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 7 (cont..3) Analysis of Continuous beams using coefficients

In the previous section we did the detailed analysis of the continuous beam ABCDE. Now we will do the analysis using the method of coefficients. For using this method, the beam should satisfy certain conditions that we saw before. If these conditions are satisfied, we can use the coefficients given in the tables 12 and 13 of the code (cl.22.5.1) for obtaining the bending moment and shear forces for the design. Let us see whether our beam satisfies those conditions:

• The section is uniform for all the spans. Because, the width of cross section of the beam is 230 mm, and the depth of cross section is 400mm through out the length of the beam ABCDE.
• The loading type is 'Uniformly distributed'
• No. of spans in our case is four
• Length of the longest span is 4.23m. 15 % of this is 0.6345m. 
     ♦ Difference in length between the second span and the longest is 4.23 -4.165=0.065m < 0.6345m.
     ♦ Difference in length between the third span and the longest is 4.23 -4.065=0.165m < 0.6345m.
     ♦ Difference in length between the fourth span and the longest is 4.23-4.23=0.0m < 0.6345m.

• We are not doing moment redistribution for our beam.


So all the conditions of cl.22.5.1 are satisfied. Now let us see the details of table 12 of the code. We have already seen that the Bending moment can be obtained as:
BM = coefficient x w_u l². 
The 'coefficient' is obtained from table 12.
Using this information, let us now calculate the BM at various points:

From the second column (which is under 'span moments') in table 12, we can obtain the following bending moments:

Bending moment near the middle of end span (span AB) due to DL:
Coefficient = 1/12; w_u =24.94; l=4.23
So BM  = 1/12 x 24.94 x 4.23² = 37.19 kNm

Bending moment near the middle of end span (span AB) due to LL:
Coefficient = 1/10; w_u =6.0; l=4.5
So BM  = 1/10 x 10.23 x 4.23² = 18.30 kNm

So the total BM due to DL and LL = 37.19 +18.30 =55.49 kNm
This value is close to the value of 28.44 kNm which we obtained earlier in fig.7.21

In this way, we can obtain the BM at the other points of the beam. The tables in the following fig.7.22  show the calculation steps:

Fig.7.22

Note that, the coefficients for spans AB and DE are the same because they are both end spans. Similarly, the coefficients for spans BC and CD are the same because they are both interior spans.

Now we will see the negative (hogging) BM at supports. The BM at end supports for this problem will be zero because they are simply supported. So we need to calculate the BM at intermediate supports only. At any intermediate support, the BM will be influenced by the magnitude and type of loads, and also the length of the span on either side of the support. So if the two spans are not equally loaded and/or they have unequal spans, we will get two different values. One from the left side span, and the other from the right side span. It is stated in cl.22.5.1 of the code that: “For moments at supports where two unequal spans meet or in case where the spans are not equally loaded, the average of the two values for the negative moment at the support may be taken for design.” Based on this, the calculations at the supports are shown in the fig.7.23 below:

Fig.7.23

Both the supports B and D use the same coefficients because they are both 'supports next to end support'. B is the support next to the end support A, and D is the support next to the end support E. In this problem, we have yet another type of support, which is the 'interior support' C. It uses different coefficients from B and D.

Now we will see the application of the other set of coefficients given in table 13, for calculating the SF. We have already seen that the Shear force can be obtained as:
SF = coefficient x w_u l.  
The 'coefficient' is obtained from table 13.

Using this information, let us now calculate the SF at various points:
From the second column ('At end supports') in table 13, we can obtain the following SF:

SF at support A due to DL:
Coefficient = 0.4; w_u =24.94; l=4.23
So SF  = 0.4 x 24.94 x 4.23 = 42.20 kN

SF at support A due to DL:
Coefficient = 0.45; w_u =10.23; l=4.23
So SF  = 0.45 x 10.23 x 4.23 = 19.47 kN

So the total SF due to DL and LL = 42.20 +19.47 =61.67 kN
This value is close to the value of 30.36 kN which we obtained earlier in fig.7.21

In this way, we can obtain the SF at the other supports of the beam. The tables in the following fig.7.24  show the calculation steps:

Fig.7.24

It may be noted that, the above values of BM and SF, that we obtained using the 'Method of coefficients' are close to the values obtained using actual structural analysis, the results of which were shown in fig.7.21 in the previous section.

