# Velocity to knock over a barrier

1. Apr 2, 2015

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. Apr 2, 2015

### PWiz

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.

3. Apr 3, 2015

### jack action

More complicated than you think.

First, you have to identify the force $F$ 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 $\delta$ with respect to the base of the barrier.

This means the stiffness of the barrier is $K = \frac{F}{\delta}$.

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

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

So, assuming the total energy of the vehicle is absorbed by the barrier, the velocity to achieve the necessary force is $v = \sqrt{\frac{F\delta}{m}}$. 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 ($\delta \rightarrow 0$), 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 $F$ and $m$, all that is missing is $\delta$.

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

Or, putting it all together:

$$v = F \sqrt{\frac{ L^3}{3EIm}}$$​

4. Apr 7, 2015