Is the Boundary Value Problem affected by shear force or moments?

[Nusc]

Suppose we have this rectangle that is stretched equally on both sides with some force, F.
Neglect shear force or moments and assuming transverse waves,

is the solution still

ε = Ae^(i(wt-kx))+Be^(i(wt+kx))

With boundary conditions:

X = +L/2, ∂ε/∂x = 0

and

X = -L/2, ∂ε/∂x = 0

?

 

[Simon Bridge]

There will be a set of solutions that will correspond with the situation you described, not just one.
If I read the description correctly, the boundary conditions are sort-of OK but the opposite sides of the rectangle, at the same x, will be oscillating 180deg out of phase.

 

[Nusc]

There will be a set of solutions that will correspond with the situation you described, not just one.
If I read the description correctly, the boundary conditions are sort-of OK but the opposite sides of the rectangle, at the same x, will be oscillating 180deg out of phase.

This is a standard BVP that can be found in a textbook no?

I have to review this material - I haven't touched BVP's for several years now. How can I derive the other solutions?

I don't understand when you say, "...the opposite sides of the rectangle, at the same x, will be oscillating 180deg out of phase. "

Once I review the material I think I'll know what you mean.

 

[Nusc]

First, I need to be sure on what the boundary conditions are, can you state algebraically what you mean by that statement?

Second, how can I determine what the other possible solutions are?

 

[Simon Bridge]

This is a standard BVP that can be found in a textbook no?

I havn't seen that specific description in a textbook - but it follows a common form.

The trick is usually to translate what you understand into maths.
Here's how I'm reading what you wrote:

There is a rectangle length L and width W (thickness we don't care about), made out of some unspecified material, aligned: $-\frac{L}{2} \leq x \leq \frac{L}{2}$.

We can represent the width at some position x measured at time t by a function w(x,t).

At $t=0$ the ends are distorted so that the wiidth there is: $w(-\frac{L}{2},0)=w(\frac{L}{2},0)=W+\epsilon_0$, i.e. both ends are initially stretched by the same amount.

The material properties of the rectangle mean there is a continuous variation of w with x.
But notice that the specific variation can be anything so long as those initial conditions are satisfied.
If we let go of the ends, we can expect the width of the rectangle at some position x at t>0 to vary according to:

$w(x,t)=W+\epsilon(x,t):\epsilon(-\frac{L}{2},0)=\epsilon(\frac{L}{2},0)=\epsilon_0$

... we can usually say more about the initial conditions than that though ... since we were the ones causing the initial distortion, we also know the precise shape of the distorted rectangle close to the ends ... i.e. we know, at least, how the width varies with x close to the ends. This can be anything we like - we usually pick:

$w'(-\frac{L}{2},0)=w'(\frac{L}{2},0)=0$ ... the primed notation indicates differentiation wrt x.

I have to review this material - I haven't touched BVP's for several years now. How can I derive the other solutions?

By solving the DE normally - the DE in question should be chosen using your knowledge of the physics involved.
For your example, it is probably fair to expect the wave equation to be useful.

It is very common to be asked to look for standing wave solutions and express other situations as a superposition of modes. This is what it looks like you were trying for with your proposed solution - but that's unlikely unless the ends are continuously being distorted (i.e. not just at t=0). Your setup could result in $w(x,t)=W+\epsilon_0\cos\omega t$ or in two pulses running into each other, depending on the material properties.

I don't understand when you say, "...the opposite sides of the rectangle, at the same x, will be oscillating 180deg out of phase. "

If the top edge of the rectangle follows $y_+(x,t)$ and the bottom edge follows $y_-(x,t)$, then $w(x,t)=y_+(x,t)-y_-(x,t)$ ...

$y_+(x,t)=\frac{W}{2}+\frac{1}{2}\epsilon(x,t)\implies y_-(x,t)=y_+(x,t)=\frac{W}{2}-\frac{1}{2}\epsilon(x,t)$

... i.e. when one goes up the other goes down: $y_+$ and $y_-$ are 180deg out of phase.

 

[Nusc]

Hey Simon,

I incorrectly stated the problem:

Suppose we have this rectangle that is stretched equally on both sides with some force, F.
Neglect shear force resistance or moments and assume vertical displacements - I'm assuming that this means that the tension on the rectangle will only be displaced in the vertical direction?

How does this change what you have written?

 

[Simon Bridge]

Well it becomes totally ambiguous what is meant by "both sides".
A rectangle has four sides.

What you wrote made it look like the short sides were stretched, and your subsequent attempts suggested that was your understanding also. Now it could mean the long sides are stretched.

Look at the way I wrote my description in post #5. See how I described what I meant by "side" and what I meant by "stretched"? This needs to be cleared up. Normally, when a question is that ambiguous, you have some coursework examples that you are supposed to be reminded of.

 

[Nusc]

What I meant was equally in all directions.

The rectangle can only have vertical displacements.

 

[Nusc]

The fact that it only has vertical displacements, despite being pulled on each side with tension,T, is this is a 1D or 2D problem?

 

[Simon Bridge]

The fact that it only has vertical displacements, despite being pulled on each side with tension,T, is this is a 1D or 2D problem?

It's your problem - don't you know?

What I meant was equally in all directions.
The rectangle can only have vertical displacements.

This revised description makes no sense at all to me.

If the rectangle is expanded equally in all directions, then it's overall size gets bigger - upon release, all it's dimensions will just oscillate in and out with the same period.

If it can only have vertical displacements, then it cannot be expanded equally in all directions.
... unless you mean the center of mass is constrained to move only vertically? In which case, the rectangle will not have any displacements as a result of the expansion.

Please take some time to work out exactly what you mean.