Wind Damage

Enough to bend the hinges....yuk yuk

You have to add in the momentum of the swinging door....kinetic energy....

"p = 0.0032 * v^2 where pressure p is in pounds per square foot and v is miles per hour. For pressure in psi divide by 144 to get; p = .00002222 * v^2 So if v = 60 mph, then p = 0.08 psi. "

2880 sq inches * .08 psi = 230.4 total lbs force ....probably not uniform though...

https://www.physicsforums.com/threads/what-psi-can-be-achieved-from-a-60mph-wind.698861/

This was the values I came up with, but because the door was on hinges, what was the torque acting on the hinges. And I also wanted to consider the impact torque when the 30# door swinging at a peripheral velocity of 60 MPH came to a sudden stop.

F=mv^2/r

Here in Michigan, we had issues with the wind during that storm. We had 800,000 people lose power over the wind, some people lost power for 7 days...

You may not want to hear this, but recent CR4 posts about torque, I calculate, 60mph winds hitting an approximately 20 square foot door are equal to the torque needed to remove the scales of about 17 fish in the 18 to 22 inch range.

Seriously, it was close to 250lbs as Solar Eagle said, my storm door is aluminum, if yours is too, then 250lbs is enough to rip it off.

Didn't rip it off, but bent the crap out of the hinges, and turned the rods and mounts of the door closer into pretzels. Lucky, that was all the damage we got. Now all I have to deal with is 14" of slush that froze into concrete the following night.

"Here in Michigan, we had issues with the wind during that storm. We had 800,000 people lose power over the wind, some people lost power for 7 days..."

Uneventful day here. Light winds 40 - 70 MPH. Damn near tolerable to be outside flying a kite on a day like that in these parts!

Small kite, big twine.

It was uneventful for me too, my electricity never even flickered. I considered going fishing, with 60mph winds behind me, I can cast a spoon 1/2 a mile.

Wind is the #1 reason the majority of our local scale utility power system feed lines are underground now.

Farmers with big field equipment not watching where they are going is #2.

Unless the hinge is seized, the torque on it is near zero. (That being the whole point of a hinge.)

You are right. I should have said the moment given that the hinge is about 1/4 inch from the point of the frame that limits the 36" door's swing.

BINGO!!!!!

It's actually worse than that. 234 lb was just the wind force when it came to a stop.

As the door opens, the wind force increases as the area exposed to the wind increases (proportional to the sine of the angle), applying a torque to the opening door equal to force times half the projected width of the door (also proportional to the sine of the angle). The amount of angular kinetic energy imparted to the door is this torque integrated over the angle, 0 to 90 degrees. (Rotational kinetic energy equals torque times angle.)

When the hinges and door closer attempt to stop it, they have to apply a counter torque to the swinging door to stop it within a small angle. Again, rotational kinetic energy equals torque times angle. The smaller the angle, the greater the torque. This inertial torque is in addition to the torque from the wind force.

Since the door closer and hinges are so close to the axis of rotation, the forces on these are very large. If you estimate the stopping angle of the door, you can calculate the total torque required to stop the door and resist the wind force. If you divide this torque by the distance from the hinge to the door closer, you can calculate the forces on the hinge and door closer.

Around here most people (who know better) use double heavy duty closer units. One top, one bottom with the stationary brackets mounted to have maximum leverage plus anchored with 3 - 4 inch screws all the way into the wall framing.

On the door the screws go all the way though and either have pan heads or are part of a decorative steel plate.

Also with the catch chain that's anchored as far out on the door and frame as it can be placed without being in the way and like the closers well anchored at each end as well.

Basically they'll blow the glass out or fold the door in half before they let go!

Depends on the quality of the door and hinges.

Say I do a head on collision between my Ford F250 pickup and a geo Ford Festiva at 60 MPH.

Which one sustains more damage?

Depends on how new your f250 is.....but generally an f250 is likely to sustain far more damage.

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Really.

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Okay, calm down. No need for hostility.

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A ford Festiva typically caps out at less than $2000 total value. Can't damage value that isn't there. There are many many f250s on the road easily worth more than $2000 which would sustain more than $2000 in damage running over a Ford Festiva at 60 mph,

Physical damage. Financial value has no meaning in physics.

Oh. Physics, huh? I was pretending we were considering it in an engineering context.

Fair enough. What are your units of damage for comparison?

