What Are the Tensions in Each Cable of a Suspension Bridge?

[hendrix7]

Homework Statement

[/B]
A simple suspension bridge consists of a uniform plane horizontal rectangular roadway of mass M, freely supported by two sets of four equally spaced light vertical cables. Each set of four cables lies in the same vertical plane and is attached to one edge of the roadway. These cables are in turn attached to a single light cable ABCDEF as shown below. A, B, C are on the same horizontal level as F, E, D respectively.

A and F are 8 m and C and D are 2 m above the level of the roadway which has length 20 m. AF = 20 m. Given that the tensions in all vertical cables are the same, show that the tension in CD is Mg/4 and find the tensions in each of the parts AB, BC, DE, EF of the connecting cable.

Homework Equations

The Attempt at a Solution

resolving vertically: Mg = 4T (T = tension in all vertical cables).
Actually, I'm not sure of this. What about the reactions at A and F?

Let tension in BC = T₁
Let a = acute angle BC makes with CD

tension in CD = T₁ cos a
now I'm stuck. How do I find cos of a?

any help greatly appreciated,
Jimi

 

[andrevdh]

You can get a with the 4 m and the 2 m

 

[andrewkirk]

You can get a with the 4 m and the 2 m

Not unless it says somewhere that the height of B and E above the road is 4m. So far as I can see, it doesn't give any heights for B or E. that seems like an omission in the problem specification, unless it may be that the angle a does not affect the answer because of some cancellation phenomenon.

Hendrix are you sure the problem does not give heights for B or E?

Edit: Why not try writing out the equations for all the forces in terms of a as an unknown, then see if a cancels out in the final formulas.

 

[hendrix7]

andrevh - I don't see how since I don't know how high the vertical cable is from the road to B.

 

[hendrix7]

andrewkirk - no. I've checked the problem and this is all there is. Was thinking maybe there's a way of using the height of 8 m but I can't see how.

 

[andrewkirk]

The problem statement is definitely missing something, because you are asked to find out the tension in BC, which is $\frac{Mg}{4\sin a}$. There is no possibility for the $a$ to cancel out in that formula, even if it cancels out in some of the other tensions (which I doubt it will).

I suggest you ask your lecturer whether:

(1) there is a piece of info missing from the problem statement, such as the height of B; or

(2) you are expected to give the answer in terms of some unknown quantity such as $a$.

 

[andrevdh]

The vector sum of the tensions T and T_CD need to cancel T_BC - draw a vector diagram.

 

[andrevdh]

With multiple equations you can solve for more than one unknown quantity.

 

[insightful]

Try cutting the cable between C and D and summing torques on half the "U" cable.

 

[andrewkirk]

I just realized that the 2m distance from C and D to the roadway is irrelevant to the problem, given that the usual assumption of massless cables is used. It doesn't affect any of the tensions. Why would they give a useless piece of data, while omitting a necessary piece? Because the problem-setter made the mistake of thinking that the datum given would enable us to work out the angle a. I give odds 20 to 1 that the 2m was meant to give the angle a, and the problem-setter didn't realize that it didn't (even uni lecturers make mistakes sometimes).

Given that observation, I suggest you begin your answer as follows.

'The information needed to determine the angle of BC to the horizontal, and hence the tension in cable section BC, is missing from the problem statement, yet an unusable piece of information is given that the distance from B to the roadway is 2m. Hence I will instead assume that the vertical height of B above C is 2m. ...

 

[insightful]

I was able to solve the problem completely with the data given. And yes, the 2m distance is irrelevant.

 

[andrewkirk]

I was able to solve the problem completely with the data given.

Really? how?

If B is 1m above C the tension in BC will have to be greater than if it is 2.5m above it. Yet it seems to me that a bridge conforming to the problem statement can be built with either height of B.

 

[insightful]

Start with Post #9.

 

[andrewkirk]

I don't understand post 9, as torques don't seem to me to be a natural way to approach the problem.

However since my last post I've realized that the problem is solvable as stated, because the forces in the cable have to align with the cables. So we solve it as follows.

There are Five unknown forces:
ABv : the vertical component of tension in AB
ABh : the horizontal component of tension in AB
BCv : the vertical component of tension in BC
BCh : the horizontal component of tension in BC
CD : the tension in CD, which is horizontal

We have the following five equations:

ABv = Mg/4 (each end has to support one quarter of the bridge)
ABh=BCh (horizontal equilibrium at B)
ABv=BCv+Mg/8 (vertical equilibrium at B)
BCh=CD (horizontal equilibrium at C)
BCv=Mg/8 (vertical equilibrium at C)

[Note: corrections were made to 1st, 3rd & 5th eqns per insightful's post below]

So we can solve the equations to find the vertical and horizontal components of tension in each cable section.
We then find the the total tension in each segment by Pythagoras.

