How to Calculate Forces in Bridge Points?

[Mister V]

Homework Statement

The problem statement:

Bridge of length: 200m
Construction can expand horizontally to allow thermic expansion
Masses of beams: small in comparison to the car
Mass of the car: 1360 kg.

The answers: Na= 11,0 kN, Ne= 3,67 kN, Fab = Fde = 7,35 kN, Fac = Fce = 6,37 kN, Fbc=Fcd= 4,24 kN, Fbd = 8,49 kN

Question: Calculate the forces in points A,B,C,D,E

Homework Equations

We've already solved a similar bridge, but there was not the possibility of horizontal expansion.

In the similar bridge we stated that, as the beams were in static equilibrium, the forces should be parallel to the beams, either pointing outwards or inwards:

The Attempt at a Solution

My problem is I haven't got a clue how to incorporate the ability to move horizontally. Do I have to allow the forces to be non parallel to the beams?
What I find hard to understand is the fact that the normal force in point A (Na) is different form the normal force in point E (Ne)? I thought that, since the massa is equally distributed along the bridge, the normal forces would be the weight of the car divided by 2.

I have made a drawing with the forces, knowing that when I found a negative force, I assumed the wrong direction.

I thought to calculate Fab by saying that there are only forces having a vertical component, so those two should be zero when added.

I stated Fn is = 6670,80. , so Fab . sin(40°) = - 6670,80, Fab = -10377 N, which apparently is wrong ?

 

[PhanthomJay]

[
In the similar bridge we stated that, as the beams were in static equilibrium, the forces should be parallel to the beams, either pointing outwards or inwards:

That's correct, as long as you have a pure truss with loads applied at the joints.

My problem is I haven't got a clue how to incorporate the ability to move horizontally. Do I have to allow the forces to be non parallel to the beams?

No, this just means that there can be no horizontal reactions at A and E, under the vertically applied load.

What I find hard to understand is the fact that the normal force in point A (Na) is different form the normal force in point E (Ne)? I thought that, since the massa is equally distributed along the bridge, the normal forces would be the weight of the car divided by 2.

me too.

I have made a drawing with the forces, knowing that when I found a negative force, I assumed the wrong direction.

I thought to calculate Fab by saying that there are only forces having a vertical component, so those two should be zero when added.

I stated Fn is = 6670,80. , so Fab . sin(40°) = - 6670,80, Fab = -10377 N, which apparently is wrong ?

Looks right to me. The answer key looks wrong.

 

[Mister V]

Thanks for helping me out!