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The formula is:

$$ f = \frac{3.5161}{2\pi L^2}\sqrt{\frac{EI}{\rho A}} $$

$A$ = area, $\rho$ = density, $I$ = second moment of area cross section, $E$ = Young's modulus, and $L$ = length. Can anyone help transpose this equation so that A is the unknown subject?

I can't find my way around it, and the people I've asked are stumped?

So i have some sprung steel, I'm calculating the frequency to make a SPECIFIC note.... but i want to know what size I need my steel (rectangular rod) to be to create the desired note/frequency.

Am I looking at it all wrong?

Mark H
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Dale Mass
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1 Answers1

0

$$ f = \frac{3.5161}{2\pi L^2}\sqrt{\frac{EI}{\rho A}} $$ $$ f^2 = \left(\frac{3.5161}{2\pi L^2}\right)^2\frac{EI}{\rho A} $$ $$ Af^2 = \left(\frac{3.5161}{2\pi L^2}\right)^2\frac{EI}{\rho} $$ $$ A = \left(\frac{3.5161}{2\pi L^2}\right)^2\frac{EI}{\rho f^2} $$ or $$ A = \left(\frac{3.5161}{2\pi L^2f}\right)^2\frac{EI}{\rho} $$

Mark H
  • 25,556