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Dynamic load of a free fall
7

Dynamic load of a free fall

Dynamic load of a free fall

(OP)
Dear colleagues,

I like to to know how to calculate the dynamic load on a steel platform due to free fall of a concrete block 4800 lbs?
And how we can protect the steel platform (already constructed -1970s)

Thank you for you direction

RE: Dynamic load of a free fall

F=Ma

RE: Dynamic load of a free fall

This is brought up time and time again. It has a ton to do with how stiff the steel platform is, how far the fall was. It's an iterative process because you have to assume a deflection of the steel platform when it has successfully stopped the mass from moving. That distance is used to determine the acceleration (in this case it's negative because the mass is slowing down) and then you solve for the force, then recheck the actual deflection under that load. Repeat steps until solution converges.

If you have an infinitely rigid platform, then you would in theory get an infinite force because you'd be stopping the mass from some velocity right before impact over zero distance meaning your acceleration would be approaching negative infinity.

And a 5000lb chunk of concrete isn't being stopped by much more than the ground, or a much larger chunk of concrete.

RE: Dynamic load of a free fall

My previous employer had a project like this (unloading stand for a logging truck) and we spent almost a month on it. Just setup a test area with some load cells and test it to find the impact force, don't try to do it by calculations alone because it's going to be almost impossible.

Professional and Structural Engineer (ME, NH)
American Concrete Industries
www.americanconcrete.com

RE: Dynamic load of a free fall

(OP)
TehMightyEngineer (Structural),

I would really appreciate if you can explain more what test needs to be done (what the test is called)?
Is it like they drop the 4800 lbs to test it???

I have a canteleved platform in 120ft up in the air which is cantelevered by the open structure.
I have to check if the platform can withstand the drop of this conctere block (part of a counterweight system)
The free fall is 12.5 feet the dimension of the platform is 20 ft by 7.00 feet and covered by grating

Please if you have any documents pics, information on similar situation direct me.

I am unclue where to find some helpful information

Thank you so much

RE: Dynamic load of a free fall

I`ve always been curious about this type of problem and never reached a solution - iterative or otherwise.

Jayrod - you say you assume a deflection/distance, and that distance is used to determine the acceleration. Position, velocity, and acceleration are all related by time and if we're assuming the distance, we cannot also assume the time, right? How do you get time?
If you have a weight falling unrestricted, you could easily determine how long it takes the weight to pass through the deflection, but in reality, the beam pushes back and it will take longer to travel this distance. I think this change in time will depend on the stiffness of the beam, and hence, you can't easily calculate it.
What am I missing?

RE: Dynamic load of a free fall

this is an impact problem. The loads generated depend on the stiffness of the structures, the time of impact/deceleration; that's why it's very difficult to get a real answer from analysis.

We do similar analysis (bird strike) with FEA (LS-Dyna) but modelling the structures realistically for stiffness (particularly at joints) is time-consuming. And testing to validate the model.

I'd take a lot of care before I started lobbing 2+ tons lumps of concrete around.

it's easy to figure out the velocity of the concrete block prior to impact, right?

another day in paradise, or is paradise one day closer ?

RE: Dynamic load of a free fall

of course, the easy (and inaccurate) way is to use an impact factor of 2 (so that the force on the structure is, near enough, 10,000 lbs).

another day in paradise, or is paradise one day closer ?

RE: Dynamic load of a free fall

(OP)

You guys keep saying we do tests. Can some body please explain what test? what company needs to be hired to do the test?
I appreciate somebody kindly explain to me what test?
Hope you are trying to help rather than mudding the water.

This is not a new problem and I thought it is a solved engineering subject

RE: Dynamic load of a free fall


To test this - get a really strong scale, take it out into the parking lot, and drop a 4800 lb chunk of concrete on it from 12.5' up. Repeat the test a couple times, and look at the force that the scale measured.
This won't accurately account for the deflection of the beam, but will capture the force dissipation associated with the crushing of the concrete and is a pretty reasonable starting point.

If this is an existing structure that wasn't designed for this criteria, I'd bet a donut that it can't take the load.
If you're designing it, it's going to be some *serious* steel.

Either way - please share the scale readings. I think we'd all be interested.

