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.

 

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Chapter 7 (cont..2)

In the previous section we completed the analysis of a continuous slab. In this section we will do the analysis of a continuous beam. A view of a four span continuous beam ABCDE is given in fig 7.17 below:

Fig.7.17
View of a continuous beam ABCDE

Total depth of the beam = 400 mm
Width of the beam = 230 mm
Width of all outer supports A,B, D & E = 230 mm
Width of inner support C = 300 mm

In the previous example of a continuous slab, we were required to calculate the loads acting per unit length of the slab (unit length of a 1000mm wide strip). But here, the loads acting per meter length of the beam is given with the problem: 

Span	DL	LL	Factored DL	Factored LL
AB	16.63	6.82	24.94	10.23
BC	16.48	6.75	24.72	10.13
CD	16.25	6.64	24.37	9.96
DE	16.63	6.82	24.94	10.23

Effective spans: But unlike in the previous example, here, only the clear spans and the widths of supports are given. We have to calculate the effective span of the various spans.

We will calculate the effective spans by the two different methods: The one based on Eurocode-2, and the other based on IS 456. For this problem, it is convenient to mention before hand that, each of the spans have their support widths less than it's ln/12. So while using the cl.22.2(b) of IS 456, we will not have to look to the portion below the magenta colored dashed line of the chart. (see fig.7a.4)

Also assume dia. of bottom bars = 16 mm, dia. of links =8mm and Cc = 30 mm
So effective depth d = 400 -30 -8 -8 = 354 mm

First we will consider span AB. The calculations based on Eurocode-2 is shown below: (fig.7a.1)

Span AB, l_n =4000

	Support A	Support B
Type of support	Non-continuous support	Continuous support
Fig. to use	Fig.(a)	Fig.(b)
h	400	400
t	230	230
ai = lesser of {h/2; t/2}	115	115

l_eff = l_n + a₁ +a₂ =4000 +115+115=4230

The calculations based on IS 456 is given below:
Clear span l_n =2850mm.
l_n/12 = 4000/12 =333.33. So t₁ < l_n/12 & t₂ < l_n/12

As mentioned above, we only need the portion above the magenta colored dashed line for all spans of the beam. This is shown in the fig.7.18 below. This fig. is applicable to all the spans.

Fig.7.18
Application of chart to span AB

Now we calculate the following:
• c/c distance between the supports = 4000 +115 +115 =4230
• clear span + effective depth = 4000 +354 =4354
Effective span = l_eff = Lesser of the above = 4230mm

Thus we calculated l_eff of span AB using the two methods.

Now we will consider span BC. The calculations based on Eurocode-2 is shown below:

Span BC, l_n =3900

	Support B	Support C
Type of support	Continuous support	Continuous support
Fig. to use	Fig.(b)	Fig.(b)
h	400	400
t	230	300
ai = lesser of {h/2; t/2}	115	150

l_eff = l_n + a₁ +a₂ =3900 +115+150=4165

The calculations based on IS 456 is given below:
Clear span l_n =3300mm.
l_n/12 = 3900/12 =325.00. So t₁ < l_n/12 & t₂ < l_n/12

Fig.7.18 is applicable here also. Now we calculate the following:
• c/c distance between the supports = 3900 +115 +150 =4165
• clear span + effective depth = 3900 +354 =4254
Effective span = l_eff = Lesser of the above = 4165mm

Thus we calculated l_eff of span BC using the two methods.

Now we will consider span CD. The calculations based on Eurocode-2 is shown below:

Span CD, l_n =3800

	Support C	Support D
Type of support	Continuous support	Continuous support
Fig. to use	Fig.(b)	Fig.(b)
h	400	400
t	300	230
ai = lesser of {h/2; t/2}	150	115

l_eff = l_n + a₁ +a₂ =3800 +150+115=4065

The calculations based on IS 456 is given below:
Clear span l_n =3800mm.
l_n/12 = 3800/12 =316.67. So t₁ < l_n/12 & t₂ < l_n/12

Fig.7.18 is applicable here also. Now we calculate the following:
• c/c distance between the supports = 3800 +150 +115 =4065
• clear span + effective depth = 3800 +354 =4154
Effective span = l_eff = Lesser of the above = 4065mm

Thus we calculated l_eff of span CD using the two methods.

