Example 1 - Number of DOF influence in Natural Frequency
Contents
Example 1 - Number of DOF influence in Natural Frequency#
In this example, we use the rotor seen in Example 5.8.1 from [Friswell, 2010]. Which is a symmetric rotor with a single disk in the center. The shaft is hollow with an outside diameter of \(80 mm\), an inside diameter of \(30 mm\), and a length of \(1.2 m\) and it is modeled using Euler-Bernoulli elements, with no internal shaft damping. The bearings are rigid and short and the disk has a diameter of \(400 mm\) and a thickness of \(80 mm\). The disk and shaft elements are made of steel.
import numpy as np
import plotly.graph_objects as go
import ross as rs
import plotly.io as pio
pio.renderers.default = "notebook"
steel = rs.materials.steel
number_of_elements = [2, 4, 6, 8, 10, 12, 14, 16, 18, 20, 40, 60]
def create_rotor(n_el):
"""Create example rotor with given number of elements."""
shaft = [
rs.ShaftElement(1.2 / (n_el), idl=0.03, odl=0.08, material=steel)
for i in range(n_el)
]
disks = [
rs.DiskElement.from_geometry(
n=(n_el / 2), material=steel, width=0.08, i_d=0.08, o_d=0.4
)
]
bearings = [
rs.BearingElement(0, kxx=1e15, cxx=0),
rs.BearingElement(n_el, kxx=1e15, cxx=0),
]
return rs.Rotor(shaft, disks, bearings)
def analysis(speed):
"""Perform convergence analysis for a given speed."""
# create reference rotor with 80 elements
n_eigen = 8
rotor_80 = create_rotor(80)
modal_80 = rotor_80.run_modal(speed, num_modes=2 * n_eigen)
errors = np.zeros([len(number_of_elements), n_eigen])
for i, n_el in enumerate(number_of_elements):
rotor = create_rotor(n_el)
modal = rotor.run_modal(speed, num_modes=2 * n_eigen)
errors[i, :] = abs(
100 * (modal.wn[:n_eigen] - modal_80.wn[:n_eigen]) / modal_80.wn[:n_eigen]
)
fig = go.Figure()
for i in range(8):
fig.add_trace(
go.Scatter(x=number_of_elements, y=errors[:, i], name=f"Mode {i}")
)
fig.update_layout(
xaxis=dict(title="Number of degrees of freedom"),
yaxis=dict(type="log", title="Natural Frequency Error(%)"),
)
return fig
analysis(speed=0)