Deflection of an Elastic Beam

[AngelofMusic]

When an elastic beam AB supports a block of weight W at a given point B, the deflection y_s (static deflection) is proportional to W. Show that if the same block is dropped form a height h onto the end B of a cantilever beam, the maximum deflection y_m = y_s(1 + (1+\frac{2h}{y})^\frac{1}{2}). Neglect the weight of the beam and any energy dissipated in the impact.

I have:

y_s = kW

T1 = 0
V1 = mgh
T2 = 0 when deflection is at a max
V2 = -mgy_m

I'm pretty sure that V2 should also include the potential energy stored in the beam, but I don't know how to express that. Would it be similar to a spring? V(beam) = 1/2ky^2 ? That's my guess, but there must be a logical way of proving it.

I think I can do the rest once I find the expression for the potential energy inside the elastic beam.

[Tom Mattson]

Hey, check this out:

http://www.engin.brown.edu/courses/En31/en31design.pdf

[turin]

Originally posted by AngelofMusic
I have:

y_s = kW
...
... there must be a logical way of proving it.Yes, there is. You have identified a force law that should look familiar (if you put the k on the other side). How do you derive the potential energy from that force law (think about F dot dx).