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

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 

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'.    

PREVIOUS

NEXT

                                                         

            Copyright ©2015 limitstatelessons.blogspot.com - All Rights Reserved

Chapter 8 (cont..1) - Layout of main bars in a continuous one way slab

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:

	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

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

	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.

PREVIOUS

NEXT

                                                         

            Copyright ©2015 limitstatelessons.blogspot.com- All Rights Reserved

Chapter 8 - Design examples of continuous slabs and beams

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 C_c 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 p_t,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

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:

	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

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:

	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.

 

PREVIOUS

NEXT

                                                         

            Copyright ©2015 limitstatelessons.blogspot.com - All Rights Reserved