Dynamic load of a free fall 7

[SKJ25POL]

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

 

[Leftwow]

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

 

[Guest262653]

@Pete1919 (Structural)

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

 

[jayrod12]

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

 

[drawoh]

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

 

[Leftwow]

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.html,

this example divides it by the 1/velocity. I think it would be F=mV^2

 

[rb1957]

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 ?

 

[SKJ25POL]

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.

 

[dhengr]

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.

 

[IDS]

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/

 

[PEinc]

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?

http://www.PeirceEngineering.com

 

[BUGGAR]

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.

 

[PEinc]

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?

http://www.PeirceEngineering.com

 

[TLHS]

What if someone's on the platform?

 

[jayrod12]

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

 

[SKJ25POL]

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

 

[BUGGAR]

Yes

 

[rb1957]

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 ?