I have a large skid platform weighing about 80,000 pounds which, when unloaded from a truck, may have one end drop about 2 feet onto a concrete surface. The final velocity is 11.3 fps. How do I calculate the reaction force when it hits? The skid is on sttel runners 12 inches wide on each side.
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Reaction Force of a Dropped Object 3
- Thread starter RonMB
- Start date
HamishMcTavish
Nuclear
You know the velocity so just work out the momentum and say in the collision that momentum is conserved. You should be able to get your reaction force then work out how the material responds.
RichLeimbach
Mechanical
A rough way to do it is this: I would mathematically model the skid and concrete as compression springs using the info you can find in most any strengths of materials book. Then calculate approximate deflection and reaction force using your known kinetic energy. This should get you in the ballpark.
Approximate:
Sum of F = ma
ads = vdv => v^2 = u^2 + 2as; u =0
a = v^2/2s => a = 11.3^2/2(2) => a = 31.92 ft/s^2
Sum of F = ma
=> Reaction = m*31.92 Aproximatelly
I can't recall English unit; Substitute m as the mass of that thingo to find the approximate reaction.
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- #5
vtl,
Your calculation appears to derive the force acting on the 'thingo' during its 2 foot fall, not the impact load when it strikes the concrete.
The derived value of 31.92 is basically a back-calculation of the the acceleration due to gravity (which RonMB presumably took as 32 ft/sec/sec), and m*31.92 is merely the self-weight of the thingo.
Your calculation appears to derive the force acting on the 'thingo' during its 2 foot fall, not the impact load when it strikes the concrete.
The derived value of 31.92 is basically a back-calculation of the the acceleration due to gravity (which RonMB presumably took as 32 ft/sec/sec), and m*31.92 is merely the self-weight of the thingo.
Opsss... Did I missed reading that the final velocity was 11.3 fps not 0 fps.
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Well if you know the delta time when and after the impact then you can use the impulse equation, then multiply by delta time to get the answer.
Else you can do it this way
Sum (F) = ma
a = deceleration
v = u + at
t = approximately the time between skid or between contacting the surface then comes to a halt. You have to know this value.
a = v/t
Sum (F) = ma
=> Reaction = ma (approximately)
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Well if you know the delta time when and after the impact then you can use the impulse equation, then multiply by delta time to get the answer.
Else you can do it this way
Sum (F) = ma
a = deceleration
v = u + at
t = approximately the time between skid or between contacting the surface then comes to a halt. You have to know this value.
a = v/t
Sum (F) = ma
=> Reaction = ma (approximately)
RonMB,
Put it this way, if everything is perfectly elastic, the skip should bounce back up to it's original height. The crux of the problem is therefore concerned with the amount of energy absorbed at impact. This will be even effected by the load in the skip, if it was sand for example (ever dropped a bag of sand on the ground?). As mentioned above the max impact force will be related to the hardness of the concrete but also the stiffness of the skip. The softer both materials are, the greater the compression distance at impact and so the lower the max loads are. The area under the load deflection cure for the initial compression will equal the kinetic energy at impact.
Hope this helps.
Speedy
"Tell a man there are 300 billion stars in the universe and he'll believe you. Tell him a bench has wet paint on it and he'll have to touch to be sure."
Put it this way, if everything is perfectly elastic, the skip should bounce back up to it's original height. The crux of the problem is therefore concerned with the amount of energy absorbed at impact. This will be even effected by the load in the skip, if it was sand for example (ever dropped a bag of sand on the ground?). As mentioned above the max impact force will be related to the hardness of the concrete but also the stiffness of the skip. The softer both materials are, the greater the compression distance at impact and so the lower the max loads are. The area under the load deflection cure for the initial compression will equal the kinetic energy at impact.
Hope this helps.
Speedy
"Tell a man there are 300 billion stars in the universe and he'll believe you. Tell him a bench has wet paint on it and he'll have to touch to be sure."
...in other words, nobody knows.
Speedy is correct; this depends upon the reaction of the different parts of your mechanical system. This is an impulse problem:
impulse force = (final momentum)/(duration of impact)
That is simple to state, but difficult to solve. The difficult part is to determine the duration of impact. It is a (potentially complex) function of the interactions of load-skid and skid-ground.
