#CONTINUOUS BEAM 3 UNEQUAL SPANS CODE#

So if the two spans are not equally loaded and/or they have unequal spans, we will get two different values. 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 we need to calculate the BM at intermediate supports only. The BM at end supports for this problem will be zero because they are simply supported. Now we will see the negative (hogging) BM at supports. Similarly, the coefficients for spans BC and CD are the same because they are both interior spans. Note that, the coefficients for spans AB and DE are the same because they are both end spans. The tables in the following fig.7.22 show the calculation steps: In this way, we can obtain the BM at the other points of the beam. This value is close to the value of 28.44 kNm which we obtained earlier in fig.7.21 So BM = 1/12 x 24.94 x 4.23 2 = 37.19 kNmД«ending moment near the middle of end span (span AB) due to LL: Using this information, let us now calculate the BM at various points:ДЇrom the second column (which is under 'span moments') in table 12, we can obtain the following bending moments:Д«ending moment near the middle of end span (span AB) due to DL: The 'coefficient' is obtained from table 12. We have already seen that the Bending moment can be obtained as: Now let us see the details of table 12 of the code. So all the conditions of cl.22.5.1 are satisfied. We are not doing moment redistribution for our beam.в™¦ Difference in length between the fourth span and the longest is 4.23-4.23=0.0m < 0.6345m. в™¦ Difference in length between the third span and the longest is 4.23 -4.065=0.165m < 0.6345m. в™¦ Difference in length between the second span and the longest is 4.23 -4.165=0.065m < 0.6345m. The loading type is 'Uniformly distributed'.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 section is uniform for all the spans.Let us see whether our beam satisfies those conditions:

#CONTINUOUS BEAM 3 UNEQUAL SPANS CODE#

For using this method, the beam should satisfy certain conditions that we saw before.В 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. Now we will do the analysis using the method of coefficients. The previous section we did the detailed analysis of the continuous beam ABCDE.

Continuous beam 3 unequal spans