Now we will consider span DE. The calculations based on Eurocode-2 is shown below:

Span DE, l_n =4000

	Support D	Support E
Type of support	Continuous support	Non-Continuous support
Fig. to use	Fig.(b)	Fig.(a)
h	400	400
t	230	230
ai = lesser of {h/2; t/2}	115	115

l_eff = l_n + a₁ +a₂ =4000 +115+115=4230

The calculations based on IS 456 is given below:
Clear span l_n =4000mm.
l_n/12 = 4000/12 =333.33. So t₁ < l_n/12 & t₂ < l_n/12

Fig.7.18 is applicable here also. Now we calculate the following:
• c/c distance between the supports = 4000 +115 +115 =4230
• clear span + effective depth = 4000 +354 =4354
Effective span = l_eff = Lesser of the above = 4230mm

Thus we calculated l_eff of span DE using the two methods. All the results from the two methods are tabulated below:

Effective spans:

Name of span	Based on Euro code	Based on IS456
AB	4230	4230
BC	4165	4165
CD	4065	4065
DE	4230	4230

Now the beam is analysed for the ten possible combinations of the loads, using Kani's method. The first combination (where LL is applied on span AB only) is denoted as case1, and it's line diagram is shown in fig.7.19 below:

Fig.7.19
Load arrangement for case 1

In this way, a total of ten combinations are analysed. The results are shown in the fig.7.20 below:

Fig.7.20
Results of analysis

Extreme values are picked from the above fig., and are tabulated below in fig.7.21:

Fig.7.21
Extreme values of BM and SF

In the above figs., '1' denotes the maximum BM at supports; '2' denotes the maximum BM at midspans, and '3' denotes the maximum SF at supports.

*2.a denotes the minimum BM at midspans. We have to take special care about minimum BM in BC and CD because, for some load combinations (cases 8 and 9), these spans are bending upwards. 

So now we have the BM and SF at the various points of the continuous beam. Just as in the previous example, let us calculate them again using the coefficients given in the code. We will do this in the next section

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Chapter 7 (cont..1)

In the previous section we saw the detailed procedure of analysis of a continuous one-way slab. By doing the analysis, we obtained the BM and SF at various points of the slab.  In this section we will see an alternate method to obtain these BM and SF. This method is called the 'Method of coefficients'. In this method, we do not have to make a complete structural analysis using Slope deflection method, Moment distribution method, Kani's method etc., We can obtain BM and SF of any span just by multiplying the length of the span and the load with some coefficients. So this is an easy method. But to use this method, the continuous one-way slab (or beam) should satisfy certain conditions:

• The beam section for all the spans should be uniform
• The loading should be substantially uniformly distributed load.
• There must be three or more spans.
• The lengths of the spans should not differ by more than 15% of the longest.
• When we use this method, we must not carry out 'moment redistribution'

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 if the above conditions are satisfied in our case:
• The section is uniform for all the spans. Because, the width of cross section of the beam is 1000 mm, and the depth of cross section is 200mm through out the length of the slab.
• The loading type is 'Uniformly distributed'
• No. of spans in our case is three
• Length of the longest span is 4.5m. 15 % of this is 0.675m. 
     ♦ Difference in length between the second span and the longest is 4.5-4.0=0.5m < 0.675m.
     ♦ Difference in length between the third span and the longest is 4.5-4.2=0.3m < 0.675m.
• 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.

According to the note below table 12, "to get the bending moment, we must multiply the coefficient by the 'total design load' and the effective span". - - -(1) 

The total design load is equal to the 'design load per meter' multiplied by the effective span.
That is., Total design load =w_u l - - - (2)

According to (1), we must multiply (2) again by l. So the bending moment can be obtained as follows: By multiplying the coefficient by the design load per meter and the square of the effective span. That is., 
BM = coefficient x w_u l². - - - (3)

Using the above 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 =9.75; l=4.5
So BM  = 1/12 x 9.75 x 4.5² = 16.45 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 6.0 x 4.5² = 12.15 kNm

So the total BM due to DL and LL = 16.45 +12.15 =28.6 kNm
This value is close to the value of 28.44 kNm which we obtained earlier in fig.7.11

In this way, we can obtain the BM at the other points of the beam. The tables in the following fig.7.14  show the calculation steps:

Fig.7.14

Note that, the coefficients for Spans AB and CD are the same because they are both end 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.7.15 below:

Fig.7.15

Both the supports B and C use the same coefficients because they are both 'supports next to end support'. B is the support next to the end support A, and C is the support next to the end support D. If in our problem, there were four spans, there would have been a support between B and C. Then that support would have been an 'interior support'. Such a support uses different coefficients than those used for B and C above. 

Now we will see the application of the other set of coefficients given in table 13, for calculating the SF.

According to the note below table 13, "to get the Shear force, we must multiply the coefficient by the total design load". - - -(1) 

The total design load is equal to the 'design load per meter' multiplied by the effective span.
That is., Total design load =w_u l - - - (2)

According to (1), we must multiply (2) by the coefficient. So we get: 
SF = coefficient x w_u l. - - - (3)

Using the above 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 =9.75; l=4.5
So SF  = 0.4 x 9.75 x 4.5 = 17.55 kN

SF at support A due to DL:
Coefficient = 0.45; w_u =6.0; l=4.5
So SF  = 0.45 x 6.0 x 4.5 = 12.15 kN

So the total SF due to DL and LL = 17.55 +12.15 =29.7 kN
This value is close to the value of 30.36 kN which we obtained earlier in fig.7.11

In this way, we can obtain the SF at the other supports of the beam. The tables in the following fig.7.16  show the calculation steps:

Fig.7.16

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.7.11 in the previous section.

In the next section we will see the analysis of a continuous beam. 

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