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.

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Chapter 6 - Design of One-way slabs

In the previous section we completed the analysis of one way slabs. Now we will see the design process. As in the case of beams, here also, we will first see the procedure for fixing up the 'preliminary dimensions'. We have seen that for analysis and design, the slabs are considered as strips, and we can take any one strip. As we are dealing with 1m wide strips, the width of the beams that we have to design is fixed at 1000mm. We have to fix up only the preliminary depth. Let us see how this is done:

We know that the depth provided to a beam must be sufficient for it to resist the bending moment acting on it. In addition to this, the depth must be sufficient to control the deflection also. So the depth depends on two criteria:
• To resist the bending moment, and
• To control deflection. 

In the case of slabs, where the depth is relatively low, the criterion for deflection control becomes more critical. That is., a slab with a certain depth may effectively resist the bending moment acting on it. But this same depth may not be sufficient to control the deflection. In other words, the slab generally requires more depth to control it's deflection than it requires to resist bending. 


So to fix up the preliminary depth, we must use the deflection control criterion. Let us take the case of a simply supported slab (whose effective span is less than 10m). We have already discussed the procedure for deflection control here. The effective depth that the slab actually have in the final structure should satisfy the following relation:

(l/d)_actual  ≤  [(l/d)_basic] α k_t  - - - (1) 

• For simply supported spans, (l/d)basic = 20
• As we are considering spans less than 10m, α need not be taken into account. 
Thus we get :
(l/d)_actual  ≤  20 x k_t - - - (2)

From this expression, we want to calculate 'd'. But k_t is also an unknown. So we assume a value for k_t. The assumed value should be as accurate as possible. For this we look at the numerous design examples of simply supported slabs that are done in the past by different designers. By examining them, we can find that the percentage of reinforcement p_t generally falls in the range 0.4–0.5. 

Sample calculation:
Let a slab have the following properties:
D =150mm; d =115mm; Ф =10mm; s =150mm
Then from Eq.5.2,

= 490.87mm². So p_t = 100A_st /bd = [490.87 x100]/[1000 x115] = 0.427 

Based on this information, we adopt the following procedure:
• Assume that the area of steel provided is equal to the area of steel required,
• Then f_st will be equal to 0.58f_y. 
Take the value of f_y that is going to be used. If it is 415, then f_st = 240.7
So we have the values required to find k_t: They are:
(a) p_t in the range 0.4–0.5, and
(b) f_st = 240.7

Corresponding to these values, we will get k_t = 1.25 from Fig.4 of the code.

So substituting in (2) we get
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. 

For continuous slabs, we can write:
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

From the above expressions, 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. 

As in the case of beams, we must do all the design checks to ensure that the final section satisfies all requirements. If the section fails to do so, the whole process should be repeated with improved dimensions.

So we have fixed the preliminary dimensions. The next step is to find the effective depth required from bending moment considerations. Once this is calculated, we must compare it with the d that we have from the preliminary depth. If the required value is less, we can proceed to find the steel required to resist this bending moment. These two steps are same as that for a beam.


Now we must know the 'rules' for distributing the calculated steel into the slab.

Concrete cover and grade of concrete:
The details about the concrete cover that has to be provided to the bars of a slab are same as those of a beam. They were discussed in the section about the design of beams, and we saw table 4.1.


So, if we are to design a slab which will be subjected to 'moderate' exposure conditions, from table 4.1,we get the value of C_c to be provided as 30 mm , and the minimum grade of concrete to be used as M25

Minimum spacing to be provided between the bars of a slab:
The clear space provided between parallel reinforcing bars should not be less than the minimum value specified in cl 26.3.2 of the code. We have seen the details of this clause when we discussed about the minimum spacing between bars of beams here. The same is applicable to slabs also. So we can show the application of this clause to slabs as shown in the fig.6.1 below:

Fig.6.1
Minimum spacing between bars of slab

Maximum spacing allowable between bars of a slab:
When we design the slab we will get the area of steel that should be provided. Sometimes, when we convert it into 'spacing of bars of a particular diameter', we may get a large spacing s. This often happens when the bending moment that the slab has to resist, is low. Large values of s may be obtained on other occasions also: When we choose to provide large diameter bars. This can be explained as follows: From 5.3 we know that

In the above equation, for the same value of A_st, we will get a larger s for a larger Ф.  

It may seem to be economical to provide bars at large spacing. But bars at large spacing will not be able to control cracks effectively. Also the bond between steel and concrete will become lesser when bars are at a large distance apart.

