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 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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Chapter 7 - Analysis of continuous one-way slabs

In the previous chapters we saw the analysis and design of simply supported beams and slabs. In this chapter we will discuss about the analysis of continuous one-way slabs. When we complete this discussion about continuous one way slabs, we will be able to do the analysis of continuous beams also.

A video of a small scale experiment, showing the behaviour of continuous beams can be seen here 

A 3D view of a continuous beam is given in fig.7.1 below. A view of a continuous slab can be seen in fig.7.2

Fig.7.1
View of a continuous beam

Before we proceed with the discussion on continuous members, we must know how to calculate the 'Effective span' of the various spans. A detailed discussion is given here.

We have already discussed about the effective span of simply supported members here.


To design a member like a slab or beam, we have to first find the bending moments (BM) and shear forces (SF) at the various points (supports, midpoints of spans etc.,) of the member. In the case of simply supported members, these can be found out easily using the three equations of equilibrium. But in the case of continuous members, we have to use one of the methods like Moment distribution method, Slope deflection method or kani's method.

Following solved example demonstrates the method of obtaining the BM and SF at various points of a continuous slab.

Solved example 7.1:
Consider a continuous one-way slab ABCD shown in the fig.7.2 below. It has a thickness of 200mm. It carries a Dead load of 1.5 kN/m² in addition to it's self weight. Also there is a LL of 4 kN/m². The effective spans of AB, BC and CD are 4.5, 4.0 and 4.2 m respectively. Our aim is to find the bending moments and shear forces at different points of the slab.

Fig.7.2
View of continuous slab ABCD

Solution:
Just as we saw in the case of a simply supported one-way slab, in the case of a continuous one-way slab also, we consider a strip of 1m width. This strip extends from one end support 'A' to the other end support 'D', as shown in the fig.7.3 below:

Fig.7.3
Strip of 1m width in a continuous slab

Now it has become a continuous beam of width 1m, and total depth D which is equal to the thickness of the slab. We know how to calculate the load per meter length of this beam: It is the load per meter area of the slab. (details here) So the next step is to calculate this load per unit area.

DL per unit area:
• Self weight = 25D = 25 x0.2 = 5.0 kN/m²
• Additional DL (given) = 1.5 kN/m²
• Total DL, w_DL = 6.5 kN/m²
Thus w_u,DL = 1.5 x6.5 = 9.75 kN/m²

• LL per unit area (given) w_LL = 4.0 kN/m²
Thus w_u,LL = 1.5 x4.0 = 6.0 kN/m²

We have computed DL and LL separately. The reason for this will be explained soon.

These loads per square area on the slab is same as the load per meter length of the 1m wide beam. So we can write:

• w_u,DL per meter length of the beam = 9.75 kN/m
• w_u,LL per meter length of the beam = 6.0 kN/m


With the above information, we can draw the line diagram for analysis. This is shown in the fig.7.4 below:

Fig.7.4
Line diagram for analysis

Based on the above fig.7.4, we can do the structural analysis of the beam ABCD. We can use methods like slope deflection method, moment distribution method, or kani's method. (The steps of the structural analysis are not shown here). Based on the results of the analysis, we can draw the BM and SF diagrams. These are shown in the figs.7.5 and 7.6 below.

Fig.7.5
Bending moment diagram

Fig.7.6
Shear force diagram

We can tabulate the values at different points as shown in the fig.7.7 below:

Fig.7.7
Tabulated results

So from the above table, we will get the required values of BM and SF at the various points. It may seem that we can now start the design process. But we are not there yet. The values in the above table, are not the 'maximum values'. Even though the whole beam ABCD has been loaded with all the maximum LL and DL, there are some 'possibilities' by which, the BM and SF at the various points along the beam will exceed the values shown in the table. Let us now examine such possibilities:

Consider a situation in which the first span AB is loaded with maximum possible DL and LL. But it's adjacent span BC and the span CD after that does not carry any LL. The line diagram of this situation is shown in the fig.7.8 below:

Fig.7.8
LL on span AB only

Note that in the above fig., the DL on spans BC and CD, are the 'unfactored' DL. This means they are the characteristic DL. The reason for this can be explained as follows: We want to know the result when 'full load' is applied on AB and 'reduced load' is applied on BC and CD. So we make a reduction on the loads on BC and CD. While making such a reduction, we must make a 'maximum possible reduction'. To achieve this, we must remove not only the LL, but also the 'Extra DL that can possibly occur'. So we avoid the application of the 'Load factor' to the DL, thus using only it's characteristic value. In short, we are using the 'lowest possible loads' on BC and CD.

