Need help with a torsional vibration problem

[hanson]

[Urgent] Need help with a torsional vibration problem

Hi all!
I am doing a problem on torsional vibration as follow:
"Two shafts A and B are geared togather such that the speed of B is 2.5 times that of the speed of A. The shaft A, of 50mm diammeter and 600mm length carries a wheel of moment of inertia of 1.7 kgm2 at the free end. The shaft B is of diameter 38mm and length 760mm, carries a wheel of moment of inertia of 0.64 kgm2 at its free end. By finding the equivalent length of shaft A, determine the natural frequency of torsional vibration if the inertia effect of the gears and shafts may be ignored. Modulus of rigifity of shaft material = 80x10^3 MN/m2.

I don't really understand the meaning of "By finding the equivalent length of shaft A". As I know from geared torsional vibration problem, the system is transform into an equivalent system of different stiffness, k, but with the same length, right? So how to do this question actually?

I tried determining the equivalent stiffness of shaft B by multipying it by 2.5^2 and the equivalent moment of inertia of the wheel on shaft B by 2.5^2, also. By doing so, I find the new diameter of shaft B and find the equivalent length of A if it is to have the same diameter of B.
But the answer I got is different from the one provided...

Anyone please kindly help?

 

[Astronuc]

Each axle has a 'wheel' (disc) at the free end, and I imagine that the moment of inertia for the disc is different than the shaft to which it is attached.

http://en.wikibooks.org/wiki/Solid_Mechanics#Angle_of_Twist

Then perhaps there is an equivalent length of shaft corresponding to the wheel.

 

[piercas]

Hi there,

Thank you for reaching out for help with your torsional vibration problem. Torsional vibration can be a complex and challenging topic, so it's understandable that you are struggling with this question. Let's break it down and try to understand it better together.

Firstly, the problem states that the inertia effects of the gears and shafts can be ignored. This means that we can assume that the gears and shafts have no impact on the natural frequency of torsional vibration. This simplifies the problem and allows us to focus on the two shafts and their respective wheels.

Now, let's look at the phrase "By finding the equivalent length of shaft A". This means that we need to find a single shaft that has the same stiffness and moment of inertia as both shafts A and B combined. This equivalent shaft will have the same natural frequency of torsional vibration as the original system.

To find the equivalent length of shaft A, we need to use the concept of equivalent stiffness. This is where we transform the original system into an equivalent system with a different stiffness, but the same length. In this case, we need to find the stiffness of the equivalent shaft that would produce the same natural frequency as the original system.

To find the equivalent stiffness, we can use the formula:

keq = k1 + k2 + k3 + ...

where k1, k2, k3, etc. are the stiffnesses of each component in the system. In this case, we have two components - shaft A and shaft B. So our formula becomes:

keq = kA + kB

We know that the stiffness of a shaft is proportional to its length and the modulus of rigidity of the material. So we can write:

keq = kA + kB = (lA x 80x10^3) + (lB x 80x10^3)

where lA and lB are the lengths of shaft A and B, respectively.

Now, we can use the formula for the natural frequency of torsional vibration:

ωn = √(keq/Ieq)

where ωn is the natural frequency, keq is the equivalent stiffness, and Ieq is the equivalent moment of inertia.

From this, we can find the equivalent length of shaft A by rearranging the formula:

lA = [ωn^2 x Ieq - (lB x 80x10^3)] / (80x