How to find cross sectional dimension of beam

Click For Summary
SUMMARY

This discussion focuses on calculating the necessary cross-sectional dimensions of a steel cantilever beam, 5 meters in length, to limit vibration amplitude to 0.5 cm when an electric motor (75 kg, operating at 1200 RPM) is mounted at its free end. Key equations include the moment of inertia \(I = \frac{1}{12}bh^3\), the effective stiffness \(k_{eq} = \frac{3EI}{\ell^3}\), and the natural frequency \(\omega_n = \sqrt{\frac{k}{m}}\). The challenge lies in determining the mass of the beam and the damping coefficient, which are critical for accurate calculations.

PREREQUISITES
  • Understanding of cantilever beam mechanics
  • Familiarity with vibration analysis and damping coefficients
  • Knowledge of moment of inertia calculations for different cross-sectional shapes
  • Ability to convert RPM to radians per second
NEXT STEPS
  • Calculate the mass of the beam based on its material and dimensions
  • Determine the damping coefficient for the system
  • Explore the effects of different cross-sectional shapes on moment of inertia
  • Learn about dynamic loading conditions on cantilever beams
USEFUL FOR

Mechanical engineers, structural engineers, and students studying dynamics and vibration analysis will benefit from this discussion, particularly those involved in designing and analyzing cantilever beam systems.

Dustinsfl
Messages
2,217
Reaction score
5

Homework Statement


If the electric motor of problem 3.66 is to be mounted at the free end of a steel cantilever beam of length ##5## m and the amplitude of vibration is to be limited to ##0.5## cm, find the necessary cross-sectional dimension of the beam. Include the weight of the beam in the computations.

The electric motor has mass of ##75## kg and speed ##1200## rpm at its mid spin. A rotating force of magnitude ##F_0 = 5000## N

Homework Equations


\begin{align}
I &= \frac{1}{12}bh^3\\
E &= \frac{m\omega_n^2\ell^3}{3I}\\
k_{eq} &= \frac{3EI}{\ell^3}\\
\omega_n &= \sqrt{\frac{k}{m}}
\end{align}

The Attempt at a Solution


Since the amplitude of vibration is limited to 0.5cm,
$$
\frac{1}{200} = \sqrt{1 + (2\zeta r)^2}\lvert H(i\omega)\rvert
$$
where
$$
\lvert H(i\omega)\rvert = \frac{1}{\sqrt{(1 - r^2)^2 + (2\zeta r)^2}}
$$
where ##r = \frac{\omega}{\omega_n}##, ##2\zeta\omega_n = \frac{c}{m}##
I solved for zeta in terms of r but I don't think that is leading any where
$$
\zeta = \pm\frac{\sqrt{\omega^4 - 2\omega^2\omega_n^2 - 39999\omega_n^4}}{2\omega\omega_n\sqrt{39999}}
$$
We can convert rpm to radians per sec by
$$
1200rpm\frac{2\pi}{1rev}\frac{1m}{60s} = 40\pi
$$
That is ##\omega = 40\pi## and our general equation of motion is
$$
m\ddot{x} + c\dot{x} + kx = 5000e^{40\pi it}
$$
The mass in the EOM is mass of motor plus mass of beam. How would I find the mass of the beam and the damping coefficient? If I can get these two pieces of information, I should be good to go.
 
Last edited:
Physics news on Phys.org
Dustinsfl said:

Homework Statement


If the electric motor of problem 3.66 is to be mounted at the free end of a steel cantilever beam of length ##5## m and the amplitude of vibration is to be limited to ##0.5## cm, find the necessary cross-sectional dimension of the beam. Include the weight of the beam in the computations.

The electric motor has mass of ##75## kg and speed ##1200## rpm at its mid spin. A rotating force of magnitude ##F_0 = 5000## N

Homework Equations


\begin{align}
I &= \frac{1}{12}bh^3\\
E &= \frac{m\omega_n^2\ell^3}{3I}\\
k_{eq} &= \frac{3EI}{\ell^3}\\
\omega_n &= \sqrt{\frac{k}{m}}
\end{align}

The Attempt at a Solution


Since the amplitude of vibration is limited to 0.5cm,
$$
0.5 = \sqrt{1 + (2\zeta r)^2}\lvert H(i\omega)\rvert
$$
where
$$
\lvert H(i\omega)\rvert = \frac{1}{\sqrt{(1 - r^2)^2 + (2\zeta r)^2}}
$$
where ##r = \frac{\omega}{\omega_n}##, ##2\zeta\omega_n = \frac{c}{m}##

I need to know ##\omega## and ##\omega_n##

I solved for zeta in terms of r but I don't think that is leading any where
$$
\zeta = \pm\frac{\sqrt{\omega^4 - 2\omega^2\omega_n^2 - 39999\omega_n^4}}{2\omega\omega_n\sqrt{39999}}
$$
You need to understand what the variables in your formulas represent.

##\omega_n## is the natural frequency of the beam, and it is a function of E and the moment of inertia of the beam. The problem statement doesn't specify a beam material or a cross section shape, so E and I have to be determined using some subjective criteria. (The length of the beam is already specified, so it is not considered here). You should be aware that I = bh3/12 is correct only for beams having a rectangular cross section.

ω is the frequency of excitation, and the problem statement has already stated that the electric motor operates at 1200 RPM. Can you calculate what ω is at this speed?

k is a little trickier. You have the motor placed at mid-span of the beam, which implies that there is a certain static deflection produced in the beam. The problem also wants you to consider the mass of the beam in solving this problem. The k you have specified is that for a beam with a uniform distributed load, but what about the effect of the motor on the deflection of the beam?
 
SteamKing said:
ω is the frequency of excitation, and the problem statement has already stated that the electric motor operates at 1200 RPM. Can you calculate what ω is at this speed?

k is a little trickier. You have the motor placed at mid-span of the beam, which implies that there is a certain static deflection produced in the beam. The problem also wants you to consider the mass of the beam in solving this problem. The k you have specified is that for a beam with a uniform distributed load, but what about the effect of the motor on the deflection of the beam?

omega was added to the OP 11minutes ago and the motor is not a mid span it is at the free end which is stated in the question. The k I specified comes from the book for a cantilever beam with end load which is the scenario we have.
 

Similar threads

  • · Replies 4 ·
Replies
4
Views
3K
  • · Replies 13 ·
Replies
13
Views
6K
Replies
7
Views
4K
Replies
1
Views
743
  • · Replies 26 ·
Replies
26
Views
2K
  • · Replies 4 ·
Replies
4
Views
3K
Replies
11
Views
2K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 1 ·
Replies
1
Views
2K