## equivalent torsional stiffness

## equivalent torsional stiffness

(OP)

My original problem is to find the torsional frequency for a rotor cage. I first looked into the torsional frequency for 2 discs (J1, J2) and 1 massless rod in between (standard textbook problem). Using torsional stiffness equation, kT = G*J/L, I was able to find my torsional frequency of the system. For a single rod (with diameter = Do) on top of the rotational axis, J orig = (pi*Do^4/32). I then tried to find an equivalent system with multiple rods (that is made up of multiple N parallel rods (with diameter = Dn) located at R away from center of rotation). I used parallel axis theorem. (pi*Dn^4/32+ R^2*(area))*N = J total. Since rods are parallel, the total k should be the sum of each k.

For an equivalent system, J orig should = J total, I was able to find Dn. The problem is when I tried to verify it using FEA, my natural frequency is way off.

Did I make wrong assumptions somewhere in there?

Given for standard text book calc: J1 = 5.66 kg-m2, J2 = 22.6 kg-m2, G = 7.69e10 kg/m2, L = 0.6m, Do = 0.1m

Standard text book calc: J orig = 9.817e-6 m^4, kT = 1.26e6 Nm/rad and freq ~ 84 Hz

FEA text book calc: freq ~ 82 Hz

For equivalent system, my calc for 4 rods with Dn = 0.0176m, R = 0.1m, same J and kT, freq ~ 84 Hz

FEA for 4 rods: freq ~ 6.2 Hz

For an equivalent system, J orig should = J total, I was able to find Dn. The problem is when I tried to verify it using FEA, my natural frequency is way off.

Did I make wrong assumptions somewhere in there?

Given for standard text book calc: J1 = 5.66 kg-m2, J2 = 22.6 kg-m2, G = 7.69e10 kg/m2, L = 0.6m, Do = 0.1m

Standard text book calc: J orig = 9.817e-6 m^4, kT = 1.26e6 Nm/rad and freq ~ 84 Hz

FEA text book calc: freq ~ 82 Hz

For equivalent system, my calc for 4 rods with Dn = 0.0176m, R = 0.1m, same J and kT, freq ~ 84 Hz

FEA for 4 rods: freq ~ 6.2 Hz

## RE: equivalent torsional stiffness

like this, but different -

http://chalet-shop.com/wp-content/uploads/2013/11/...

Can you animate the FEA results?

depending how the rods are affixed to the end disks and how thick the disks are the disk may be bending at each rod too.

## RE: equivalent torsional stiffness

http://files.engineering.com/getfile.aspx?folder=8...

http://files.engineering.com/getfile.aspx?folder=f...

The BC's for textbook case (case 1) is cylindrical support (axial and radial: fixed. tangential: free) on outer surfaces of both discs and the rod.

The BC's for multirod case (case 2) is cylindrical support (axial and radial: fixed. tangential: free) on outer surfaces of both discs.

The rods are rigidly bonded to the discs. I also tried free-free condition with no BC's. The result is the same.

I've been scratching my head over this for many days wondering what did I miss?

## RE: equivalent torsional stiffness

## RE: equivalent torsional stiffness

http://ruina.mae.cornell.edu/Courses/ME4735-2012/R...

Apply a load, solve for the displacement, look at the result as a rotation/chord then angle out at the rod radius on disk 2 .