Based on the above line diagram, we can do a structural analysis. The results are shown in the table in fig.7.9 below:

Fig.7.9
Tabulated results

We can see that the sagging moment at the midspan in AB is now 27.13. But in the fig.7.7, it was 24.95. So the sagging moment has increased due to the reduction in loads on other spans. In this way, the moments at other points also may increase depending on the pattern of loading. We must consider all possibilities. That is., we must consider all possible load combinations. We must do the structural analysis for each of these combinations, and then note down the maximum value at various points. All possible combinations are analysed, and the results are shown in tabular form in the fig.7.10 below:

Fig.7.10
Possible combinations and results

Now from the above fig., we have to take out the maximum values at each point along the length of the slab. This is shown in the fig.7.11 below:

Fig.7.11
Extreme values

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 because, for one particular combination (case 6), this mid span is bending upwards. This usually happens when a lightly loaded short span comes in between two heavily loaded long spans.

We can examine the various extreme values in the above fig. and make the following conclusion:

(a) Extreme values of the negative (hogging) BM at supports:

• -32.45 kNm for the first interior support 'B' is obtained from case 4 where LL is applied on the spans on either side of that support.
• -29.08 kNm for the second interior support 'C' is obtained from case 5 where LL is applied on the spans on either side of that support.


In general, the extreme values of the negative BM at a support occurs when the LL is applied on the spans on either side of that support and also on every alternate span thereafter.

'and also on every alternate span thereafter' in the above statement cannot be demonstrated in our slab because it has only three spans. On a beam or slab which has more spans, it can be shown as in fig.7.12 below:

Fig.7.12
Loading arrangement for maximum hogging moment at support B

(b) Extreme values of positive BM at center of spans:

• 28.44 kNm at the center of span AB is obtained from case 6 where the LL is applied on span AB itself and also on span CD which is an alternate span.
• 12.9 kNm at the center of span BC is obtained from case 2 where LL is applied on span BC only.
• 25.45 kNm at the center of span CD is obtained from case 6 where the LL is applied on span CD itself and also on span AB which is an alternate span.


In general, The maximum positive (sagging) moment in a span occurs when LL is applied on that span and every other alternate span. On a beam or slab which has more spans, it can be shown as in fig.7.13 below:

Fig.7.13
Loading arrangement for maximum sagging moment at midspan BC

(c)Minimum value of positive BM at center of spans:
• 6.8 kNm at the center of span AB is obtained from case2 where
1. LL is applied on the adjacent span which is BC and 
2. No LL is applied on AB & CD.

• -7.7 kNm at the center of span BC is obtained from case6 where
1. LL is applied on the adjacent spans which are AB and CD 
2. No LL is applied on BC.

• 5.33 kNm at the center of span CD is obtained from case2 where
1. LL is applied on the adjacent span which is BC and 
2. No LL is applied on AB & CD.

In general, The minimum positive moment in a span occurs when

1. No LL is applied on that span but
2. LL is applied on the adjoining spans and every alternate span thereafter.


On a beam or slab which has more spans, it can be shown as in fig.7.13 below:

Fig.7.13
Loading arrangement for minimum sagging moment at midspan BC

It should be noted that this minimum positive moment may turn out to be the maximum negative moment at the midspan. That is, the beam will be bending upwards. This has occurred in case 6 above where a lightly loaded short span is placed in between two heavily loaded long spans

(d) Extreme values of total shear force at supports:
• 30.36 kN for the first end support 'A' is obtained from case6 where:
1. LL is applied on the span AB of which this support is a part and 
2.  LL is applied on the alternate span CD

• 78.56 kN (42.65 from the outer side plus 35.91 from the inner side) for the first interior support 'B' is obtained from case4 where LL is applied on the spans on either side of that support.

• 74.6 kN (40.0 from the outer side plus 34.6 from the inner side) for the second interior support 'C' is obtained from case5 where LL is applied on the spans on either side of that support.

• 28.66 kN for the last end support 'D' is obtained from case6 where
1. LL is applied on the span CD of which this support is a part and 
2.  LL is applied on the alternate span.

In general, the extreme values of the shear force at a support occurs when the LL is applied on the spans on either side of that support and also on every alternate span thereafter. This is the same condition for obtaining the extreme values of negative BM at supports (fig.7.12).

So we obtained some general conditions for the extreme values of BM and SF. We did this using an example. But these general conditions can be obtained by using the principles of 'Influence lines' also.

In the next section we will discuss a simplified procedure given by the code.

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