First natural frequency for bending, axial and torsion modes

AI Thread Summary
The discussion focuses on calculating the first natural frequency for bending, axial, and torsion modes of a mass attached to a massless rod fixed at one end. The bending frequency is derived using the equation mx'' = kx - mg, leading to the natural frequency formula Wn=sqrt(k/m). For the axial mode, the rod behaves like a spring under the mass's weight, while the torsional mode requires adjusting the stiffness constant k to account for torsional properties. Participants clarify that the mass should have a moment of inertia for accurate calculations, and they discuss the formulas for stiffness in bending, axial, and torsional modes. The conversation emphasizes understanding the basic principles and formulas necessary for solving the problem.
Feodalherren
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Homework Statement


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.

Homework Equations

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 let's 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?
 
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Feodalherren said:
For bending you can set up the problem as
mx'' = kx - mg
Sure, but the interesting part is the formula for determining k from the dimensions of the rod and the properties of the material.
Feodalherren said:
for the torsional one would I just adjust my k to be the torsional value,
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.
Feodalherren said:
not sure what they mean by the axial one.
That would be longitudinal vibration of the rod, like a spring.
 
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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?
 
Bending k = EA/L
Axial load k = EA/L
Torsional k = πGD4 / 32L
 
Feodalherren said:
Bending k = EA/L
Not sure what EA is here. (I am not an engineer.) I am familiar with a formula like 3EI/L3.
 
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Kindly see the attached pdf. My attempt to solve it, is in it. I'm wondering if my solution is right. My idea is this: At any point of time, the ball may be assumed to be at an incline which is at an angle of θ(kindly see both the pics in the pdf file). The value of θ will continuously change and so will the value of friction. I'm not able to figure out, why my solution is wrong, if it is wrong .
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