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Say we have the following indeterminate beam:

enter image description here

We can make it determinate by adding a pin at B, and then adding back the bending moment at B which was destroyed by the pin:

enter image description here

My lecture handouts then go to say that we can consider the beam as two separate spans, and then use a compatibility condition to calculate $M_B$:

enter image description here

However, there are several things about this step that are not clear to me. How/why are we able to a split a multi-span beam into its individual spans? Is there any sort of rule that we need to follow when doing so?

And also, what is going on with the supports? We start with B being a pin over a roller support, and then we split into two spans, we get a roller support at B (looking at AB) and a normal pinned support at B (looking at BC).

I've tried to understand by drawing a free body diagram, but I don't really know how to proceed after this: enter image description here

Any clarity on how exactly and when we can consider a multi-span beam as separate spans would be very much appreciated.

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1 Answer 1

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1 - "How/why are we able to a split a multi-span beam into its individual spans?"

This continuous beam constitutes a structural indeterminate system to the first degree, so there is a need to release a restrain and apply restoring force at the same location afterward to satisfy the structural compatibility (deflection or angle of rotation) conditions. In this case, the continuity of the beam at support $B$ was selected to be released (note that there are other ways of releasing restrains), which resulted in two independent, yet stable, beam systems.

2 - " And also, what is going on with the supports?...."

As addressed above, these two beam systems must be "stable", and the common support must be "comparable/compatible". Without in-span horizontal load, the change of roller to pin support at $B$ for the right side beam system has no effect on the results, only satisfying the global equilibrium requirement (a beam is unstable when supported by two rollers).

3 - The sketch below represents the free body diagrams after the release. It is your job to find the rotation angles at $B$ that satisfy the compatibility condition.

enter image description here

Example:

What you can say about the structural compatibility at support $B$?

enter image description here

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  • $\begingroup$ Just to make sure I'm understanding this correctly. Before adding the pin, the beam had to be continuous over the roller. Once we added the pin over the roller at B, we destroyed the bending moment there and the resistance to rotation, which meant that the left hand side and right hand side could rotate independently, no longer limited by the need for continuity - hence the two independent beam systems. And the supports on the two independent systems are chosen to ensure we have stable determinate beams, and not unstable mechanisms. That sound about right? $\endgroup$
    – VJ123
    Apr 7 at 0:26
  • $\begingroup$ That sounds right, though "adding a pin" is not necessary (and confusing). You only need to know separating the beam and adding the results that satisfy the compatibility condition are based on the concept of "superposition". $\endgroup$
    – r13
    Apr 7 at 0:44
  • $\begingroup$ Got you, thanks! $\endgroup$
    – VJ123
    Apr 7 at 8:53
  • $\begingroup$ You are welcome. The analytical procedure is - 1) Release continuity (moment capacity) of the beam at support B to separate the beam segments. 2) Reimpose the released restrain (ML=MR=MB) at the released end of each segment. 3) At the released end, the left side segment will have a rotation angle equal to the sum of rotations caused by the span loading (W) and the imposed moment (MB), the resulting rotation angle should be identical to the rotation angle of the right-side segment caused by the imposed moment (MB). Write the equation for this condition, then solve for the unknown moment (MB). $\endgroup$
    – r13
    Apr 7 at 15:49
  • $\begingroup$ Ah that's a useful summary, cheers. $\endgroup$
    – VJ123
    Apr 8 at 11:31

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