Calculating the spring displacement response from an acceleration input

[Raddy13]

TL;DR

Determine the displacement response of a mass on a spring when subject to an acceleration impulse

I'm working on a project where we have a mass (50 kg) sitting on a spring (350 N/mm) and are subjecting it to a sudden impulse (20g) along the spring axis to simulate a shock. We have the profile of the acceleration defined as:

$a(t) = x''(t) = P\cdot \sin^2 (\pi \cdot t / T)$

Where P (peak acceleration) and T (pulse width) are known and fixed. I know the normal equation is:

$F(t) + m \cdot g = m \cdot x'' + b \cdot x' + k \cdot x$

The b term drops out since there is no damper in this system, but that's as far as I can make it and I'm about a decade removed from my differential equations class. Is it possible to determine the function x(t) response from the information I have?

 

[BvU]

are subjecting it to a sudden impulse (20g) along the spring axis to simulate a shock.

If you are subjecting the mass to an acceleration $a(t)$, all you have to do is integrate the accelration twice to ge the displacement .

This is probably not what you had in mind. 'a sudden impulse' suggests you want to apply a force and give the spring a chance to counter that. Right ?

 

[Raddy13]

That would be ideal. The mass is a piece of equipment that sits on a shock mount in the bed of a mining vehicle and the vehicle is constantly hitting potholes and ruts and an acceleration measurement found that the worst case acceleration at the mounting location is 20gs. The equation listed above is from a study I found on approximating acceleration profiles in shock analysis.

 

[BvU]

Consider as $x$ the deviation from steady state (so that $mg$ also disappears from the equation)
and instead of $a(t) = P\sin^2\left ({t\pi\over T}\right)$ write $F(t) = mP\sin^2\left ({t\pi\over T}\right)$, then you are left with $$m\ddot x + kx = mP\sin^2\left ({t\pi\over T}\right)$$

My DE class is 4 decades back, so I use wolframalpha and type in

d^2x/dt^2 = -x +sin^2(t);x(0)=0, dx/dt(0)=0

and they come back with

which looks OK, and a 'solution'

which I don't fully grok now (late and Heineken), but maybe will tomorrow. They even show a plot (for periodic bumps):Advantage: gets you started and easy to re-introduce some damping

 

[Raddy13]

Awesome, thank you for the reply!

 

[JBA]

To start you need to determine actual dynamic displacement of your mass vs the static load displacement amplitude Xdyn/Xstatic ; and. for that you must first determine the β ratio between the natural frequency of your spring and the input frequency of your shock profile. The equation for the natural frequency of your spring/mass is: f (sprg &mass) = √(k / m) ÷ 2π, where m = mass mounted on the top of the spring + 1/2 the mass of the spring and k is your spring rate.
Next, the time of your shock determines the 1/t apparent frequency of your shock input; and then β = f spring & mass / (1/t).

From there, the actual dynamic mass displacement X dyn / X static = 1 / sqrt[ (1-β^2)^2 + (2*ζ*β)^2]; where for your undamped case the ζ (damping factor) = 0 and the formula reduces to: X dyn = X static / sqrt[ (1-β^2)^2].

Once the impulse displacement is completed your load and spring will then oscillate at the above calculated f sprg &mass frequency in a sin wave configuration; and, even with no external damping, the amplitude of the successive oscillations will decline due to work energy consumption of the spring and aerodynamic drag on the mass.

(PS The 1/2 mass of the spring a way of representing the average spring displacement between the x=0 at the input end and the x=max at the output ends of the spring)