Design Beam for Impact Absorption

[jmggks]

I am trying to figure out how to design a beam to absorb the impact from a falling object. A conveyor carries 13,000 lb. parts over an industrial enclosure that we are building. The customer wants the roof beams to protect the people inside if a part falls from the conveyor. The fall distance is only 1 ft. It's okay if the beams are damaged in the process as long as they do not give way.

Using Blodgett's Design of Welded Structures 2.8-5, I can assume a beam size, calculate the elastic spring rate for a simply supported beam, and solve for the applied force. In this case, 16 ft. span, assume W10x49 beam, and this solves to P=143 kips. But it would only take 82 kips to produce FuZ for this beam, so this would seem to indicate that I need a bigger beam ... which will give me a stiffer spring rate ... which will give me a higher force ... which will require an even larger beam ... and so on and so on never reaching a solution.

I know the problem with this analysis is that the spring rate of the beam will drop by orders of magnitude once it reaches yield, which will lessen the force. Using an elastic spring rate is the problem. I think what I need to do is to equate the potential energy delivered by the dropped part to the energy absorbing capacity of the beam at rupture. Can anyone point me to a method for doing this?

Thanks for your input.

 

[phamENG]

The number you're looking for is essentially the integral of the stress strain curve from 0 to rupture.

I'd be careful about doing that, though. You need to make sure your static load (once the impulse has subsided) is less than yield, or your plastic hinge will continue to deform to failure.

What about using a concrete roof and tighter beam spacing to spread the load? Or running infill beams perpendicular to your W10s that are moment connected across them or a set of perpendicular grillage type beams sitting on top. That will help with load sharing, too, I should think.

 

[Tomfh]

I think what I need to do is to equate the potential energy delivered by the dropped part to the energy absorbing capacity of the beam at rupture. Can anyone point me to a method for doing this?

Calculate the internal work your beam does reaching yield. Call this A.

Calculate the internal work your beam does yielding say another 2 inches. Call this B.

Calculate the change in potential energy of the object between the initial position, and the final position (including yield deflection). Call this C.

If A+B > C, your beam will yield less than 2 inches.



Make sure the peak load the beam will experience won't break anything else. The stiffer the beam the higher the shock loading.

 

[HTURKAK]

We just know the wt of impacting object 13,000 lb and fall height 1 ft.

Provide more info. to get more useful responds..What is the shape , material of impacting mass ? is that a steel sphere ? Why a simply supported single steel beam resisting to impact? A sketch will be helpful to see the case.

 

[Mohanlal0488]

Hi Jmggks

I have done a similar design on numerous occasions.
Previously I have also developed a linear time interval over which the force acts and thereafter used my own FEA code to perform the analysis transiently as this also gives a decent indication of what happens after the impact. This type of analysis only proved to be useful for certain circumstances.

I am not sure of the process requirements, but why don't you look into abrasion resistant steel and high yield strength steel, these are used widely for the design of static screens/grizzly's.

 

[r6155]

Be careful with CHARPY impact test, absorbed energy 20 ftlbf (27J) a test temperature of 70ºF (21ºC).
See ASTM A6

Regards

 

[jmggks]

Tomfh - can you elaborate on how to calculate the internal work, especially past yield?

For a W10x49 spanning 16 ft. modeled as simply supported, I calculate an elastic spring rate of 53494 lb./in., a deflection at yield of 1.063 in., so energy absorbed to initial yield = 0.5 k x² = 0.5*(53494 lb./in)*(1.063 in)² = 30,234 in.lb.

(13000 lb.) * (12 in. + 1.063 in.) = 169,819 in.lb., so at yield, the beam has not absorbed enough energy to stop the part.

How do you calculate the internal work to deflect another 2 inches? Is that simply the load to cause initial yield * 2 inches, or are you considering a tangent modulus for steel once you are past yield? Also, how do you know that another 2 inches of deflection has not ruptured the beam?

Thanks!

 

[bones206]

I posted a derivation on this thread that might be useful: thread507-461666

 

[jmggks]

Thanks bones206. One question - the deflection formula is based on elastic behavior over the entire cross section, so it is only valid to initial yield, right? I am thinking to do an analysis similar to what you linked except using S instead of Z. The end result doesn't change much.

 

[Tomfh]

How do you calculate the internal work to deflect another 2 inches?

It’s your yield Force (57 kips?) multiplied by 2 inches. 114000 inch pounds.


It’s the same concept as 1/2 kx^2, the only difference is you are now calculating the area under the post yield rectangle, vs the pre yield triangle.

 

[bones206]

In my simplified assumption of elastic-perfectly plastic behavior, the assumption is elastic behavior right up to the point of full plastic behavior.

