User`s guide
97
Solution for displacement is:
Where:
For a cantilever beam, the displacement and slope are zero at the fixed end, and the
moment and shear are zero at the free end. Thus the boundary conditions are:
when x = 0, y = 0,
.
when x=L,
,
.
Applying the boundary conditions yields
The equation for time is
So the exact expression of
natural frequency in rad/sec is
where E is Young's modulus of elasticity, I is moment of inertia of cross section, L is
effective length of beam, and is the density, A is the area of cross section. The dimensionless
wave number = 2/wavelength.
values for cantilever beams are: β1L = 1.8751=
, β2L =
4.6941=
, β3L = 7.8548=
, β4L = 10.99557=
, β5L = 14.1372=
, β6L = 17.279=
.
Therefore, the natural frequency of cantilever beam with a rectangular cross section is
Eq.22
Eq.17
Eq.18
Eq.19
Eq.20
Eq.21