So the code specifies some upper limits to  the spacing. The c/c spacing provided between parallel reinforcing bars of a slab should not be greater than that specified in cl 26.3.3 (b) of the code. The requirements stated in this clause can be shown as in fig 6.2 given below:

Fig.6.2
Maximum spacing between bars of slab

In the above fig., it is indicated that the diameter of the main bars should not exceed D/8. This is because 'small diameter bars spaced closer together' is effective in reducing cracks and in improving bond than 'large diameter bars spaced at larger distances'. However, bars smaller than 8mm in diameter are generally not used as main bars of slabs. 

Minimum Area of reinforcement for slabs:
The area of reinforcement provided in slabs should not be less than that specified in cl 26.5.2 of the code. The requirements stated in this clause can be written as follows:

6.3: For Fe 250 steel:
A_st ≮ 0.0015A_g


6.4: For Fe 415 steel:
A_st ≮ 0.0012A_g

Where A_st is the area of steel provided in the slab, and A_g is the gross area of cross section of the slab. A_g is equal to bD where b is the width of the slab and D is the total depth of the slab. When we consider a strip of slab for the design, b is equal to the width of the strip.

Transverse reinforcement for one way slabs :
We have seen in the previous chapter that transverse reinforcement is required for one-way slabs. These bars are also called distribution bars of a One-way slab. The quantity of this reinforcement that we have to provide, is given by the code, and is same as that given by 6.3 and 6.4 above.

The c/c spacing between the distribution bars should not exceed the smallest of the following (cl 26.3.3.b (2)):

• 5 times the effective depth d of the slab
• 450 mm 

This is shown in fig. 6.1 above

Deflection control for one-way slabs
We have seen some details about deflection control checks in one way slabs when we discussed about the 'preliminary depth' at the beginning of this chapter. There we assumed a value for p_t. But now we are discussing the actual 'check'. This is the 'last but one check' that we have to perform after the design of a one way slab (The last one being the check for M_uR). So here we will be using the actual p_t that is provided. The procedure is same as that for a singly reinforced beam. So we can write the same expressions that we wrote for the beam:

4.19
For singly reinforced rectangular slabs with span less than 10m,

(l/d)_actual  ≤  [(l/d)_basic] k_t

4.20
For singly reinforced rectangular slabs with span greater than 10m,

(l/d)_actual  ≤  [(l/d)_basic] α k_t

Partial fixity at supports:
Another topic that we must consider, is the design at supports. If the slab is simply supported, the bending moment will be zero at the supports. The details to be considered in such a situation is demonstrated in the following presentation:


(The first five slides are same as the ones we saw in an earlier presentation about the behaviour of slabs in chapter 5)




This completes the discussion about the design of a simply supported one way slab. We will now see a solved example which demonstrates the actual process of design.

Solved example 6.1

In the above solved example, we have determined the reinforcement requirements of a simply supported one way slab. Based on that design we must draw a plan and a sectional view, showing the layout of bars. This is given in the fig.6.3 and 6.4 below:

Fig.6.3
Plan view showing reinforcement details of slab

Fig.6.4
Sectional elevation

In the plan view, we can see that alternate bars are bent up at supports. The reason for this can be explained as follows: We have designed the slab as 'simply supported'. So the bending moments at the support will be equal to zero. But if in the future, partial fixity (shown in the presentation above) is introduced at any of the supports, bending moments will develop at that support. These bending moments will be hogging in nature. So tension will develop near the top surface of the slab at such supports. Thus we have to provide top steel. This can be achieved by bending up the bars. The requirements for this top reinforcement is given by cl.D-1.6 of the code. According to this clause, the bars must extend a distance of 0.1l from the support into the slab. Also, these bars should have an area equal to 50% of that provided at midspan. Thus, when we bend up alternate bars, we will get 50% at the top.

Technically, we need to show only two bars of a particular set. But if we show more bars, the symmetry and pattern will become more clear. Such a plan view is given here. 

In the sectional view in fig.6.4 above, the two bars are shown separately only for clarity. The two bars are in fact provided in the same layer as indicated by the '0'mm dimension.

This completes the design and detailing of a 'simply supported' one way slab. Next we will see 'continuous' beams and one way slabs. But to analyse continuous beams and slabs, we must know how to calculate the 'effective spans' in continuous members. So in the next chapter we will see the details about this effective span also.

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