How is the load distributed on the skid? Sounds like one end of your skid is already resting on the ground; so the final velocity is greatest at the far end, and the impact is therefore related to the load distribution (i.e. how far is the load C.G. moving?).
What are the contact points/surfaces of the skid-ground?
The mass/elastic characteristics of the load (e.g. a rubber tired load will be different than a solid block of metal)
Etc.
There ought to be a "reasonable" approximation for the impact duration so as to establish some maximum loads, but I can't rationalize any (delta-time) values in my own mind.
It also seems like this kind of "problem" is encountered a lot in: the military, the construction industry, transportation,... At the very least, try posting your question on some of the other forums of this site.
Speedy is correct; this depends upon the reaction of the different parts of your mechanical system. This is an impulse problem:
impulse force = (final momentum)/(duration of impact)
That is simple to state, but difficult to solve. The difficult part is to determine the duration of impact. It is a (potentially complex) function of the interactions of load-skid and skid-ground.
How is the load distributed on the skid? Sounds like one end of your skid is already resting on the ground; so the final velocity is greatest at the far end, and the impact is therefore related to the load distribution (i.e. how far is the load C.G. moving?).
What are the contact points/surfaces of the skid-ground?
The mass/elastic characteristics of the load (e.g. a rubber tired load will be different than a solid block of metal)
Etc.
There ought to be a "reasonable" approximation for the impact duration so as to establish some maximum loads, but I can't rationalize any (delta-time) values in my own mind.
It also seems like this kind of "problem" is encountered a lot in: the military, the construction industry, transportation,... At the very least, try posting your question on some of the other forums of this site.
In most cases of impact loading, it is virtually impossible to predict loads, deflections, or stresses, because the distribution of stress and strain is not the same as it is in static loading. That said, the following approximation works pretty well for a relatively heavy body at relatively low velocity, which appears to be your case.
First, calculate the elastic deflection in your structure at rest after impact. The impact load will cause deflection greater than that static deflection by a factor of approximately 1 + (1 + 2h / d)^.5,
where h is the drop height, and d is the deformation which would be present if the load were gradually applied. This should allow you to back-calculate the approximate impact load and reaction.
Use a generous Safety Factor, and consider that it is unlikely that the load will be evenly divided between skids at impact.
This approximation is for a dropped object, while your problem involves rotation to impact, but it should serve to give you a feel for the forces involved. Calculate the free-fall drop height which results in the same terminal velocity and use that as h in the foregoing equation.
You might find helpful information in test requirements for shipping containers. I do not have any such specs at hand, but I have used them in the past, and they deal specifically with objects rotating to impact. I do not recall whether they offer a method of calculating loads at impact.
HTH
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- #10
poetix99 correctly identifies this as a complex problem, one that few people really understand because most people don't require it for their jobs. I have limited knowledge, but I know that the governing equations for impact require using a factor known as the restitution coefficient, which relates how much energy is lost (inelastic collision) during impact. A small description is available at:
but there is more to it than this. An article in the September 2001 issue of Advanced Materials and Processes titled "Materials in Sports: Designing for Ball Impacts" related restitution coefficient to deflection, stress, etc. for impacts. An Internet search at Google would be a place to start.
but there is more to it than this. An article in the September 2001 issue of Advanced Materials and Processes titled "Materials in Sports: Designing for Ball Impacts" related restitution coefficient to deflection, stress, etc. for impacts. An Internet search at Google would be a place to start.
I agree with the other posts that this problem is simple in principle, but difficult to solve in practice. The main reason is that it will be very difficult for you to obtain two things that you need: the coefficient of restitution and the time the skid is initially in contact with the floor.
The coefficient of restitution (e) is a measure of how elastic (e=1) or inelastic (e=0) a collision is. This number will help you calculate how the skid will bounce when it hits the floor. You already have the velocity of the skid just before it hits the ground. Using the coefficient of restitution, you can calculate the velocity as it bounces upward. You need this velocity to properly calculate the change in momentum during the collision, which you can then use to calculate the impulse of the force.
