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I'm considering the problem below.

It's a setup with a roller joint and a pinned joint with a hinge between the two rigid links.

Since there are now 3 unknowns for 3 equations and a zero moment at the hinge the system is onefold statically underdeterminate. If I would add an external moment over the hinge would that make the structure statically determinate?

enter image description here

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  • $\begingroup$ No but you can do a quasistatic analysis of a mechanism as a function of a variable say the angle between the beams. Then assume the link is rigid. It would be a sufficient way to analyze a slow moving system with no shock loads. $\endgroup$ – joojaa Nov 4 '20 at 4:55
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if you randomly put a moment at the center hinge, then the problem would still be indeterminate.

  • Make the moment large the hinge will close up.
  • Make the moment small the hinge will open up (if you consider weight).

You would need to put a rotational constraint at the hinge point to make it stable. (I am not certain is 'statically determinate' would be an applicable term for this)

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If you think about it, a support is just an applied force or moment, right? If you have a simply supported beam with a force $F$ in the midspan, the supports will be two forces, each $F/2$ in the opposite direction.

But then, what if, instead of supports, you simply had two forces $F/2$ applied at the ends? That'd be exactly the same, right?

Wrong.

The difference between supports and forces is that supports are "flexible": they generate the forces needed to keep a structure stable. An equivalent force is only valid for a specific loading case.

So sure, if you have $F=100$ and apply two forces at the ends, each equal to $S=50$, that'll be identical to having supports at those ends. But if you then have $F=200$, those applied forces won't cut it. Meanwhile, if you have supports, well, then they'll generate forces of $S = 100$ each.

Basically, think in terms of order of operations: your structure suffers some loading configuration $q$, and then from those you calculate your supports. From a procedural perspective, supports are different from forces in that supports are a function of forces.

In your case, if you apply a bending moment at the hinge (it'd have to be on both sides of the hinge), then sure, you could make a balanced structure for a specific loading criteria. But you'd just be mimicking the behavior you'd get if you simply didn't have a hinge at all, and simply had a continuous (though bumpy) beam.

So, what you need isn't a bending moment at the hinge. What you need is to get rid of the hinge.

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  • $\begingroup$ First of all, You explained it a lot clearer than me. I was caught by your perspective that supports are a function of forces. Would you consider that it may be more specific than that i.e. that supports are a function of structural loads? $\endgroup$ – NMech Nov 3 '20 at 17:09
  • $\begingroup$ @NMech: oh yeah, I was using "forces" as a general term for loading, likely because I forgot the word "loads" exists... >.> $\endgroup$ – Wasabi Nov 4 '20 at 3:42
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I add an alternative answer.

If you add a moment to the hing but your are willing to accept suspended configuration, it will be a determinat structure.

Consider the hinge at lower level than the supports, then apply the moment between the beams in a way that it wants to open them.

Let L, be the length of each beam and W its weight. And the angle between the beams a.

$M/L*sina=1/2* 2W= W$

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