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Calculating impact force, different results from different equations.

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GrouchyStressedTensePoor

GrouchyStressedTensePoor

Mechanical
Aug 2, 2018
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Hi all,

I was hoping my first post here wouldn't be something that I expect will turn out to be a something simple but here we go.

I am trying to calculate impact loading due to a mass with some initial velocity/kinnetic energy striking a cantilevered beam. Therefore the main method I have been using has been to equate kinnetic energy with strain energy. For the moment we can forget the cantilever bit.

The first method is:

K.E = ½mv²

F[sub]peak[/sub] = kδ

S.E = ½Fδ = ½(F²/k)

∴ ½(F²/k) = ½mv²
F²/k = mv²
F² = kmv²
F = √(kmv²)

The second method requires knowledge of the impact duration, which is why I am not using it as my main method since I do not know the duration of the impact (whereas the stiffness is more easily estimated). To calculate the impact duration I have assumed that the system behaves as an undamped second order mass-spring system, subject to free vibration. The point of impact would be the point of zero displacement and acceleration with maximum velocity (the impact velocity) and the "end" of the impact would be the subsequent point of maximum displacement and acceleration with zero velocity. This means that the impact duration in seconds would be ¼ of the time period of the second order mass spring system.

The natural frequency for a second order mass-spring system is: ω[sub]n[/sub] = √(k/m)
∴ ƒ = (1/2π)√(k/m)
∴ T = 2π√(m/k)
∴ t[sub]impact[/sub] = (π/2)√(m/k)

This can then be input into the following equation for impact force: F = 2mv/t

The above equation is something my boss found online, but looking at it it seems to be based on the assumptions that the impact force will be double the average force (which makes sense for a linear force beginning from zero like an elastic force), mv/t being a form of F=ma that assumes that the velocity drops to zero in time t.

I would expect then that using the same values for initial velocity, stiffness, and mass the two methods should give the same answer, but they dont seem to? Am I being dense? Where does the disparity come from?

e.g. v=3, k=5, m=20
F = √(kmv²) = √(5*20*9) = √900 = 30N
t = (π/2)√(20/5) = πs
F = 2mv/t = (2*20*3)/π = 38.2N

It would make some sense to me if the answers were off by a factor of 2 or 4, but I don't understand why one method returns an answer that is 1.27 (4/π) times the other.

Thanks in advance.
 
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It's a first order spring mass system.

I prefer your first approach, SEmax in the spring=KEmax of the mass. The assumptions you are making are that the cantilever is massless, there is no internal damping, and that no energy is lost at the point of impact. These are reasonable in some cases. This is an energy method, so it is intrinsically more robust than time based methods. In your second approach you have to make more assumptions so I'd expect more errors.





Cheers

Greg Locock


New here? Try reading these, they might help FAQ731-376
 
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wouldn't deflection and duration of impact be related to initial velocity (maybe under a constant (sinusoidal?) deceleration ?)

another day in paradise, or is paradise one day closer ?
 
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The impulse/momentum relationship is impulse, F*t, equals the change in momemtum, m(v2-v1). V2 is the final velocity of mass, m, and v1 is the initial velocity.
The impact force equalling 2*w, w being dropped object weight, comes from the special case of suddenly applied load of a weight dropped from zero height.
Since the duration of impact,t, is difficult to know, the energy approach seems the better one.

Ted
 
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"The second method requires knowledge of the impact duration, which is why I am not using it as my main method since I do not know the duration of the impact" Time lapse camera can help you in this case. Time lapse cameras were used by NJ DOT to determine impacts between vehicles and New Jersey barrier at least 30 years ago.
 
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The natural frequency for a second order mass-spring system is: ωn = √(k/m)
∴ ƒ = (1/2π)√(k/m)
∴ T = 2π√(m/k)
∴ timpact = (π/2)√(m/k)

This can then be input into the following equation for impact force: F = 2mv/t
Click to expand...
I haven't read the whole thread, but...
How did you decide ttimpact is 1/4 of T.
I would have imagined we approximate timpact as 1/2 of T. i.e. one half of a cycle (velocity goes from +Vmax thru zero to -Vmax)


=====================================
(2B)+(2B)' ?
 
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GrouchyStressedTensePoor,

Using kinetic energy and strain energy is good physics, and engineering. You start your calculations with good numbers. From the resulting strain energy, you get deflection of your cantilever beam. Since you know the average velocity, you can work out the time, and feed that back into your second equation, and see how that works out.

--
JHG
 
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This reference suggests that the difference is in the definition of average force. Time average vs. distance average.


What do you mean by average force?
The net external force on a constant mass object obeys Newton's second law, Fnet external=ma. The most straightforward way to approach the concept of average force is to multiply the constant mass times the average acceleration, and in that approach the average force is an average over time. Another approach that is valuable for assessing the impact force in high speed collisions is to use the impulse of force. When you strike a golf ball with a club, if you can measure the momentum of the golf ball and also measure the time of impact, you can divide the momentum change by the time to get the average force of impact. This average force is also an average over time.


