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force due to falling objects 2
- Thread starter FrustEng
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The equations you described are for Kinetic Energy and Potential Energy. Perhaps you can elaborate upon your paricular design problem. Is your falling object going to strike a beam or some other structural element? Lets give you an example:
Potential Energy from a suspended object will convert to Kinetic Energy when the object falls and then convert to Spring Energy when it strikes the beam.
Spring Energy=0.5kx^2
x=deflection
k=48EI/L^3 (Spring Constant for a simple beam with concentrated load at the center)
P=kx (Concentrated Load)
You can refer to Design Of Welded Structures by Omer Blodgett for more info.
Potential Energy from a suspended object will convert to Kinetic Energy when the object falls and then convert to Spring Energy when it strikes the beam.
Spring Energy=0.5kx^2
x=deflection
k=48EI/L^3 (Spring Constant for a simple beam with concentrated load at the center)
P=kx (Concentrated Load)
You can refer to Design Of Welded Structures by Omer Blodgett for more info.
whilst you can't disagree with the previous post (watch, someone will ! ... actually someone already has, einstein), anyways, i think the OP is asking about the impulse of the mas ?
Frusteng, over to you, 'cause the force is independent of the height. are you dropping something onto something else ? the force felt by the receipent will depend on the height (the velocity achieved) and the time it takes to stop it.
good luck
Frusteng, over to you, 'cause the force is independent of the height. are you dropping something onto something else ? the force felt by the receipent will depend on the height (the velocity achieved) and the time it takes to stop it.
good luck
motorcity. thats for an object that isn't falling. Impact forces resulting from falling objects is a different story.
Impact equations are not an easy thing to figure out. The defining factor is how long does it take the object to come to a complete rest once it hits the stationary object. There is a finite, however usually quite large deceleration of the object when it hits the ground. It will all depend on how much of a distance the object is going to decelerate. The longer the distance, the less the deceleration and thus the less the force that is a result.
In reality, it is going to be impossible to calculate the impulse force, without guessing at some point. There is surely some guide out there with something along the lines of multiplying times the weight (F=ma*3), but I don't know of it.
Impact equations are not an easy thing to figure out. The defining factor is how long does it take the object to come to a complete rest once it hits the stationary object. There is a finite, however usually quite large deceleration of the object when it hits the ground. It will all depend on how much of a distance the object is going to decelerate. The longer the distance, the less the deceleration and thus the less the force that is a result.
In reality, it is going to be impossible to calculate the impulse force, without guessing at some point. There is surely some guide out there with something along the lines of multiplying times the weight (F=ma*3), but I don't know of it.
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- #6
Well, I rememer that F=ma is the force needed to keep (m) moves at a rate of acceleration (a). This force obviously is independant of the height.
My problem is that I am required to account for the load applied on a beam due to the fall of the object,( 30kg from 5m). what is that load in Newton or kN or Pound...
Thanks
My problem is that I am required to account for the load applied on a beam due to the fall of the object,( 30kg from 5m). what is that load in Newton or kN or Pound...
Thanks
FrustEng,
You have defined your case further. You know the following:
a) I (moment of inertia)
b) E (youngs modulus)
c) L (beam span)
d) h (height of object above beam)
e) g (gravity)
f) m (mass of falling object)
Now calculate:
k (Spring Constant of your beam) = 48EI/L^3
Now calculate:
x (deflection of beam) = (2mgh/k)^0.5
Now calculate:
P (Load due to falling object hitting beam) = kx
There you go.
You have defined your case further. You know the following:
a) I (moment of inertia)
b) E (youngs modulus)
c) L (beam span)
d) h (height of object above beam)
e) g (gravity)
f) m (mass of falling object)
Now calculate:
k (Spring Constant of your beam) = 48EI/L^3
Now calculate:
x (deflection of beam) = (2mgh/k)^0.5
Now calculate:
P (Load due to falling object hitting beam) = kx
There you go.
