Structural Engineering Stress Question

[012anonymousx]

Homework Statement

Piano wire weighs 77kN/m^3 and breaks at a stress of 3000MPa. Calculate the length of piano wire that can be lowered down a very deep mine before it breaks.

Homework Equations

stress = force/area

The Attempt at a Solution

Piano wire = 77 000N/m^3
stress = 3000 000 000Pa(N/M^2)
Area of cross section = (pi)(d^2)/4

Force exerted by x m^3 of piano wire is (x)(77000)

Therefore

3 000 000 000 = (4)(77000)(x) /(pi)(d^2)

ASSUME d = 0.0002m

x = 0.122399m^3

Volume = (pi)(r)^2 * h
0.122399 = (pi)(0.001)^2 * h
h= 38 960

This seems a bit too much.
My professor is known for throwing out questions that require "engineering intuition for making assumptions."
Not sure if my solution is right or I can do it without making an assumption.

This is a problem set, no solutions please but only tell me if I'm way off (or close and where i made the mistake).

I appreciate it.

 

[rock.freak667]

That is correct, from what I've read online, the free-breaking length around 30 km depending on the UTS you are given.

Also, you didn't need to assume any diameter. Instead of finding force, you could have just said that at the breaking length 'x', the stress would be 77x kN/m² or 77x kPa = 0.077x MPa.

Thus 0.077x = 3000 => x= 38,961 m.

 

[SteamKing]

Rather a lot, really.

 

[012anonymousx]

Wait... whaa...?
How can you not assume any diameter..?
What is going on here...what formula is that. Isn't the length you can go down heavily dependent on the diameter of the piano wire? That formula doesn't assume anything.

You could lower a 0.1cm diameter wire or a 1m wire...

But if you plug these diameters into the original formula for stress, you get huge differences

Stress = (77000)(x)(4) / (pi)(d^2)

Rearrange for x and the x is dependant on d.

So what is going on...?

 

[rock.freak667]

Wait... whaa...?
How can you not assume any diameter..?
What is going on here...what formula is that. Isn't the length you can go down heavily dependent on the diameter of the piano wire? That formula doesn't assume anything.

You could lower a 0.1cm diameter wire or a 1m wire...

But if you plug these diameters into the original formula for stress, you get huge differences

Stress = (77000)(x)(4) / (pi)(d^2)

Rearrange for x and the x is dependant on d.

So what is going on...?

They gave you the Force per unit volume.

So

$$77 \frac{kN}{m^3} = \frac{Force}{Volume} = \frac{F}{Area \times x}$$

$$77x \frac{kN}{m^2} = \frac{Force}{Area} = Stress = 3000 MPa$$

 

[pongo38]

I understand that the strength of cotton is sometimes expressed as the length that would just break under its self-weight, even though is not carried out that way.

 

[012anonymousx]

Interesting...

Sure, my solution was really close, but only because I appeared to have picked a good diameter.

I could easily have picked a bad diameter and got it wrong.

So what was fundamentally wrong with my solution?

Also, I notice that one cannot solve for the area (diameter) or the force using this solution.

Because you have 77kN/m^3= Force / (area * x)

You only know force/area together, but its two variables still.

Is your solution implying that one can lower any diameter of piano wire (theoretically)? It sort of makes sense because the wider it is, the less stress, but more weight. A straight relationship.

But more importantly, please, someone tell me what was fundamentally wrong with my solution. I cannot see any gap in the logic but it can turn deadly because it is obviously wrong and might mess me during exam.

 

[PhanthomJay]

There was nothing wrong with your solution, even though you used d =.002 m but wrote it down as .0002 m.

There was something wrong with your other solutions, however, when you chose different diameters . It should give you the same result for length. The breaking length is independent of the diameter.

Stress = F/A
and since F = (weight density)(volume), then
F = (weight density)(A)(L)
Stress = (weight density)(A)(L)/A
The Areas cancel, so
Stress = (weight density)(L), or
L = Stress/weight density
which is independent of the diameter.
As Rockfreak pointed out twice.

 

[012anonymousx]

Aha! Thank you... that clarified a lot.

Okay, I have another really difficult question: again, this is problem set so please push in the right direction.

A mercedes-Benz weighing 4000kg and traveling 160km/h runs into a concrete post. Assuming all kinetic energy is used to crush the low allow steel structure at the front of the car, how many kg of steel will be crushed before the car is stopped?

E = 0.5 m v^2

E = (0.5)(stress)(strain)(volume)

Information about low allow steel from a quick fact sheet I got:
Weight: 77000N/m^3
E (young's modulus) = 200 000MPa
Compressive Str = 420Mpa
(I thought these were the only relevant things)

So I began by finding the kinetic energy:
Ek = 3 950 617.284J

But immediately after I'm stuck.

E = (0.5)(stress)(strain)(volume)

The only real thing I know here is the volume, based on the kg of steel specified in the question. But my thinking was to SOLVE for that volume to find the amount of steel.
I could resolve stress into force/area and volume into area * h to cancel h, but then two problems present themselves:
what the heck is the height. And the force is way too complicated to solve. You need a lot more data to find the acceleration during that 0.2s collision.

I feel like this is a similar problem to the first one where the dimensions shouldn't matter one bit. Afterall, the wider it is, the less distance it will be crushed but more kg will be crushed (because of more area). And the thinner it is, the more distance but since it is thinner... well you get the idea.

But I cannot cancel them out.

Any ideas?

 

[pongo38]

As the car crushes it goes beyond the elastic limit into plastic crumpling. Therefore Young's modulus E is of no relevance. Try thinking about energy dissipation over a short period of time.

 

[012anonymousx]

Ah, I appreciate it. I got it, thanks a lot.