How do I calculate the damage on a lightning pole from a car impact?

[Dalit]

Hey,
I would like to design a lightning pole to be installed on the street (or in parking lots).
The height of the pole is about 1m.

I would like to calculate the impact of a car heating the pole in order to evaluate the damage to the pole (and to the car).

I know the car mass (1000-2500[kg]), and the car velocity (10-30 [km/hr]).
Therefore I can calculate the kinetic energy at impact.

I would like to know how do I calculate the damage on the pole, or what are the forces and energies it feels.

TIA, D

 

[Dr.D]

You have to be a lot more specific about the nature of the impact to even start making reasonable guesses.

 

[Dalit]

The scenario is a car heating a steel pole.
The Pole height is about 1m. The car's speed is between 10-30km/h.
I would like to know the impact on the pole.
The pole is fixed to the pavement on a concrete tile with 4 screws.

Does that make more sense?

 

[Claws]

The scenario is a car heating a steel pole.
The Pole height is about 1m. The car's speed is between 10-30km/h.
I would like to know the impact on the pole.
The pole is fixed to the pavement on a concrete tile with 4 screws.

Does that make more sense?

Before you can begin to assume a realistic scenario, you'd want to know how much of the energy is absorbed, by the car crumbling. Then you could probably guestimate the forces acting on a beam (the light post) then moving on to the forces exerted on the four bolts.

 

[Dalit]

How do I know the energy absorbed by the car?
I guess each fender have a different energy absorbtion.

 

[Dr.D]

You need to specify how high off the ground the contact occurs, what the mass distribution of the steel pole is, the stiffness and energy absorption of the front end of the car (as mentioned by Claws), etc.

Does the car strike the pole on a fender, on the bumper, is it head on or at an angle, etc.?

 

[Dalit]

The car will hit the pole by it's fender, can I find an estimation of how much energy the fender absorb?
The pole geomentry is a rectangular, but it's installed in such a way that the fender hits the corner of the rectangular.
The pole material is SAE1020, how can I calculae the profile stiffness and energy absorption?

 

[Dr.D]

You still did not mention how high off the ground this impact occurs.

Why don't you go push on a few fenders and see how stiff they are? I think you will find that some are stiffer than others, so which fender is going to participate in this impact? The energy absorption depends on the structure supporting the fender as well as the fender itself; have you identified that?

The whole point is this: The problem you pose is not a simple problem at all. Consultants get paid very, very well to solve this problem for specific cases where all of the conditions are very fully specified. They do lots of FEA modeling and employ much proprietary data on energy absorption. You will get no general solution to an ill defined problem; it does not exist.

 

[Dalit]

the impact occurs between 35-60 cm above ground

 

[Dalit]

Dr. D
When I install a pole in a parking lot, there are many types of cars going in and out.
I don't know which will tackle my pole, and I am aware of that the fender energy absorption differs from one another.
However the material and geomatry of the pole is my design which I can calculate.
I would like to do some guestimate to have a general idea of what the pole will absorb.
The model should be same as a lamppost.
btw - do you know any o fthe experts you have mentioned?

 

[Dr.D]

Look under Consulting Engineers - Accident Reconstruction. They will burn through your design budget in a flash. Mostly they work for lawyers in settling law suits.

 

[Dalit]

I need engineering assistance.
Gladly I still don't have a law suit standing against me.
I was wondering with which tool I can module this problem.

 

[nvn]

Dalit: What are the cross-sectional dimensions (mm) and wall thickness of your lighting pole rectangular tube? What is the mass (kg) of the lighting assembly on the top end of your pole? What are the dimensions and thickness of your pole rectangular base plate, and what are the bolt hole spacing dimensions?

 

[xxChrisxx]

Just go for worst case scenario and assume an infinitely stiff impact, all the impact energy is absorbed by the pole and acts at the point of impact. That simplifies the problem, you can work out the stresses in the beam and then those at the bolts. This will give an over estimate on the damage to the pole.

From what I've seen of people running into parking bollards at relatively low speed (5mph) is that the concrete fails before the bolts do. Above that speed, well I just don't know what would happen.

 

[Dalit]

Hi nvn,
The rectangular external dimensions are 18X18cm. Wall thickness of the pole is 2.5mm.
The overall weight of the unit is about 20Kg.

