When an elastic beam AB supports a block of weight W at a given point B, the deflection [tex]y_s[/tex] (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 [tex]y_m = y_s(1 + (1+\frac{2h}{y})^\frac{1}{2}).[/tex] Neglect the weight of the beam and any energy dissipated in the impact.(adsbygoogle = window.adsbygoogle || []).push({});

I have:

[tex]y_s = kW[/tex]

T1 = 0

V1 = mgh

T2 = 0 when deflection is at a max

[tex]V2 = -mgy_m[/tex]

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.

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# Homework Help: Deflection of an Elastic Beam

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