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- Thread starter MMCS
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I go to infinity.The massless bar case is also a limiting case of a small mass, m, located at the end of the bar (3L, down by the dark thingies) and the inertial effects (or lack thereof) of this mass are indistinguishable from the massless bar limit as m goes to zero.I see what you're saying. My thoughts are (for the moment) that a massless bar presents something of a pathological case. I am thinking of treating the problem as though the rod did have finite mass, ergo rotational inertia I, then solve for theta... and then let I go to infinity.The massless bar case is also a limiting case of a small mass,

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EDIT -- And what do the distances "L" at the bottom of the figure have to do with the problem? Are there transmission line considerations?

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berkeman said:

EDIT -- And what do the distances "L" at the bottom of the figure have to do with the problem? Are there transmission line considerations?

I suspect that it's a schematic of a mechanical problem of some form. Maybe a beam fixed at one end and with springs and damping located at certain distances? A model for a beam?

If the OP will enlighten us (or someone familiar with the format can elucidate all the technical nuances encoded in the diagram), maybe we can help.

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berkeman said:

EDIT -- And what do the distances "L" at the bottom of the figure have to do with the problem? Are there transmission line considerations?

The 'thing on the left ' is presumably part of a wall. A massless bar or board runs from x = 0 to x = 3L. If the wall ends at x = -d, then the pivot point is at x = 0, the springs exert a force at x = L , the dashpot about x = 2L and the displacement output y(t) occurs at x = 3L.

The problem deals with torques about the pivot point. x(t) is the input displacement at the top of the top spring. y(t) is the output displacement at x = 3L.

So, OP, write Ʃ torques = angular acceleration and solve for y(t)/x(t). Since it asks for the

I don't know why they ask for the transfer function

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rude man said:The 'thing on the left ' is presumably part of a wall. A massless bar or board runs from x = 0 to x = 3L. If the wall ends at x = -d, then the pivot point is at x = 0, the springs exert a force at x = L , the dashpot about x = 2L and the displacement output y(t) occurs at x = 3L.

The problem deals with torques about the pivot point. x(t) is the input displacement at the top of the top spring. y(t) is the output displacement at x = 3L.

So, OP, write Ʃ torques = angular acceleration and solve for y(t)/x(t). Since it asks for thetransfer functionI assume you've been exposed to the Laplace transform or the Fourier.

I don't know why they ask for the transfer functionand time constant. The transfer function by itself contains all the available information including time constants.

Rude man: To expand on this, it looks like those dark rectangles dangling from the bottom are not masses. It seems to me they are rigid constraints (i.e., with zero displacement). So, in my judgement, the system consists purely of springs and dashpots. x(t) is an arbitrary time-dependent forcing displacement, and y(t) is the time-dependent response of the spring dashpot system. It doesn't seem like the angular accelerations will be required, although the vertical velocities and displacements are important, and, as you said, a balance of moments is required. Also there is a kinematic condition between y(t), and the displacements at other key locations along the plank.

Chet

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Chestermiller said:Rude man: To expand on this, it looks like those dark rectangles dangling from the bottom are not masses. It seems to me they are rigid constraints (i.e., with zero displacement). So, in my judgement, the system consists purely of springs and dashpots. x(t) is an arbitrary time-dependent forcing displacement, and y(t) is the time-dependent response of the spring dashpot system. It doesn't seem like the angular accelerations will be required, although the vertical velocities and displacements are important, and, as you said, a balance of moments is required. Also there is a kinematic condition between y(t), and the displacements at other key locations along the plank.

Chet

I agree, those dark thingies "hanging" at the bottom are additional "walls" i.e. fixed boundaries.

You agree or not, the thing connecting the springs and dashpot is a massless rod or similar? And it's pivoted at that triangular extension away from the wall on the left?

The equation of motion is one of summing torques = angular acceleration of the rod = d

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Agree.rude man said:I agree, those dark thingies "hanging" at the bottom are additional "walls" i.e. fixed boundaries.

You agree or not, the thing connecting the springs and dashpot is a massless rod or similar?

Yes.And it's pivoted at that triangular extension away from the wall on the left?

Not exactly. If the rod is massless, its moment of inertia is zero. The moment balance is then:The equation of motion is one of summing torques = angular acceleration of the rod = d^{2}θ/dt^{2}.

sum of torques = 0

Since θ is small, we can work exclusively with y(t). The downward displacements at L and 2L are y/3 and 2y/3.y(t) is then = 3L*θ(t).

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K (y/3) + K (y/3) + C dy/dt (2y/3) = y(t)

**i have ommitted the x from Kx because of no initial displacement?? **

K(2y/3) + C dy/dt (2y/3) = y(t)

Laplace Transform

K(2y/3) + C(2y/3) S = y(t)

Is this correct? What would be the next step to get the transfer function?

