# How to calculate the tension in a beam of a bridge?

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1. Apr 28, 2017

### Wimpalot

1. The problem statement, all variables and given/known data
A bridge, constructed of 11 beams of equal length L and negligible mass, supports an object of mass M as shown in the picture.
Assuming that the bridge segments are free to pivot at each intersection point, what is the tension T in the horizontal segment directly above the point where the object is attached (see image)? If you find that the horizontal segment directly above the point where the object is attached is being stretched, indicate this with a positive value for T. If the segment is being compressed, indicate this with a negative value for T.

2. Relevant equations
tnet = 0 at all points
t = Fr

3. The attempt at a solution
So I worked out that:
FP = 2/3 Mg
FQ=1/3 Mg

But after that I have no idea what to do. I am completely lost. I was told to try considering the torques on single points but I don't understand how to do that in this case.

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2. Apr 28, 2017

### jbriggs444

The trick in many problems involving torques and rotations is to start by choosing an appropriate axis.

3. Apr 28, 2017

### Wimpalot

So not just the usual x-y?

4. Apr 28, 2017

### jbriggs444

Not the direction of the coordinate axes. The origin of the coordinate system -- the axis of rotation about which you compute torques.

5. Apr 28, 2017

### Nidum

Hint : Look for two self stable substructures .

6. Apr 28, 2017

### Wimpalot

Okay then in that case I am not entirely sure. The point next to the beam?

Sorry, not sure what you mean/what to do with that information

7. Apr 28, 2017

### jbriggs444

There are a lot of beams in that truss. Which one did you have in mind?

8. Apr 28, 2017

### Wimpalot

The one I am trying to calculate the tension of. The point on the left?

9. Apr 28, 2017

### jbriggs444

If you write an equation for the torques about that point, you will have contributions from the supporting force from the left pier, the supporting force from the right pier and the downward force from mass M. The tension force T has a line of action that passes through your chosen axis. So it drops out of the equation.

This particular choice of axis does not let you write down a simple equation involving T. It is a poor choice.

Nidum's hint is equally (or perhaps more) important. You want to be computing torques applied to some particular sub-structure. If you think about dividing the truss up into two relevant parts, where might you put the dividing line?

10. Apr 28, 2017

### Wimpalot

I'm not sure then. I have no idea how to do this question. Maybe the left-most triangle?

11. Apr 28, 2017

### jbriggs444

OK.
OK. Let's say you use the left-most triangle. Now pick an axis of rotation.

12. Apr 28, 2017

### Wimpalot

The point that the mass hangs from?

13. Apr 28, 2017

### jbriggs444

OK. Now what external torques act on that triangle?

14. Apr 28, 2017

### Wimpalot

The reaction force FP and maybe the tension force in the beam T?

15. Apr 28, 2017

### jbriggs444

Those are the two of the external forces, yes. To get external torques you would have to multiply by the length of their moment arms, right?

Can you tell us why you ignored the external force from the dangling mass M?

16. Apr 28, 2017

### Wimpalot

And are both moment arms just L? And what about the directions?

Because that way the torque due to the mass is 0

17. Apr 28, 2017

### jbriggs444

Let's start with the torque from the left hand supporting point. Is it clockwise or counter-clockwise? If the left hand supporting force is $F_P$ and the length of the beams is $L$, what is the torque from that left hand supporting force?

Proceed to the torque from the [tension? compression?] T in the horizontal beam. Must it be clockwise or counter-clockwise? Does that mean that it is tension? Or compression? How large is the torque? Be careful. It's not all right angles for this one.
Good. That is why choosing that point for the axis of rotation was wise. Neither the weight of the dangling mass M nor any forces from the beams to the right contribute to the torque around that axis.

18. Apr 28, 2017

### Wimpalot

Clockwise so negative and thus equal to FP*L
Is it equal to the tension and L*sin(60)? I am not sure. It would have to be tension right to counteract the other torque?

19. Apr 28, 2017

### jbriggs444

Yes, the torque from the left hand pier is clockwise and its magnitude is FP*L. Yes, if you adopt the convention that clockwise is negative, that means that it is negative.
You do not sound very sure. But yes, the moment arm is given by L*sin(60).

Let's go slowly. We can do this with mathematics (positives and negatives). Or we can do it with physical intuition (balancing clockwise and counter-clockwise torques).

Let's try the mathematical way. The triangle is both massless and is continuously at rest. The sum of the torques on it must be equal to _____?

20. Apr 28, 2017

Zero