GrouchyStressedTensePoor
Mechanical
- Aug 2, 2018
- 32
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
Fpeak = 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: 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
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.
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
Fpeak = 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: 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
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.