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I have an intermediate shaft that drives my spindle with pitch (3). On the spindle there is a slide which is moved translationally by the spindle. My intermediate shaft is rubbery and has a very low torsional stiffness. Is it possible to convert this torsional stiffness by the pitch of the spindle to a translational stiffness?

r/EngineeringStudents - Convert torsional stiffness to translational stiffness
Similar to this picture except that an intermediate shaft is connected to reduce the stiffness of the system

I am relatively new to the subject and would appreciate any help.

Many thanks

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  • $\begingroup$ Are you trying to model this system as a sdof oscillator in order to predict its vibrational behaviour? $\endgroup$ – NMech Dec 26 '20 at 13:17
  • $\begingroup$ Hello Nmech, yes thats the reason, but i cant figure it out how $\endgroup$ – thestudentneedhelp Dec 27 '20 at 10:35
  • $\begingroup$ IMHO, if you want to go down that road the best way is through the equivalent systems modelling using the kinetic energy (translational/rotational) and the potential energy (torsional springs and bending spring in this case). I could give you an example, however, at end of the day I don't think it would be at all accurate to model this system. if you want I can write up a simpler example for torsional gears of the equivalent systems method (see link for a linear example ) $\endgroup$ – NMech Dec 28 '20 at 8:36
  • $\begingroup$ Dear Nmech, I would be very grateful if you could write a simple example since I am not so familiar with the subject matte $\endgroup$ – thestudentneedhelp Dec 28 '20 at 10:00
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Lets assume you have the following system:

A motor supplies a torque, which turns a shaft. The shaft has a gear on it (with $z_1$ teeth), which meshes to a second gear (with $z_2$ teeth) designed such that the speed of the second shaft is greater than the first.

enter image description here

The shafts are mounted on bearings each with a torsional damping coefficient $c_t$.

The angular displacement of the shaft 1 (directly connected to the motor) is denoted with $\theta_1$. The relationship between the angular velocity of shaft 1 and shaft 2 is given by:

$$\frac{z_1}{z_2}=\frac{\dot{\theta}_2}{\dot{\theta}_1}$$

1. kinematic relationship

Seemingly, the problem here is that you have two different angular displacements. However, in truth you only have one, because $\theta_2$ is proportionally dependent to $\theta_1$. The exact equation is :

$$\theta_2 = \frac{z_1}{z_2} \theta_1$$

2. derivation of equivalent system.

The main idea is that you come up with parameters $I_{eq}$, $c_{eq}$ and $k_{eq}$ that for a given angular displacement have the same energy with the system.

2.1 Kinetic energy of shafts.

The kinetic energy of a rotating object is given by:

$$T_i = \frac{1}{2} I_i \dot{\theta}_i^2$

where:

  • $J_i$ is the mass moment of inertia $[kg m^2]$

because both the gear (g) and the shaft (s) are rotating the above can be written for a shaft with a gear as:

$$T_i = \frac{1}{2} (I_{s,i} + I_{g,i}) \dot{\theta}_i^2$

so the total energy is :

$$T_{total} = T_1 + T_2 = \frac{1}{2} (I_{s,1} + I_{g,1}) \dot{\theta}_1^2 + \frac{1}{2} (I_{s,2} + I_{g,2}) \dot{\theta}_2^2$$

this can be simplified by substituting $\dot{\theta}_2 = \frac{z_1}{z_2} \dot{\theta}_1$ to:

$$T_{total} = T_1 + T_2 = \frac{1}{2} (I_{s,1} + I_{g,1}) \dot{\theta}_1^2 + \frac{1}{2}(I_{s,2} + I_{g,2})\left( \frac{z_1}{z_2} \dot{\theta}_1\right)^2$$

$$T_{total} = \frac{1}{2} \color{red}{\left( (I_{s,1} + I_{g,1})+ (I_{s,2} + I_{g,2}) \left( \frac{z_1}{z_2} \right)^2 \right)}\dot{\theta}_1^2 $$

So the equivalent mass moment of inertia $I_{eq}$ to simulate the system which is rotating with $\dot{\theta}_1$ is:

$$ I_{eq} = (I_{s,1} + I_{g,1}) + (I_{s,2} + I_{g,2}) \left( \frac{z_1}{z_2} \right)^2 $$

2.2. Damping

The work done by the viscous dampers from time a to time b is equivalent to the rotational speed.

$$W_{ab} = -\int_{\theta_{1,a}}^{\theta_{1,b}} c_t \dot\theta_1 d\dot{\theta}_1 -\int_{\theta_{2,a}}^{\theta_{2,b}} c_t \dot\theta_2 d\theta_2 $$

Again, this can be simplified by substituting $\dot{\theta}_2 = \frac{z_1}{z_2} \dot{\theta}_1$ and $\theta_2 = \frac{z_1}{z_2} \theta_1$ to:

$$W_{ab} = -\int_{\theta_{1,a}}^{\theta_{1,b}} \color{red}{c_t \left(1 + \left(\frac{z_1}{z_2} \right)^2\right)} \dot\theta_1 d\dot\theta_1 $$

So the equivalent damping coeffient is the red part:

$$c_{eq} =c_t \left(1 + \left(\frac{z_1}{z_2} \right)^2\right)$$

2.3. Dynamic energy.

In this system, some energy is stored in the shafts as torsional energy. However, we won't calculate the $k_{eq}$ because in this system A) the angular displacement related to the stored torsional energy is orders of magnitude smaller to that of $\theta_1$, B) the stored energy - usually (however probably not in your example)- is normally comparatively small, and C) because the elastic energy remains constant if the applied torque remains the constant.

final comments

This is the general idea of a modelling method to turn a system like this to an SDOF. In your particular problem you would have other things, like bending of the beam, etc. At the end of the day, it might not be possible to model it accurately with a SDOF, but you'd need to go to a MDOF System.

Additionally, there are other issues of concern (like how to calcucate the damping coefficients etc). As I said, in my comment, I don't think this is particularly useful for what you are intending to do.

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Your machine is already doing what you ask.

The torque in the shaft needed to move the collar is, approximately:

$$\tau=\frac{Fd_m}{2}tan(\phi +\lambda)$$

  • T = torque

  • F = load on the screw

  • dm = mean diameter

  • $\mu $, = coefficient of friction

  • l = lead

  • $\phi $, = angle of friction

  • $\lambda $, = lead angle

If the shaft can not take this torque it will fail and break.

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  • $\begingroup$ Hello Kamran, that's right, but I want to calculate the translational stiffness to get to the natural frequency of my system, so I want to convert the rotational stiffness into a translational stiffness. My basic idea was that it is somehow possible to do this with the help of the spindle pitch. Do you have any ideas? $\endgroup$ – thestudentneedhelp Dec 25 '20 at 8:22
  • $\begingroup$ I doubt it. usually the friction is so high that the system will not vibrate. many times it will lock up. $\endgroup$ – kamran Dec 25 '20 at 10:03
  • $\begingroup$ Hi Kamran, for this reason my intermediate shaft is made of elastomer. Very sad I really thought there was a connection. $\endgroup$ – thestudentneedhelp Dec 25 '20 at 10:27

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