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

  • $\begingroup$ 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, 2020 at 4:55

3 Answers 3


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)


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?


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.

  • $\begingroup$ 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, 2020 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, 2020 at 3:42

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.

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


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.