Bridge Question - Stress / Strain

[Dethrone]

View attachment 3269

I have gathered all the information:

Span: $1991$ m
Breaking Stress: $1770$ MPa
Load: $370$ KN/m
Drape: $201.2$ m

So far, I've found the allowable stress by dividing the breaking stress by the safety factor, which is $787$MPa. Then, I calculated the total load of the bridge, by multiplying the span, with the load: $737 000$ KN.
Then I divided it by two, because each tower is responsible for 1/2 of the load. Now I am stuck, any ideas?

 

[I like Serena]

Hey Rido! ;)

The given breaking stress is the one for a single wire.
Were you planning to suspend the bridge on a single thread?

 

[Dethrone]

Hi ILS! (Wave)

So right now, what I did is finding the vertical force of the bridge, by $\frac{wl}{2}=\frac{370 \cdot 1991}{2}$

The horizontal force is given by $\frac{wl^2}{8h}$, which can be calculated by examining the moments on half the bridge...

Then I calculated the net force by applying pythagorean theorem. I know that stress = $\frac{F_{total}}{nA}$, so that is dividing the total force by the number of wires, and each of them should be equal to the allowable stress. Now, is that right? I don't think the bridge is still held by a single thread (Giggle)

 

[I like Serena]

Actually, I misread the problem statement originally, so my comment about a single thread is a off.
Sorry for that. Before we get confused, let's define a couple of symbols.

\begin{array}{|c|c|l|}\hline
\text{Symbol} & \text{Value} & \text{Description}\\
\hline
l & 1991 \text{ m} & \text{Span}\\
\sigma_{max} & 1770 \text{ MPa}& \text{Breaking stress}\\
w & 370 \text{ kN/m} & \text{Load (usually called $q$, but I'll follow your choice)}\\
h & 201.2 \text{ m} & \text{Drape}\\
d & 5.23 \text{ mm} & \text{Diameter of a wire} \\
FoS & 2.25 & \text{Factor of safety} \\
W & & \text{Weight of the bridge} \\
\alpha & & \text{Angle of the cable with the horizontal where it is connected to the tower} \\
V & & \text{Total vertical force on 1 tower cause by the bridge section} \\
H & & \text{Total horizontal force on 1 tower} \\
F_{total} & & \text{Combined tensional force in all wires that are connected to a tower} \\
F_{max} & & \text{Breaking tensional force of all wires together} \\
n & & \text{Number of wires needed for both cables} \\
\hline \end{array}

So right now, what I did is finding the vertical force of the bridge, by $\frac{wl}{2}=\frac{370 \cdot 1991}{2}$

The weight of the bridge section is:
$$W=wl$$
And indeed, this gets divided over the 2 towers.
So that leaves us with a vertical force on a tower of:
$$V=\frac 12 W = \frac{wl}{2}$$

Good!

The horizontal force is given by $\frac{wl^2}{8h}$, which can be calculated by examining the moments on half the bridge...

Then I calculated the net force by applying pythagorean theorem.

I'm not sure how you examined those moments, but the horizontal force is indeed:
$$H=\frac{wl^2}{8h}$$

So we get indeed that:
$$F_{total}=\sqrt{H^2+V^2}$$

I know that stress = $\frac{F_{total}}{nA}$, so that is dividing the total force by the number of wires,

The definition of stress is:
$$\sigma = \frac{F}{A}$$
So yes, the breaking stress of all wires together is:
$$\sigma_{max} = \frac{F_{max}}{n\cdot \pi d^2}$$

and each of them should be equal to the allowable stress.

Huh?
The breaking stress is only related to the material and not to a wire with some specific diameter.

Now, is that right? I don't think the bridge is still held by a single thread (Giggle)

Yep! It looks right! ;)

... except that you didn't get to the final answer yet.

 

[Dethrone]

Okay, I just realized that there are two main cables in the bridge, so should I also divide by two again?

Using $\frac{wl}{2}$, I get 368335 kN on each each tower, and 184167.5 kN on each cable.

Applying

$$H=\frac{wl^2}{8h}$$

The load is split between two cables? so do I apply

$$H=\frac{(w/2)l^2}{8h}$$

to find the horizontal component of force in each cable? or do I divide the resulting horizontal component by two?

 

[I like Serena]

Okay, I just realized that there are two main cables in the bridge, so should I also divide by two again?

Using $\frac{wl}{2}$, I get 368335 kN on each each tower, and 184167.5 kN on each cable.

Applying

$$H=\frac{wl^2}{8h}$$

The load is split between two cables? so do I apply

$$H=\frac{(w/2)l^2}{8h}$$

to find the horizontal component of force in each cable?

There's no need. You can treat both cables, consisting of many wires, as one.
Or rather as $n$ for the number of wires in both cables combined.

So you have a total force that is divided over the combined wires of both both cables.

 

[Dethrone]

I'm going to proceed by steps, just to make it is all correct.

$$V=\frac{wl}{2}=\frac{370 \cdot 1991}{2}=368335 \text{ kN}$$
$$H=\frac{wl^2}{8h}=\frac{370 \cdot (1991)^2}{8 \cdot 201.2}=911226.373 \text{ kN}$$

$$F_{net}=\sqrt{V^2+H^2}=\sqrt{368335^2+911226.373^2}=982855.114\text{ kN}$$

I'm going to be very careful with my units in the next step:

$$\sigma_{max} = \frac{F_{max}}{nA}=\frac{982855.114 \text{ kN}}{\pi \left(\frac{5.23}{2}\right)^2\cdot 787 \text{ MPa}}=58132.84 \text{ wires}$$

Remembering this the the number of wires in both cables...we can divide by two to arrive at $29067$ wires. Is this right :D?