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- #8
EddyC,
Sounds too complicated especially that I don't know the size of the beam. A trial and error is required I guess. Any practical conservative solution?
This is a temporary overhead canopy for the enterance of a building where some constuction is being done. The structure is wood.
Thanks
Sounds too complicated especially that I don't know the size of the beam. A trial and error is required I guess. Any practical conservative solution?
This is a temporary overhead canopy for the enterance of a building where some constuction is being done. The structure is wood.
Thanks
"Sounds too complicated especially that I don't know the size of the beam. A trial and error is required I guess. Any practical conservative solution?"
I don't think a simple conservative solution exists. I think you have three choices:
1. Use your judgement to determine that you shouldn't worry about 30kg from 5m. I hate monkeymeters, but I think that's 66 lbf falling from about 15 ft. I don't know enough about your roof system to know if that's a lot. Intuitively, I think I'd be worried about it landing between two beams and crashing through the deck a lot more than the effect of landing on a beam.
2. Tell them they can't do it.
3. Calculate the force imposed on the beam from this load as suggested earlier. Break out that physics book. This calculation is not difficult. Calculate the kinetic energy of the object when it contacts the beam. The kinetic energy is transferred to spring potential energy, allowing you to calculate how far the beam deflects. From this deflection, you should be able to get the beam internal forces.
DBD
I don't think a simple conservative solution exists. I think you have three choices:
1. Use your judgement to determine that you shouldn't worry about 30kg from 5m. I hate monkeymeters, but I think that's 66 lbf falling from about 15 ft. I don't know enough about your roof system to know if that's a lot. Intuitively, I think I'd be worried about it landing between two beams and crashing through the deck a lot more than the effect of landing on a beam.
2. Tell them they can't do it.
3. Calculate the force imposed on the beam from this load as suggested earlier. Break out that physics book. This calculation is not difficult. Calculate the kinetic energy of the object when it contacts the beam. The kinetic energy is transferred to spring potential energy, allowing you to calculate how far the beam deflects. From this deflection, you should be able to get the beam internal forces.
DBD
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Hi Frusteng
Whilst I agree with the majority of the previous posts here is my ten pennies worth.
Firstly a collision which involves a fixed surface falls outside the theory of "conservation of linear momentum" as
the fixed surface would be an external impulse.
So in the case of a moving and fixed object collision Newtons law of Restitution should be used.
This however would be difficult as stated previously as the time of contact between the two objects would be difficult
if not impossible to determine, as would whether there was
any bounce of the moving object from the fixed surface.
However if we assume that there is no elasticity in the collision (no bounce of object from fixed surface)and therefore the coefficient of restitution is zero then we can
say that :-
Impulse (J)= final momentum - initial momentum
The final momentum will be zero as the coefficient of restitution is zero however the initial momentum is that of the falling object which can be calculated as follows:-
conservation of energy:- mgh= 0.5*m*V^2
30*9.81*5=0.5*30*V^2
V= 9.9 m/s
therefore Impulse= final momentum - initial momentum
= 5*9.9-0 = 49.52Ns (impulse force)
regards desertfox
Whilst I agree with the majority of the previous posts here is my ten pennies worth.
Firstly a collision which involves a fixed surface falls outside the theory of "conservation of linear momentum" as
the fixed surface would be an external impulse.
So in the case of a moving and fixed object collision Newtons law of Restitution should be used.
This however would be difficult as stated previously as the time of contact between the two objects would be difficult
if not impossible to determine, as would whether there was
any bounce of the moving object from the fixed surface.
However if we assume that there is no elasticity in the collision (no bounce of object from fixed surface)and therefore the coefficient of restitution is zero then we can
say that :-
Impulse (J)= final momentum - initial momentum
The final momentum will be zero as the coefficient of restitution is zero however the initial momentum is that of the falling object which can be calculated as follows:-
conservation of energy:- mgh= 0.5*m*V^2
30*9.81*5=0.5*30*V^2
V= 9.9 m/s
therefore Impulse= final momentum - initial momentum
= 5*9.9-0 = 49.52Ns (impulse force)
regards desertfox
GregLocock
Automotive
That's probably fine as far as it goes, but that is not a conservative solution.
