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 wu 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; wu =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; wu =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 wu 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; wu =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; wu =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.
• 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 wu 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; wu =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; wu =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
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 wu 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; wu =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; wu =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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