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Dynamic load of a free fall 7
- Thread starter SKJ25POL
- Start date
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...
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.
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.
SlideRuleEra
Structural
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.
![[idea]](/data/assets/smilies/idea.gif "[idea] [idea]")
![[r2d2]](/data/assets/smilies/r2d2.gif "[r2d2] [r2d2]")
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.
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.
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....
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
SlideRuleEra
Structural
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.
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.
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
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
SlideRuleEra
Structural
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.
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.
at 150 lb/ft^3, 4800 lb would be a 3.2-ft cube
TTFN
I can do absolutely anything. I'm an expert!
homework forum: //faq731-376 forum1529
TTFN
I can do absolutely anything. I'm an expert!
homework forum: //faq731-376 forum1529
Guest262653
Geotechnical
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:
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.
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.
- Thread starter
- #34
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
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
-
1
- #35
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.
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.
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.
SlideRuleEra
Structural
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.
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.
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
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
- Thread starter
- #40
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
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
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