 

[Dethrone]

I'm going to head to bed now, so I'll read the response in the morning, but why is $58133$ wires the number of wires in the whole bridge, and not just the half section of the bridge? The way I got the vertical and horizontal force, I examined only half of the bridge...(Wondering)

 

[I like Serena]

I'm going to proceed by steps, just to make it is all correct.

$$V=\frac{wl}{2}=\frac{370 \cdot 1991}{2}=368335 \text{ kN}$$
$$H=\frac{wl^2}{8h}=\frac{370 \cdot (1991)^2}{8 \cdot 201.2}=911226.373 \text{ kN}$$

$$F_{net}=\sqrt{V^2+H^2}=\sqrt{368335^2+911226.373^2}=982855.114\text{ kN}$$

Yep!

I'm going to be very careful with my units in the next step:

$$\sigma_{max} = \frac{F_{max}}{nA}=\frac{982855.114 \text{ kN}}{\pi \left(\frac{5.23}{2}\right)^2\cdot 787 \text{ MPa}}=58132.84 \text{ wires}$$

Strain does not have $\text{wires}$ as unit.

Did you mix up $\sigma_{max}$ and $n$? (Wondering)

Remembering this the the number of wires in both cables...we can divide by two to arrive at $29067$ wires. Is this right :D?

Since the problem statement asks for the number of wires you need for each cable, yes, you can divide by two. (Nod)

For the record, I'm getting $29079$.
Did you make a rounding error? (Crying)

Since we have the first 3 digits in common, where did you round an intermediate result to 3 digits?

I'm going to head to bed now, so I'll read the response in the morning, but why is $58133$ wires the number of wires in the whole bridge, and not just the half section of the bridge? The way I got the vertical and horizontal force, I examined only half of the bridge...(Wondering)

There are 2 cables that connect 2 towers.
That means that weight of the bridge section is divided over 4 connection points.

Due to symmetry you can look at only 1 tower that carries half of the bridge section.

And conceptually, you can merge the 2 cables into 1 cable without changing the total forces, since we're only talking about the weight of the bridge, and not about stability.

 

[Dethrone]

Yep, I did mix up both stress and number of wires by accident, but I did recheck my calculations, and I did indeed get 29066.41799 wires.

$$n_{wires}=\frac{982855.114 \text{ kN}}{\pi \left(\frac{5.23}{2}\right)^2\cdot 787 \text{ MPa}}=58132.84 \text{ wires}$$

I do understand that the bridge is symmetrical and everything, but it still confuses me. If we only look at half of the bridge, and examine only the vertical and horizontal forces acting on half the bridge, we determine the net force acting on half the bridge. Shouldn't the number of wires be reflecting the number of wires on only half the bridge? So if the net force was $982855.114$, then the net force of the whole bridge is that times 2...

 

[I like Serena]

Yep, I did mix up both stress and number of wires by accident, but I did recheck my calculations, and I did indeed get 29066.41799 wires.

$$n_{wires}=\frac{982855.114 \text{ kN}}{\pi \left(\frac{5.23}{2}\right)^2\cdot 787 \text{ MPa}}=58132.84 \text{ wires}$$

Your rounding error is most likely in $787\text{ MPa}$, which you calculated and rounded down to 3 digits prematurely.
Anyway, it's not a big deal. (Wink)

I do understand that the bridge is symmetrical and everything, but it still confuses me. If we only look at half of the bridge, and examine only the vertical and horizontal forces acting on half the bridge, we determine the net force acting on half the bridge. Shouldn't the number of wires be reflecting the number of wires on only half the bridge? So if the net force was $982855.114$, then the net force of the whole bridge is that times 2...

The tensional force in the cable is not equally divided along its length.
The closer you get to a tower, the greater the tensional force, since it carries more of the bridge below it.
If the cable would break, it would be close to a tower, where the force is greatest.

More to the point, in the free body diagram (that I hope you have drawn ;)), we have a weight going down, and a vertical force at each of the towers, that are both half of the weight.

The weight of complete bridge section is carried where the wires are connected to the left tower and where they are connected to the right tower.
So where the wires are connected to the left tower, they have to carry half the weight of the bridge, and where the (same) wires are connected to the right tower, they again have to carry half the weight of the bridge. (Nerd)

 

[Dethrone]

Thanks, I get it now! I just noticed your excellent table many posts ago, and you listed $\alpha$ to be the angle at the tower made with the horizontal. I've seen that technique used to answer this question before...what is the difference? Is using $\alpha$ less accurate?

 

[I like Serena]

Thanks, I get it now! I just noticed your excellent table many posts ago, and you listed $\alpha$ to be the angle at the tower made with the horizontal. I've seen that technique used to answer this question before...what is the difference? Is using $\alpha$ less accurate?

No such thing.
I only defined $\alpha$ to be able to refer to that particular angle. (Nerd)

We have that:
$$\tan \alpha = \frac VH = \frac {4h}{l}$$
after which we don't need it anymore.

It appears you had already figured that out, so it never came up. (Mmm)

 

[Dethrone]

Oh, and there is something else that still isn't connecting, it is stressing me out. You tried to explain it before, but it is straining my brain.

Assume that there is only one main cable, not two. First we divide the load by two, where each tower supports half of the load. Then, by analyzing half the bridge, we are able to get the net tension at one tower. If we use that tension which supports half the load, are we not getting half the wires of that cable...? How does the tension give us all the wires in the cable? i.e if right tower tension gives us 335 wires, then isn't the total number of wires 670?

Now extending this into two main cable, we looked at half the bridge. Similarly, the load gets divide by two among both towers, WL/2, and we get the net tension in both (half) cables at the right tower. If we were to calculate the number of cables, why isn't it only for half the bridge...i.e two half cables connecting to the right tower, which is one complete cable assuming symmetry? (Wondering)