 

[dhengr]

Jmggks:
The idea of a grillage of beams, as suggested by PhamENG (6MAY21, 22:10) is an important concept. You obviously have to distribute this impact load to 5 or 6 reasonably sized beams to start to make it work, and that second spanning layer will help do that. You haven’t said anything about the size and shape of the 13k piece and how it might fall off the conveyor, 12”, as a point load on one corner, as a line load, or fairly much as a distributed load, etc. You do want fairly large deflections in the extreme case, and the start of yielding, because this reduces the magnitude of the impact loading and absorbs the energy over time and distance. One layer can yield, as long as the other layer holds things up, and together, as deflection progresses; and of course, at each node the two crossing beams have the same deflection. Now, analyze this as a grillage of beams acting in two directions, and you can imagine the system starting to take that loading.

We know nothing about the conveyor or the room below it which is being protected. I would put one beam layer on either side of the conveyor and parallel to it. This could be a/the higher layer and would reduce the fall distance from the conveyor. The second layer would be immediately under the conveyor. You might also look at some sort of a guard railing system as part of the conveyor structure, which would prevent the load from moving laterally, and falling off the conveyor, a guiding/centering system. Have you ever had one of these pieces fall of the conveyor and onto the shop/facility floor? If not, what’s the worry, we don’t normally put a second major load carrying ceiling structure below the bldg. roof, to protect people, but sometimes bldg. roofs do fail.

 

[jmggks]

There have been some posts asking for more detail about the falling part, the conveyor etc. The conveyor carries finished weldments in a major ag/construction manufacturing facility. As such, there is a wide variety of shapes, sizes, weights, dimensions etc. The conveyor is a overhead power & free conveyor. Parts can fall, but it is extremely unlikely that they would fall above our structure - the conveyor there is level with no diverge or converge points. Problems are much more likely to occur at elevation changes or diverge / converge points.

The most likely scenario for a falling part above our structure is that a large part, suspended by 2 chains from the overhead conveyor, comes unhooked on one end and rotates downward, delivering essentially a point load where the "corner" of the part hits our structure. The customer has specified the magnitude of the load and 12" clearance / fall distance to our structure.

We only have 10" of height to work with for our structure because we are supporting another conveyor below from our booth roof. So I appreciate the ideas about multiple layers of beams but in this case the total depth cannot increase and the fall distance cannot be decreased.

My plan at this point is based on the suggestions by Tomfh and bones206. A W10x49 spanning 16' takes 57k to reach initial yield and deflects 1.06" in the process. If I assume perfectly plastic behavior from that point and another 3.19" deflection, then the total energy absorbed is (0.5)(57k)(1.06")+(57k)(3.19")= 212 in-k. The energy delivered by the falling part is (13k)(12" + 1.06" + 3.19")= 212 in-k. So at 4.25" deflection, the beam has stopped the falling part. Calculating the strain in a 16' long W10x49 deflected 4.25", it something like 0.005 so not even close to the strain limit of steel (around 0.2 from what I can find). The columns and beam end connections can easily be designed to take 57k / 2 = 28.5k elastically with normal safety factors.

BTW by this analysis a much lighter beam would work, but the budget includes W10x49s and the customer wants something confidence inspiring, so W10x49s it is. Much of the conveyor fall guard in the plant is supported by much lighter members, and this analysis suggests why that is workable.

Our roof will also have C6 on 2' centers running between the W10x49s (parallel to conveyor travel) and delta wire conveyor guard (

https://deltawire.com/products/conveyorguardmesh/)

on top of all of it, so the assumption of a point load on a single beam is conservative, and there is no way for a falling part to miss the fall containment.

Objections? Suggestions?

Thanks to all who responded.

 

[Tomfh]

The columns and beam end connections can easily be designed to take 57k / 2 = 28.5k elastically with normal safety factors.

You should design the connections and columns to resist the beam’s REAL strength, which will be more than the 57 kips nominal yield load.

Eg you said your beam is good for 82 kips at ultimate. In reality it might be 100+ kips (real ultimate is often a lot more than published ultimate). So it’s more like 100 kips (plus safety factors) that the connections and columns need to resist.

 

[bones206]

Yea the dynamic reactions are going to be greater than static weight divided by 2. Since the impact factor (aka dynamic load factor, or DLF) is unknown, it’s good practice, as Tomfh suggested, to design the connections to develop the full beam capacity, with due consideration of material overstrength (SIF or ASF) and strain-rate effects (DIF).

These factors for various materials can be found in the literature. A good one that is freely available is the UFC 3-340-02 blast design manual.