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I had come across a question on shear force and bending moment diagram where a cantilever beam with a roller support at point C is given and the portion beyond point C in overhang. Does this hinge act as a internal release point in the structure? If yes then the does this mean bending moment about point B is zero.enter image description here

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A hinge is a point where there is no restriction on rotation. For other points on a beam, the rotation to the left of a point must be equal to the rotation to the right of that same point; that is, there can't be a discontinuity in the rotations along a beam.

Hinges, however, don't have this restriction and therefore allow for discontinuities of rotation. And since bending moment is generated by a beam trying to resist changes to its curvature, we can conclude that there is no internal bending moment at a hinge.

does this mean bending moment about point B is zero

To be clear, a stable structure will have zero bending moment about any imaginable point. We tend to perform our $\sum M = 0$ calculations around supports because it eliminates some variables, but that equilibrium equation is valid about any point in the universe.

But if that's the case, how do we obtain non-zero bending moments along a beam? Well, that's because what we're calculating there is internal bending moment. And internal bending moment is calculated as the sum of bending moments to one side of the relevant point.

So, at midspan in a beam, for example, you calculate the bending moment generated by all the loads to the left (or right) of the beam, which will (usually) result in a non-zero value, representing the beam's internal reaction to the curvature being felt at that point.

But hinges don't resist rotation, so we know that the internal bending moment at the hinge is zero. Therefore, the bending moment to either side of the hinge is zero. That's what makes hinges different from other points on a beam. Any other point has a guaranteed null external bending moment (sum of all loads throughout the structure), but only hinges have guaranteed null internal bending moments (sum of all loads to the left of the hinge equal to zero).

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  • $\begingroup$ Thanks a lot for the answer. $\endgroup$
    – Ashmark
    Commented Jun 30, 2020 at 8:58
  • $\begingroup$ @AshishMutekar if this answered your question, feel free to click the checkmark to the left to mark it as the accepted answer. $\endgroup$
    – Wasabi
    Commented Jun 30, 2020 at 12:04
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If yes then the does this mean bending moment about point B (the hinge) is zero.

You should re-read your notes, or look up "hinge" in your book's index, and start reading up, starting with the first mention.

How can the bending moment around a hinge be anything but zero? What happens if you have a hinge and it has bending moment? What does it do?

Does this hinge act as a internal release point in the structure?

You phrase that as if you're really saying "why would a designer put a hinge there?". The answer is -- they probably wouldn't. This looks like a question in a first course in statics, where all beams are rigid and straight. If so, then the textbook author is avoiding a situation where the beam is overconstrained. By putting a hinge in the middle, they turn the problem from one where the cantilever and the roller support "fight" one another* for control of the beam position into one where the position of the left portion of the beam** is controlled by the cantilever attachment, and the position of the right portion of the beam is controlled by the hinge and the roller support.

* Such a situation is called "overconstrained"; determining the forces on an overconstrained member gets into a lot of issues that I know intuitively but never studied. Basically, you have to know how the member bends in response to forces, and you have to track the interplay between those forces and the bending of the member. When you start to consider the fact that everything bends (the member you care about, the thing it's fastened to, the fasteners, etc., etc.) then the solution gets very complicated, which is why old graybeards will just eyeball the assembly and tell you what to do, without using a computer.

** I say "left portion" and "right portion", but really, you could treat this as two beams.

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