biz
Guest
looks like 419,8 kn.
calculates the "spinta" exercised by a wild boar falling - resumption of physics 2 (vengement)
Legs
Guest
I wanted, by scruple, to perform the same analysis by making direct use of the response to the impulse always in the hypothesis of mechanical linearity (of which it is necessary to verify the validity).
the solution can be obtained thus: u(t) = i * h(t)
u displacement, impulse and response to the unitary impulse.
i = variation amount of motion which is mcing*vcing = 100x6,26-0 = 626 kg*m/s
setting to zero the damping hmax = 1 / (m * omega)
omega = square root(k/m) = square root(4,717e7/105) = 670,25 rad/s
umax = 626 / (105*670,25) = 0,0089m = 8.9mm
this value is to be considered the maximum oscillation around the value of the static case.
ustat = (100*9,8 / 4,717e7)*1000 = 0.02mm
the numerical solution therefore coincides with that of @f_ingrasciotta (the mass of the beam affects little. therefore consider it or not leads to significant differences).
Obviously in reality this solution is not acceptable.
during fall tests with mountain equipment (with ropes ripping around 12kn) reached 100kg mass amplifications thrown into the vacuum (on the ground, then with air friction) at values between 8 and 9 (measure made with dynmeter); Note that the minimum strength of 9kn for a single operator of 100kg is used in lifeline checks.
if you use the solution for the case to the heaviside (i.e. the force remains applied) you get to a factor 2. Therefore, in reality, I expect a solution between 2 and 9, certainly not 400.
in fact the hypothesis of linearity is not satisfied. the plasticization of the beam must be considered.
with a beam of so modest stiffness and resistance you can not expect to remain in the elastic field. but of course I understand that somewhere you have to leave.
only move to the case of non-linearity mechanics or use a beam with characteristics such as to remain in the linear field.
the solution can be obtained thus: u(t) = i * h(t)
u displacement, impulse and response to the unitary impulse.
i = variation amount of motion which is mcing*vcing = 100x6,26-0 = 626 kg*m/s
setting to zero the damping hmax = 1 / (m * omega)
omega = square root(k/m) = square root(4,717e7/105) = 670,25 rad/s
umax = 626 / (105*670,25) = 0,0089m = 8.9mm
this value is to be considered the maximum oscillation around the value of the static case.
ustat = (100*9,8 / 4,717e7)*1000 = 0.02mm
the numerical solution therefore coincides with that of @f_ingrasciotta (the mass of the beam affects little. therefore consider it or not leads to significant differences).
Obviously in reality this solution is not acceptable.
during fall tests with mountain equipment (with ropes ripping around 12kn) reached 100kg mass amplifications thrown into the vacuum (on the ground, then with air friction) at values between 8 and 9 (measure made with dynmeter); Note that the minimum strength of 9kn for a single operator of 100kg is used in lifeline checks.
if you use the solution for the case to the heaviside (i.e. the force remains applied) you get to a factor 2. Therefore, in reality, I expect a solution between 2 and 9, certainly not 400.
in fact the hypothesis of linearity is not satisfied. the plasticization of the beam must be considered.
with a beam of so modest stiffness and resistance you can not expect to remain in the elastic field. but of course I understand that somewhere you have to leave.
only move to the case of non-linearity mechanics or use a beam with characteristics such as to remain in the linear field.
F_Ingrasciotta
Guest
I said a good shit. with the value of 419 kn of load you would have a moment in the zone of half-carry pairs to 419kn*l/2=209 knm (with l=500mm). this value leads to "spatial" tensions so you do not remain in the elastic/linear field.
@legs Can you post a couple of images of the simulation you did? I am very curious to see how everything was set [il mondo FEA mi affascina sempre molto]
mamobono
Guest
In fact, in the eye, the forces found are too high to be true.
the equation of the conservation of kinetic energy cannot be used because there will be its dispersion in the shock, as it says here
www.youmath.it
So it is possible to have an idea of the forces in play only experimentally?
I follow the discussion ..