RE: Dynamic load of a free fall

They literally mean build a platform that is exactly the same as the one you want to know about, then drop masses on it from heights to determine what kind of force modifier to use.

In response to Once20036. You're right, it's a combination thing, I've never officially ran a number using this method, so I would have to spend more time fine-tuning it, but you have to assume something, a distance it is stopping over, the time it takes it to stop, something. If you have the distance it stops over and the speed prior to impact, you can determine the acceleration without caring about time.

I see it sort of the following procedure:
1) determine speed of mass right before impact
2) Assume a stopping distance (deflection) x
3) Use conservation of energy to determine the spring constant K that would result from said stopping distance x
4) determine force in spring using distance x and spring constant
5) apply force to model to determine actual deflection
6) Substitute calculated deflection for assumed deflection from step 2
7) repeat steps 2 through 6 until the solution converges.

Again that's just how I understand it being possible to determine, I don't know if it would ever converge, I also don't have a feel for how the localized damage from impact may affect the results.

SKJ25POL
In all seriousness, your platform is likely only designed for a total load between 7000 and 14000 lbs. That's gives you, at best, a dynamic impact factor of 2.8. Your 5000 lb hunk of concrete will be moving at 20 MPH by the time it gets to the platform, if you flipped your cantilever platform vertical would it stop a car driving 20MPH? My money is no chance. Or at least not without some serious deformation happening.

RE: Dynamic load of a free fall

a 5,000lb car ? bigsmile and I think a car will crumple more than a concrete block ! mind you the concrete will shear at some stage, so ... ??

I would not use a scale in a car park, though it would give you the worst case load (the ground is much stiffer than your structure, i'd expect.

The best test is build a copy of your structure and drop your concrete block on it ... but, of course, no one's going to do that ! (way too expensive, and way too risky with some analysis to give you confidence that it'll work)

Personally, I'd go with analysis that assumes an impact load factor ... 2, 3, ... and goes from there. Remember the impact force is only part of the problem, what about the impact area (is the block landing face on, or corner first ?) and where ? (the middle of the panel is probably the most compliant, at the support the structure is very stiff)

another day in paradise, or is paradise one day closer ?

RE: Dynamic load of a free fall

Build a short spring loaded platform on top of existing structure, so you can control the deceleration to some degree?

RE: Dynamic load of a free fall

rb1957,

Couldn't you just assume some elastic deformation of the structure, and work out the resulting accelerations, forces and resulting stresses? Obviously, elastic deformation will be very small, accelerations and forces will be high, and there is an excellent chance this thing will not work. The OP can iterate if he wants more accurate results. It should not be that difficult to estimate roughly the maximum elastic deformation.

I would wonder about pieces of shattered concrete falling one hundred twenty feet to the ground? Given that the deformation is elastic, what is to stop the concrete block from bouncing off in some random direction?

--
JHG

RE: Dynamic load of a free fall

I'm just throwing darts here:
Start getting a handle on the deflection, do a static analysis of the platform support structure, find out at what load it will yield, and what the deflection is at that load. Then figure out what de-acceleration will produce that force, then figure out what time that process takes over the deflection available. Somewhere in there you may find out it either works or does not work. Your end result may be that the structure has to deflect over a long period of time, you may be able relate the strain energy of the structure to this time frame, maybe not. If the time frame is an hour, that's unrealistic. If it is on the order of a millisecond, I think you may be in the ball park. I'd be interested in seeing the data. Use MathCAD or some other software package so you can iterate, change things, make plots, etc...

RE: Dynamic load of a free fall

2
"of course, the easy (and inaccurate) way is to use an impact factor of 2 (so that the force on the structure is, near enough, 10,000 lbs)." No don't do that. Check out how that factor of 2 is derived. For a start you can see it gives an odd answer, since it ignores the drop height.

You could build an accurate scale model of the setup and instrument that and test it. Scaling it is not going to be straightforward.

A non linear FEA model such as LS Dyna would be used in the automotive industry to simulate crashing cars into concrete blocks to some tremendous accuracy. I think that is overkill, and you probably haven't got the budget and you certainly don't have the correlation data.

jayrod12 suggested this is an impact problem, which he's simplified down to a single degree of freedom spring mass system. That is the way I'd approach it. You know the stiffness of the platform, you know the velocity at initial impact, you know the mass. First year dynamics. k*x2=m*v2 gives you x, then k*x is your maximum force. This assumes linear elastic, in real life you might prefer to absorb the energy in a plastic hinge in which case you need to know the moment of plasticity of your platform.

