Should I consider internal forces for this hinged structure?

AI Thread Summary
For a hinged structure under a load, the moment-equations for both scenarios presented are deemed correct, but the direction of the force F must be accurately determined. If F is aligned with rod DE, the additional rod does not contribute any forces, effectively making it a 0-rod. Internal forces do not need to be considered for the calculations in this context. To find the required force F, applying a moment-equation about point A is appropriate. The discussion emphasizes the importance of accurately defining force directions in structural analysis.
Mech_LS24
Messages
148
Reaction score
16
Summary:: A hinged structure applied with a load, should the structure totally been calculated in order to find the resultant force?

Hello,

Two situations, the first is a simple structure with a 20 N load at point C. The moment-equation is composed next to it.

Second situation:
A hinged structure is again loaded with 20 N. A moment-equation can also been found next to it.

For both situations,
  • Are the moment-equations correct? Or should I take internal forces also in consideration?
  • Would a 'extra' rod drawn in situation 2 affect the moment-equation?
  • Could situation 2 be simplified? I was thinking about kind of a 0-rod (see sketch). That is in line with the force so it actually does nothing more than passing a force.

1624214599916.jpeg


Mentor note: Moved from engineering forum, so no template.
 
Last edited by a moderator:
Physics news on Phys.org
The momentum equation is fine, but how dit you determine the direction of F? If E is a pin joint, DE will rotate until it is in the line of F.

Same question for situation 2. If F is in the same direction as DE, the new rod has no forces. if E is a pin joint this should happen. If you don't know what α is, you can't solve it.
 
willem2 said:
The momentum equation is fine, but how dit you determine the direction of F?
I see, I think I made a mistake there. Force F should be vertically down instead, this would make more sense.

willem2 said:
If F is in the same direction as DE, the new rod has no forces. if E is a pin joint this should happen.
This makes sense :). That would make rod AE a 0-rod.

With this information, I think I can say the internal forces doesn't need to be considered. If I want to calculate the needed Force F, I can apply a moment-equation about point A?
 
Thread 'Have I solved this structural engineering equation correctly?'
Hi all, I have a structural engineering book from 1979. I am trying to follow it as best as I can. I have come to a formula that calculates the rotations in radians at the rigid joint that requires an iterative procedure. This equation comes in the form of: $$ x_i = \frac {Q_ih_i + Q_{i+1}h_{i+1}}{4K} + \frac {C}{K}x_{i-1} + \frac {C}{K}x_{i+1} $$ Where: ## Q ## is the horizontal storey shear ## h ## is the storey height ## K = (6G_i + C_i + C_{i+1}) ## ## G = \frac {I_g}{h} ## ## C...
Back
Top