Static structure - find a fourth relation

[Moara]

Homework Statement

As shown in the figure, we are given a static structure with articulated supports A and B. C is a point articulated too. We are asked to find the reaction forces and moments in A and B.

Relevant Equations

$F_r = ma$, $M_r = I\alpha$

First, since A and B are articulated, the moments due to A and B are zero. Now, we may call reaction forces in A, $V_A$ and $H_A$ and in the same way, call the reactions in B as $V_B$ and $H_B$. With that and Newton's third law, I managed to find three equations (equilibrium of translation in the vertical and in the horizontal plus the rotation equilibrium), but, still, I need one more equation or argument to completely find the four forces.

 

[Lnewqban]

You can remove one support at a time and calculate the forces that are needed to keep the structure from rotating about the other.
It is bassicaly an ABC triangular closed structure.

 

[Moara]

I don't think I understand how am I supposed to remove one support, could you clarify, please? Meanwhile, I tried to split the structure looking only at the branch AC, can I say that in C there will be only horizontal forces, hence finding that $V_A = 10 \ kN$ ?

 

[haruspex]

plus the rotation equilibrium

There are two components that, in principle, can rotate independently, so you should have two rotation equilibrium equations.

 

[Moara]

There are two components that, in principle, can rotate independently, so you should have two rotation equilibrium equations.

with that in mind, I did the equilibrium of rotation looking to C, from bar AC, getting another equation and finding $H_A = \frac{90}{7}$, is it correct doing that ?

 

[Moara]

with that in mind, I did the equilibrium of rotation looking to C, from bar AC, getting another equation and finding $H_A = \frac{90}{7}$, is it correct doing that ?

It appears that I have to look AC and BC separately right

 

[Moara]

with that in mind, I did the equilibrium of rotation looking to C, from bar AC, getting another equation and finding $H_A = \frac{90}{7}$, is it correct doing that ?

It appears that I have to look AC and BC separately right

 

[haruspex]

with that in mind, I did the equilibrium of rotation looking to C, from bar AC, getting another equation and finding $H_A = \frac{90}{7}$, is it correct doing that ?

How do you get 90/7?

 

[Moara]

How do you get 90/7?

I probably did a mistake in my calculations

 

[Moara]

I probably did a mistake in my calculations

but I think that horizontal and vertical equilibrium plus rotation equilibrium of AC and BC around C is correct

 

[haruspex]

but I think that horizontal and vertical equilibrium plus rotation equilibrium of AC and BC around C is correct

Yes, that should work.

 

[Lnewqban]

I don't think I understand how am I supposed to remove one support, could you clarify, please? Meanwhile, I tried to split the structure looking only at the branch AC, can I say that in C there will be only horizontal forces, hence finding that $V_A = 10 \ kN$ ?

Please, see example 3.14 in following link:
https://eng.libretexts.org/Bookshel...Determinacy_and_Stability_of_Beams_and_Frames

:)