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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Effective span example 5

In this section we will calculate the effective spans of the continuous slab shown in the fig.7e.1 below:

Fig.7e.1
Plan and elevation of continuous slab

From the fig., we can see that, in this problem, the supports have different widths. The outer two walls on either sides have a width of 200mm and the middle wall have a width of 230mm.

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. (Fig.7a.4)

Also assume dia. of bottom bars = 10 mm and Cc = 30 mm
So effective depth d = 150 -30 -5 = 115 mm

First we will consider span AB. The calculations based on Eurocode-2 is shown in Table 7e.1 below:

Table 7e.1
Span AB, l_n =2850

	Support A	Support B
Type of support	Non-continuous support	Continuous support
Fig. to use	Fig.(a)	Fig.(b)
h	150	150
t	200	200
ai = lesser of {h/2; t/2}	75	75

l_eff = l_n + a₁ +a₂ =2850 +75 +75 =3000

The calculations based on IS 456 is given below:
Clear span l_n =2850mm.
l_n/12 = 2850/12 =237.50. 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.7e.2 below. This fig. is applicable to all the spans.

Fig.7e.2
Application of chart to span AB

Now we calculate the following:
• c/c distance between the supports = 2850 +100 +100 =3050
• clear span + effective depth = 2850 +115 =2965
Effective span = l_eff = Lesser of the above = 2965mm

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 in Table 7e.2 below:

Table 7e.2
Span BC, l_n =3300

	Support B	Support C
Type of support	Continuous support	Continuous support
Fig. to use	Fig.(b)	Fig.(b)
h	150	150
t	200	230
ai = lesser of {h/2; t/2}	75	75

l_eff = l_n + a₁ +a₂ =3300 +75 +75 =3450

The calculations based on IS 456 is given below:
Clear span l_n =3300mm.
l_n/12 = 3300/12 =275.00. So t₁ < l_n/12 & t₂ < l_n/12

Fig.7b.2 is applicable here also. Now we calculate the following:
• c/c distance between the supports = 3300 +100 +115 =3515
• clear span + effective depth = 3300 +115 =3415
Effective span = l_eff = Lesser of the above = 3415mm

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 in Table 7e.3 below:

Table 7e.3
Span CD, l_n =3300

	Support C	Support D
Type of support	Continuous support	Continuous support
Fig. to use	Fig.(b)	Fig.(b)
h	150	150
t	230	200
ai = lesser of {h/2; t/2}	75	75

l_eff = l_n + a₁ +a₂ =3300 +75 +75 =3450

The calculations based on IS 456 is given below:
Clear span l_n =3300mm.
l_n/12 = 3300/12 =275.00. So t₁ < l_n/12 & t₂ < l_n/12

Fig.7e.2 is applicable here also. Now we calculate the following:
• c/c distance between the supports = 3300 +115 +100 =3515
• clear span + effective depth = 3300 +115 =3415
Effective span = l_eff = Lesser of the above = 3415mm

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 in Table 7e.4 below:

Table 7e.4
Span DE, l_n =2850

	Support D	Support E
Type of support	Continuous support	Non-Continuous support
Fig. to use	Fig.(b)	Fig.(a)
h	150	150
t	200	200
ai = lesser of {h/2; t/2}	75	75

l_eff = l_n + a₁ +a₂ =2850 +75 +75 =3000

The calculations based on IS 456 is given below:
Clear span l_n =2850mm.
l_n/12 = 2850/12 =237.50. So t₁ < l_n/12 & t₂ < l_n/12

Fig.7e.2 is applicable here also. Now we calculate the following:
• c/c distance between the supports = 2850 +100 +100 =3050
• clear span + effective depth = 2850 +115 =2965
Effective span = l_eff = Lesser of the above = 2965mm

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

Table 7e.5: Effective spans

Name of span	Based on Euro code	Based on IS456
AB	3000	2965
BC	3450	3415
CD	3450	3415
DE	3000	2965

We have branched off from the main discussion. The layout map given below will help us to navigate easily between the various sections. Links to some more examples are also given in the layout map.


                                                         

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Effective span example 4

In this section we will calculate the effective spans of the continuous slab shown in the fig.7d.1 below:

Fig.7d.1
Plan and elevation of continuous beam

From the fig., we can see that, in this problem, all the supports have equal width. We will calculate the effective spans by the two different methods: The one based on Eurocode-2, and the other based on IS 456.

