Calc force from mass striking an object

[trainguy]

Hi all.

This one's simple for you dynamic types.

I have a mass, and its distance to a fixed object. This mass is subjected to an acceleration that varies in time, from 0 to a peak back to 0, triangular pulse.

How do I calculate the force exerted on the object?

I do not have any dynamics texts here at home.

Any help would be appreciated, this is for evaluation of a SS bracket on a train, not a school assignment...

Thanks in advance.

tg

 

[IRstuff]

Your question and post title are inconsistent with each other. Do you want the force on the moving object while it's moving? Or do you want the impact force?

In the latter case, there is no closed-form solution, since the force is a function of how long or how far it takes to complete the collision.

TTFN

 

[rmetzger]

from what I gather the triangular acceleration curve is the result of a mass striking the bracket and imparting motion to it. The up side of the curve as the component absorbs the energy of the impact and accelerates - and the down side of the curve once the object has absorbed most of the impact and is near its max speed. You are looking for the impact force that caused the acceleration curve? Is this correct?

 

[ivymike]

do you know the width of the base of the triangle (duration of event)?

 

[TheTick]

Impact force is not a problem readily solved by classic dynamics. Such problems are best solved from an energy-transfer and momentum-transfer perspective.

Much depends on the shape and material properties of the objects. Questions that can't be answered by kinematic equations:[ul][li]How much energy is absorbed in deformation?[/li][li]What is the peak instantaneous force and/or acceleration?[/li][/ul]

"Customer satisfaction, while theoretically possible, is neither guaranteed nor statistically likely.--E.L. Kersten

http://www.EsoxRepublic.com

 

[IRstuff]

If the acceleration curve is given, then the force is essentially f=m*a.

TTFN

 

[barbless]

Trainguy,

It sounds like you might be describing a pile driver. If so then you are looking for an "average force of blow" Which =(W*S)/d where W is the weight of the pile driver; S is (distance that the weight falls plus the distance the pile is driven) and d is the distance the pile is driven. The impact force is not accurately known, thus "average force of blow" is the solvable term. This, of course neglects losses due to dissipated heat and strain energy.

 

[MintJulep]

Barbless,

Your equation results in units of mass per time squared. Doesn't seem to have any physically possible interpretation.

Trainguy,

Not nearly enough information to solve the problem.

A mass subjected to a time-variant acceleration implies a time-variant force as an external input, as noted by IRstuff.

Differentiate the acceleration vs time curve twice wrt time to arrive at position vs time. Does the mass close the distance to the fixed object before or after the input force drops back to zero?

In either case, the force on the fixed object is not constant wrt time. It is depentant upon the elastiscity of the mass and fixed object. How long does it take to absorb the energy? This will define the deceleration rate of the collision. If the input pulse falls to zero before the start of the collision, then the average force on the object is simply F=ma. If the input pulse is still driving durring the collision, you need to add that force in as well.

As noted by The Tic, it seems as if this problem could be better solved by looking at energy.

 

[zekeman]

I believe you are talking abount a collision between the mass that has an initial velocity, v ,and a final velocity of zero after impact.And you are assuming a tranglar pulse of force that stops the mass.
The only thing that is certain is that the impulse (i.e., the area under the force time curve is equal to the change of momentum by Newton's second law, viz.
mv=area under ma-time curve= impulse
Dividing by m the both sides of the equation , the equation becomes
v=area under the acc-time curve. If the assumption that the pulse is triangular and you call the peak acc A,and the impulse time,T, then the area under the Acc-time curve is
AT/2=v.
The impulse time, T, is a function of the wall/mass material and is not easily determined.

 

[barbless]

Mintjulip,

I admit that my pile driver analogy doesn't exactly describe his problem as stated but as for not having any "physically possible interpretation", maybe you should have directed that criticism toward Machinery Handbook so that they can remove their discussion and formulas dealing with pile drivers from their book.

 

[FredGarvin]

Barbless' equation units work out to be a measure of force since the other two items (S/D) cancel each other out because they are both units of distance.

I am wondering where in Machinery's Handbook they discuss pile drivers...I thought I knew that book pretty well. There are so many things in it it's easy to miss something.

 

[barbless]

FredGarvin,

Re. Machinerys Handbook.
The discussion of pile drivers can be found in the MECHANICS section under the subtitle "Force of a Blow".

 

[trainguy]

Thank you all for your input. This is beyond my capacities at the moment, so we have retained the services of a consultant who will be able to help us.

By the way, I was talking about a railway passenger impacting a fixed table on a railway car during a crash. The objective is to determine whether or not a bolted table bracket will fail or not for a certain crash pulse.

Cheers, and happy holidays to all.

PS:

Fred Garvin,
Are you THE Fred Garvin, from SNL? Welcome back... If not, please disregard this.

tg

 

[thorngulch]

trainguy:

You stated that the acceleration was a triangular pulse, so the force would be, for no energy absorbed by anything, a similar triangular pulse. The maximum acceleration would provide a maximum force. The integral of the either would be equivalent to the kinetic energy of course. Now, the problem is whether or not there was energy conversion due to deformation etc. As you seem to be interested in a shear problem, then solve for the load as you don't need an exact answer but one with a sufficient safety factor. If it deals with litigation, and you need to know if you designed with a sufficient safety factor, you will need to do real world testing to demonstrate safe design. As there would be energy absorbed by materials in the impact, the ideal values would yield a higher value than in the real world.