Once you have the impulse of the force, you need to know how long the skid is in contact with the floor. This can be done with modeling, as pointed out above. That won't be easy, though, because then you'll need even more parameters that you probably don't have. You could estimate how long the contact period is, and then divide that time into the impulse to get the force; however, this will be a time average force. For simplicity, you may want to assume the force is triangular to give you a better estimation of the maximum force.
Calculating an accurate force-time history is nearly impossible. I've seen experimental accelerometer data of collisions and I don't see any way that you could accurately repeat those traces using a theoretical/numerical model, unless you turn it into a PhD thesis (and do testing to get needed parameters).
Haf
The coefficient of restitution (e) is a measure of how elastic (e=1) or inelastic (e=0) a collision is. This number will help you calculate how the skid will bounce when it hits the floor. You already have the velocity of the skid just before it hits the ground. Using the coefficient of restitution, you can calculate the velocity as it bounces upward. You need this velocity to properly calculate the change in momentum during the collision, which you can then use to calculate the impulse of the force.
Once you have the impulse of the force, you need to know how long the skid is in contact with the floor. This can be done with modeling, as pointed out above. That won't be easy, though, because then you'll need even more parameters that you probably don't have. You could estimate how long the contact period is, and then divide that time into the impulse to get the force; however, this will be a time average force. For simplicity, you may want to assume the force is triangular to give you a better estimation of the maximum force.
Calculating an accurate force-time history is nearly impossible. I've seen experimental accelerometer data of collisions and I don't see any way that you could accurately repeat those traces using a theoretical/numerical model, unless you turn it into a PhD thesis (and do testing to get needed parameters).
Haf
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- #13
haf is spot on about the number of variables you would need to acquire in order to good numbers on this. the first thing that comes to mind for me is the concentration of the force at impact i.e - will the skid hit perfectly flat (the force is spread out over many square inches) or will the load cause it to initially hit on a corner of the skid (the load is concentrated- and the concrete becomes gravel). i would seriously consider massey's post as the prudent solution.
You have to calculate the impact coeficient for this case, for this you need to know, If I remember correctly the strain of the two objects in contact and at rest, that means, how much deflection you have when you put the slab slowly on the floor, then the force is the resting force times the impact coefficient.
If you lower a big object until it is at zero distance from the floor, but the load on the floor is still zero, and you suddenly let it drop, that means letting all the weight on the floor and not held by a crane for example, the IC is 2, that means that if the object weight 1000 pounds, the impact will be equivalent to put a 2000 pounds slowly, so, at the very least, you are way adobe a IC of 2, which for a 80K pound slab is quite a lot.
An impact is defined as any aplication of a load that lasts less that half the natural period of any of the objects.
If you lower a big object until it is at zero distance from the floor, but the load on the floor is still zero, and you suddenly let it drop, that means letting all the weight on the floor and not held by a crane for example, the IC is 2, that means that if the object weight 1000 pounds, the impact will be equivalent to put a 2000 pounds slowly, so, at the very least, you are way adobe a IC of 2, which for a 80K pound slab is quite a lot.
An impact is defined as any aplication of a load that lasts less that half the natural period of any of the objects.
There's the problem:
velocity is known-I forget what it is and it doesn't matter.
mass is known-I forget what it is and it doesn't matter.
what does matter is that these values are multiplied and probably = a fairly large momentum.
what matters even more is that time is very, very, very small (because both objects are very rigid) which makes the force very, very, very large.
Doesn't sound too techy and it's 2 AM and I'm too lazy to look up the values above for an example - but the smaller value for time = that much larger a force.
You know questions like this arise from people who want to unload something like this, and an engineer might say "hey, hold on a minute, you can't just drop that thing off the trailor".
Then the boss says "why not we're only dropping it 2 feet, what's the problem?
To which the engineer replies "well there will be a tremoundous amount of force generated when that skid impacts the ground, you may damage the equipment or the hard deck".
To which the boss replies "is that right? well how much force is going to be generated?"
To which the engineer replies "I can tell you in a couple of minutes". Then a couple of minutes later the engineer remembers that in college all those assumptions we used to make on a regular basis don't really apply here. And he begins to say things that cause people who hear to question whether or not he really knows what he is talking about.
