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Air Velocity/Pressure on Wrecked Auto

  1. Aug 2, 2006 #1
    Can any one help?
    I recently transported a wrecked vehicle on the interstate on a flat bed trailer. The individuals that I transported this vehicle for are claiming that they should not have to pay me due to damage done by a 4" strap used to hold the vehicle on the trailer.
    I claim that the strap was the only way to safely transport the vehicle.
    I believe that air being forced through a hole the size of a basketball would have blown the hood off the vehicle on the road.
    They claim it could have been held on with a bungee strap or rope tied inside the engine compartment.
    I believe that there would have been enough force going through the hole by the hood and the drivers side fender that the hood would have been blown off the vehicle.
    I used that 4" strap for safety reasons.
    They are refusing to pay me for the work due to the dents caused in the passenger side front fender by the strap.
    I am trying to determine the amount of pressure created by freeway speeds blowing into the hole that may have caused the hood to come off the vehicle.
    I feel like they are trying to extort the money from me and could really use some guidance in determining if their strap would have held the hood down.
    I have pictures of the vehicle. Link to Pictures.
    I believe both the freight broker and the owner of the auto body shop who bought this vehicle for salvage are trying to rip me off.
    Can anyone help?
    I am tired of being taken advantage of.
  2. jcsd
  3. Aug 3, 2006 #2
    The necessary conditions to determine the strength needed to hold down the hood can be split up into 3 areas.

    First we need to model 4 forces: the drag pushing the hood towards the rear of the car (which is small and I will consider almost neglegible) The lift force trying to flip the hood up, the weight ofthe hood, and the force exerted by the restraint to hold the hood down.

    Next we need to determine an expression for the lift. The lift is the integral of the pressure distribution over the top and bottom of the hood. In the worst case scenario lets assume the air is stagnant under the hood and that we a constant free-stream flow over the top. The pressure gradient is then found by using bernoulis equation (potential flow should be ok here, this car is not traveling near an significant fraction of a mach number). This means delta P = 1/2*density of air*velocity^2. Let the density of air be 1.1kg/m^3 and lets assume we are traveling 60 mph which is approximately 27 m/s.

    Then the pressure is 0.55*27^2 which is about 391 Pa (pascals) these units are in Newtons per square meter which is the force per unit area. Assume the hood's dimenions are 1.6m x 1.6m

    Well for a constant force distribution F = P*A = 391*1.6^2 = 1000 N or 1 kN.

    This means the vertical reaction force at the pivot added to the lifting force must equal the restraining force in addition to the weight. However we can eliminate the reaction force by summing moments about the pivot.

    If the lift distribution is constant over the whole hood its effective moment arm is located at the centroid of the force distribution which in this case is 0.8m from the pivot. (Since this is not the case, it'll actually be closer to the front since you have significantly more pressure where those gaps are, and for simplicity lets assume the moment arm is 1m) this means its located about 62% down the length of the hood as opposed to 50%, but thats a rough estimate. So anyways
    Lift*Lift moment arm = Restraint*Restraint moment arm + Weight*Weight moment arm.

    lets assume the restraint is placed 1.2 meters from the pivot - the closer to the free end of hood the less stress it will experience. Also lets assume the mass of the hood is 15 kg (about 35 pounds) which is approximately 150 N, located 1/2 the length means its moment is 150N*0.8m = 120N-m

    so 1000*1 = R*1.2 + 120

    R = (1000-120)/1.2 = 733 N

    so the force needed to hold the hood down is 733 N

    assume 1/2 this load is carried on the left side of the belt and the other 1/2 is carried on the right side of the belt.

    To measure the stress we are simply looking for the pressure.

    so the stress on a side of the belt is 367N/area of belt
    assume the thickness of the belt is 4mm and that its width is 5cm

    that has an area of 0.004m*0.05m = 0.0002m^2

    367/0.0002m = 1835000 Pa which isnt terrible, the yield stresses of most metals are in the order of 10^8-10^9 (the giga pascal range). Here the stress is 1.8 MPa (mega-pascals). To find how close you are the envelopes simply take 1.8Mpa or whatever number u get doing a similar analysis and save the measre of stress (pressure).

    Now the part I don't know how to do - find the yield stress of your material. For common metals/alloys/construction materials these are all listed. Fibers and composites are a little tougher to find. If you can find the yield stress of your material then you have to do it though a tensile test. Anyways lets says the the yield stress is 3.6 MPa, and our calculuted maximum stress is about 1.8Mpa then the factor of saftey is given by (yield stress/max allowed stress) which in this case is 2. Meaning those belts could handle twice the load, you could then make variations to your numbers in the analysis (moment arm, areas, speeds) to see what conditions yield that envelope. But thats the kind of analysis you need to do.

