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Force of a 2x4
2

Force of a 2x4

Force of a 2x4

(OP)
In Florida, for the large missile impact test, a 2x4 (between 7' and 9' long), weighing between 9 lbs. and 9.5 lbs., is shot from a cannon and strikes the surface of a hurricane shutter.  The distance from the end of the cannon to the hurricane shutter is 9' plus the length of the 2x4.  The 2x4's speed at impact is 50 feet per second.  What is the force with which the 2x4 strikes the hurricane shutter?

Using a 2x4 length of 9' and a weight of 9 lbs., I come up with 19.42 pounds of force, which does not seem correct.

Thanks!

RE: Force of a 2x4

It seems to me the force would be dependent upon the deflection characteristics of the 2x4 and of the shutter at impact, and there would be no way to calculate it with the information given.

RE: Force of a 2x4

JStephen is right.  The force will be much higher if you're hitting a brick wall, than if you're hitting a pillow (just for extreme ends of the spectrum).

F=MA, you know the mass, but the acceleration is dependent upon the stiffness of whatever it's hitting.

RE: Force of a 2x4

You have to assume some type of de-acceleration meaning you need to guess how much the shutter will deflect.  That will give you the -A you need.  Try 1'' then 3'' and then 6''  See what you get..

The actual answer can only be found through actual testing.  Thats why Ford and Chevy and BMW wreck cars all the time....IF they could figure it out through computers or even hand calcs - they wouldn't be wrecking multi-thousand dollar cars.

RE: Force of a 2x4

(OP)
JStephen, PAStructuralPE, MiketheEngineer:

Again -- thanks!

RE: Force of a 2x4

your calc (of 2g) does look light.

as you say, you know the mass, and have a pretty good idea of the velocity at impact, but you need to estimate the duration of the impact which is influenced by the stiffness of the impact surface ... a window shutter sounds like it would deflect some (absorbing some of the impact energy) and increase the duration of the impact

RE: Force of a 2x4

I have checked for impact force before, and I usually pick a very small impluse interval so that the force seems reasonable like a tenth or twentieth of a second. Also, I just assume that the final velocity is zero (just hits the wall and drops, not accurate but gotta use something maybe if it is a hard wall I would give it a backwards velocity but it would still be just a guess)

I usually get around 250 lbs and than I double it because it is an impact load (so I can use it with a static analysis).

RE: Force of a 2x4

Well, unless this is some professor's idea of a (useless without approximations and assumptions!) exam question, the real answer has to come from testing.

But a useable velocity doesn't need to come from a cannon:  A 2x4 dropped vertically should reach terminal velocity that in about 80 - 100 feet.  You'd need a clear span and a safely restricted area below to prevent injuries.  Then test a few - back-derive the assumptions (stiffness, movement/deflection/times you need to make the real calc's with the legally-required numbers.

RE: Force of a 2x4

You going to need calculus.  Of course, you still remember how to do integration by parts, right?

RE: Force of a 2x4

It is going to be different depending on whether it hits at a support or in the most flexible spot.
For the most flexible spot, calculate the "spring rate". The energy equals the spring rate by the deflection squared divided by 2. The maximum force equals the spring rate by the deflection.

For the most rigid point, it may well be the crushing strength of the 2x4.

Michael.
Timing has a lot to do with the outcome of a rain dance.

RE: Force of a 2x4

ash060

250 pounds equiv static load seems very very low to me.

I imagine that an equivalent static load is several hundred psi over the 2x4 surface area.

  

RE: Force of a 2x4

I usually double the static load I get for impact, so I would use 500 lbs.

Again it is just an estimate, but something is better than nothing and I used some numbers to get to an answer.

RE: Force of a 2x4

Using conservation of momentum and assuming the time it takes for the 2x4 to come to a complete stop = 0.1 sec, I came out with a force of 148 lbs. Since it is a dynamic force, a Dynamic Load Factor would have to be applied [and that could be as high as 2.0].

 

RE: Force of a 2x4

The problem is determining the time it takes to stop.  Where did you get 0.1 sec?  If it takes 0.05 seconds, which seems just as arbitrary, your force is much greater.

Seem like a test is the sure fire way to get an answer.

RE: Force of a 2x4

Quite true IceNine. My guess comes from previous experience with impact....but it may not be applicable here. In fact I have seen delta t as low as 0.0001 sec. So you are right: testing is the way to go. I just wanted to respond and make note that 19.42 lbs does seem way off (as the OP suggests).

RE: Force of a 2x4

I have yet to encounter a client that will pay for such a test.

RE: Force of a 2x4

If it was only 20lbs you wouldn't need shudders...

Someone said IMPULSE, which is the key term for impact forces. I am too rusty at my mechanics but like others have said you would need to know the exact time of the deceleration. I would think from watching Myth Busters you would need a high speed camera with a timer on it, then you could determine the time from initial impact to the point the 2x4 stops moving forward. Then you could plug this time into your impulse formula to determine the force.