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MMCS said:

K (y/3) + K (y/3) + C dy/dt (2y/3) = y(t)

**i have ommitted the x from Kx because of no initial displacement?? **

K(2y/3) + C dy/dt (2y/3) = y(t)

Laplace Transform

K(2y/3) + C(2y/3) S = y(t)

Is this correct? What would be the next step to get the transfer function?

This is not what I get. Instead of

K(2y/3) + C dy/dt (2y/3) = y(t)

I get

K(2y/3) + C dy/dt (

From this, you should be able to take the Laplace Transform and get the transfer function.

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How do you get 4/3 for the moment of C?

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Chestermiller said:This is not what I get. Instead of

K(2y/3) + C dy/dt (2y/3) = y(t)

I get

K(2y/3) + C dy/dt (4/3) =kx(t)

From this, you should be able to take the Laplace Transform and get the transfer function.

How did you get 4/3 for the moment for c?

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The displacement is 2y/3, and the moment arm is 2L.MMCS said:How did you get 4/3 for the moment for c?

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K*2/3*y(s) + C*2/3*y(s) = Kx(S)

y(s)/x(s) + K/ ( K*2/3 ) + (C*4/3*S)

3/2 * K/( K + 2*C*S)

divide through by K

3/2 * 1/ 1 + 2*C/K

3/2 * 1 / 1 + 2τ

Thanks for the help. i don't want to flood the forum but i am stuck on a similar type question, I am going to post it on here in a minute, it involves unit impulse

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MMCS said:

K*2/3*y(s) + C*2/3*y(s) = Kx(S)

y(s)/x(s) + K/ ( K*2/3 ) + (C*4/3*S)

3/2 * K/( K + 2*C*S)

divide through by K

3/2 * 1/ 1 + 2*C/K

3/2 * 1 / 1 + 2τ

Thanks for the help. i don't want to flood the forum but i am stuck on a similar type question, I am going to post it on here in a minute, it involves unit impulse

There seem to be a lot of equal signs missing and some s's missing, so none of this looks right. Please reconsider your algebra. For the transfer function, I got

[tex]\frac{(3/2)}{(1+sτ)}[/tex]

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Ok let me check...

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K*2/3*y(s) + C*2/3*y(s) = Kx(S)

y(s)/x(s) = K/ ( K*2/3 ) + (C*4/3*S)

y(s)/x(s) = 3/2 * K/( K + 2*C*S)

divide fraction by K

3/2 * 1/ 1 + 2*(C/K) * S

2*C/K = time constant = τ

3/2 * 1 / 1 + τS

I think that's it, we have the same answer now

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Chestermiller said:This is not what I get. Instead of

K(2y/3) + C dy/dt (2y/3) = y(t)

I get

K(2y/3) + C dy/dt (4/3) =kx(t)

From this, you should be able to take the Laplace Transform and get the transfer function.

Assuming a finite rotational inertia I for which my torque-summing equation is

Iθ'' = kL(x - Lθ) - kL

θ'' + 2kL

and after Laplace etc. I got

θ(s)/X(s) = kL/(Is

and if we take I → 0,

θ(s)/X(s) = (k/2L)/(cs + k)

and Y(s)/X(s) = (3k/2)/(cs + k)

which looks a lot like what you and the OP got.

Last edited:

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rude man said:For the spring force I get kx/2 = ky/6, not 2ky/3.

here's my reasoning:

Apply displacement x. Then in the absence of any restraining forces, the mid-point of the springs displaces to x/2.

So now, to get force times distance needed to pull the midpoint back to zero, you can compute the difference in potential energies E before and after the pullback:

E_{before}= 1/2 k (x/2)^{2}+ 1/2 k (x/2)^{2}= kx^{2}/4

E_{after}= 1/2 k x^{2}+ 0 (top & bottom spring respectively) = kx^{2}/2

delta E = Fx/2 = kx^{2}/4 or F = kx/2.

I now assumed finite I for which my torque-summing equation is

Iθ'' = kLx/2 - 2cL^{2}θ'

and after Laplace etc. I got θ(s)/X(s) = (kL/2) / s(sI + 2cL^{2})

and if we let I → 0 we get the surprising result that

θ(s)/X(s) = k/4cLs

i.e. an integral relationship between x and y: since y = 3Lθ, Y(s)/X(s) = 3k/4cs.

BTW: in your approach using only forces, what happened to the force exerted by the hinge?

Hi Rude Man,

I haven't been able to follow in detail yet what you have written, but I'll try some more. With regard to your BTW question, we took moments around the hinge.

Chet

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Chestermiller said:Hi Rude Man,

I haven't been able to follow in detail yet what you have written, but I'll try some more. With regard to your BTW question, we took moments around the hinge.

Chet

OK. Don't look too hard at it. I need to revise the net spring force for the third time.

See my final effort as post # 19, edited twice!

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