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Velocity to knock over a barrier

  1. Apr 2, 2015 #1
    I have a problem of a vehicle that impacts a barrier and knocks it over. At what minimum velocity does the barrier tip? (assuming it did not slide and there is no damage absorbing energy in the collision), sort like

    Barrier 2 ft wide, 2 ft tall, 6 ft long, 3600 lbs.
    Vehicle 2000 lbs, bumper height 1.5 ft.

    If we use energy KE of vehicle stopped by the impact = PE of barrier to move the CG up to the tipping point, 1/2 m v^2 = m g h so, v = sqrt (2 m g h / m) = sqrt (2 * 3600 * 32.2 * (sqrt 2 -1)/ 2000) = 7 fps
    the vehicle bumper height is not a factor and seems incorrect?
     
  2. jcsd
  3. Apr 2, 2015 #2
    You don't have to use kinetic energy here. What's important is the average force exerted on the wall upon collision. I think you need to know the duration of the impact, and then see if the force knocks the wall over it's center of mass. You'll have to use the impulse equation.
     
  4. Apr 3, 2015 #3

    jack action

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    More complicated than you think.

    First, you have to identify the force [itex]F[/itex] needed to tip the barrier from the free body diagram of the barrier.

    Of course, we will assume that this force is less than the friction force needed to initiate sliding of the barrier and less than the force needed to break or permanently deform the barrier. Otherwise you have to use the smallest of these forces as your critical force.

    Under that force, the barrier will bend in such a way that, at the bumper's height, the barrier will displace by a distance [itex]\delta[/itex] with respect to the base of the barrier.

    This means the stiffness of the barrier is [itex]K = \frac{F}{\delta}[/itex].

    The energy needed to bend the barrier is then [itex]\frac{1}{2}K\delta^2[/itex] or [itex]\frac{1}{2}F\delta[/itex].

    For a vehicle of mass [itex]m[/itex], coming at velocity [itex]v[/itex], the total energy of the system is [itex]\frac{1}{2}mv^2[/itex].

    So, assuming the total energy of the vehicle is absorbed by the barrier, the velocity to achieve the necessary force is [itex]v = \sqrt{\frac{F\delta}{m}}[/itex]. If some energy is absorbed by the vehicle (and it has too, as it will deform just like the barrier), the vehicle's velocity will need to be higher to transmit enough force to tip the barrier. So this velocity is the absolute minimum velocity needed.

    What the previous equation means is that the stiffer the barrier is ([itex]\delta \rightarrow 0[/itex]), the shorter will be the time duration of the impulse that PWiz mentioned. So the force created by the change in momentum will be larger under the same vehicle speed.

    So, now we know [itex]F[/itex] and [itex]m[/itex], all that is missing is [itex]\delta[/itex].

    For this, you have to model the barrier as a cantilever beam and then the deflection [itex]\delta = \frac{FL^3}{3EI}[/itex]. [itex]L[/itex] is the distance between the ground and the bumper, [itex]E[/itex] is the Young's modulus of the barrier and [itex]I[/itex] is the area moment of inertia of the barrier.

    spring1.gif

    Or, putting it all together:

    [tex]v = F \sqrt{\frac{ L^3}{3EIm}}[/tex]​
     
  5. Apr 7, 2015 #4
    From Jack we have velocity = F * (material and geometric constant). On the force side of the equation above we need acceleration, which requires knowing velocity? That does not work when you process it through. I think the beam analogy is not right, we have a pivot and that equation is for deformation related to the young modules of the material, not movement. deformation does approach 0 though.

    Measuring the force applied to tip a cinder block - it is maximum at the start and decreases to 0 just before it tips. I performed scaled testing on a cinder block, using a 7.5x7.5x15.5 inch block, 41 lbs and tipped it over pushing on a force meter, and tipped it over with a 10.5 lb sledge hammer pendulum. The force varied and showed a consistent moment - with a slow push the force measured 10 at the top, 13.5 at 3/4, and 20 lbf at the middle height - moment of about 155 lb-in. For the 10.5 lb swing, it knocked over when the height of the hammer was 30 inches (impacting at the 3/4 height of the block 11.625 in, so it fell18.35 inches), so v= sqrt( 2 g h) for a speed of about 9.9 fps or 6.77 mph at impact.
    Using the energy balance calculation I had v = sqrt (2 g h/ m) , the velocity would have been only 2.89 mph, so it could be the average velocity, which would double it to 5.8 mph and still calculate low vs. the 6.8 real world testing. I don't think I can just scale the value - its the wrong calc and not necessarily a linear multiple.

    Looking at the force values that were tested each several times. There was no impulse and its not the right sensor for that measurement.
    We know from typical crash testing of vehicle to barrier impacts, that the duration of the impulse is 0.05 to 0.1 seconds (airbag control module data).
    F = m a = m v / t = impulse v = F t / m, that does not make sense either and would result in a very small number as I do not have the impulse force, but the constant force.

    We know the force decreases as we push on the block, so as the mass does not change the acceleration must change. Our impact impulse accelerates the block with a max acceleration at impact which decreases to nearly 0 at the tipping point. Finding that induced acceleration I think is key.
     
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