I Which direction do the loads and reactions go in a Ferris wheel?

AI Thread Summary
In a Ferris wheel with pretensioned cable spokes, the loads primarily act downward due to gravity, while the reactions from the structure counteract these loads. The outer ring of the Ferris wheel hangs from the bottom spokes, which transfer tension loads from the hub to the rim as the wheel rotates. The load at the hub and main axle is directed downward and is transferred to the ground through A-shaped frames. Understanding these load and reaction dynamics is crucial for structural integrity in construction consultancy. This analysis highlights the importance of tension and support in the design of rotating structures like Ferris wheels.
Smilemore
Messages
3
Reaction score
1
Considering a ferris wheel with the spokes being pretensioned cables. Which directions do the loads vs reactions go?
 
Physics news on Phys.org
Smilemore said:
Considering a ferris wheel with the spokes being pretensioned cables. Which directions do the loads vs reactions go?
Can you produce for us a drawing? And then identify points where you think there is a load/reaction pair?
 
Smilemore said:
Considering a ferris wheel with the spokes being pretensioned cables. Which directions do the loads vs reactions go?
Welcome to PF. Is this question for schoolwork?
 
berkeman said:
Welcome to PF. Is this question for schoolwork?
No. I am in construction consultancy but just interested
 
Smilemore said:
Considering a ferris wheel with the spokes being pretensioned cables. Which directions do the loads vs reactions go?
Welcome!
If the ferry wheel is like in a bicycle wheel, the outer ring is hanging from the bottom spokes, which transfer tension loads between the hub and the bottom arc of the rim in a sequence, as they are relocated by rotation respect to the vertical.
The load at the hub and main axle is transferred down onto the ground by two A-shaped frames.

basic-force-diagram.png
 
Last edited:
I have recently been really interested in the derivation of Hamiltons Principle. On my research I found that with the term ##m \cdot \frac{d}{dt} (\frac{dr}{dt} \cdot \delta r) = 0## (1) one may derivate ##\delta \int (T - V) dt = 0## (2). The derivation itself I understood quiet good, but what I don't understand is where the equation (1) came from, because in my research it was just given and not derived from anywhere. Does anybody know where (1) comes from or why from it the...
Back
Top