The pole base is a concrete tile. It is assembled to the floor using 4 - 8X0.5inch nuts.
spacing dimensions are 30cm

Attached top view of the impact scenario as I see it.

 

[Dalit]

Chris,
I can calculate the energy at impact at different speeds.
I having trouble converting the energy into force that impact the pole.
considering worst case scenario that no energy is absorb by the car's fender.

 

[xxChrisxx]

You need to know how long the impact event takes. You can then use this along with the momentum to find the force.

Its called the impulse momentum theorem

F dt = d(mv)

 

[Dalit]

I can only guess that the impact will take 1 or 2 seonds.
I guess the velocity of the car will be zero after the impact.
So I have the force that hits the pole.
Can I use that and convert the problem to a static force problem with the calculated force?

 

[nvn]

Dalit: You say the overall mass is 20 kg, but what is the mass of only the lighting assembly, attached to the top of the pole?

In post 1, you said the pole height is 1 m. Is that correct?

You said the pole base is a concrete tile, but is the pole cast (encastre, encased, embedded) into the concrete? Or does the pole have a welded steel base plate, and the steel base plate is bolted to the concrete tile with four nuts? What are the dimensions and thickness of the concrete tile? What are the dimensions and thickness of the steel base plate? How is the steel base plate attached to your square pole, and the dimensions of this attachment?

 

[Dalit]

The pole mass, total weigth, include external case and electrical components is about 20Kg.
The pole height is about 75-80cm, a bit less than 1m, that is correct.
The pole is not casted it's bolted by 4 screws, the concrete tile is secured to the pavement by 4 5/8" nuts.
The concrete tile thickness is about 8cm and the steel base is 2.5mm.
The plate is soldered to the pole. The pole is attached by 4 screws to the concrete plate.

 

[nvn]

Dalit: What are the dimensions of the 2.5-mm-thick steel base plate, and what are the steel plate bolt hole spacing dimensions? What is the bolt size (shank diameter) of the four bolts in the steel base plate?

 

[Dalit]

The plate is 178X178mm.
Bolt spacing is 130X130mm (4 bolts)
shank diameter is 15mm.

 

[Dalit]

Attach is the pole design and the hitting scenario

 

[xxChrisxx]

I can only guess that the impact will take 1 or 2 seonds.
I guess the velocity of the car will be zero after the impact.
So I have the force that hits the pole.
Can I use that and convert the problem to a static force problem with the calculated force?

The impact time would be measured in tenths of seconds say .2-.5s (this needs to be measured empirically if you are doing it properly). As we've simplified the model to a rigid body elastic collision the car would transfer all of its meaningful energy almost the instant it hits the pole.

 

[Dr.D]

Dalit, the reason you are having trouble with this is that there is no definite relation between the initial energy and the maximum force developed. It all depends upon the details of the impact. One of the key factors is how long does the impact last. Is it 0.1 milliseconds or is it 10 milliseconds, or something in between? It makes a huge difference as to what the maximum force is. Remember that the impulse, the integral of the force of interaction between your post and the car taken over the length of the impact time, must be enough to bring the car to rest.

Let the speed of the car be V before impact, M be the mass of the car, and let the impact interval be deltaT. Then the average (not the maximum, but the average) force that acts between the post and the car is
Favg*deltaT = Mcar*v
Then
Favg = (Mcar*V)/deltaT

By GUESSING any number we like for deltaT, we can get Favg as large or as small as we want. Until you know something about the nature of the impact and in particular the time duration of the impact, you are flying blind on the force estimate.

 

[nvn]

Dalit: If you apply a horizontal static force to the pole (cantilever) equal to the vehicle weight, the rectangular steel tube is already greatly overstressed, and yielding. But your vehicle has an impact velocity of v = 8.33 m/s. Therefore, I currently estimate the dynamic amplification factor (daf) as daf = 43.22. Therefore, the equivalent static load, applied transversely to the cantilever at a height of 600 mm, currently appears to be P = 43.22(2500 kg)(9.8067 m/s^2) = 1060 kN. In other words, the results indicate the lighting pole will be annihilated, and this impact will probably cause very little decrease in the vehicle velocity.

Secondly, I currently predict the damage to the vehicle will be "relatively minor"; however, even minor damage to vehicles can be quite expensive to repair, because after repair of bent parts, usually local repainting is required.

