Energy lost due to bushing friction

[George Zucas]

EDIT: I accidentally put this thread into Engineering instead of Mechanical Engineering section, please move!

Hello all,

I am designing a system which will basically just stand in two different positions all the time by a compressed spring. The spring is pushing the system while the forces on the other side keep the system in place. There will be a sensor which will detect the time when the force is not there. This will work since when the force is not there, the spring will push the system up and the sensor will be triggered. Now I need to select a suitable spring.

I've approached the problem by calculating the potential energy change between the normal position and the position in which the sensor will be triggered ( IF there are any alternatives I'd be glad to learn it). It is about 10000 J (it is a small system). Now I need to find the energy lost to the bushings in the system (there are 8 bushings). The coefficient of friction is 0.19 and it is a H7-h9 fitting.

I am kinda at a loss to how to determine this. I don't know how the fitting comes into play for example or the roundness of the contact. What I would do is to take frictional force as F=mu*mg=0.19*0.25*9.81=0.47 N (0.25 is the shaft weight, 10 mm is radius) and therefore the frictional moment is 0.47*10=4.75Nmm. Though I don't where to go after that or how wrong it is up to this point.

Any help is appreciated.

 

[Baluncore]

Do the shafts slide or rotate in the bushings ?
Is the spring force carried by the bushings ?

I think we need a diagram or a picture that shows the arrangement of components.

 

[George Zucas]

I'd gladly post the picture of the entire system if it wasn't for the fact that I can't delete it later. I am cautious in case any problem occurs at workplace due to that. Though I can provide a hand-drawn picture. In the picture the ones marked with B shows bushings. It is a simple but close representation of the system.

The shafts rotate in the bushings.

And thanks for the reply.

 

[Baluncore]

The surface finish detail and area of contact should not be needed if the coefficient of friction is specified.
Note that depending on the lever lengths, the force on the lower pin and bush will be twice the applied force.

Energy lost to bushes per cycle will depend on the pin diameter and the angle of rotation.
That is the total linear distance moved by the pin–bush contact multiplied by the force and by the coefficient of friction.

 

[George Zucas]

Thanks for the answer. I made some silly errors, even typing out the problem makes one understand the problem better.

The pins have a diameter of 20 mm and the angle of rotation is about 30 degrees. So (2*pi*(20/2))*(30/360)=5.24 mm (but this is not linear distance, do you mean the direct distance between the first and the second position of a point which is 5.18 mm? Not that there is much of a difference but for the sake of understanding the concept).

The force on the lower bushing will include the weights and double the force as you said , considering your note and the weights, the force is approximately 1000 N on the bushings.

So, from what I understand:

The movement distance of the contact is 5.2 mm,
The force is approximately 1000 N,
Coefficient of friction is 0.19,
Energy loss is then (5.2mm)*(1000N)(0.19)=1045 Nmm=1.05 Joule
Assuming each bushing cause the same loss (top ones are much lower)= 8.4 Joules

 

[Baluncore]

It seems reasonable to me. Build a prototype and measure the temperature rise while in operation.
To lower the energy loss further, lubricate the bush, lengthen the rocker arm or reduce the pin diameter.
Don't forget it will rock back again, that doubles the distance per full cycle.

 

[George Zucas]

Thanks a lot. A rough estimation is what I needed since I have to add a huge safety factor to ensure that the system is working even 10 years from now anyway (it'll stay in the same position for decades, waiting in case that force no longer exists :) ). Though it seems finding a spring strong enough to push the system while having a small size is quite a challenge.