Static structure - find a fourth relation

  • Thread starter Moara
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  • #1
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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##
Captura de tela 2021-09-10 140458.png

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.
 

Answers and Replies

  • #2
Lnewqban
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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.
 
  • #3
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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## ?
 
  • #4
haruspex
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plus the rotation equilibrium
There are two components that, in principle, can rotate independently, so you should have two rotation equilibrium equations.
 
  • #5
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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 ?
 
  • #6
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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
 
  • #7
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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
 
  • #8
haruspex
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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?
 
  • #10
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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
 
  • #11
haruspex
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but I think that horizontal and vertical equilibrium plus rotation equilibrium of AC and BC around C is correct
Yes, that should work.
 
  • #12
Lnewqban
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