# Homework Help: First natural frequency for bending, axial and torsion modes

1. Dec 5, 2017

### Feodalherren

1. The problem statement, all variables and given/known data
We have a rod of length L fixed to a rigid support. At the end of the rod there is a mass, m. Assume that the rod has no mass. Find the first natural frequency for the bending, axial and torsion modes.

2. Relevant equations

3. The attempt at a solution
So I'm reviewing some stuff from my undergraduate degree as I will be taking night classes for my master's in systems and controls next year. I'm doing some very basic stuff but wowzie is it difficult to remember some of this stuff that I haven't seen since I graduated.

So lets take the first part, that I think that I remember. For bending you can set up the problem as

mx'' = kx - mg

This is of a familiar form and we can see that for a simple mass-spring system the natural frequency is

Wn=sqrt(k/m)

So for a 1 DOF system we have found the 1st natural frequency. Is this correct?

I'm not sure what they mean by the axial one. Would I simply assume that the rod is acting as a spring and that the mass m is pushing down on it?

And for the torsional one would I just adjust my k to be the torsional value, but where does the supposed torque come from?

2. Dec 5, 2017

### haruspex

Sure, but the interesting part is the formula for determining k from the dimensions of the rod and the properties of the material.
This is a strange one to ask for a point mass on the end of a rod. There is no rotational inertia. Is that what you meant by no torque? Assume the mass has some moment of inertia and replace m with I. Again, the interesting part is the formula for k.
That would be longitudinal vibration of the rod, like a spring.

3. Dec 5, 2017

### Feodalherren

Thanks. I know how to find the formulas for k, it's not a problem. I just wanted to make sure that I had the basic problem down. It's not really a point mass. I was just simplifying it to see if what I was doing was accurate.

The mass is an AISI 1005 steel ball with diameter of 8 mm. So it's just as simple as using the torsional k?

4. Dec 5, 2017

### Feodalherren

Bending k = EA/L
Torsional k = πGD4 / 32L

5. Dec 5, 2017

### haruspex

Not sure what EA is here. (I am not an engineer.) I am familiar with a formula like 3EI/L3.

6. Dec 7, 2017

### Feodalherren

Last edited: Dec 7, 2017