Free body diagram for two-force members

In summary, the armature shown is not appropriate to be a two-force member because it does not have a rigid link between points 2 and 3.
  • #1
kps
5
1
Hoping someone can chip in and help me understand a problem when it comes to two-force members.

I have a structure as shown below. Essentially its a single body attached at 3 points with pin joints. My question is, is it appropriate to treat each of the left and right side as individual members, making them single two-force members and therefore have a reaction force in x AND y, at both point 2 and 3. Or is the fact that it's a single rigid body mean it can't be a two force member and there will only be a y reaction force at 2 and 3.

Many thanks!
Screenshot 2023-04-22 120537.png
 
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  • #2
Hello @kps ,
:welcome: ##\qquad## !​

Here at PF the guidelines require you post your attempt at solution before we are allowed to help.
kps said:
at both point 2 and 3

In the mean time: ask yourself if the members exercise a horizontal force on each other at point 1 ...
(do the experiment if in doubt !)

##\ ##
 
  • #3
Well I don't get how this would be required to be a two-force member because we have 3 points that can counter any moments. So assuming point 1 is right in the middle between 2 and 3, there would only be a vertical (y) reaction at both points of 0.5F, no horizontal force?
 
  • #4
I take it you did not perform the experiment ?

What about this not-so-subtle hint ? Still no horizontal forces at point 1 ?

1682199056667.png


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  • #5
Can two force members have the force along their axis?
 
  • #6
BvU said:
I take it you did not perform the experiment ?

What about this not-so-subtle hint ? Still no horizontal forces at point 1 ?

View attachment 325347

##\ ##
I get what you're trying to show, and correct me if I'm wrong, is that you're implying that the steep angle means there is a horizontal force from both sides acting on 1. However, that only applies if I already assumed that there are axial forces only. My question is why? and how does this scenario differ from for example a bike shown from side view, with a cyclist sat right in the middle of the two wheels providing a vertical reaction force of 0.5*mg at each wheel, with no horizontal component.
 
  • #7
kps said:
Essentially its a single body attached at 3 points with pin joints.
What does this mean??? A rigid body is rigid (by itself, from internal forces). The drawing is not at all clear to me (but very little is lately). Are the bars infinitely strong? Is there a lower tension member?
 
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  • #8
kps said:
and how does this scenario differ
the wheels of the bike are not pivot points like 2 and 3, but, well, rollers. If your points 2 and 3 were wheels or sliding points they would move apart because point 1 is a hinge

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  • #9
kps said:
Well I don't get how this would be required to be a two-force member because we have 3 points that can counter any moments. So assuming point 1 is right in the middle between 2 and 3, there would only be a vertical (y) reaction at both points of 0.5F, no horizontal force?
I share your confusion because you have described a single rigid body, but your diagram represents an armature.
Each of those have different distribution of internal forces.

The main difference comes from the way in which points 2 and 3 stay put and keep the distance between them.

If the armature has a 2-3 rigid link, then links 1-2 and 1-3 can be each a two-force member and there will only be y-reaction forces at 2 and 3.

If that 2-3 rigid link does not exist, then supports 2 and 3 will need to have each x-reaction forces
that prevent the armature from spreading links 1-2 and 1-3 and from collapsing.
Normally, those x-reaction forces are possible due to friction or anchoring respect to a solid substrate such as ground.
 
  • #10
Lnewqban said:
I share your confusion because you have described a single rigid body, but your diagram represents an armature.
Each of those have different distribution of internal forces.

The main difference comes from the way in which points 2 and 3 stay put and keep the distance between them.

If the armature has a 2-3 rigid link, then links 1-2 and 1-3 can be each a two-force member and there will only be y-reaction forces at 2 and 3.

If that 2-3 rigid link does not exist, then supports 2 and 3 will need to have each x-reaction forces
that prevent the armature from spreading links 1-2 and 1-3 and from collapsing.
Normally, those x-reaction forces are possible due to friction or anchoring respect to a solid substrate such as ground.
Maybe my initial post wasn't clear so I apologise. But what I was actually intended the links to be is a single body. Imagine a suspension control arm with 3 attachments (spherical bearings). So joint 1 has a solid connection that joins 1-2 and 1-3 together, it's all one rigid body, it's just the force is applied through a spherical bearing that would have a bolt going through it for example, hence the pin joint at 1. I think a typical truss example would have 1-2 and 1-3 be separate links, so with no joints at 2 and 3 any vertical load would split them. Wouldn't my case have internal bending preventing them from spreading and collapsing? Would this change the reaction forces?

Screenshot 2023-04-23 0052092.png
 
  • #11
kps said:
... what I was actually intended the links to be is a single body..... Wouldn't my case have internal bending preventing them from spreading and collapsing? Would this change the reaction forces?

View attachment 325349
Confusion solved, then.
No apologies needed.

You are correct, that rigid body must have internal moments that try to bend it, just like the frame of the bicycle that was mentioned before.
Therefore, x-forces will not arise at points 2 and 3, if the body is supported as shown in your original diagram.

If the arm pivots about point 1, then input and output forces at points 2 and 3 should be considered as acting perpendicular to the arm distances 1-2 and 1-3.

Please, see:
http://507movements.com/mm_157.html

http://507movements.com/mm_385.html
 
Last edited:
  • #12
kps said:
Would this change the reaction forces?
Only if the body deforms. Which rigid bodies do not, by fiat. So the static rigid body is a just a triangle (or any shape with attachment at 1,2,3 and appropriate center of mass. I do not understand the elaborate nomenclature (two-force member????....reaction forces?...input-output forces? ) For me they are obfuscatory.
 
  • #13
Lnewqban said:
Confusion solved, then.
No apologies needed.

You are correct, that rigid body must have internal moments that try to bend it, just like the frame of the bicycle that was mentioned before.
Therefore, x-forces will not arise at points 2 and 3, if the body is supported as shown in your original diagram.

If the arm pivots about point 1, then input and output forces at points 2 and 3 should be considered as acting perpendicular to the arm distances 1-2 and 1-3.

Please, see:
http://507movements.com/mm_157.html

http://507movements.com/mm_385.html
The bell crank is analogous to my FBD it seems.

So to clarify, for the second link you posted (movement 385), forces will only be reacted axially, as each link is a separate member?
Whereas the bell crank (movement 157), although has the same joints, can have reaction forces in any direction, dictated by applied load, this is because its one solid member and not two
 
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  • #14
kps said:
The bell crank is analogous to my FBD it seems.

So to clarify, for the second link you posted (movement 385), forces will only be reacted axially, as each link is a separate member?
Whereas the bell crank (movement 157), although has the same joints, can have reaction forces in any direction, dictated by applied load, this is because its one solid member and not two
Exactly.
For the second link I posted (movement 385), which is a mechanism, forces will be oriented axially, respect to each link.
Those links or members can only transfer traction or compression efforts; there is no moment to talk about.

Unlikely, the bell crank (movement 157), has one rigid joint, which loads the links with moment efforts, which value is always maximum around that pivot.
 

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