    Now by no means am i a materials guy. Since this belt is more like a hoop our method for finding the stress is only an estimation, what we really need to find to get a precise answer is something called the hoop stress, but I can't remember the formula for that off the top of my head, either way it should yield a similar result.

    If you do this analysis for the tape or whatever they suggested you use I would to do the following to construct an argument. Find the factor of safety first for the tape, if its below 1 that means there is failure, if its close to one, the find what variation is perhaps speed is required to decrease the factor of safety to 1. Remember, increased speed means increased lift which means increased force in the tape which means increased stress. If you only need a few mph to cause the factor of safety to appraoch 1 you have a good argument. Likewise I'd show that the belt has a higher factor of safet and perhaps you'd need to go 200 mph to cause the factor of safety to drop to 1 (which is beyond the performance limit of the vehicle - therefore the belt cannot fail due to variations in speed only).

    Either ways this is just a run through, sorry its long but it is somewhat of an involved problem. There is no way you'll be able to model the flow over the hood exactly due to the geometry, the only way to get a good measure would be to set the thing up in a big wind tunnel which would be massive overkill.

    The following equation should give you the factor of safety- make sure u are consistent with units

    2*yield stress*Belt cross sectional area*Length of hood*fraction of hood moment arm where the belt is located

    what i mean by the last term is if the hood is 2 meters long and the fraction of hood MA is 0.8 this would correspond the belt being located 1.6 meters towards the front of the hood. Anyways multiply those and thats your numerator

    the denominator is (2*Pressure*Area of hood*center of pressure length - Weight of hood*Length of hood)

    the pressure of the hood is given by 1/2*density of air*speed^2) so

    denominator is (density of air*speed^2*hood area*Length of hood/2 - weight of hood*length of hood)
    this is with the constant pressure distribution assumed, so.....

    FOS = num/den where
    num = (2*yield stress*Belt cross sectional area*Length of hood*fraction of hood moment arm where the belt is located)
    den = (density of air*speed^2*hood area*Length of hood/2 - weight of hood*length of hood)

    make sure your bases of units are kilograms, meters, and seconds. the end result is in pascals, using the metric system u can throw on a the prefix (I would assume this would be in the mega pascals 10^6) but in reality this is a high estimate, theres no way the pressure distribution is this high. I would assume the real answer is maybe 0.6 - 0.7 multiplied by the calculated answer. but it could be on the order of a couple megapascals.

    Show them this analysis and then ask them to tell you under what speed (without a strap) the hood would fly up smashing their windshield costing them even more. I don't want to rant here but sometimes people need to appreciate the level of analysis needed to actually quantify some result. Its better to be safe than sorry especially when you're estimating some condition that is wayyyyy too complex to analyze. Personally, being an aero engineer, I would strap the hood down like crazy, never would i use some cheap form of tape, especially if that thing is on the highway. That thing looks like it could fly off and hit another car even, I believe what you did was what any other considerate person should have done.
    Last edited: Aug 3, 2006
  4. Aug 3, 2006 #3
    Thank you for your response.
    I am still trying to digest it all but seem to be understanding everything.
    It is just taking a little while to sink in.
    The strap that they said that I should have used has a a rating of 5000 lbs holding strength and a 3500 lb holding strength during quick tension release situations.
    I believe enough air would flow into the engine compartment and create pressure that would lift the already damaged hood off the vehicle. Micro burst wind conditions on the highway are always prevalent out here in the west.
    This vehicle was to travel on a trailer from Las Vegas, NV to Idaho Falls, ID.
    My case is that even attaching the ratchet strap to the underside of the hood and to the frame would not guarantee secure holding power of the strap due to "Possible" damage sustained in the wreck prior to me even getting the vehicle.
    Micro bursts of wind in addition to the steady pressure created by constant wind velocity caused me to believe that the hood would fly off. Not might fly off.
    I travel at interstate speeds of between 60-75 miles an hour and will work with your suggested direction to determine the amount of pressure that would be created by wind velocity inside the vehicle engine compartment.
    I do know that some will escape the bottom bottom but I believe it would do nothing more than create negative pressure and increase the possibility of lifting the hood.
    I will work with this and I really appreciate the help.
    Thank you in advance.
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