I think ;)

RE: Force of a 2x4

'cept the time of the impact is affected by the stiffness of the support (ie the shutter).

if you're trying to pass a design, then testing would do it (and be some fun as well).

I imagine your code says something like "the shutter has to withstand the impact of a 10lbs 2x4 with a velocity of ?? ft/sec", a bit like our birdstrike requirements.

RE: Force of a 2x4

50 fps sounds a bit low, that is not even close to hurricane force winds. Even a Cat 1 has sustained wind speeds up to 95 mph. I would expect something over 100 fps.

RE: Force of a 2x4

I'll take a stab at it.  Let me know if my reasoning is flawed.

If you know how far the shutter deflects, you can figure the time. Assuming constant deceleration(probably wrong), the average velocity is 50 x 12/2 = 300 in/sec. So t = x/300 sec. Your acceleration is 50ft/sec / t.  For 1" deflection would take 1/300 sec.  Acceleration a = v/t = 50 ft/sec / 1/300 sec = 15000 ft/sec^2.  F = weight x 15000/32.2 = 4192 lb.  If it deflects 10in, F=about 420 lb.  If the shutter breaks, who knows?
I guess if you know the max allowable deflection of the shutter, you can guess the allowable speed or wind speed it can withstand (for a given wt. of projectile).  If the shutter behaves like a beam (or a spring), the accelleration would be variable. Your max. acceleration and force would be twice that calculated.

RE: Force of a 2x4

olineemg -

Since you are in Florida and mentioned an "air cannon", there must be a concern about debris penetration and not structural since most of the life-safety requirements are based on the testing conducted in Texas (Texas Tech?) where they have been testing wall and some window/door/shutter assemblies for use within the FEMA guidelines that are for life safety and not based on numeric analysis. There is a volume of test that have been performed for by groups and associations that want approval for standards inclusion. The standards are very strict - not penetration for walls after a 12' long 2x4 fired from an air cannon at 145 mph. For a long period only 2 wall assemblies were approved, but finally, a laminated wall of layers of 3/4" plywood on 2x6 studs and steel plate was included.

The basis is practical performance and not assumed structural calculations.

The earlier post about the fun of the testing was very accurate. the fun was not an exaggeration. The people testing verified the construction, assemblies, anchorage and took great pride in separating those that met the criteria from the others, but I imagine all test information is test available. None of the failures were disclosed publicly.

Other testing was done for doors and windows with a lower criteria, but could be valuable.

If I could take a month or two off to participate, it would be a real learning experience.

Much of the testing is the basis for coastal code requirements that are part of the regional/local codes and the heavy insurance discounts in addition the FEMA requirements.

Dick

Engineer and international traveler interested in construction techniques, problems and proper design.

RE: Force of a 2x4

Dick:
Shudders or impact resistant glazing is required in Florida for all wind-borne debris regions, see attached map.

I would not disagree with your assertion, but I would think the bigger reason to protect the openings is to maintain the envelope so the structure remains enclosed rather than partially enclosed. Garage doors, windows, and doors are the achiles' heel of structures during wind events, especially residential structures. You could add localized roof failures to the list also, once a couple of jack trusses fail then the roof is in trouble.

If people listen to the authorities, they should not be in the wind-borne debris region when a hurricane hits. That is a big IF.

I think you are referring to a "hardened" structure that meets specific FEMA guidelines for use as a shelter. I assume the OP's shudders are for general use in wind-borne debris regions, and Miami-Dade is usually the litmus test for standard wind-related product approval, though maybe now Florida has adopted S FL's standards state-wide when it comes to product approvals. That is a whole specialty area of engineering I am not involved with...

And I am in the same boat as you, I would love to do this type of testing!

RE: Force of a 2x4

The Texas tests were the basis showing up in the hardened FEMA structure standards for zero penetration. They were altered downward slightly for other situations/materials in the coastal code areas, because the conditions in Florida were not as severe.

The insurance discounts (40-60%) offered in selected Florida areas were for structures built beyond the minimum code standards that are slow to change. One interesting thing was a weak point in the codes for the gable end of structures was found in damage assessment. This was almost as critical as the openings. Obviously, they also require some sort of additional anchorage of the garage door, which is major cause of roofs going and opening up everything to further damage.

It is interesting to see the window testing. While attending a door and window installation certification class they were testing windows for penetration in the same lab and I saw the same happy smile was on the face of the technicians there as I saw in Texas. It is great to get behind the scenes to see the basis for the codes and standards.

Dick

Engineer and international traveler interested in construction techniques, problems and proper design.

RE: Force of a 2x4

(OP)
Thanks all for the time and thought.

Ash060, in this case, you are right, in a sense:  the client has no plans to pay US for the testing.  Our goal is to assist in the design of shutters that will pass muster when the client pays a testing facility to do the test.

Concretemasonry, you are right -- would be fun.  Kind of reminds me of that great old Letterman segment, "Dropping Stuff from a Five-Story Building."  Maybe we could get Letterman to do it and save everyone a bunch of bucks.

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