Cheers

Greg Locock


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RE: Dynamic load of a free fall

It's a simple energy problem. mgh = spring energy of platform. You'll have to figure the spring constant of the platform. From this, you compute the deflection of the platform and the corresponding forces in the platform.
As a check, a suddenly applied load applied to a non-deflected platform will cause 2g's as it deflects the platform.

RE: Dynamic load of a free fall

I had a much smaller, but illustrative project. It was a metal frame with and electronic instrument inside; part of a stack that clipped like LEGO bricks. One of the tests was a 3 foot onto plywood. The G forces recorded for bottom down were less than for top down. At first the test guys were puzzled, but it turned out the engaging pins on the bottom embedded themselves in the plywood, increasing the impact duration and lowering the peak load.

tl;dr - prediction is very inaccurate because collision deforms and damages the items in ways that are expensive to evaluate. Test it in the condition you expect and see if works afterwards.

RE: Dynamic load of a free fall

Could you put a bunch of high density foam in the area of concern? It'd be one time use only, but you could significantly increase the time of deceleration.

RE: Dynamic load of a free fall

A 4800 lb concrete block is probably much smaller than 7x20 ft in planform, so it will likely punch right through the grating and keep on going, impacting on whatever is under the grating.

If there's any structure between grating and ground, e.g. sway bracing of the cantilevered platform, and/or perimeter and intermediate support of the grating, and that structure is at all efficient, it will probably buckle, the block will fall farther, and progressive buckling or subsequent impacts by the block in its now deflected trajectory will have a fair chance of bringing down the whole damn thing.

Could be an expensive test.

Could also be an expensive analysis.

Analysis gets interesting, even more so when you try to account for 35+ years of neglected maintenance and corrosion of the tower and uncertainty about the properties of the material actually used.

Have you considered a new tower, designed from the start for the subject load case?
Or perhaps with redundant load paths for the counterweight, to mitigate the chance of it falling?



Mike Halloran
Pembroke Pines, FL, USA

RE: Dynamic load of a free fall

Omer Boldgett has some examples of shock loading in his "Solutions to Design of Weldments" a jewal of a little book that can be obtained from Lincoln Electric for a few dollars...best money you will ever spend ...he also has other great books at great prices...

RE: Dynamic load of a free fall

that should be Omer Blodgett....

RE: Dynamic load of a free fall

1. Please look into the design of "dolphins" to resist the impact load of ships. Similar procedure would apply here. Essentially it is based on the spring constant discussed by many on the forum here. However, the spring constant of dolphin's structure is well known/defined based on experiments and experience.
2. The rough and dirty method is to multiply by two. But the equation derived to arrive at the factor of 2 is itself grossly rough and flawed. It all boils down to the time of contact during impact which is not taken into consideration in deriving the factor of 2.
3. I liked 1Gibson's simple and practical approach of spring loading/padding the platform in some way and form to increase the time of contact/impact and hence reduce the dynamic load; and also double the static load.

I hope this helps.

RE: Dynamic load of a free fall

Don't guess at an impact factor, work the problem backwards to determine at what static load (positioned at the area of impact) the platform will fail. Divide this load by 4800 lb. to find the impact factor at which failure occurs. Based on the description of a grating covered cantilevered platform, I expect the calculated impact factor will be too low to be reasonable - a value less than 1 (failure below 4800 lb) would not surprise me. Keep in mind that the entire platform does not have to fail, just the area of impact.

To get an idea of the size of the area of impact, calculated that a (hypothetical) 4800 lb. sphere of concrete is just under 4 feet in diameter. Of course it could be any shape.

www.SlideRuleEra.net idea
www.VacuumTubeEra.net r2d2

RE: Dynamic load of a free fall

The potential energy of the concrete block is its weight times the height from which it falls. The beam/structure kinetic energy is 0.5(k)(x^2) where x is the deflection and k is the stiffness of the beam. The stiffness is dependent on the load case and beam type (uniform load, concentrated load; cantilever, simply supported, etc.). By setting the concrete's potential energy equal to the beam's kinetic energy, you can calculate the deflection. You then can back out the equivalent load.