Hi.
the equation of the conservation of kinetic energy cannot be used because there will be its dispersion in the shock, as it says here
Urto anelastico
Cos'è un urto anelastico? Spiegazione con definizione, formule e leggi di conservazione degli urti anelastici, con esempi ed esercizi svolti.
www.youmath.it
I follow the discussion ..
Hi.
Legs
Guest
@f_ingrasciottaTo tell the truth, I've done two accounts by hand.
in this document (but in any dynamic book): find the formulation used.
I simply reset the damping coefficient and took the maximum value of the response to the unit pulse.
in this document (but in any dynamic book): find the formulation used.
I simply reset the damping coefficient and took the maximum value of the response to the unit pulse.
F_Ingrasciotta
Guest
the energy is not preserved, as you have rightly observed, but the amount of motion is (if you consider an anaelastic shock where after the impact the two bodies remain united).In fact, in the eye, the forces found are too high to be true.
the equation of the conservation of kinetic energy cannot be used because there will be its dispersion in the shock, as it says hereSo it is possible to have an idea of the forces in play only experimentally?Urto anelastico
Cos'è un urto anelastico? Spiegazione con definizione, formule e leggi di conservazione degli urti anelastici, con esempi ed esercizi svolti.
I follow the discussion ..
Hi.
Okay, I don't know why but I imagined a fem simulation.
biz
Guest
I'm coming.
I had omitted the term g in writing the values:
_It is 6.4 mm with my data
_It is 9.2 mm with data @f_ingrasciottatherefore there is no serious error (3.4%), but the calculation is more accurate @f_ingrasciotta
incorrect courier.
I had omitted the term g in writing the values:
_It is 6.4 mm with my data
_It is 9.2 mm with data @f_ingrasciottatherefore there is no serious error (3.4%), but the calculation is more accurate @f_ingrasciotta
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biz
Guest
energy is also converted into thermal energy and plastic deformation.
Legs
Guest
as rightly points out News part of the energy is dissipated in the plastic deformations of the beam and part in heat.
plastic deformations are concentrated in plastic hinges.
but in an isostatic beam the attainment of the force that leads to the plasticization of the beam also leads to its collapse.
It follows that this particular problem directly reaches the collapse without allowing any type of plastic dissipation.
therefore it is only necessary to foresee two solutions: change the constraints in ways to have a hyperstatic beam or increase stiffness and resistance so as to maintain the phenomenon in linear elastic field.
@f_ingrasciotta the energy budget is still applicable also in the non-linear field. it is enough to take into account the elastic and plastic quota (I pass the thermal quota for simplicity).
Then, wanting to simplify even more we could consider rigid-plastic behaviour (elastic deformations are usually negligible compared to plastics) and calculate all the energy dissipated as the simple product of the force for moving. where the force is the one that produces the plasticization of the beam. As already mentioned, however, it is not acceptable that the beam becomes labile. That is why we must increase the constraints.
In practice, I will write that the energy supplied to the system (potential or kinetic) must be equal to the energy dissipated.
plastic deformations are concentrated in plastic hinges.
but in an isostatic beam the attainment of the force that leads to the plasticization of the beam also leads to its collapse.
It follows that this particular problem directly reaches the collapse without allowing any type of plastic dissipation.
therefore it is only necessary to foresee two solutions: change the constraints in ways to have a hyperstatic beam or increase stiffness and resistance so as to maintain the phenomenon in linear elastic field.
@f_ingrasciotta the energy budget is still applicable also in the non-linear field. it is enough to take into account the elastic and plastic quota (I pass the thermal quota for simplicity).
Then, wanting to simplify even more we could consider rigid-plastic behaviour (elastic deformations are usually negligible compared to plastics) and calculate all the energy dissipated as the simple product of the force for moving. where the force is the one that produces the plasticization of the beam. As already mentioned, however, it is not acceptable that the beam becomes labile. That is why we must increase the constraints.
In practice, I will write that the energy supplied to the system (potential or kinetic) must be equal to the energy dissipated.