# Homework Help: Reaction force in a beam

Tags:
1. Mar 27, 2016

### Jonski

1. The problem statement, all variables and given/known data

Calculate the reaction force at A?
2. Relevant equations
ΣMb = 0

3. 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.

2. Mar 27, 2016

### Simon Bridge

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.

3. Mar 27, 2016

### Jonski

So would the FBD look like this:

The question says consider the reactions at A and D to be rollers, so thus there is no horizontal force

4. Mar 27, 2016

### SteamKing

Staff Emeritus
The pin at point B can develop a force, but it cannot sustain a moment. That's what's missing from your FBD above.

5. Mar 27, 2016

### Jonski

So would I add a vertical and horizontal force at point B and then use the equations of equilibrium to solve?

6. Mar 27, 2016

### SteamKing

Staff Emeritus
You can add a horizontal force at pin B, but it looks like all of the applied loads are vertical.

7. Mar 27, 2016

### Jonski

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.

8. Mar 27, 2016

### SteamKing

Staff Emeritus
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.

9. Mar 27, 2016

### Jonski

So would the FBD look more like this:

#### Attached Files:

• ###### Screen Shot 2016-03-28 at 2.02.04 pm.png
File size:
150.3 KB
Views:
81
10. Mar 28, 2016

### SteamKing

Staff Emeritus
Yes, that's more like it.

11. Mar 28, 2016

### PhanthomJay

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.