Impact Loading 1

[andyfabian]

I have to design a cage for a tiger in a local zoo. The cage fencing is suppose to be some sort of wire mesh. I'm trying to determine what loads to use for a 600# tiger. The assumption is the tiger will run at the fence and hit it a specified velocity.

My boss wants me to use Fd = 1/2mv^2. He said assume a deflection of 2" in the fence on impact.

After starting the job, I realized that something didn't seem right. I worked it out on paper and determined that simply utilizing Fd = 1/2mv^2 is wrong. The distance simply can't be arbitrary.

Therefore, I used the Work-Energy relation to develop the following:

1/2k(X2^2 - X1^2) = 1/2 mv1^2

This now gave me a relation between the tiger's mass and velocity with respect to some stiffness (k) in the wire mess. The deflection would be simply = X2 - X1.

The problem now is two things:

1) The left side of the equation is developed from F = kx, which is NOT valid for deceleration effects. Therefore, the equation is more complex, and I do not know this solution.

2) More importantly, my boss is becoming loosy goosy about it and is throwing around arbitrary numbers to use. I am not comfortable with this at all.

This now leads to my question. Is there a code that simply outlines the requirements for cages or containment structures similar to this? There has to be something that helps an engineer not have to re-invent the wheel each time this comes up.

Thanks for help in advance.

 

[MotorCity]

andyfabian,

I would take a different approach to the design.

First, you may want to investigate loads other than the animal that might control the design. For example, is the cage exposed to wind, snow, etc. Also, do vehicles or other mobile equipment frequently work in or near the cage?

If it turns out that the controlling load is the tiger, I would apply an impact factor, say a 25% or 30% increase.

I am not aware of any design aid or code that governs this type of design.

Good Luck

 

[IRstuff]

Your boss gave you a reasonable approach.

The tiger's kinetic energy is constrained by the spring behavior of the fence, hence the integral of -kx*dx is set equal to the kinetic energy.

TTFN

 

[andyfabian]

IRstuff - please clarify. If you re-read what i wrote, i was the one that came up with the work-energy relation. The idea of the fence disapating the energy as a spring would, is my idea. I think it makes sense, except for the fact that the equation assumes that there is no change in acceleration.

I know that sounded a bit immature, but that's not how i mean it. I just think you are mixing things up; therefore, i am confused about what you mean or which idea you agree with (regardless of whose it is).

So, assuming you are actually agreeing with me and not my boss, are you certain that acceleration can be ignored?

And, if i use the work-energy relation, i still have a messy problem with a minimum budget.

Finally, what my question really pertains to, is if anyone knows of a code that would eliminate re-inventing the wheel?

I know i could simply increase the load arbitrarily by 100% for impact, but I don't know if that's correct.


Thanks.

 

[ishvaaag]

I think you can start from M·V=F·t that rules quantity of movement. You know mass, you can decide on a maximum V, and from this V, and the maximum penetration of the bars unto the body of the tiger when it jumps against the bars (say between 1 and 12 inches) you can guess t (since V diminishes from V to 0 one would calculate the time it takes to run the decided length of penetration at V/2) and so get a set of suitable Forces (following hypotheses) that dissipate the initial quantity of movement M·V in time t.

The running deflection of the bars will have more to say when yu assume a short dissipation path than long ones.

From the more conservative estimates of the forces you can proceed to design the bars.

 

[IRstuff]

I read your posting:

"My boss wants me to use Fd = 1/2mv^2. He said assume a deflection of 2" in the fence on impact"

This means setting 1/2mv^2 = 1/2kx^2 assuming a 600lb tiger and 40mph, you get k=33E6 N/m, a huge number, but a solution nonetheless. This results in a maximum force of kx=1.7E6 N over the impact area.

The acceleration term is inherent in the integration of kx*dx=1/2kx^2.


TTFN

 

[JWB46]

Hi Andyfabian. I think IRstuff and your boss are right. Let the kinetic energy be absorbed by strain energy in the cage structure. Better still if you treat the maximum possible velocity case as an extreme event, you can also take allowance for plastic deformation along with elastic strain energy. (That's how we design boat fenders on offshore platforms etc.) Or if you have the time and are allowed by other constraints, design a net-like structure that will deflect more but will easily absorb high strain energy and remain elastic.

Hope this helps. (- Wish I had some interesting jobs with tigers etc.)Good luck.

 

[ChipB]

Hey Andy,

I'm with JWB46 in wishing I had a cool design project like that.

Your acceleration is a function of the spring constant (obviously). If there are cross bars as well, I'd put 50% of the force one bar and assume the next two adjacent bars take 25% each.

To get the sprink constant k, I'd place a force at midspan and increase (or decrease) until the midspan deflection was 1". Whatever force that is required to cause this deflection, I'd use as my spring constant in kips/inch.

You then have 1/2MV^2=1/2kX^2 (I think)

 

[jheidt2543]

Lions and tigers and bears, Oh my!

You could call PETA, but that is probably not a good idea. I would call some zoos or zoo associations and ask about design standards for animal cages. In fact, I would start at the top and try for Marlin Perkins, Jack Hanna or the Crocodile Hunter - really!

That's one of the neat things with this forum, we get to hear about all the fun projects everybody else gets to work on.

 

[sacem1]

I would start from the other end of the problem, how much pressure can the paw, leg, head or jaw bones transmit to the cage, if the poor tiger crushes itself to death and breaks the cage he will be of no danger so instead of thinking the animal is suicidal, which I think up to now is reserved to us humans and whales, why not research the force that can transmit the tiger before pain stops him, now if you are going to propell the tiger with a canon at 40 mph against the cage then you have the problem of recovering all the grinded tiger meat...
SACEM1

 

[JoeTank]

Odd image... I wonder at what speed would the tiger rice itself on the screen?

Steve Braune
Tank Industry Consultants

http://www.tankindustry.com

 

[IRstuff]

I don't think that pain will be what stops an animal already moving at 40 mph.

Besides, that might be all quite moot. Given that a tiger is essentially a rather large cat, I'd be more concerned about whether the darn fence is high enough.

TTFN

 

[rijse]

The usual maximum impact factor that may be achieved is 2, i.e., if you use 2 times the load for your design you will be on the safe side. Due to the spacing of the grill, it is obvious that the tiger will strike more than one. In additon, the grill will act as a two-way structure depending upon its aspect ratio.

 

[andyfabian]

Thanks rijse,

That happens to be the exact approach i have taken.

 

[dbuzz]

What an interestnig project! It certainly breaks the mould for our usual work.

You'll probably find the design is governed by some local or state statutory authority.

Good luck! The consequence of failure in this case is scary!

 

[ThomasH]

Hi

If you could drop the tiger, freefall, into the cage grill you would get a impact factor of 2 if I ignore the tigers "deformation". I don't think the tiger can outrun freefall so this i probably a limit in some respects.

Also, just a thought, in order for the grid to "look" strong enough it will probably require a grid with such dimensions that the "numbers" might be of less importance. It's a big tiger in there after all.

But I must say it's an interesting project you got on your hands.

Good Luck

Thomas

 

[ThomasH]

Oops, I didn't see that rijse had already mentioned the "factor 2 method". Sorry about that. Instead I agree with him/her.

Regards

Thomas

 

[flamby]

What is this "factor 2" method ThomasH/rijse apart from being a crap? See Greg Locock's excellent arguments on this in the thread below.

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