I know this isn't the conventional structural question, but I haven't had any response in the vibration forum. The question is a mix of vibration and stress. I have a pendulum with a spring attached. I can determine a relationship to predict the natural frequency of the system. Where I'm having trouble is finding a relationship to predict the maximum bending moment in the beam, so that I size the beam properly. I've attached an illustration. Any help or direction would be appreciated.
Eng-Tips is the largest engineering community on the Internet
Intelligent Work Forums for Engineering Professionals
Bending moment in a Pendulum with Spring
- Thread starter meca
- Start date
I'd be designing it such that the spring and the pivot are supports and I would be calculating the force from the mass based on it's maximum velocity stopping over an assumed distance.
Then iterate a solution until the distance used to determine the force equals the deflection calculated on the beam from that force.
I think that would be an upper bound solution.
I also think you have to think of the beam as a rigid element otherwise you couldn't accurately predict the frequency of the vibration (as some of the energy would be eaten up by the beam flexing).
Then iterate a solution until the distance used to determine the force equals the deflection calculated on the beam from that force.
I think that would be an upper bound solution.
I also think you have to think of the beam as a rigid element otherwise you couldn't accurately predict the frequency of the vibration (as some of the energy would be eaten up by the beam flexing).
JeanLucPicard
Structural
Fundamentally, you have 3 main natural frequencies... I'm shooting blind now, but I think you should make a modal analysis. Read the deflections, rotation angles, and with stiffness work out the forces and moments?
Live long and prosper!
Live long and prosper!
JeanLucPicard
Structural
Can you use softwatres? Or you must do it by hand?
Live long and prosper!
Live long and prosper!
Meca:
The max. moments in the beam will occur at the instant of the max. acceleration or deceleration, of the mass, in either direction. This should be represented by a sine wave. Sounds like you should open you Structural Dynamics text book. It’s single degree of freedom problem the way you’ve described it. Right?
The max. moments in the beam will occur at the instant of the max. acceleration or deceleration, of the mass, in either direction. This should be represented by a sine wave. Sounds like you should open you Structural Dynamics text book. It’s single degree of freedom problem the way you’ve described it. Right?
Agree that it looks a lot like homework.
Summing moments about the pivot should relate angular acceleration and spring force; solution should just be sines/cosines. I would think maximum moment would be at the point where the spring attaches to the arm; summing moments about that point when accelerations are known would give you the internal moment.
Summing moments about the pivot should relate angular acceleration and spring force; solution should just be sines/cosines. I would think maximum moment would be at the point where the spring attaches to the arm; summing moments about that point when accelerations are known would give you the internal moment.
What happens if the mass is suspended on a cable instead of a beam? The cable cannot resist moment. You would still have a dampened pendulum. The cable would be straight when the mass passed directly under the suspension point (maximum velocity, zero acceleration); the cable would be kinked at the spring attachment when the mass reached its greatest horizontal distance from the suspension point (zero velocity, maximum acceleration or deceleration).
At the instant of maximum sweep, the three tensile forces at the junction of the cable and the spring must be in equilibrium.
BA
At the instant of maximum sweep, the three tensile forces at the junction of the cable and the spring must be in equilibrium.
BA
I assume the spring is just a spring, so the system is not damped.
I believe if the beam is considered massless and with small-angle approximations, the spring and the flex of the beam could be treated as one equivalent spring, so it's just a single degree of freedom spring-mass system.
I would assume maximum moment in the beam was when the mass was all the way at either side. At the limits of travel, the mass velocity will be zero, so all the kinetic energy is then transferred to potential energy in the springs. Knowing the maximum velocity, you can calculate maximum travel or vice versa.
I believe if the beam is considered massless and with small-angle approximations, the spring and the flex of the beam could be treated as one equivalent spring, so it's just a single degree of freedom spring-mass system.
I would assume maximum moment in the beam was when the mass was all the way at either side. At the limits of travel, the mass velocity will be zero, so all the kinetic energy is then transferred to potential energy in the springs. Knowing the maximum velocity, you can calculate maximum travel or vice versa.
JStephen,
I'm not sure why you say the system is not damped. If the mass was attached to the beam without a spring, it would oscillate at its natural frequency. Wouldn't the addition of a spring act as a damper?
I assume the spring is unstressed when the mass m is directly below the suspension point (point A). When m swings to the right, the spring goes into tension; when m swings left, the spring is compressed.
In the diagram, m is located a distance of z.sinθ to the right of point A and the spring has stretched h.sinθ, so it has a tension T = kh.sinθ (assuming the spring remains horizontal as it stretches which is a slight inaccuracy). If that represents the rightmost point of travel of mass m, the instantaneous bending moment in the beam is:
M = T.cosθ.h.(z-h)/z = k.sinθcosθ.h[sup]2[/sup](z-h)/z
M is maximum at the limits of the travel of mass m and is zero when m is directly under point A.
BA
I'm not sure why you say the system is not damped. If the mass was attached to the beam without a spring, it would oscillate at its natural frequency. Wouldn't the addition of a spring act as a damper?
I assume the spring is unstressed when the mass m is directly below the suspension point (point A). When m swings to the right, the spring goes into tension; when m swings left, the spring is compressed.
In the diagram, m is located a distance of z.sinθ to the right of point A and the spring has stretched h.sinθ, so it has a tension T = kh.sinθ (assuming the spring remains horizontal as it stretches which is a slight inaccuracy). If that represents the rightmost point of travel of mass m, the instantaneous bending moment in the beam is:
M = T.cosθ.h.(z-h)/z = k.sinθcosθ.h[sup]2[/sup](z-h)/z
M is maximum at the limits of the travel of mass m and is zero when m is directly under point A.