Same as anything else. Its relative to the overall value of the object or system as a whole.

Spending $50 to fix a broken $30 remote control isn't worth it but spending $50 to fix the $4000 60" commercial grade HD monitor it goes with is.

As I see things with the Oroville dam its both a active and very necessary flood control system plus ~820 Megawatt power station which in my books puts its value in today's economy and other politically, environmentally and socially driven aspects at a pretty high number.

High enough, that my views, that even at a Billion dollar major renovation and rebuild price tag that guarantees its safe operation for the next 100+ years it would be well worth it.

It needs repair improvement and it is very much a justifiably fixable issue.

Forgive me for being slow here, but didn't you just finish declaring `....financial value has no meaning....'?

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Perhaps there are some caveats you could explain.

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Which incurs more damage from 1000 rounds from a 50 bmg ...a warehouse full of fine china or a warehouse full of red plastic solo cups?

Same relativistic results. I wouldn't pay a dime to own either and thusly financially they still hold zero value when calculating the energies expended in either case.

Now however if I got to do the shooting I'd pay to be the one pulling the trigger regardless of which warehouse I was pointed at.

Fair enough.

The Ford, obviously.

OK, answer this one:

from a dead calm, a 100 mph wind gust hits a Ford F250, and also a geo Ford Festiva side-on, and both are parked at the edge of an embankment (for drama). Which one takes off rolling down the embankment, and how many times does it roll before it hits the bottom, where the embankment is 45 degrees, and is 500' hypotenuse?

Neither rolls over, the 500' hippopotamus stops them from rolling.

Ahh, that was hypotenuse not hippopotamus.......

Somebody been eating too many donuts.

Well then its the Cavalier. It rolls 23 times (amount of years in production).

Ding! Ding! Ding! Ding! Ding! Folks we already have a winner!

I even gave you one GA, before you edited that post to include the number of rolls. By the way, The Cavalier did no donuts in the making of this commercial, since you can't spin out in a Cavalier, or a Vacalier.

I beg to differ, my ex wife owned one before the deer tore it apart, it had the 2.4 liter and was a blast to do donuts in reverse... Now that we are divorced I will admit to the reverse donuts...

She never understood while even after a front end alinement, her front tires only lasted a year and a half, hmm..........

Dear Mr.gringogreg,

The pressure created by wind is given by the Equation P = (0.004) x V^2

where P = Pressure created by the blowing wind in Lbs./Sq.Inch, V = Velocity of wind in Ft/Sec.

You have mentioned that wind speed is 60 Miles/Hr., and Door Size is 36"x 80" size.

The wind speed 60 Miles/Hr. works out to 88 Ft./Sec and hence the pressure experienced by the door is P (in Pounds?Sq.In) = (0.004) x 88^2 = (0.004) x 7744 = 30.796 Lbs/Sq.Inch = 4436.624 Lbs/Sq.Ft

Door Area 36 x 80 = 2880 Sq.Inches.

Therefore Load on the Door = Area X Pressure = 2880 x 30.796 = 88,692.48 Lbs. = 39.59 Tons which is very huge force for this door.

DHAYANANDHAN.S

Haha... can I have you do my tax returns?

Are you sure about your math? I live in a mobile home. It is 70' long by about 11' high to the roof, which is about 120,960 square inches. Using your math, the 60mph winds we experienced last week put 3,414,660.48 pounds of force on my mobile home.

Umm, no, that much force and I would not be talking to you, I would not be here. My house wasn't even moving during that wind.

88,000lbs on my storm door and you might find it on the other side of the planet.

I didn't even lose a piece of skirting, at 16" x 24", with this math, it would have had over 11,000lbs on it.... Nope....

I'm thinking you are off by a factor of 144. I believe your formula calculates pounds per square foot, not psi.

I suspect you guys may be missing something. When I was in school many moons ago, we studied something called, I think, "impulse and momentum." I totally forget the equations, unless I dig out my old physics book or Google it. But I'm pretty sure it had a Δt in the denominator. As Δt gets small, the force can become huge. Thus, the quicker the storm door stops, the bigger the forces on the hinges, closer, and whatever else. I looked at this for my high jumping (in a big school I wouldn't even be on the team), and the force on my heal was tremendous. No wonder I broke several of the "unbreakable" Fiberglas heel cups!

