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

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Homework Help Overview

The discussion revolves around calculating the tension in a beam of a bridge that supports a mass. The bridge consists of multiple beams, and the problem involves analyzing forces and torques to determine the tension in a specific horizontal segment above the mass.

Discussion Character

  • Exploratory, Assumption checking, Mathematical reasoning

Approaches and Questions Raised

  • Participants explore the concept of torques and the importance of selecting an appropriate axis for calculations. There is a focus on understanding how to compute torques from various forces acting on the structure.

Discussion Status

Participants are actively engaging with the problem, discussing the implications of different choices for the axis of rotation and the resulting torques. Some guidance has been offered regarding the selection of substructures and the identification of external forces, although there is still uncertainty about the correct approach.

Contextual Notes

There is mention of the need to consider the geometry of the truss and the angles involved, as well as the significance of sign conventions in determining the nature of the forces (tension or compression).

Wimpalot
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Homework Statement


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.

Homework Equations


tnet = 0 at all points
t = Fr

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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The trick in many problems involving torques and rotations is to start by choosing an appropriate axis.
 
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jbriggs444 said:
The trick in many problems involving torques and rotations is to start by choosing an appropriate axis.
So not just the usual x-y?
 
Wimpalot said:
So not just the usual x-y?
Not the direction of the coordinate axes. The origin of the coordinate system -- the axis of rotation about which you compute torques.
 
Hint : Look for two self stable substructures .
 
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jbriggs444 said:
Not the direction of the coordinate axes. The origin of the coordinate system -- the axis of rotation about which you compute torques.
Okay then in that case I am not entirely sure. The point next to the beam?

Nidum said:
Hint : Look for two self stable substructures .
Sorry, not sure what you mean/what to do with that information
 
Wimpalot said:
Okay then in that case I am not entirely sure. The point next to the beam?
There are a lot of beams in that truss. Which one did you have in mind?
 
jbriggs444 said:
There are a lot of beams in that truss. Which one did you have in mind?

The one I am trying to calculate the tension of. The point on the left?
 
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
jbriggs444 said:
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?

I'm not sure then. I have no idea how to do this question. Maybe the left-most triangle?
 
  • #11
OK.
Wimpalot said:
Maybe the left-most triangle?
OK. Let's say you use the left-most triangle. Now pick an axis of rotation.
 
  • #12
The point that the mass hangs from?
 
  • #13
Wimpalot said:
The point that the mass hangs from?
OK. Now what external torques act on that triangle?
 
  • #14
jbriggs444 said:
OK. Now what external torques act on that triangle?

The reaction force FP and maybe the tension force in the beam T?
 
  • #15
Wimpalot said:
The reaction force FP and maybe the tension force in the beam T?
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
jbriggs444 said:
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?

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

Because that way the torque due to the mass is 0
 
  • #17
Wimpalot said:
And are both moment arms just L? And what about the directions?
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.
Because that way the torque due to the mass is 0
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
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
Wimpalot said:
Clockwise so negative and thus equal to FP*L
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.
Is it equal to the tension and L*sin(60)?
You do not sound very sure. But yes, the moment arm is given by L*sin(60).

It would have to be tension right to counteract the other torque?
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
Zero
 
  • #21
Wimpalot said:
Zero
So does that give you an equation to solve?
 
  • #22
jbriggs444 said:
So does that give you an equation to solve?

0 = -FP*L + T*L*sin(60)
Which implies:

T = FP*L/(L*sin(60))
 
  • #23
Wimpalot said:
0 = -FP*L + T*L*sin(60)
Which implies:

T = FP*L/(L*sin(60))
So the torque due to T is positive or negative?
And the force due to T must therefore be in which direction?
 
  • #24
jbriggs444 said:
So the torque due to T is positive or negative?
And the force due to T must therefore be in which direction?
Negative? Which implies compression.
 
  • #25
Wimpalot said:
Negative? Which implies compression.
What's negative?
Unless you identify what you are talking about and what sign convention you are using, "positive" and "negative" are just meaningless noises.
 
  • #26
jbriggs444 said:
What's negative?
Unless you identify what you are talking about and what sign convention you are using, "positive" and "negative" are just meaningless noises.

Sorry the torque due to T is negative.
I broke the space key on my keyboard so I am being a bit lazy with words.
 
  • #27
Wimpalot said:
Sorry the torque due to T is negative.
I broke the space key on my keyboard so I am being a bit lazy with words.
So you add a negative to a negative and get zero?

[Back in #18, you said that the torque due to FP was negative. Now you say that the torque due to T is also negative]
 
  • #28
jbriggs444 said:
So you add a negative to a negative and get zero?

[Back in #18, you said that the torque due to FP was negative. Now you say that the torque due to T is also negative]
Sorry I got mixed up the torque due to the tension should be positive.
 
  • #29
Wimpalot said:
Sorry I got mixed up the torque due to the tension should be positive.
Ok. And since your convention is that positive is counter-clockwise, we are talking about compression. If T is the "tension" in that beam then T must be negative.

So, can you now solve for T?
 
Last edited:
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  • #30
jbriggs444 said:
Ok. And since your convention is that positive is counter-clockwise, we are talking about compression. If T is the "tension" in that beam than T must be negative.

So, can you now solve for T?
Yeah just T = -FP*L/(L*sin(60)) which is correct according the site.

Thank you very much for all the help I really feel like I understand this now and can apply it all in the future so thank you very much.
 
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