If we want the angles, we use the fact that the cables must align with the force in them, and calculate the angle from horizontal as
arctan(vertical compt of tension / horiz compt of tension).

What was throwing me was that we can build the bridge with a range of different angles for AB and BC. What I didn't realize is that doing that with anything other than the angles that fall out of the above equations will cause the points B and E to move laterally as well as vertically, so that the cables going down from B and E are no longer vertical. Since the problem statement requires that they are vertical, there is only one possible set of angles that works.

An interesting practial realisation from this, for me, is that, when the load changes on a suspension bridge - eg between light and heavy traffic - there will be slight lateral displacement in the (almost) vertical cables.

 

[insightful]

Are you forgetting that there are two cable systems supporting the bridge (one on each side of the roadbed)?

 

[andrewkirk]

Good point. Yes I was. Anything involving Mg needs to be divided by another factor of 2. I'll make the corrections in the post

 

[insightful]

So, how do you solve for CD?

 

[andrewkirk]

To compensate for the 1st eqn being the sum of the 3rd and the 5th, we use the extra equation that I didn't write down :

(ABv+BCv)/(ABh+BCh)=(8-2)/(4+4)

So it seems we do need to use the given 2m after all.

 

[NascentOxygen]

So, how do you solve for CD?

Consider, say, the left half of the suspension cable, and take sum of moments about A.

 

[insightful]

Consider half of the suspension cable, and take sum of moments about A.

Exactly!

 

[insightful]

So it seems we do need to use the given 2m after all.

That can't be, because the tension in the "U" cable is independent of the length of the vertical cables.
(However, 2m does end up being one of the dimensions in the "U" cable.)

 

[hendrix7]

Thanks for all your input but I'm still not sure I fully understand this. Just to say, this isn't actually a coursework. It's an old exam question from ODLE (UK exam body). I have the answers (unfortunately not the mark scheme) if anyone wants me to post them.

 

[insightful]

I have the answers (unfortunately not the mark scheme) if anyone wants me to post them.

Yes, it would be fun to see if they match mine [(mg/4)sqrt(2) for AB and EF, and (mg/8)sqrt(5) for BC and DE].

 

[andrewkirk]

That can't be, because the tension in the "U" cable is independent of the length of the vertical cables.

Yes, but it's not independent of the vertical drop of the U cable from A to C and we need to use the 2m to calculate that drop.

 

[insightful]

Yes, but it's not independent of the vertical drop of the U cable from A to C and we need to use the 2m to calculate that drop.

Yes, quite right.
Did you successfully calculate the vertical drop from B to C with your method?

 

[andrewkirk]

My answer matches yours insightful. Let's hope they're both right.

 

[hendrix7]

The textbook says: √5Mg/2 for AB and EF; √2Mg/4 for BC and DE.

 

[insightful]

The textbook says: √5Mg/2 for AB and EF; √2Mg/4 for BC and DE.

Hmmm.. Consider point D: We know the horizontal force to the left is (Mg)/4 and so also the horizontal part of vector DE to the right must be the same. Then, the angle of DE to the horizontal must be cos^-11/sqrt(2) = 45^o. But this means the vertical part of vector DE must also be Mg/4 upward. But isn't this the only force countering the vertical weight of the bridge on point D, which is Mg/8 downward?

 

[andrewkirk]

The textbook says: √5Mg/2 for AB and EF; √2Mg/4 for BC and DE.

The textbook author has fallen into the trap I fell into, which insightful pointed out in post 15, of forgetting that there are two sets of cables, one on each side of the roadway. So the textbook answer has erroneously doubled the tensions.

 

[insightful]

I'd love to see how the textbook got their answers. Doubling doesn't appear to explain my discrepancy.

 

[andrewkirk]

I'd love to see how the textbook got their answers. Doubling doesn't appear to explain my discrepancy.

True.
The textbook answer looks ridiculous to me. Since $\frac{\sqrt{5}}{2}>1.1$ it is saying, via its value for AB and EF, that each of the four pylons is resisting a force equal to more than the entire weight of the bridge.

 

[NascentOxygen]

Yes, it would be fun to see if they match mine [(mg/4)sqrt(2) for AB and EF, and (mg/8)sqrt(5) for BC and DE].

These are also the answers I arrived at. (No saying I haven't blundered somewhere, though.)