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
First off it is impossible to calculate the reaction force for this due to not being able to calculate the initial impact time as stated from many above. You could possibly set up accelerometers or velocity transducers and collect transient data to possibly separate initial impact force and time from all off the inevitable resonance from structure and concrete but WHY ????? In a "REAL" world scenario I would have to agree with Massey. Take 2 cranes or for that matter 1 huge crane (I have worked around cranes capable of lifting 800 tons, awsome machines) to lift this skid off the truck. The safety issues, which far out weigh the damage to equipment issues, should be of the utmost importance. If you have ever seen anyone crushed by miscalculated judgement the pen and pad would not even be a part of this equation. Roy Gariepy
Maintenance and Reliability Dept.
Bayer Corporation Dorlastan Fibers Div.
Goose Creek, South Carolina USA
Maintenance and Reliability Dept.
Bayer Corporation Dorlastan Fibers Div.
Goose Creek, South Carolina USA
I agree with micjk, but I suggested two cranes because for some reason it was going to drop 2' when taken from the trailor.
It may have abeen a rigging problem or they may be lifting one end with a huge fork-truck and dragging it off the trailor.
Or they may be chaining it to the building and taking off in the truck to let it drop - beats me.
Ideally one well rigged lift with an adequate crane, second best is two cranes where the balanced rigging isn't as big of an issue. And last and least is chaining the skid to the building, better yet the neighbors building, and taking off in the truck without looking back.
It may have abeen a rigging problem or they may be lifting one end with a huge fork-truck and dragging it off the trailor.
Or they may be chaining it to the building and taking off in the truck to let it drop - beats me.
Ideally one well rigged lift with an adequate crane, second best is two cranes where the balanced rigging isn't as big of an issue. And last and least is chaining the skid to the building, better yet the neighbors building, and taking off in the truck without looking back.
GregLocock
Automotive
Agree with all the above sensible suggestions, but if you really want to design the package so that it can be dropped (eg it may be a customer spec), then why not fit a proper shock-absorbing system into the skid. This can be easily tuned so that it is the dominant compliance in the system, thereby simplifying the problem of assessing the shock on the load.
Cheers
Greg Locock
Cheers
Greg Locock
Bowey,
The sum total of your suggestions would seem to be that you would end up calculating the force acting on the body during its fall as M*g, or as we generally know it, its self weight.
That doesn't seem to be much value in estimating the impact load when the free fall is suddenly stopped.
The sum total of your suggestions would seem to be that you would end up calculating the force acting on the body during its fall as M*g, or as we generally know it, its self weight.
That doesn't seem to be much value in estimating the impact load when the free fall is suddenly stopped.
Here is link to a paper - "Calculating the impact force of a mass falling on an elastic structure"
Rich2001,
That was an interesting paper that may be helpful in some other situations and it does show that the impact force can be calculated without knowing the coefficient of restitution. However, in this particular case I don't see how it could work for the following reasons:
knowing the arresting strain isn't helpful because the modulus of elasticity of the concrete is not known (too many unknowns to use published values),
and the stress cannot be determined because the Area of the concrete that the force acts over cannot be accuratly determined.
this object is not in 'free fall', originally it was stated that one end of this object may fall two feet and hit the ground - this indicates that the CG has a moment arm about some pivot point.
the length L cannot be known - or rather I don't see how there could be one in this case. (?)
But other than that it was interesting.
That was an interesting paper that may be helpful in some other situations and it does show that the impact force can be calculated without knowing the coefficient of restitution. However, in this particular case I don't see how it could work for the following reasons:
knowing the arresting strain isn't helpful because the modulus of elasticity of the concrete is not known (too many unknowns to use published values),
and the stress cannot be determined because the Area of the concrete that the force acts over cannot be accuratly determined.
this object is not in 'free fall', originally it was stated that one end of this object may fall two feet and hit the ground - this indicates that the CG has a moment arm about some pivot point.
the length L cannot be known - or rather I don't see how there could be one in this case. (?)
But other than that it was interesting.
I am amazed that nobody can even attempt an approximation for this. I do realise that it is very complicated, but surely there are material property values available that would allow this, without having to take the experimental route.
If an FEA was applied what data would be required? Would the material yield stresses and dynamic modulii be sufficient?
Just curious!! 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."
If an FEA was applied what data would be required? Would the material yield stresses and dynamic modulii be sufficient?
Just curious!! 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."
Massey,
The method is to calculate the static load on the runners by statics, as if the skid platform had been placed gently in the tilted position. Multiply that load by 1 + (1 + 2h / d)^.5, to obtain the approximate dynamic load upon impact.
I hope I didn't say something to imply that the part had to actually be placed in position, dropped, or even built, before the approximation could be carried out.
The only Roark I have at hand is the 4th edition, in which this approach is described on page 370. I am pretty sure that the method can be found in later editions, but I don't have my copies with me just now.
Sorry I didn't write more clearly in my previous post.
Regards,
Lcubed
The method is to calculate the static load on the runners by statics, as if the skid platform had been placed gently in the tilted position. Multiply that load by 1 + (1 + 2h / d)^.5, to obtain the approximate dynamic load upon impact.