 

[Dr.D]

nvn: Two questions, please.

1) Why would you begin by considering applying the weight of the vehicle as a horizontal static load? I just don't see any rationale at all for this.

2) How did you arrive at your dynamic load fact or 43.22? That seems incredibly high to me, and I'm really curious how you got there.

 

[Dalit]

Dr.D
According to your equestion:
"Favg = (Mcar*V)/deltaT"
V is the velocity of the car at impact?

 

[xxChrisxx]

The velocity at impact is whatever you state it to be...

I think the more important question is should you be designing this when you have to ask questions like that? I'm really not trying to bash you but this is all fairly fundamental stuff.

 

[Dalit]

The velociyt is actually the before and after, I am assuming the velocity after is zero.
I guess it's obvious.

 

[xxChrisxx]

Dalit: If you apply a horizontal static force to the pole (cantilever) equal to the vehicle weight, the rectangular steel tube is already greatly overstressed, and yielding. But your vehicle has an impact velocity of v = 8.33 m/s. Therefore, I currently estimate the dynamic amplification factor (daf) as daf = 43.22. Therefore, the equivalent static load, applied transversely to the cantilever at a height of 600 mm, currently appears to be P = 43.22(2500 kg)(9.8067 m/s^2) = 1060 kN. In other words, the results indicate the lighting pole will be annihilated, and this impact will probably cause very little decrease in the vehicle velocity.

Secondly, I currently predict the damage to the vehicle will be "relatively minor"; however, even minor damage to vehicles can be quite expensive to repair, because after repair of bent parts, usually local repainting is required.

That is an utterly bogus calculation and I am going to call you on it. Where did you pluck a DAF of 43 from. The load has nothing to do with gravity even if it WAS static, last time I remembered gravity acted downwards not horizontally.

 

[Dr.D]

Dalit, in answer to your question, right before I wrote the equation, I had written,
"Let the speed of the car be V before impact...

xxChrisxx, it was nvn, not Dalit, that pointed the gravity load in the horizontal direction and produced the dynamic load factor of 43+. I asked him about this previously, but he has not answered yet. Perhaps he still may.

 

[nvn]

Following are my responses to questions in post 27. Dr.D: (1) That is just the convenient way daf is quantified. It is a factor multiplied by a mass accelerating at 1 g, which happens to be the same as the vehicle weight.

(2) If we write a conservation of energy equation for the impact, and solve for the maximum deflection of the beam, the maximum deflection divided by delta (where delta = deflection caused by applying m*g in the horizontal direction), simplifies to daf = v/(g*delta)^0.5. However, this pretends the beam remains elastic. We already know the beam can go inelastic, even for a 1 g acceleration. Therefore, I agree that the daf value I posted overestimates the equivalent static load, but underestimates the beam deflection. The beam will go inelastic near the base, and may experience sudden collapse near the base due to local buckling. I also agree that a more significant decrease in the vehicle velocity, during the impact, might occur (than what I stated earlier), especially for smaller vehicles at lower impact velocities, if the beam near the base does not experience local buckling.

 

[Dr.D]

nvn: thanks for your response regarding your approach to the dynamic load factor.

Conservation of energy most certainly does NOT apply in any impact problem, and this one in particular. The approach assumes a fully elastic system, all the way to extreme deflections, with all of the deflection occurring in the beam and none in the striking body, the car. Given the nature of automobile construction, with crushable front ends, etc., it is unreasonable to treat the car as a rigid body while taking the post as a beam deforming elastically to the extent necessary to fully absorb all of the energy of the car.

I really cannot buy into this approach to a dynamic load factor calculation at all.

 

[xxChrisxx]

Dalit, in answer to your question, right before I wrote the equation, I had written,
"Let the speed of the car be V before impact...

xxChrisxx, it was nvn, not Dalit, that pointed the gravity load in the horizontal direction and produced the dynamic load factor of 43+. I asked him about this previously, but he has not answered yet. Perhaps he still may.

Oops.. reading comprehension fail, sorry Dalit.

nvn: But in this case its safer to assume no acceleration and the car is traveling at constand speed. The gravity still doesn't apply horizontally. Like Dr.D I am not in any way convinced by the approach you are taking.

