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Using Newton's 2nd law for rotation with a moving coordinate. 2

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happygopher

Mechanical
Feb 4, 2010
14
I am working on a problem where there is a mass at the end of a slender cylindrical beam as shown in the attached figure. There is a torque load applied to the mass about the Y-axis (on mass's local coordinate system). This produces a deflection in the beam in the direction of the positive X-axis.

If I want to describe the dynamic behavior of this system, it seems to me that I can't just use Newton's second law directly because I would only be considering the mass as a rotating object. It is however also moving as the beam deflects.

How can I modify [tt]T = I*d2/dt2(theta)[/tt] to account for this?

My initial thoughts are that maybe "I" is a function of beam deflection; however, I am not exactly sure what this relationship would be.

I would appreciate any insights.
 
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Are you trying to determine a vibratory motion with this problem?
 
Sort of, but more just calculating dynamic response for variable input (torques). It will be part of a larger Simulink model. This is just the portion that I am having trouble accurately describing.
 
It seems that if you concentrate your analysis to a CoordSys located locally on the mass, then the only motion you would anticipate would be undamped torsional oscillation.

If you want the eqns of motion for "the system" then you need to assign a CS to the grounded base of the column, and resolve all motions to that. In this way it would give you eqns of motion of the mass as a function of (X, Z, <RotationAboutY>) where <RotationAboutY> is a function of Theta.

TygerDawg
Blue Technik LLC
Virtuoso Robotics Engineering
 
In that case then I would need to resolve the load placed on the mass by the beam into torque and force. The torque part I already know but does anyone know how I would obtain the force load?

I suppose I could use the equations obtained from the numerical solution for large displacement beam theory shown here (PDF):
Does that seem to make sense?
 
Guessing, but the beam acts as a cantilever spring, applies a reaction force to mass when off-center. Instaneously variable based on position. Spring constant related to beam bending formula for combined cantilever load & moment load. Identify loads on mass and resolve for instanteous position to match beam formula.

But all this for gross analysis only, doesn't take into account induced vibrations into beam by variable torque.

TygerDawg
Blue Technik LLC
Virtuoso Robotics Engineering
 
tylerdawg,

I agree with what you have proposed; however, I am not sure about the part with identifying the spring constant. Since it is a large displacement problem wouldn't the equations be nonlinear and I'm not sure that there is a readily available solution for the spring constant.
 
This problem is best done by using Lagranges equations which involve
T, the kinetic energy of the system and U, the stored energy in the massless beam

T= 0.5*m*y'^2 + 0.5 I*@'^2

where y and @ are the angular and vertical deflections and @' and y' the velocities of the free end of the beam and U which is fairly messy but is of the form

U=a*y^2+b*@^2+c*y*@

The Lagrangian is
L=T-U and

The equations of motion come from

d/dt(partial of L with respect to @')- dU/d@=T(t)

1) I@"+ 2*b*@+c*y=T(t)
Similarly for the vertical motion

2)My"+2*a*y+c*@=0

The two coupled equations describe the dynamics of putting torque, T(t) at the end on the mass at the end of the beam.

The problem is to get the stiffness matrix, F,which is messy but straightforward.


F=[K]*Y

where F=f,M, the force vector and Y=y.@ the motion vector



















 
fwiw, I'm inclined to agree with zekeman.

happygopher, three questions:
1 - are you satisifed to neglect the mass of the cylindrical beam (it certainly helps)
2 - do you have reason to reject typical small-displacement assumptions? These are big simplifiers
3 - are you interested primarily in the sinusoidal steady state response and/or resonant frequencies? This may lead to simplification along the way in both analytial and numerical solutions.

=====================================
(2B)+(2B)' ?
 
"The problem is to get the stiffness matrix, F,which is messy but straightforward."

correction, the stiffness matrix is [K].
 
@zekeman: Thank you for your post. I'll have to work through what you have put down and see if I can make sense of everything.

