Forces acting on a Walking Beam (oil rig pump) from a pivot pin

[link223]

Homework Statement

So I just want to see which forc acts on the walking beam at E but itsn't quite obvious to me from the figure.

Relevant Equations

Structural Analysis

So.. question:
- How do I know that only the pin is at work at E and not those 2 beams? my guess: It is because those 2 beams are connected to the pin whilst the pin is the one that exerts a force on that walking beam DEF?

 

[haruspex]

Homework Statement:: So I just want to see which forc acts on the walking beam at E but itsn't quite obvious to me from the figure.
Relevant Equations:: Structural Analysis

How do I know that only the pin is at work at E and not those 2 beams?

I have no idea what your question means.
What do you mean by pin E being "at work"? Do you mean it is doing work? It does not move, so it can't do work.
What two beams? DEF is a single beam.

The motor exerts a torque at A on beam AB. In addition, the counterweight exerts a force at B. These result in a force at D, which in turn exerts a torque about E on beam DEF, etc. You do not need to consider forces at E.

 

[link223]

I have no idea what your question means.
What do you mean by pin E being "at work"? Do you mean it is doing work? It does not move, so it can't do work.
What two beams? DEF is a single beam.

The motor exerts a torque at A on beam AB. In addition, the counterweight exerts a force at B. These result in a force at D, which in turn exerts a torque about E on beam DEF, etc. You do not need to consider forces at E.

Thank you for your answer!
That is indeed what I did (just solved the exercise with this exact method) but on the FBD, there will be a horizontal and vertical force exerted on the beam DEF at E because of that pin.
What i was confused with was whether it is the actual force components from the pin that needs to be drawn or also the forces that replace those two member that connect to E (the ones for stability) which is appearantly not the case.

 

[haruspex]

Thank you for your answer!
That is indeed what I did (just solved the exercise with this exact method) but on the FBD, there will be a horizontal and vertical force exerted on the beam DEF at E because of that pin.
What i was confused with was whether it is the actual force components from the pin that needs to be drawn or also the forces that replace those two member that connect to E (the ones for stability) which is appearantly not the case.

You can draw forces at E for completeness, but forces applied at E cannot exert a torque about E. So as long as you use the torque balance about E for your equation you can ignore those forces.

 

[Lnewqban]

If you are drawing the FBD of beam DEF, pin E should have represented two reactive forces that oppose the two external forces (at D and at F) at that exact frozen instant represented in the picture.

 

[link223]

You can draw forces at E for completeness, but forces applied at E cannot exert a torque about E. So as long as you use the torque balance about E for your equation you can ignore those forces.

Yes that is indeed what I did (it was for completeness :) ) thank you

 

[link223]

If you are drawing the FBD of beam DEF, pin E should have represented two reactive forces that oppose the two external forces (at D and at F) at that exact frozen instant represented in the picture.

Thank you that is indeed what I did, I was just confused whether those two member that are connected to E are actually connected to E (the pin) or actually exert some force on DEF directly which is not the case.