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(OP)
If a beam is designed with a dynamic impact factor of "2," what would the equivalent drop height?

Hi, carvinwood.

Assuming that when you refer to an impact factor of 2 you mean that you have multiplied your static stresses and deflections by 2 to allow for dynamic effects, then you may be surprised by the answer.

You will double your 'static' stresses etc by simply applying the full load 'suddenly', and you don't need to drop it at all.

For most purposes, we calculate the deflections etc on the basis that the applied load is increasing proportionally to the deflection.  That gives us a linear (triangular) graph for load/deflection.  Since the area under the graph gives us the strain energy, the final strain energy = 0.5*W1*delection, where W2 is the final load.

If a load W2 is applied at full value from the very beginning, the load/deflection graph is a horizontal line, and the area underneath is a simple rectangle, thus the strain energy is 1.0*W2*deflection.

Thus for the same strain energy, stresses and deflections, W2=W1/2.

If you were to apply the original load W1 suddenly, you would end up with 2*deflection etc.

Thus any drop at all will generate an impact factor greater than 2

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