A couple of questions though. Can you point me in the direction of something that describes the stored energy of the beam that you used? I have been having trouble finding anything of that form online.

zekeman said:
U=a*y^2+b*@^2+c*y*@

When you used the Euler-Lagrange equation, where does the T(t) come in to that? I'm reading up on it but having a little trouble understanding that part.

zekeman said:
d/dt(partial of L with respect to @')- dU/d@=T(t)


@electricpete:

1) I think this is quite a reasonable assumption for my problem.
2) I really wish that I could use small displacement analysis because then I wouldn't really have much issue with it. However, the beam will see quite large deflections. Probably up to complete retroflexion.
3) I am primarily interested in being able to accurately model the system in Simulink and locate the tip position in response to applied load. The response will actually go to a steady state value when the rest of the model (with damping) is in place.

 
Attached is an analysis using zekeman's suggested approach.

It uses Maple to do the grunt work and ASSUMES small deflections.

That assumption is not exactly what you were looking for, but maybe will help you get the idea of Lagrange.



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(2B)+(2B)' ?
 
 http://files.engineering.com/getfile.aspx?folder=2e0c433c-731f-4a1e-bfb0-a59c89136b7a&file=happy_gopherR5.pdf
electricpete said:
Attached is an analysis using zekeman's suggested approach.

Wow! Thanks a lot for taking the time to do that. I will look through it and see if I can adapt it to the large deflection case.

One thing I found a bit ago that may help (not sure yet) is a solution for beam shape under pure moment loading.

 
Hmmm. I don't think you have pure moment loading since there is force required to accelerate the mass m at the end Fy = m * d^2(y)/dt^2 (using symbols/terminology like my attachment). That force is provided by the beam. The shear V at end of the beam is not zero.

=====================================
(2B)+(2B)' ?
 
Clarification in bold:
The shear V at end of the beam is not zero.
should've been
The shear V in the beam is not zero.

(at any moment in time, shear is constant along the length of the beam for this particular problem where forces are applied only at the ends of the beam)

=====================================
(2B)+(2B)' ?
 
electricpete said:
Hmmm. I don't think you have pure moment loading since there is force required to accelerate the mass m at the end Fy = m * d^2(y)/dt^2 (using symbols/terminology like my attachment). That force is provided by the beam. The shear V at end of the beam is not zero.

Ah, yea good point. I suppose that that condition is only valid if no mass is considered.
 
I wanted to do a "double-check" that the solution I provided before was correct. Results are attached and suggest that it is correct (at least predicts the natural frequenies reasonably well).

I examined the following example problem:
Rod diameter = 0.1m
Rod length = 1m
Disk diameter = 0.5m
Disk thicknss = 0.01m
Properties (steel)
rho = 7700 kg/m^3
E = 2E11 N/M^2

In the tab entitled "overview", there is an icon containing a pdf of the Maple solution which is the same as before, except I have added at the end some calculations using the equations of motion to calculate the natural frequencies. The results are:
f1 = 69hz
f2 = 660hz

The remainder of the spreadsheet is one I have developed for solving certain vibration problems numerically. When solving the same problem, the results are:
f1 = 67.7hz
f2 = 662hz

It seems reasonably close to me, so I have a little more confidence that the previously-provided solution is correct (within stated assumptions)

=====================================
(2B)+(2B)' ?
 
 http://files.engineering.com/getfile.aspx?folder=ade544fb-847a-4a4f-82df-57af252e136d&file=HappyGopher.xls
Correcting the shaft density in section 15 (cell G15 of tab rotorsections) from 7700 to 0.1 (~0) results in improved agreement of the first mode frequency:

Solution by Maple using previously-derived equations of motion:
concludes:
f1 = 69hz
f2 = 660hz


Solution by transfer matrix method
f1 = 69.0hz
f2 = 663hz



=====================================
(2B)+(2B)' ?
 
 http://files.engineering.com/getfile.aspx?folder=9f61ad31-558b-4aa6-8f84-10889d438ac3&file=HappyGopherR1.xls
Electricpete,

Good job!

As far as I can tell you got the potential energy right and the 2 coupled equations as well; and your small motion comments were on the money.

I may add that we both ignored the gravitation field.

happygopher,

You probably know that you just add mgy to the stored energy term U if the gravity field exists. Also, a large motion solution would be prohibitively difficult and I doubt that you need it for your analysis. In fact when these large motions occur, they are associated with a resonance or near resonance condition which allows the designer to accommodate the environment.I personally have never seen the need in all my years in this.\ business.
 
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