BA
- Thread starter
- #13
Thanks to all. Even though it looks like homework, it's not. It's a type of damping device, and I'm trying to determine how to properly size the beam. I've created a finite element model that gives some stresses, but the hand calculations results are an order of magnitude higher. I'm trying to figure out if the problem is in my FEA or my hand calculations. This is complicated by the fact that it's been 25 years since I've worked a dynamics problem. I'll try some of these suggestions and see if it leads me anywhere. Thanks again.
I'm with JStephen on several points here:
1) single mass = single DOF = single Tn.
2) I expect max moment to occur at limits of travel.
3) I don't see how the spring dampens anything. I think that it's just one component of the two that comprise the flexibility of the restraint at the single DOF (spring and beam flex). The combined stiffness may or may not be linear depending on how far we're taking the small deflection business. The spring will limit movement and alter the natural period but it won't dampen in my opinion.
@meca:
1) Should the spring really be a dash pot (damper) or damper and spring combo?
2) Is the beam/mass the thing that is meant to be damped? Or is this entire assembly to be mounted to something else that requires damping?
I like to debate structural engineering theory -- a lot. If I challenge you on something, know that I'm doing so because I respect your opinion enough to either change it or adopt it.
1) single mass = single DOF = single Tn.
2) I expect max moment to occur at limits of travel.
3) I don't see how the spring dampens anything. I think that it's just one component of the two that comprise the flexibility of the restraint at the single DOF (spring and beam flex). The combined stiffness may or may not be linear depending on how far we're taking the small deflection business. The spring will limit movement and alter the natural period but it won't dampen in my opinion.
@meca:
1) Should the spring really be a dash pot (damper) or damper and spring combo?
2) Is the beam/mass the thing that is meant to be damped? Or is this entire assembly to be mounted to something else that requires damping?
I like to debate structural engineering theory -- a lot. If I challenge you on something, know that I'm doing so because I respect your opinion enough to either change it or adopt it.
FishTheStructure
Structural
Agree with KootK - the spring does not do any damping unless it has dampening characteristics (think shocks vs. springs in a car's suspension)
Great spirits have always encountered violent opposition from mediocre minds - Albert Einstein
Great spirits have always encountered violent opposition from mediocre minds - Albert Einstein
Spring- force is proportional to deflection.
Damper- Force is proportional to velocity (I think that's Coulomb damping, but it's been a while- and damping could be of other types).
Anyway, as shown, it's seems to be purely a spring, zero damping, so single-degree-of-freedom linear system.
Note that in a real case, it might be reasonable to neglect the deflection in the beam part which would simplify the problem.
Damper- Force is proportional to velocity (I think that's Coulomb damping, but it's been a while- and damping could be of other types).
Anyway, as shown, it's seems to be purely a spring, zero damping, so single-degree-of-freedom linear system.
Note that in a real case, it might be reasonable to neglect the deflection in the beam part which would simplify the problem.
Here's my Vibration Forum post of 24Feb15 @ 2:21
(The joys of double posting without terminating the first thread!!)
- - - - - - - - - - - -
I'm a bit confused.[ ] The text on your diagram states that the beam is NOT rigid, but the diagram itself implies that it IS rigid.[ ] If you have derived a formula for the natural frequency you must have assumed a rigid beam, because it's too hard if the beam is flexible.
For a rigid beam, the natural frequency is
omega = sqrt[(kh²+mgz)/(mz²)]
The equation of motion then becomes
theta = A*sin(omega*t) where A is the (arbitrary) amplitude of the motion.
Double-differentiate to get accelerations.
From here it is a simple exercise to employ d'Alembert's Principle to derive an algebraic expression for the bending moment you are seeking.
- - - - - - - - - - - -
Taking it through (now that I've had some sleep) I get the maximum moment at the time of maximum excursion, as others have correctly pointed out above.[ ] Its magnitude is
(z-h)*(kh²)/z
multiplied by the angular amplitude of the motion.
Plus or minus a few small-angle approximations, which the OP told us were acceptable, this looks remarkably like the result from BAretired above.[ ] This should surprise nobody.
(The joys of double posting without terminating the first thread!!)
- - - - - - - - - - - -
I'm a bit confused.[ ] The text on your diagram states that the beam is NOT rigid, but the diagram itself implies that it IS rigid.[ ] If you have derived a formula for the natural frequency you must have assumed a rigid beam, because it's too hard if the beam is flexible.
For a rigid beam, the natural frequency is
omega = sqrt[(kh²+mgz)/(mz²)]
The equation of motion then becomes
theta = A*sin(omega*t) where A is the (arbitrary) amplitude of the motion.
Double-differentiate to get accelerations.
From here it is a simple exercise to employ d'Alembert's Principle to derive an algebraic expression for the bending moment you are seeking.
- - - - - - - - - - - -
Taking it through (now that I've had some sleep) I get the maximum moment at the time of maximum excursion, as others have correctly pointed out above.[ ] Its magnitude is
(z-h)*(kh²)/z
multiplied by the angular amplitude of the motion.
Plus or minus a few small-angle approximations, which the OP told us were acceptable, this looks remarkably like the result from BAretired above.[ ] This should surprise nobody.
SomptingGuy
Automotive
Maximum bending moment will be related to vibration amplitude. Vibration amplitude will be controlled by the system damping, which you have not given.
- Steve
- Steve
- Status
Similar threads
- Question
- Locked
- Question