RE: Dynamic load of a free fall

Sorry Greg Locock, I missed your post before I posted my previous response.

RE: Dynamic load of a free fall

I think SlideRule has cut through the fog on this debate....his suggestion will determine wheather the platform as now designed has any chance at all of meeting the reguirements, as is ,or even if reinforced
which may lead to a whole new approach to the problem....

RE: Dynamic load of a free fall

Quote:

I think SlideRule has cut through the fog on this debate....his suggestion will determine wheather the platform as now designed has any chance at all of meeting the reguirements, as is ,or even if reinforced
which may lead to a whole new approach to the problem....

The trouble with the SlideRule approach is that calculating the "impact factor" doesn't actually tell you much about whether the structure will survive the given impact load. If it comes out less than 1 we know that it won't, but we don't know what the required level is (other than that it will be much higher than the figure of 2 that some people have mentioned).

I suggest the following approach:

1. Carry out an incremental non-linear static analysis (applying the load over an estimated contact area) until something breaks. That is until the maximum strain in any fibre of any element exceeds the rupture strain.
2. Calculate the work done by the applied force (area under the force-deflection diagram)
3. Calculate the potential energy of the falling concrete block (adding in the deflection of the impact point, if it is significant)
4. If 2 is not significantly greater than 3, review the possibilities for strengthening and/or the potential consequences of a falling concrete block falling through this part of the structure and potentially causing further damage on the way down.
5. If 2 is significantly greater 3 hire someone who does dynamic analysis for a living to review whether the dynamic effects will cause the structure to fail anyway.

Doug Jenkins
Interactive Design Services
http://newtonexcelbach.wordpress.com/

RE: Dynamic load of a free fall

Assume that a 4800 lb. block of concrete is say, 4' wide x 4' deep x 2' high. It is sitting on the grating in the ideal position - block is flat on the bottom and is applying a uniform load over the 16 sq. ft. (4' x 4') contact area.

The load on that grating, and its support steel is 300 PSF.

What is the rated live load of the grating?
40 PSF?
60 PSF?
Maybe, 100 PSF?
And 300 PSF is the static load.

Here is an analogy to get a feel for the amount of energy that platform is subjected to:
Think of the 4800 lb. block as the ram of a single-acting pile hammer. The stroke is 12.5 feet. Energy delivered at impact - 60,000 foot-pounds. There are plenty of pile hammers that deliver that much energy, and more, but a typical sized hammer for routine pile driving of say, 12" HP and 16" concrete piles develops 15,000 ft.-lb. to 25,000 ft.-lb.
Decide for yourself if a platform could survive a direct strike from a 60,000 ft.-lb. pile driver.

www.SlideRuleEra.net idea
www.VacuumTubeEra.net r2d2

RE: Dynamic load of a free fall

SlideRuleera - if the structure won't take the static load then I agree the problem is easy.

My points were:
1. If the structure will take 2x the static load, that's still nowhere near enough (with which I am sure you agree).
2. It's not that hard to get a reasonable estimate of the maximum energy the structure will absorb before collapse.
3. But even if the structure will absorb that energy (or can be strengthened to do so) you still need to get it reviewed by someone who is familiar with this stuff.

But I agree that it is most unlikely that the structure will come anywhere near taking that impact load.

Doug Jenkins
Interactive Design Services
http://newtonexcelbach.wordpress.com/

RE: Dynamic load of a free fall

Doug - No problem. We know so little about the actual platform. As SAIL3 mentioned, I'm just suggesting that the "easy" math may be enough to prove that the platform won't survive - never have to go to a detailed analysis.

IMHO, the biggest unknown: Is the platform 20' wide, cantilevered out 7'... or is it 7' wide, cantilevered out 20'? Failure mode would likely be different for the two cases, especially since the concrete block will likely be small compared to the 20' dimension.

www.SlideRuleEra.net idea
www.VacuumTubeEra.net r2d2

RE: Dynamic load of a free fall

Nice little problem and an opportunity to review some of the stuff we learned at the Mechanics of Materials classes 20 years ago.

