Forces in Truss Members (Statics)

AI Thread Summary
The discussion revolves around calculating reaction forces in a truss system with specific connections at points A and B. Point A is a pinned connection with forces in both X and Y directions, while point B is a roller connection with a force only in the Y direction. The challenge arises from the lack of dimensions in the problem, complicating the calculation of the perpendicular distance from the applied load to point A. After some back-and-forth, it is confirmed that the correct perpendicular distance is 1.5L, leading to the conclusion of the reaction force at point D being 1.5KN. The participant expresses gratitude for the assistance and acknowledges a breakthrough in understanding the problem.
cpmustang07
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Homework Statement



See attachment.
Edit: Point A is a pinned connection (has a reactionary force in X and Y direction). Point B is a roller connection (only has reactionary force in Y direction)

Homework Equations



Sum Fx=0
Sum Fy=0
Sum of Moments=0


The Attempt at a Solution




Find reaction forces at point A and D. Normally this wouldn't be an issue. I'd just find the reaction force of D by solving for the moment around point A. Then find A by knowing that the sum of forces in Y direction is 0. However, dimensions are not given in the problem. I have the answers, but I do not see how to get to them.

After finding reaction forces, isolate each joint and begin with joint with no more than two unknowns. Proceed until all forces are found.
 

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These are all equilateral triangles...each horizontal member has the same length, L. What would be the perp. distance from the applied load to A?
 
PhanthomJay said:
These are all equilateral triangles...each horizontal member has the same length, L. What would be the perp. distance from the applied load to A?

3*cos(60)*L, correct?
 
cpmustang07 said:
3*cos(60)*L, correct?
Interesting way to say 1.5L, which is correct.
 
SammyS said:
Interesting way to say 1.5L, which is correct.

No...
 
Phrak said:
No...
Why not?? If the length of each horizontal member is L, the perpendicular distance from the line of action of the applied load to A is 1.5 L.
 
PhanthomJay said:
Why not?? If the length of each horizontal member is L, the perpendicular distance from the line of action of the applied load to A is 1.5 L.

right. 1.5L to A and 1.5KN on D.
 
Thank you for your help guys. I'll work on it again today and see if this different train of thought will get me through it. :)

I got it guys. Thank you. Apparently I'm really slow. ;)
 
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