & l =4.23 m). With this value of the hogging moment, we have to do a complete design just as we do at other supports B, C and D. The results are: d required = 181.93mm, Ast required = 215.88 mm2. These results are applicable at support E also because spans AB and DE are exact mirror images. So the table 8.7 can be modified by adding these details:
Table 8.8
One row for 'support A' has been added at the beginning, and another row for 'support E' has been added at the end. Now the table is complete. In the next section we will see the arrangement of bars according to the numbers given in the above table.
Now we want to set up the preliminary dimensions of a continuous beam. Unlike for a slab (width of a strip is taken as 1000 mm), for a beam, we want to set up both the preliminary width and depth. We have already seen the procedure for doing this here, when we discussed about the design of beams. Based on that, we can use a range of 10 to 16 for l/D for the depth.
After fixing the preliminary dimensions, we can start the design process. The design process is same as that for a simply supported beam that we saw earlier here : 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 beams 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 various code requirements are satisfied. 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 8.2:
We will do the design of the continuous beam, 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.7.22 and 7.23). Those results are reproduced in the table 8.6 below:
Table 8.6
| BM | |
| Span AB | 55.49 |
| Supp. B | -63.69 |
| Span BC | 41.44 |
| Supp. C | -53.55 |
| Span CD | 38.89 |
| Supp. D | -61.76 |
| Span DE | 55.49 |
We will use the largest span, which is the end span. So l = 4230 mm. D = 400 mm. Thus we get l/D = 4230/400 =10.575. This falls between 10 and 16. So the value of D is satisfactory.
Though we found it satisfactory, after the design, we must do all the necessary checks and confirm that the beam section with overall depth 400 mm 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 8.6 above, we can see that the BM at midspan AB is 55.49 kNm. The pdf file given below shows the detailed steps of design:
Steel at midspan AB
So we got the steel as 3-#16 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 steel as 3-#16 at Support B. In a similar way we can obtain the required spacing at other points also. These are shown in the table 8.7 below:
Table 8.7
| BM | d req. | Ast req. | dia. | No. | Ast pr. | |
| Span AB | 55.49 | 263.62 | 481.39 | 16 | 3 | 603.19 |
| Supp. B | -63.69 | 282.42 | 562.9 | 16 | 3 | 603.19 |
| Span BC | 41.44 | 228.58 | 349.16 | 16 | 2 | 402.12 |
| Supp. C | -53.55 | 259.81 | 462.61 | 16 | 3 | 603.19 |
| Span CD | 38.89 | 221.38 | 325.84 | 16 | 2 | 402.12 |
| Supp. D | -61.76 | 279.02 | 543.42 | 16 | 3 | 603.19 |
| Span DE | 55.49 | 263.62 | 481.39 | 16 | 3 | 603.19 |
The above table 8.2 is some what incomplete. This is because, we have not considered the end supports A and E. 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 beam ABCDE 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. But for beams, we have to apply cl.22.5.2. According to this clause, we have to design the supports A and E for a hogging moment whose magnitude is given by Wl/24. Where W is the total load, and l is the effective span. So we will first calculate this moment:
Here W is the total load. The total load is obtained as W =wl. where w is the total factored load per unit length. So the hogging moment can be obtained as wl2/24. This is equal to = 26.22kNm (w =24.94 +10.23 & l =4.23 m). With this value of the hogging moment, we have to do a complete design just as we do at other supports B, C and D. The results are: d required = 181.93mm, Ast required = 215.88 mm2. These results are applicable at support E also because spans AB and DE are exact mirror images. So the table 8.7 can be modified by adding these details:
Table 8.8
| BM | d req. | Ast req. | dia. | No. | Ast pr. | |
| Supp. A | -26.22 | 181.93 | 215.88 | 16 | 2 | 402.12 |
| Span AB | 55.49 | 263.62 | 481.39 | 16 | 3 | 603.19 |
| Supp. B | -63.69 | 282.42 | 562.9 | 16 | 3 | 603.19 |
| Span BC | 41.44 | 228.58 | 349.16 | 16 | 2 | 402.12 |
| Supp. C | -53.55 | 259.81 | 462.61 | 16 | 3 | 603.19 |
| Span CD | 38.89 | 221.38 | 325.84 | 16 | 2 | 402.12 |
| Supp. D | -61.76 | 279.02 | 543.42 | 16 | 3 | 603.19 |
| Span DE | 55.49 | 263.62 | 481.39 | 16 | 3 | 603.19 |
| Supp. E | -26.22 | 181.93 | 215.88 | 16 | 2 | 402.12 |
One row for 'support A' has been added at the beginning, and another row for 'support E' has been added at the end. Now the table is complete. In the next section we will see the arrangement of bars according to the numbers given in the above table.
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