In the next section we will see the design of continuous beams and slabs.

 

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Chapter 4 (cont..5) - Requirements for Deflection control

In the previous section we completed the checks and 'rules' for distributing the bars in a beam. Now we will discuss about the check for deflection control. When loads act on the beam, it deflects. If this deflection is excessive, following undesirable effects may be produced:

• Appearance and efficiency of the structure is affected
• Appearance and efficiency of the partitions and equipments that the structure supports is affected
• Psychological discomfort is caused to the occupants
• Slopes and levels of the floor surfaces are affected
• Wider cracks are formed on the under side of beams and slabs, which will affect their durability
• Deflection of a roof slab will lead to ponding of water over the roof

So we must make sure that the deflection is kept within specified limits. Let us see how this can be done:

The total depth D of a beam or a slab has a major role to play in controlling deflection. We can prove this by taking the example of a beam made completely with a linear elastic material like steel. We will later extend our discussion to reinforced concrete which is a composite material.

Consider one such beam. Let it be simply supported and carry a uniformly distributed load of w per unit length. It has a length of l. We know that the deflection Δ at midspan of such a beam is given by

Eq.4.12

Where l = length of the beam
E = Young's modulus of elasticity of the material of the beam
w = Load per unit length of the beam, and
I = Moment of inertia of the beam section = bd³ /12

We also know that the maximum bending moment M_max occurs at the midspan, and is given by 

Eq.4.13
M_max = wl²/8

From the above Eq.4.13, we get 

Eq.4.14
w = 8M_max / l²

Now we take the basic bending equation:
M/I = f/y = E/R . Here we need only the first two ratios. We get:

Eq.4.15
M = (f/y)I

Let us apply Eq.4.15 to the midspan section of the beam:
M =M_max; f = Maximum tensile stress at extreme fibre; y = half the total depth of the beam = D/2; I = bd³ /12

So we get
Eq.4.16  

Substituting this value of M_max in 4.14, we get
Eq.4.17

Substituting this value of w in the basic equation for deflection (Eq.4.12), we get

This can be simplified as
Eq.4.18

[²⁴⁰ /₁₁₅₂][ ^f/_E] is taken as a constant because both f and E are constants.

So we get a relation between Δ and D. Δ is in the numerator on the left side and D is the denominator on the right side. So, if we increase D, Δ will decrease. The length l cannot change because it depends on the size and shape of the rooms, or the distance between supporting columns.

We derived the above Eq.4.18 for a simply supported beam, carrying a uniformly distributed load. The expression for Δ (Eq.4.12), and the expression for M_max (Eq.4.13) are for simply supported beams. We can do similar derivation for other types like continuous beams, fixed beams etc., and also other types of loads such as point loads, uniformly varying loads etc.,. In all those cases, we will get the same equation as in 4.18. The only differences will be in the value of the 'constant'. This means that in all cases, when D increases, Δ decreases.

The Eq.4.18 cannot be applied directly to reinforced concrete because it is not a linearly elastic material. And also the values of f, I and E are dependent on the extend of cracking, percentage of reinforcement and on the long term effects like creep and shrinkage. So how do we apply it to reinforced concrete beams? The answer is that we follow the procedure given in the code. The code adopts the concept that we discussed above for a linearly elastic material, makes some suitable approximations, and prescribes some limiting l/d ratios. So the next topic that we have to discuss is:

Code procedure for deflection control

Cl.23.2.1 of the code gives us some limiting l/d (span/Effective depth) ratios. Note that here d is the effective depth, not the total depth D. These ratios, which are given by the code are the 'limiting ratios'. We have to calculate our own ratio (denoted as (l/d)_actual) with the actual length and effective depth of the beam that we are designing, and compare it with that given by the code. The ratio that we calculate should not be greater than the ratio that is given by the code. Supposing that the value we obtain from the code for a particular problem is '22'. Then the condition can be represented as (l/d)_actual ≤ 22.  d is in the denominator. So we can say that when the total depth D increases (then the effective depth d will also increase), we have a better chance for satisfying the condition.

