Response to impact pulse 3

[davidmandis]

My problem involves a spring mass system with a triangular impulse acceleration as the input on impact. I have simplified the force-deflection response as two linear behaviors as shown in attached figure.
I will need to model two separate responses for the input as two ramp loads, and using the results of the first one as the initial conditions to the second. The two ramp inputs will be,

f(t)=2P(t/tp) 0<t<tp/2
f(t)=2P(1-(t/tp)) tp/2<t<tp

I also will have to calculate two responses for the two different spring behaviors,
F=kx x<3750
F=k x>3750

Now, I cannot seem to figure out how to superimpose all the results. Should I just add all the responses (by response I mean the output displacements) or is there some other way to superimpose the results?

Thanks,
David

 

[electricpete]

IF the acceleration represents ths input to the system AND you had a linear system, you could decompose the input acceleration as you have done and add the "responses" to the acceleration. But that does not apply for a non-linear system which you have.

It also seems unusual to me to have acceleration an an input. Is there a mass in the system? If not, displacement is second integral of acceleration and force is known function of displacement.. seems like a pretty trivial problem. More details of the total problem to be solved would help.


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[davidmandis]

The system actually represents a landing gear with a composite crash tube to absorb the energy of impact. Now the input would be a force since I know the rate at which the system (system includes mass and spring) deccelerates, so the mass times the g force would give the force applied which would be a function of time and will have the same time step as the acceleration.
I have converted the non-linear spring behavior into a linear response. But I am not sure how to combine the results of the steps as shown in first post.

 

[electricpete]

Just trying to follow the logic.

If everything you said is true ( you know the acceleration and therefore the spring must absorb force F(t)=m*a(t) where a(t) is as given in the jpg file above), then you can trace F(t) and determine a displacement. But displacement of what? You told me you already know acceleraiton.

How can you know a(t) ahead of time if you haven't factored the non-linear spring into the problem yet? Doesn't it affect the acceleration?

I would not consider that representing the non-linear curves as two straight lines makes it linear. A linear system has output proportional to input (and that type of linearity is a prerequisite to using superposition). It would be represented by a straight line that goes through 0.

I think if you can break the problem into separate time intervals, then it could be linear within each time interval. In the time when the spring is in the linear region, using your linear spring constant F = k1. In the time when the spring is in the flat region, model it as a static applied force F = k2.


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[electricpete]

Correction:
"...then you can trace F(t) and determine a displacement"
Should have been:
"...then you can double integrate F(t)/m and determine a displacement"

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[davidmandis]

You are right about the non-linearity...the spring is indeed non-linear and i meant it is linear if we consider it in two different sections but I dont think that is the right approach.
What I mean by acceleration as input is that the whole system including the mass and the spring is deccelerating at the rate shown in the figure but I do not know the acceleration of the mass in the system itself. In other words the input to the system is acceleration which can be converted into a force. I have shown the spring mass system in the figure attached.
A transient analysis would have worked out if the spring was liner but I don't know how to include the non-linearity of the spring in the system.

David

 

[GregLocock]

I strongly recommend that you build a time based model that just builds up the displacements from the velocities from the accelerations from the forces.


Cheers

Greg Locock

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[electricpete]

Yes. Break it up into regions. The breakpoints between the regions are tp/2, tp, and anytime the spring changes between it's two regions.

If you have numerical values AND units, it might be a little easier to see where the breakpoint might fall.

In your file impact_dynamics.jpg, the horizontal axis is labeled displacement in one figure and load in the other. Supposed to be displacement on horizontal axis in both?


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[GregLocock]

I meant cruder than that. Every 0.01 seconds at a rough guess. It takes a while to write your first spreadsheet like that but it is a very useful technique , and can be extended to including multidof systems.



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[electricpete]

Yes, a numerical integration of the initial value problem would be easier and save a lot of algebra.

for Dave's benefit, here's one way to set it up:

State variables:
dm = displacement of mass
vm = velocity of mass

Derivatives of state variables
d/dt(dm) = vm
d/dt(vm) = (1/M) * f(dg - dm)

where f is the nonelinear spring function which gives force as function of displacement
dg is displacement of the ground node which is a known function of time that we can determine ahead of time from two integrations of the function a(t) plotted above. (actually of course ground doesn't move but we can consider the initial inertial frame of the mass where the ground does move).

The simplest numerical integration is Euler's method
xk = xk-1 + deltaT * dx/dt (xk-1)

I have a spreadsheet at home where I programmed in vba a variable step size Runge-Kutta Feldberg 4th order algorithm with 5th order correction of step size.... seems to work just as good as Matlab ODE45. If you provide the numerical values and units I will set it up and post it.


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[electricpete]

By the way, it seems like one assumption needs checking. Is the spring force really a function of displacement only with no dependence on time history... or does it act differently if you apply the force slowly or suddenly (i.e. some damping)?

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[electricpete]

If you throw out the 5th order variable step size correction and stick with 4th order fixed step size, the fourth order Runge Kutta (RK4) is very very easy. It is a helluva lot easier than RKF45 and a helluva lot more accurate than Euler... RK4 is O(dt^4) while Euler is O(dt^1) where dt is step size.

RK4 is as follows:

Let f(x,t) be the computed derivative

f1 = f(x,t)
f2 = f(x+ f1*dt/2, t+dt/2)
f3 = f(x+ f2*dt/2, t+dt/2)
f4 = f(x+ f3*dt, t+dt)

estimate of x(t+dt) is
x(t) + dt*(1/6) (f1+2*f2+2*f3+f4)


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[davidmandis]

Thank you Pete and Greg for your replies, I really appreciate it. I think I will solve it first using Greg's method and use the results as validation for the MATLAB results.
Pete, I am trying to figure from your directions how to solve my problem in MATLAB using ode45. Do you know of any reference I could use and would love to see how you solve it using a spreadsheet. I have attached the problem statement with all the units and other details I could gather. Let me know if I missed out any details.

Thanks again,
David

 

[davidmandis]

Sorry I forgot the attachment. Here it is,

 

[electricpete]

What is the value and units of M?

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[electricpete]

Also can you respond to this query:

In your file impact_dynamics.jpg, the horizontal axis is labeled displacement in one figure and load in the other. Supposed to be displacement on horizontal axis in both?

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[davidmandis]

M is in lbs and it is deflection on the horizontal axis for both. I have made this correction in a pdf attachment in my last post.

 

[electricpete]

Sorry, I didn't see your last attachment. Got it now. Thanks.

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[electricpete]

You show a single degree of freedom system but list two masses.

Is the spring located between the aircraft and the occupant and the aircraft plays the role shown in the SODF diagram as ground (except that it's accelerating with the curve given)?

Or are the aircraft and occupant masses lumped together as the mass in the SDOF and the ground accelerating with respect to their initial inertial reference frame?


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[davidmandis]

yes the spring is located in between the aircraft floor and the occupant and the aircraft floor is accelerating according to the given curve. So when the aircraft crashes, the ground imparts a force of 48 G's to the aircraft floor. one thing I missed out is that the force should be multiplied by 42 to get the max force Fm. So that will make,
Fm = 4738230 N = 1075197 lbs-force

David