2 DOF oscillator max force response

AI Thread Summary
The discussion focuses on determining the maximum force response on a component in a two-degree-of-freedom oscillator system. The user seeks to express the force applied to mass m2 as a function of system parameters (m1, m2, c1, c2, k1, k2, y) without involving positions x1 and x2. They have derived a matrix system of equations but struggle to eliminate x1 and x2 from their final expression. The conversation emphasizes that for steady-state responses, initial conditions are not necessary, and the nominal length of the springs can be disregarded in oscillatory analysis. The user ultimately aims to find a force frequency response function for the system.
saxymon
Messages
4
Reaction score
0
Hello,

I am mounting a component onto structure and I need to determine the maximum force input into the component.

My system can be represented by a base driven two degree of freedom oscillator:

http://www.freeimagehosting.net/newuploads/xowmr.png

I need to determine the force applied to m2:

F = k2*(x2-x1) + c2*(x2-x1)

This force needs to be a function of (m1,m2,c1,c2,k1,k2,y) and not of (x1,x2).
Basically, for a given input y, what will the be force response on m2.

Every time I solve the system of equations, my result is a function of x1 and/or x2.


Thank you in advance for your help!

p.s. If it is easier, feel free to remove the dampers from the system. An un-damped system will work for my purposes.
 
Engineering news on Phys.org
This is somewhat unfamiliar territory for me. A real spring has some length, right? Let that be l2 or l1 when the springs are unloaded.F2 = -k2d
d = x2 - x1 - l2
F2 = m2dx2/dt = k2(x1 + l2 - x2)

Then the force on m1 would be the negation of that plus the force from the spring connecting it to the base

F1 = m1dx1/dt = k2(x2- x1 - l2) + k1(y + l1 - x1)

This should be a system of differential equations which has solutions dependent on initial values of x1 and x2 as well as the entire input function y(t). You must further specify the problem by placing constraints on the initial positions, and the input function y(t). Then you may still be left with a seemingly large space of y(t) functions over which to maximize the force.
 
Last edited:
Thanks for the reply MisterX. A couple comments...

The nominal length of the spring can be ignored as this is an oscillatory system (and as long as the stiffness is the same, the response will be the same, regardless of length).

Also, I am seeing the force on m2 (not m1), which is defined as
F = k2*(x2-x1)

I am looking for a steady state response due to a stationary random input which means initial conditions are not needed (they effect only the transient response). Think of this as a "Force" Frequency Response Function


Here is what I've done...

my matrix system of equations is...

[m1 0 ; 0 m2] [x''1 ; x''2] + [k1+k2 -k2 ; -k2 k2] [x1 ; x2] = [0 ; k1] [0 y]

Interestingly enough, the force I need to recover is naturally in the lower portion of the equation giving:

F = k1*y - m2*x''2

I make the reduction to

F = k1*y + Wn^2*m2*x2

but still was not able to remove all of the x2's and x1's (after making a number of substitutions).
 
Posted June 2024 - 15 years after starting this class. I have learned a whole lot. To get to the short course on making your stock car, late model, hobby stock E-mod handle, look at the index below. Read all posts on Roll Center, Jacking effect and Why does car drive straight to the wall when I gas it? Also read You really have two race cars. This will cover 90% of problems you have. Simply put, the car pushes going in and is loose coming out. You do not have enuff downforce on the right...
I'm trying to decide what size and type of galvanized steel I need for 2 cantilever extensions. The cantilever is 5 ft. The space between the two cantilever arms is a 17 ft Gap the center 7 ft of the 17 ft Gap we'll need to Bear approximately 17,000 lb spread evenly from the front of the cantilever to the back of the cantilever over 5 ft. I will put support beams across these cantilever arms to support the load evenly
Thread 'Physics of Stretch: What pressure does a band apply on a cylinder?'
Scenario 1 (figure 1) A continuous loop of elastic material is stretched around two metal bars. The top bar is attached to a load cell that reads force. The lower bar can be moved downwards to stretch the elastic material. The lower bar is moved downwards until the two bars are 1190mm apart, stretching the elastic material. The bars are 5mm thick, so the total internal loop length is 1200mm (1190mm + 5mm + 5mm). At this level of stretch, the load cell reads 45N tensile force. Key numbers...

Similar threads

Back
Top