Assuming an elastic collision, we can have an impact force estimate following what others have already said in this thread:
- Potential energy of the falling mass = mg(h+delta_tip)
- Delta_tip = PL^3/(3EI)
- Strain energy on the cantilever bar loaded at the tip = P/2 * PL^3/(3EI), by Clapeyron's theorem
- From an energy conservation assumption, we get a 2nd degree polynomial equation which is easily solvable for P:
P^2.L^3/(6EI) - PmgL^3/(3EI) -mgh = 0
Additionally, we can throw an efficieny factor to the equation to simulate energy losses (and the quantification of this factor is the real problem...).
I've attached a simple spreadsheet to perform these calculations automatically.

This approach doesn't consider the following factors...
- aerodynamic drag on the falling mass;
- initital velocity of the mass;
- energy dissipation at contact (noise, heat, plastic deformation);
- stiffness of the falling mass;
- non-linear behaviour of the cantilever material;
- shear deformation of the cantilever;
- bouncing of the weight and reloading of the beam;
- (...)
... but is an estimate anyway.



RE: Dynamic load of a free fall

(OP)
Thank you everyone for your input and time, I appreciate it.
Sliderule,
Thank you for your response and I have attached the steel framing of the platform with the location of the concrete block above the platform.
Please if the clears some mud, I would like to know if any of your comments or anybody else"s comment will change.

I appreciate continuing your advice and direction.

Sincerely
SKJ25POL

RE: Dynamic load of a free fall

I might:

1) Assume that the counter wight touches down on it's north end. The north load path looks stiffer than the south.

2) Make a SAP model to work out the stiffness of the system relative to the point load mentioned in #1.

3) With stiffness in hand, work out the dynamic amplification factor as others have declined above.

4) Start checking stuff.

There appears to be a fair bit of flexibility built into the system with those beams cantilevered from the columns like that. Might not be too bad.

I like to debate structural engineering theory -- a lot. If I challenge you on something, know that I'm doing so because I respect your opinion enough to either change it or adopt it.

RE: Dynamic load of a free fall

I bet you can't get enough fasteners in the W8's. Are you welding the connections? I continue to push for the energy analysis. Avscorreia gets it. Use SAP or whatever to get the platform spring constant at the point of impact, solve for deflection from the falling mass, and back-compute the equivalent load in all the members from that load.

RE: Dynamic load of a free fall

avscorreia (Geotechnical),
Good approach for a simple cantilever problem. Your equations are correct. However, in your spread sheet you have coded the second term of the quadratic equation incorrectly. Should have removed mu (0.95) from there.

RE: Dynamic load of a free fall

SKJ25POL - Since the counterweight has a flat bottom and is positioned directly over a beam it is not going to "punch through" the grating. The W8x17 will easily support the 4800 lb. static load - time to move on the a more detailed analysis of the type discussed by our colleagues.

However, I would not assume an elastic collision. Suggest taking the F=MA approach mentioned by jrisebo.
One way is to assume a deflection under the impact location (for discussion say 3").
Also assume that the platform survives the impact, more or less intact with the block sitting on it.
The 4800 lb. block has to decelerate from its 28.4 ft/sec impact velocity to zero ft/sec in a distance of 3" (or whatever distance you assume).
Compute the force (exerted by the platform) to make that happen.
Apply that force (as a static value) to platform at the location of the impact and see if the platform can actually withstand it. IMHO... I doubt it.

www.SlideRuleEra.net idea
www.VacuumTubeEra.net r2d2

RE: Dynamic load of a free fall

SKJ25POL,

What is it you are trying to accomplish here? Are you trying to determine whether or not it is safe to drop the concrete block? Is this a one-time event followed by extensive repairs, or will this happen regularly?

Could you place some sort of crush structure on your platform to control the deceleration of your block? If the crush is plastic rather than elastic, the block won't bounce and land somewhere else.

--
JHG

RE: Dynamic load of a free fall

(OP)
This is to avoid or prevent platform failure and stopping the concrete block to fall and injure anybody underneath.

drawoh,
you stated "Could you place some sort of crush structure on your platform to control the deceleration of your block? If the crush is plastic rather than elastic, the block won't bounce and land somewhere else."

Answer, yes. What's out there? What will you suggest?

