Formula for Natural Frequency of Cantilever Beam

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SUMMARY

The discussion centers on two equations for calculating the natural frequency of a cantilever beam. The first equation is F1 = k² * sqrt(E * I / (m * L⁴)) / (2 * π), while the second is F1 = k² * sqrt(E * I / (m * L⁴)). The key distinction is that the first equation provides frequency in Hertz (Hz), whereas the second yields angular frequency in radians per second. The constant k is defined as 1.875 for the first natural frequency, and the moment of inertia I is calculated as b * d³ / 12.

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  • Understanding of cantilever beam mechanics
  • Familiarity with the concepts of natural frequency and angular frequency
  • Knowledge of the moment of inertia calculation
  • Basic proficiency in differential equations
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Oscar6330
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I found two different versions of equations to find natural frequency of a cantilever beam. I am not sure which one is right. I would appreciate if someone could make things a bit clear here


F1= k^2*sqrt(E*I/(mpl*L^4))/(2*pi) where k=1.875 for first natural freq and I= b*d^3/12;

OR is it

F1= k^2*sqrt(E*I/(mpl*L^4)) where k=1.875 for first natural freq and I= b*d^3/12;

Basically I am not sure why some equations have /(2*pi) while others do not and which one is correct
 
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Angular frequency , \omega radians /second = 2\pif cycles per second ?
 
Studiot said:
Angular frequency , \omega radians /second = 2\pif cycles per second ?

Thanks. So the first equation gives freq in Hz while other one gives answer in radians per second?
 
Yes the solution to the governing differential equation for say a cantilever with mass m at the end is

m\ddot x + kx = 0

which has solution

\omega = \sqrt {\frac{k}{m}}

where omega is in rads/second
 
Last edited:
Studiot said:
Yes the solution to the governing differential equation for say a cantilever with mass m at the end is

m\ddot x + kx = 0

which has solution

\omega = \sqrt {\frac{k}{m}}

where omega is in rads/second

thanks for your kind reply.

So

F1= k^2*sqrt(E*I/(mpl*L^4))/(2*pi)


w1=k^2*sqrt(E*I/(mpl*L^4))
 

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