Reinforced Concrete Design: August 2015

Saturday, August 29, 2015

Chapter 8 (cont..5) Design of a continuous beam

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:

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

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

	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, A_st 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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Chapter 8 (cont..4) - Distribution bars and other details

In the previous section we completed the arrangement of the required reinforcement using straight bars. Now we will see the details about distribution bars. We have seen that the area of distribution bars depend only on A_g the gross area of cross section. So the same steps that we did while using bent-up bars are applicable here also. Thus we can provide #8 @200 for the distribution bars.

So we have completed the design of slab using straight bars. 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 (straight bars) final checks

Curtailment of bars
Now we will discuss about the curtailment of bars. We have seen the curtailment details while using bent-up bars here. Now we will see the details while using straight 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.

When we discussed about the curtailment of bent-up bars, we saw 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. This is applicable for straight bars also.

We will first see the curtailment of top bars at supports. We will take support B as an example. The fig.8.18 given below shows the details.

Fig.8.18
Curtailment of top bars

Here, bar types b'' and c'' 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 A_st,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. In this length, all of b'' and c'' are working together. But beyond this distance, all bars are not present. The b'' bars are not travelling beyond this distance on either side of the support. But c'' bars are continuing their travel for a distance of 0.15l more. So in effect, within a distance of 0.15l, full capacity is present. After this distance, the capacity is reduced because all bars are not present.

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 0.5A_st,sup.B. This will be readily achieved because all the alternate bars in the group are continuing their travel. 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.

The above discussion about 'curtailment of top bars at an interior support' can be simply written in a few steps as follows:

• At the interior support of a continuous one way slab, there will be two types of top bars.
• The longer bars have a length of 0.30l₁ + 0.30l₂ + width of the support
• The shorter bars have a length of 0.15l₁ + 0.15l₂ + width of the support.
• These two bars are provided alternately
• Care should be taken to see that the required lengths are provided on the concerned spans. (0.15l₁ and 0.30l₁ on the left side and 0.15l₂ and 0.30l₂ on the right side). The required areas should also be satisfied

Now we will see the curtailment of bottom bars in an interior span. We will take span BC as an example. Fig.8.19 given below shows the details.

Fig.8.19
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 A_st,mid.BC. Beyond this portion, on either sides, the BM is of lower magnitude. So c' stops at 0.25l₂.

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 0.5A_st,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.

The above discussion about 'curtailment of bottom bars at an interior midspan' can be simply written in a few steps as follows:

• At the interior span of a continuous one way slab, there will be two types of bottom bars.
• The longer bars extend from support to support
• The shorter bars have a length of l₂ -0.50l₂ =0.50l₂
• These two bars are provided alternately
• Care should be taken to see that the shorter bars are provided at the exact mid portion. Because 0.25l₂ is marked from the center line of either supports. The required areas should also be satisfied

So we have completed the discussion on the curtailment of bars at interior supports and spans. Now we will see the curtailment at an end span. Together, we will see the curtailment at an end support also. Fig.8.20 below shows the details.

Fig.8.20
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 all the a'' have a length of 0.1l from the face of the support. These bars are not part of any group. They are working alone to resist the hogging moment at support A. 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.8.18 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=0.60l₁. 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.

So the discussion about 'curtailment of bottom bars at an end span' can be simply written in a few steps as follows:

• At the end span of a continuous one way slab, there will be two types of bottom bars.
• The longer bars extend from support to support
• The shorter bars have a length of l₁ -0.15l₁ -0.25l₁=0.60l₁
• These two bars are provided alternately
• Care should be taken to see that the shorter bars are provided at the exact required portion. (0.15l from the end support and 0.25l from the interior support). The required areas should also be satisfied
This completes the details of the arrangement using 'straight bars'. In the next section we will see the design of a continuous beam.

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Tuesday, August 25, 2015

Chapter 8 (cont..3) - Straight bars for one way continuous slabs

In the previous section we completed the arrangement of the required reinforcement using bent-up bars. In this section we will see how the same required reinforcement can be achieved by using 'straight bars'.

We have to begin our work from the point where we obtained table 8.3. This table gives us the spacing required at various points. When we used bent-up bars, we saw that the spacing in one span or support will have 'influence' on the adjacent span or support. We were able to give only two spacing: 140mm at all spans and interior supports, and 280 mm at the end supports.

