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force due to falling objects
2

force due to falling objects

force due to falling objects

(OP)
My physics information are not that fresh so I need some help please.
I’d like to know how to calculate the force of a falling 30 kg object from 5m high.
Is it 1/2*m*v^2 where v^2=g.h


Thanks

RE: force due to falling objects

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.

RE: force due to falling objects

Some guy named Newton suggested F=ma

RE: force due to falling objects

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

RE: force due to falling objects

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.

RE: force due to falling objects

(OP)
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

RE: force due to falling objects

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.

RE: force due to falling objects

(OP)
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

RE: force due to falling objects

"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

RE: force due to falling objects

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

RE: force due to falling objects

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.

RE: force due to falling objects

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.

RE: force due to falling objects

"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

RE: force due to falling objects

(OP)
Desertfox,

I think you made a small mistake in the last line of your calculation. Did you mean (Impulse,Ns,)=m.s=30*9.81*9.9=2913Ns?

RE: force due to falling objects

Hi FrustEng

The last line is correct there is no need to introduce the gravity constant.
The impulse force is based on momentum and its units are
Newton-seconds.

 kg= Ns^2/(m)

so mass * velocity = (Ns^2/(m))* (m/s) = Ns

Regards Desertfox

RE: force due to falling objects

Hi DBD

If the guys having trouble with the concepts you are talking about how is he going to calculate the deflection and stiffness of the beam and then iterate,

regards desertfox

RE: force due to falling objects

(OP)
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.

RE: force due to falling objects

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

RE: force due to falling objects

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.

RE: force due to falling objects

FrustEng,

I get the same numbers that you do.

RE: force due to falling objects

I find the equation can be made simpler by equating the energies.

The energy of the falling object as it strikes the beam (at the centre to be conservative)is 'mgh'.

Neglecting the losses due to impact (again to be conservative), this energy is stored as the strain energy in the beam while the beam deflects. The strain energy of the beam is given by     (1/2EI) 0L?(Mx)2 dx

If P is the dynamic force causing the bending of the beam, the strain energy can be written in terms of P and the same can be equated to the potential energy of the falling mass. Solving for P the force can be obtained.

As Desertfox has stated in his post, we assume no bouncing during the clollision. Further, this impact will also make the beam vibrate and dissipate the additional energy before coming to rest in static equillibrium.

Trilinga

RE: force due to falling objects

FrustEng,

I dug a little further on this subject. The book Design Of Welded Structures by Omer Blodgett covers exactly the problem that you descibed. He uses the following formula:

U (Allowable Energy Storage Of Beam) = [(L)(I)((Fy)^2)]/[(6E)(c^2)]

You can set U=mgh (Potential Energy)

RE: force due to falling objects

Frusteng,

going back to your post of the 23rd ... g = 9.81 m/sec2 ... but your other dimensions are in mm ... did you remember the conversion ?

btw, i don't like the expression "impulse force" ... there's something called "impulse" and there's something called "force", and forces have units of "N".

going back over some earlier posts, i agree with DBDavis, back on the 22nd; isn't the critical question how the cover (between the studs) is going to react ?  this'll deflect a lot more than the stud and absorb the energy this way.  you've dealing with wood ... maybe structures/civil guys are more comfortable with calculating stresses, but i don't know ... maybe if you've got good material properties.  i'd prefer to answer the question along judgement lines.

going back over the analysis responses, we're assuming that the mass hits on one stud which reacts the load in isolation.  how much load is distributed through the canopy to other beams ?  if not much, 'cause they're too far apart or the canopy is too flexible, then you've got a whole different problem !  if the studs are closely pitched or the canopy is stiff (again judgement ... "close" or "stiff") then the load will be distributed through more than one stud, which isn't a problem if you are comfortable being conservative. ... but then if the (one stud) structure is insufficient, are you going to say it's no good, when it might be ?

as an analysis question, what you're doing is transferring KE from the mass to strain energy in the structure (be it a stud or a membrane/plate cover.

btw, could you test the set-up (off-shift?) ?  and why 30kgs and 5m ? regulation requirements ?

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