 

[nvn]

xxChrisxx: I agree, the vehicle is traveling at a constant velocity, v, prior to impact. Acceleration/deceleration occurs during impact.

The symbol g can also be a unit of acceleration, equal to 9.806 65 m/s^2; it does not always refer to gravity.

 

[xxChrisxx]

If you are talking about a deceleration, its going to be much much higher than that for a crash. If you are talkign an acceleration then what you are describing is a supercar at maximum acceleration plowing into the post.

Either way the number is bollocks. There is a reason why gravity is given the symbol 'g' and any other acceleration is given 'a'. You clearly used the acceleration 'g' as gravity which has no bearning what so ever. I am not getting at you using a or g the value '9.81m/s/s' is simply not sensible to use in this case.

 

[FredGarvin]

The "g" usage is simply a kin to the aero guys saying they are pulling g's. It is a unit used in measuring accelerations. The unit implys an acceleration equivalent to the acceleration due to gravity. It's used very widely in vibration analysis.

 

[xxChrisxx]

I've reread nvn's post and it does say 1g deceleration, another fail for me reading stuff correctly!

However I still believe the 1g deceleration is still bogus in this case. It seems to be a number plucked out of thin air.

 

[Dalit]

So what should I take into account for calculating the impact?

I have the car mass (M), car velocity at impact (V=5-25kmph), impact duration 10 millisecond.
I can calculate the momentum or kinetic energy apply at impact.
I would like to know how much is absorb by the car and how much is absorb by the pole, or how do I calculate energy at each after impact?
Hitting the pole at 5kmph gives something of 300,000N, it seems outrages.

 

[xxChrisxx]

F dt = m dv

Integrate for F (this applies to the simplified case only)

So dv = starting v - 0 (as we assume it stansfers all energy so speed is 0).

 

[nvn]

Dalit wrote:[/color] "What should I take into account for calculating the impact?"[/color]

Dalit: It depends on your objective. Is your objective to make the beam strong enough to not yield or rupture? Or is your objective to make the beam weak enough to not damage the vehicle? Also, keep in mind, some older vehicles have heavy steel bumpers mounted directly to solid steel supports. In your case, these would act almost like a rigid body. Do you want to include these vehicles in your analysis, or exclude them?

 

[Dalit]

Chris,
Car mass is 2500Kg (going on the large ones). M=2500
Velocity at impact is 10 kmph (2.78m/sec), dv=2.78m/sec
Impact duration Is is 0.1 milliseconds or is it 10 milliseconds, according to Dr. D
t=10milisecond
F=~700,000N
that's the force hitting the pole.
If I calculate the energy for the same data
Ek=9960 [Kg*m2/sec2], that's the energy at impact.
After impact car doesn't fully stops, it keeps advancing due to inertion, how do I had that in the calculation?

 

[xxChrisxx]

If you want to do that, you'll need to do tests for the impact time and coefficient of restitution of the materials.

700KN is high, but you;ve got to remember you are making a lot of assumptions that drives this figure upwards. 10 miliseconds is a short amount of time even for a crash, the car will also not go to 0, the car (and most likely the beam) will plastically deform (further increasing contact time).

Also you've got to remember a 2500Kg car is on the heavy side, you are talking larger 4 x 4's.

 

[Dalit]

nvn,
My pole have electricity inside. My objective is to protect the electricity so I don't have to call technician every other day.
I am less concern what will happen to the car. I would, however, like to consider worst case scenario.
I also have to design an experiment to back it up.

 

[xxChrisxx]

Also I think 10ms is too uncraslistic a time for this crash. It'll probably be more like 100ms or 150ms.

 

[Dalit]

I am going on the worst case, since the pole have a PCB and electrical components.
If the external beam will have big deformation it my break the inner components.

 

[nvn]

Dalit: Is your beam an actual, continuous, square steel tube cross section? Or is the metal shell shown in your attached picture connected as multiple steel panels? I cannot tell from the picture you attached. If connected as multiple steel panels, are the steel panels continuously welded? Or are they connected by only a few fasteners?

 

[Dalit]

The pole is a steal tube, rolled forming welded at one joint line.
there is an additional small brackets welded at the corners to install the unit, they are not designed as extra strength although I think they have a small contribution.
However it is neglected.

 

[xxChrisxx]

What electronics are in it? and what are they for?