Thank you

RE: Dynamic load of a free fall

This is how I would do it:

You have to figure out your velocity first.
take your acceleration of 32.2 ft/sec^2

set equation up Velocity = 32.2 * time

Take your kinematic equations for displacement to solve for time
(change in height/0.5*32.2)^1/2=time

Plug in time for velocity equation

Set up equation for Momentum
P=mass*velocity

Then add F=P+mg

Your momentum will be The force applied. Add Impact safety factor from asce 7-10


RE: Dynamic load of a free fall

@Pete1919 (Structural)

You're right. As I used a*2 for the calculation of b I should've multiplied b by mu. Thanks!

RE: Dynamic load of a free fall

Leftwow, how is momentum a force directly the way you've shown?

RE: Dynamic load of a free fall

SKJ25POL,

How about those barrels and boxes they place in front of highway bullnoses? Those are designed to stop things weighing around 4800lb. If you don't want people hit by your concrete block, bouncing is every bit as bad as platform failure.

Alternately, you could design a sheet metal or plastic box that would crush, control deceleration and force, and contain your dropping concrete block. The analysis for this is way beyond my abilities.

Have you any control over the orientation of the block when it hits stuff?

--
JHG

RE: Dynamic load of a free fall

Well if mass has acceleration then force exists. If mass does not have acceleration, then there is no force.
In either case, momentum will be present.

If m is constant

F=ma=m*(dv/dt)

I guess I need to still divide my velocity by time. dang

http://hyperphysics.phy-astr.gsu.edu/hbase/impulse..., this example divides it by the 1/velocity. I think it would be F=mV^2

RE: Dynamic load of a free fall

SKJ25POL,

I assume we're looking at a plan view, with the counterweight suspended above the beam (and not the grating). btw, the red box is NTS compared to the 4' dim'n nearby.

Is the beam built-in at the ends (ie with some moment connection) ? to help distribute the load through more of the structure.

Is it possible to erect support struts under the beam, to provide a better out-of-plan loadpath ?

how far is the counterweight above the beam ?

what else will happen if the counterweight breaks free ? (what is it balancing ?)

Is it easier to double-up on the cable controlling the counter-weight, to provide a secondary loadpath ?

The beam will deflect into an arc, so the block load will concentrate on the outer edges. will the web buckle ? will the upper cap split ?? It makes sense to reinforce the web over this span, at least the likely impact points.

another day in paradise, or is paradise one day closer ?

RE: Dynamic load of a free fall

(OP)

rb1957 (Aerospace,

I assume we're looking at a plan view, with the counterweight suspended above the beam (and not the grating). btw, the red box is NTS compared to the 4' dim'n nearby. YES IT IS PLANVIEW

Is the beam built-in at the ends (ie with some moment connection) ? to help distribute the load through more of the structure. I ASSUME IS SIMPLE CONNECTION. HAVENT CHECKED MYSELF.

Is it possible to erect support struts under the beam, to provide a better out-of-plan loadpath ? IF THAT'S THE SOLUTION, YES

how far is the counterweight above the beam ? 12.5ft

what else will happen if the counterweight breaks free ? (what is it balancing ?) NOT SRURE

Is it easier to double-up on the cable controlling the counter-weight, to provide a secondary loadpath ? THAT'S NOT MY CALL

The beam will deflect into an arc, so the block load will concentrate on the outer edges. will the web buckle ? will the upper cap split ?? It makes sense to reinforce the web over this span, at least the likely impact points.

RE: Dynamic load of a free fall

SKJ25POL:
You’ve got a bunch of smart and serious people trying to help you (wasting their time, guessing at what you are trying to do), and it has only taken you about 30 and 40 posts to actually start to explain/show your problem in enough detail so we can start to see it and understand what is going on. Why is the counterweight (2' x 4' x 4' high = 4800lbs. of conc.) likely to fall? Can you do something up at the counterweight to prevent that, safety cables as Rb1957 suggests, for example? Why not build a light braced frame up 12.33' off the existing platform and tie this framing into the bldg. so the ct.wt. can’t fall but a few inches? Either of the above are easier to do than to try to justify that block of conc. falling 12.5' to the canti. platform, 120' up in the air, over people below. How do you know that something will break and allow the conc. blk. to fall straight down to the platform, without rotating or swinging?