Assume dia. of bottom bars = 10 mm, and Cc =30mm.  So effective depth d = 150 -30 -5 = 115 mm

First we will consider span AB. The calculations based on Eurocode-2 is shown in Table 7d.1 below:

Table 7d.1
Span AB, l_n =2850

	Support A	Support B
Type of support	Non-continuous support	Continuous support
Fig. to use	Fig.(a)	Fig.(b)
h	150	150
t	300	300
ai = lesser of {h/2; t/2}	75	75

l_eff = l_n + a₁ +a₂ =2850 +75 +75 =3000

The calculations based on IS 456 is given below:
Clear span l_n =3500mm.
l_n/12 = 2850/12 =237.5. So t₁ ≮ l_n/12 & t₂ ≮ l_n/12

Note the 'Not less than' symbol in the above step. So we cannot turn left. Let us see if we can turn right: For this, we have to calculate the lesser of {ln/12; 600} That is., the lesser of {237.50 ; 600}, which is 237.50. So we can write:
t₁ > lesser of {ln/12 ; 600} & t₂ > lesser of {ln/12 ; 600}

Thus we take the deviation to the right. The path taken by the calculations is shown in the fig.7d.1 below:

Fig.7d.1
Application of chart to span AB

At 'A', the deviation is towards the right. If at 'A', the deviation is to the right, we will be needing the whole portion of the chart. So proceeding downwards, at 'B', the deviation is towards the left because, the span AB that we are considering, is an End span. Next, at 'C', the deviation is towards the right because, span AB has one end free, and the other end continuous. Thus we reach a 'blue box' which gives us the steps in the final calculations that have to be made.

So we calculate the following:
• clear span + 0.5 x effective depth =2850 + 0.5 x115 =2907.5
• clear span + 0.5 x width of discontinuous support =2850 + 0.5 x300 =3000 
Effective span = l_eff = Lesser of the above = 2907.5mm

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 in Table 7d.2 below:

Table 7d.2
Span BC, l_n =3300

	Support B	Support C
Type of support	Continuous support	Continuous support
Fig. to use	Fig.(b)	Fig.(b)
h	150	150
t	300	300
ai = lesser of {h/2; t/2}	75	75

l_eff = l_n + a₁ +a₂ =3300 +75 +75 =3450

The calculations based on IS 456 is given below:
Clear span l_n =3300mm.
l_n/12 = 3300/12 =275.00. So t₁ ≮ l_n/12 & t₂ ≮ l_n/12

Note the 'Not less than' symbol in the above step. So we cannot turn left. Let us see if we can turn right: For this, we have to calculate the lesser of {ln/12; 600} That is., the lesser of {275.00 ; 600}, which is 275.00. So we can write:
t₁ > lesser of {ln/12 ; 600} & t₂ > lesser of {ln/12 ; 600}

Thus we take the deviation to the right. The path taken by the calculations is shown in the fig.7d.2 below:

Fig.7d.2
Application of the chart to span BC

The path taken is different from that in the previous fig.7d.1. This is because the previous span AB is an end span. The present span BC is an intermediate span. So here, the deviation at 'B' is towards the right. After this point, there is no other deviations, and we reach a 'blue box'. Thus the effective span is the clear span between the supports. So we can write:
l_eff of span BC = 3300

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 in Table 7d.3 below:

Table 7d.3
Span CD, l_n =3450

	Support C	Support D
Type of support	Continuous support	Continuous support
Fig. to use	Fig.(b)	Fig.(b)
h	150	150
t	300	300
ai = lesser of {h/2; t/2}	75	75

l_eff = l_n + a₁ +a₂ =3300 +75 +75 =3450

The calculations based on IS 456 is given below:
Clear span l_n =3300mm.
l_n/12 = 3300/12 =275. So t₁ ≮ l_n/12 & t₂ ≮ l_n/12


Note the 'Not less than' symbol in the above step. So we cannot turn left. Let us see if we can turn right: For this, we have to calculate the lesser of {ln/12; 600} That is., the lesser of {275 ; 600}, which is 275. So we can write:
t₁ > lesser of {ln/12 ; 600} & t₂ > lesser of {ln/12 ; 600}

The fig.7d.2 is applicable here also because CD is also an intermediate span just like BC. Thus the effective span is the clear span between the supports. So we can write:
l_eff of span CD = 3300

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 in Table 7d.4 below:

Table 7d.4
Span DE, l_n =2850

	Support D	Support E
Type of support	Continuous support	Non-Continuous support
Fig. to use	Fig.(b)	Fig.(a)
h	150	150
t	300	300
ai = lesser of {h/2; t/2}	75	75

l_eff = l_n + a₁ +a₂ =2850 +75 +75 =3000

The calculations based on IS 456 is given below:
Clear span l_n =2850mm.
l_n/12 = 2850/12 =237.50. So t₁ ≮ l_n/12 & t₂ ≮ l_n/12

Note the 'Not less than' symbol in the above step. So we cannot turn left. Let us see if we can turn right: For this, we have to calculate the lesser of {ln/12; 600} That is., the lesser of {237.5 ; 600}, which is 237.5. So we can write:
t₁ > lesser of {ln/12 ; 600} & t₂ > lesser of {ln/12 ; 600}

The first fig.7d.1 is applicable here also because DE is also an end span just like AB

So we calculate the following:
• clear span + 0.5 x effective depth =2850 + 0.5 x115 =2907.5
• clear span + 0.5 x width of discontinuous support =2850 + 0.5 x300 =3000 
Effective span = l_eff = Lesser of the above = 2907.5mm

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

Table 7d.5: Effective spans

Name of span	Based on Euro code	Based on IS456
AB	3000	2907.5
BC	3450	3300
CD	3450	3300
DE	3000	2907.5

We have branched off from the main discussion. The layout map given below will help us to navigate easily between the various sections. Links to some more examples are also given in the layout map.


                                                         

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