He does know what he is talking about, but he can't deliver a confident answer because it's not possible. Not without knowing things that can't be known at the time. For example how long a time period it is from the point of contact with the ground until it stops moving.
Don't you all remember the huge values of force generated from impact-momentum problems from dynamics class? The more rigid the components - the shorter the time period - the larger value of force generated at impact.
Make a gutsy call - cross your fingers and drop it or use the suggestion I made on 11/14/02.
Please let us know what you decided and what happened. By now you may have the machinery installed in your plant for production already - but then again you may looking for a job. Good night.
velocity is known-I forget what it is and it doesn't matter.
mass is known-I forget what it is and it doesn't matter.
what does matter is that these values are multiplied and probably = a fairly large momentum.
what matters even more is that time is very, very, very small (because both objects are very rigid) which makes the force very, very, very large.
Doesn't sound too techy and it's 2 AM and I'm too lazy to look up the values above for an example - but the smaller value for time = that much larger a force.
You know questions like this arise from people who want to unload something like this, and an engineer might say "hey, hold on a minute, you can't just drop that thing off the trailor".
Then the boss says "why not we're only dropping it 2 feet, what's the problem?
To which the engineer replies "well there will be a tremoundous amount of force generated when that skid impacts the ground, you may damage the equipment or the hard deck".
To which the boss replies "is that right? well how much force is going to be generated?"
To which the engineer replies "I can tell you in a couple of minutes". Then a couple of minutes later the engineer remembers that in college all those assumptions we used to make on a regular basis don't really apply here. And he begins to say things that cause people who hear to question whether or not he really knows what he is talking about.
He does know what he is talking about, but he can't deliver a confident answer because it's not possible. Not without knowing things that can't be known at the time. For example how long a time period it is from the point of contact with the ground until it stops moving.
Don't you all remember the huge values of force generated from impact-momentum problems from dynamics class? The more rigid the components - the shorter the time period - the larger value of force generated at impact.
Make a gutsy call - cross your fingers and drop it or use the suggestion I made on 11/14/02.
Please let us know what you decided and what happened. By now you may have the machinery installed in your plant for production already - but then again you may looking for a job. Good night.
bowey,
The "T" mentioned isn't how long it takes to get to the ground (that is elementary) the "T" we are concerned with is how long it takes to stop moving down once it hits the ground (that is not possible to calculate with mass, distance, velocity, and a whole list of other knowns}.
In fact it is not too far out to just say that it is impossible to calculate because you have to make assumptions as to the "stiffness" of the concrete, the re-enforcement in the concrete, the sub-structure beneath the concrete. You may be able to use some standard rule of thumb as to the stiffness and restitution of the concrete.
But then you have to consider the skid and for this there no assumption or rule of thumb so to speak. Because this depends on the skid material, how it is made, how it is re-enforced, how the load is concentrated within the skid, how the machinery is secured within the skid, how the machinery is built with its' joints and pivot points, and the material it is made of...etc... suffice to say that you cannot "calculate" the co-efficient of restitution therefore you cannot calculate the amount of "T" it takes to stop moving vertically once the load begins to be applied to the hard-deck.
The "T" mentioned isn't how long it takes to get to the ground (that is elementary) the "T" we are concerned with is how long it takes to stop moving down once it hits the ground (that is not possible to calculate with mass, distance, velocity, and a whole list of other knowns}.
In fact it is not too far out to just say that it is impossible to calculate because you have to make assumptions as to the "stiffness" of the concrete, the re-enforcement in the concrete, the sub-structure beneath the concrete. You may be able to use some standard rule of thumb as to the stiffness and restitution of the concrete.
But then you have to consider the skid and for this there no assumption or rule of thumb so to speak. Because this depends on the skid material, how it is made, how it is re-enforced, how the load is concentrated within the skid, how the machinery is secured within the skid, how the machinery is built with its' joints and pivot points, and the material it is made of...etc... suffice to say that you cannot "calculate" the co-efficient of restitution therefore you cannot calculate the amount of "T" it takes to stop moving vertically once the load begins to be applied to the hard-deck.
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