Cheers
Greg Locock
Please see FAQ731-376 for tips on how to make the best use of Eng-Tips.
Cheers
Greg Locock
Please see FAQ731-376 for tips on how to make the best use of Eng-Tips.
FrustEng,
I stand fully behind my method. And you are right, it does require trial & error iteration. You have to first pick the beam size in order to get the Moment Of Inertia. You will find that there is no code-based "handbook" solution for this kind of problem. Its all based upon physics & dynamics.
I stand fully behind my method. And you are right, it does require trial & error iteration. You have to first pick the beam size in order to get the Moment Of Inertia. You will find that there is no code-based "handbook" solution for this kind of problem. Its all based upon physics & dynamics.
"I stand fully behind my method. And you are right, it does require trial & error iteration. You have to first pick the beam size in order to get the Moment Of Inertia. You will find that there is no code-based "handbook" solution for this kind of problem. Its all based upon physics & dynamics."
His idea requires trial and error but this is not so bad. For the forces we're talking about, my bet is that if it'll work for any decent LL, it'll work for this case. You might do an interation or two, but come on, like you've never had to iterate before!
And about the "Newtons law of Restitution" post earlier. The guy's having trouble with kinetic energy of a falling object and potential energy of a spring. You might as well have told him to consider relativistic effects. LOL.
DBD
His idea requires trial and error but this is not so bad. For the forces we're talking about, my bet is that if it'll work for any decent LL, it'll work for this case. You might do an interation or two, but come on, like you've never had to iterate before!
And about the "Newtons law of Restitution" post earlier. The guy's having trouble with kinetic energy of a falling object and potential energy of a spring. You might as well have told him to consider relativistic effects. LOL.
DBD
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EddyC,
Given E=200000N/mm^2, I=37e6mm^4, L=3000mm, h=5000, g=9.81, and m=30kg,
k=13155 N/mm
x=14.9 mm, and
P=196766N or 197kN
Means the force of falling 30kg object is equavalent to 197kN standing still on the baem? or I am having units probem?
Desertfox,
your last line of calc. yielded 49.52 Ns Impulse force. How that translates into an equavelent point force, because the 30kg object, if standing still on the beam, would mean a 30*9.81=300 N force.
Sorry, My physics are really not that good.
Given E=200000N/mm^2, I=37e6mm^4, L=3000mm, h=5000, g=9.81, and m=30kg,
k=13155 N/mm
x=14.9 mm, and
P=196766N or 197kN
Means the force of falling 30kg object is equavalent to 197kN standing still on the baem? or I am having units probem?
Desertfox,
your last line of calc. yielded 49.52 Ns Impulse force. How that translates into an equavelent point force, because the 30kg object, if standing still on the beam, would mean a 30*9.81=300 N force.
Sorry, My physics are really not that good.
Hi FrustEng
The last line of my calculation should of read:-
30kg * 9.9 m/s= 297Ns not 49.52Ns.
The point load you are looking for would depend on how long the impact lasted ie:- if the impact lasted 0.5 seconds the
average force would be 297Ns/0.5s =594N.
Note there is no need to introduce the gravity constant
which you appear to have done in your last 2 posts.
regards desertfox
The last line of my calculation should of read:-
30kg * 9.9 m/s= 297Ns not 49.52Ns.
The point load you are looking for would depend on how long the impact lasted ie:- if the impact lasted 0.5 seconds the
average force would be 297Ns/0.5s =594N.
Note there is no need to introduce the gravity constant
which you appear to have done in your last 2 posts.
regards desertfox
essentially this is an impact loading problem, therefore all you have to do is calculate the static deflection due to your 30kg object as delta=(pl^3/48EI).Next you amplify the static deflection to account for the distance it falls delta x((1+(1+2(h/delta))^.5. Then max load is 30kg x 9.81 x amplified static deflection.
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