Frame question?

  • Thread starter Chappu
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  • #1
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Hi! I am having trouble with this frame. To make a long story short, I did the problem and got the right answer. The issue is that when I wanted to try solving the problem by exploding the structure into individual members, I couldn't get an answer anymore and it's bothering me because I do not understand why it would not work. Please refer to the attachment.

Homework Statement





Homework Equations





The Attempt at a Solution


I did not get very far. I simply exploded the structure and looked at the 2 force members DE and CD (I put the 700N on one of them) and tried solving ∑Fx=0 and ∑Fy=0
For member CD:
∑Fx: Cx-Dx=0 -> Cx=Dx -> C(cos45°)=D(cos45°) -> C=D
∑Fy: Cy-Dy-700N=0 -> C(sin45°)-D(sin45°)=700N -> (replace above equation here)
C(sin45°)-C(sin45°)=700N -> 0=700N ?
 

Attachments

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Answers and Replies

  • #2
907
88
You incorrectly assumed that the force D acts parallel to the member. The force D should act parallel to member ED.
 
  • #3
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You incorrectly assumed that the force D acts parallel to the member. The force D should act parallel to member ED.

Alright that makes sense but if we were looking at member ED without the 700N why would the force be parallel to it in that case? I know it would have to be in order to keep it static but I am having trouble understanding whats the difference between these two situations.
Also, if the 700N was split up into lets say 350N on each, wouldn't that also change things?
 
  • #4
907
88
Are you talking about the force D acting on member ED? I would say similarly that this force must be parallel with member CD.
 
  • #5
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Are you talking about the force D acting on member ED? I would say similarly that this force must be parallel with member CD.

Wouldn't the member not be in static equilibrium then? Since its a 2 FM and even if the force at E counteracts it there would still be a net torque?
 
  • #6
907
88
2 FM's can only transfer forces axially. There is no other way. Without the force at E there can be no forces generated in the members.
 
  • #7
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Sorry I am having a really hard time understanding this. Are my diagrams fine? How would that even work? How could the member without the 700N be in static equilibrium in that case? The forces at D wouldn't even be equal and opposite.
Without the 700N, it would not be in equilibrium then? and the 700N can't just be on one of the 2 FMs?
 

Attachments

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  • #8
907
88
Your diagrams are not correct.

Maybe another way to think about it is to say there is a resultant force R acting at each end of a 2FM. The components are the other truss members and any nodal forces.
 
  • #9
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Your diagrams are not correct.

Maybe another way to think about it is to say there is a resultant force R acting at each end of a 2FM. The components are the other truss members and any nodal forces.

Do you think it would be possible for you to provide the correct diagram so I can visualize it a little better? Sorry for it taking so long to sink in its just I thought I understood it before so I am not sure what parts how I was looking at it are incorrect
 
  • #10
907
88
I don't think I need to draw it.

For member CD, there is a force R at point C and also another force at point D that is equal but opposite. The components of R at point D are 700N and the reaction force from member ED.

Now these components are not 90° to one another. So you need to work out their relationship.
 
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