I hope I didn't say something to imply that the part had to actually be placed in position, dropped, or even built, before the approximation could be carried out.
The only Roark I have at hand is the 4th edition, in which this approach is described on page 370. I am pretty sure that the method can be found in later editions, but I don't have my copies with me just now.
Sorry I didn't write more clearly in my previous post.
Regards,
Lcubed
PeterCharles
Mechanical
MASSEY (Mechanical) wrote :-
"Or they may be chaining it to the building and taking off in the truck to let it drop - beats me."
In the late 1970's the company I worked for was doing work in Iran. There weren't any buildings around, but there was this tree ................
I believe damage was minimised by the splintering of the wooden packing cases.
"Or they may be chaining it to the building and taking off in the truck to let it drop - beats me."
In the late 1970's the company I worked for was doing work in Iran. There weren't any buildings around, but there was this tree ................
I believe damage was minimised by the splintering of the wooden packing cases.
RonMB
Many have answered on how to analyze the problem using momentum and elasticity. It is a bit more difficult than most admit and as some have suggested. We do solve such problems every day in analysis of granular mechanics related to ore flow in chutes and bins, during comminution action in grinding mills, crushers, in rock mass failure action in mining, et.al. The tools we use, we developed in house. These are derived from DEM (Discrete Element Method) All forces can be obtained. See our website and an animation:
The dropped object must be characterized with respect to: a) center of spin in 3-D and inertia in 3-D on impact.
b) irregular shape of the surface and orientation on impact
c) structural failure criteria of object and surface (non-linear elastic, elasto-plastic, or plastic rheology - ie. strain rate dependence and exceeding yield prop.), and surface energy properties (crack tip growth) to assess whether the contact point will carry the load or fracture and the results from the fracture
d) restitution properties - linear; nonlinear differential-integral function
e) and so on
Itasca Consulting sell such a code developed by Peter Cundall at web site:
The method is developed from discontinum mechanics a super set of continum.
Lawrence Nordell
Conveyor Dynamics, Inc.
Many have answered on how to analyze the problem using momentum and elasticity. It is a bit more difficult than most admit and as some have suggested. We do solve such problems every day in analysis of granular mechanics related to ore flow in chutes and bins, during comminution action in grinding mills, crushers, in rock mass failure action in mining, et.al. The tools we use, we developed in house. These are derived from DEM (Discrete Element Method) All forces can be obtained. See our website and an animation:
The dropped object must be characterized with respect to: a) center of spin in 3-D and inertia in 3-D on impact.
b) irregular shape of the surface and orientation on impact
c) structural failure criteria of object and surface (non-linear elastic, elasto-plastic, or plastic rheology - ie. strain rate dependence and exceeding yield prop.), and surface energy properties (crack tip growth) to assess whether the contact point will carry the load or fracture and the results from the fracture
d) restitution properties - linear; nonlinear differential-integral function
e) and so on
Itasca Consulting sell such a code developed by Peter Cundall at web site:
The method is developed from discontinum mechanics a super set of continum.
Lawrence Nordell
Conveyor Dynamics, Inc.
Perhaps I am too late with my suggestion, but, I think that since you are applying a very big force (at least 6 times 80k pounds), on two thin skids, you should consider that you will damage your concrete floor (a fragile material, which will crack locally at least), your skids (which will be subjected to a very big distributed load perhaps yielding them), and whatever is attached on top.
Around here, when we have to do something like that, we use old tires, lying on their side, and we drop the object over that. We have the advantage of having it separated from the floor when trying to forklift it later.
But 80k pounds is a lot!, several tires would be needed...
sancat
Around here, when we have to do something like that, we use old tires, lying on their side, and we drop the object over that. We have the advantage of having it separated from the floor when trying to forklift it later.
But 80k pounds is a lot!, several tires would be needed...
sancat
The impact question is a frustrating one. I am currently working on redesigning a train derailer. The derailer, essentially a steel bar clamped over the rail head at an angle, must withstand the impact of a locomotive and succeeding train cars going a given speed.
Many questions arose from this seemingly simple problem. What masses should be considered: the wheel, the truck, the locomotive, or the whole train? What elasticity can be expected from a derailer braced against railroad ties? What is the estimated length of impact? What is the momentum of the locomotive/wheel/train after impact? Friction forces? Etc etc...
Back in the bad old days I think they just kept beefing the rerailers up until they didn't break. I may have to do the same!
Many questions arose from this seemingly simple problem. What masses should be considered: the wheel, the truck, the locomotive, or the whole train? What elasticity can be expected from a derailer braced against railroad ties? What is the estimated length of impact? What is the momentum of the locomotive/wheel/train after impact? Friction forces? Etc etc...
Back in the bad old days I think they just kept beefing the rerailers up until they didn't break. I may have to do the same!
diamondjim
Mechanical
If this is put on a slope and as it is pulled off
the truck, wont the truck over turn by the weight
of the 80,000 pounds? Use a crane.
the truck, wont the truck over turn by the weight
of the 80,000 pounds? Use a crane.
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