Alternatively, I would add 2 more W8x17's, 8"-7" long (north & south) about 16" either side of the one centered under the ct.wt. You don’t want the ct.wt. tipping off that one centered W8. I would put some sort of a 3' high railing around the impact area to prevent tipping. This would enclose an area of about 8'-7" by 4' (east & west). Your crush system could be as simple as a 2.5- 3' deep cribbing of pine 2x4's laid up in such a way that it would break and crush on impact and absorb a bunch of energy as the conc. blk. continued to break successive layers and finally stop. Or, it could be 8 or 10 layers of stacked soda cans, standing up and reasonably tightly fitted, into containing boxes constructed of a crushable material. The layers of can should be separated by a horiz. layer of material which causes a layer at a time to crush. This too, would absorb energy just like the front crush regions on today’s cars. It slows the load, over a period of time, which allows the impact load to be something short of infinite.

RE: Dynamic load of a free fall

Quote:

The beam will deflect into an arc, so the block load will concentrate on the outer edges. will the web buckle ? will the upper cap split ?? It makes sense to reinforce the web over this span, at least the likely impact points.

The ends of the counterweight are only 2 feet from the connection to the transverse beams, so almost instantly after impact it will generate a plastic hinge at each end of the beam, with an associated shear force of Mp/2. That will rapidly reduce, because the transverse beams have the same plastic moment, with a lever arm of 4 feet.

Knowing the shear force at the points of impact (if the beams will take this load without failure) you can calculate the deflection required to bring the falling weight to a stop, and from that calculate the required plastic strain at the cantilever supports. I haven't done the numbers, but I suspect the two transverse beams will have nowhere near enough rotation capacity, and you will need to find a way to effectively distribute the load, or find another way of dealing with a dropped counterweight.

Note that as this is just an order of magnitude calculation, and the required deflection will be much greater than the elastic limit, you can get an acceptable first approximation calculation by assuming a constant force equal to the shear force that will cause plastic hinging at the point of support (Mp/L). Then:
deflection = mgh/(Mp/L)

Doug Jenkins
Interactive Design Services
http://newtonexcelbach.wordpress.com/

RE: Dynamic load of a free fall

Would it be easier or possible to design some additional (redundant) support up at the counterweight that would prevent the counterweight from falling rather than letting it drop and smash into the platform?

www.PeirceEngineering.com

RE: Dynamic load of a free fall

Like someone said before, you can pick a marine fender out of a catalog that will absorb this kind of energy. The catalog will give you the force vs. deflection vs. energy absorption so you can figger everything nice and tight. Just lay that fender down on your platform and strap it down. Problem solved! The bill is in the mail.

RE: Dynamic load of a free fall

I may be wrong, but I assume that the platform was built for and is currently being used for a different purpose than catching a heavy hunk of concrete. If so, wouldn't a marine fender interfere with the platform's normal use?

www.PeirceEngineering.com

RE: Dynamic load of a free fall

What if someone's on the platform?

RE: Dynamic load of a free fall

TLHS that's where the dodgeball training comes in handy. Dodge, duck, dip, dive and dodge.

RE: Dynamic load of a free fall

(OP)
BUGGAR (Structural,

Sive you brought up the marine fender and I am not familiar with the product, could you please attach a catolog that you see the product best suits the problem?

The ones that I saw on internet are cone shape and a plate on top, are u suggesting these ones?

Thanks

RE: Dynamic load of a free fall

Yes

RE: Dynamic load of a free fall

now that we're describing the lump of concrete with a useful name (counterweight) that suggests more questions ...

does it move in normal operation ? (so that 12' is the highest it could be above the platform, and <1' the lowest)

there are lots of "fender" designs that wouls be more suitable here ... I'm thinking of flat 2D shapes (like a bed mattress).

i don't like the idea of saying "assume the counterweight falls" and not answering the logical question "what happens to the thing bereft of the counterweight's support ?" of course it could be a regulatory question "assume the 'hand of god' sunders the cable supporting the counterweight above your platform, will your platform fail ?" I'm assuming this platform has been erected for some time now, and now someone's asking "what happens if ..."; maybe the answer is "the crane (or whatever), that relied on the counterweight, falls over causing mayhem and destruction; oh, and the counterweight crushes the platform".

another day in paradise, or is paradise one day closer ?

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