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

[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?

 

[Dalit]

PCB, contactor, RCD...
It's a smart pole.
The Pole is wired to a main voltage source.

 

[nvn]

Dalit: If you want to include older vehicles having bare steel bumpers connected directly to solid steel supports, then it is acceptable, due to your stated objective and scenario, to assume the vehicle is a rigid body. Also, the maximum strength of your beam that we could hope for is if we pretend your beam cross section does not experience local buckling near the base (even though local buckling near the base is likely, but has not been checked).

For an impact at a height of 500 mm, and pretending no local buckling of your beam cross section occurs near the base, your beam can absorb a total strain energy of U = 2540 N*m before rupturing. And the beam inelastic deflection (at a height of 500 mm) is 34.9 mm when your beam ruptures.

Therefore, for vehicle mass m = 2500 kg, and impact velocity v1 = 2.78 m/s, the vehicle velocity after your cantilever ruptures is v2 = 2.39 m/s. The impact duration is roughly 13.5 ms. If the vehicle mass is instead m = 1000 kg, and v1 = 2.78 m/s, then v2 = 1.15 m/s. Either way, this analysis indicates your beam is annihilated.

 

[xxChrisxx]

That's a really good response nvn.

It there any possibility of you fitting the main controls into the base Dalit?

 

[Dr.D]

nvn, you said, "The impact duration is roughly 13.5 ms." How did you come up with this value?

 

[nvn]

Dr.D: As a rough approximation, deflection divided by average velocity gives (0.0349 m)/[0.5(2.78 + 2.39) m/s] = 13.5 ms.

 

[Dalit]

nvn,
I don't follow your calculations.
you said that "strain energy of U = 2540 N*m".
you need the force at impact, Please advise how do you calculate it.

you also mentioned that "the beam inelastic deflection (at a height of 500 mm) is 34.9 mm when your beam ruptures." How did you calculate that?

In the third section, your calculations are based on velocity at impact 2.78 m/s (that's given), you assume that velocity after imact is v2 = 2.39 m/s. how come?

 

[nvn]

Dalit: I don't have time to try to explain nonlinear analysis here on the forum. You will need to study your favorite mechanics of materials or strength of materials textbooks, for a long time. You also need to be a good programmer. Even then, it will be a gross approximation. To get more accurate results, you could run a simulation of the impact using very advanced software. Perhaps LS-Dyna would do, but it is probably quite expensive and time-consuming. I only have time to give the results I obtained for your beam, but not further explanation.

The force versus deflection curve I obtained for your beam was P(y) = (y<1.26)(43083*y + 599.4) + (y>1.26)[37787(y + 3.589)^0.2261], for y = 0 to 35 mm, where P = load (in units of N) applied to cantilever by vehicle at a height of 500 mm, and y = cantilever deflection (at a height of 500 mm) in units of mm. Integrate the area under this curve to obtain the beam strain energy, U, to the point of extreme fiber rupture (y = 35 mm). This is based on mild steel. Once you obtain U in N*mm, convert it to N*m. Furthermore, although I didn't do this, you can probably divide the above U by 0.70 or 0.60, or maybe even 0.55, to try to make U also account for energy lost to heat and other forms. After you obtain U, write a conservation of energy equation for the impact, and solve it for v2 to obtain the vehicle final velocity.

Similarly, here is the force versus deflection curve I obtained if you increase your beam wall thickness to 6.0 mm. P(y) = (y<1.26)(99388*y + 362.4) + (y>1.26)[83513(y + 4.434)^0.2357], for y = 0 to 36 mm, where y = 36 mm is the point of extreme fiber rupture.