ADJUSTING INTERNAL FORCE OF GIRDER WITH DIFFERENT SPAN
When designing and constructing reinforced concrete structures, especially floor beams, the treatment of different spans is an important factor to ensure the safety and efficiency of the structure. Beams with different spans can lead to uneven distribution of internal forces, affecting the bearing capacity and stability of the entire structure. Therefore, adjusting the internal forces for beams with different spans is necessary to optimize the performance of the structure and ensure safety.
1. Internal force adjustment method
When the adjacent spans differ by more than 10%, the existing results can be used with the necessary adjustments. It is usually necessary to adjust the positive moment (internal force) at the larger span and the negative moment range of action at the smaller span.
1.2 Consider a 2-span beam
When the 2 spans are equal, we have positive moments at the spans M1 and M2 and negative moments at the support is Mb.
Next, keep the BC span as l and reduce the AB span and set it as x.
We see that when decreasing x from l to zero, M1 decreases to M1′, Mb decreases to Mb’ and M2 increases to M2’=kM2. Until M1′ decreases to zero, from here on in the AB span there is only negative moment.
When x = 0, the beam becomes 1 span and Mb’=0; M2′ reaches its maximum value.
From the above analysis, we can draw the following conclusions:
- For safety in design, we can take Mb as when calculating with a beam of equal span, take M1′ calculated with span x, take M2’ = k1.M2 with k1=1.375 - (0.375x/l).
1.2 Consider the 3-span beam in case 1
When the 3 spans are equal, we have positive moments at the spans M1 and M2, M3 and negative moments at the supports are Mb, Mc.
Next, keep the middle span equal to l, reduce 1 or both edge spans to x1, x3, then the positive moments M1, M3 gradually decrease (as with M1 of a two-span beam).
We see that when x is gradually reduced, the negative moments Mb, Mc also decrease in the beam according to x1, x3. The positive moment M2 increases to M2’=k2.M2 with:
For safe design, do not reduce Mb and Mc. Take M1′ and M3; according to the span x1, x3.
1.3 Consider the 3-span beam in case 2
Keep the first span and the third span equal to l, reduce the middle span BC to x2 < l (figure c above).
When reducing x2, Mb’ and Mc’ initially decrease, but at some point (x2 < 0.5.l), Mb’ and Mc’ gradually increase. When x2 = 0, it will become a 2-span beam.
When Mb’ and Mc’ decrease, M1′ and M3′ increase. For safe design, take Mb and Mc as when the span is even, take M2′ calculated according to x2, take M1’=k.M1 and M3’=k.M3 with k given in the table below:
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Source: Compiled from the Internet