So our next aim is to obtain the value of l/d given by the code. It is not very easy to obtain it from the code. What the code gives us is the (l/d)_basic . We have to apply some 'modification factors' to it, to finally obtain the l/d ratio. The 'modification factors' depend upon the particular problem that we are considering. We will now see the values of (l/d)_basic and the modification factors:

Code recommendations for (l/d)_basic:
For reinforced concrete beams of rectangular cross section and slabs of uniform thickness, cl 23.2.1(a) gives the following values:

4.19:
(l/d)_basic for spans upto 10m:

Cantilever       7
Simply supported  20
Continuous     26

Modification factor α 
According to section (b) of the above clause, when the span is greater than 10m, for simply supported and continuous members, we must find the ratio 'α  = span/10' and multiply it to the above basic values. 

But for cantilevers, when the span is greater than 10m, actual deflection calculations should be made.

Modification factor for tension steel k_t 
According to section (c) of the above clause, we must calculate a modification factor k_t from fig.4. of the code. This modification factor depends upon the area of the tension reinforcement and also the stress in the tension reinforcement. So the first step is to find f_st, the stress in the tension steel. In Limit state method, the stresses at ultimate state are considered. So we take the stress when the ultimate load (factored load) is applied on the beam. A newly designed beam will be under reinforced, and so the stress in steel will be 0.87f_y. But for deflection control checks, we consider the working loads. So f_st is the stress in steel when the working loads are applied on the beam. When we discussed 'Working stress method', we learned how to determine the stresses when any given load is applied on the beam. We can determine it whether the beam is under reinforced or over reinforced. But it involves lengthy calculations. So the code gives us a formula to calculate f_st. It is given along with the fig.4 of the code:

The second step is to find the percentage of tension reinforcement. It is given by: 

Where A_st,p is the actual area of steel provided

Now we can find the modification factor k_t from fig.4 of the code. A presentation demonstrating the general procedure for obtaining required values from graphs is given here.

Modification factor k_c
According to section (d) of the above clause, we must calculate a 'modification factor' k_c from fig.5. of the code. This modification factor depends upon the area of the compression reinforcement. So we will discuss about it when we take up the design of doubly reinforced sections. This factor need not be calculated when we design singly reinforced sections.

Modification factor k_f
According to section (e) of the above clause, we must calculate a 'reduction factor' k_f from fig.6. This reduction factor is for flanged sections. So we will discuss about it when we take up the design of flanged sections. This factor need not be calculated when we design singly reinforced sections.

In the above discussion we saw three l/d ratios. They are:
• (l/d)_basic  which is the basic value given by the code. (4.19 given above)
• l/d which is obtained by applying the required modification factors to (l/d)_basic .
• (l/d)_actual which is obtained using the finalized values of l and d of the beam which we are designing.

Their application in three steps can be shown in the form of a flowchart as in the fig.4.15 below:

Fig.4.15
Application of l/d ratios

So the above discussions can be summarized as follows:

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

(l/d)_actual  ≤  [(l/d)_basic] k_t

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

(l/d)_actual  ≤  [(l/d)_basic] α k_t

If the beam is a cantilever with span greater than 10 m, actual deflection calculations should be made.

Cl.23.2 (a) of the code specifies that the final deflection due to all loads including the effects of temperature, creep and shrinkage and measured from the as-cast level of the ,supports of floors, roofs and all other horizontal members, should not normally exceed span/250.

The method of using 'limiting l/d ratios' that we discussed above is expected to satisfy this requirement.

It should be noted that in cases where the span is large, or loading is heavy, or creep and shrinkage effects are more, or when strict deflection control is required, we must calculate the actual deflection, the method of which is discussed in a later chapter.

So we have completed all the checks that have to be done after the design process. We will now see some solved examples which demonstrate the design procedure and various checks:

Solved example 4.1

Solved example 4.2

Solved example 4.3

So we have completed the design and detailing (for flexure) of simply supported beams. Next we will see the analysis of simply supported one way slabs.

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