Calculating Reaction Force at A for a Beam with Pin Joint | Beam Homework

AI Thread Summary
To calculate the reaction force at point A of the beam, the equilibrium equation ΣMb = 0 was applied, leading to an initial calculation of Ay = -40 kN. However, the user questioned the accuracy of this result, suspecting that the sum of vertical forces may not equal zero due to an upward force at point By of 60 kN. It was clarified that the pin at B can exert a force but cannot sustain a moment, which is crucial for the free body diagram (FBD). The discussion emphasized that the beam is only resting against the wall at point E, allowing for an axial force but no moment at that location. Overall, careful consideration of forces and moments is essential for accurately solving the problem.
Jonski
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Homework Statement


Screen Shot 2016-03-27 at 5.21.42 pm.png

Calculate the reaction force at A?

Homework Equations


ΣMb = 0

The Attempt at a Solution


As there is a pin joint at B it is possible to consider the bar only from A-B.
From here I applied
ΣMb = 0
120 + 20*2 + Ay*4 = 0
This gives Ay = -40kn
Also since there are no forces in the x direction Ax = 0
and Hence A = -40kn or 40n downwards
However this is wrong. I am not sure what I am doing wrong unless it is something that the sum of the y forces don't equal 0, but I think there would be a force at By= 60kn up to combat this.
Any help would be appreciated, thanks.
 
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120kN.m is a moment about A isn't it?
What is the reasoning that suggests that you need only consider the A-B length?
The pin connection at B does not stop rotation about B caused by the force distribution on B-E - just prevents translation of that point.
 
So would the FBD look like this:
Screen Shot 2016-03-28 at 12.40.31 pm.png

The question says consider the reactions at A and D to be rollers, so thus there is no horizontal force
 
Jonski said:
So would the FBD look like this:
View attachment 98083
The question says consider the reactions at A and D to be rollers, so thus there is no horizontal force
The pin at point B can develop a force, but it cannot sustain a moment. That's what's missing from your FBD above.
 
SteamKing said:
The pin at point B can develop a force, but it cannot sustain a moment. That's what's missing from your FBD above.
So would I add a vertical and horizontal force at point B and then use the equations of equilibrium to solve?
 
Jonski said:
So would I add a vertical and horizontal force at point B and then use the equations of equilibrium to solve?
You can add a horizontal force at pin B, but it looks like all of the applied loads are vertical.
 
SteamKing said:
You can add a horizontal force at pin B, but it looks like all of the applied loads are vertical.
So you're saying it is there, but it would just be zero. Also I'm assuming then that Ex would be 0?
If i do that won't I end up with more unknowns than equations, in which case it would be unsolvable.
 
Jonski said:
So you're saying it is there, but it would just be zero. Also I'm assuming then that Ex would be 0?
If i do that won't I end up with more unknowns than equations, in which case it would be unsolvable.
That's unlikely, since there are only two supports for the entire beam.

BTW, the beam is only resting against the wall at E. It can't develop a moment at that location, only an axial force.
 
SteamKing said:
That's unlikely, since there are only two supports for the entire beam.

BTW, the beam is only resting against the wall at E. It can't develop a moment at that location, only an axial force.

So would the FBD look more like this:
 

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  • Screen Shot 2016-03-28 at 2.02.04 pm.png
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  • #10
Jonski said:
So would the FBD look more like this:
Yes, that's more like it.
 
  • #11
In my estimation, I believe that the wording in this problem is not very good. When it says "resting against a wall", I think it implies "resting atop a wall", otherwise, with the hinge at B, the beam would be unstable.

The force from the pin at B is internal, so when drawing the FBD of the entire beam system, that reaction doesn't enter into the equilibrium equation. I would proceeded by isolating AB first, but you have to be careful with your plus and minus signs, and interpretation of them.
 
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