There are, however, situations in which the distance traveled in a collision is readily measured while the time of the collision is not. In a case like a car crash with a tree, one can measure the distance of the collision and use the work-energy principle to evaluate the average of the force over the distance. Such an average over distance is not the same as the time average of force, but it is nevertheless useful in a case like evaluating the wisdom of wearing your seatbelt at all times in your automobile because without it, your stopping distance in a collision will be much shorter, and the average force on you much greater.

In the example illustrated above, a ball is brought to rest by impact with a spring. This provides an opportunity to evaluate the force average over time and the force average over distance. To make the comparison, the nature of simple harmonic motion is used, and the time to bring the ball to rest is one-fourth of a period of the periodic motion. The elastic motion would follow Hooke's law with a force constant k. The work-energy principle may also be applied, using the fact that in stopping the ball, the kinetic energy would be converted to spring potential energy. It is clear that the two approaches to an average force are not the same, although in this physical system they are proportional.

To provide a numerical example, consider a 1 kg mass traveling at 10 m/s that strikes a spring with spring constant k=10 N/m. This system would have a natural oscillation period of 1.99 seconds. So the time to stop the mass would be about a half second and the time average of force would be 20.1 Newtons. Using the work-energy principle to evaluate the distance average gives 15.8 Newtons.

Ted
 
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the frequency response solution is assuming a sinusodial force, and the energy solution a linear force, no?

another day in paradise, or is paradise one day closer ?
 
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Hi all,

Thanks for all the responses!

Greg - Is it not a second order system since it includes both inertia and spring force, so the characteristic equation includes displacement and the second derivative of displacement?

rb1957 - I admit I find it counter intuitive that the duration does not depend on initial velocity, but I believe it doesn't (like a pendulums time period is the same no matter what height its swing begins at). The deflection absolutely will be since the initial velocity implies a higher kinnetic energy, which means that the beam will have to deform further as it absorbs a greater amount of strain energy (and the force would also be higher, proportional to the deflection through the stiffness of the beam). The reason I do not believe the acceleration is constant is that the force will increase linearly with deflection due to Hooke's law, and so through F=ma the acceleration should also be linear.

Your second reply - I don't think the frequency response assumes a sinusoidal force, since I am assuming free vibrations with no externally applied force. In dynamic system/vibration terms F=0. The force I am talking about is the restoring spring force intrinsic to the system itself, in dynamic systems/vibration terms it would be kx(t). Hope that made sense, I think I didn't explain that very well.

Ted - The impact force being given for the suddenly applied case is something that made me a bit wary of the second method, as I have seen a lot of people claim that any impact force can only ever be twice the static force, which is clearly false. However mathematically I think the second method should still return the same answer as the first in the case I have used above?

Your second reply was very helpful thank you! I feel like that was the "Ohhh" moment I was hoping for! Did you find that information in a reference? I will try to find it by googling the text.

Chicopee - Unfortunately I do not see a way of setting up a test in a sufficiently controlled environment to film the impact, having said that it is my preferred option and I will continue looking into it. Fortunately for the moment this is just a first estimate, and hopefully worse case scenario.

Pete - The reason I have chosen ¼T is that this will be the time over which the original kinnetic energy of the mass will be fully transferred into strain energy (Vmax to V=0). From V=0 to -Vmax the strain energy in the beam will be transferred back into kinnetic energy of the mass as the beam unloads and pushes the mass away. To me this portion of the movement is not impact, more like a recovery phase.

drawoh -

If the average velocity is v = 3/2,
The deflection is, δ = F/k
δ = 30/5
δ = 6
∴ t = δ/v
t = 6/(3/2) = 12/3
t = 4
if F = 2mv/t
F = 2*20*(3/4)
F = 30N

Ok well that appears to have worked, thank you. I was sure I had tried that earlier and it didn't!
 
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maybe the small difference in force is because energy methods assume a linearly increasing force, a parabolic velocity, and the natural frequency approach assumes a sinusoidal velocity (due to sinusoidal displacement) ?

another day in paradise, or is paradise one day closer ?
 
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The first method is based on energy conservation principle. It also assumed that there is no energy loss during the impact, therefore the derived impact force is larger than the real value.
the second method is based on momentum conservative principle. The tough point is to estimate the impact time.

Personally, I prefer method 1.
 
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IDS said:
Surely the answer is that:

Quote:
This can then be input into the following equation for impact force: F = 2mv/t

is just wrong.
Click to expand...

I think that is the conclusion I came to also after reading through the replies.

Out of curiosity where did you get the equation:

IDS said:
F = pi.mv/2t = 2mv/t * pi/4
Click to expand...

Since that pi/4 is similar to the factor that the answers disagreed by initially.

GSTP

Graduate Mechanical Design Engineer
UK
 
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