But while using straight bars, there is no such influence. This is because the bottom bars in the spans will not become top bars at supports. So we are free to give any convenient spacing at different points. The only points to note is that the spacing that we provide should be a multiple of 5 or 10mm, and it should be less than the required spacing. Based on this we will prepare the final table which will show the actual spacing given at various points. This is given as table 8.5 below:

Table 8.5

	BM	d req.	Ast req.	S req.	s pr.	Ast pr.
Supp. A			252.91	310.54	200	392.70
Span AB	28.60	90.76	505.82	155.27	150	523.60
Supp. B	-29.76	92.59	527.55	148.88	140	560.71
Span BC	17.75	71.79	309.32	253.78	200	392.70
Supp. C	-27.61	89.53	490.47	160.05	150	523.60
Span CD	24.92	85.06	440.32	178.28	170	462.00
Supp. D			220.16	356.56	200	392.70

The above table gives the final spacing that we intend to give while using 'straight bars'. Note that this table corresponds to the table 8.4 which we prepared for bent-up bars. In span BC, we could have given a spacing of 250 mm (because the required spacing is 253.75). But it is decided to give a spacing of 200 mm. This is based on a common design practice of not giving a spacing greater than 200 mm. Now we will see how the bars are arranged in these spacing using straight bars. The following figs. gives the details.

Fig.8.13
Midspan AB and support B

Reinforcement details of one way continuous slab using straight bars

The above fig. shows all the bars required for the following purposes:
• Resisting the sagging moment at midspan AB
• Resisting the hogging moment at support B
• Resisting the hogging moment due to possible partial fixity at support A

Let us first see the sagging moment at midspan AB. For this, two types of bars are shown in the fig.: a' and b'. Among these, a' are short bars. They do not reach the supports. b' are longer bars. They extend into both the supports A and B. Both these bars should be provided together and alternately. They will work together to resist the sagging moment at midspan AB. This is shown in the plan view given below:

Fig.8.14
Plan view

From the plan view it is clear that, the distance between any two adjacent a' type bars will be equal. If we denote this distance as 2s₁, then the distance between any two b' type bars will also be the same 2s₁. So the 'net' distance between any two adjacent bars will be equal to s₁. From table 8.5 above, it can be seen that we have given a value of 150 mm for s₁. This is less than the required 155.27 mm.

Now we will see the hogging moment at support. For this, two types of bars are shown in the sectional view in the previous fig.8.13: b'' and c''. Among these two types, c'' are longer. Both these bars should be provided together and alternately. They will work together to resist the hogging moment at support B. This is shown in the plan view given below:

Fig.8.15
Plan view at support B

Plan view showing top bars at intermediate support of a continuous one way slab

From the plan view it is clear that, the distance between any two adjacent b'' type bars will be equal. If we denote this distance as 2s₂, then the distance between any two c'' type bars will also be the same 2s₂. So the 'net' distance between any two adjacent bars will be equal to s₂. From table, it can be seen that we have given a value of 140 mm for s₂. This is less than the required 148.88 mm.

Next we will see the details at midspan BC. The arrangement is similar to midspan AB. The fig.8.16 below shows the sectional elevation:

Fig.8.16
Details at midspan BC

Two types of bars are shown in the fig.: c' and d'. Among these, c' are short bars. They do not reach the supports. d' are longer bars. They extend into both the supports B and C. Both these bars should be provided together and alternately. They will work together to resist the sagging moment at midspan BC.

So we have seen the following:
• A typical end support (support A)
• A typical end span (span AB)
• A typical intermediate support (support B)
• A typical intermediate span (span BC)

Based on this we can give a satisfactory arrangement of straight bars for a continuous one way slab with any number of spans. The complete arrangement for the whole of our slab ABCD is given below:

Fig.8.17
Sectional Elevation

Sectional elevation showing reinforcement details of a continuous one way slab

So we have completed the arrangement using straight bars. In the next section, we will see the distribution bars and other details.

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Tuesday, August 18, 2015

Chapter 8 (cont..2) - Checks for continuous slab

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 A_g, the gross area of cross section of the slab, which is obtained as A_g= 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 = 0.0012A_g = 0.0012 x bD = 0.0012 x1000 x200 = 240mm²

Now, spacing required can be determined using Eq.5.3

Where Φ= 8mm and A_st = 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 A_st 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.

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

Details of the curtailment of top bars at an intermediate support of a one way continuous slab.

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 A_st,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 0.5A_st,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

Curtailment of bottom bars in the interior spans of a continuous slab

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 A_st,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 0.5A_st,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'.