A rock is falling from an embankment of mass $m_a$ at a height of $h$. Below, a fall protection fence catches the rock, dissipating the energy through tension cables to steel posts and then into concrete piles.

An example from a RTA standard drawing

How would you calculate the "impact" capacity of the tension cables, steel posts and concrete foundations?

EDIT: At this height, the rock initially has a potential energy of $PE=mgh$. Assuming there are no losses due to friction, bouncing, heat, sound, etc (a very conservative assumption), the potential energy will be mostly converted to kinetic energy, $KE=0.5mv^2$. The fence will absorb this complete energy.

Say the rock hits in the middle of the bay between the two posts. The cables will deflect with the impact of the rock. How do you calculate how much energy the cables absorb and whether the cables have sufficient capacity? How is this different from the applied force?

  • $\begingroup$ First calculate the energy of the rock... Show what you have done so far - this is not a free homework completion site. $\endgroup$ – Solar Mike Jan 21 '19 at 7:19
  • $\begingroup$ Or, start from what is the maximum the concrete foundations can support... $\endgroup$ – Solar Mike Jan 21 '19 at 7:28
  • $\begingroup$ I have edited the question to show that I understand the set-up of the problem. $\endgroup$ – lukeweatherstone Jan 21 '19 at 21:55
  • $\begingroup$ So, did you research the elastic and yield values of the chain mesh and cables? That need to be compared to the energy to be dissipated... $\endgroup$ – Solar Mike Jan 21 '19 at 21:59

Say speed of the rock when it hits the fence is.

$V = \sqrt {2gh}\quad$ and its kinetic energy is $E = 1/2 mV^2 $

This energy will be trapped by the fence and transferred to the posts. Assume inelastic collision with the post, or for simplicity ignore the mass of the post'

Elastic energy in a deflected cantilever post should be equated with the kinetic energy of the falling rock. From there you can calculate all other reactions and moments.

Commercial chain-link usually is strong enough for heights of up to 5 feet at spans of 6-8 feet.

if you show your effort we can help you.

  • $\begingroup$ You don't have to be concerned with the velocity. Substituting V into the kinetic energy equation results in the Potential Energy equation U = mgh. At the top of its fall, a rock will have potential energy of mgh & at the bottom of the fall the potential energy is fully converted to kinetic energy. $\endgroup$ – Fred Jan 21 '19 at 14:01
  • $\begingroup$ I have edited the question to show that I understand the set-up of the problem. $\endgroup$ – lukeweatherstone Jan 21 '19 at 21:57

Theoretically, impact energy will partially be absorbed by deformation of the rock and partially by deformation of the structure. However, the deformation of the rock will be very small, so just assume all energy will be absorbed by the structure. And since you don't know the exact shape of the rock, it is not practical to do any calculations on it anyway.

You will want to calculate the contact force as a function of the deformation of the structure, $F(u)$. I recommend for an initial approximation to assume either linear-elastic behaviour or perfectly plastic behaviour depending on the stiffness of the component the rock hits. The energy absorbed by the structure is the work of the contact force, so you solve $\int_0^{u_{max}} F(u)du = mgh$ for the deformation.

This will give you an equivalent static impact force and a deformation to check the structure for. Just remember that the load on the foundation varies a lot depending on whether the rock hits a post or a cable midspan between posts.


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