Deriving Potential Energy from a Force Law

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SUMMARY

The discussion centers on deriving the maximum deflection of a cantilever beam when a block of weight W is dropped from a height h. The relationship established is that the maximum deflection y_m is given by the formula y_m = y_s(1 + (1 + (2h/y))^0.5), where y_s represents the static deflection proportional to W. Participants explore the potential energy stored in the beam during deflection, drawing parallels to spring mechanics and referencing the formula for potential energy in elastic materials, V(beam) = 1/2ky^2. The conversation emphasizes the importance of understanding force laws to derive potential energy accurately.

PREREQUISITES
  • Understanding of static deflection in elastic beams
  • Familiarity with potential energy concepts in mechanics
  • Knowledge of force laws and their applications
  • Basic principles of energy conservation in mechanical systems
NEXT STEPS
  • Study the derivation of potential energy from force laws using F dot dx
  • Explore the mechanics of elastic beams and their deflection characteristics
  • Learn about energy conservation principles in impact scenarios
  • Investigate the mathematical modeling of cantilever beams under dynamic loads
USEFUL FOR

Mechanical engineers, physics students, and anyone involved in structural analysis or materials science will benefit from this discussion, particularly those interested in the mechanics of elastic materials and energy transfer during impacts.

AngelofMusic
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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.

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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Originally posted by AngelofMusic
I have:

[tex]y_s = kW[/tex]
...
... 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).
 

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