# Impact capacity of tension cables, columns and foundations

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.

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?

• First calculate the energy of the rock... Show what you have done so far - this is not a free homework completion site. – Solar Mike Jan 21 '19 at 7:19
• Or, start from what is the maximum the concrete foundations can support... – Solar Mike Jan 21 '19 at 7:28
• I have edited the question to show that I understand the set-up of the problem. – lukeweatherstone Jan 21 '19 at 21:55
• 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... – 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.

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.