You can look at momentum and time (momentum/time = force) or energy and distance (energy/distance = force). Most of the time, after the fact, you don't know about stopping time, but you can figure (or guess) stopping distance.

The swinging door gets kinetic energy from the wind:

E = ∫τ dΘ, where Θ is the angle of the door, and τ is the torque from the force of the wind on the door which is pivoting about its hinges.

The torque is: τ = F x r where F is the wind force and r is the distance it is applied relative to the hinges. If the door is width w and is open angle Θ, r will be

r = w/2 sin(Θ).

Force, F = P w h sin(Θ), where P is the wind pressure, w is the width of the door, h is the height of the door, and Θ is the angle the door is open.

So torque, τ = F x r = P w h² sin²(Θ)/2 and kinetic energy is the integral of this from Θ = 0 to π/2

E = .5 P w h² ∫sin²(Θ) dΘ = (π/4) P w h²/2

∫sin²(Θ) dΘ = (π/4)

E = (π/4) P w h²/2

Now to get the torque applied by the hinges and door closer to stop the swinging door, divide Kinetic energy by the stopping angle (in radians). The smaller the stopping angle, the greater the torque required to stop the swinging door.

Add this stopping torque to the torque from the wind force to get the total torque. To translate this torque to force on these components, divide total torque by their separation.

Now would this be enough torque to scale a fish?

Rixter: Thanks; shows how much physics/math goes away in 50+ years when not used much.

I wish to thank you all for this fine lesson in mechanics. I now have a better appreciation for the destructive power of the wind. In the future, given the calculations you have so generously provided, I will make sure that my storm door is closed and tightly latched when the next storm blows in... That is, once I have it replaced or, hopefully repaired (if the hinges can be replaced on this model).

Go full length....

Oh, wind its actually complicated if you would be so conservative thinker. You can draw a free body diagram on this.

Vertical load 50 lbs

Horizontal =? You can actually tie a spring scale and a rope measure displacement and wind velocity simultaneously. Now you have a better experimental empirical formula to use, assuming relationship of wind speed and force at considerable amount is proportional.

Sounds like the real answer is $239.95 for a new storm door...

That's my next problem The door itself is not damaged, just the hinges. I don't have any information about the make or model of the door (probably Anderson from 1985 or so when this condo was built. This model has a screen that rolls up into the top of the door). The hinges can be removed from the door, but I can't tell if they can be removed from the frame. There are two screws through the frame to each hinge, but there are also what appears to be two rivet heads.

I'm going to do an internet search, and perhaps a trip to Home Depot in the hope that I can replace the hinges.

Nah! do not replace the hinges. Just take the door off, and install it on an outhouse, after you lay it down and hammer the crap out of the hinge area until it lies flat.

Hint: make sure to underlay the door with a plate of steel to keep from mashing the hinges flat, or hammering some new pattern into the door surface you don't want.

Either that, or shell out for a whole new door, go cheap or go home.

Most storm doors are in fact 100% recyclable by weight.

The hinges on mine broke years ago, being integrated in the door and frame were not replaceable. Think frame with rolled edge with 3 slots cut in it and door with 3 rolled spots with dowels in them in the rolled part of the frame.

I put shims in the door and frame to line it up, cut the old dowels out, and zip screwed 3 heavy duty hinges in and painted them flat white. Good enough for an older mobile home, only other option was replacement of frame and door.

New doors on a trailer house. Like putting a square peg in parallelogram shaped hole.

I'm planing to do 3 - 4 new windows on mine this week before the new siding goes on and I have every reason to suspect that every one will require at least one shim or spacer board on one or two sides hand planed to fit followed by most of a tube of caulking to hide/seal it properly anyway. Everything else that has been replaced required it and no two items were skewed the same direction and amounts either.

"...only other option was replacement of frame and door."

When you consider the time and labor invested, that's probably the cheapest way to go.

Where is Doorman when I really need him?

Just a follow up.

I was able to track down the door's manufacturer and went to their web site. lo and behold the web site had a service page where they explained how to find the model and serial number for my door. Entering this information and description of the damaged components, they promptly replied via e-mail with the correct part numbers for my specific door, and instructions how to to perform the repair, and directed me to a page where I could order the parts.

Total cost $60 vs. $300 for a new door

The internet can be a wonderful tool if one makes the effort to use it.

Good on you!

Reminds me I need a new drive belt for my 45 year old Snapper self-propelled (right now it is manual propelled) lawn mower!